Discrete Geometry (Pure and Applied Mathematics (Marcel Dekker))

z^M

H -- --

2v^M

< ————— < sm 7

INI-lkll"

where the inequality in the middle is equivalent to

This follows from |b||,||g|| < p - {/r/10. n Lemma 11. For all p 6 ST, C2(p) n ^Pr C Z2. Proof. By Lemma 10 the edge (or vertex) [a, b] corresponding to p 6 Sr is contained in C\(p}. Let c be the integer on dPr immediately after b in anticlockwise order. Then c <E Lk(p) for some fc > 1. We will show k — 1. There are two cases.

LATTICE POINTS ON THE BOUNDARY

43

Case 2

Case 1 Figure 2

Case 1. k is odd, k = It + 1 (I > 1). Set x = \(b + c] and let &', x', c' denote the last points on the chords d(p] (i - 1, t + 1, It + 1). As x ± |p 6 Z2, |x' — x| < ^||p|| must hold. Then, a fortiori,

1

> \x'-x\ >

It is easy to see geometrically that the last expression is minimal on the interval a 6 [0,1/8] at a = 1/8, and when a = 1/8, it is minimal for t > 1 if I = 1. Now with a = 1/8 and t = 1

1 ey I \lr\ I

^

But this contradicts

l/\/10 < 0.3163.

Case 2. k is even, k = 21 and £ > 2. Write x (resp. y) for the intersection of the segment [6, c] with CV(p), CV+i(p). The same way as above and with analogous notation \xf — x\ + \y' — y\ < \\p\\ as otherwise either [x',x] or [y',y] would contain an integer. Thus

\\p\\

x -x

\y' -y\> \Ce+l(p)\ + \Ct(p)\ -

- C2i(p}\.

The right hand side is the smallest when a = 1/8, again, and then when t = 2. A contradiction follows after a short computation. D According to Lemmas 10 and 12, Ci(p) n dPr = {c,d} C Z2 for every p € Sr. Assume c, 6, a, d come in this order on dPr. Define T(p] = {[d,o],[o,6],[6 >C ]} and Tr =

T(p).

BARANY AND BOROCZKY, JR.

44

By Lemma 10, T(p) and T(q) are disjoint for distinct p, q G Sr. The boundary of Pr consists of three parts: the first contains the segments in Tr, the second the edges corresponding to p G Er, and the third is the rest. This last group consists of edges [a t -,6j] corresponding to p. While the first and second group may have some overlap they are both disjoint from the third.

Figure 3

Lemma 12. For large r (along the sequence rn) 3.8671r

4.2602r.

Proof. The perimeter of Pr is 2r?r( 1+0(1)) and the second group contributes o(r) to it (in view of Lemma 8). So it suffices to estimate ^"L^ \x\\. It is not hard to see that, for p E 5V,

because a < 1/8. Then, by (9),

24 / i —7= ( 1 + - 1 r < 2.4158r. TTx/TO V r

This proves the lower bound. For the upper bound notice that when p £ Sr \d-a\

-c| > \C2(p) C\ Pr\ > \C2(p)\ - 2\\p\\

LATTICE POINTS ON THE BOUNDARY

Thus

45

^ E l(p^ = E l(p^ - E ^

The Fact shows the second sum that it is o(r). So, again by (9), —= ) (1 + o(l))r > 2.0228r. D V5 57T/

9. PROOF OF (6) AND (7) Next we prove (6). The number of edges (or segments) in the first group is at most o 18 / r \ 2 /3 o/o 315VI < 3|Pn/>5 2 | < — ( —] (l + o(l)) < 1.2345r2/3. \10/

7T

The contribution of the second group is o(r 2 / 3 ). Finally, in the third group, m

Ell p. Under these conditions m is maximal if the g,- are as short as possible. So choose the smallest p satisfying m < |Pn (pB2\pB2}\. Then = i(p 3 -/> 3 )(l + o(l)).

We get p < \/3A46Qr showing that m < \Tr](pB2\pB2)\ = -(p2 - p2)(l + o(l)) < 3.9458r2/3, 7T

proving (6). Finally we prove (7). Given [i^v] C rB2, the missed area, A( / u,v), beyond this segment is defined as A(u, v) = Areajx e rB2 : [0, x] D [M, v] ^ 0}. It is easy to see that ,

where w, v are close to the boundary of rB2 and M, v are almost orthogonal to 11 — v. For p 6 Sr let [d',c'] = C*2(p) with dr, d, c, c' in this order. It is easy to see (we omit the proof) that A(d',a) < A(d,a) and A(6,c') < A(b, c). Then

46

BARANY AND BOROCZKY, JR.

the missed area beyond the chord C2(p) is

A(d, a) + A(a, b) + A(6, c) > A(d', a] + A(a, 6) + A(6, c') >

where o(l) is uniform in p. Summing this for all p £ 5V shows that the missed area beyond the segments in Tr is 2 3 2 3 ^ z_-/ L ( p , r ) = V L(p,r) - o(r / ) > 2.2478r / HP||

The sum is minimal, under the condition that the p and ^ | qi\ > 3.8671r, if the qi are as short as possible. The by now routine computation reveals that m 3

> 0.4267r 2 / 3 . D

Remark. This proof gives a very small value for the implied constants in Theorems 4 and 5. Computational experience (and [3] as well) shows Nr w 3.45r 2 / 3 and \dPr n Z2| w 12r2/3.

REFERENCES [1] G.E. Andrews: A lower bound for the volumes of strictly convex bodies with many boundary points, Trans. Amer. Math. 5oc., 106 (1965), 270273. [2] A. Balog and I. Barany: On the convex hull of the integer points in a disk, DIMACS Series in Discrete Math., 6 (1991), 39-44. [3] A. Balog and J.-M. Deshouliiers: On some convex lattice polytopes, Number theory in progress, Vol. 2 (Zakopane-Koscielisko, 1997), 591606, de Gruyter, Berlin, 1999. [4] I. Barany and D. Larman: The convex hull of the integer points in a large ball, Math. Annalen, 312 (1998), 167-181. [5] L. Fejes Toth: Lagerungen in der Ebene, auf dem Kugel und im Raum, Springer, Berlin (1953). [6] E. Grosswald: Representation of integers as sums of squares, Springer, 1985.

LATTICE POINTS ON THE BOUNDARY

47

[7] P.M. Gruber and C.G. Lekkerkerker: Geometry of Numbers, North Holland, Amsterdam (1987). [8] S. B. Konyagin and K. A. Sevastyanov: Estimation of the number of vertices of a convex integral polyhedron in terms of its volume, Funk. Anal. PriL, 18 (1984), no. 1, 13-15. (in Russian) [9] G.H. Hardy and E.M. Wright: Introduction to the theory of numbers, Oxford, Clarendon (1954). [10] V. Jarnik: Uber Gitterpunkte und konvex Kurven, Math. Z., 24 (1925) 500-518. [11] R. Kannan and L. Lovasz: Covering minima and lattice point free bodies, Annals. Math., 128 (1988) 577-602. [12] L. Kuipers and H. Niederreiter: Uniform Distribution of Sequences, New York, 1974. [13] W. Schmidt: Integral points on surfaces and curves, Monats. Math., 99 (1985) 45-82. E-mail address: [email protected] ''Imre Barany'' E-mail address: carlosfirenyi.hu ''Karoly Boroczky,

Jr.''

THE ERDOS-SZEKERES PROBLEM FOR PLANAR POINTS IN ARBITRARY POSITION

TIBOR BISZTRICZKY1 Department of Mathematics and Statistics, University of Calgary, Calgary, AB, Canada, T2N 1N4 GABOR FEJES TOTH2 Alfred Renyi Institute of Mathematics, Hungarian Academy of Sciences, Budapest, Hungary, H-1053

ABSTRACT. Let A; and / be integers 2 < k < I. We consider the ErdosSzekeres convex n-gon problem for sets S of points in the plane with the property that the convex hull of any k points of S contains at most / points of S.

1. INTRODUCTION Recall the classical theorem of Erdos and Szekeres [2] according to which, for every natural number n > 4, any set of points in general position in the plane with sufficiently large cardinality contains some n points in convex position. A set of points in the plane is in general position if no three points of the set are on a line, and it is in convex position if no element of it is contained in the convex hull of the rest of the points. Let Pn denote the property that a set does not contain any n points in convex position. Another way of formulating the theorem of Erdos and Szekeres is that the cardinality \S\ of a set S of points in general position in the plane satisfying property Pn is bounded. Let f(ri) denote the maximum cardinality of such a set. 1

Supported in paxt by NSERC Operating grant. Part of this research was done while the author was visiting at Western Washington University in Bellingham, WA. Research was also supported by OTKA Grants no. 26358 and 30012. 2

49

50

BISZTRICZKY AND FEJES TOTH

Erdos and Szekeres showed that

The upper bound has been improved recently in a series of papers by Chung and Graham [1], Kleitman and Pachter [7] and finally by Toth and Valtr [10] who established the presently best known upper bound, namely

There still remains a large gap between this bound and the lower bound 2 n ~ 2 , which is conjectured to be sharp for all n. The theorem of Erdos and Szekeres has been generalized in various directions (see [9] for an extensive survey on the subject). Here we give yet another generalization, the starting point of which is the following reformulation of the condition that the points are in general position. A set S in the plane is in general position if the convex hull of any two points from S contains just these two points of 5*. What happens if we weaken the premises in the theorem of Erdos and Szekeres and instead of requiring the points to be in general position, we assume only that the convex hull of any two points from S contains at most I points of 5, where I is some integer greater than two? More generally, for integers k and /, 2 < k < /, we consider sets S with the property 'Ptkj) that the convex hull of any k points from S contains not more than / points of S. The property that a set satisfies both Pn and P(k,i) is denoted by P2 ^ . Is the cardinality of a set of points in the plane with property P?k ^ bounded? As expected, the answer is yes. So it is legitimate to define //J. ^ as the maximum cardinality of a set with property P?k ^. Of course, we have /(n) = /," 2y We shall give upper and lower bounds for f?k ^, some of which turn out to be surprisingly strong. Obviously, //],. n = I for n < k + 1, so we deal only with the case n > k + 1. We need to introduce additional notation in order to formulate some of our bounds. For /, i, j > 2, we define the numbers 9 l ( i i j ) recursively as follows: (1)

0(2,j)=sitt,2) = l

(

(2)

if j is odd,

;2 +1

if j is even,

0i(3,j)=00',3) = < [

(3)

fc^

J>*

9i(i,J)=9i(i-l,J)+9i(iJ-l)

for *, j > 3.

Now we are in the position to formulate our main result.

THE ERDOS-SZEKERES PROBLEM

51

Theorem. We have n

(4)

(5)

£]fl(t,n + 2 - i) < /(»2>0 < 0,(n - l,n) + 1

/(» jZ) = n - 1 + J -

fc

/or 3 < fc < / <

/or / > 2, n > 3;

- 1, / < n;

and (6)

n _i

+ (n-

/or 3 < k < I and k + I < n.

We conjecture that the lower bound in (4) is sharp. It is easy to give the general term of the sequence g^ij}- We have

To see this we note that #2(2, j) = 1 = P'"^4) an^ #2(3, j) = j — I = Pt-2 4 )' that is the equation holds for i = 2 and i — 3. Recalling that (™) = (^i*) + (m~1} we see that also the recurrence relation (3) is satisfied by the numbers g z ( i , j ) = (J^!.^4)- We have

i-5N 1=2

i=

thus the bounds in (4) for //^ 2^ = /(n) are those of Erdos and Szekeres, and Toth and Valtr. We observe that l

-(j - 2) + 1 < 9l(3,j) < l-^(j

- 1),

j > 3.

Using these initial values instead of those given in (2), we get that 2\\ i-2 J for j > 3. Hence (/ + 2)2-4 -

i-3

< /(n2)0 <

-

«-3

s

+X

/«- ^ > 3, n > 3.

As an immediate consequence of (6), we get the following Corollary. If for the integers k, I, and n with 3 < k < I and n > k + 1, both |^5 and ^-^f are integers, then

52

BISZTRICZKY AND FEJES TOTH

We note that there is equality in (6) also when k = I, yielding that f?k k^ = n — 1 for k > 3. This case is covered by (5) as well. We recall the problem of Erdos asking whether sufficiently large cardinality of a set of points S in general position in the plane guarantees that S contains the vertices of an empty convex n-gon; that is n points in convex position such that their convex hull does not contain any other points of S. It turned out that the answer to this question is yes for n = 4 and 5 (see [4]) and no for n > 7 [5]. The problem remains undecided for n = 6. However, Karolyi, Pach and Toth in [6] showed that property 7^(3,4) and sufficiently large cardinality implies for any set that it contains an empty convex n-gon for any n. This result was extended by Valtr in [11] to any k < I.

2. UPPER BOUNDS FOR fjl ^ The upper bound in (4) is easily obtained through a slight modification of the original methods of Erdos and Szekeres [2]. We say that the points (xi, yi), (0:2,1/2), • • • ^ (xrm ym)i xi < X2 < • • • < xmi form an m-cap if

y-2 - yi <"y»o

JL,

—.

nr> -

*^ 1

Vm - ym-i

ya -3/2 T*o

**-' O

'T'o

**-> 2,

T*

^TTl

T*

i

*^77l— 1

Similarly, they form an m-cup if

y2 ~ yi
yi *c ... <, ym

ym—i .

A set of two points is a 2-cap and at the same time a 2- cup. We consider sets S in the plane satisfying property P(2,i) such that no line through two points from S is parallel to the x- or y-axis and, furthermore, S contains neither an ^-cap nor a j-cup. Let gi(i:j) be the maximum number of points in such a set. It is not a mistake that we denote these numbers with symbols already used in the introduction, as they are equal to those defined by the relations (1), (2), and (3). Erdos and Szekeres proved that the numbers defined here satisfy the recurrence relation (3) if / = 2; however, their argument applies without any modification for all / > 2. Obviously, if a set of points contains neither a 2-cap, nor a 2-cup, then it consists of a single point, so the initial value conditions (1) are satisfied. It remains to show that the initial value conditions (2) hold as well. Let S be a set satisfying property "P(2,n which contains neither a 3-cap nor a j-cup. Then all elements of S lie on the boundary of conv S. We claim that ( l-^fl

(7)

if j is odd,

\S\
j>3 ls even

>

THE ERDOS-SZEKERES PROBLEM

53

This holds obviously when 1 = 2. Next we investigate the case / = 3. Consider a set R of 2s + 1 points of S lying on s consecutive sides on the boundary of conv S with each side containing exactly one additional point of S in its relative interior. From R, we choose each second vertex and each point in the relative interior of a side. This gives t = ["^p"] + s points which, when arranged with increasing re-coordinates, form a t-cup. We observe that the ratio r^r is greater than or equal to | with equality only if s = 1, that is \R\=3. We consider the subsets Ri of S consisting of points in a maximal sequence of consecutive sides on the boundary of conv S each side containing an additional point of S in its relative interior. Each Ri contains a tj-cup, Ol 7? . I where ti > -^ unless \Ri\ = 3. These points, together with the remaining 9l 91 vertices of convS1 form a it-cup, where t > -L-1. Here strict inequality holds unless the sets Ri exhaust all vertices of conv S and each Ri contains exactly three points. 2| c| Since S contains no j-cup, we have -^-L < j - — 1, and consequently

151 ^^ 3(J ~ 1) PIS Equality can hold here only if |5| is a multiple of 3 and j is odd. If \S\ is ., 2|S - . even, then -^ < j — 1,-.1hence 2

Finally, we assume that I > 4. Let M be the set of sides of conv S containing at least two points of 5" in their relative interior and let V be the set of points of S which do not belong to a side from M. Then the sides belonging to M, together with the vertices from V, cover the set S. Since each side of conv 5 contains at most / points from S, it follows that

We observe that the set V satisfies property "P(2,3), so it contains a i-cup with t > -^-. We augment this set with |M| pairs of points chosen from the relative interior of the sides from M. The resulting set is a t'-cup with t' > -^ + 2|M|. Since S contains no j-cup, 3

As / > 4, it follows that ?-_i_21H J

\S\
^~

f ^^r1 + \V\<<

(, ^\ ' +1

if j is odd,

if j is even.

54

BISZTRICZKY AND FEJES TOTH

This completes the proof of (7). It is easily seen that this bound is sharp. This is obvious for 1 = 1. If / > 2, then the bound can be achieved if, with the possible exception of one point, all points of S lie on distinct segments, each containing I points. This shows that the numbers defined here indeed coincide with the ones defined by the relations (1), (2), and (3). Obviously, f(2,l)

<9l(nin}-

Using the idea of Toth and Valtr [10], we get that

For the proof of (5), we need the following generalization of Caratheodory's theorem. Lemma. Let S be a set of points in the plane and X a set of i > I points contained in conv 5. Then there is a subset SQ of S containing at most 2i points such that X C conv SoThe proof follows by induction on i. Let p\ and p% be two points in conv S. Let the line p\pi intersect the boundary of conv S in q\ and q%. Then there are points r\, r2, r3, and r\ of S on the boundary of conv 5 such that qi 6 conv{ri,r 2 } and 2 points in conv S. By the inductive hypothesis, there are, not necessarily distinct, 2(i — 1) points r i , . . . , r 2 ^_ 2 of S such that {pi,... ,pi-i} C conv{ri,... ,r 2i _ 2 }. Let the ray pi-ipi intersect the boundary of conv S in qi. Then there are two points r^i-i and r^i of 5 the convex hull of which contains %. It follows that {pi,... ,pi} C conv{n,... ,r 2 i}. It is easily seen that the bound in the lemma is sharp: If S consists of the vertices of a convex n-gon with n > 2i and X is the set of the midpoints of i mutually disjoint closed sides of conv 5", then X is not contained in the convex hull of any 2i — 1 points from S. We note that the proof above carries over with obvious modifications to Ed yielding the following statement. // in Ed a set X of i points is contained in the convex hull of a set S, then there is a subset SQ of S containing at most di points such that X C conv SQ. The bound di cannot be replaced by a smaller one. We prove (5) for the case 4 < k < I < [^jf-\ — 1. As we noted above, the case k = /, including the case k = I = 3, is covered also by (6) and will be treated there. Let 4 < k < I < [4£J — 1, I < n and consider a set S of n + I — k points in the plane satisfying property P(k,i)- We note that then S also satisfies

THE ERDOS-SZEKERES PROBLEM

55

property P(kij_f~+k') f°r all k', 2 < k' < k. We have to show that 5 contains n points in convex position. Suppose the contrary. Then, in particular, conv 5 has at most n — I vertices. Therefore, we can pick / — k + 1 points of S which are not vertices of conv S. According the lemma, there are 1(1 — k + 1) points among the vertices of conv 5, the convex hull of which contains these points. Thus 5 does not satisfy property P(2(i-k+i),3(i-k+i))Since 3(/ - k + 1) - 1 - 2(1 - k + 1) = I - k and in view of / < [f J - 1 we have 2(1 — k + 1) < fc, this contradicts our assumption. Thus S does contain some n points in convex position. This shows that / J^ ^ < n — I +1 — k. The example showing that the bound in the lemma is sharp shows that this bound is sharp, as well. Finally, we turn to the proof of the upper bound in (6). Let 3 < k < /, n > k + 1 and let S be a set of points in the plane satisfying property P?k ». Let m be the number of vertices of conv S. Then m < n — I. If m < A;, then 5 has at most / elements which is, for n > k+1, smaller than the upper bound in (6). If m > k, then we decompose conv S into a maximum number of kgons and possibly one convex polygon with less than k vertices. It is easily seen that such a decomposition contains [^zf 1 polygons. Each polygon in this decomposition can contain at most / — A; points from 5" different from its vertices. Thus S contains at most m + (I — k) C^rrf ] < n — 1 + (I — k) [frf 1 points. 3. LOWER BOUNDS FOR f?k » The proof of the lower bound in (4) also follows the original ideas of Erdos and Szekeres [3]. There is a very transparent description of the construction of Erdos and Szekeres in the book of Lovasz [8]. We shall follow the description there. Consider a regular 4(n — 2)-gon inscribed in a circle of radius r centered at the origin. Let v % , . . . :vn be the vertices of it lying in the region: x > 0 and \y\ < re, enumerated in counter-clockwise direction. We note that any line ViVj, 2 < i < j < n, forms an angle greater than 45° with the re-axis. It follows that if r is sufficiently large, then the unit circular discs Ci centered at Vi, 2 < i < n, have the property that any line Pipj, Pi 6 (7j, PJ 6 (7j, 2 < i < 3' < n, forms an angle greater than 45° with the re-axis. We place a set Si of gi(i,n + 2 — i) points into Ci such that it satisfies property P(2,i) such that 5Z- contains neither an i-cap nor a (n + 2 — z)-cup. We assume that no line through two points from Si is parallel to the re-axis and, furthermore, that any line through two points of Si forms an angle smaller than 45° with the re-axis. Clearly, the last condition can be achieved by applying a suitable affinity. We claim that the set 5 = Uf_ 2 5j with Sr=29i(i>n + 2 — i) points satisfies property P?2z^.

56

BISZTRICZKY AND FEJES TOTH

It is clear from the construction that a line connecting two points of S belonging to different sets Si and Sj can contain no further points of S. Since each of the sets <%, 2 < i < j < n, satisfies property P (2,1)1 it follows that so does S. Let S be a subset of S in convex position. We have to show that S contains at most n points. Consider first the case that S is contained in one of the sets Si, 2 < i < n. Let p\ and q\ be the points of S with the least and largest re-coordinate, respectively. Let ^2, • • • ,pj and q^-, • • • ,qk be the points of S above and below the linep\q\, respectively. Then the pointspi^pi, • • • ,Pji
THE ERDOS-SZEKERES PROBLEM

57

FIGURE 1 We note that for any natural number i property T^s+i) implies P(k,k+(k-2)i)Indeed, the convex hull of any A; points from a set S can be decomposed into at most k — 2 triangles. If 5 satisfies property T-^a+j), then each of these triangles contains at most i points of S different from its vertices. Thus, the total number of points of 5" contained in the convex hull of k points is at most k + (k — 2)i. Using the observation above, the general case of the lower bound in (6) follows by constructing a set of n— l + (n— 3) j^ points satisfying property

REFERENCES [1] Chung, F.R.K. and Graham, R.L., Forced convex n-gons in the plane, Discrete Comput. Geom., 19 (1998), 367-371. [2] Erdos, P. and Szekeres, G., A combinatorial problem in geometry, Compositio Mathematica, 2 (1935), 463-470. [3] Erdos, P. and Szekeres, G., On some extremum problems in elementary geometry, Annales Universitatis Scienciarum Budapestiensis, Eotvos, Sectio Maihematica, 3-4 (1960-61), 53-62. [4] Harborth, H., Konvexe Fiinfecke in ebenen Punktmengen, Elem. Math. 33 (1978), 116-118.

58

BISZTRICZKY AND FEJES TOTH

[5] Horton, J.D., Sets with no empty convex 7-gons, Canad. Math. Bull. 26 (1983), 482-484. [6] Karolyi, Gy., Pach, J. and Toth, G., A modular version of the ErdosSzekeres theorem, Studio, Sci. Math. Hungar., 38 (2001), 245-259. [7] Kleitman, D.J. and Pachter L., Finding convex sets among points in the plane, Discrete Comput. Geom., 19 (1998), 457-459. [8] Lovasz, L., Combinatorial problems and exercises, North-Holland, 1979. [9] Morris, W. and Soltan V., The Erdos-Szekeres problem on points in convex position—a survey. Bull. Amer. Math. Soc. (N.S.), 37 (2000), 437-458. [10] Toth, G. and Valtr, P., Note on the Erdos-Szekeres theorem, Discrete Comput. Geom., 19 (1998), 457-459. [11] Valtr, P., A sufficient condition for the existence of large empty convex polygons, Discrete Comput. Geom., to appear. E-mail address: E-mail address:

gfejesOrenyi.hu ''Gabor Fejes T o t h ' ' [email protected] ''Tibor Bisztriczky''

SEPARATION IN TOTALLY-SEWN 4-POLYTOPES

TIBOR BISZTRICZKY Department of Mathematics and Statistics, University of Calgary, Calgary, AB, Canada, T2N 1N4 DEBORAH OLIVEROS Department of Mathematics and Statistics, University of Calgary, Calgary, AB, Canada, T2N 1N4

ABSTRACT. For a totally-sewn neighbourly 4-polytope P, we examine the problem of determining a minimal set Ti. of hyperplanes with the property that for any facet F of P, there is an H 6 H that strictly separates F from a fixed interior point of P. I. INTRODUCTION Let O be an interior point of a convex o?-polytope P C Ed. The Separation Problem is to determine the smallest number s(O,P) of hyperplanes of Ed that are needed to strictly separate any facet of P from O. The importance of the Problem is partly due to the fact that if O is the origin of Ed then s(O,P) is the Gohberg-Markus-Hadwiger covering number for the polar of P; cf. [2] and [6]. In [3] and [4], the Problem was solved in the case that P is cyclic. In [5] and [7], partial solutions were presented in the case that P is neighbourly and four-dimensional. Specifically, it was shown that s(O,P) < 9 in the case that P is a particular type of totally-sewn neighbourly 4-polytope, called semi-cyclic (see definition in Section 2). An edge E of a neighbourly 4-polytope P C E4 with n vertices is universal if the quotient polytope P/E has n—2 vertices. Since P/E is a plane polygon, it follows that E is contained in n — 1 facets, of P. say, PI,P2,... , Pn-2- For any of these P^, there is a point x G E4 with the property that it is beneath each facet P of P (P is a facet of conv(P U {x}}} with the exception of 59

60

BISZTRICZKY AND OLIVEROS

the other n — 3 Pj's, and it is beyond each of those (Fj is not a facet of conv(P U { x } ) } . The polytope P — conv(P U {x}} is again neighbourly and it is said to be obtained by sewing x onto P. If P has u universal edges then there are u(n — 2)P's obtained from P by sewing. If P itself is obtained from a 4-simplex by a series of sewings then we say that P is totally-sewn. If P is cyclic or semi-cyclic then P is totally-sewn. In case of the former (latter), exactly one (n — 2) of the u(n — 2)P's is again cyclic (semi-cyclic). Both a cyclic and a semi-cyclic P is obtained from a 4-simplex by sewings via the same set of universal edges, and so a cyclic 4-polytope is also semi-cyclic. In the following, we show that the vertices of a semi-cyclic P determine certain special pencils of hyperplanes (Theorem 3). Next, the existence of these special pencils allow us to bound from above s(0,P) for each of the u(n — 2)P's (Theorem 4). Finally, we determine which of the u(n — 2) sewings preserve the special pencils (Theorem 5). Specifically, there are at least two such sewings via any universal edge that has a common vertex with an other universal edge. In summary, we establish that s(O,P) < 9 for a class of totally-sewn 4polytopes P that, for the first time, includes polytopes for which there is no explicit description of the facial structure. A solution for all totally-sewn 4-polytopes P will likely require a positive answer to the following: Is each P a subpolytope of a P sewn as above?

2. SEPARATION PROPERTIES We restrict our attention to E4, and assume familiarity with cyclic and neighbourly polytopes (cf. [8] and [9]). As usual, the convex and affine hull of a set X of points is denoted, respectively, by conv(X) and aff(X). For sets X\,X-2,... , Xn and a point x, let [Xi,X 2 ,... ,Xn] = (X^Xz,...

conv(XlUX2(J---(JXn},

,X n ) = aff(Xi U X2 U • ..*„),

[*] = [{*}] and (*) = <{*}}. Henceforth, P denotes a convex neighbourly 4-polytope in J54 with the set V(P) of vertices, the set £(P) of edges, the set U(P] of universal edges and the set F(P) of facets. We recall that P is simplicial, [x,y] € £(P) for any x ^ y in V(P) and (cf. [10]) if |V(P)| > 7 and [x,y] € £(P) then (1) [x,y] € U(P) if, and only if, x and y lie on the same side of every hyperplane determined by a subset of V(P). Let P = [P, x] be obtained by sewing x onto P. Then for some E = [x, y] € U(P] and F = [E,z,w] e ^(P), x is beyond (beneath) each F € .F(P) that

SEPARATION IN TOTALLY-SEWN 4-POLYTOPES

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is (is not) in

, P, P) =

We say that x is sewn through T(E, F, P), and note that V(P) = V(P) U{x} and (2)

where U°(P) = {E G U(P] \ E n F = 0 or |£n {z,w}l = 1}. If P is totally-sewn then P is totally-sewn, and we may assume that V(P) = {xi, X2, ... , xn}, Pi = [xi, £2, • • • , #i] and P;+i is obtained by sewing Xj + i onto Pi for « = 5, . . . , n — 1. If P = Pn is cyclic then Xj+i is sewn through .F([xi,Xi], [xi,Xi,X2,Xi_i],Pi), and if P is semi-cyclic then Xi+i is sewn through .F([xi,x.;],Pi,Pi) for some PJ G J~(Pi) that contains [xi,Xi], As noted above, we are interested in pencils of hyperplanes generated by the vertices of P. For mutually distinct XJ,XK and x^ and V(P), let

We determine first the connection between the facial structure of P and the arrangements of hyperplanes in H(P, Xj, xjt, Xi) and ?i(P, x^, x^, xm) depicted in Figure 1.

Lemma 1. Let Q = [xj,xk,X£,xu,xv,xw],

Q' = [xj,Xk,xi,xm,xu,xw]

and

1.1. If (A) then [xj,Xjfc,X£J and [xu,xv,xw] are not faces ofQ. 1.2. // (B) and [xk,xi] £ U(Q] then [xk,xt,xv] and [xj,xu,xw] are not faces ofQ. 1.3. If (C) and [xj,xm] €U(Q] then (D). Proof. We recall that [V(Q)j = 6 implies that Q is cyclic, and hence, three vertices of Q span a face if, and only if, the remaining three span one. In c?=e of (A). [xj,Xk,xg] is clearly not a face. In case of (B). [x'K.z".:,x.;] is a face for xr £ {xj,xu,xw}. Thus if [xfc,x^] ^ U{Q) then [xk,xi,xv] is not a face.

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(B)

(D)

(C)

Figure 1 Assume (C) and that [xj,x m ] G U(Q}. Then [xj,x m ] E U(Q'}, and [xj,xm,xw] is a face of Q'. Hence, [xfc,X£,x u ] is a face of Q', and it follows by 1.2 that [xfc,X£J E U(Q'}. Then QY[xfc,x^] is a quadrilateral with the diagonals determined by {x/^x^x^Xu} and (xfc,X£,x m ,x w ). Thus, we need only to determine the position of xv in (D). We note that Q is also cyclic. Relabel its vertices so that Gale's Evenness Condition is satisfied by the vertex array c\ < c% < • • • < cj < c§ = GI . Then [cj,Cj + i] G U(Q] for i = 1,... , 7, and each 2-face of Q contains some [Ci,Cj + i].

Since [xj,x m ] G l^(Q}^ [xm^v^w] and [xmxv,xw] are not faces of Q and (xj,X)-,X£,xm} separates xv from xu and x^,, it is easy to check that the vertex array is Xm < Xj < xw < x^ < xv < xc. < z.u or x.,, < x.j < xw < x^ < xv < Xfc < xu. In either case, (x m ,Xfc,x^,X£) separates xu from Xj and xw. D

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Let E — [x0,a;ft] £ U(P}- Lemma 1.3 states that H(P,xa,Xk,xi) completely determines 7Y(P,xj,,x^^xi). We simplify our notation by letting "H(P,E,Xk,xi) denote either one. The T-t(P,E,Xk,xe) that is of particular interest is the one depicted in Figure 2, and called a pentagram. We are assuming that V(P)\{xa,Xb,Xk,xt} is the disjoint union of (some possibly empty) clusters R, 5, T, U and V. We note that a cluster X is always on one side of (xa,Xk,X£,Xi) for any Xi £ X, and for simplicity, we denote X usually by a single x £ X. We say that P has the pentagram property if "H(P,E, Xk,Xi) is a pentagram for each E £ U(P] and all Xfc ^ x^ in V(P)\E. Clearly, if P has at most nine vertices then it has the pentagram property.

Figure 2 Lemma 2. Let E — [xc,Xd] £ M(P), P = [P,x] be obtained by sewing x through JF(£, F,P), E = [xa,xb] £ U(P] nU(P) and H(P,E,xk,xe) be a pentagram. ThenT-L(P,E,Xk-,xe] is a pentagram. Proof. Since E £ U(P], it follows that E ^ E and, say, xt, <£ {xc,Xd}. If [xa,Xk,xi] is a 2-face of P then there is at most one non-empty cluster of 1-i(P,xa,Xk,Xi) on each side of ( x a , X k , x i , X b ) , and the assertion is immediate. If [xa,Xk-,xe] is not a face of P then, say, xc is contained in one of the clusters of 7Y(P, xa,Xk-,Xi^Xb}. Since [x,xc] €&/(P), it follows from (1) that if, say, xc £ S1 then H(-P, 2;a5 ^fc, ^) is a pentagram with S = SU{x} replacing S. D As noted in the Introduction, Theorems 3, 4 and 5 are our key results, and each focuses on pentagrams. The proofs of Theorem 3 and 5 involve

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deducing the existence of new pentagrams from known ones, and are based upon the following simple observation: (3) // "H(P, [x a ,Xb],Xfc,x^) is a pentagram (cf. Figure 2) and xm 6 V(P) such that [x£,x m ] € U(P] then 7i(P, [x£,x m ],x a ,Xfc) is a pentagram whenever (x a ,Xfc,X£,x m ) supports R, 5, [7, V or T(J{xb}. Theorem 3. A semi-cyclic 4:-polytope has the pentagram property. Proof. We assume that P = [zi, # 2 , . . . , xn] C E4" is semi-cyclic with respect to the vertex array x\ < x<± < -. - < xn; that is, PQ — [xi,X2,... , zg] is cyclic with respect to x\ < X2 < • • • < zg, and for m = 6 , . . . , n — 1, Pm+i = [Pm, Xm+i] is obtained by sewing xm+1 through J^([xi, x m ],F m , Pm) for some Fm G ^(Pm) containing [xi,x m ]. Then (2) yields that {[xi,x n ], [x n ,x n _i]} C U(P) and U(P}\{[xi,xn}} C {[x m _i,x m ] | m = 2 , . . . ,n}. In [7], it was shown that for x\ < x& < x^ < xn, 7i(P, [xi,x n ],Xfe,X£) is the pentagram depicted in Figure 3 and with the following properties: (4) X; < xi for each Xi £ S U C/, (5) Xj > Xi for each Xj 6 T U V", and (6) if xr < xu and {x r ,x u } ts contained in T or V then (xi,Xfc,X£,x u ) separates xr and xn.

Figure 3 Let E 6 U(P] and xp ^ xg in V(P)\E. We may assume that n > 6 and that 7i(P n _i,£",Xp,x^) is a pentagram for any E' 6 U(Pn-i) and any x^ ^ x^ in V(Pn-i)\E'. Since the assertion is

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immediate if E — [x\,xn] or [xp,xg] e M(P), we assume that E = [x m _i,x m ] for some 2 < m < n and [xp,x9] ^ ^(P)- Finally, we may assume that Xp "C. Xq.

Case 1. E — [x n _i,x n j. Then x\ < xp < xq < x n _2. If xp — x\ then the assertion follows from H(P,xn-i,xi,xq) = H(P,xi,xk,xt} with (a*,a:*) = (a; g ,a; n _i) in Figure 3. If xp ^ x\ then it follows from "H(P, x n ,Xfc,X£) with (x^x^) = (x p ,x g ), and the fact that (xfc,X£,x n ,x n _i) supports T or V by (6).

Case 1. E- [xm-i,xm] 6 U(Pn-\). Then we may assume by Lemma 2 that {xp,xq} (£ V(P n -i), and hence that xq = xn. Since we have assumed that [xp,x9] ^ £^(P), xp > x\. If xp < Xm-i then H(P,xm-i,xp,xn) = H(P,xn,xk,xi} with (xk,x£) = (x p ,x m _i) and (5) yield that, say, xm e T. By (6), {x m _i,x p ,x n ,x m } supports T and hence the assertion follows via (3). If xm < xp then H(P, xm, xp, xn) = H(P,xn,xk,xe) with (xk,xe) = (x m ,x p ) and (4) yield that S U U C Pp = [ x i , X 2 , . . . ,xp] and that, say, x m _i 6 S. We note that {[x m _i,x m ], [XI,X P ]} C U(Pp] and so, (x T O _i,x m ,xi,x p ) = (xi,Xk,xi,xm-i) supports S. Then (x n ,Xfc,x^,x m _i) = (x m ,x p ,x n , x m _i) supports 5, and the assertion follows. D Theorem 4. Let P = Pn C E4 be totally-sewn with |V(P)| = n > 6, s(P) = max{s(O',P) | O' G intP} < 9 and t/te pentagram property. Let P = [P, x] 6e obtained by sewing x onto P, and O e intP. T/ien s(O,P) < 9. Proo/. Let x be sewn through F(E*,F*,P] with P = [xa, xfc, yi, 2/2, • • • > yn-2, 2/n-i =yi], ^* =fca,a:&],[x a ,x 6 ,yi,y i+ i] e ^"(P) for i = 1,2,... , n - 2 and F* = [x a ,x b ,y n _ 2 ,yi]. Thus, with y\ := [xa,xb,yi]/£;*, P* = P/£* is an (n — 2)-gon with the cyclically labelled vertices y^y^, • • • ^Vn-2We refer to [4] and [7] for solutions to the Separation Problem in the cyclic and semi-cyclic cases, and note that we may assume that O E int P and that there exist l
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66

y\ Figure 4 From the sewing of x and (2), we have that x is beyond exactly [xa,Xb,yi,yi+i\ for i = 1,... ,n - 3 and {[x0,x], [x,x6]} C U(P). It is easy to check that F(P}\F(P} consists of [ x , x a , X b , y i ] , [x,x a ,Xfc,y n _2] and, [x,z a ,y;,y m ] and [x,x b ,^,y i+ i] for i = 1,... ,n - 3. Then with P* = P/[x,xa], P£ = P/[x,Xb] and the self-evident notation, the indicated positions of O* in Figure 5 follow from (11).

Figure 5 For the position of O in P, we use (7), (11) and the fact that 7i(P, [x0, xj,], Uk,yi) is a pentagram. We assume the worst case scenario that none of the

SEPARATION IN TOTALLY-SEWN 4-POLYTOPES

67

clusters R, S1, T, U and V of the pentagram are empty, and as a simplification, we represent R, S and T by single elements.

Figure 6 We verify that s(0,P) < 9 by determining nine hyperplanes HI, ... :Hg in E4 such that for any F e -F(P), there is an # e (#1,... , #9} such that O £ H and jff separates F and O. We note that either x € .F or F G ^(P), and if P e .F(P) then either F C Q or FH (17 U V] = 0 or F intersects both U U F and R U 5 U T. In the case that F contains x, we refer to Figure 5 and choose H 6 {#i,#2,#3,#4} with #1 = (x,xa,yk,yi), HI = (x,xb,yk,yi), H3 = (x,xa,xb,yk) and #4 = (x,xa,xb,yi). In the case that F G ^(P), we refer to Figure 6 and choose H = H$ = (xa,Xb,yk,yt) ifP C Q, andfT e {^1,^2} if Fn(C/"UF) = 0. Let P intersect both C^UV and RuSUT. Then 1.1 yields the choice of H € {^e, #7, #8, ^9} with Jf6 = (xa,yk,yi,y'v), H7 = (xa,yk,yi,y'u), H8 = (xbjyk,ye,yv) and HQ = (xb,yk,yi,yu). If some of the clusters are empty then some of the Hi are superfluous or identical. D Since s(Po) < 9 and Pn has the pentagram property for n < 9, it follows from Theorem 4 that s(Pio) < 9. Our final task is to determine which sewings (onto a Pg or a semi-cyclic Pn) preserve the pentagram property. The following example shows that not all sewings work. Let Pg be the 4-polytope ]V|3 described in [1]. Then U(Po) — {[xs,^], [379,2:4], [xsjXe], [x7,a*8]} and Pg is totally-sewn. Let P = [Pg,.^] be obtained by sewing x through ^"([2:3, x9], [x3, x9, x 6 ,077], Pg). Then E= [x3, x] e U(P) and it is a simple exercise to check that H(P,E,X5,xs] is not a pentagram.

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BISZTRICZKY AND OLIVEROS

Theorem 5. Let P have the pentagram property, {[x0, X{,], [x&, xc]} C U(P], [x a ,Xb,x c ] C F G ^(P) andP = [P, x] 6e obtained by x through F([xa,Xb\,F, P) . Then P has the pentagram property. Proof. We recall W(P) from (2), choose an ?i(P,£',Xfc,x^) and note that we may assume that [xfc,X£,z] is not a 2-face of P for either vertex z of E1. Let E1 = [x,x a j. Then z = x and the preceding yield that x& ^ {x/^x^}, and so 7Y(P, [x a ,x b ],x f c ,X£) is a pentagram (cf. Figure 2). Since {[x,X(,], [xb,x c ]} C £/(P), it follows from (1) that neither x& and x, nor x& and xc, are separated by any H G H(P, xa,X]-,xg). Hence, either xc G T and (x a ,Xfc,X£,x) supports J RuS'UTU{xb} or x c G {xfc,X£J and 7i(P, x a ,Xfc,X£) has at most two non-empty clusters. In either case, 7i(P, E,Xk,xe) is a pentagram. Since ?i(P, [x a ,x],Xfc,X£) is a pentagram, it follows from [x, x^,] G W(P] that (3) is applicable. Hence, 7Y(P, [x,X{,],Xfc,X£) is a pentagram. Let E = [x u ,x v ] G U(P] HZY(P). By Lemma 2, we may assume that, say, Xfc = x. From above, we are assuming that [x u ,x, X(\ and [x v ,x,X£J are not 2-faces of P; that is, {xa,x;,} D {x u ,x u ,X£} = 0. Since H(P, [xu,xv],xa,X£), 7i(P, [x,x a ],x u ,X£) and ?i(P, [x, xa],xv,xe) are pentagrams, it is easy to check that 7Y(P, [x u ,x v ],x, X£J is one as well. D

REFERENCES [1] Altshuler, A. and Steinberg L., Neighborly 4-polytopes with 9 vertices, J. Combin. Theory Ser. A 15 (1973), 270-287. [2] Bezdek, K., The problem of illumination of a convex body by afnne subspaces, Mathematika 38 (1991), 362-375. [3] Bezdek, K., and Bisztriczky, T., Hadwiger's covering conjecture and low dimensional dual cyclic polytopes, Geom. Ded. 46 (1993), 276-286. [4] Bezdek, K., and Bisztriczky, T., A proof of Hadwiger's covering conjecture for dual cyclic polytopes, Geom. Ded. 68 (1997), 29-41. [5] Bisztriczky, T., Separation in neighbourly 4-polytopes, to appear. [6] Boltyanski, V., Martini, H., and Soltan, P.S., Excursions into combinatorial geometry, Springer, Berlin, 1997. [7] Finbow, W. and Oliveros, D., Separation in semi-cyclic 4-polytopes, to appear. [8] Grtinbaum, B., Convex polytopes, Interscience, New York, 1967. [9] Shemer, L, Neighborly polytopes, Israel J. Math. 43 (1982), 291-314. [10] Shemer, I., How many cyclic subpolytopes can a non-cyclic polytope have?, Israel J. Math. 49 (1984), 331-342. E-mail address: E-mail address:

[email protected]. ca '"Tiber Bisztriczky'' [email protected] ; 'Deborah Oliveros''

ON A CLASS OF EQUIFACETTED POLYTOPES

GERD BLIND Math. Institut B, University of Stuttgart, Pfaffenwaldring 57, D 70550 Stuttgart, Germany ROSWITHA BLIND Waldburgstr. 88, D 70563 Stuttgart, Germany

ABSTRACT. A convex d-polytope is called equifacetted if all its facets are combinatorially isomorphic. Let Bd be the class of those equifacetted dpolytopes where all the facets are isomorphic to the pyramid over the combinatorial (d — 2)-cube (d > 4). We determine the polytopes of Bd with minimal vertex number and we look for gaps in the vertex numbers of polytopes of Bd. It turns out that the situation is quite different for d = 4 and

I. INTRODUCTION A convex d-polytope is called equifacetted if all its facets are combinatorially isomorphic. Common examples are the simplicial and the cubical polytopes. Further classes of equifacetted polytopes (or of equifacetted spheres) have been constructed and studied under various aspects ( see e.g. [6], [7], [8], [9]). Let d > 4. We consider a class Bd of equifacetted d-polytopes "between" simplicial and cubcial polytopes, namely their facets are pyramids over combinatorial (d — 2)-cubes, so in the most interesting case d = 4, the facets are pyramids over quadrangles. Such polytopes are called b- equifacetted. Simple examples are the bipyramids over cubical (d — l)-polytopes. In dimension d = 4, further examples can be constructed as "degenerate bipyramids" as follows. Take as base. e.g.. the 3-polytope K of Figure 1, whose boundary complex is that of a cube with one 2-face divided into two triangles by an additional edge e. Let / be a segment such that I intersects 69

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the affine hull of K in a single point belonging to the relative interiors of both I and e, that is, / and e are diagonals of the quadrangle spanned by them. Then the convex hull of I and K is a 6-equifacetted polytope, which we call PQ. Note that the boundary complex of PQ is closely related to that of the bipyramid over the 3-cube: Their 1-skeletons are isomorphic, and their 2-skeletons only differ in the position of one quadrangular 2-face. An interesting problem for convex polytopes is as follows. Given a class Pd of convex d-polytopes, determine the minimal vertex number of all polytopes of Pd, and characterize the polytopes with minimal vertex number. Within the class of all polytopes, e. g., exactly the simplices have minimal vertex number, and within the class of cubical polytopes, exactly the combinatorial cube has minimal vertex number (see [I]}. In the following, combinatorially isomorphic polytopes are considered as equal. We have Theorem 1. A b-equifacetted 4-polytope has at least 10 vertices. There are precisely two b-equifacetted ^-polytopes with 10 vertices, namely the bipyramid over the 3-cube and the polytope PQ. Thus in contrast to the situation for simplicial and cubical polytopes, the minimal 5-equifacetted 4-polytope is not unique.

Figure 1: The polytopes K and L. Moreover, whereas gaps exist in the possible vertex numbers of cubical 4-polytopes, namely only even vertex numbers occur (see [2]), this is not

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71

the case for 6-equifacetted 4-polytopes. In fact, such a polytope with 11 vertices can be constructed as a degenerate bipyramid over the 3-polytope L in Figure 1, and 6-equifacetted 4-polytopes with n vertices, n > 12, can also be constructed as bipyramids or degenerate bipyramids. However, for d > 5, the 6-equifacetted d-poly topes with few vertices strongly reflect the properties of cubical polytopes. Let Cd be the d-cube and let Td be the cubical polytope arising from pasting together two d-cubes along a common facet. For d > 4, within the class of cubical polytopes, exactly Cd has minimal vertex number and exactly Td has the next larger vertex number (see [3], [4] and [5]), and this is transfered to 6-equifacetted d-polytopes: Theorem 2. For d > 5, a b-equifacetted d-polytope has at least 2d~l + 2 vertices and is then the bipyramid over Cd~l. The next larger vertex number is 2d~l + 2d~2 + 2 and occurs only for the bipyramid over Td~l. The two theorems are proved separately. For the proof of Theorem 1, we must change boundary complexes of polytopes. Such changes, however, in general do not again yield a polytope. So we must pass to 6-equifacetted 3-spheres (see Section 2.1). In fact, we prove Theorem 1 for 6-equifacetted 3-spheres. Theorem Ib. A b-equifacetted 3-sphere has at least 10 vertices. There are precisely two b-equifacetted 3-spheres with 10 vertices, namely the boundary complex of the bipyramid over the 3-cube and the boundary complex of PQ.

2. PROOF OF THEOREM IB 2.1. Vertex figures and basic figures. We start with the definition of a 6-equifacetted 3-sphere. A topological pyramid over a quadrangle is the topological image of a pyramid over a quadrangle together with its faces. A finite family P of topological pyramids over a quadrangle is called a topological b-equifacetted complex if the following conditions of a complex are satisfied. (1) Each face of a member of P is itself a member of P. (2) The intersection of any two members of P is a face of each of them. As usual, a member of P is called a face of P, and a facet is a face of dimension 3. Notions using only the inclusion of faces are defined as for usual complexes, e.g., the valence of a vertex or the isomorphism of two complexes. We consider isomorphic complexes as equal.

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P is a b-equifacetted 3-sphere if in addition to (1) and (2) it satisfies (3) set P is a topological 3-sphere. Let v be a vertex of P. We define the vertex figure P/v of P at v as follows. As usual, let star v be the complex spanned py the facets of P containing v. By (3), set(star v) is a topological 3-ball, so that its boundary is a topological 2-sphere, which is divided up by its intersections with those faces of star v containing v. The resulting family of vertices, edges and 2-faces is a complex and is called the vertex figure P/v of P at v. The fact that P is 5-equifacetted has consequences on P/v. An edge of P/v either stems from a triangular or from a quadrangular face and in the latter case it is called a D-edge. The 2-faces of P/v are quadrangles and triangles, and the triangles fall into pairs with a common D-edge. Removing all the D-edges from P/v, we obtain a net B/v on the 2-sphere, which we call basic figure. It has the property (*) The 1-skeleton of B/v is a complex, and the 2-faces of B/v are quadrangles. In order to prove Theorem Ib, let P be a 6-equifacetted 3-sphere with at most 10 vertices. Let v be a vertex of P and let n be the valence of v, so that 4 < n < 9, and n is the vertex number of P/v and B/v.Then all the possible B/v and P/v are obtained as follows; the results are collected in Table I (In the diagrams of B/v, their base is also a 2-face of B/v. In the diagrams of P/v, their base is either a qudadrangular face of P/v or consists of two triangular faces with a common D-edge, which is indicated by a broken line. The other D-edges of P/v are marked by D.). Since P has at most 9 vertices apart from v, the number of D-edges of P/v is at most 9 — n. Thus for n — 9, the complex P/v has no D-edge and so all its faces are quadrangles, which is easily seen to be impossible. Thus the case n = 9 does not occur. Next, for each n (4 < n < 8) we determine all possible B/v. By Euler's theorem, the number of quadrangles of B/v is n — 2, and then it is easily seen that B/v has a 2-valent vertex for 4 < n < 1. For n = 4, B/v consists of two quadrangles and so is uniquely detemined. For n > 5, by property (*), B/v cannot have an edge with both vertices 2-valent. Thus, if B/v has a 2-valent vertex vj, removing w and the two incident edges yields again a net on the 2-sphere satisfying (*). So for n — 5, 6, and 7, all such nets and so all possible B/v can be constructed inductively. For n = 8, B/v could also have a 2-valent vertex, but the nets so obtained cannot occur as B/v, since each 2-valent vertex of B/v must lie in a D-edge of P/v, of which P/v has at most one. Thus for n = 8, all the vertices of L>IV are o-valent, which uniquely determines B/v.

ON A CLASS OF EQUIFACETTED POLYTOPES

3 <1> S?£ x *o g
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i_

C

o> ro E ro CD 3 D -4c cr o

•° -n i

(U 0) £i en

Table 1: The possible basic figures B/v and vertex figures P/v.

73

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GERD BLIND AND ROSWITHA BLIND

From all possible B/v we immediately get all possible P/v by adding Dedges - not more than indicated in Table 1 - in such a way that a complex arises. Note that each vertex w of P/v is also a vertex of P. Our next aim is to estimate for w its valence as a vertex of P. Comparing P/v with star v we see that the non n-edges of P/v are also edges of P, that w lies in an edge containing v, and that for every quadrangle of B/v containing w and divided by a D-edge of P/v there is one more edge of P containing w. In this way, the fat drawn vertices of P/v in Table 1 have valence at least 6, 7 or 8, respectively, as stated. 2.2. The case n — 8. We will prove Lemma 2.2. // P has an S-valent vertex, then P is as stated in Theorem 16. Proof. According to Table 1, there are two possibilities for the vertex figure P/v. In the left case, v is the apex of all the facets containing v, and star v contains 9 vertices. Since P has at most 10 vertices, all the facets of P not belonging to star v but containing a quadrangular 2-face of star v have a common apex. Thus P is isomorphic to the boundary complex of a bipyramid over the 3-cube. In the right case it is shown analogously that P is isomorphic to the boundary complex of PQ, which proves Lemma 2.2. D 2.3. The case n — 4. From now on we assume that P has minimal vertex number. Then we have Lemma 2.3. P has no ^.-valent vertex. Proof. Assume to the contrary that P has a 4-valent vertex v. Then Table 1 shows that the 1-skeleton of B/v is a quadrangle; we consecutively call its vertices 91,92,9s, 94- The shape of P/v shows that v lies in two quadrangular 2-faces, and we denote their vertices opposite to v by p\ respectively p2- Then these vertices pi, j>2,
ON A CLASS OF EQUIFACETTED POLYTOPES

75

their base contains both q\ and 93 or both q
2.4. The number of 7-valent vertices. By Lemmas 2.2 and 2.3, we may assume 5 < n < 7 for the valence n of each vertex of P. Thus in Table 1, that vertex figure containing an at least 8-valent vertex is not possible, and all the other fat drawn vertices have valence at most 7. Hence P has either no or at least two 7-valent vertices. Moreover Lemma 2.4. P has either no or at least four 7-valent vertices. Proof. Note that the facets of P fall into pairs with a common quadrangular 2-face. Removing these 2-faces, we may consider each such pair of facets as a topological octahedron. The list of all (remaining) possible P/v in Table 1 shows that a vertex of P is 7-valent if and only if it lies in an edge lying in precisely 4 such octahedrons. Thus if an edge of P lies in four octahedrons, then both its vertices are 7-valent, and since a 7-valent vertex lies in only one such edge, 7-valent vertices occur as pairs. For a 7-valent vertex v, the two possible P/v lead to two possible linked complexes link v o f v , see Figure 2; note that the basic face of these diagrams is a quadrangular face / of link v. Let pi and fi (i = 1,2) be vertices and 2-faces, respectively, as indicated in Figure 2.

Figure 2: The possible linked complexes of a 7-valent vertex.

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GERD BLIND AND ROSWITHA BLIND

Now each fa and / is a base of two pyramidal facets, one with apex v, and the other with apex one of the 9 vertices of link v, namely p\ or p2, since P is a complex. By assumption, P has no 4-valent vertex, so the facet with base fi must have apexp^. It follows that pi or p2 is 7-valent. Thus if P has a 7-valent vertex, it has at least three of them and by the first part of the proof at least four. D

2.5. Completing the proof. As in the proof of Lemma 2.4, noticing that the facets of P fall into pairs with a common quadrangular 2-face, we may consider each such pair as a topological octahedron. Letting x be the number of such octahedra, we determine the number of incidences of vertices of P and octahedra. On the one hand, since an octahedron has 6 vertices, this number is 6x. On the other hand, note that the number of octahedra incident with an n-valent vertex v equals the number of 2-faces of -B/t>, which is n — 2 (see Table 1). Letting vn be the number of n-valent vertices of P (n = 5,6, 7), we thus have (1)

3^5 + 4t>6 + 5i>7 = 6x

Moreover, x is the number of quadrangular 2-faces of P. The list of all (remaining) possible P/v in Table I shows that every 6-valent vertex is the apex of at least one facet of P and every 7-valent vertex is the apex of precisely three facets. Hence (2)

w6 + 3w7 < 2x

According to Lemma 2.4 we now distinguish the cases ^7 = 0 and v? > 4. In case v-j = 0, Table 1 shows that the vertex figures of the 5- and 6-valent vertices are unique and have one respectively two quadrangular faces. Thus instead of (2) we have the equality (3)

v5 + 2v6 = 2x,

which together with (1) yields VQ = 0. So all the vertices of P are 5-valent, which is impossible because of the fat vertex of the corresponding vertex figure. In case v-j > 4, since for a 7-valent vertex v the complex star v has 10 vertices, P has precisely 10 vertices. Hence (4)

v5 + v& + v7 = 10

ON A CLASS OF EQUIFACETTED POLYTOPES

77

Subtracting (2) from (1) and taking into account (4) yields v-j < 30 — 4rr. Together with v-j > 4 this yields 4x < 26 or x < 6. On the other hand, because of (4) and v? > 4, the left side of (1) takes its minimum for vj = 4, VQ = 0 and ^5 = 6, so that 38 < 6x or x > 7, which is a contradiction, so that the case V7 > 4 is also not possible, which proves Theorem Ib.

3. PROOF OF THEOREM 2 Let d > 5, let P be a 6-equifacetted d-polytope and let mo := 2d~l +2 d ~ 2 . Assuming that P has at most mo + 2 vertices we must show that P is the bipyramid over Cd~l or Td~l. Let F be a facet of P and let v be the apex of F. Then v is also a vertex of P. It is easy to see that the facets of P containing v can be labelled F = Fi,...,Fk so that each F;,2 < i < k, has a common (d-2)-face with some Fj, j < i. In particular, the facets F — F\ and FI have a common (d-2)-face containing v, which as face of F is a pyramid with apex v over a (d-S)-cube, and because of d > 5 it follows that v is also the apex of F^. Iterating this argument shows that v is the apex of every Fj-,1 < i < k. Hence the linked complex link v of v is a cubical (d-2)-complex. Thus, by [3], [4] and [5], link v is either the boundary complex dCd~l of some Cd~l, or it is isomorphic to the boundary complex dTd~l of some Trf-1, or it has more than mo vertices. Let / be a (d — 2)-face of link v. Then / lies in two facets of dP, one with apex v and the other with some apex w ^ v. The above argument shows that link w is also a cubical (d — 2)-complex , and that w is not a vertex of link v and v is not a vertex of link w. Thus P has as vertices the vertices of link v and of link w;, which may partly coincide, and at least two additional vertices v and w. Moreover, if link v has a (d — 2)-face not belonging to link w, these arguments also show that then P must have even one more vertex. Since P is assumed to have at most mo + 2 vertices, it follows that both link v and link w have at most mo vertices and so are isomorphic to dCd~l or dTd~l. We first consider the case that link v is isomorphic to dTd~l and so has mo vertices. By the above, since P is assumed to have at most mo + 2 vertices, every (d — 2)-face of link v also belongs to link w, so that link v = link w, and P is the bipyramid over some Td~l, as claimed. It remains the case that link v is some dCd~l, and without loss of generality we may assume the same for link w. Thus link v spans a hyperplane Hv and the vertices of link v are the vertices of the polytope Hv fl P. In the same way, link w spans a hyperplane Hw and the vertices of link w are the vertices of the polytope Hw D P. Recall that link v and link w have

78

GERD BLIND AND ROSWITHA BLIND

a common (d — 2)-face /. Thus, if link v and link w have more common vertices than those of /, then Hv — Hw and P is the bipyramid over some Cd~l, as claimed. However, if link v and link w have only the vertices of / in common, then link v and link w together have mo vertices and, by the above, P has at least mo + 3 vertices, which contradicts our assumption on P.

4. REMARK It can be shown that all 6-equifacetted 3-spheres with 11 vertices arise from those with 10 vertices by a simple operation, namely the inverse operation to that used in the proof of Lemma 2.3

REFERENCES [1] G. Blind and R. Blind, Convex polytopes without triangular faces, Israel J. Math. 71 (1990), 129 - 134. [2] G. Blind and R. Blind , Gaps in the numbers of vertices of cubical polytopes I, Discrete Comput. Geom. 11 (1994), 351 - 356. [3] G. Blind and R. Blind, Cubical d-polytopes with few vertices for d > 4, Geom. Dedicata 65 (1997), 247 - 255. [4] G. Blind and R. Blind, Cubical 4-polytopes with few vertices, Geom. Dedicata 66 (1997), 223 - 231. [5] G. Blind and R. Blind, The almost simple cubical polytopes, Discrete Math. 184 (1998), 25 - 48. [6] J. Bokowski and P. Schuchert, Equifacetted 3-spheres as topes of nonpolytopal matroid polytopes, Discrete Comput. Geom. 13 (1995), 347 361. [7] J. Bokowski, On the construction of equifacetted 3-spheres. Invariant methods in discrete and computational geometry (Curagao, 1994), 301 - 312, Kluwer Acad. Publ. Dordrecht, 1995. [8] J. Bokowski, P. Cara and S. Mock, On a self dual 3-sphere of Peter McMullen, Discrete geometry and rigidity (Budapest 1999). Period. Math. Hungar. 39 (1999), 17 - 32. [9] M. A. Perles and G. C. Shephard, facets and nonfacets of convex polytopes, Acta Math. 119 (1967), 113 - 145. E-mail address: blindQmathematik.uni-stuttgart.de ''Gerd Blind'' E-mail address: Roswitha.Blind9spd-online.de ''Roswitha Blind''

CHESSBOARD RAMSEY NUMBERS

JENS-P. BODE Abteilung fur Diskrete Mathematik, Technische Universitat Braunschweig, 38023 Braunschweig, Germany HEIKO HARBORTH Abteilung fur Diskrete Mathematik, Technische Universitat Braunschweig, 38023 Braunschweig, Germany STEFAN KRAUSE Abteilung fur Mathematische Optimierung, Technische Universitat Braunschweig, 38023 Braunschweig, Germany

ABSTRACT. The chessboard Ramsey number r — r(G,H) is the minimum number r such that every two-coloring (green and red) of the sides of the squares of the rxr-chessboard contains a green G or a red H. The existence is possible only for G being a path P$ or P± and for H being a certain tree. The exact values of r(P 3 ,Pfc) and the values for r(P 3 ,#) and r(P±,H] for all small graphs H are determined and the existence of r(P±,Pk) is proved.

1. INTRODUCTION The n x n-chessboard Bn many times has been the object of problems in combinatorial geometry. Here we will consider the set of sides of the squares of Bn. Then we may ask, for example, for the largest n for which it is possible to remove vertex disjoint edges from Bn such that no path of given length remains. More general, we partition the set of all sides into two color classes and then ask for the existence of certain monochromatic configurations. If Bn is interpreted as the graph Bn with the vertex points of the squares as vertices, and with the sides as edges then we will ask for the minimum number r = r(G,H] such that every two-coloring (green and red) of the edges of BT contains a given subgraph G in green or a given subgraph H in red. That is, we ask for a Ramsey like number where instead of the complete graphs Kn the chessboard graphs Bn serve as 'host graphs'. In 79

BODE, HARBORTH, AND KRAUSE

80

the literature, besides the classical case of Kn [4], for example, complete bipartite graphs [1], cube graphs [2] and octahedron graphs [3] are discussed as other sequences of host graphs.

2. EXISTENCE For the existence of r(G,H) it is sufficient to consider connected graphs G and H, both having at least two edges. A coloring of Bn is called critical if it neither contains a green G nor a red H. In Figures 1 to 4 critical colorings are shown, which can be extended to Bn in general. The chessboard Ramsey number r(P3,Cfc) does not exist, where Pi denotes a path with i vertices and Cj a cycle of length j. If k is odd, Bn does not contain Ck- Otherwise, if k = 2a -f 2 then we have the following critical colorings of Bn. All vertical edges are colored red. In every row the horizontal edges are alternating green and red such that in the columns always exactly a consecutive edges are of the same color (see Figure 1 for n — 5 and a — 2). In this coloring any red Cj enclosing a green edge implies j > 2a + 4 and any other red Cj implies j < la. Hence, a red Ck = C*2a+2 does not occur.

Figure 1. r(P3,C6] does not exist.

Figure 2. r(G7H) exists only if G C C^ or H C C4.

The coloring of vertex disjoint green and red squares (see Figure 2 for n — 5) proves that either G or H has to be a subgraph of €4 if r(G,H) exists. In Figure 3 the vertical edges of Bn are colored red and the horizontal edges alternating green and red, starting with green in the first column. This is a critical coloring of any Bn for r(Pz, H) if H is not a subgraph of L in Figure 5. Critical colorings of any Bn as in Figure 4 for n = 7 prove that r(P±,H} cannot exist unless H is a subgraph of Z in Figure 6. In these colorings the top row of horizontal edges is periodically colored red-green-red-green-redred starting at the left upper vertex. Then in the following rows of horizontal

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CHESSBOARD RAMSEY NUMBERS

Figure 3. r(P 3 ,#) exists only if H is a subgraph of L.

Figure 4. r(P^^H] exists only if H is a subgraph of Z.

Figure 5. Subgraph L.

Figure 6. Subgraph Z.

edges the periods row by row are shifted two edges to the right. The first two columns of vertical edges are periodically green-red-red. And the periods every second column are shifted down by one. The critical colorings as in Figures 1 to 4 imply the following theorem. Theorem 1. For the existence of the chessboard Ramsey number it is necessary that one graph is PS and the other one is a subtree of L (Figure 5), or one graph is P± and the other one is a subtree of Z (Figure 6). 3. PATHS For paths Pk we determine the exact values of r(Ps,Pfc). Theorem 2.

{

2 3 ff 1 - 1

for k = 3,4,5, for k = 6, for k > 7.

Proof. It suffices to prove the asserted values as lower bounds for k = 3,6, and k = 1i -\- 1, i > 4 and as upper bounds for k = 5 and k = 2i + 2, i > 3. The lower bounds for k = 3 and 6 follow from Figures 7 and 8. The colorings of P»;_i, i > 4, as in Figure 3 for i = 8, neither contain a green Pa nor a red P?i+i so that r(Ps,P2i+i) > i. For the upper bound in the case k = 5 every two-coloring of P>2 without ?. green PS contains a red PS of two horizontal or two vertical edges inside B
BODE, HARBORTH, AND KRAUSE

82

Figure 7. r(P3,P3) > 2.

Figure 8. r(P3,P6)>3.

For k = 1i -\- 2, i > 3, we prove that any two-coloring of B{ without a green P^ contains a red P^i+2- We distinguish the two cases that the red subgraph is connected or not. If the red subgraph is connected then there exists a path from the lower left to the upper right corner of B{ having at least 2i edges. If this path has more than 1i edges then we have a red Pii+i- It remains that every such path has exactly 2z edges which then only uses edges upwards and rightwards. By symmetry we may assume that the first edge of this path is upwards. Then we consider the first horizontal red edge from the bottom in the leftmost column. If this edge does not belong to the path then, using this edge, there is a path with 2i + 1 red edges from the second vertex in the lowest row of vertices to the vertex in the upper right corner (see Figure 9). Otherwise, we obtain a path of 2z + 1 red edges starting at the fourth vertex in the last row or at the third vertex in the second last row of vertices as indicated in Figure 10. This is not possible only if the original path uses the third vertex in the second last row and the third horizontal edge in the last row is green. Then, however, we use a path surrounding the second square in the last row of squares and this path has 1i -\- 2 red edges (see Figure 11).

Figure 9.

Figure 10.

Figure 11.

If the red subgraph is not connected then we can assume by symmetry that all horizontal edges of one column are green since any green edge separating two red components force both parallel neighboring edges within this column to be also green (see Figure 12). We consider one of these columns of separa.rting green oHges and the largest of the two remaining parts of -0Z, say the right one, having a width of at least l^} edges. At the left side of this part we have a red path Pi+s of i

CHESSBOARD RAMSEY NUMBERS

83

vertical red edges and at both ends one horizontal red edge. We can extend this P;+s at both ends by red edges either rightwards or upwards and either rightwards or downwards, respectively. If at each vertex vertical edges are prefered then both extensions of P;+s may meet each other to close a red cycle of at least 2i + 2 red edges, that is, a red P2i+2 exists. This is always the case if another column of separating green edges exists in the chosen part of Bi. Otherwise both extensions end on the right border of P>;. If these extensions use together at least two vertical red edges then we have at least i + 2 [^] + 2, that is, at least 2i + 1 red edges, and a red P2i+2 exists. If at most one vertical red edge occurs then one extension, say the lower one, has to end in a vertex at the corner of Bi. In this case the dashed edges in Figure 13 can be used to obtain a path having the two desired additional red edges.

Figure 12.

Figure 13.

For paths Pj. we prove the existence of r(P^^Pk). Theorem 3. The chessboard Ramsey numbers r(P^Pk) do exist. Proof. A sufficiently large two-colored board Bn without a green Pj and without a red P/- contains green and red components. We choose a red component having the largest geometrical diameter d (which is less than A;). This red component is enclosed by a separating set of green components. This separating set determines a separating curve C through edge-to-edge squares of Bn such that only green edges are intersected. No square can contain a starting point of three separating curves since three green edges of the square determine a green PJ. Thus the red edges outside C and adjacent to the edges of the separating set of green components belong to the same red component. Since Bn is large enough this red component has a diameter larger than d, a contradiction. 4. EXACT VALUES FOR SMALL GRAPHS Besides the exact values of r(Pa, Pjt) and the existence of r(P±, Pk) we used a computer to determine all values r(Ps, H) and r(P4, H] for subtrees H of the

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graphs L and Z, respectively, with up to 8 vertices for H (see Table 1). In this case, both sets of subtrees are identical. In general, we do not know how to characterize the sets of subtrees of L and Z and whether the chessboard Ramsey numbers exist for all these subtrees.

G

H

ft

ft

G

H

r

i r

4

Table 1. Values of r(G,H) for G = P3, G = ft, and all H with up to 8 vertices.

REFERENCES [1] F.Harary, H.Harborth, and L Mengersen: Generalized Ramsey theory for graphs. XII. Bipartite Ramsey sets. Glasgow Math. J. 22 (1981), 31-41. [2] A.Bencziir, H.Harborth, and L. K. J0rgensen: Cube graph Ramsey numbers. Abh. Braunschweig. Wiss. Ges. 47 (1996), 151-157. [3] H.Harborth and I. Mengersen: Ramsey numbers in octahedron graphs. Discrete Math. 231 (2001), 241-246. [4] S. Radziszowski: Small Ramsey numbers. Electronic Journal of Combinatorics (Dynamic surveys), www.combinatorics.org E-mail address: E-mail address: E-mail address:

jp.bode9tu-bs.de "Jens-P. Bode" h.harborthfitu-bs.de "Heiko Harborth" stefan.krause9tu-bs.de "Stefan Krause"

MAXIMAL PRIMITIVE FIXING SYSTEMS FOR CONVEX FIGURES

VLADIMIR BOLTYANSKI Centre de Investigation en Matematicas, A.P. 402, 36000 Guanajuato, GTO, MEXICO. HERNAN GONZALEZ-AGUILAR Centre de Investigation en Matematicas, A.P. 402, 36000 Guanajuato, GTO, MEXICO.

ABSTRACT. We give here a complete classification of compact, convex figures with respect to the maximal cardinality of their fixing systems. This classification generalizes and makes more precise the results obtained by L. Fejes Toth [9], B. Tomor [12], and S. Fudali [10]. Some examples illustrate the text.

1. INTRODUCTION n

Let M C M be a compact, convex body and F be a subset of its boundary bdM. (We use the term body for a convex set if its interior in Rn is nonempty.) A nonzero vector v 6 W1 is said to move the interior intM of the body M off the set F if for every translation of M in the direction of v the interior of the translated body has no point in common with F, i.e., (Xv + intM) n F = 0 for every

A > 0.

Definition 1. Let M C Mn be a compact, convex body. A set F C bdM is a fixing system for the body M if there is no nonzero vector that moves the interior of M off the set F. The fixing system is primitive if no proper subset is a fixing system for M. In other words, F is a fixing system for M if for every vector v ^ 0 there exists a number A > 0 such that (Xv + intM) n F ^ 0. Visually, F is a fixing 85

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GONZALEZ-AGUILAR

system if, drifting 'nails without friction' at all points of F, we eliminate the possibility to translate M in any direction. The concepts of fixing system and primitive fixing system were introduced by famous Hungarian mathematician Laslo Fejes Toth [9]. He established (cf. also [12]) that for every compact, convex body M C K2 the maximal cardinality £> max M of its primitive fixing systems satisfies the inequality £> max M < 6, the equality being attained if M is a hexagon with pairwise parallel opposite sides (Fig. 1). Furthermore, S. Fudali [10] showed that these hexagons are the only planar figures for which the indicated upper bound is attained. Evidently, every minimal fixing system F for the body M C Mn (i.e., a fixing system of the least cardinality) is primitive. B. Griinbaum [11] proved that the cardinality £ m i n M of minimal fixing system for M satisfies the inequalities The lower bound is attained, in particular, for n-dimensional ball (see Example 3 below), the upper one being attained for n-dimensional parallelotope (Example 1). In this connection, Griinbaum formulated the following problem: To prove that the upper bound is attained only for n-dimensional parallelotopes.

Fig. 1 Fig. 2 Being based on the result of Fejes Toth, L. Danzer [8] conjectured that for any three-dimensional compact, convex body M the maximal cardinality of primitive fixing system is not greater than 14 where the upper bound is attained for rhombic dodecahedron (i.e., a three-dimensional zonotope analogous to the regular hexagon, Fig. 2). But this conjecture was disproved by B. Bollobas [1] who constructed for any integer k > 4 a three-dimensional compact, convex body M\~ with £> max A/& = k. A simpler construction was offered in [4] (see Example 4 below). To formulate next results, we recall the definition of the functional m,d introduced in [2]. Let M C Mn be a compact, convex body. By mdM denote the greatest integer m such that the unit outward normals po,pi, ...,pm of

MAXIMAL PRIMITIVE FIXING SYSTEMS

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M at some regular boundary points ao,ai,...,a m are minimally dependent, i.e., (i) there are positive numbers Ao,Ai,...,A m such that \opo + ^ipi + ... + Xmpm = Q(M) any m of the vectors po,pi, ...,pm are linearly independent. In other words, the vectors PQ,PI, ...,pm (emanating from the origin 0 G Kn) are the vertices of an m-dimensional simplex that contains 0 in its relative interior. Remark that 1 < mdM < n for any compact, convex body M C Mn and mdM = 1 if and only if M is an n-dimensional parallelotope. Using the functional mdM, in articles [3] and [6] (cf. also Section 44 in monograph [5]) the following estimates were proved: nH

—- < g . M < 2n + 1 - mdM. mdM ~ min The upper one contains the positive solution of the above Griinbaum's problem. Indeed, if £ min M = n, then mdM = 1, i.e., M is an n-dimensional parallelotope. For n > 3 and 2 < m < n there are compact, convex bodies M C Kn with mdM = m and any large (integer) cardinality of primitive fixing system for M (this was established in [4], see Example 4 below). Even if mdM = 2, the following exact upper estimate is established in [7]: 9 £max^ < ^n + an

where an = 0 an = -3

if n = 0(mod4), if n = 2(mod4),

an = —-

if n = 1 (mod 4),

an =

if n = 3(mod4).

2. MAIN RESULTS In this article we give a complete classification of compact, convex planar figures with respect to the cardinality of their maximal primitive fixing systems. Definition 2. Let M C Mn be a compact, convex body. Boundary points ai,a2 of M are antipodal if there are two parallel (and distinct) support hyperplanes Fi,F2 of M with ai G FI and a 2 G F2. The points ai,a2 are strictly antipodal if the hyperplanes can be chosen in such a way that every hyperplane Fi,F2 has only one point in common with M. The antipodal points £-..02 are semistrictly antipodal if they are not strictly antipodal, but the above support hyperplanes Fi,F2 can be chosen in such a way that one of them has only one point in common with M.

V. BOLTYANSKI AND H.

Fig.

GONZALEZ-AGUILAR

4

In Fig 3, the points a, 6 £ bdM are strictly antipodal. Furthermore, the boundary points a,c are semistrictly antipodal. Finally, the points 6,c are antipodal, not being neither strictly antipodal nor semistrictly antipodal. Definition 3. Let M C En be a compact, convex body and v ^ 0 be a vector. A boundary point a of M is illuminated by the direction of the vector v (or, for brevity, a is illuminated by the vector v) if a+ Xv 6 int M for A > 0 small enough. In Fig. 4 the point a is illuminated by u, whereas 6 and c are not illuminated. Theorem 1. A set F C bd M is a fixing system for a compact, convex body M C Mn if and only if each direction illuminates at least one point of the set F. This Theorem is known (see Theorem 44.1 in [5]). The following three Theorems contain a complete classification of compact, convex planar figures with respect to £> max M. More detailed, Theorems 2, 3, and 4 give a necessary and sufficient condition under which the equality £> max M = k holds for k = 6, 5 and 3, respectively. In other cases £ max M = 4. Theorem 2 (S. Fudali [10]). Let M C K2 be a compact, convex body. The equality £> max M = 6 holds if and only if M is a hexagon with pairwise parallel opposite sides. Before to formulate the next Theorem, we say several words about antipodal vertices of convex pentagons. Definition 4. A convex pentagon P is flattened if there are t.\vo its neighboring vertices which are antipodal points of P.

MAXIMAL PRIMITIVE FIXING SYSTEMS

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d

Fig. 5

Fig. 6

In Fig. 5 the convex pentagon P has two neighboring vertices a, b for which the sum of corresponding interior angles is less than TT; these points are strictly antipodal, i.e., P is flattened. In Fig. 6 the vertices a and b have the sum TT of the interior angles; these points are antipodal (not being strictly antipodal), and hence P is flattened. Even if for every two neighboring vertices of P the sum of corresponding interior angles is greater than TT, then no two neighboring vertices of P are antipodal, i.e., P is non-flattened. Theorem 3. Let M C R2 be a compact, convex body with £ max Af < 6. The equality £>max M = 5 holds if and only if there is a non-flattened convex pentagon P inscribed in M such that every two non- neighboring vertices of P are antipodal points of the body M. The requirement £ max Af < 6 in Theorem 3 is illustrated by Example 5. Theorem 4. Let M C R2 be a compact, convex body. The equality £ max M = 3 holds if and only if M satisfies at least one of the following two conditions: 1) there exist three boundary points of M every two of which are strictly or semistrictly antipodal points of M; 2) there exists a segment [a, b] C bdM such that a and b are strictly or semistrictly antipodal points of M . The following two Theorems give £ max Af for two interesting classes of planar figures. Theorem 5. Let M be a figure of constant width h. Then 5 Theorem 6. Let

if M is the Realeaux triangle, if M is a Realeaux pentagon, in all other cases.

be the regular polygon with k vertices. Then 3 if k = 3, 4 if k = 4.7,9 or k > 10, 5 if fc = 5, 8, 10, 6 */ Jfc = 6.

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Example 1. Let M C M be an n-dimensional parallelotope. Choosing one relatively interior point in every facet of M, we obtain a fixing system F for M that consists of 2n points. This fixing system is primitive, and every primitive fixing system for M has that form. Thus QminM = Qm^M = In. Example 2. Let M C R2 be a hexagon with pairwise parallel opposite sides (Fig. 1), and F be its vertex set. Each direction illuminates at least one vertex of M. Hence (by Theorem 1) F is a fixing system for M. This fixing system is primitive. Moreover, F is the only primitive fixing system for M consisting of six points. Example 3. Let M C Mn be a compact, convex body that is smooth] this means that every boundary point a of M is regular, i.e., there is only one support hyperplane of M through a. Furthermore, let po?Pi>-",Pn be minimally dependent unit vectors in Rn. Denote by T C Mn the circumscribed n-dimensional simplex of M with outward normals pQ,pi,...,pn of its facets. Let ao,eti, ...,a n be common points of the facets of T with M (if a facet has more than one point in common with M, we take an arbitrary one of them). Each direction illuminates at least one facet of T, and hence it illuminates at least one of the points ao, ai, ...,a n . Theorem 1 shows that F = {ao,ai,...,a n } is a fixing system for M. This fixing system consists of n + 1 points. Consequently, by Griinbaum's estimates, this fixing system has the minimal cardinality, i.e., £> min M = n + 1. Thus every smooth, compact, convex body M C Mn (in particular, every n-dimensional ball) satisfies Grimbaum's lower estimate. At the same time £» max M = In. Indeed, let P be a circumscribed parallelotope for M and Z/ z -, i = 1, ...,2n, be the facets of P. For every i — 1, ...,2n choose a point a; 6 L{ fl bdM. We obtain a set F = {ai,...,«2n} C bdM. Every nonzero vector illuminates at least one facet of P, and hence it illuminates at least one of the points ai,...,a2 n - Thus F is a fixing system for M. This system is primitive. Consequently £> max M > In. Let now F' = {61,..., 6^} be an arbitrary primitive fixing system for M and Pi, ...,pk be unit outward normals of M at the points 61, ...,6fc, respectively. Then {pi,...,pk} is a positive basis of Rn, i.e., every vector is representable as a linear combination of the vectors pi, ...,pk with nonnegative coefficients (otherwise there exists a vector that doesn't illuminate any point of F'). But every primitive positive basis of Mn consists of no more than 2n vectors (cf., for example, [11]), i.e., k < 2n, and hence £ma][M = In.

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Fig. 7 Example 4. Let M = conv(B U {a}) C M3 where B is a ball and a is its exterior point. In [4] the equality £ max M = oo is proved; this means that for any integer k > 4 there is a primitive fixing system for M consisting of k points. Moreover [4], for every A; > 4 there exists a zonotope Z^ C R3 with £ max Zfc > k. These examples can be generalized for greater dimensions. In particular, for every n > m > 3 there is a compact, convex body M C Rn with mdM = m and £max.M = oo.

Fig. 10

Fig. 9 2

Example 5. Let M C R be a regular hexagon with vertices a, 6, c, d, e, f (in cyclic order). Denote by p a relatively interior point of the segment [e,/]. Then P = conv{a,6, c,cf,p} is a non-flattened convex pentagon inscribed in M and satisfying condition of Theorem 3. Nevertheless, Qma,xM = 6, i.e., conclusion of Theorem 3 doesn't hold. This shows that the condition £ max M < 6 in Theorem 3 is essential. Let now II be a closed half-plane with {a, 6, c,d, e,p} C intll and / ^ II (Fig. 9). Then M' = M n II is a compact, convex body in R2 with Qm^M' = 5. Indeed, P is an inscribed pentagon for M' satisfying condition of Theorem 3 and, moreover, 0 max M' < 6 (by Theorem 2). Example 6. Let M be the Realeaux triangle (Fig. 7). Every two of the vertices a, 6, c are strictly antipodal points of M. By Theorem 4, £ max M = 3.

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If II is a closed half-plane containing a, 6, c in its interior, but not containing M (Fig. 10), then M' = M n II is another figure with £ max M' = 3 (again by Theorem 4). However, if we cut the point c by a line parallel to belli (Fig.11), then we obtain a figure N with 6m^N = 4 (see Lemma 4 below or use Theorems 2, 3, and 4). Example 7. Let N C R2 be a semicircle with diameter [a,m]. Then ^max-^ — 3 by Theorem 4. Let now b be a relatively interior point of the segment [a,m] and / be the line through b that is perpendicular to [a, raj. Denote by II the closed half-plane with bdll = / that contains a, and let M — TV D II. Then £> max M = 3 (again by Theorem 4), since the segment [a, b] is contained in bdAf and a, b are semistrictly antipodal boundary points of M.

M Fig. 11

Fig. 12 2

Example 8. For the body M C M in Fig. 12 we have £ max Af = 5. Indeed, the inscribed pentagon P with the vertices a, 6, c, d, e satisfies the condition of Theorem 3.

4. AUXILIARY PROPOSITIONS In the proofs below we assume that the plane R2 is Euclidean, i.e., a scalar product is defined in R2. In particular, it is possible to consider the length /(/) of any segment / C R2 and the angle a between any two nonzero vectors in R2 (with 0 < a < TT). Lemma 1. Let M C R2 be a compact, convex body and F be a primitive fixing system for M. Then F is the vertex set of an inscribed polygon P of M, i.e., no three points of F are situated in the same line. Moreover, if p and q are non-neighboring vertices of P, then they are antipodal points of M. Proof. Assume that three points a, 6, c € F are situated in / n bdM where / is a support line of M. Let b be situated between a and c in /. If a vector e ^ 0 illuminates the point a, then it illuminates b. In other words, there is no direction that illuminates only the point a £ F, contradicting the

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primitivity of the fixing system F. Thus no three points of F are situated in a line. Let p and q be non-neighboring vertices of P. They divide bdM into two open arcs A and B. There exists a vertex a of P situated in A (since p and q are not neighboring). Denote by g and h the rays emanating from p and <7, respectively, and touching B. Let g and h be their opposite rays. If g and h have a common point (Fig. 13), then it is impossible to illuminate only the point a £ F, contradicting the primitivity of the fixing system F. Thus g n h = 0, i.e., the arc B is situated in a strip with parallel sides through p and
K

Fig. 13

Fig. 14

Lemma 2. Let G C K2 be a strip with parallel sides K\,K-2. Let, further, a G KI, 6 G KI\ c G G, e G intG where c and e are situated in an open half-plane with the boundary line ab. Then there is no strip H C M2 with parallel sides L\,L-2 such that c G L\,e G L?, and a,6 G #• Proof. Introduce coordinate system x,y with a as the origin, JiTi as the x-axis, and the ray emanating from a through b as positive y-axis. We may suppose that c and e have positive abscissa (Fig. 14). Assume that the strip H exists, and denote by v a vector with positive abscissa that is parallel to LI,LI- Since a G H, ordinate yv of the vector v is nonnegative. Similarly yv is nonpositive, since b G H- Hence yv = 0. But in this case the strip H cannot contain the points a and 6, since e G intG. • Lemma 3. Let M C K.2 be a compact, convex body with £ max M = 5 and F = {a, 6, c, d, e} be a primitive fixing system for M, the points being situated in cyclic order. Then the pentagon P = convF is non-flattened. Proof. Assume that two neighboring vertices of P are antipodal points, i.e., P has the form as in Figs. 5 or 6. In the case of Fig. 5, there are three vertices of P inside a strip with the sides through a and b. respectively. In this case, by Lemma 2, c and e cannot be antipodal points of M, contradicting Lemma 1. In the case of Fig. 6, since c and e are antipodal (by Lemma

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1), M is situated in the strip with the boundary lines be and ae. Hence the vector v — c — b doesn't illuminate any point of _F, contradicting Theorem 1. • Let M C M2 be a compact, convex body and a be an its non-regular boundary point. Denote by / i , / 2 the rays which emanate from a and touch M (Fig. 15). The angle between the rays is the interior angle of M at the point a. The adjacent angle is the exterior angle of M at the point a. If a is the value of the interior angle at a, then the exterior angle has the value (3 = TT — a. It is known that the set of all non-regular boundary points of M is at most countable, and the sum of values of corresponding exterior angles is not greater than 2?r. m2

Fig. 15

Fig. 16 2

Lemma 4. Let M C R be a compact, convex body. Assume that there are two parallel segments /i, /2 in its boundary with 0 < /(/i) < ^(^2)- If there exists a chord I of M parallel to I\ and I l(Ii), then £> max M > 4. Proof. Let [a, b] be a chord of M that is parallel to I\ and has the maximal length (among the parallel chords). Then there are two parallel support lines m\,m<2 of M passing through a and 6, respectively (Fig. 16). Let c and d be relatively interior points of I\ and /2, respectively. Then F = {a,6,c, d} is a primitive fixing system for M, and hence £> max M > 4. • Lemma 5. Let M C R2 be a compact, convex body. Assume that there exists a segment [a, 6] C bdM such that a and b are strictly or semistrictly antipodal points of M. Then £> max Af = 3. Proof. Let / and m be parallel support lines of M passing through a and 6, respectively. We may suppose that / has only the point a in common with M (since a and b are strictly or semistrictly antipodal points of M). Denote by p the unit vector that is parallel to I and illuminates the relatively interior points of the segment [a, 6] (Fig. 17). Furthermore, let n be the support line of M that is parallel to ab. Denote by q the unit vector such that b = a + Xq with A > 0. Denote by g C bdM the open arc consisting of all points

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which are illuminated by the vector q. The endpoints of g are a and c G n. Analogously, denote by h C bdM the open arc consisting of all points which are illuminated by the vector — q. The endpoints of h are b and d £ n. It is possible that the points c and d coincide. Let now F be a primitive fixing system for M. Then there is a relatively interior point x of the segment [a, b] that belongs to F (otherwise the vector p doesn't illuminate any point of F}. Furthermore, there is a point y 6 F that belongs to the open arc g (otherwise the vector q doesn't illuminate any point of F}. Analogously, there is a point z £ F D h. We show that the points x,y,z form a fixing system for M. Indeed, if a nonzero vector is situated between p and q or between p and —
Fig. 18 Lemma 6. Let M C K2 be a compact, convex body. If there are boundary points a,6,c of M every two of which are strictly or semistrictly antipodal, then 0 max M = 3. Proof. Denote by ia& the open arc of bdM with the endpoints a and b which doesn't contain c. The open arcs tj,c and tca are denned analogously. If at least one of the segments [a, 6], [6,c][c,a] is contained in bdM, then £ max M = 3 by Lemma 5. Thus we may suppose that none of the segments is contained in bdM. Since the points a and 6 are strictly or semistrictly antipodal, there are two parallel support lines lajb of M with a £ /a, b £ /j, such that at least one of the lines / a ,/& has only one point in common with M. By assumption, c ^ /a, c £ /&. Let vc be the unit vector that is parallel to

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lajb and illuminates the arc i a &. Then the opposite vector —vc illuminates the point c and at least one of the arcs tic,tca (more detailed, if the line la contains only the point a in common with M, then — vc illuminates the arc £ ca , and analogously for the arc i& c ). Similarly we define the vectors va and Vh (Fig. 18). Thus each of the vectors va,Vb,vc, — va, — Vj,, — vc illuminates at least one of the arcs £ a &,£& c , tca. If a vector p ^ 0 is situated strictly between va and — fc, then p illuminates the arc 2&c (and similarly for vectors situated in the interiors of the other five angles). Thus every nonzero vector illuminates at least one of the arcs tab-i thai tea-

Let now F be a primitive fixing system for M. Since the vector va illuminates only the points of the open arc 2& c , the set tbc fi F is nonempty. We choose an arbitrary point a of this set. Similarly, choose points b 6 tca D F and c € tat, n F. Then the points a, 6, c form a fixing system for M (since every nonzero vector illuminates at least one of the arcs 2 a &?4c7*ca and hence illuminates at least one of the points a,6, c). Since F is primitive, it coincides with {a,6,c}. Thus every primitive fixing system for M consists of three points, i.e., £> max M = 3. •

5. PROOF OF MAIN RESULTS Proof of the Fejes Toth estimate. Assume that a primitive fixing system F C bdM consists of seven (or more) points, and let ai,...,a7 be successive vertices of the convex polygon P = convF (Fig. 19). By Lemma 1, ai and 03 are antipodal points of M. Let G D M be a strip with parallel sides Ki^K^ such that a\ € A'i,a-3 G KI- The point 02 is situated in an open half-plane HI with the boundary line 0,1013, and 04, 05, ae, 07 are situated in the opposite open half-plane 112. Maybe 0,7 £ K\ and 0,4 € A'2, but the points 05,05 are situated in the open set 11% n mtG. Then 04 and ag are non-neighboring vertices situated in 112 and, by Lemma 1, they are antipodal points of M. This means that there is a strip H D M with boundary lines through 0,4 and ae, respectively, contradicting Lemma 2. This contradiction shows that £ m a x M < 6. • Proof of Theorem 2. Let £ max M = 6. Assume that F is a primitive fixing system for M consisting of six points a\, a^, cts, 04, as, fle m cyclic order. Let G D M be a strip with the boundary lines K\ and KI through the antipodal points 01,03. If two of the points 04,05,05 are situated in the interior of G, we obtain the contradiction, as in the previous proof. Hence ric. and n-4 should be situated in the lines K\,K
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i.e., M = P is a hexagon with parallel opposite sides (the primitive fixing system F being the vertex set of the body M). Conversely, if M is a hexagon with pairwise parallel opposite sides, its vertex set is a primitive fixing system for M, and hence £ max Af = 6. • Proof of Theorem 3. The part "only if is contained in Lemmas 1, 3. We prove the part "if. Let P be a non-flattened inscribed convex pentagon of M such that every two non-neighboring vertices of P are antipodal point of M. Denote by F the vertex set of P. Any direction illuminates at least one vertex of P. All the more, any direction illuminates at least one point of the set F C bdM, i.e., F is a fixing system for M. Furthermore, let a be an arbitrary point of F and 6, e be its neighboring vertices in P. Then b and e are antipodal points of M, i.e., there is a strip with the boundary lines through b and e that contains M. Hence there is a direction that illuminates only the point a 6 F. Similarly for other points of F. Thus the fixing system F is primitive. •

'6

Fig. 19

ll

Fig. 20

Proof of Theorem 4. The part "if follows from Lemmas 5 and 6. We prove the part "only if. Let Qm^xM = 3. Then the body M contains nonregular boundary points (otherwise £ max M = 4, see Example 3). Let a be the non-regular boundary point with the greatest exterior angle (or one of the points with the greatest exterior angle if such a point is not unique). Denote by /i,/2 the tangential rays of M emanating from a and by m\,m
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semistrictly antipodal. Then at least one of the points c, d belongs to M (otherwise £> max M = 4 by Lemma 4). Assume for definiteness that c € M, i.e., b and c are antipodal. If they are neither strictly nor semistrictly antipodal, then the points x, y, z, w as in Fig. 20 form a primitive fixing system for M, contradicting the condition £> max M = 3. Hence b and c are strictly or semistrictly antipodal, i.e., again M satisfies condition 2) of Theorem 4. Thus if b G M, then Theorem 4 holds. Analogously, if d 6 M, Theorem 4 holds. Case 2. None of the points b,d belongs to M and, moreover, a and c are the unique common points of M with the boundary of the parallelogram abed. Denote by /j, 1'2 the tangential rays of M emanating from c. Then l'i C mi and 1'2 C TII^ (otherwise the exterior angle of M at c is greater than the exterior angle at a, contradicting the choice of the point a). Let P be a circumscribed parallelogram of M that contains the points a,c in its interior and x,y,z,w be common points of M with the boundary of P (Fig. 21). We assume P to be so close to abed that x,w are illuminated by the vector c — a and y, z are illuminated by the vector a — c. (Remark that the set of all directions for which there is a support line, having a segment in common with M, is no more than countable, and therefore we may suppose that each side of P has only one point in common with M.) The set F = {x,y,z,w} is a fixing system for M. Indeed, if a nonzero vector is situated between c — a and b — d (inclusively), then it illuminates the point w. Furthermore, if a nonzero vector is situated between a — c and 6 — rf, it illuminates the point z. Similarly for vectors situated between d — b and ±(a — c). Hence any direction illuminates at least one of the points z , y , 2 , w , i.e., F is a fixing system. Moreover, this fixing system is primitive (since each side of P contains only one point belonging to F), contradicting the condition > M = 3. Thus case 2 cannot be realized.

Fig. 21

Fig. 22

Case 3. None of the points 6, d belongs to M: moreover, c, £ M and M has one segment in common with the boundary of abed. For definiteness, assume that the side [a, b] has a segment [a, a;] in common with the boundary of

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the parallelogram abed. Let P be a circumscribed parallelogram of M that contains the points a, c in its interior and x'', y, 2, it; be common points of M with the boundary of P (Fig. 22). We remark that the point x' may coincide with x or differ from it. As in the previous case, {x',t/,2;, w} is a primitive fixing system for M, contradicting the condition £ max M = 3. Thus case 3 cannot be realized, too. Case 4: b £ M, d £ M, c £ M, and M has more than one segment in common with the boundary of abed. By Lemma 4, M cannot have two segments in common with two opposite sides of abed. For defmiteness, assume that the side [a, b] has a segment [a, x] in common with the boundary of abed and either [a, d] or [6, c] has a segment in common with the boundary of abed. As earlier, we construct a primitive fixing system for M that consists of four points, contradicting the condition £ max M = 3. Thus case 4 also cannot be realized. Case 5. None of the points 6,c,d belongs to M. Denote by b' the point of the set m<2, n M nearest to c and by d' the point of the set m\ D M nearest to c. If b' and d' are not antipodal points of M (Fig. 23), then it is possible to construct a primitive fixing system consisting of four points (with the help of a circumscribed parallelogram P) as in the previous cases, i.e., £> max M > 4, contradicting the condition of the Theorem. Even if b' and d' are antipodal points of M, not being neither strictly antipodal nor semistrictly antipodal (Fig. 24), then Qm!LXM > 4, by Lemma 4, again contradicting the condition of the Theorem. Thus b' and d' are strictly or semistrictly antipodal. The points a and b' are strictly or semistrictly antipodal and, analogously, a and d' are strictly or semistrictly antipodal. Now the points a, b', and d' satisfy condition 1) of Theorem 4. •

Proof of Theorem 5. Let M C R2 be a figure of constant width h. If 6, c are antipodal points of M then they are strictly antipodal and the segment [6, c] has the length h. Furthermore, let a,p, q be three boundary

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points such that a,p are antipodal and a,g are antipodal, too. Then each of the segments [a,p] and [a,g] has the length h and the boundary arc pq (not containing a) is the circular arc of radius h. Now, according to Theorem 4, if £max^ — 3, then M is the Realeaux triangle, i.e., each of the segments [a, 6], [6, c], [c, a] in Fig. 7 has the length h and every arc ab, be, ca is a circular arc of radius h. Furthermore, if Qm^M = 5 and P — conv{a,6,c,d,e] is the inscribed pentagon as in Theorem 3, then each diagonal of P has the length h and every boundary arc ab,bc,cd,de,ea is the circular arc of radius h. This means that M is a Realeaux pentagon of constant width h (Fig. 8). Even if a body M C K2 is a figure of constant width h distinct from the above ones, then £>max.M — 4 (since the equality Qm^M = 6 is impossible by Theorem 2). Proof of Theorem 6. Let Mk C M2 be a regular polygon with k = 2n+l vertices, n > 2. By Theorem 2, ^ max M^ < 6. Assume £> max M/j = 5. Then there is a non-flattened convex pentagon P inscribed in Mk as in Theorem 3. Let a,b,c,d,e be successive vertices of P. The center of Mk denote by g. Since a and c are antipodal points of Mk, there are two parallel, noncoincident support lines / a ,/ c of Mk with a G la,c € lc- Maybe one of the sides of Mk is situated in la or lc, but 2n (or all 2n + 1) sides are situated in the strip between la and lc. Namely, n (or n + 1) of the sides of Mk are situated in one half-plane with the boundary line ac, and n (or w+1) of them are situated in the other half-plane. This implies that the angle between the vectors a — q and c — q is not less than 2™r\ '^ • Similarly, the angle between a — q and d — q is > 2™ 1 • 27T. Hence the angle between c — q and d — q is — 2n+i ' ^ (independently, whether c and d are situated on different sides of the line ag or not), i.e., every side of P is visible at the angle < 2n+i ' ^ from q. Consequently the chord ac is visible at the angle < 2n2+1 • 2?r from g, i.e., the angle a between the vectors a — g and c — g is < 2 2 l • 2yr. Thus 2 n • 2?r > a > • 27T, 2n + 1 ~ - 2n + 1

contradicting n > 2. This contradiction shows that £> max Af/; 7^ 5. Similarly Qm^Mk ^ 3. Thus if n > 2, then £> max Af2 n +i = 4. Let now Mk C K2 be a regular polygon with k = 2n vertices, n > 5. Then £> max Mfc < 6. Assume that £>max Mk = 5. Then there is a non-flattened convex pentagon P inscribed in Mk as in Theorem 3. Let a,b,c,d,e be successive vertices of P. The center of Mk denote by q. Since a and c are antipodal points of Mk, there are two parallel, noncoincident support lines / a ,/ c of Mk with a G / e ,c G lc- Maybe two opposite sides of Mk are situated in la U lc, but In — 2 (or all In] sides are situated in the strip between la and lc. Namely, n — 1 (or n) of the sides of M\. are

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situated in one half-plane with the boundary line ac, and n — I (or n) of them are situated in the other half-plane. This follows that the angle between the vectors a — q and c — q is not less than -i^ • 27T. Similarly, the angle between a — q and d — q is > ^^ • 2?r. Hence the angle between c — q and d — q is < ^ • 2?r, i.e., every side of P is visible at the angle < ^ • 2?r from q. Consequently the chord ac is visible at the angle < ^ • 2?r from #, i.e., the angle (3 between the vectors a — q and c — q is < ^ • 27T. Thus 27T > /? > ^—— • 2?r, 2n - ^ ~ 2n contradicting n > 5. This contradiction shows that Qm^Mk ^ 5. Similarly £ max Mfc ^ 3. Thus if n > 5, then £>max-M2n = 4.

It follows from above that£ max M7 = 4, ^ max Mg = 4 and gmAXMk = 4 for A; > 10. It remains to consider the value QmiLXMk for k = 3,4,5,6,8,10. It is easily shown that £ max -Mfc = k for k — 3,4,5,6. Furthermore, let ai,a 2 , ...,aio be successive vertices of MIQ. Then P = conv{a 2 ,a 4 ,a6,a8,aio} is a non-flattened inscribed pentagon of M\Q that satisfies the condition of Theorem 3. Hence £> max Mio = 5. Finally, let ai,a2,...,as be successive vertices of MS. Then P — conv{a2,a4,a5,a7,as} is a non-flattened inscribed pentagon of MS that satisfies the condition of Theorem 3. Hence gm!iXMs — 5.

REFERENCES [1] B. Bollobas, Fixing systems for convex bodies, Studio, Sci. Math. Hungar., 2 (1967), 351-354. [2] V. Boltyanski, Kelly's theorem for .ff-convex sets, Soviet. Math. Doklady, 17 (1976), no. 1, 78-81. [3] V. Boltyanski and H. Martini, Combinatorial geometry of belt bodies, Results Math., 28 (1995), 224-249. [4] V. Boltyanski and H. Martini, On maximal primitive fixing systems, Beitrdge zur Algebra und Geometric, 37 (1996), 199-207. [5] V. Boltyanski, H. Martini, P. Soltan, Excursions into Combinatorial Geometry, Springer- Verlag, Berlin-Heidelberg-New York, 1997. xiv + 419 pp. [6] V. Boltyanski and E. Morales Amaya, Minimal fixing systems for convex bodies, Journal of Applied Analysis, 1 (1995), no. 1, 1-13. [7] V. Boltyanski and E. Morales Amaya, Cardinalities of primitive fixing systems for convex bodies, Discrete and Computational Geometry, 24 (2000), no. 2-3, 209-218. [8] L. Danzer, Review 2942, Math Reviews, 26 (1963), 569-570. [9] L. Fejes Toth, On primitive polyhedra, Acta Math. Sci. Hungar., 13 (1962), 379-383.

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[10] S. Fudali, Fixing systems and homothetic covering, Acta Math. Sci. Hungar., 50 (3-4) (1987), 203-225. [11] B. Griinbaum, Fixing systems and inner illumination, Acta Math. Sci. Hungar., 15 (1964), 161-168. [12] B. Tomor, The fixing problem for convex figures, Mat. Lapok, 14 (1963), 120-123. E-mail address: E-mail address:

boltianOcimat .mx ''Vladimir Boltyanski ' ' hernanQcimat .mx ''Hernan Gonzalez-Aguilar' '

THE NEWTON-GREGORY PROBLEM REVISITED

KAROLY BOROCZKY1 Eotvos University, Department of Geometry, Pazmany Peter setany 1/c, Budapest, Hungary, 1117

ABSTRACT. The Newton-Gregory problem; namely, whether thirteen points can be placed on the sphere S2 in a way that the spherical distance between any two is at least |, has received much attention since 1694. The first rigorous solution is due to K. Schiitte & B.L. van der Waerden [12], who used their method developed earlier in order to deal with packings of any number of equal spherical circles (see K. Schiitte & B.L. van der Waerden [11]). A simple and elegant solution for the Newton-Gregory problem was given by J. Leech [10]; more precisely, he provided the sketch of his argument. Here we present a complete proof along different lines.

1. INTRODUCTION For k > 4, we write a^ to denote the maximum number with the property that one can place A; points on the sphere S2 so that the spherical distance between any two different points is at least a^. In case of thirteen points on S2, it seems to be out of reach to determine 013. On the other hand, the question whether 013 > | has attracted much attention. The reason is an equivalent formulation; namely, whether thirteen unit balls in E3 can touch a given unit ball without overlapping. Isaac Newton and David Gregory discussed this problem in 1694, and Newton stated that the maximal number of touching balls should be twelve. He was right, but the first and still not complete solutions were provided only in 1874/75 by the physicists C. Bender [4], S. Giinter [8] and R. Hoppe [9]. The first "rigorous proof was given by K. Schiitte &; B.L. van der Waerden [12], and this proof was further simplified by J. Leech [10] to an elegant argument. 1

Supported by OTKA T 029786 103

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The proof by Leech considers the network of segments of pairs of points whose length are less than a prescribed constant (arccos ^), and estimates the area of the resulting possibly non-convex spherical polygons. There is one apriori possible arrangement where the estimates do not work, but this case is eliminated by graph theoretic arguments. The sketch [10] of J. Leech is reproduced in the books M. Aigner and G. Ziegler [1], and Ch. Zong [13]. The paper by J. Leech is only two pages, but it is partially due to the fact that he does not provide proofs for his estimates for the area of spherical polygons. This was acknowledged by M. Aigner and G. Ziegler, say in the preface of the second edition [2] of their beautiful book. In this paper, we prove using a different line of argument THEOREM 1.1. Among thirteen points on the sphere S2, there always exist two whose spherical distance is less than ^. We note that f = 1.047197551 The paper K. Boroczky and L. Szabo [3] in this volume verifies that even a\3 < 1.02746114, while K. Schiitte and B.L. van der Waerden [11] exhibited an arrangement showing that 013 > 0.997223592....

2. THE PROOF OF THEOREM 1.1 We prove Theorem 1.1 by contradiction, so suppose that (1)

«13 > |-

We write | • | to denote the area on the sphere, and d(-, •) to denote spherical distance. In this paper, distance, triangle, eic, always mean spherical distance, spherical triangle, etc. We use the spherical Laws of Sine and Cosine in order to calculate of areas of polygons with given sidelengths and angles. First we recall the notion of Delone triangulations. Let x\,... ,xn be a family of points on S2 that intersects any open hemisphere. We say that a circle is a supporting circle if its interior does not contain any of the points, and the boundary contains at least three of the points of the family x\,... ,x n . The spherical convex hull of the points in the boundary of a supporting circle is called a Delone cell. Now the Delone cells are the radial projections of the faces of the convex hull of the points x\,... , xn as a subset of the Euclidean three space E3. It follows that the Delone cells form a cell decomposition of S2. Any triangulation of this cell complex whose vertices are x i , . . . ,xn is called a Delone triangulation. In addition, the family x\,... ,rr n is called saturated with respect to a distance a < | if on the one hand, the distance between any two of the n points is at least a, and on the other hand, any

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x E S2 is of distance less than a from at least one of #1,... ,xn. We observe that if xi,... ,xn is saturated with respect to a distance a then the circumradius of any Delone cell or triangle is less than a. We write T(a) to denote the regular triangle of side length a. Let us quote the well-known triangle bound due to L. Fejes Toth [5] (see also L. Fejes Toth [6] p. 244): LEMMA 2.1. If T is a Delone triangle for a saturated arrangement of points with respect to a distance a < f then \T\ > \T(a)\. 2

Since a triangulation of 5 with at least twelve vertices contain at least 20 triangles, we deduce that a\2 is the spherical distance between neighbouring vertices of the regular icosahedron (see L. Fejes Toth [5], or L. Fejes Toth [6] p. 227), and hence (2)

ai3 < 012.

Actually ai2 can be characterized with the property that the circumradius of the regular pentagon with side length 012 is also a\<2In order to verify Theorem 1.1, we will prove that if a\z > | then the total area of the Delone triangles is larger than 4?r, what is readily absurd. It is natural to check what the triangle bound yields directly: Let us consider a Delone triangulation for some saturated family of n > 13 points on 52 such that the mutual distances between the points are at least 013. The area of each Delone triangle T satisfies \T\ > \T(al3)\ > T(f )| = 0.551285... according to Lemma 2.1. Since the number of Delone triangles is 2n — 4 by the Euler formula, and 24|T(|)| > 47r, we deduce that n — 13. On the other hand, 22 |T(^)| < 47r, therefore we search for six Delone triangles whose total area is at least 3.8. We can not ensure this fact for any extremal arrangement, but for a one where the number of pairwise distances that are ais attains its possible minimum. We will apply the triangle bound to the 16 remaining Delone triangles, and obtain a contradiction with (1) by (3)

16 |T(|)|+3.8 > 12.62 >47r.

Let pi,... ,pis € S2 satisfy that the pairwise distances are at least aia, and number of pairwise distances that are 013 attains its possible minimum. We colour the spherical segment pipj red if the distance of pi and PJ is a\^. The geometric graph on the thirteen points resulting this way is called the red graph. The maximality of 013 yields that there exist some red edges. Any circular disc whose diameter is a red edge contain no other points from the family than the endpoints, therefore the red edges are edges also in

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the Delone triangulation. Since the number of red edges is minimal for Pi, • • • 5 Pi3> we deduce that • at least three red edges meet at each of the vertices, and • the red edges divide S2 into convex polygons whose sides are of length 013-

Any such convex polygon is either a regular triangle, a rhomb, or a pentagon because a\^ > |. Next we claim that the red graph has no isolated vertex. Otherwise, let P be a red fc-gon, k < 5, which contains an isolated vertex of the red graph. Now P contains at least k triangles in any Delone triagulation, and hence \P\ > k - \ T ( a \ % ) \ . On the other hand, the area of P is at most the area of the regular Ar-gon of edge length 013 according to the isoperimetric inequality for polygons (see say A. Florian [7] p. 183). This is absurd because a\^ < a\^ yields that the circumradius of the fc-gon with edge length aia is less than 013. In turn, the claim follows. For any red rhombi or pentagon, one or two diagonals, respectively, occur as edges of the Delone triangulation. We call these diagonals Delone diagonals. We write S((p) to denote the square of side length
«3(f)+<*4(f)=7T.

We recall two well-known statements (Facts 1 and 2 below), which are consequences of the following elementary fact: Let pqr be a spherical triangle. Then the set of those points r', which are not separated from r by the line pq, and for which \pqr\ = \pqr'\ is a circular arc that passes through r and the antipodals of p and q. This circle is usually called the Lexell circle of the triangle pqr. We note the last property of the Lexell circle, Fact 1 and Fact 2 are stated as Proposition 2 (vii), Proposition 2 (viii), and Proposition 3 (i), respectively, in the paper K. Boroczky and L. Szabo [3] in this volume. Fact 1. Assume that two sides of a triangle is given, and the angle enclosed by these two sides decreases from TT to zero. Then the third side decreases, and the area of the triangle increases until the center of the circle passing through the vertices is separated by the third side, and the area decrease?, after that position.

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Fact 2. Given a circular arc 7 connecting p and q, moving r along 7 towards the midpoint of 7 increases the area of the triangle pqr. We need estimates on the area of red pentagons, and first we consider certain quadrilaterals: PROPOSITION 2.2. Let Q = pqrs be a convex quadrilateral Leaving the sides of Q unchanged, if we increase /.qps from its minimum then the area of Q increases until the points p, q, r, 5 lie on a circle, and the area of Q decreases after that position. Proof: Fact 1 yields that the angle /.qrs and the diagonal qs increase, while the angles /.pqr and Zjpsr, and the diagonal pr decrease. Thus by symmetry, it is sufficient to verify the following statement: Let us assume that the circular disc whose boundary passes through p, q and s contains r in its interior. If we increase /.qps by any small amount then the area of the quadrilateral pqrs decreases. We write P0)<70) r o ? s o to denote the new positions of p, g,r, s, and QQ to denote the new quadrilateral. Let t be a point on the circle such that r lies in the interior of the quadrilateral P — spqt, and together with Q, we deform P into the quadrilateral PQ = sopoqoto keeping the lengths of the sides of P. Let us consider the points rq and rs in PQ where the triangle qorqtQ (the triangle SQ^O) is congruent to the triangle qrt (the triangle srt). Now Fact 1 yields that d(qo,rs) > d(qo,r0) and d(so,rq} > d(s 0 ,^o) 5 and hence we deduce again by Fact 1 that the interiors of QQ and the triangles qorqtQ and SQrsto are disjoint. In particular, it is sufficient to prove that \PQ\ < \P\. Finally, we write g to denote the radius of the circle passing through p, q and s, and place the circular cap of radius Q on top of each side of PQ. According to the isoperimetric inequality (see say A. Florian [7] p. 183), the union of these four arcs encloses an area which is less than the area of the circle with radius Q, and hence \PQ\ < \P\. QED PROPOSITION 2.3. Let P be a red pentagon. Then (i): | P | > | T ( f ) | + |S(f)| = 1.91...; (ii): if the angle at the common endpoint of the Delone diagonals is at most 3 then \P\ > 2.16. Proof: We denote the vertices by q\,... ,95 where q\ is the common endpoint of the Delone diagonals. Our argument is based on deformation using Proposition 2.2. During the deformations, we keep the properties that side length are of y? = 013, and the diagonals are of length at lerist co. In addition, we keep the Delone diagonals, and keep that the angle at q\ and 9s (at qi and 94) is at least the angle at #2 (at 95).

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We may assume that P is not the regular pentagon since that has the maximal area according to the isoperimetric inequality for pentagons. First we consider (ii). If then we obtain a decreasing family of sets. Therefore the area is minimal if ip = |, and hence P\ = 2.169... . For (i), we allow that an angle of P is possibly TT. If (p is given then the minimal P is a triangle whose two sides are 2ip and one side is (p. Now Fact 1. yields that among these triangles, the area is minimal for (p = |, and in this case P is the union of a regular triangle and a square. OED Fact 1. yields that the Delone diagonal of a red rhombi R connects two vertices where the angle a is at least 0:4(013), and the area of .R is a decreasing function of a. PROPOSITION 2.4. Let y>\, and let the rhombi RI,... ,Rk and the triangles T\,... , T/ form a neighbourhood of a common vertex p without overlapping. In addition, assume that 2k + I > 6, the sides of the rhombi and the triangles meeting atp are of length (p, and the angle atp of a rhombus (triangle) is at least a±((p) (a^(ip)). Then 2k + I = 6, and |fli| + ... + | J R fc | + |Ti| + ... T , | > 2 - T ( f ) | + 2 - 5 ( f ) | = 3 . 8 2 . . . .

Remark: A triangle T{ in this lemma will be either a red triangle, or a triangle that is cut off by a Delone diagonal from a red rhombus or a red pentagon. Proof: We observe that 0:3(). First assume that k < 2. Then the sum of the angles at p is at least 20:3(1) + 2a4(|) = 2?r, therefore we have two squares 5"(f) and two regular triangles T ( f ) . So let k > 3, and hence considering the sum of the angles at p show that k = 3 and I — 0. We cut each Ri, i — 1,2,3, into half by the diagonal that does not contain p, and write T( to denote the triangle that is the half, which contains p. It follows by Fact 2 that the sum of the areas of T{, T^ and Tg is minimal then the angle of the two of the triangles at p are the possibly minimum. In particular, two of the rhombi are squares S((p). Now decreasing ip decreases the areas of the rhombi R\, R<2 and R^. Thus
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LEMMA 2.5. Let us consider a Delone triangulation of some 13 points on S2 such that the mutual distances are at least 0^3. If ais > ^ then there exist six triangles of the Delone triangulation meeting at a common vertex whose total area is larger than 3.8. Proof: The Delone triangulation on thirteen vertices has a vertex p\ of degree at least six by the Euler formula. We will find the six triangles for Lemma 2.5 at p\. Case I. There exists no red pentagon such that some Delone diagonal ends at pi. If pi is the vertex of a red rhombus or a red pentagon then this red polygon has a Delone diagonal that cuts off p\\ in particular, cuts off a Delone triangle that is of the same kind as the triangles in Proposition 2.4. Therefore Proposition 2.4 yields Lemma 2.5. Case II. There exist a Delone diagonal of some red pentagon, which ends at pi. If at least two red pentagons meet at p\ such that the Delone diagonals of both pentagon meet at p\ then Proposition 2.3 (i) provides six Delone triangles with total area at least 3.82. Therefore let P be the only red pentagon, such that p\ is a vertex, and has a Delone diagonal that ends at pi. We write p\,... , p^ to denote the vertices of P in cyclic order in a way that the diagonal pips is a Delone diagonal. If pi is the end point of exactly one Delone diagonal then p5p3 is the other Delone diagonal. We write p'3 to denote the vertex of the rhombus R — pip2p'sp5 opposite to p\. Since pipzPz is a Delone triangle, p$ is not contained in the interior of the circular disc determined by P1,P2 5 P3, and hence the point p% lies in the circular disc determined by pi,p2,P5- We deduce that /.p2pip$ > ^4(^13). Next, we observe that p± is not contained in the interior of the circular disc determined by P2,P3,P5 because p^psps is a Delone triangle, which in turn yields that the centre c of the circular disc determined by P2,P3,P5 lies on the same side of the great circle psps as P2. Therefore the triangle P2P3P5 has larger area than the triangle according to Fact 1., and hence the total area of the Delone triangles and PIP3P5 is at least \R\. Now if we replace the quarilateral P1P2P3P5 by the rhombus R then we decrease the total area of the Delone triangles at pi, and R can be one of the rhombi Rj in Proposition 2.4. Therefore we will consider only the rhombus R instead of P, and the analogous argument as in Case I based on Proposition 2.4 yields Lemma 2.5. Finally, we assume that pi is the vertex of exactly one red pentagon P, and Delone diagonals of P end at p\. If the angle of P at p\ is at most 3 then the total area of the three Delone triangles inside together with any other three Delone triangles is greater than 2.16 + 3|T(|)| > 3.81 according to

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Proposition 2.3 (ii), confirming Lemma 2.5. If the angle of P at p\ is larger than 3 then the number of Delone triangles which are not contained in P and meet at p\ is three, and two of these triangles form a red rhombus RQ whose larger angle is at p\. Now the angle of RQ at p\ is at least 2?r — a(|) — 3, and hence the possible minimum of \Ro\ is obtained when the angle at p\ is minimal and all side lengths are |. We deduce that \Ro\ > 1.34. On the other hand, we have already seen that |P| > 1.91, and the area of the sixth triangle at p\ is larger than 0.55, which in turn yield that the total area of the six Delone triangles at p\ is larger than 3.8. Therefore the proof of Lemma 2.5, and in turn of Theorem 1.1, is now complete.

REFERENCES [1] M. Aigner, G. Ziegler: Proofs from THE BOOK. First edition, SpringerVerlag, 1999. [2] M. Aigner, G. Ziegler: Proofs from THE BOOK. Second edition, Springer-Verlag, 2001. [3] K. Boroczky, L. Szabo: Arrangements of thirteen points on a sphere. (see this volume) [4] C. Bender: Bestimmung der grossten Anzahl gleicher Kugeln, welche sich auf eine Kugel von demselben Radius, wie die iibrigen, auflegen lassen. Arch. Math. Phys. (Grunert), 56 (1874), 302-306. [5] L. Fejes Toth: On the densest packing of spherical caps. Amer. Math. Monthly, 56, (1949), 330-331. [6] L. Fejes Toth: Regular Figures. Pergamon Press, 1964. [7] A. Florian: Extremum problems for convex discs and polyhedra. In: P.M. Gruber and J.M. Wills, editors, Handbook of convex geometry, 179-221, North-Holland, Amsterdam, 1993. [8] S. Giinther: Bin stereometrisches Problem. Arch. Math. Phys. J (Grunert), 57 (1875), 209-215. [9] R. Hoppe: Bemerkung der Redaction. Arch. Math. Phys. (Grunert), 56 (1874), 307-312. [10] J. Leech: The problem of the thirteen spheres. Math. Gazette, 40 (1953), 22-23. [11] K. Schiitte, B.L. van der Waerden: Auf welcher Kugel haben 5, 6, 7, 8 oder 9 Punkte mit Mindestabstand eins Platz? Math. Ann., 123 (1951), 96-124. [12] K. Schiitte, B.L. van der Waerden: Das Problem der dreizehn Kugeln. Math. Ann., 53 (1953), 325-334. [13] Ch. Zong: Sphere packings. Springer-Verlag, 1999. E-mail address: boroczkyScs.elte.hu ''Karoly Boroczky''

ARRANGEMENTS OF 13 POINTS ON A SPHERE

KAROLY BOROCZKY 1 Department of Geometry, Eotvos Lorand University, H-1117 Budapest, Pazmany Peter setany 1/c LASZLO SZABO 2 Computer and Automation Research Institute, Hungarian Academy of Sciences, H-llll Budapest, Lagymanyosi utca 11

ABSTRACT. Let a^ denote the maximum number with the property that one can place k points on the unit sphere S2 so that the spherical distance between any two different points is at least a^. The exact value of a^ is determined only for some small values of fc, namely, for k < 12 and k — 24. In this paper we are concerned with the first unsolved case k = 13 which is particularly interesting because of its close relation to several famous problems in discrete geometry. A certain arrangement of 13 points on S2 shows that ai3 > 0.997223592.... On the other hand K. Schiitte and B. L. van der Waerden proved that 013 < ^ = 1.04719755.... In this paper we prove that a13 < 1.02746114. 1. INTRODUCTION Let a,k denote the maximum number with the property that one can place k points on the unit sphere S2 so that the spherical distance between any two different points is at least a/-. The problem of determining the exact value of a/j together with the corresponding extremal arrangements of points was posed by the Dutch biologist Tammes [16] who was led to this problem by examining the distribution of the openings on the pollen grains of different flowers. The problem of Tammes is solved only for some small values of A;, for k = 3,4,6,12 by L. Fejes Toth [4], for k = 5,7,8,9 by K. Schiitte and B. 1

The work was supported by Hungarian Scientific Research Fund No. T029786. The work was supported by Hungarian Scientific Research Fund No. F035117. Ill

2

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L. van der Waerden [14], for k = 10,11 by L. Danzer [3] (and, independently, for k = 10 by L. Ears [10] and for k = 11 by K. Boroczky [1]), and for k = 24 by R. M. Robinson [13]. The first unsolved case is k = 13 which is particularly interesting because of its relation to the so called Newton-Gregory problem. It is known that there was a famous controversy between David Gregory and Isaac Newton in 1694 concerning the following question: How many unit spheres can simultaneously touch a given unit sphere without overlapping each other. Newton conjectured that the answer was 12 while Gregory thought 13 was also possible. Surprisingly, it took more than two hundred years before K. Schiitte and B. L. van der Waerden [15] proved in 1953 that Newton's guess, the '12', was correct. For a more complete picture we note that several "solutions" were given in the physics literature in 1874/75 [8, 11] but these arguments are not accepted as mathematical proofs by the experts. We also note that J. Leech [12] sketched an elegant 2-page-long proof in 1956. Unfortunately, most of the details of that proof are left to the readers and a complete version of the proof has not been published yet. Recently, K. Boroczky [2] has given a new simple proof along a completely different line. The case k — 13 is also interesting from the following point of view. Consider a packing of congruent copies of a convex body K. Two members of the packing are called neighbours if their boundaries have a point in common. Following the terminology of L. Fejes Toth, the maximum possible number N(K] of neighbours of K in a packing of congruent copies of K is called the Newton number of K. A packing of congruent copies of K in which each body has exactly N(K) neighbours is called a maximal packing. Observe that the maximal density lattice packing of unit balls in E3 is maximal. A conjecture of L. Fejes Toth [6] states that any maximal packing of unit balls in E3 is essentially unique, i.e., it is composed of parallel layers of the same hexagonal structure as the layers in the densest lattice packing, and thus must have density —P=. According to L. Fejes Toth, the first step in proving vl8 this conjecture might be to give a "good" upper bound for 013. This paper is intended to complete this first step, namely we prove Theorem 1. a13 < a0 = 1.02746114. We note that one can construct an arrangement of 13 points on S2 such that the distance between any two points of the arrangement is at least 0.997223592... (see [7]). We state two interesting consequences of Theorem 1 for ball packings in E3. Theorem 2. Let B\, f?2, • • • •> -#14 be fourteen different members of a packing of unit balls in E3. Assume that each of the balls B
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the ball B\. Then the distance between the centres of B\ and B\<± is at least 2.205279217705. Corollary 1. The distance between the centres of any two non-neighbouring balls in a maximal packing of unit balls in E3 is at least 2.205279217705.

2. PROOF OF THEOREM 1 For technical reason we will use the result of K. Schiitte and B. L. van der Waerden [15] which says that a\z < -|. We will assume, for contradiction, that ao < «i3 < f. First we recall some definitions. Let (p < ^. We say that a point set 'P on S2 is a (^-saturated point set if the distance between any two different points of 'P is at least (p and for any point q on S2 there exists a point of fJ* whose distance from q is smaller than 7 and (p < ^. Consider a set 7 of k points such that the distance between any two different points of IP is at least ip. Let us join two points of y by a segment if their distance is exactly (p. In this way we obtain a graph which is called the graph of 9. Among the sets of k points in which the distance between any two different points is at least

the edges do not cross each other, the edges form a connected system, the edges divide S2 into strictly convex polygons, each vertex is either isolated or of degree at least three.

By the indirect assumption there exists a non-trivial

7aQ > 27T. It is not hard to see that triangles, rhombi, or pentagons of the >p graph cannot contain an isolated point of the

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interior of a hexagon of the (p graph, then this hexagon cannot contain other vertices of the (p graph in its interior and the degree of the isolated point is 5 or 6 in the Delone triangulation of the corresponding (p point set. Suppose that an isolated point of the (p graph is contained in the interior of a hexagon of the

, arrange the isolated point in the hexagon, if it is possible, in such a way that its degree is 6 in the Delone triangulation. If it is not possible, then arrange the isolated point so that its distance from the vertex of the hexagon not adjacent with it in the Delone triangulation is as small as possible. Repeat the above process for each isolated point in the (f> graph. Then we obtain a point set which will be called a <^-final point set. Note that the , and R( 1. For each point p in the interior of the circle C of centre o and radius - on S2 let p' denote the point of the segment o*p on S2 for which op' — [i • op. Then for any two interior points p, q of the circle C on S2 we have \L • pq > p'q'.

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We will prove that if a curve 7 joining the points p and q on S2 is differentiable and its length is s, then the image 7' of 7 under the above map is also differentiable and for its length s' we have fis > s'. Fix the point p and the angle £opq = a and examine limg_>p 2_2__ If o,p, q are collinear, then, obviously, 2-2_ = fj,. Now assume that 0 < a < TT. Let r denote the point on the half line emanating from o and going through q for which or = op = x. Clearly, lim g _p £prq — lim?_,p £p'r'q' — |. Hence lim ? _ p 2J1 = if^f — v. Since the function sin x is strictly concave on the range 0 < x < TT, therefore sin/ux < yusinx which implies that v < [i. At the same time 2-£- = ^ and lim g _ p £qpr = \a — -|| = /3. Thus n.

lim

tan 2 (3 + 1

From this, using the differentiability property of 7, the assertion follows.

D

Turning to the proof of Theorem 2, let 01, o 2 , . . . , o14 denote the centres of the balls B\,Bi,... ,-814, respectively. Suppose, for contradiction, that oio14 < 2.205279217705. For 2 < i < 14 let pi denote the intersection point of the segment o\O{ with the boundary of the ball B\. Apply the map described in Proposition 1 with o* — p\± and A = ^- for the points Pi,pz,-.. ,^13. Now easy calculation shows that the distance between any two different points of {p^iP^i • • • iPi^iPi^} is at least CQ, a contradiction.

4. GENERAL POLYGONS In this section we prove some properties of spherical polygons. First we introduce some notations. For n = 3,4,5,6, let T(<^), S(ip), P^}-, and H((p] denote the regular spherical rc-gons of side length
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(ii) Let pqr be a spherical triangle. Now pq = qr if and only if /.prq = /.qpr. On the other hand pq < qr if and only if Lprq < £qpr. (iii) Let pqr be a spherical triangle. Then pq + qr > rp. (iv) // we fix the lengths of two sides of a spherical triangle and we increase the angle between them, then the third side increases. (v) Let o denote the centre of the circumscribed circle of a spherical triangle pqr. Then £rpq -f- A.rqp — £prq — '2£opq if pq does not separate r and o, while £rpq -+- £rqp — /(.prq = — lA-Opq if pq separates r and o in the circumscribed circle. Thus, if we fix p and q and we move r on the circle arc ^pq", then £rpq + /.rqp — £prq remains constant. (vi) The area of a spherical triangle with angles a,/3,7 is a + Q + 7 — TT. (vii) Let pqr be a spherical triangle. Then the set of those points r1 which are not separated from r by the line pq and for which \pqr\ = \pqr'\ is the fixed circle arc p'q' going through r where p' and q' denote the antipodals of p and q on S2, respectively. This circle arc is usually called the Lexell circle of the triangle pqr. (viii) Let pqr be a spherical triangle and let C denote its circumscribed circle. Let t denote the midpoint of the arc "pg" containing r. If we move r toward t on the circle C, then both \pqr\ and £prq strictly increase. (ix) Let pqr be a spherical triangle. Now pq + qr — TT if and only ij £qpr = TT. On the other hand pq-\- qr < TT if and only if £prq-\TT. Especially, if pq,qr < ^, then Lprq < ir — Lqpr. (x) Let pq be a great circle arc and let s be its perpendicular bisector great circle. For any point r of s we have /Lprq > pq with equality if and only if pr — ^. Furthermore £prq strictly increases if pr increases or decreases from -|. We continue with proving some further properties of spherical polygons. Proposition 3. (i) Fix the lengths of the sides pr and rq of a spherical triangle pqr and increase £prq from zero. Then \pqr\ strictly increases while the centre of the circumscribed circle of pqr lies in the interior of the hemisphere bounded by the line pq and containing pqr (i.e. £rpq -f Lrqp — Lprq > Q). The area of pqr attains its maximum if the centre of the circumscribed circle of pqr lies on the side pq (i.e. /Lrpq + A.rqp — £prq = 0). Finally, \pqr\ strictly decreases while the centre of the circumscribed circle of pqr lies in the interior of the hemisphere bounded by the line pq and not containing pqr (i.e. £rpq + £rqp — Lprq < Q). (ii) Let pqr be a spherical triangle which contains the centre of its circumscribed circle. If pq > p'q', qr > q'r', rp > r'p' for some spherical triangle p'q'r', then \pqr\ > \p'q'r'\ unless pqr and p'q'r' are congruent.

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Proof, (i) The assertion can be proved easily by Proposition 2 (vii) and the argument used in its proof by the method of decomposition. (ii) By (i), one can decrease the area of a triangle • by decreasing an arbitrary side of the triangle if the centre of the circumscribed circle lies in the triangle, • by decreasing one of the two smaller sides of the triangle. Fix the triangle A' of side lengths a',b',c': and among the triangles A of side lengths a,6,c satisfying the conditions described in (ii) let A be such one whose area is as small as possible. Let a, 6, c denote the sides of A. If A is congruent with A', then we are done. If A is not congruent with A' and the area of A can be decreased by none of the two ways described at the beginning of the proof, then the centre of the circumscribed circle of A cannot lie in the interior of the triangle. Indeed, in this case only the lower bounds for the sides of A prevent the decrease of the area of A, which implies that the corresponding sides of A and A' are equal, i.e., A and A' are congruent, a contradiction. Thus the centre of the circumscribed circle of A lie on the (unique) greatest side, say on a, of A. Since a also cannot be decreased, therefore a = a'. This implies that a' > max(6, c) > max(6',c') and b > &', c > c'. Hence we can obtain a congruent copy of A' by decreasing the two smaller sides of A. From this |A| > |A'| follows. D Proposition 4. A polygon of maximal area among polygons of given side lengths is inscribed in a spherical circle. Proof. This is an immediate consequence of the well-known isoperimetric inequality. D Proposition 5. (i) // <7i 03(9391) with equality if and only if the triangle is regular. (iii) If q\qi + 9293 < it in a triangle 919293 and q{ and q'3 are such points of the sides q\q<2 and q^q-2, respectively, for which qiq( = g3g3, then (iv) Suppose that the vertices q\ and q? and the sum ^.q-^qiq^ + are fixed in a triangle K> then s = 9193+9392 attains its minimum when q\qz = q^q? and the function s(^.q^q\q2] changes monotonously

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from the symmetric position. We note that if A.q?,q\qi + then s — TT, a constant.

Proof, (i) Consider a triangle gig2gs in which qiq% < q\qi and . Let ^2 and 92 denote the images of gi under the reflections with respect to g2 and g2, respectively. Since gigs = q'2q3 and gig3 = q'2q3 in the triangles q\q3q2 and 9ig3g 2 , respectively, and q\q'2 < q\q'2, therefore £q\q3q2 < ^qiqsq? and thus /gig392 < ^q\q^qi- Recall that qiq3 < f , hence f/i Z.q3qiq2. Suppose, for contradiction, that £q$q\qi < ^gsgig2- Let g denote the intersection point of the line gsg~2 and the segment q\q qi is minimal if and only if q^qz = q$q\- Thus the minimum possible value of £qiq3q2 is a3(q-3qi). (iii) Set Zr/ 2 gir/ 3 = fti and /f/29s 0 where qiq3 - q(q'3 « (cos 0i + cos/33)£ means that lim E _o 9 l g 3 ~' 7 i g 3 _ CQS^J + cos ^3. ^.From this the assertion follows. (iv) We deal only with the case £q3q\q2 + ^ 71" can be settled similarly. Now gigs + gsg2 < TT. Suppose that < ^q?>qiq\- Leaving the vertices q\ and q2 and the sum £.q$q\q2 + unchanged, increase £qzq\q
g2gs) = (gig 3 - gigs) + (^q3 - g2gs) ~ -(ggs - gg3 cos «) + (993 - 99s cos a) =

(99s - 993)(1 + cos a). Since gig3 + g3g2 < TT, therefore singig 3 > sing 2 gs (do not forget gigs > 9293) which implies that qq'3 > ggs. On the other hand limj^o^ = ^9i9392 ^ K- Thus (gg3 - gg3)(l + cose) > 0 from which the assertion follows. This completes the proof of Proposition 5. D

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The special case of the next lemma when q\qi > q^qz = (73(74 = q^q\ was proved by W. Habicht and B. L. van der Waerden [9]. Lemma 1. Let 91929394 be a convex quadrangle. Leaving the side lengths of the quadrangle unchanged, if we increase A.q±q\qi from its minimum, then the area of the quadrangle increases until the points 91,92,9s, 94 become cocircular and then the area decreases. (The angle ^929394 and the diagonal 9294 increase, while the angles £q\qiq?>, ^9s949i> and the diagonal 9193 decrease, as Lq\q\qi increases.) Proof. The statement in brackets is obvious. By symmetry, it is enough to show that if the circle going through 91,92,#4 contains 93 in its interior and we increase A.q^q\qi by a sufficiently small amount, then the area of the quadrangle decreases. Let rs denote the intersection point of the circle going through 91,92,94 and the angle bisector of the (concave) complement of ^929394. Consider the quadrangle 9192^394- Leaving the vertices 91,92 and the lengths of the sides 9i92,92 r 3, r s94,949i unchanged increase £q\q\q Lq±q\92, hence ^q(r'3q2 > /94^s92). Then £q'4r'3q2 = £q(r'3q% + £q3r'3ql + ^9^392 and thus Zg|r3g2 > ^93^392 and £q3r'3q4 > Z.q3r3q4. This yields that q^q3 > 9293 and q'4q3 > q4q3. Let ki^7"^!- Subtracting the sum of the areas of the triangles r^q^q^ and r^q^q-^ from both sides of this inequality, we obtain [91929394! > |9i9293r39s94l- At the same time I9i929|7>39394l > I9i929 3 9 4 l from which I9i929s94| > |9i929394l follows. D We will frequently use the following consequence of Lemma 1. Remark 1. Let 91929394 be a convex quadrangle whose side lengths q\q<2, 9293?93945949i are fixed and q\qi — 9394. Then \q\929394) is maximal when = ^929394 (and /949i92 = ^9i949s^- If ^9i929s < ^929394, then < ^949i92 and q\qz (< 9294) is the Delone diagonal 0/51929394-

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The special case of the next lemma when P is a pentagon of side length (p was proved by W. Schiitte and B. L. van der Waerden [14] using a result of W. Habicht and B. L. van der Waerden [9]. Lemma 2. Let (p < 012- If P is a convex pentagon of side lengths at most if, then P contains no point whose distances from its vertices are at least (p. Proof. (Indirect.) Let ge be a point in P for which qiq§ > <£, 1 < i < 5. Split P into the five triangles qiq^qe, #293 |T( 4 and let qiqiqi+i, 2 < i < n — I, be n—2 triangles of the Delone triangulation of a point set. IfY^i=2 ^-QiQiQi+i < TT, then the points qi, 3 < i < n — 1, lie in the circle going through qn,q\,q<2. Proof. Obvious.

D

Proposition 7. Let

<*n((p). Proof. Since the radius of the circle going through qn-,qi-,q2 is a strictly increasing function of £qnq\q2, therefore, by Proposition 6, it is enough to show that we can arrange in only one way the points in the circumscribed circle of S(
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Proposition 8. Let (p < a\2- If 9i92 = in a triangle 919293 of the Delone triangulation of a (p -saturated point set, then ^919392 < ^(v?) with equality if and only if 919293 is regular. Proof. Let q'3 be the midpoint of the circle arc q\q2 containing #3. Then ^9i9s92 < ^9i9s92- We know from the regular icosahedron that 2m(ai2) + ai2 = TT. This implies that 2m(<^) + y? < TT for 99 < 012- Hence, by Proposition 2 (x), we have ^919392 < QsC^) unless the triangle 919293 is regular. D Proposition 9. Let (p < Ij- one? consider the Delone triangulation of a (psaturated point set Q = {91,92, • • • ?9fc}(i) If 919293 is a Delone triangle, then ^919293 < 20:3(92). (ii) /fgi9293 and q\q^q^ are Delone triangles and q±q\ = (p, then ^9491 92 > ^9i929s wz£/j equality if and only 2/9293 = y? one? gigs = 9294(iii) If 9192 < d((p), then q\q^ is a Delone edge unless there exist two points of Q, say q% and q±, for which qiq^q^q^ is congruent with S( d((p) with equality if and only if there exist two points of Q, say q$ and q±, for which gi#3g2 / ^-qiqiqz- On the other hand, if <7i ?2?3, then consider the supporting circle going through £q\q\qi = £q\qiqz(iii) First note that four points can be arranged in the circumscribed circle of S((p) only at the vertices of a square. Let h denote the midpoint of the segment 9192- If 9i92 < d((f), then the circle of centre h and diameter d((p) can contain no point of Q other than q\ and q?. Therefore there exist two points hi and h^ on the perpendicular bisector of the segment q\q, and ^949i9s + ^9s9i92 < K, then the centre c of the circumscribed circle of the triangle 929394 lies in the interior 0/^929394.

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Proof. Assume that 9493 > 9293. Since only the greatest side of a triangle can separate the triangle from the centre of its circumscribed circle, therefore it is enough to show that 9493 cannot separate c from the triangle 929394 (if c G 9s94i then 9493 separates c from 929394, by definition). First examine the case A.q^q\q<2 < 20:3(92). Consider the circle going through 94,91,92- The radius of this circle is at most (p. Let c\ denote the centre of this circle. By Proposition 6, this circle contains the point 93. Since c\q\ < 9491, therefore, by Proposition 2 (x), we have £q$ciqi — ^9iCi92 > asCy). The diameter of the circle sector q\c\qi is 92 and 9293 > 92, hence ^930192 > 0:3(92). Thus A.q^c\q\ + ^910192 + ^920193 > 803(92) > TT which implies that c\ lies in the interior of ^939491. Since c\q\ > 091, it follows that c also lies in the interior of ^939491.

FIGURE 1. The case ^949192 > 2a 3 (<^). Next we deal with the case £.q±q\qi > 1a^((p). Suppose, for contradiction, that 9394 separates c from 929394- Let h be the midpoint of the segment 9493 and let c-2 denote the centre of the supporting circle going through 91,93,94. The situation is shown in Figure 1. Since 92 does not lie in the interior of this circle, therefore hc^ > /ic, and thus (p > 0294 > 094 = 092- This and Proposition 2 (x) imply that ^94092 > /949i92 > 20:3(92). Since 9293 > <£>, we have ^92093 > 0:3(9?), hence Aq^cq^ + ^.q^cq^ > 80:3(9?) > TT, a contradiction. This completes the proof of Proposition 10. D 6. (p GRAPHS Let k > 13 and V3 < f. Consider a set IP of k points in which the distance between any two different points is at least 99. Let us join two points of f by a segment if their distance is exactly
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distance between any two different points is at least (p consider one whose graph has as few edges as possible. This graph will be called a (f> graph of k vertices and the corresponding point set will be called a (p point set of k points. It is easy to see that in a non-trivial

13 and

5. Then in a

, then qqn < q^qs, then If qlq3 = q^qs, then all four of the above inequalities become equalities and gi?2?394?5?6 is centrally symmetric.

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(ii) Let (p < -|. Let 9i929s94959e be a convex hexagon of side length (p and diagonal lengths not less than (p. In each pair of opposite angles of the hexagon consider that one which is not smaller than the other one. Then these three angles are either at the vertices 91,93,95 or at the vertices g 2 ,94,96(iii) Let ip < ^. Let 919293949595 be a convex hexagon of side length (p. If the diagonals q3q& and 9092 of this hexagon are Delone diagonals, then (iv) Let (p < -|. Let 91939495 be a convex quadrangle where 9394 — q&q\ = 9 < 9i94 < TT - Lp and 9193 > q±q&. Then ^.qiq3q4 < Zg 4 g 6 9i. If 91 93 = 9496, then £qiq3q4 = /949e9iProof, (iv) The case of equality is trivial, therefore assume that ^496- The diagonal q\q± splits the quadrangle into two triangles. For the side lengths of these two triangles q^q\ = 9394 and 9194 = 9491 hold, thus /969i 94 < ^939491If we rotate the triangle 959194 in such a way that qe and q\ are carried to 93 and 94, respectively, then g4 is carried to an interior point q\ of the region / V otherwise £q3q6qi > /Lq\q<2q3 and £q5q6qi > £q3qeqi. Now, using the fact that f an(i 9s9s > 9-,we can apply Proposition 12 (iv) for the quadrangles. Hence ^91^293 < ^9i9e9s an(^ ^9s949s ^ ^939695- Since the greatest angle of the hexagon is ^959691, therefore, by Proposition 12 (ii), we have ^929394 < ^9s9e9i, ^949596 < ^919293, and /9e9i92 < /9s9496- Taking these inequalities together with ^959691 = ^.q^q'6qi into account we obtain

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that the sum of angles of the hexagon 9i929s949s96 is at most 4?r, a contradiction. D In the next five lemmas we will examine the structure of the Delone triangulations of (f> point sets of k points for a&+i < (p < ^. We note that these (p point sets are also ^-saturated point sets. This implies among others that the radii of the circumscribed circles of the Delone triangles of these point sets are at most 9. Lemma 4. Let k > 13 and max(|-,afc+i) < graph of k vertices does not exist an empty convex hexagon q\929394 959e of side length (p with the property that one of the vertices of the hexagon, say q§, is adjacent to all Delone diagonals of the hexagon. Proof. (Indirect.) Suppose that 9692,9693,9694 are Delone diagonals in 9i929394959e- It follows from Remark 1 and Proposition 12 that the greatest angle of the hexagon is ^959591 • Let q'6 denote the image of q6 under the reflection with respect to the diagonal 9195. Proposition 13 implies that 9g9s > (p. We show that min(9692,9694) > <£. By Proposition 11, this proves the lemma. We may assume that 9592 < 9e94- Then ^959192 < ^919594 and thus 9692 < 9694- Therefore it is enough to deal with 9692We prove that, with the given structure of the Delone diagonals, q'6q2 attains its minimum if 929s949s96 is an inscribed pentagon. First assume that 9e929s94 is not an inscribed quadrangle. Fix the points 94,95,96,91 and leaving its side lengths unchanged deform the quadrangle 9i929s94 in such a way that ^929194 decreases. Then 9692 decreases, furthermore 9296 decreases, while 9395 and 9395 increase. This implies that the structure of the Delone diagonals remains unchanged until 9e929s94 becomes an inscribed quadrangle. Thus q'6q2 can be minimal only if 9e929s94 is an inscribed quadrangle. Next, if 9s949s96 is not an inscribed quadrangle, then 9594 < 9395- Fix the points 96,9i,92,93 and leaving its side lengths unchanged deform the quadrangle 93949596 in such a way that ^959693 decreases. Then 9692 decreases, furthermore 9395 decreases, while 9495 and 9492 increase. This implies that the structure of the Delone diagonals remains unchanged until 93949595 becomes an inscribed quadrangle. Thus 9692 can be minimal only if 93949596 is an inscribed quadrangle. Since q'6q<2 attains its minimum, therefore in this case both of the above quadrangles are inscribed quadrangles, i.e., 9293949596 is an inscribed pentagon.

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We have to show that Lq'&q\q 0:3(9?) provided that q^qsq^ — j, then 0:3(9?) + 0:5(99) = TT (cf. the tessellation (3,5,3,5)), thus 0:3(9?) + a 5 ( f ) > K f°r 9? > f- This implies that gluing the polygons T(9?) = q'lq^qQ and P((p) — ^5(9?) and q^qe > q'2q'6. Hence /g39i9s = ^qiqe > 0:3(9?) in #o from which q2q'6 > 9? if < f 5 then, by Proposition 2 (x), A.q^q\q-^ > qsqs- Consider again the hexagon H' = T(9?) U -P(9?). Obviously, ^3^5 > 93^5. Set 93
.

27T

.9?

.

. 7T

smm — sm 7-5(9?) sin — and sm —- = sin rs(9?) sin — o ^ o from which sin m = 2 cos fO sin ^. At the same time ^ (

. 03(9?) sm 2

1 2 cos f

Now the desired inequality 2m > 0:3(9?) follows from the inequality \/2

v/5- 1 _

This completes the proof of Lemma 4.

1

D

Lemma 5. Let k > 13 and max(|-,ajt+i) < 9? < f • Then in a (p graph of k vertices does not exist an empty convex hexagon
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Let q'6 denote the image of q& under the reflection with respect to the diagonal q\q*>. We show, as in the proof of Lemma 4, that g;

q±qi in such a way that X^2 (p. Thus we may assume that during the deformation we reach the position where q§q± = q±q<2 before q% becomes adjacent to the circle going through 92?
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FIGURE 2. Illustration to the proof of Proposition 14. such a way that for the new positions d[ and d'2 of di and d^-, respectively, /Ld\cd\ — Zc/2cd2 holds. The situation is shown in Figure 2. For i — 1,2, set /.diad'j = Si and /.dibd'j — £;. Then Lad\b — /.ad'fi = /.cadi + £cbdi — A.cad\ — /Lc\)d\ — Si — £i for i — 1,2. To prove Proposition 14 it is enough to show that &\— £\ < S? — e?. Since the arcs did^ and d^d^ are congruent and adi < ae?2, ad\ < ad'2, bd\ > bd%, bd\ > bd'2, therefore, by Proposition 2 (viii), Si < 6-2 and e\ > £2 from which the assertion follows. D Lemma 6. (i) Let (p < ^. There exists a uniquely determined convex hexagon Hn = of side length (p such that in^n^ = ^.n^n^nQ — ^.n^nQUi and = (f> holds for some point n'6 ^ UQ. Setting ip(p) — £n$nQn\ and i^*((p} = A.riQn\n<2,, the function V'Cv) zs strictly increasing while i^*((f) is strictly decreasing. (ii) Let k > 13 and max(^,afc + i) <

^.q^q^q^, then ^
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For 04(9?), hence from the triangle n^n'^n'^. This implies that lA.n-2.n'^n\ + 0:3(9?) < + 0:3(9?) < TT = /rijWgni. Thus 2/712^1^1 — /T^n^ni < — 0:3(9?). For ^5 = n'2 we have ^ < ^.n^nin'^ = /n27i27ii + 0-3(9?) since in the triangle 7iin 2 7Z2 the sum of the two smaller angles is equal to Ln^n^'^- This implies that S2£n-2n'2n\ > TT — 2013(9?). Therefore 2/n2« 2 n i ~ ^n'^n'&n\ — 2/7i27i'2ni - TT + 0:3(9?) > -03(9?). By Proposition 14, the function 2/712^5711 — /dsngTii attains the value — 0:3(93) exactly once. Let 715 denote the position of d$ corresponding to this value. ___ ^ Note that if 0^5 lies in the interior of the smaller arc n27i5, then /.d$n'&n\ < ^.n^n'6n\ and the corresponding value of the function is greater than — a 3 (<^). Now we have 2/711715712 + 0:3(92) = /Ln^n'&n\ for the point 715. Let HQ ^ n'6 denote the point for which HQU\ = UQU^, = 9? and let 714 and 713 be the images of the points n\ and ne, respectively, under the reflection with respect to the midpoint of the segment 712715. It is easy to see that in this way we obtain a hexagon Hn described in Lemma 6 (i). Observe that if we fix the points 7ii,ri2,7i 6 , the hexagon Hn is uniquely determined. Indeed, the greatest angle of the hexagon Hn is at the vertex n6, hence 4.ninf6n5 = /nin 6 7i 5 > 0:6(9?) > 203(9?). Thus /n57i67i2 < 2?r 80:3(9?) < TT. Therefore it is enough to move d$ on the smaller arc n^n^. To complete the proof of (i) it remains to show the monotonicity of ^(9?) and V;*(v?)Since ^(9?) = £n$n6n\ = /TisTigTZi, therefore we will examine the position of the points HI , 712 , 715 , n'Q . Suppose, for contradiction, that ^(9? + ^9?) < t/>(9?) for a suitable small positive A(p. We will change the position of the four points in three steps to reach the position corresponding to 9? + A(f> from the position corresponding to 9?. If ^(9? + Z\9?) < ^(9?), then first rotate 715 around n'6 until ^.n^n'&n\ = ^(9?+Z\9?). Applying Proposition 14 with a = n?, b = HI, c — 7Z6, and d = 715 we obtain for the new position of 715 that 2/711715712 + #3(9?) > If ^(9? + A
holds for the new point n5. Then rotate HI around n'6 until /7ii7i67i2 = 03(9? + ^9?). Now increases, therefore the left hand side of inequality (*) becomes (strictly) greater than the right hand side.

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Leaving the point n'6 and the directions n 6 ni,n 6 n2,n 6 W5 unchanged, increase the lengths of the segments n 6 ni,n 6 n2,n 6 W5 to (p-\-A(f>. Then Ln\n$n'6 and Zn 6 n5iZ2 increase, hence ^nin^n^ also increases. Thus the left hand side of inequality (*) increases again. Now observe that the last position of the points fti,n2,n5,7i6 corresponds to . We know from the previous argument that A.n^n'6n\ < Ln$n'&n\ and a 3 (^) < a3((f>-\- A Zn 5 n 6 n 2 . On the other hand Oi^(ip} < a^((p + A(p) < TT — 0:3 (y> + A(fi) < A.n§n'&ni. Therefore the area of the triangle n$n'&n
This proves that V;*(v?) ^s a strictly decreasing function. (ii) Consider an empty convex hexagon 9i(?2 jLq^qy,^- Let q'6 ^ q@ denote the point for which q'6q5 = q'6qi = p. Since ^q^qsqi > ^.q^q^q^-, then, by Proposition 12 (ii), we have Also £q4q5q6 > ^.q3q4q5, by Proposition 12 (iii). Hence Now £q4qsq'6 > £q'6qiq2 also holds, thus q±q'& > If 92^6 ^ V7; then, taking Proposition 13 into account as well. Proposition 11 shows that the number of edges of the ip graph is not as small as possible, a contradiction.

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If / ^ 92 and leaving its side lengths unchanged deform the quadrangle q& remains unchanged and the distance q<2q'6 and the angle £q$q§q\ also do not change. Furthermore ^.q^q^qi remains the greatest angle and the the hexagon becomes centrally symmetric. Consider the hexagon Hn with the point n'Q and arrange them in such a way that n'6 = q'6, HI — q\, and the points n2 and q2 are not separated by the line q'6q\. We will concentrate on the points 715, n'6 = q'6, n\ = qi, and n-2 of Hn. Suppose, for contradiction, that ^>(<^) > /Lq$q§q\. Then

hence £n5n'6ni > Consider the point 55 as the point d$ in the construction of the hexagon Hn in (i) which, therefore, lies in the interior of the smaller arc n'^n^ and thus 2Zn 2 -a3(<^). This implies that l^n^q^qi + c*3(v?) > Let 94,93, 9g denote the images of £q$qQq\. In the centrally symmetric hexagon 9i<72 ^.q^q^qQ. At the same time, by the centralsymmetry, gsgi = q'4n-2 and ^5^1 = V? a Thus £q$qQqi > t/>((/?). If ^Lq^q&q\ coincides with ^n. in the hexagon 9i
contradiction as we have already seen. = tl>(<£>) and 52^6 ^ ^5 then the hexagon Further, since max(/gig2?3?^949596) < therefore gi
n-

Now the proof of Lemma 6 is complete.

D

Let j < (f < -|. Let ^ = ii^2^3^4^5^6 denote the degenerated convex hexagon of side length (f> for which trfe =

13 anc? max(|-,afc + i) < in which q^qe^q^q^^qeq^ are Delone diagonals, then min(Z<72

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If(f>> 0.805153655, then max(Zg6^g3)- The function $((p) is strictly increasing and TJ)((p) < We note that ^(1.02746114) = 3.092711117 .... Proof. Consider a convex hexagon H = gi ^O Fix a position of Ht and arrange the hexagons Hn and 77 in such a way that ti — HI = ^-t^t^t^ = ^(v 7 )We examine what happens if we move the point 93 / £3 by a sufficiently small amount toward £3 on ^£3. We concentrate on the points ii, qi, 93, 94. For technical reason we change the conditions concerning #4. Instead of requiring that q4 G M4 we require that A-qiq^q^ remains unchanged. Observe that the fact that /
+ 3 cos (p + 2 cos2 (p is a strictly increasing function. We know that as(f) = 1.1070... < 1.2566... = 3p and a 3 (^) = TT < ^, therefore a 3 (<^) < 2^ for f < 9? < ^,

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q^ — Q3v^' <

iQr69i- Then /ge9i92 £ -^939495 by Proposition 12 (ii). We want to prove that ^929394 > ^(v7) f°r V < f an(i ^939495 < ^"(v7) f°r i < < •

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Let q'6 denote the image of qQ under the reflection with respect to the diagonal q\q$. As we have seen at the beginning of the proof of Lemma 6 (ii), if 929e > fi then, by Proposition 11, the number of edges of the

4>((p}- Indeed, if Lq^q&q\ — ^(v?), then A29394 = ^(V>) > ^(V 7 )-

We deform the hexagon 9192 9s 94 9s 9e in such a way that the structure of the Delone triangulation remains unchanged, ^q^q^qi remains the greatest angle, and Z ^959193Consider the triangles 9193*74 and q\qsq4. We know that q\q% = 9193, 9394 = 9394, As9i94 < ^9s9i94, and £q3q4qi, £q~3q4qi < |, hence /9i9s94 > ^9i9s94- Thus Z(?2g394 > ^929394 > ^(v] and Zg39495 < -^939495 < ^'"(v)If 9s9e < 9295 in 9i929s949596, then fix the quadrangle qiq^qsqe and leaving the side lengths unchanged deform the quadrangle 92939495 in such a way that ^929594 decreases. Then Z(?i y, a contradiction. Now we are at one of the previously discussed cases where we have already showed that Zg29s94 > $(). Since Z#29394 decreases while d-q^q^qc, increases during the deformation, the first part of the lemma follows. It remains to prove that 4>(^>) is a strictly increasing function. Consider the hexagon Ht, and let t denote the intersection point of the diagonals t\t^ and £3^5. Set a = Zt^i, ft — /^sti, 7 = /(.tt^. Considering the triangle

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we obtain that sin 7 =

cos (p COS j

I 2 cos2 y 1 -f COS (£>

from which it easily follows that 7 is a strictly decreasing function of (p. Fixing the hypotenuse of the right angled triangle tt^t^ if 7 decreases, then /3 increases. After that, fixing 7, if the hypotenuse ^3^4 increases, then f3 increases again. This implies that ft is a strictly increasing function of (p. Finally, considering the triangle tt\t^ we obtain that v/2 cos (f> + 1 tan a = --2 -- 2 cos (p — I

from which it is easy to see that a is a strictly increasing function of (p. Thus $(¥>) = a + (3 is a strictly increasing function of (p. This completes the proof of Lemma 7. D Lemma 8. Let k > 13 and a^ + i < (p < -|. If in a (p graph of k vertices a convex hexagon (p for 1 < i < 6, then the degree of q? in the Delone triangulation of the corresponding y point set is 5 or 6. Proof. Obviously, the degree of q-r can be 6 and it also can be 5 if

024. Note that if, say, the segment q?qQ can be a Delone edge in no Delone triangulation of the point set, then, by Proposition 9 (iii), we have Obviously, the point q-r is of degree at least three. Suppose that only three Delone triangles are adjacent to 97. If the vertices different from q? of one of these triangles are consecutive vertices, say q\ and 92? of the hexagon and ^.q^q-rq^ and /.qrqsqe > ^qsqrqi from which 4.q4q5q6 > 2?r On the other hand, Lq^q^q\ < 2as(<^) < TT, by Proposition 9 (i),

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if TT, a contradiction. It remains to deal with the case when q7 is of degree 4 and either 97 is adjacent to four consecutive vertices of the hexagon or q7 is not adjacent to a pair of opposite vertices of the hexagon. If, say, the vertices 95, ge are n°t adjacent to 57, then, by Proposition 8, we have Lq^q-iqi + Z^rtfs + /gs?794 < 3a3(v?). Thus £qiq7q4 > 2yr - 3a3(<£>) > 2«3((p) in the Delone triangle 919754, a contradiction by Proposition 9 (i). Now assume that, say, 93 and q@ are not adjacent to 97. Obviously, one of the two regions /.q^qiqQ is convex. Let ft denote the other one. We may assume, without loss of generality, that q4 and 95 are contained in ft. Let q'5 and q'4 denote the points in ft with the property q'5q$ = q'5q7 — q'4q7 = q'4qs = if. Since q7q5 > q7q'5 and g794 > qiq'4, therefore /.q5q&q7 > A^etf? and ^ ^?4?3 d(ip) which implies that ry i [ (Q j

-

Therefore £q4q7q'5 > TT — a4((f>} > a^((p] and thus q'4q'5 > (/?, a contradiction again. D We turn to the preparation for the proof of the main result, Lemma 9, of this section concerning the number of isolated points in convex hexagons of (p graphs. First we prove some technical propositions. Proposition 15. Let qiq^q^qi be a convex quadrangle such that q\q^ = #394 and a = Zgi^gs < ^?2z,qz, the side lengths q\q^ — ?3 K- If we fix the length of the side q\q4 instead of a + ft, then the function a + ft of a is increasing if qq\ + qq4 > TT and is decreasing if

qq\ + qq* < K. Proof. The two statements of Proposition 15 are equivalent, therefore we prove only the latter one. Let a < ft and increase a by a sufficiently small Act. Then ft decreases, say, by Aft. Now q is the instantaneous centre of the motion of the segment q\q4. Let Aq\ and Aq4 denote the amounts of the corresponding moves of qi and 94, respectively. Then Aq\ : Aq4 w ^ing^ : sin qq4. If qq\ + qq4 < TT, then Aq\ < Aq,i a.nd thus Aa < A3 taking 91^2 — 93^4 into account. Therefore the sum a-}- ft decreases. Similar argument shows that if qq\ + qq4 > TT, then a + ft increases. D

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139

Let z\ denote the unique real root of the equation 8z3 — z2 — 4z — 1 = 0. Set 2 arccos Zl = fa = 1.041344073 .... Proposition 16. Let v7 < f ana let P = 9i929s949s be a convex pentagon of side length in which a = ^Lqiq^qz < ^929394 — ft(i) If (p < fa, then there does not exist a pentagon P described above in which a + ft > 27T — o^y?). If (p — fa, then there exists exactly one pentagon P — PI described above in which a + ft > 2?r — a^((p]. Now q\q± = 2 fa and a = ft = T r - ^ M inPlf (ii) If fa < ct( 2<^. This implies that if a -f ft > 2?r — a.^((f>}^ then q\q± > 2(p even holds for the polygonal line <7i"^3 are in this order on ) - 2<5 in P. Let Ps = 51^2935455 be a degenerated convex pentagon of side length (p in which 3i5~4 = 2y> and 9355 = (p and which is not separated from P by the line 9293- Then ^35455 is a regular triangle and 51^29355 is a rhombus in which ^Si92) + 8. Set Z555i 8. Now Zm3<7354 = 6 + #1, hence a + ft = 2?r - a3(v?) - 8 - #1 in Ps- The situation is shown in Figure 6.

BOROCZKY AND SZABO

140

m2

FIGURE 6. The pentagons P and Ps.

Let P = in which a-\-ft — 2?r — 0:3 (v?) and which is not separated from P by the line q^qs- If q\q$ = 2

TT - 6. Then /.q5q3m3 < Z^s^ms, and thus Z a3(
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141

Let Q = riq2q3r4 be a convex quadrangle in which r\q2 — 9293 — 937-4 = Sis4 = 2<^ in this quadrangle Q' = r(q2q3r'4. We will use the notions a = ^-Tiq2q3 and ft = ^92^3^4 in the quadrangle Q as well. We note that the quadrangle Q (and thus the pentagon P) does not exist for (p < 0.74614 since for these values /.r(q2q3 > ^49392, although a < ft. This follows from the monotonicity of as() and a4((p). We will not use the above numerical value in the course of the proof. We will use, however, that if 4.r(q2q3 > ^54^3^2, then in the case of strict inequality no quadrangle Q exists, while in the case of equality Q is congruent with Q' for which there does not exist a corresponding pentagon P. Thus we may assume that ^.r(q2q3 < 4.s4q3q2. Let Q° — r^q2q3r^ be the convex quadrangle in which r^q2 = q2q3 = 93^4 = <£>, ^r^q2qs = ^^293^4 = TT - aY ', and which is not separated from Q by the line q^qz- Let q denote the intersection point of the lines r\q<2 and r4 < ^7-4 in the whole range, hence the pentagon P also does not exist. If r^7-4 < 2). If we decrease ^.r.\q^q2 from the position corresponding to Ps, then ri 7*4 becomes smaller than 1(p and thus we obtain the pentagons P =

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If ^949392 decreases in the pentagon P = 9192*739495 (where r$ = 94), then the diagonal 9194 (= 7^4) decreases. (3) qr^ + q$4 > TT. Then r^ decreases in the whole range as ^49392 decreases. The argument is the same as in (2). We obtained that appropriate convex pentagons P — 9i92
Thus r^r^ < 2ip holds if and only if . . sin (p >^ cos 99 sin -- .\- sin (p cos — cos

Set z = cos ^. Using the fact

2

\r

4 cos2 f'

we obtain that 0 > 8z3-z2-4z-l = f ( z ) . The equation 8z3-z2-4z-l = 0 has a unique real root z\ and f ( z ) < 0 for z < z\. This implies that r\r\ < 2ip if (p > tpi and r^r^ ~ 2(f> if (p = tp\ (recall that ^i = 2arccos^i). To complete the proof of Proposition 16 it remains to prove that the function cx((p) — As\q<2(b introduced in (ii) is strictly decreasing. For a given ty\ < (p < ^, let Ps = 519293*455 be a degenerated convex pentagon described in (ii). Consider the convex quadrangle Q — 9192*7394 for which 9192 = 9293 = 9394 = (p + Atp with a sufficiently small Z\<£, the points 5i and 93 lie on the segments 9291 and 9293, respectively, and ^929354 = We will show that 9194 — 5^52 < 1.7 A^p < 2Atp which implies that a(y>) is a decreasing function. Let (3\ — ^92^154 and /?4 — £.q-$s±s\. Since min(9255,93S5) > 9?, hence Let 9 denote the point for which 939 = (p-\- A(p and 54 lies on the segment Examining the increase of s^s^ at the points si and 54 we obtain that 919- 5i54 w (cos/3i+cos/3 4 )Z\<£ < 2cosa3((p0}A(p < Q.7A
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143

Z. We construct the degenerated convex pentagon P* = s^^qss^s^ of side length (f> + A(f described in (ii). Start with the quadrangle Q — . In this way we have constructed the desired pentagon P* and Zsi<72] is a strictly decreasing function. This completes the proof of Proposition 16. D Proposition 17. Let ip < ^. If a convex hexagon H — qiq^q^q^q^Q of side length (p contains at least two points qi, i > 7, in its interior for which qiqj >y,i>l,j>l,i^j, then • the diagonals of H are greater than (p, • any line connecting two interior points qi, i > 1', goes through interior points of two opposite sides of H, • any segment connecting two interior points qi, i > 1, is a Delone edge of the Delone triangulation in the hexagon. Proof. If a diagonal of H is not greater than 7 (for pentagons cf. Lemma 2). Thus the diagonals of H are greater than d(tp}. We will examine q'rq'8. It is clear that q'7q'8 < 3
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BOROCZKY AND SZABO

The distance of #7 from the line q\q-z is at least m(<^), hence qrq'7 > m((p}. We obtain similarly that qsq'8 > ^(92). Thus q'7q'8 > 2ra(<^) + d( 3
it is enough to prove that —^ and -^ 1 Then

are increasing functions. Set x =

l 1+ x andA J( d ((£>)\ = ,_ V4 - : which are increasing functions. This implies that m((p] and d((p) function from which the assertion follows. This completes the proof of Proposition 17.

are convex D

Proposition 18. Let(p < ^. Consider a convex pentagon P — \P,\, (ii) |Ps| is a strictly increasing function of (p.

Proof, (i) According to Lemma 1, we can deform P to Ps by such kind of deformations in which three consecutive vertices are fixed and the remaining two change in such a way that P remains convex, its sides remain of length 92, its diagonals remain of length at least , then the angle of P at their common endpoints is 80:3(92) > TT, a contradiction. Also observe that if P becomes a triangle, say the triangle (p, I < i < 8, 7 < j < 8, i ^ j.

ARRANGEMENTS OF 13 POINTS ON A SPHERE

(ii)

(iii)

145

Let

13 aracf

We note that assertions (i) and (ii) remain true with qiqj > (f for (f < ty\ and

V, 1 < « < 8, 7 < j < 8, i ^ j. Suppose, for contradiction, that H contains 2 + / points, / > 0, in its interior in such a way that this 2 + / points together with the vertices of H form a (^-saturated point set in H. Then H consists of 8 + 2/ Del one triangles. We will show that the sum of the area of 8 triangles among them is greater that \H(
Now/(z) = 48arcsmf-12arcsin^-47r. I f O < y < f , then 1 < x < -^. Since

24 therefore /'(x 2 ) = 0 for a:2 = y 52/45 while f'(x) > 0 for x < x2 and f'(x} < 0 for x > x-2- This implies that as x (and 1, and = 2 arccos ± = 0.94912607 .... This yields that f ( x } > 0 for 1 < x < x?, which proves that the hexagon H cannot contain two points with the required properties for

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BOROCZKY AND SZABO

and from the vertices of H are at least

< | and let # = \ (> (/?) denote this maximum value. Let us join two points by a segment if and only if their distance is exactly tpi. In this way we obtain a graph. Among these graphs consider one in which the number of edges is as small as possible. This graph will be called a scarlet graph (with scarlet edges). Obviously, an interior point is isolated or is of degree at least three in the scarlet graph and at least one of the interior points is not isolated. It is also obvious that the edges of the scarlet graph divide H into convex polygons of side number at most five. This implies, by Lemma 2, that no interior point can be isolated and no diagonal of H can be a scarlet edge. Observe that two interior points, say q-j and gs, cannot be joined to the same vertex of H by a scarlet edge. Indeed, if, say, q\qj and q\qs are scarlet edges, then the scarlet edges adjacent to q\ divide /g2 |, by Proposition 5 (ii), a contradiction. According to Proposition 17, let q'7 and q'8 denote the interior points of opposite sides, say q\q ^49893 in the triangle q^qs, since Ar^s in the triangle qzqiqs, since q^qs > ^q^q^qi. At the same time 0:3(99) > Zg4g8 ^4<M8> by Proposition 8, and / 0:3(^1), by Proposition 5 (ii). Hence Z ^-q^q&q's taking 0:3(^1) > o^v3) into account. This implies that TT > /^2?394 > ^
147

ARRANGEMENTS OF 13 POINTS ON A SPHERE

Note that the point qz is not adjacent to qg (if it exists) in the scarlet graph, since q-^qs is a Delone edge, by Proposition 17, each scarlet edge is a Delone edge, and the Delone edges do not cross each other. Hence the polygon of the scarlet graph containing q% must be the pentagon 98?45
q4<

q5

FIGURE 7. Scarlet graphs for two and three isolated points. Note that (f>i < ai2 = arccos 4= = 1.1071 ---- Indeed, since q'7q7 > m(ipi) and q'8q8 > m(y?i), therefore q'7q'8 = q'7q7 + q?qs + qsq's > 2m( 2ra(ai2) + a\2 = TT if y>\ > a\2 which contradicts to the fact that H is contained in an open hemisphere. We prove that the hexagon H cannot contain three isolated points from which Lemma 9 (ii) follows. Let V's < ^ < f • Suppose, for contradiction, that there exists a convex hexagon H = qiqiq^q^qsqe, of side length \ in its interior corresponding to a scarlet graph. Let n denote the centre of #7\ and the points ti,\ and the points £3, 97,^9,^7 are in this order. Let /4,/n be the points on
BOROCZKY AND SZABO

148

and angle 013(^1). The situation is shown in Figure 8. By Proposition 17, the vertices
t6

t4

t8

t9

tl

FIGURE 8. Illustration to the case ^3 < f , then nt\ > ^ and thus min(ng2, wg4,n -| which implies that the triangle
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149

(1) Rotate ge around 99 to t$. Then q2q4 remains unchanged, q4qs decreases, and #2i) + v?i > ^3(^3) + ^3 = 1.6473... > f. Next assume that each of the pairs (^3,^5), (Me? Ms), (^8^95*1^2) contains exactly one point among g6^8 and set q'4 € 94^5, q'4ts = ^e^s- Rotate g4 around q8 to g^. Then i. Thus, in what follows, let q6 = q'6. (2) If q2t2 = q4t5 = qet8 does not hold, then q2t2 < q4t5 = q&t8. Set q'2 € M25 ?2^2 — r3(?/'3) + ?i-3 > |, This proves that for

which can contain three points in its interior with the property that the

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BOROCZKY AND SZABO

distances of these three points from each other and from the vertices of the hexagon are at least (p. We continue the proof of Lemma 9 with the examination of hexagons with two isolated points. Recall that such hexagons do not exist for (p < ^2 = 0.949126. Suppose, for contradiction, that there exists a convex hexagon H of side length tp with two isolated points for some (p > fa. For a fixed
9i97, 92 97, 97 q&, 9s 94, 9s 95 We will distinguish two cases: (p and (p\ — (p. First assume that (p. Then we can decrease the area of Hm unless Hm is a regular triangle of side length 2

, <£,2)• Then Zg2949e = ^(K). Obviously, O^K) is an increasing function of ip, therefore S(K) = ^q^q^qz = \(K — OL^(K}} is a decreasing function. Let us define the kernel of the triangle qiqsqs as that part of the triangle q\q^q^ which is covered by none of the interiors of the circles of radius

arccos -4= let #1,93,95 denote those points in the interior of the triangle q2q4q& for which q{q2 = q(q& = 9s96 = 9s94 = 9s94 = 9392 = VThe kernel of the triangle 919395 is contained in the regular triangle 9^9395 whose diameter is 9^93. If /.q(q2q'3 < ^(v) then q(q'3 < 9?, hence Hm exists only for greater value of (p. Since 8 is a decreasing function, therefore Z.q'lq2q'3 = TT — 4^ is an increasing function of
ARRANGEMENTS OF 13 POINTS ON A SPHERE

151

Let fa = 1.0257. Then S(fa) > 0.486955. Hence, for arccos -4 for fa < < faNext consider the case fa <

, 1 < i < 6. Since nsi = nsj, 1 < i,j < 6, therefore the circle of centre n and radius ns\ contains the kernel. Thus it is enough to show that nsi < ^. Since q^n < q\n, therefore £s\nq\ < \^q^nq\ = f • At the same time £s\q\n — az'2^' — as(y) < f ~ f — f? hence £ns\qi > ^. Let s denote the point of the segment s\q\ for which sqi = fa. Since £.ns\q\ > ^j- and nq\ < f, therefore nsi < ns. We prove that ns = u() = ^((f>] is an increasing function. Since 1 — cos < 1 + 3 cos v? + 2 cos2 (f>

is a strictly increasing function, hence the value °^2 ' — a'3(fa) > 2.5 bounds 77(v?) from below on the interval fa < y> < fa. Now, as

) < u;(Vi) < 0.499561 < 0.51285 = \ < f, which implies that it is not possible that Hm is congruent with T(2<^) for ^>^ < ^> < fa since the diameter of the kernel of T(2<^) is smaller than |T(2)| > \T(2fa)\ + 14|TOi)| > 13 > 4?r, a contradiction. In the remaining part of the proof we deal with the case when (p > fa, the value
152

BOROCZKY AND SZABO

adjacent to the side qrqs is at least 2?r — «3(v) and thus, by Proposition 16 (i), we have (p > t^i. This completes the proof of Lemma 9 (i). Now, let -01 < V < f • Consider the degenerated convex hexagon Hr — of side length tp which contains the points 97,93 and for which = 2tp, r\r<2q-r and r^r^q^ are regular triangles of side length ), and thus the pentagons PS and Pj are congruent with the pentagon Ps denned in Proposition 16 (ii). We may assume that a((p) = £riq7q8 — /.r4q8q7 < £q8q7r2 = ^9798^5 =fl(<+>)•We may also assume that for the pentagon PI in Hm we have ^.qiqjqs+^qrqsqs > 2?r - a3((f>) and Z £riq7q8. If qi = ri, then q5 - r5, q4 = r4, and q2 = r2, hence Hm - Hr. If q\ ^ n, i-e., ^.qiq-rqs > £riq7q8, then qir5 > r^s = 2y, which implies that /?598?7 < £r5q8qr. Let P[ = #iq& with the degenerated quadrangle qiqsq'sq'e- This implies, by Remark 1, that |Pi| > \P{\. Also by Remark 1, we have \P{\ > |P4|, since these degenerated pentagons are quadrangles. Hence 2|Pi| > 2|Pi| from which \Hm\ > \Hr\, a contradiction. Thus Hm = Hr if £q\q7q8 + £q7q8qs = 2?r - a 3 (<^)Next assume that Z.q\q7q8 + /.q7q8qs > 2?r — a^((p) in the pentagon PI. Deform PI in such a way that it remains a pentagon of side length y>, the sum ^ and ^9^9798 < Applying Proposition 16 (iii) for Pj% we obtain that /.r\q7q^ < hence /.riq7q8 < £qiq7q8. Then girs > firs = 2<£>, which implies that ^9s9s97 < -^^59897- At the same time /Lr^q^q-r < £q
ARRANGEMENTS OF 13 POINTS ON A SPHERE

153

Since /gi#7 \P^\ > |P4| and |P2| > IP^I > I A | where P{ = #i<77. This yields that |Pi| -f \Pi\ > 2|Pi| from which \Hm\ > \Hr\, a contradiction again. This proves that Hm = Hr. Let ^5 = 1.045 and let v]>\ < (f> < ^5. We give a lower bound for the area of the pentagon Ps(vO - r^r^r^q^q-r. Let Ps = 5152535455 be the degenerated convex pentagon of side length

and 5355 = (p. Then /5i525s = TT — as(v?) < TT — as('0i) < 1.9131.... Let P3(} = ^2r3r4^8?7 denote the degenerated convex pentagon of side length (f> in which r'2r'4 = 2(p and Lr'^q^q-i = a(tps) = 2.1345 . . . (cf. Proposition 16 (ii)). Since the corresponding sides of the quadrangles 51525354 and T^q^q^r'^ are equal and ^515253 < Z.r'4qsq7, therefore 0:3(93) = ^535455 < 4.q7r'2r'3 from which (p < qrr'3. This means that the median q?r'3 corresponding to the greatest side r'2r'4 of the triangle r'2r'4q7 is greater than the half of this side. This implies that the triangle r'2r'4q7 contains the centre of its circumscribed circle. We show that IP^y?)! is an increasing function of (p. Fix the point q$ and the directions q$q7 and qsr'4 in P3( by a sufficiently small Ay>- Let P^((f + A(p) = T2r3rf4q8q'7. Obviously, the triangle r'4qsq'7 contains the triangle r'4q%qi and r'4q'7 > r'4q-r since

\r4q7r'2\. This implies that \P3(

|P^(v)| from which \P3(p}\ > \P3(^i}\ follows. At the same time ^.r^q^qj = a(tp) > a(tp^), by Proposition 16 (ii). Thus |Ps(^)l > 1^(^)1, by Remark 1, which implies that |P3(^)| > |Ps(^i)|. This yields that \Hm\ = 2|T(^)| + 2|P3(v)l > 2|r(^)| + 2|P^)| > 5.2995. Consider a point set described in Lemma 9 (iii). Then there are at least 14 Delone triangles outside H. Hence for the area of the whole sphere we have 4?r > \H\ + 14|T( |J7m| + 14|T(^)| > 12.8775 > 4?r, a contradiction. Finally, let ^5 < (p < ^. Now Hm is divided into two regular triangles of side length

\Ps(ip)\ > |PS(^5)|, i — 1,2. Thus \Hm\ = 2|T(V)| + |Pi| + |P2| >

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BOROCZKY AND SZABO

Consider again a point set described in Lemma 9 (iii). Then there are at least 14 Delone triangles outside H. Hence for the area of the whole sphere we have 4w > \H\ + 14\T()\ > \Hm\ + 14|T(t/>5)| > 2|P S (V> 5 )| + 16|T(^5)| > 12.5791 > 47r, a contradiction. This completes the proof of Lemma 9. D 7. (/P-FINAL GRAPHS

Let k > 13 and max(^r,a/; +1 ) < (p < ^ and consider a (f> graph of k vertices. If the (p graph is non-trivial, then the faces of the graph are regular triangles, rhombi, pentagons, and hexagons. Polygons of side number greater than six cannot occur. The triangles, rhombi, or pentagons of the

ARRANGEMENTS OF 13 POINTS ON A SPHERE

155

If 9i97 = 9597 — ¥>> then on the one hand ^959691 = /.qiqrqe < ot^( V9? tnen again /.q7qiq2 < 7T-a 4 (v?) < a 4 ( graph with some Delone triangles inside. We claim that ^.q^q^q^ < a 4 (v?). First, leaving gi, 97,94, 9s, 96 unchanged, decrease #297 until #297 — ¥> by moving qi and q$ in such a way that q\q-2 = - Then /Lqtq^qz increases. Secondly, leaving 91,92,93,94,97 unchanged, decrease q^q-r until 95959197 becomes an inscribed quadrangle by moving 95 and qe in such a way that 94^5 = q^q^ = ge^i = graph together with the new red edges (of length 13 and max(^,afc + i) <

-final point set of k points where and q^q^ are edges of the corresponding if -final graph. Set ft = (i) If (3 is the angle of a quadrangle in the if -final graph, then ft < ot^ip}. (ii) If ft is the angle of a convex pentagon in the (p-final graph, then ft < (iii) If ft is the angle of an empty convex hexagon in the ip- final graph, then ft < <*(0.805153654) < a 4 (y>). (iv) If ft is an angle in the ip- final graph not listed in (i)-(iii), then ft <

Proof, (i) Obvious. (ii) The statement is true for P((p). Suppose that a convex pentagon 7192935455 m the
156

BOROCZKY AND SZABO

whose symmetry axis goes through q?. Since /Lq\q^q4 < £q\q?,q'4, therefore the points q'4 and q'5 lie outside the circle going through 9i,?2)?3Increase ^qiq^q^ in the symmetric convex pentagon of side length *(y) < ^(0-805153654) < a4(0.70767526) < a4( 13 and max( ^r,a/j + i) < (p < ^. Let q\q^q^ be a triangle in the Delone triangulation of a ip-final point set of k points where q\qi and q^q?, are edges of the corresponding (p-final graph. Set ft = (i) If ft < a^((f] ana #i\ > ki^^^^sl where qiq^q^q'^ is a symmetric convex pentagon of side length (p and its symmetry axis goes through either q'4 or q'5. (iii) For a fixed ft, the functions |#i V? then and qq'4 — q\qy,i i.e. for which the triangles gi TT since the left hand side equals to the sum of the angles of the triangle \qiq3q'4\ with equality if and only if q4 = q'4. (ii) We have seen in Lemma 10 that ft > cx.4(tfi) may occur only if the Delone triangle qiq^q^ is the part of a convex pentagon <7i
ARRANGEMENTS OF 13 POINTS ON A SPHERE

157

<£>-finaI graph. We may assume that (73*75 is a Delone diagonal in the quadrangle 91 939495. Then the point 95 does not lie in the interior of the circle going through 91,92,93- Fix the points 91,92,93 and deform the quadrangle 9i 939495 in such a way that q3q4 = q4q5 = q5qi = (p until 9s, 9i, 92, 9s become cocircular. Let q'4 and q'5 denote the new positions of q4 and 55, respectively. Then #295 = 9i9s > <£ and q3q'5 > (p since ft > a4( 9294- Now, by Remark 1, the area of 91939495 decreases, from which |9i 9293949s! > |9i 92939495! where the symmetry axis of qi 92939X goes through q'4. Equality may occur in the above inequality only if q4 = q'4 and 95 = 95(iii) First we examine the function 1 9192 9394) • Let 91 929394 be a rhombus of side length

) < ft < a4(, If ft > a3((p), then increase

a3(

) < ft < a^((p). If ft = 0:4 (<£>), then (p cannot be increased and the pentagon is divided into a regular triangle and a square of side length 04(^7), then increase

a4((p + A + Arcas^, hence qiq~2 > q3q'4. Let ^4 be the point on the ray emanating from c and going through q'4 such that 9s94 = 9i92- Then q4q > q'4q. Since the triangle 9i92C is contained in the triangle 9i92C, therefore | = £cqiq2 < A.cq\q
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BOROCZKY AND SZABO

g~5 around g~2 and g~i, respectively, toward each other until the angles at the vertices q^ and q\ in the symmetric convex pentagon of side length

g^g. Arrange the pentagons 9192939495 and q^q^q^q^q" in such a way that 9i92 C gig^ 9 ~ g' and both pentagons are contained in the same hemisphere bounded by the line gig2- Let 5 denote the centre of this hemisphere. Consider the points gi? and q^ on the segments g^g" and g2ga, respectively, such that g^gg = g^g^ = (p. Then sq'5 = sq% — sq£ — sgg, since the angles at the vertices 91,91,92,92 *n ^ne pentagons are ft. On the other hand ^9559s < ^95593, hence the distance of q'5q^ from s is greater than the distance of q^q^ from s which is greater than the distance of g^g^ from 5. This implies that 91929395 C gi929s9s- Taking the inequality 994 < gg" into account we obtain that 9192939495 C gig^s^'^s which proves the monotonicity of the function Iqiq^qsq^q'slThis completes the proof of Lemma 11. D For further reference we summarize some properties of y-final graphs in the next remark. Remark 3. Let k > 13 and max(^y : ,a; c +i) <

then the pentagon q\qiqzq^q^ is either a convex pentagon of side length (p, or it can be obtained from a convex hexagon of side length (p by cutting off a Delone triangle from this hexagon. Lemma 12. Let k > 13 and max(^ L ,afc + i) <

<£, therefore the radius of the supporting circle going through g2,96,93 is greater than the radius of the supporting circle going through 92,ge,9i- Hence the segment g29e separates the point q\ and the centre of the supporting circle going through 92,96,93- The radius of the circle going through g2,ge,9s is greater than the radius of the supporting circle going through g2,ge,gs, hence the segment

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before q^qe becomes (p. Note that the greatest angle of the hexagon is ^q5q&qi or Zg2?304First suppose that Lq$q§q\ > ^.q^q^q^. Let q'6 denote the image of the point ^>(y) > ^(1.02746114) = 3.11663278... > 2.461918835... = 2a 3 (f) > 2a3(v?) for 1.02746114 < (p < f and Z$20394 > V'(^) > V'(f) = 2.455117551... > 2.446054045... = 2a3(1.02746114) > 2a3(^) for f < ip < 1.02746114. Thus Z(72 2a 3 (y>) for | < if < | from which ^.q^q^q'^ < «3 ( «3(<^)Consider the semicircle on the circle of radius 9 and centre 94 whose one endpoint is #3 and which contains q$. If a point q moves on this semicircle from 93, then the area of the triangle
= ^-5

<

which yields that ^2^4 < V- Thus ^g lies on the arc qeq'6 of centre 95 again from which q£qs > q'6qz > ty follows in this case as well. This completes the proof of Lemma 12. D Lemma 13. Let ip < |. Let q\qiq?, and q\ q^q^ be two triangles of the Delone triangulation of a (^-saturated point set in which q\q {qiqzQsq^ where (Mi = 9392 = 9 and A.qzq^q\ < as(v?) for the point q3.

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(iii) // 91929394 is an inscribed quadrangle then [91929394! > | ^(y)! where q^ ^ qi, q^q^ = , and qiq^q^q^ is an inscribed quadrangle. The function \q\929394! is a strictly increasing function 0/^949192Proof. Let c and r denote the centre and the radius, respectively, of the circle going through 94,91,92- By Proposition 6, the point 93 lies in the circle going through 94,91,92- Fix the points 94,91,92- We may assume that 9493 > 9392(i) In this case r < (£>. Let 93 ^ q\ and 93 ^ q\ denote the points for which 9394 = 9392 = Sotsfa) > TT. Hence, by Proposition 3, we have 1^492931 > 19402£31 which proves (i). (ii) In this case r > (p. The vertex 93 of the triangle 949293 of minimum area lies either on the circle of radius if and centre 91 or on the circle of radius

q3a

FIGURE 9. Illustration to the proof of Lemma 13 (ii). If 93 lies on the circle of centre 91 and radius c^, then, obviously, [9492931 > |949293| with equality only if 93 = 93 since 9493 > 9293If 93 lies on the circle of centre 92 and radius cr. then it is enough to show that the area of the triangle 949293 decreases as 93 moves from 93 to 93. By Proposition 3, this follows from the inequality ^939492 + ^949^92 > ^949293-

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If O3(9p)-f/ TT for the point GI, then, using the fact that qi lies outside the circle of centre c\ and radius 92, we have ^.c\q^q^ > /.ciq^q^ /Lc\q^q-2 > Zci 0, from which the assertion follows. Now let 03(92) + ^ TT and /3 < TT — 03(9?), therefore a > 03(92). We also know that ft > 03(92). In the triangle q^c\q^ we have 9401 = c\q^ = 92, and /.q^ciq^ > 203(9?), thus, by Proposition 3, the area of this triangle is less than the area of T((p). This implies that 26 + a + 03(92) < 303(92) < /? + 203(92) from which a - £ < (ft — 8} + (03(95) — 6 — e) follows. This proves (ii). (iii) In this case q^c — r. Since r is the radius of a supporting circle, therefore r < 9?. With fixed qi,q2,Q4 the quadrangle gi A.q±cq\ — / 03(9?). We increase /g4gig2 and ^.q\qiq^ in the inscribed quadrangle 13 anc?max(^r,afc+i) < 9? < -|. Let P — q\qiq-$q±q$ be a convex pentagon of side length 9? in a 92 -final graph of k vertices in which q,q^q^ of minimum area. (i) If / 03(9?) + 04(92), then the smaller Delone diagonal of P is of length 9?. (iii) The area of P is a strictly decreasing function of £q±q*>q\. (iv) Let Ps = si 52533455 denote the degenerated convex pentagon of side length 9? in which 3481 = 2y and 5355 = (p. Then \P\ > \PS\(v) With fixed ^.q^q^q\ the area of P is a strictly increasing function of (p. Proof. We may assume that £q*>q\qi > ^.q^q^q^. It follows from Remark 1 that #3
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(iii) First assume that Aq^q^q\ < 013(y>) + a4((p). Then P is symmetric to the perpendicular bisector of 9192- Fix the points #3, #4,95 and deform the quadrangle 93959192 in such a way that ^.q^q^qi increases but q$qi — q\q^ — 929s — <Pand ^94959i < ttsC'tO + Q^^) remain true. Then, by Remark 1, the area of the pentagon decreases. Next fix the points 94,95, q\ and deform the quadrangle 94919293 in such a way that §1^2 — #293 = 9s94 — f remains true and ^q^q^qi increases to £q4q5qi. By Remark 1, the area of the pentagon decreases again. Secondly assume that A.q^q^q\ > 03(9?) + ct4((p). Then q^q^ = 9 in P. This means that 93^4^5 is a regular triangle and 93959192 is a rhombus. If ^949s9i increases, then the greater angle of the rhombus increases, thus its area decreases. (iv) is an immediate consequence of (i), (ii), and (iii). (v) First assume that Zg4 be a sufficiently small positive number and consider the pentagon P( + Atp] = 9i92939495 °^ s^e length (f + Ap in which £q'4q'5q'l - £q'4q'3q'2 = ^qsQi- Obviously, (9495911 = | tnus 7 by Proposition 3 (ii), we have |<7i + A(p)\. Secondly assume that A.q^q^,q\ > as((/?) + 0:4(). Then 93^5 = (p in Let Z\ a Zi(^) + a4((^+Zi^) also holds. Consider the pentagon P((p+A(p) = of side length (/? + A(f> in which 4.q'4q'5q'i = Lq±q$q\ and ^3^5 = 4.q'3q'5q{. Fix the side length in the rhombus 93959194 and decrease ^939591 to ^.q'^q'^q'^. Since ^q^q^qi is the greater angle of the rhombus, hence the area of the rhombus increases. Next fix /.qzq^qi in the (new) rhombus 93959194 and increase the length of its sides to (f> + A(p. Then the area of the rhombus increases again. This implies that |P(y)| < |P(v + A ^939295- If we fix the vertices 91,92,95 and increase ^919293, then the sum of the areas of the rhombi decreases.

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We note that Remark 4 can be proved in a much simpler way as well. Indeed, it is enough to observe that the area of the triangle q\q$qz decreases as /) > ^?293?4 ^ ^97 o-nd decrease then the sum of the areas of the polygons decreases. Proof. If /9i<72. Split this pentagon into the triangles q^q^q'^ q'^q&i q^qeqr- Let q denote the midpoint of q^qs. Fix the points q$ and q§ and increase /#3 ?3?5?65 ^ kiftM^ + l^^kslSecondly deform the rhombus kiMa&l + Ift^Msl follows. This completes the proof of Lemma 16. D Lemma 17. Let <-p < ^. Let P = <7i
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function of £q4q$qi while q\q4 > | (for 0.90455689431 < (p < | this always holds). Proof. Let q denote the midpoint of q3q4. Set £q4q*,q\ = a, £q$q4q\ — A49i?5 = ft, £q4qiq = 1, £qiq±q - & and qiq4 = a > |. As we have seen in the proof of Lemma 14 (iii), the area of P is a strictly decreasing function of a. Furthermore, it is easy to see that ^qsqiqz = 2((3 + 7) and /Lq^qj,q4 = £q?,q4q5 = ft + 8 decrease as a increases in P. We will prove that both (3 + 7 and ft -\- 6 are strictly concave function of a while a > |-. (i) Fix the vertices #3 and q4 and decrease 4.qsq4qs = ft+8 in the symmetric convex pentagon P by a sufficiently small positive A. Then a increases by, say, Aa. Let P' = |, therefore 7 increases but ft -f 7 decreases. We determine the limit lim^_+o -

qi

FIGURE 10. The rotation of the triangle q^q\q4 around q4. Rotate the triangle q$q\q4 around q4 until q$ = q'5. Let q\ denote the new position of q\. The situation is shown in Figure 10. Then £qiq'5q{ = Aa. Since £q\q4q\ = A, therefore q\q\ w Zisina, q$q'5 w A sirup, and q\q( w Aa sirup. Now A.q\q\q^ w ^ and the angle of qiq[ and ^1^5 also tends to | as zA tends to zero, hence ^.qiqiq'-^ w /?, 4.qiq[qi ~ ^ — (/? + 7), and 77 + 7 in the triangle q\q\q'i. Thus cos(ft + 7) cos 7 from which lim

7)

sm a cos 7

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To prove that /3 + 6 is a strictly concave function of a it is enough to show that si^ocos *s a strictly increasing function of a which is obvious by the previously mentioned monotonicity properties of ft + 7, a, and 7. (ii) Fix the vertex q\ and the symmetry axis and decrease ^.qq\q^ = /3 + 7 in the symmetric convex pentagon P by a sufficiently small positive A. Then a increases by, say, Aa. Let P' = <7i ^, therefore 6 > .

qi

q4

FIGURE 11. The rotation of the triangle q$q\q4 around q\. Rotate the triangle q4q$q\ around q\ until q$ — q'5. Let q4 denote the new position of q4. The situation is shown in Figure 11. Then £q4q'5q'4 = Aa and q~4 lies in the region ^.q\q^q. Since ^.q^q\q^ = A, therefore q±q± w zAsina, q$q'5 ~ Asiiap, and 94^4 w Z\o;sin(^. Now 4.qq4q'4 w ^ because the distance of the point q'4 from the line g#i is j. At the same time £q4q4q\ w ^ and the angle of 94^4 and 94^5 also tends to ^ as A tends to zero. Hence £q4q4q'4 « /?, TT — (^3 + 7), and £q4q4q'4 w 6 in the triangle q±q4q'4. Thus sn <

from which lim

7)

sin a sin 6

To prove that ft + 7 is a strictly concave function of a it is enough to show that g|"qj^g is a strictly increasing function of a which is obvious by the previously mentioned monotonicity properties of ft + <5, a, and 6 (do not forget a > f and ft + tf > <5 > \).

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This proves Lemma 17 for a > ~. Since a > 05(9) and sin | — sin 9 sin ^, therefore a > | for sin2 9 > ^1, i.e., for 9 > 0.90455689430.... D Proposition 19. Let k > 13 and max(^ L ,a/,. +1 ) < (p < |. // 603(9) > 2?r around this point, or at least one blue edge is adjacent to this point in which case 2?r > 04(92) + 403(9) > 04(024) + 403(024) = 2?r around this point, a contradiction in both cases. D Let R((p) denote the rhombus of side length 9 which has an angle 2ir — 20:3(9) - 0:4(9)- Arrange the polygons S() in this order around a common vertex of them without overlapping each other such that the angle of R( 13 and max(0.90455689431,aA+1) < 9 < f. // the degree of a point q is six in the Delone triangulation of a (p-final point set of k points, then the sum of the areas of the six triangles adjacent to q is at least \Hs(
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We know that the angle between two consecutive red edges is at least 03(9) and we also know from Proposition 7 that the angle between two red edges in the pattern red-blue-red is at least 04(9). At the same time the angle between two consecutive red edges is less than 04(9) and the angle between two red edges in the pattern red-blue-red is less than 203(9). Indeed, if the angle between two consecutive red edges is at least 04(9), then 2?r > 03(9) + 304(9) > ^3(024) + 3o 4 (a 2 4) > 6.3343 > 2?r, while if the angle between two red edges in the pattern red-blue-red is at least 203(9), then 2?r > 4a3() > 40:3(024) + 0-4(024) = 2?r around 9, a contradiction in both cases. Leaving the colour pattern of the edges adjacent to q unchanged we will deform the hexagon q\929s94959e in order to decrease its area. Let q'4 and q'& denote the images of 9 under the reflections with respect to the lines 9395 and 9591, respectively. Since max(/93995,/9599i) < 203(9), therefore, by Lemma 13 (i), we have |993949s| > |993949s| and |99s969i| > 1995964i I-

In what follows we examine the hexagon 9i9293949596 instead of the hexagon 919293949596Decrease ^91992 and ^92993 to 03(9) in the hexagon 9i929394959e m such a way that none of ^93995 and ^95991 decreases. Obviously, 04(9) < ^93995 < 203(9) and a4()l(b) Assume that the two blue edges are consecutive edges in the cyclic order of the edges around 9 in the Delone triangulation. Without loss of generality we may assume that 995 and 995 are the blue edges. Then, by Remark 3, the pentagon 994959591 is either a convex pentagon of side length 9, or it can be obtained from a convex hexagon of side length 9 by cutting off a Delone triangle from this hexagon. In the latter case if 994959e9i is not a convex pentagon of side length 9, then replace 994959591 with a convex pentagon of side length 9 whose area is less than the area of 994959e9i and in which the Delone diagonals are also adjacent to 9 (cf. Lemma 12). Thus we may assume that 9495 = 9595 = 9e9i = 9We know that ^91994 < 27r-3o3(9) < 27r-3o3(a24) = O3(a 2 4)+O'4(a24) < <23(<^) + a4((t>) in the pentagon 994959591 of side length 9. Since ^91994 > 05(9), therefore the angle between two consecutive red edges is less than 04(9). Indeed, if the angle between two consecutive red edges is at least 04(9) for some 9 > 9* = 0.84814272, then 2?r > 203(9) + «4(v) + ^(v) > 203(9*) + Ct4( 6.28318534 > 2?r around 9, a contradiction.

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Replace qq^q^q&qi with a symmetric convex pentagon qq^q^q&q\ of side length if. We know from the proof of Lemma 14 (i) that the area of qq^q^q^qi is not greater than the area of qq^q^q&q\. By Lemma 14 (iii), the area of the pentagon qq^q^qi increases if £q\qq± increases. Now decrease £q\qq-, ?3??4, respectively, to a^ip). Then the areas of these triangles and the above pentagon decrease. This implies that the sum of the areas of the six triangles adjacent to q is at least 3 T((p)\ + |P C (<£>)| where Pc(v) 'ls the symmetric convex pentagon of side length (p whose two greatest angles are 2?r — 3a3(<^). Arrange the polygons P C ((^),T((/?),T(}\ > \Rs((p)\. As a first step, we show that both \Hp() in the symmetric convex pentagon P c (v)- Then, by Lemma 14 (iii), the area of the pentagon increases. Secondly, leaving the greatest angle 2vr — 303(1^ + Atp} unchanged increase the side length (p to

\Hs((p)\ follows from the next table.

V 7T/3

1.01091770670 0.98181291240 0.95785375082 0.93773682084 0.92058198917 0.90577007057 0.89285051676 0.8814861579ft 0.87141864083 0.86244598312

\H. \HP 3.82126647249803 . . .4.25023082951690 . . . 3.50146649194088 . . .3.82126647256370 ... 3.25456862549919 ... 3.50146649194355 . . . 3.05848758575130 ... 3.25456862555406 . . . 2.89916943157644 . . .3.05848758580918 . . . 2.76729350927521 ... 2.89916943164899 ... 2.65644890911128 . . .2.76729350929594 ... 2.56208489037588 . . .2.65644890915333 . . . 2.48088210916607 . . . 2.56208489041388 . . . 2.41036177505207 ... 2.48088210921272 ... 2.34863451519861 ... 2.41036177510203 ...

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(ii) Assume that three red and three blue edges are adjacent to q in the Delone triangulation. By Remark 3, the three blue edges are not consecutive edges in the cyclic order of the edges around q in the Delone triangulation. (a) Assume that no two blue edges are consecutive edges in the cyclic order of the edges around q in the Delone triangulation. Without loss of generality we may assume that qq\, qq$, and qq$ are the red edges. If the angle between two red edges, say Lq\qq^^ is at least 203(92), then, by Lemma 11 (ii), we have \qq\qiq-3\ > Iqqiq^sl where q'2q = q'2q\ = <£> and the line q\q% separates q and q'2. Hence |qiq?q%q±q$q&\ > | I-^X^)! from which the assertion follows. Thus assume that the angle between any two red edges is smaller than 203(99). Replace the points q2,q4,q& with the points q'2,q'4,qQ, respectively, such that qiq'2 - q3q'2 = q3q'4 = q5q'4 = q5q'6 = qiq'6 = ki^s^s^l- Applying Remark 4 to the rhombi separated by the red edges, we obtain that the sum of the area of these three rhombi is minimal if the greatest one is 203(99) while the smallest one is 04(92) among the angles ^)\. (b) Assume that two blue edges and two red edges are consecutive edges in the cyclic order of the edges around q in the Delone triangulation. Without loss of generality we may assume that qq\, 593, and qq$ are the red edges. We will examine the triangle qqeqi, the quadrangle qqiq^qz, and the pentagon 203(92) cannot occur, because, by Proposition 7, in this case 2-rr > 2as((/?) + 04(99) + a>$(i£>) > 2a3( 2?r around the point 203(99), then, by the same argument we used at the beginning of case (ii)(a), we are also done. If ^.q^qqe > 03(99) + #4(92)? then, by Remark 3, Lemma 12, and Lemma 14 (ii), we can replace the pentagon qq^q^q^qQ with a convex pentagon qq3q'4q'5q6 of side length 99 whose area is smaller than the area of qq^q^q^q^ and in which the Delone diameters are qq'4 and qq'5 one of them is of length 99. In this way we are again in a situation examined in (i).

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Thus, in what follows, assume that ^91993 < 2a3(p) and ^93995 < Let q' and q'2 denote the images of the point q under the reflections with respect to the lines q\ qe and 9193, respectively. By Lemma 13 (i), the area of the rhombus 99i929s is not greater than the area of the quadrangle g Ai993 (> «4(v)), then replace the triangles q&qqi,q\ qq'2 ,92993 with the triangles q6qq',q'qqi-,qiqq3- The six triangles around q obtained in this way remain of type described in (ii)(b), the only difference is that now qq' is the blue edge instead of qq'2. At the same time, this replacement decrease the sum of the areas of the triangles around q since |9i9293| = I9i99s| > I9i99e| = \qiq'qe\- Thus we may assume that / 0:4(92)) increases. At the same time |9i939e| also decreases since q\ moves on the circle of centre q going through the points 93,9i,ge and £q6qqi < ^9i99sWe may rotate q\ around q toward q@ until either £q\qq3 = 2a3(ip) which has already been examined in (i), or q\qQ = if. Next consider the pentagon qq^q^q^q^. By Remark 3, Lemma 12, and Lemma 14 (i), there exists a symmetric convex pentagon qqsq'^qe of side length
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171

the Delone edges remained unchanged. ^From this I follows. (iii) Assume that six blue edges are adjacent to q in the Delone triangulation. Then (p, I < i < 6, and qqi, 1 < i < 6, are the Delone edges in q\q2q^q^q?>q&- We prove that \Hm\ = \H.()\. Since Hm is of minimum area, therefore, fixing four consecutive vertices and the side lengths of Hm, we cannot move the remaining two vertices of Hm in such a way that the greater one of the angles at these two vertices increases. Only the point q can prevent this increasing, in such a way that either the distance of q from some vertices of Hm are (p or some triplets of consecutive vertices of Hm together with q form inscribed quadrangles. If Hm is of minimum area, then it is clear that these conditions fix the position of q in Hm . We know from Proposition 9 (iii) that qiqj > d((p), 1 < « < .7 < 6. We will distinguish four essentially different cases. For convenience, if qi-iqiqi+iq, 1 < i < 6 (go — <7e> 2|%>)| + 2|r(y>)| > |%»)| + |fl(vO| + 2|r(y>)| = l#.(v»)l(b) The point q is a vertex of two inscribed quadrangles corresponding to two consecutive vertices, say q\ and q-z, of Hm. Then ) < TT. Therefore the move of q into the region ^.q^qqe is prevented by the condition qq$ — qq@ = (p. This implies that ^qsqqe = otsfo). Thus we have 2?r < a5((/?) + 3a3(vp) < a 5 (ai 2 ) + 80:3(012) = 2?r around the point q, a contradiction. (c) There are at least three vertices of Hm whose distance from q is

) < TT, therefore the move of q into the region £q<2qq& is not prevented by the presence of ^3,^4,95 themselves. Thus at least one of qq^ = (p and qqe — ^ hold. Al the same time the move of the points 93, #4, 95 is prevented

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by at least one of the conditions qqs =

. We distinguish two subcases. (dl) Assume that qq% = qq§ — (p. Then q&qiqiq is congruent with S(<£>). If 994 = <£>, then we have a situation already examined in (i)(a) or (ii)(a). On the other hand, if qq$ > 92, then qq% — qq$ —

(f> (the case 992 —

-> then we can rotate q around q$ into the interior of the circle going through 9e?9i,92 5 a contradiction. Thus 9e9i92939 is a symmetric convex pentagon of side length (f>, hence /9e99s < a 3 (<^) + « 4 (^) < TT. On the other hand, the move of the points 94,95 is prevented by at least one of the conditions qq^ = (p and qq5 — (f>. Therefore we have a situation already examined in (i)(b) or (ii)(b). This completes the proof of Lemma 18. D We close this section with two lemmas which give information about the possible vertex degrees in Delone triangulations of <^-final point sets. Lemma 19. Let k > 13 and max(1.01427543570,afc+i) < y < f. There exists no point whose degree is at least seven in the Delone triangulation of a (p-final point set of k points. Proof. Set (f>0 = 1.01427543570. A point of degree at least seven in the Delone triangulation is a point of degree at least three in the <^-fmal graph. By Remark 3, it can be only at most two blue edges between two consecutive red edges. Thus, by Proposition 7, the average of the angles of the Delone triangles at their common vertex between two consecutive red edges is at least 0:3(y), a4.^ , or Q5^ if the number of blue edges between the two red edges is 0, 1, or 2, respectively. Since o^v?) > 0:3(^0) > 1.217864400 > |, oiM > 2i^£ol > 0.942310624 > f, and ^M > ^1 > Q.788239891 > \, therefore there does not exist a point of degree at least eight in the Delone triangulation. Suppose, for contradiction, that there exists a point, say gg, of degree seven in the Delone triangulation. Let 9i9892,929893,939894,949895,959s96,969897, 9?989i be the Delone triangles adjacent to gg. By Proposition 19, the degree of qs, in the final graph is three or four. (i) Assume that qs is adjacent to four red and three blue edges. If no two of the blue edges are consecutive, then, by Proposition 7, we have 2?r > as(<^) + 30:4(92) > QJ3((£>0) + 30:4(^0) > 6.871728145 > 2?r around gg, a contradiction. If there are two consecutive blue edges, then, again by Proposition 7, wo hove 2;r > 2a3((f>) + a4() > 2a3(^o) + «4(^o) + a 5 (^o) > 6.806628942 > 2?r around gg, a contradiction again.

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173

(ii) Assume that q$ is adjacent to three red and four blue edges. Note that three blue edges cannot be consecutive. We distinguish two cases. (a) Assume that no two of the red edges are consecutive. We may assume, without loss of generality, that q\q%, as( 2?r, therefore £qiqsq$ < 2?r — Thus, in view of Remark 3, Lemma 12, and Lemma 14 (i), we have Iqiqsqsqeqrl > ^iq&qsq'M where q\q8q$q'&q'7 is a symmetric convex pentagon of side length

)-\-Oi^((f]-\- Oi$((p) > 2 ?r, hence ^.qiqsqs < 2?r — a 5 (y?) < 2a3(v?). Similarly, Z^stfs < 2a3(p). Hence, by Lemma 13 (i), we have \q\qiqzqs\ > |<7i<72<73<78| and Iqsq'^qsqsl where the quadrangles qiq^qsqs and qsq'^qsqs are rhombi. This implies that \qi q-2q3q^q^q&q7\ > ^iq^^^'MWe may assume that 4.qiq$q3 < ^.q^qsqs. Our lower and upper bounds for the angles between the red edges remain true while the degree of q$ remains seven in the Delone triangulation, i.e., the structure of the Delone edges around q$ remains unchanged. Fix the pentagon qiq^q^q^q'-^ and the structure of the (/7-final graph, and increase ^.q^q^q^ until the quadrangle qiq^qsqs becomes congruent with S(tp). By Remark 4, the sum of the areas of the rhombi qiq^qsqs and qsq'^qsqs decreases during this deformation. Also V remains true since ^.q^q^q^ < 2a^((p) holds during the deformation. Let qi&Qsqs and q^q^q^qs denote the new rhombi (here <7i)). Then Iqiq'^qsq'^q'M > kiMs^s^T-lFix the square qi&qsqs and move the point q$ in such a way that qs^qsqs remains a rhombus and )Let S(S,S,p) = <7i).

174

BOROCZKY AND SZABO

Thus |9i929394959697l > min(|5'(5', S,p}\ , \S(S, r, P)|). At the end of the proof we will show that both \S(S,S,p)\ + 15|T(»| > 4?r and |5(5,r,P)| + 15|T(y>)| > 4?r hold, a contradiction. (b) Assume that there exist two consecutive red edges. We may assume, without loss of generality, that 9193, 929s, and 9593 are the red edges. By Proposition 7, we have ^919392 > «3(^) and min(Zg 2 |9293949598| and |g596979i98| > 1 9596 ^ki 9s | where q2q'3q'4q5q8 and q5q'6q'7qiqs are symmetric convex pentagon of side length y? whose symmetry axes go through q5. Thus Iqiqiqsqiqsqsqrl > ^iqiq'^h^'MOur lower and upper bounds for the angles between the red edges remain true while the degree of gg remains seven in the Delone triangulation, i.e., the structure of the Delone edges around q$ remains unchanged. Fix the triangle q\q^qs and increase ^q^q&qi in such a way that q^q^q^qsqs and qsqQq'^qiqs remain symmetric convex pentagon of side length if> whose symmetry axes goes through 95. By Lemma 17, the sum of the areas of the two pentagons is a strictly concave function of /.q^qsqi- The sum of the areas of the two pentagons is maximal in the symmetric position, i.e., where /.q^qsqs — ^.qsqgqi, hence the sum of the areas of the two pentagons decreases as /.q5qsq\ increase and attains its minimum where ^q^q^qi is maximal, i.e., where 929s949598 is congruent with P( \9i 929.394 959697 1-

Now /9i9892 < 2?r - 2a 5 (). If 9192 > ), and 9596979i98 is a symmetric convex pentagon of side length (£> whose symmetry axis goes through q'5. We know that 1 91929394 9s 9697 1 > \S(T,P,p)\. We will show that |5(T,P,p)| + 15|T(^)| > 4?r, a contradiction. To finish the proof we show that 4?r — 15|T(<^)| < min(|5'(5<, 5*,p)|, \S(S,r,P)\,\S(T,P,p)\) for ^ <

ARRANGEMENTS OF 13 POINTS ON A SPHERE

175

First we show that \S(S, S»|, \S(S, r, P)\,\S(T, P,p)\ are increasing functions of (p. We know that |T(<^)|, |6*(<^)|, |P(<^)| are increasing functions of The greater angle ^.q^qsq^ = 27r — &$(<£) — as(^) is a strictly decreasing function while the sides are strictly increasing functions of

D

Lemma 20. Let k > 13 and max(l. 02029547044, ak+l) < y < f . There exists no point whose degree is three in the Delone triangulation of a p-final point set of k points. Proof. Set ^9s9i94 > ^949192- Set Zg29i9s = fti, ^9s9i94 = #2, ^949i92 = fa- We know from Lemma 10 that ft < a 5 (y?) < a5(|), hence /?3 > 27r-2a 5 (^) > 27T-2a 5 (|). On the other hand /32 > 2x - ft - f32 from which / 3 2 > 7 r - § - > 7 r > 1.93566 > a4(§) > a 4 (v») (> By Lemma 11 (ii), for j = 1,2, there exists a symmetric convex pentagon j(/3j,q>) of side length

176

BOROCZKY AND SZABO

Now / | Mj(/3j, |M 3 G3 3 ,¥>)|. If /?3 > 0:4(9?), then the quadrangle qiq^q^q^ is a part of a convex pentagon PS of side length 9?. Then, by Lemma 11 (ii), there exists a symmetric convex pentagon MS ($3, 9?) of side length 9? in which two adjacent angles are fa and Let Ti,T2,... ,T 2 /c_4 denote the triangles of the Delone triangulation of the 9>final point set. We may assume that 1 < n < 8 for the triangles Tn contained in the polygons Pi,Pz,Q and that 1 < n < 9 for the triangles Tn contained in the polygons Pi^P^^P^For j = 1,2,3, if we fix /9j, then the function \Mj(flj,(p)\ is a strictly increasing function of 9? while Mj(flj, 04(97) holds, then Mj(/3j,ipi) also exists since 04(97) > 04(9?!). If we fix /?3, then the function \M'z(f3^^(f>}\ is a strictly increasing function of 9? while M'^(ft^^(p) is a rhombus, i.e., while fa < 04(97). We will decrease 97 to (pi. Note that M^(fa^(f>] remains a rhombus in the whole range if fa 5: 04(97}). If fa > 04(971), then there exists a value 971 < 97' < 97 such that fa = 0.4(97'). Hence \M^(fa^}\ + |T(y»)| > |M^(/3 3 V)I + I?V)I = Thus we obtain that (i) if fa > 04(97), then

n=10

(ii) if 04(921) < fa < 04(97), then

ARRANGEMENTS OF 13 POINTS ON A SPHERE

177

(iii) if 03 < 0:4(921), then 2

2Ar-4 n=9

The function jMg (,$3,921)1 is a strictly increasing function of $3 since if we decrease the smaller angle of a rhombus, then the area of the rhombus decreases. For j — 1,2,3, the function |Mj(/3j, 0:4(951) must hold. We have seen that f3\ > 02 > 0-4(921). If we decrease (3% to such a 0'3 for which #, > o 4 (y?i) > P'3, then |M 3 (/3 3 , l^3(«4(Vi)>Vi)l = 15(^)1 + ^(^)1 - |^(o 4 (^),^)| + |T(^i)| > IM^^OI + ir^!)!. Now we distinguish two cases. (a) Assume that fti>^-. We have seen at the beginning of the proof that & > ft = 2?r - 2a 5 (i). Hence IMi^,^)! + \M3(^^\ > |M3(^,^i)| > |M^,v»i)| + |T(v>i)|, and \M3(03,¥i)\ > Thus in all cases (i)-(iii) above we have 4?r > 2|Mi(^,9?i)| + \M3(03,if>i)\ + 14|T( 12.74423472838 > 4?r, a contradiction. (b) Assume that f}2 < %j-- We have seen at the beginning of the proof that fa > ftZ = TT - a5 (^/ 3 ). At the same time 83 = IK - fa - fa > ,#T = f -a 5 (f). Hence |M1(A,9l)| + |M 2 (/9 2 ,^i)| > (M^f and M3

178

BOROCZKY AND SZABO

Thus in all cases (i)-(iii) above we have 4?r > |Mi(3f, 12.56637061459 > 4?r, a contradiction again. This completes the proof of Lemma 20. D 8. A GRAPH LEMMA

In this section we concentrate on the combinatorial structure of Delone triangulations of (/>final point sets of exactly 13 points. Lemma 21. Let G be the graph of the Delone triangulation of a set of 13 points on S2. Suppose that each vertex of G is of degree four, five, or six. Then G contains two non-adjacent vertices of degree six. Proof. Let {pi,p2> • • • 5 £>is} denote the vertex set of G. As usual, the cycle formed by the edges pilpi2iPi2Pi3i • • • iPii-iPinPiiPii °f G will be denoted by (pi1,pi2,... ,pi,). If (piiPjiPi) bounds a face of G, then this will be denoted by \pitPjiPi]- If (pi,PjiPi) does not bound a face of G, then this will be denoted by ] p i , p j , p i [ . First we prove that every cycle of length three in G bounds a face of G. Suppose, for contradiction, that, say, }PI,P2,P3[- Split G into the graphs GI and G% of vertex number v and 16 — v, respectively, along the cycle (pi,p2,Ps) in such a way that (P\,P2,P3) belongs to both G\ and GI, i.e., \P\iPiiPz] holds in both G\ and G-2- We may assume, without loss of generality, that 4 < v < 8. The cycle (pi,p2,p3) will be called the boundary of GI and G<2, the remaining parts of G\ and G? will be called the interiors of G\ and 6*2, respectively. It is clear that only the boundary can contain vertices of degree three, the interior vertices are of degree at least four. This immediately implies that v = 4,5 cannot occur, otherwise we have an interior vertex of G\ of degree three. Thus we may assume that 6 < v < 8. For i = 1,2, at most one of the vertices P1,P2-,P3 can be of degree three in Gi since a triangulated planar graph of at least five vertices cannot contain two adjacent vertices of degree three. Hence, for i = 1,2, at most one of the vertices pi,p2->P3 can be of degree three and at most one of the vertices P\,P2,P3 can be of degree five in G;, the other vertices among pi,p2,Ps are of degree four in G;. Assume that the degree of p\ is as large as the degrees of p% and p$ in GI. Simple case analysis show that • for v = 6 there are two essentially different triangulated simple planar graphs, exactly one of them is suitable as GI, • for v = 1 there are five essentially different triangulated simple planar graphs, exactly one of them is suitable as GI, • for v = 8 there are fourteen essentially different triangulated simple planar graphs, exactly one of them is suitable as G\.

ARRANGEMENTS OF 13 POINTS ON A SPHERE

FIGURE 12. The triangulated simple planar graphs for v = 6,7,8.

179

BOROCZKY AND SZABO

180

Figure 12 shows the possible triangulated simple planar graphs for v = 6,7,8. In this figure each graph is stereographically projected from one of its vertex of maximum degree. The graphs suitable as G\ are boxed in Figure 12. For v — 8, the graph GI also has eight vertices, hence it is identical - up to a cyclic permutation of the vertices P1,P2-,P3 - to GI. This means that Pi is of degree at least seven in G, a contradiction. For v = 7, the vertex p\ is of degree three. Let p4 denote the third neighbour of p\ different from p2 and p^ in G<2- Now the cycle (P4,P2,P3) splits G into two graphs each of which is of eight vertices. But this is impossible as we have already seen. For v — 6, if a vertex, say p1? is of degree three in 6*2, then, denoting by p4 the third neighbour of p\ different from p2 and ^3 in G^-, the cycle (P4,P2,P3] splits G into two graphs of seven and nine vertices, respectively, which is impossible. On the other hand, if each oip\,p2,p$ is of degree four, then let Pi^-, Pi2 •, Piz denote the interior vertices of GI for which \pi-L,Pi,p?\, \Pi2iPiiP3\-, and \piz,Pi,p2\ hold in (72. Now Pil,pi2,piz are pairwise distinct and they form a cycle which splits G into two graphs of seven and nine vertices, respectively, a contradiction again. This proves that every cycle of length three in G bounds a face of G. Next we prove that if G contains a vertex of degree five whose neighbours are also of degree five, then G is uniquely determined. Namely, G contains exactly one vertex of degree four, two of its neighbours are non-adjacent vertices of degree six, and the remaining ten vertices are of degree five.

p9

plO

FIGURE 13. The (first and second) neighbours of p\ in G. Let pi be the vertex of degree five whose neighbours are also of degree five. I- t P2,p3iP4iP5iP6 be the five neighbours of p\ in the cyclic order of vertices around p\ in G. Finally, let pil-,pi2-,Piz-lPi±-,Pi$ be the vertices of G different from pi for which lpil,p2,p3],lpi2,P3,P4],[pi3,P4,p5},[pii,p5,P6},[pi5,P6,Pi] p

ARRANGEMENTS OF 13 POINTS ON A SPHERE

181

hold. It is easy to see that 21,^2,23,^4,^5 are pairwise distinct and each of them is at least seven (do not forget that each cycle of length three in G bounds a face of G}. Set ij = j + 6 for 1 < j' < 5. Now the vertices P7,P8,.P9,P10,P11 form a cycle (p?,.P8,P9,Pio,Pii) in G. The situation is shown in Figure 13. The remaining two vertices, pi2 and ^13, lie in the pentagon bounded by (p7iP8,P9,PiOiPu)- Since p^ and ^13 are of degree at least four, therefore there exists a vertex among p7,pg,;P9,Pio,Pii, say p?, which is adjacent to both pi2 and ^13. The degree of p-? is at most six, hence we have - up to symmetry - that [pr,p8,Pi3],[p7,Pi3,Pi2],b7,Pi2,Pn]- Now consider the triangulated empty hexagon bounded by (p8,P9,Pio,Pii,Pi2,Pi3) in G. Then either [^9,^13,^12] or bio, Pis, ^12] holds otherwise p12 or p^3 is of degree three. We may assume, again by symmetry, that bio,Pis,^12] holds. Since pio is of degree at most six, therefore p\s is adjacent to p$. From this the assertion follows. Now Lemma 21 can be proved as follows. The last observation implies that G contains at least two vertices of degree six. Indeed, by Euler theorem, we have CQ — €4 — 1 where cn denotes the number of vertices of degree n in G. Hence, if CQ = 1, then 05 = 12. However, in this case G contains a vertex of degree five whose neighbours are also of degree five, thus G contains two vertices of degree six, a contradiction. First assume, for contradiction, that CQ = 2 and the two vertices, say Pi and £>2, of degree six are neighbours. Now the neighbours of p\ and p% form a cycle (p3,p4,p5,pe,P7,P8,P9,Pio) of length eight where the vertices Ps,P4,P5,P6,P7 are incident to p% and the vertices p7,Ps,P9,Pio,Ps are incident to pi. The situation is shown in Figure 14. The octagon bounded by (P3,P4,P5,P6,P7,P8,P9,P10) contains the vertices pii,pi2,Pis- We know that G contains a vertex pi of degree four. We will distinguish three different cases. p4

p3

p7

p8

FIGURE 14. The neighbours of p\ and p? in G. (i) i = 3,7. We may assume, without loss of generality, that i = 3. Then P4 and PIQ are adjacent in G. Now pu is of degree five and its neighbours

182

BOROCZKY AND SZABO

are also of degree five, thus G contains two non-adjacent vertices of degree six, a contradiction. (ii) i — 4,5,6,8,9,10. We may assume, by symmetry, that i — 8 or i — 9. It is easy to see that p4 is adjacent to one of pn,pi2,Pi3, say to pn. If i = 9, then pii is adjacent to p8 and pw and the cycle (p3,P4,P5,P6-,P7,p8,Pii,Pio) separates the vertices PI,P2,P9 from the vertex p\^- Now p\2 is of degree five and its neighbours are also of degree five, thus G contains two non-adjacent vertices of degree six, a contradiction again. (iii) i = 11,12,13. Consider the octagon bounded by the cycle (^3,^4,^5, P6-,P7,P8->P9iPio) and containing the vertices pn-,Pi2,Pi3- Let Pj denote the fifth neighbour of p^ different from pi,p2,P4,Pio- Since G is simple and the vertices P3,J>4,£>5,P6,P7,P8,P9,£>10 are of degree five, therefore j > 11. Set j = 11. Similar argument shows that the fifth neighbour of pi different from pi,p2,p6,p8 is either p\i or pis, say p^. Also, the fifth neighbour of P4 different from p^^Ps^Ps^Pn must be ^13. Now consider the triangulated empty octagon bounded by the cycle (p5,p6,Pi2-,P8,P9,Pw,Pu-,Pi3)- Note that from each vertex different from pn of this octagon starts at least one diagonal of the octagon belonging to the triangulation. However, this is impossible since in any triangulation of a polygon we can always find at least two vertices which are incident to none of the diagonals belonging to the triangulation. Next assume, for contradiction, that c& — 3 and the three vertices, say pi, P2-, and ps, of degree six are pairwise neighbours. Now the neighbours of pi, p2, and p3 form a cycle C = (p 4 ,P5,P6,J>7,P8,P9,Pio,Pi 1,^12) of length nine where the vertices P4,p$,p6,p7 are incident to ^2, the vertices p7,ps,.P9?Pio are incident to pa, and the vertices £>io,;Pii,Pi2,P4 are incident to p\. The situation is shown in Figure 15. The nine-gon bounded by C contains the vertex 13.

p6

p9

p5

plO

pll

FIGURE 15. The neighbours of pi,p2,Ps in G.

ARRANGEMENTS OF 13 POINTS ON A SPHERE

183

The neighbours of pis form a cycle C' = (pinPi2, • • • ,p 2i ) of length either four or five in G. Observe that if a vertex pi of C belongs to C", then at least one of the two neighbours of pi in C also belongs to C", otherwise Pi is of degree at least six in G. This implies that either p^, p; 2 ,... ,pi{ are consecutive vertices in C or p^,^, • • • ,p;( can be partitioned into two groups of cardinalities 2 and / - 2, respectively, so that the vertices in each of these groups are consecutive vertices in C (do not forget / < 5). Now the remaining 9 — / vertices of C are also either consecutive vertices in C or they can be partitioned into two groups so that the vertices in each of these groups are consecutive vertices in C. Thus there exist at least two but at most five consecutive vertices in C which do not belong to C'. By symmetry, we may assume that these points are p e ? - - - ,Pm where 7 < m < 10. We may also assume that p$ = p;2 and pm+i — Pi3 • This means that there exists a triangulated empty polygon P in G bounded by the cycle (ps,... ,p m+ i) of length at least four. Since both p$ and p m +i are of degree at most five, therefore they can be adjacent to none of the diagonals belonging to the triangulation of P. However, p$ and pm+i are neighbouring vertices of P, therefore at least one of them must be adjacent to a diagonal belonging to the triangulation of P, a contradiction. This proves that if CQ — 3, then there exist two non-adjacent vertices of degree six in G. Finally, if CQ > 3, then trivially exist two non-adjacent vertices of degree six in G. This completes the proof of Lemma 21. D Lemma 19, Lemma 20, and Lemma 21 imply Corollary 2. Let max(1.02029547044,ai4) <

REFERENCES [1] Boroczky, K.: The problem of Tammes for w = 11. Studia Sci. Math. Hungar. 18 (1983), 165-171. [2] Boroczky, K.: The Newton-Gregory problem revisited. This volume. [3] Danzer, L.: Endliche Punktmengen auf der 2-Sphare mit moglichst grofiem Minimalabstand. Habilitationsschrift, Universitat Gottingen, 1963. (English translation: Finite point-sets on S2 with minimum distance as large as possible. Discrete Math. 60 (1986), 3-66.) [4] Fejes Toth, L.: Tiber die Abschatzung des kiirzesten Abstandes zweier Punkte eines auf einer Kugelflache liegenden Punktsystems. Jber. Deutsch. Math. Verein. 53 (1943), 66-68.

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[5] Fejes Toth, L.: On the densest packing of spherical caps. Amer. Math. Monthly 56 (1949), 330-331. [6] Fejes Toth, L.: Remarks on a theorem of R. M. Robinson. Studio, Sci. Math. Hungar. 4 (1969), 441-445. [7] Fejes Toth, L.: Lagerungen in der Ebene, auf der Kugel und im Raum. Zweite Auflage. Springer Verlag, Berlin-Heidelberg-New York, 1972. [8] Giinther, S.: Ein stereometrisches Problem. Archiv Math. Physik (Grunert) 57 (1875), 209-215. [9] Habicht, W., van der Waerden, B. L.: Lagerung von Punkten auf der Kugel. Math. Ann. 123 (1951), 223-234. [10] Hars, L.: The Tammes problem for n — 10. Studia Sci. Math. Hungar. 21 (1986), 439-451. [11] Hoppe, R.: Bemerkung der Redaktion zu 'Bender, C.: Bestimmung der grossten Anzahl gleicher Kugeln, welche sich auf eine Kugel von demselben Radius, wie die iibrigen, auflegen lassen. Archiv Math. Physik (Grunert) 56 (1874), 302-306.' Ibid. 307-312. [12] Leech, J.: The problem of the thirteen spheres. Math. Gazette 40 (1956), 22-23. [13] Robinson, R. M.: Arrangement of 24 points on a sphere. Math. Ann. 144 (1961), 17-48. [14] Schiitte, K., van der Waerden, B. L.: Auf welcher Kugel haben 5, 6, 7, 8 oder 9 Punkte mit Mindestabstand Eins Platz? Math. Ann. 123 (1951), 96-124. [15] Schiitte, K., van der Waerden, B. L.: Das Problem der dreizehn Kugeln. Math. Ann. 125 (1953), 325-334. [16] Tammes, P. M. L.: On the origin of number and arrangement of the places of exit on the surface of pollen grains. Rec. Trav. Bot. Neerl. 27 (1930), 1-84. E-mail address: [email protected] ''Karoly Boroczky' 1 E-mail address: lszaboQilab.sztaki.hu ''Laszlo Szabo''

ON POINT SETS WITHOUT k COLLINEAR POINTS

PETER BRASS1 Institut fur Informatik, Freie Universitat Berlin, TakustraBe 9, D-14195 Berlin, Germany

ABSTRACT. In the following we discuss two problems on sets of n points in the plane, in which no k points are collinear: • the maximum number of k — 1-point lines (generalized orchard problem), and • the largest cardinality of a subset with no k — 1 points collinear. For both problems we present a slight improvement of the current bounds. We also give a simpler construction that reaches the same asymptotic bounds as Ismailescu's recent construction, which in turn improved Grunbaum's long-standing lower bound.

1. INTRODUCTION A classical general-position restriction for point sets is 'no three points on a line'. Many problems and results exist for such point sets, especially all problems dealing with 'order types' like the Erdos-Szekeres convex polygon problem, the halving lines problem, the maximum crossmatching problem etc. There is now even a complete list of all possible order types of point sets of up to ten points with no three points collinear [1, 2]. Much less is known about the point-line incidence structure in sets 'no four points on a line' or more generally 'no k points on a line'. Several problems were stated in literature for such sets, but not much progress has been made up to now, and it is even nontrivial to construct interesting examples of such sets. It is the aim of this note to collect the results and point out several small improvements for two such problems, the generalized orchard problem, and the general position subset selection problem. Supported by DFG Heisenberg grant Br 1465/5-2. 185

186

BRASS 2.

THE GENERALIZED ORCHARD PROBLEM

The classical orchard problem asks for the maximum number of lines containing three points that can be spanned by a set of n points with no four points on a line. This problem has quite a long history, which is surveyed in [4], but is traditionally known as Sylvester's orchard problem, although earlier references exist. There is a trivial upper bound of

since each pair of points can belong to at most one three-point line, and each three-point line contains three pairs; and it was shown, using a construction based on points selected from a cubic curve [4, 14], that this trivial bound is asymptotically best possible. The currently best bounds are

where the upper bound follows from the existence of at least -^n two-point lines ('ordinary lines') [5]. Exact values were determined in [4] and again in

[16]. /(n)\

5 6 7 8 9 10 11 12 13 14 15 16 37 6 8 10 13 16 20

Thus the classical orchard problem is, up to a small gap, essentially solved; it might still be interesting to look for a characterization of the extremal sets (are they necessarily on a cubic curve?), but we do understand the asymptotic behaviour of the extremal numbers. The situation is quite different if we ask for the maximum number i£ rchard (n) of Appoint lines in a set with no k + 1-points collinear, for k > 4. This seems to be a natural generalization, but although it was mentioned many times by Erdos [6, 7, 12, 13], it received less attention since it is much more difficult. This problem was first stated by Erdos and Karteszi in two papers written in Hungarian 1962/63. Some progress was made by a nice construction of Griinbaum [17], and very recently, an even better construction of Ismailescu [18]= There is again a trivial upper bound ££ rchard (n) < ^-i) O(n2}, by counting pairs, and in the 'Handbook'-article [13] Erdos and Purdy even wrote 'For k > 3 there are no nontrivial upper bounds. For example, tf chard (n) < ^n2 is trivial, but not even ^ rchard (n) < ^(1 - e)n 2 is known.' But indeed this can be reached by a very simple improvement of the counting of pairs: Proposition. £f chard (n) < —n2 + O(n).

POINT SETS WITHOUT k COLLINEAR POINTS

187

Proof: Melchior's inequality [19] (a simple consequence of the Euler formula) for the numbers ti of z'-point lines states <2 > 3 + *4 + 2*5 + 3t6 + • • • if tn = 0.

Thus *2 > *4, so 7*4 «2

+ 6*4 < (2) and *4 < ±n2.

D

Thus by adding a very small amount of geometry (Melchior's inequality holds also for pseudoline arrangements) we can improve the upper bound. But the true challenge is to find a subquadratic upper bound, or a good lower bound. It is simple to construct an Q(nlogn) lower bound by iterated Minkowski sums. The lattice cube {!,... , k}d (or any projection of this set into the plane) is a set of n = kd points, no k + 1 -of them collinear, which spans | ((A: + 2)d —fcd)lines with k points each, which gives a lower bound of

But real progress was made only by Griinbaum [17], who found recursive construction, which obtains from a set X of n points, no k + 1 of them collinear, and containing ra lines of k points each, and an integer parameter <7, a set of qn + m points, no k + 2 of them collinear, and containing qm lines of A; + 1 points each. For this he chooses a family ( t £ [0,1], intersect in a single point. The family <^(z,y) : = (t,+(l_y^l~^x(l_t}r t*+(1J'$+%~t+I<,_t})) has this property. Then the new set consists of the union of all projective copies , i = 0,... ,
BRASS

188

point of the image lines ( p 0 ( l ) , . . . , ^ and choosing q = n^^ gives Grunbaum's lower bound t°,n, TchaTd (n) > ckn1+^. ^ ' rv

It is simple to see that one can improve the lattice cubes by adding some further points, but Ismailescu found a tricky combination of several cubes that even has a better exponent. He reached with n — kd — 1 points a total of k I

^

'

I

V

v

'

/

"

'

k +1

2

k,

,

collinear fc-tuples without k + 1 collinear points. So t°k a (n) > c^n lo « fc 3 which gives for k = 4 the same order ^l(n^) as Grunbaum's bound, but for 5 < k < 35 it is better; and since Grunbaum's bound is a recursion in fc, this improves the lower bound for all k > 5. Thus the currently best lower ,

,

,

,

bounds are if chard (n) > c 4 n2, ^ rchard (n) > ckn orchard

>

for k

for 4 < k < 18 and

>

Ismailescu's set 5(4,3) and extended cubes: 63/70/80 points with 59/100/130 4-lines A different construction which gives the same asymptotic performance as Ismailescu's is as follows: denote for any set X of points in d-dimensional space and real number a by (a, X) the set

Then we construct from two sets (oti, Xi) U (02,^2) U . . . U (ak,Xk) and («!, YI) U (o;2, ^2) U . . . U (afc, Yfc) in dx + 1 and dy + 1-dimensional space (with the same oti, . . . ,0;^) a 'hyperplane-wise' cartesian product (ai, (Xl x YI)) U (a2, (X2 x Y 2 ) ) U . . . U (a fc , (Xk x y fc )). If both original sets do not contain fc + 1 collinear points, then this 'product' set also does not contain k + 1 collinear points, as can be seen by a projection on the 'factor' sets. Also, if we count only those fc-lines not contained in a

POINT SETS WITHOUT k COLLINEAR POINTS

189

hyperplane (aj,]R dx ), (aj,IR dr ), their number is multiplicative. Thus if we have some sets Xi,... ,Xk C lRd, each containing at most a points, and not containing k + l collinear points, and numbers ai , . . . , a/y such that the d + 1-dimensional set (ai,X\) U (a-^iX^} U . . . U (ctk,Xk) contains b lines that intersect each of the sets (oti,Xi) in a single point (fc-lines), then the cartesian D-th power set (ai, Xf* ) U (0:2, X%} U . . . U (ajfc, X®) is a set of at most n = kaD points in JRdD+l, no k + l of them collinear, and containing at logfe

least bD = (f ) loga lines of k points each. If we use (ai, 0:2, • • • , ctk-i> ak)

=

1 < i < k (similar to Ismailescu's construction) we find a = k, b = k + 4; thus <£rchard(A;D+1) > (k + 4)D.

Basis case and first product of the new construction for k — 4 In this simplest case, this leads to another problem already studied under the title of 'probing lines by lines': if there is an unknown set of lines, of which we only observed the intersection points with a set of k parallel 'probe lines', how can we find the lines of that hidden set, and what is the maximum number /(a, k) of lines that it can contain, if on each of the k probe lines there are only a intersection points [3]. The above examples shows that f ( k , k ) > k + 4, and at least for k < 18, any improvement of that lower bound would also give a better construction for i£rchard(n). 3. THE GENERAL POSITION SUBSET SELECTION PROBLEM

Another related problem, also asked many times by Erdos [9, 10, 11], and still no nearer to a solution, is whether a set of n points in the plane, no k + 1 of them collinear, contains a large subset in which no k points are collinear. This is wide open for each A; > 3. It is always possible that there actually is a subset containing some positive fraction of the points, but the best lower bound is by a greedy or randomized selection, giving a subset of

190

BRASS

cn ^ fc-i1 points in which no A; are collinear. If we denote this number, the minimum cardinality of the maximum independent subset (containing no k collinear points) by Sfc(n), then in the simplest case (A; = 3, so no 4 points collinear, but many collinear triples) this gives a lower bound of c\/n and an O(n) upper bound. A slight improvement of the lower bound follows from a result of [20] on independent sets in partial Steiner systems: i

1

Theorem 1. s/e(n) > Proof: The collinear A>tuples form a partial Steiner A>tuple system, i.e. a system of A;-element subsets such that any two of them intersect in at most one element. It was shown by [20] (for k — 3), [21] (all A;) that any such system of subsets contains an independent set of size at least c^n k~l (logn) k~l, where an independent set is a set that does not contain any of the A;-tuples. D Unfortunately the same authors also showed that this bound is best possible in the class of partial Steiner systems, so for a further improvement we need more geometry beyond the fact that any two lines intersect in at most one point. The upper bound also admits a very slight improvement to o(n) by use of the density version of the Hales-Jewett theorem [15]. Theorem 2. Sfc(n Proof: We consider a general plane projection of the lattice cube {!,... , k]d. This is a set of n — kd points, no k + 1 of them collinear. Suppose we always select a positive fraction en of the points without selecting a complete line, then we can also do that to the lattice cube. But the density version of the Hales-Jewett theorem [15] says that for each e > 0 there is a cfo(A;, e) such that any subset of ekd points of a lattice cube kd with d > do(k, e) will contain a 'combinatorial line', which is a line of A; points obtained by keeping some coordinates at a arbitrary fixed values, and taking the same parameter t = 1,... , k for all the other coordinates. D The 'combinatorial lines' used in the Hales-Jewett theorem are really only a subset of the Ar-point lines occuring in the kd cube, so this upper bound is too large. But for small dimensions the kd cube is really a bad example, since it almost reaches the ^r-n upper bound. Better examples can be obtained by deleting many points from the cube.

POINT SETS WITHOUT k COLLINEAR POINTS

191

No-4-collinear sets with no-3-collinear subsets

REFERENCES [1] O. Aichholzer, H. Krasser: The point set order type data base: a collection of applications and results, in: 'CCCG 01', 13th Canadian Conf. Comput. Geom. (2001). [2] O. Aichholzer, F. Aurenhammer, H. Krasser: Enumerating order types for small point sets with applications, in: 'SCG 01', 17th ACM Symp. Comput. Geom. (2001) 11-18. [3] M. de Berg, P. Bose, D. Bremner, W. Evans, L Narayanan: Recovering lines with fixed linear probes, in: 'CCCG 98' 10th Canadian Conf. Comput. Geom. (1998), see extended version at http://www.cs.arizona.edu/people/will/papers/probe.ps.gz [4] S.A. Burr, B. Griinbaum, N.J.A. Sloane: The orchard problem, Geometriae dedicata 2 (1974) 397-424. [5] J. Csima, E.T. Sawyer: There exist 6n/13 ordinary points, Discrete Comput. Geom. 9 (1993) 187-202. [6] P. Erdos: Research Problems 36, Periodica Math. Hungarica 15 (1984) 101-103. [7] P. Erdos: Problems and results in combinatorial geometry, in: 'Discrete Geometry and Convexity', Annals New York Acad. Sci. 440 (1985): 111. [8] P. Erdos: Some of my old and new problems in elementary number theory and geometry, Congressus Numerantium 50 (1985) 97-106. [9] P. Erdos: On some metric and combinatorial geometric problems, Discrete Math. 60 (1986) 147-153. [10] P. Erdos: Some old and new problems in combinatorial geometry, in: 'Applications of Discrete Mathematics', C.D. Ringeisen et al., eds., SIAM, Philadelphia 1988, 32-37. [11] P. Erdos: Problems and results on some extremal problems in number theory, geometry and combinatorics, Rostock Math. Kolloq. 38 (1989) 6-14. [12] P. Erdos: Some of my favorite unsolved problems, in: 'A Tribute to Paul Erdos', A. Baker et al., eds., Cambridge Univ. Press 1990, 467-478.

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[13] P. Erdos, G. Purdy: Extremal problems in combinatorial geometry, in: 'Handbook of combinatorics. Vol. 1', R.L. Graham et al., eds., North-Holland/Elsevier 1995, 809-874. [14] Z. Fiiredi, I. Palasti: Arrangements of lines with large number of triangles, Proc. Amer. Math. Soc. 92 (1984) 561-566. [15] H. Furstenberg, Y. Katznelson: A density version of the Hales-Jewett theorem, J. Anal. Math. 57 (1991) 64-119. [16] O. Giering: Zum Problem von Sylvester Punktmengen mit A;-Tripeln, Sitzungsber., Abt. II, Osterr. Akad. Wiss., Math.-Naturwiss. Kl. 204 (1995) 119-143. [17] B. Griinbaum: New views of some old questions of combinatorial geometry, in: 'Colloq. Int. Theorie Comb., Roma 1973, Tomo I' (1976) 451-468. [18] D. Ismailescu: Restricted point configurations with many collinear ktuples, manuscript 2001, to appear in Discrete Comput. Geom.. [19] E. Melchior: Uber Vielseite der projektiven Ebene, Deutsche Math. 5 (1941) 461-475. [20] K.T. Phelps, V. Rodl: Steiner Triple Systems with minimum independence number, Ars Combinatoria 21 (1986) 167-172. [21] V. Rodl, E. Sinajova: Note on independent sets in Steiner Systems, Random Struct. Algor. 5 (1994) 183-190. E-mail address:

brassQinf.fu-berlin.de

''Peter Brafi''

THE BECKMAN-QUARLES THEOREM FOR RATIONAL d-SPACES, d EVEN AND d > 6

ROBERT CONNELLY Department of Mathematics, Cornell University, Ithaca, NY JOSEPH ZAKS 1 Department of Mathematics, University of Haifa, Israel

ABSTRACT. A. Tyszka [8] proved that every unit-distance preserving mapping of Q8 to Q8 is an isometry. The purpose of this paper is to extend this property to all even d, d > 6, by using the result of Zaks [10]. A mapping / : Qd —>• Qd is called unit-distance preserving if \\x — y\\ = I implies ||/(^) — /(y)|| = 1. The Beckman-Quarles Theorem ([1], see also [2]) states that every unit-distance preserving mapping of Rd to Rd, d > 2, is an isometry. Tyszka [8] (see also [6, 7]) proved that every unit-distance preserving mapping of the rational 8-space Q8 to Q8 is an isometry. Zaks [10] extended Tyszka's result for unit-distance preserving mappings / : Qd —> Qd, for all even d of the form d — 4k(k + 1), as well as for all odd d of the form 2k2 — 1 which are also complete squares. The purpose of this paper is to extend the results of Tyszka and Zaks to all even dimensions d, d > 6.

1 This paper was written while J. Zaks was visiting R. Connelly at Cornell University. The Zaks wish to express their gratitude to Gail Dennis and Bob Connelly for their superb hospitality. 193

194

CONNELLY AND ZAKS

For d even, let Ai and J3,, 1 < i < d, be the Id points in Qd, given by

1,..

2z - 1,

= (0,-. • ?n"5 - (o,.. • ?U, = (o,.. n = (o,.. • ?U, •

5

U

5

1

2z, 1 2' 1 2'

o,

-I

1 2'

o,

2>

1 2'

o,

2' — 2, 2 '

o, ,0),

Let Vd denote the set (Ai, #1, A?,B2, ... ,A^,J5^} in Qd. The polytope Pd — conv(Vd) is, of course, a cross-polytope in Qd. The set Vd is a twodistances set, since \\Ai — _0;|| = A/2 for all z, 1 < z < e/, and \\Ai — A j \ \ = \Bi — Bj \ = \\A.i - Bj | — 1 for all z < j. All the points of Vd lie on the (d — l)-sphere of radius l/A/2, centered at the origin. Every two points of Vd which are at distance A/2 apart are diametrically opposite on that sphere, so let us call them anitpodal pair. Lemma 1. If d is even, then Qd contains a regular d-simplex of edge length one if, and only if, d-\-\ is a complete square. Moreover, the maximal number of rational points in Qd, d even, of mutual distance 1 is either d or d + 1. Lemma 1 is part of the following result, due to Chilakamarri [4]. Let G'(Qd, 1) denote the unit-distance graph, in which the vertices are the points of Qd, and two vertices x and y are connected by an edge if, and only if, jjx — 7/11 = 1. The maximum clique u j ( G ( Q d , I ) ) of the unit-distance graph G(Qd,l)isgivenby d + 1, d, d -f 1, (/, d — 1,

if d is even and d — 4k(k + 1); if d is even and d / 4k(k -f 1); if d is odd and d = 2k2 - 1; if d satisfies condition A; otherwise,

where the condition A is that d is odd, d is riot of the form 2k2 — 1 and there exists a solution to the Diophantine equation dx2 — 2(d — l)y2 = z2 with x ^ 0. We have slightly changed the presentation of the formula for u(G(Qd, 1)), from the one given in [4], due to the fact that if d = 2k2 — 1, then d satisfies the condition A. (Take, for example, x — 4, y — 1 and z — 2k.} Lemma 2. // d > 5, then Qd contains a triangle T having edge lengths 1, A/T and ^/r, for any rational number r, for which 1 < PROOF OF LEMMA 2. Let d > 5, and let r be any rational number satisfying 1 < 2-y/f. The vertices of T can be chosen to be the points (±1/2,0,... ,0) and ( 0 , z , y , z , t , 0 , . . . ,0), provided (±l/2) 2 + x2 + y2 + z2 + t2 = r, where

THE BECKMAN-QUARLES THEOREM

195

x, y, z and t are rational numbers. Therefore x2 + y2 + z2 + t2 = r — 1/4, where r — 1/4 > 0. The existence of such four rational numbers is given by Lagrange's Four Squares Theorem. D Lemma 3. If x, y, z and t are points in Qd, such that \\x-y\\ = \\z-t\\, then there exists a congruence f : Qd —»• Qd, such that f ( x } = z and f ( y ) = t. PROOF OF LEMMA 3. A suitable rational congruence / can be given by a translation h by the vector z — x, followed by a reflection with respect to the perpendicular bisector of the edge [h(y),t]. For more details, see [9]. D Lemma 4. // d is even, d > 4 and if f : Vj —>• Qd is a unit-distance preserving mapping, then either f(A{] = /(#;) holds for some i, 1 < i < d, or else \\f(Ai) - /(£;)! I = V2"holds f°r al1 i, 1 < i < d. PROOF OF LEMMA 4. Let d > 4 be even and let / : V& ->• Qd be any unit-distance preserving mapping. Suppose that f(Ai) ^ /(-#;) holds for all l 4 and if f : Vj —> Rd if- a unit- distance preserving mapping, then either f(A{} = f(Bi) holds for some i, 1 < i < d, or else ||/(A t -) - /(#;)! I = V^ holds for all i, 1 < i < d.

CONNELLY AND ZAKS

196

Lemma 5. If d is even, d> 6, and if\\f(x] — f(y}\\ = 1 for some antipodal pairs x and y ofVj and some unit-distance preserving mapping f : Qd —» Qd, then f(Vd) contains a regular d-dimensional simplex of edge length one. PROOF OF LEMMA 5. If, say, ||/(Ai) - /(#i)|| = 1, then /(Ai), f ( B l ] and f(A-2) form an equilateral triangle of edge length one, hence the (d + l)-set f(B\) U {/(A z -) | 1 < i < d} forms the vertices of a regular rf-simplex of edge length one in Qd. D Lemma 6. If d is even, d > 6 and if d + 1 «s every unit-distance preserving mapping f : Qd

i a complete square, then Qd preserves the distance

PROOF OF LEMMA 6. Suppose d is even, d > 6 and d+ 1 is not a complete square, and let / : Qd —> Qd be any unit-distance preserving mapping. The Moser-Moser spindle (see [8, 9, 10]), as shown in Figure 1, is the amalgamation of a triangle and two pairs of adjacent equilateral triangles. (The dotted lines are not part of the configuration.) It was used to show that the chromatic number of the unit-distance graph in the plane is at least 4.

C B B C Figure 1. The Moser-Moser Spindle as an amalgamation as follows. It consists of d + 1 congruent Define the configuration copies TO, TI, ... , Td of Vrf. For every i, 1 < i < d, consider the z'th antipodal pair of vertices A;, i of TO together with some antipodal pair of vertices A* and B* of T;. Let S = {U,Vi,V2} be a triangle of edge lengths \\U - Vi\\ = \\U - V2\ = \/2 and Jl^i - V2\\ = 1. The existence of the triangle S follows from Lemma 2. There exists a rational congruence transformation
197

THE BECKMAN-QUARLES THEOREM

hf

inT;

x

R

U

Figure 2. The Amalgamations of the T,-'s to get Mj The configuration Md is taken as the union TO Ui<; 6 and if d + 1 is not a complete square, then every unit-distance preserving mapping f : Qd —» Qd preserves the distance 2. PROOF OF LEMMA 7. Let d be even, d > 6 such that d-\-1 is not a complete square, and let / : Qd —> Qd be any unit-distance preserving mapping. By Lemma 6 the mapping / preserves the distance \/2. Let the mapping g : Qd —» Qd be given by the following dx d block-matrix 5, in which d/2 copies of the matrix A appear along the main diagonal:

A=

B=

/A 0 0 A

0 \ 0

^ 0 0

The mapping g sends the usual d unit vectors to mutually orthogonal vectors of length \/2, hence it represents a blow-up mapping by the factor

198

CONNELLY AND ZAKS

of ^/2. It follows that a proof, similar to the one given for Lemma 6, implies that the mapping / preserves the distance 2. This completes the proof of Lemma 7. D We thank M. Perles for mentioning to one of us the property of such a construction, which enables us to expand Qd by any factor of the form ^/f, provided r is the sum of the squares of two rational numbers. Theorem 1. For every even d, d > 6, every unit-distance preserving mapping f : Qd —* Qd is an isometry. PROOF OF THEOREM 1. Let d be an even number, d > 6. If d -f 1 is a complete square, then the assertion of Theorem 1 holds by [10]. If d+1 is not a complete square, then every unit-distance preserving mapping / : Qd —> Qd preserves the distance 2, by Lemma 7. It is a well-known result of Benz [3] (see also [5, 8, 10]), that every mapping / : Qd —»• Qd that preserves the distances 1 and 2 is an isometry. This completes the proof of Theorem 1. D Based on the idea of Tyszka [8], it follows from both [10] and Theorem 1 that the following discrete version holds. Corollary 1. For every even d, d > 6, and for every two points x and y in Qd, there exists a finite set Sx,y in Qd, containing x and y, for which ||/(x) — f ( y ) \ \ = \\x — i/H holds for every unit-distance preserving mapping 9 —»• Od • Jf •• °x,y ^ V Corollary 2. For every even d, d > 6, and for every t — \/a2 -f b2 in which a and b are rationals, every mapping f : Qd —> Qd which preserves the distance t is an isometry. REMARK Notice that in Theorem 1 the target space is assumed to be Qd and not Rd. In some related results, such as when the mapping / is a function from Qd, where d is an even number of the form 4k(k + 1), then the target space can be taken to be Rd instead of Qd, and it is still possible to conclude that it is a congruence into Rd. For our result, where d is even, d > 6, and d ^ 4k(k + 1), we need the target space to be Qd in order to use the property that the target space does not contain a unit-distance d-dimensional simplex. REMARK J. Zaks extended recently Tyszka's result, by showing that every unit-distance preserving mapping of Qd to Qd is an isometry for all odd d of the form 2k2 — 1 ; this will appear soon in the Journal of Discrete Mathematics.

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REFERENCES [1] F. S. Beckman and D. A. Quarles Jr., On isometrics of Euclidean spaces, Proc. Amer. Math. Soc. 4 (1953), 810-815. [2] W. Benz, An elementary proof of the theorem of Beckman and Quarles, Elemente der Math. 42 (1987), 4-9. [3] W. Benz, Geometric Transformations (in German), Biblio. Inst. Mannheim, 1992, 322 pp. [4] Kiran B. Chilakamarri, Unit-distance graphs in rational spaces, Discrete Math. 69 (1988), 213-218. [5] B. Farrahi, A characterization of isometrics of rational spaces, J. Geom. 12 (1979), 65-68. [6] A. Tyszka, A discrete form of the Beckman-Quarles theorem, Amer. Math. Monthly 104 (1997), 757-761. [7] A. Tyszka, Discrete version of the Beckman-Quarles theorem, Aequat. Math. 59 (2000), 124-133. [8] A. Tyszka, A discrete form of the Beckman-Quarles theorem for rational eight-space, Aequat. Math. 62 no. 1-2 (2001), 85-93. [9] J. Zaks, On the chromatic number of some rational spaces, Ars Comb. 33 (1992), 253-256. [10] J. Zaks, A discrete form of the Beckman-Quarles theorem for rational spaces, J. of Geometry 72 no. 1-2 (2002), 199-205. E-mail address: E-mail address:

connellyfimath.cornell.edu ''Robert Connelly'' jzaksfimath. half a. ac.il ''Joseph Zaks''

EDGE-ANTIPODAL CONVEX POLYTOPES - A PROOF OF TALATA'S CONJECTURE

BALAZS CSIKOS1 Department of Geometry, Eotvos University, Budapest, P.O.B. 120, Hungary, H-1518

ABSTRACT. A convex n-dimensional polytope in Rn is called edge-antipodal if for any edge of the polytope there is a linear function on Rrf whose restriction onto the polytope attains its minimum and maximum at the endpoints of the edge. We prove that an edge-antipodal 3-polytope can not have more than 12 vertices. This theorem gives a positive answer to Conjecture 6 of [8] posed by I. Talata.

1. INTRODUCTION 1

A subset S of W is called antipodal if for every pair of points x,y 6 £, one can find a linear function / on Rn such that

/(x) £ /(y) and /(x) < /(z) < /(y)

Vz € S.

n

It is known that an antipodal set in R has at most 2n elements. In particular, its convex hull is a polytope. The definition implies that an antipodal set coincides with the vertex set of its convex hull. We shall call a convex polytope P antipodal if its vertex set is antipodal. Thus, there is a oneto-one correspondence between antipodal sets and antipodal polytopes. For reference and related results see [3], [6], [7]. A polytope P is edge-antipodal or shortly e-antipodal if for any edge xy of P, there is a linear function / on W1 such that

/(x) ^ /(y) and /(x) < /(z) < /(y)

Vz € P-

1 Supported by the Hungarian National Science and Research Foundation OTKA T032478. 201

202

CSIKOS

An equivalent definition appears in [8]. Edge-antipodality seems to be a weaker condition on polytopes than antipodality, however, the author does not know any edge-antipodal 3-polytope which is not antipodal. The aim of this paper is to prove Talata's Conjecture 6 of [8], predicting the existence of a number m such that no edge-antipodal 3-polytope has more than m vertices. Namely, we shall prove the following theorem. Theorem 1. An edge-antipodal polytope in E3 cannot have more than 12 vertices. Remarks. (i) The upper bound 12 on the number of vertices is probably not the best one. It requires further ideas to find the sharp upper bound. (ii) A. For has recently proved that Talata's conjecture is true in any dimension (private communication). His proof, however, does not yield any explicit upper bound on the number of vertices of an edge-antipodal polytope. (iii) The following construction of I. Talata shows that in dimension n > 4, edge-antipodality of a polytope does not imply antipodality of its vertex set. Let 61,62,... ,en be a basis of W1 and denote by 0 the origin. The convex hull P of the points 0, 61,62,...,e n _i and p = ^ZY(CI -f 62 + • • • + e n _i) is the union of two (n — l)-dimensional simplices. P is strictly edge-antipodal, i.e., for any edge qr of P, there is a linear function on W1 whose restriction onto P attains its minimum and maximum only at q and r respectively. Translate the diagonal [Op] by en and stretch it slightly from en by a factor of A > 1. The segment we obtain that way has endpoints en and q = Ap + en. It is not difficult to verify that if A > 1 is not too large (in fact A < (n — l)/2 is enough), then the convex hull of P U {en,q} is edge-antipodal, but its vertex set is not antipodal.

2. PROOF OF THEOREM 1 Recall that for A, B C W1 and A € R, we define the scalar multiplication XA of the set A and the Minkowski sum A + B by XA = (Ax | x e A},

A + B = {x + y |xe A, y € B}.

The following well-known result due to F. John (see [4] or Theorem 8 on p. 13 of [2]) will be essential. Lemma 1. Let C — —C be a centrally symmetric convex body in Mn. Then there exists a solid ellipsoid E centered at the origin such that

4=£ C C C E. \/n

EDGE-ANTIPODAL POLYTOPES

203

Recall that if C = —C is a centrally symmetric convex body in Rn, then the Minkowski metric dc induced by C is the metric on Rn defined by d c (x,y) = inf{A | x - y 6 AC)

Vx,y € Rn.

C can be obtained from the metric dc as the unit ball centered at the origin, i.e., C = {x 6 Rn | dc(x.,Q) < 1}, thus, C and dc determine one another uniquely. A Minkowski metric is said to be a Euclidean metric if the metric space (Kn,dc) is isometric to Rn equipped with the standard Euclidean distance function. A Minkowski metric dc is Euclidean if and only if C is a solid ellipsoid. Corollary 1. If C — —C is a centrally symmetric convex body in W1, and dc is the Minkowski metric induced by C, then there is a Euclidean metric d in W1 such that d/^/n < dc < d. Proof. Choose E as in the above lemma, and let d be the Euclidean metric whose unit ball is l^nE. D Let P be an edge- antipodal n-dimensional polytope in Rn. Consider the difference body C — P -\- ( — P) of P and the Minkowski metric dc induced by it. The diameter of P with respect to dc is equal to 1. Indeed, we have x - y 6 P + (-P) — C for any x,y £ P and this yields
(/(x) - /(y)) - A(/(x') - /(y')) < A(/(x) - /(y)).

From (1) we get A > 1 and dc-(x,y) > 1. On the other hand x — y 6 C, thus dc(x,y) = 1. Assume that P is an edge-antipodal 3-polytope, and apply Corollary 1 to C = P+(-P). We obtain a Euclidean metric d such that (l/^/3}d < dc < d. Our previous observations imply that the diameter of P with respect to d is at most V^ while the length of any edge of P with respect to d is at least 1 . Talata's conjecture can be derived from the following theorem proved in [I]Theorem 2. Denote by 8*j the infimum of the diameters of?,-polytopes with N vertices having no edge length less than 1. Then limn_oo $N — 3. (Lengths and diameters are taken with respect to a fixed Euclidean metric.)

204

CSIKOS

Indeed, if NQ is a number such that % > \/3 for N > NO, then an edgeantipodal 3-polytope cannot have more than NQ vertices. The idea to attack Talata's conjecture with the help of Theorem 2 was suggested to the author by I. Talata. Sharpening slightly the methods used in [1] one can prove that an edge-antipodal 3-polytope can not have more than 176 vertices. This upper bound is much weaker than the bound 12 in Theorem 1. To prove Theorem 1 we use a simple consequence of edge-antipodality of P. The faces of an edge-antipodal polytope are edge-antipodal themselves. Since an edge-antipodal polygon in the plane is either a triangle or a parallelogram, faces of an edge-antipodal 3-polytope are triangles or parallelograms. Thus, Theorem 1 follows from the theorem below. Theorem 3. If a convex 3-polytope in the Euclidean 3-space has the following properties: (i) the edges of P are not shorter than 1; (ii) the diameter of P is not greater than \/3; (in) the faces of P are triangles or parallelograms; then P has at most 12 vertices. Proof. If we add to the edge graph of the polytope the longer diagonal of each quadrilateral face (in the case of equal diagonals we choose any of the diagonals), we obtain a triangulation of the boundary of P, whose edges have length > 1. We claim that the area of any triangle in this triangulation is at least \/3/4. Indeed, if for the sides of the triangle we have 1 < a < b < c < \/3, then for the angle 7 opposite to c we get that 7 is in the interval [7r/3,27r/3], because being c is a largest side otherwise would imply 7 > 27T/3, with further implication c2 = a2 + b2 - lab cos 7 > I2 + I2 + 2 • 1 • 1 • (1/2) = 3,

contradicting assumption (ii). However, if 7 is in the interval [7r/3,27T/3], then we have (!/2)a6sin7 > v/3/4. We conclude that if the triangulation constructed above consists of / triangles, then the surface area of P is at least /\/3/4. On the other hand, we know that the surface area of a convex body of average width w is not more than the surface area of a sphere of radius w/2 (Theorem 22.2 in [5]) and the average width of a convex body does not exceed its diameter, therefore, the surface area of P is at most 4 • (\/3/2) 2 7r = 3:r. Comparing the lower and upper bounds on the surface area of P, we obtain that the number of triangles in the triangulation is at most 4\/37r & 21.7656. Since the number of edges in the triangulation is e = 3//2, / must be an even natural number, hence / < 20.

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The number v of vertices can be expressed with the help of / using Euler's formula and this proves the theorem.

D

3. ACKNOWLEDGEMENTS The author is obliged to I. Talata for his thought-provoking remark, stimulating the research presented in this paper.

REFERENCES [1] Boroczky, K. and Csikos, B., Small Convex Poly topes with Long Edges and Many Vertices, Discrete Comput. Geom., to appear. [2] Gruber, P.M. and Lekkerkerker, C.G., Geometry of numbers, 2nd ed., North- Holland, Amsterdam, 1987. [3] Griinbaum, B., Convex Polytopes, Wiley-Interscience, London - New York - Sydney, 1967. [4] John, F., Extremum problems with inequalities as subsidiary conditions, Studies and essays presented to R. Courant, Interscience, New York, 1948, 187-204. [5] Leichtweifi, K., Konvexe Mengen, VEB Deutcher Verlag der Wissenschaften, Berlin, 1980. [6] Makai, E.Jr. and Martini, H., On the number of antipodal or strictly antipodal pairs of points in finite subsets of R.d, Applied Geometry and Discrete Mathematics, 457-470, DIMACS Ser. Discrete Math. Theoret. Comput. Sci., 4, Amer. Math. Soc., Providence, RI, (1991). [7] Makai, E.Jr. and Martini, H., On the number of antipodal or strictly antipodal pairs of points in finite subsets of Rd. II., Period. Math. Hung., 27 No. 3. (1993), 185-198. [8] Talata, I., On extensive subsets of convex bodies, Period. Math. Hung., 38 No. 3. (1999), 231-246. E-mail address:

csikosQcs.elte.hu ' 'Balazs Csikos''

SINGLE-SPLIT TILINGS OF THE SPHERE WITH RIGHT TRIANGLES

ROBERT J. MACG.DAWSON l Department of Mathematics and Computing Science, Saint Mary's University, Halifax, NS, Canada, B3H 3C3

ABSTRACT Sommerville [5] gave a classification of the isosceles triangles that can tile the sphere (subject to certain additional conditions) in 1923 , and Davies [1] completed the classification in 1965. However, if the edge-to-edge restriction is relaxed, there are other such triangles; such triangles must have combinations of angles adding to 180°. Here, we examine the case in which one angle is right, but no combination of the other two angles adds to 180°. We show that there is a unique triangle of this type which tiles, and examine a couple of "near misses". 1. INTRODUCTION In 1923, D.M.Y. Sommerville [5] classified the edge-to-edge tilings of the sphere with congruent isosceles spherical triangles, with directly congruent spherical triangles, and with congruent spherical triangles subject to the condition that the angles meeting at any vertex of the tiling are congruent. This last condition forces all adjacent pairs of triangles to be mirror images, so that the triangle tiles by repeated reflection. H.L. Davies completed the classification of edge-to-edge tilings with triangles in 1965 [1], adding two more continuous families of tilings that did not satisfy Sommerville's conditions. There are also, however, non-edge-to-edge tilings, including some using tiles that cannot tile in an edge-to-edge fashion [2, 3]. If a triangle tiles in a non-edge-to-edge fashion, there must exist some combination of its angles that adds to 180°. 1

Supported in part by NSERC operating grant. 207

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In [2] the isosceles spherical triangles that tile the sphere with congruent copies were completely classified. Most of these - four infinite families and five sporadic cases - tile edge-to-edge and were already known. One of these infinite families, the semilunes, consists of the triangles with angles 360°/™,360°/n, and 180° - 3600/™ for n even. When n is odd, the triangle does not tile in an edge-to-edge fashion, but nevertheless tiles. Three more sporadic cases were found, with angles (150°, 60°, 60°), (100°, 60°, 60°), and (80°, 60°, 60°). These too tile only in a non-edge-to-edge fashion, the lastmentioned in three distinct ways [3] . Each isosceles triangle that tiles the sphere can be bisected to yield two congruent right angled triangles; these, the (90°, 75°, 60°), (90°, 60°, 50°), and (90°, 60°, 40°) triangles respectively, therefore also tile the sphere. The right-angled triangles form a particularly interesting case, because every right triangle has at least one combination of angles - two right angles that adds to 180°. In a paper now in preparation [4], B. Doyle and the author will exhibit a full classification of right triangles that tile the sphere. In most cases, a right triangle that tiles the sphere in a non-edge-toedge fashion has other combinations of angles that add to 180° as well, and examining these combinations is a useful tool in the classification. This paper considers the exceptional case, in which no other combination of angles adds to 180°; we shall see that there is a unique such tile.

2. CLASSIFICATION OF TILINGS It follows from the classifications in [2] and [5] that every isosceles triangle that tiles the sphere has either two right angles, or some other collection of angles adding to 180°. Given any scalene right triangle, we will represent the measure of the larger of the two non-right angles of the triangle by (3 and that of the smaller by 7. The lengths of the edges opposite these angles will be B and C respectively, with H as the length of the hypotenuse. (Note that it may be that (3 > 90° and B > H .) We will denote the number of tiles by N . By considering the spherical excess, we see that (1)

AT = 72

Let

(2)

V = {(a, 6, c) € Z3 : a(90°) + b/3 + 07 = 360°, a, b, c > 0} .

We call the triples (a, b, c) the vertex vectors of the triangle and the equations vertex equations. The vertex vectors represent the possible (unordered) ways to surround a vertex with the available angles. A vertex which is in the middle of an edge of one of the adjoining triangles is called split If a right angles, b (3 angles, and c 7 angles meet at a split

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vertex, then (2a, 26,2c) is a vertex vector. This combination of angles will be represented by the split vector (2a,26,2c)/2 In this section, we will determine which right triangles with the single split (4,0,0)/2 tile the sphere. One special case can be dismissed quickly. A triangle with two right angles tiles the sphere if and only if 7 = (360/n)° (see [2]). Thus, while these triangles can tile in a non-edge-to-edge fashion (e.g., by tiling the Northern and Southern hemispheres each with n triangles, and rotating one slightly with respect to the other), they can always tile edgeto-edge. Lemma 1. If a right triangle has no split vectors except for (4, 0, 0)/2, and it tiles in a non-edge-to-edge fashion, then either 1C = B or 1C = H. Proof: By assumption, there is a split vertex O, with two right angles, at which the tiling fails to be edge-to-edge. If there is no split vertex configuration including the angles (3 or 7, then the extended edge containing O cannot extend past either adjacent vertex (Figure la), so consists of exactly two edges. It cannot be of length B + C (lb); and if it is of length 2B, then the other side must be covered in the same way and the tiling is edge-to-edge at O (lc). Thus the extended edge consists exactly of two short edges. The only combinations of edges that can possibly add to 1C are 1C itself, H, and J5; and in thefirstcase the tiling is edge-to-edge at O. |

O

a

Ob

FIGURE 1. The neighborhood of a split vertex Proposition 1. The only right triangle that tiles the sphere, has no split vector except for (4,0,0), and has 1C = H is the (90°, 108°, 54°) triangle. Proof: If every right angle is part of a (4,0,0) vertex or the corresponding (4,0,0)/2 split, the triangles are grouped into pairs whose unions are equilateral triangles. It is easily seen that any equilateral triangle that tiles the sphere does so in an edge-to-edge fashion; the triangles that do so are the (120°,120 0 ,120 0 ),(90 0 ,90°,90 0 ), and (72°, 72°, 72°) triangles, yielding the spherical tetrahedron, octahedron, and icosahedron respectively. The triangles obtained by bisecting these are the (90°. 120°., 60°), (90°, 90°, 45°), and (90°, 72°, 36°) triangles, all of which have other splits. Thus the tiling has a vertex of some other type that includes a right angle.

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FIGURE 2. One of the three tilings with the (90°, 108°, 54°) triangle Consider such a right angle (a\ in triangle 1, Figure 3) The edge A\B\ must be shared with the corresponding leg of another triangle (2), with the opposite orientation. Now consider the right angle of triangle 2. It cannot be part of a split, so the leg B\A^ must in turn be paired with a similar leg; and this band of triangles continues by induction until it closes or overlaps itself.

FIGURE 3. The equatorial belt If it closes, the vertices Ai form a regular n-gon about a point O and the vertices BI form a congruent n-gon about the antipodal point. If n = 2, we have the degenerate (90°, 180°, 90°) triangle. If n = 3, two more triangles (arranged into an equilateral triangle) fill each polar region; we have the vertex formula (1,2,1) which, with the relation (3 — 27, gives the (90°, 108°, 54°) triangle as the unique possibility. (It follows that this triangle tiles the sphere in exactly three ways, excluding reflections, depending upon whether the distinguished vertices of the two polar equilateral triangles are connected by a short edge, a long edge, or no edge in the equatorial belt.) If n > 3, the angles of the polar n-gons must be greater than /3, and two adjacent edges must be covered by distinct triangles. If an edge is covered by two short legs, the triangles form an equilateral (/3, /?, /?) triangle; if by a hypotenuse, the exposed long leg must be paired with another such and

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their union is again a (/?, /?, (3) triangle. We conclude that each of the vertices Ai must have the vertex vector (1,3,1). The third vertices of the ((3,P,/3) triangles surround a point 0, so that (0,2m, 0) must be a vertex vector as well.

FIGURE 4. The vertices A{ are equidistant from O. With the relation /3 — 27, this forces the triangle to have angles (90°, 77y , 381 ) This yields a contradiction, as /? does not divide 360°. | However, there are multiple covers in which 84 (90°, 77'j , 38| ) triangles cover the sphere three times (Figure 5). As each of the 28 equilateral triangles in the polar polygons can be divided into two tiles in three distinct ways, there are billions of distinct tilings in this family.

FIGURE 5. A multiple cover with the (90°,77y°,38|°) triangle It is also noteworthy that the (90°, 108°, 54°) triangle can be dissected into three (90°, 60°, 54°) triangles; this triangle therefore also tiles the sphere. (An historical note: the (90°, 60°, 54°) tiling was actually discovered first. B. Thomas noticed that some of the small triangles fitted together into larger ones; subsequent investigation showed that in fact the entire tiling was a subdivision of a larger one.)

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FIGURE 6. A tiling with the (90°, 60°, 54°) triangle Proposition 2. If a right triangle has no split vector except for (4,0,0)/2, and has 2C = B, it does not tile the sphere. If such a triangle is to tile in a non-edge-to-edge fashion, the configuration of triangles 1-3 in Figure 7 must occur. We note that if a triangle has no split vector except for (4, 0, 0)/2, it has no vertex vector (a, 6, c) with a = 2, for otherwise (0, 2fo,2c) would be a vertex vector as well, and (0,26,2c)/2 would be a split vector. A similar argument rules out a = 3. We also note that the edge PQ of triangle 1 in Figure 7 cannot have an overhang at either end; as there cannot be a second right angle at Q, we conclude that the triangle 4 must be as shown . We now consider what other angles might fill the gap at Q. Any answer to this question, along with the equation 1C — B, gives sufficient information for the numerical determination of the angles f3 and 7; we shall see that the only case in which these yield an integer N corresponds to a known tile that has other splits.

FIGURE 7. Configuration near a split with 1C = B We cannot have b + c > 6 and b > 3, as this forces (3 + 7 < 90°. Nor can we have (l,2n + 1,0), (1,3,2) or (1,2,2™ + 1), as every hypotenuse

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must be paired with another. Moreover, 7 > arctan(l/2) ~ 26.5651° and /3 > arctan(2) ^ 63.4349°, so (l,2,c) is impossible for c > 6. This leaves only (1,2,2), (1,2,4), (1,3,1), and (1,4,0) as possible candidates. It is not hard to see that the vertex vector (1,2,2) corresponds to the (90°,90°,45°) triangle. This, as mentioned above, tiles edge-to-edge, and was listed by Sommerville and Davies. Numerical solutions appear to be required for the other cases. (1,2,4) gives 0 « 70.1281°, 7 w 32.4355°, and equation (1) yields N « 57.3042. Similarly, (1,3,1) gives j3 « 77.6906°, 7 w 36.9282°, and N w 29.2459. Finally, the vertex vector (1,4,0) gives /3 = 67.5°, 7 w 29.3001°, and N « 105.8801. As N is not an integer in any of these cases, the triangle cannot tile the sphere. | If (as seems likely) N is irrational, there is not even a finite-density multiple cover. However, it should be noted that the (90°, 77.69...° , 36.92 ...°) triangle does have a rather elegant "near miss", which fails to close up exactly (Figure 8).

FIGURE 8. A near tiling

3. CONCLUSIONS While most spherical triangles that tile in a non-edge-to-edge fashion have split vectors using non-right angles, the (90°, 108°, 54°) triangle tiles (with 10 copies) using only the (4,0, 0)/2 split, and is the only spherical triangle that does so but does not tile edge-to-edge. This is an important preliminary step in the classification of all right triangles that tile the sphere [4].

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REFERENCES [1] Davies, H. L., Packings of spherical triangles and tetrahedra. Proc. Colloquium on Convexity (Copenhagen, 1965) 42-51 [2] Dawson, R. J. MacG., Tilings of the sphere with isosceles triangles, (to appear, Disc. Comp. Geom.) [3] Dawson, R. J. MacG., A triangle that tiles the sphere in exactly three ways, (to appear, Disc. Comp. Geom.) [4] Dawson, R. J. MacG., Doyle, B., Right triangles that tile the sphere, in preparation [5] Sommerville, D. M. Y., Division of space by congruent triangles and tetrahedra, Proc. Roy. Soc. Edinburgh 43 (1923), 85-116 E-mail address: [email protected]

VERTEX-UNFOLDINGS OF SIMPLICIAL MANIFOLDS

ERIK D. DEMAINE MIT Laboratory for Computer Science, 200 Technology Square, Cambridge, MA 02139, USA. DAVID EPPSTEIN1 Department of Information and Computer Science, University of California, Irvine CA 92697-3425, USA. JEFF ERICKSON2 Department of Computer Science, University of Illinois at Urbana-Champaign, 1304 W. Springfield Ave, Urbana, IL 61801, USA. GEORGE W. HART SUNY Stony Brook, Stony Brook, New York JOSEPH O'ROURKE3 Department of Computer Science, Smith College, Northampton, MA 01063, USA.

ABSTRACT. We present an algorithm to unfold any triangulated 2-manifold (in particular, any simplicial polyhedron) into a non-overlapping, connected planar layout in linear time. The manifold is cut only along its edges. The resulting layout is connected, but it may have a disconnected interior; the triangles are connected at vertices, but not necessarily joined along edges. We extend our algorithm to establish a similar result for simplicial manifolds of arbitrary dimension.

1. INTRODUCTION It is a long-standing open problem to determine whether every convex polyhedron can be cut along its edges and unfolded flat in one piece without overlap, that is, into a simple polygon. This type of unfolding has been

Martially supported by NSF grant CCR-9912338. Partially supported by a Sloan Fellowship and NSF CAREER award CCR-0093348. Supported by NSF Distinguished Teaching Scholars award DUE-0123154. 215 2

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termed an edge-unfolding] the unfolding consists of the facets of the polyhedron joined along edges. In contrast, unfolding via arbitrary cuts easily leads to non-overlap. See [11] for a history of the edge-unfolding problem and its applications to manufacturing. Recently it was established that not every nonconvex polyhedron can be edge-unfolded, even if the polyhedron is simplicial, that is, all of its faces are triangles [2, 3]. In this paper we loosen the meaning of "in one piece" to permit a nonoverlapping connected region that (in general) does not form a simple polygon, because its interior is disconnected. We call such an unfolding a vertexunfolding: facets of the polyhedron are connected in the unfolding at common vertices, but not necessarily along edges. With this easier goal we obtain a positive result: Theorem 1. Every connected triangulated 2-manifold (possibly with boundary) has a vertex-unfolding, which can be computed in linear time. This result includes simplicial polyhedra of any genus, manifolds with any number of boundary components, and even manifolds like the Klein bottle that cannot be topologically embedded in 3-space. Our proof relies crucially on the restriction that every face is a triangle. The problem remains open for nonsimplicial polyhedra with simply connected or even convex faces; see Section 5. We extend this result in the natural way to higher dimensions in Section 4. 2. ALGORITHM OVERVIEW Let M be a triangulated 2-manifold, possibly with boundary. Following polyhedral terminology, we refer to the triangles of M. as facets. The (vertexfacet) incidence graph of M. is the bipartite graph whose nodes are the facets and vertices of A4, with an arc (i>, /) whenever v is a vertex of facet /. A facet path is a trail4 (i>o, /i, vi, /2,1*2, • • • > /fc> vk) m the incidence graph of M. that includes each facet node exactly once, but may repeat vertex nodes. In any facet path, Vi-i and Vi are distinct vertices of facet /j for all i. Because each facet node appears only once, no arc is repeated. A facet cycle is a facet path that is also a circuit, that is, where VQ = v^. Our algorithm relies on this simple observation: Lemma 2. If M has a facet path, then M has a vertex-unfolding in which each triangle of the path occupies an otherwise empty vertical strip of the plane. A trail is a walk in which no arc is repeated. A walk in a graph is an alternating series of nodes and arcs with each arc incident to the surrounding nodes.

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VERTEX-UNFOLDINGS OF SIMPLICIAL MANIFOLDS

Proof. Let p be a facet path of M. Suppose inductively that a facet path p has been laid out in strips up to facet /j_i, with all triangles left of vertex Vi, the rightmost vertex of /i_i. Let (vi, fi, fi+i) be the next few nodes in p; recall that Vi ^ Vi+i- Rotate facet /j about vertex Vi so that vi is leftmost and Vi+i rightmost, and the third vertex of fi lies horizontally between. Such rotations exist because fi is a triangle. Place fi in a vertical strip with Vi and Vi+i on its left and right boundaries. Repeating this process for all facets in p produces a non-overlapping vertex-unfolding. D

A, C, B, D

A, B, C, D

FIGURE 1. Vertex-unfolding the top four triangles of the regular octahedron and making the connections planar. Thus to prove Theorem 1 it suffices to prove that every connected triangulated 2-manifold has a facet path. It might be more pleasing to obtain a vertex-unfolding based on a noncrossing facet path, one that does not include a pattern (..., A, v, C,..., B, v, D,...) with the facets incident to the vertex v appearing in the cyclic order A,B,C,D. Because a facet path has either no or at most two odd nodes (its endpoints), and because any such planar graph has a noncrossing Eulerian trail, we can convert any facet path into a noncrossing facet path. Specifically, we can replace each crossing pair of vertex-to-vertex connections with one of the two alternate pair of connections, whichever alternate pair keeps the vertex-unfolding connected [4, Lem. 1]. See Figure 1. The existence of facet paths has two further applications: (1) The vertex-unfolding resulting from a noncrossing facet path can be viewed as a hinged dissection [7] of the surface. Thus we demonstrate a hinged dissection for any triangulated 2-manifold. (2) A facet path also yields an "ideal rendering" of any triangulated surface on a computer graphics system with a 1-vertex cache: each triangle shares one vertex with the previous triangle in the graphics pipeline. This result is in some sense best possible: an ideal rendering for a 2-vertex cache in which every adjacent pair of triangles shares two vertices is not always achievable, because there are triangulations whose dual graphs have no Hamiltonian path [1].

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Figure 2(a) shows a vertex-unfolding of the triangulated surface of a cube, obtained from a facet path by our algorithm. Figure 2(b) shows a less regular vertex-unfolding. Note that the vertices do not necessarily lie on a line. Several more complex examples are shown in Figure 3. In our examples, we permit the triangles to touch along segments at the strip boundaries (as in (a) of the figure), but this could easily be avoided if desired so that each strip boundary contains just the one vertex shared between the adjacent triangles. (a)

•

•

•.

i

;

;

i

i

;

•

;

;

(b)\

FIGURE 2. Laying out facet paths in vertical strips: (a) cube; (b) 16-facet convex polyhedron.

3. FACET PATHS IN 2-MANiFOLDS In this section, we prove that every triangulated 2-manifold has a facet path, which, by Lemma 2, yields Theorem 1. 3.1. Notation. We first establish terminology for arbitrary dimensions, anticipating Section 4, but specializing to dimension 2 for this section. A d-manifold is a topological space such that every point has a neighborhood homeomorphic to the neighborhood of some point in a closed halfspace in IRd. Interior points of a 2-manifold have neighborhoods homeomorphic to a disk, and boundary points have neighborhoods homeomorphic to a half-disk. (Thus, the vertex-unfoldings in Figures 2 and 3 are not manifolds.) A simplicial d-manifold is a simplicial complex5 homeomorphic to a A simplex is the convex hull of d + 1 independent points: a triangle, tetrahedron, etc. A simpicial complex is a collection of simplices such that every pair of simplices that intersect do so in exactly one face of each. The proofs in this section proofs only require abstract simplicial complexes; the geometry of a simplex is only used in Lemma 2.

VERTEX-UNFOLDINGS OF SIMPLICIAL MANIFOLDS

219

76

0/5 jU A A/vl. ,.MN A yo pi»w v^'^/vf^-Y/J

746 -T-^JV-VV-

-A/lip^Tv^Y^ ^"^V^

^'-f1-

FIGURE 3. Vertex-unfoldings of random convex polyhedra (generated by code from [10]). The number of triangles is indicated to the left of each unfolding. The unfoldings were constructed using an earlier, less general version [6] of the algorithm in Section 3.

d-dimensional manifold, possibly with boundary. For example, any simplicial convex d-polytope is a simplicial manifold homeomorphic to the sphere §d-1. We call the full d-dimensional simplices facets and the codimension-1 simplices (i.e., the simplices of dimension d — 1) ridges. When d = 2, ridges are also called edges. A triangulated 2-manifold is an edge-to-edge gluing of (topological) triangles. Triangulated 2-manifolds are not necessarily simplicial complexes; see Figure 8. The dual l-skeleton M.* of a simplicial manifold Ai is a simple graph, with a node for each facet and an arc between any two facets that share a ridge. We call a simplicial manifold M a tree manifold if its dual 1-skeleton M* is a tree, or equivalently, if it is connected and its codimension-2 simplices all lie on the boundary. Every tree manifold T is homeomorphic to a ball. A tree 2-manifold is a simplicial complex with the topology of a triangulated polygon with no interior vertices. A tree 3-manifold has the topology of a triangulated polyhedron in IR3 with no diagonals or interior vertices. Given a simplicial manifold .M, we take an arbitrary spanning tree of M* to yield a tree manifold T corresponding to M. We will find our facet paths in T rather than in M, which is no loss because any facet path or facet cycle of T can be mapped to a facet path or facet cycle of M.; recall that a facet path may repeat vertex nodes.

DEMAINE, EPPSTEIN, ERICKSON, HART, O'ROURKE

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(e)

FIGURE 4. Tree manifold (a) to scaffold (b) to connected scaffold (c) to facet cycle (d) to strip layout (e). The layout is based on a different facet path than used in Figure 2(a). As in Section 2, the (vertex-facet) incidence graph of a simplicial manifold has a node for every vertex and every facet, and an arc (v, f) whenever v is a vertex of facet /. A facet path is a trail that includes each facet node in the incidence graph exactly once, and a facet cycle is a circuit in that passes through every facet exactly once. We now generalize facet paths and cycles to more general subgraphs that span the facets. A scaffold is a subgraph of the incidence graph in which every facet appears and has degree 2, and at most two vertices have odd degree; if every vertex has even degree, we call it an even scaffold. (See Figure 5(d) below for an example of a scaffold.) Any facet path is a scaffold, and any Euler walk through a connected scaffold is a facet path. Thus, our goal is to find a connected scaffold for T, which yields a facet path for T, which yields a facet path for Ai. The full process is illustrated for a cube in Figure 4. 3.2. Connected Scaffolds. First we establish a slightly weaker result: Lemma 3. Every triangulated polygon with no interior vertices has a (possibly disconnected) scaffold. Proof. Let T be a triangulated polygon with no interior vertices. We prove the lemma by induction on the number of triangles, with two base cases. If T is empty, we are done. If T is a single triangle, then a path between any two vertices is a scaffold. Henceforth, assume that T has at least two triangles.

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VERTEX-UNFOLDINGS OF SIMPLICIAL MANIFOLDS

(d)

FIGURE 5. (a) A polygon triangulation T. Lightly shaded triangles are ears; darker shaded triangles are hats, (b) Removing a Mickey House hat. (c) Removing a dunce cap. (d) A scaffold produced by our algorithm. An ear in T is a triangle that is adjacent to at most one other triangle. We call a triangle in T a hat if it is adjacent to at least one ear and at most one non-ear. If we remove all the ears from T, we obtain a new triangulated polygon T'. See Figure 5(a). If T' is non-empty, then it has at least one ear, and every ear in T' is a hat in T. On the other hand, if T' is empty, then T consists of exactly two triangles, which are both ears and hats. In either case, T contains at least one hat. To perform the induction, we choose a hat in T, find a cycle in the facetvertex incidence graph of that hat and (at most two of) its adjacent ears, and recursively construct a scaffold for the remaining triangulation. We have two inductive cases. Suppose T has a hat H = qrs with at least two ears E = pqr and F = rst (a "Mickey Mouse hat"). See Figure 5(b). We construct a cycle (r, E, q, H, s, F, r) through the facet-vertex of these three triangles, and recursively construct a scaffold for the smaller triangulation T \ {H, E, F}. Otherwise, let H = qrs be a hat with just one adjacent ear E = pqr (a "dunce cap"). See Figure 5(c). We construct a cycle (q,H,r,E,q) through

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the facet-vertex of those two triangles, and recursively construct a scaffold for the smaller triangulation T \ {H, E}. D Following the proof gives us an easy linear-time algorithm, consisting of a simple depth-first traversal of the input triangulation's dual tree. Figure 5(d) shows a scaffold computed by our algorithm. The scaffold we construct may be disconnected, but we now show how to make it connected using a series of local operations. Lemma 4. Every triangulated polygon with no interior vertices has a connected scaffold. Proof. Let S be a scaffold with more than one component. If S is not actually an even scaffold, its two odd vertices must be in the same component; every other component has an Euler circuit and so must be 2-edge-connected. Choose a pair of triangles A = pqr and B — qrs that lie in different components of S. Without loss of generality, suppose S contains the edges (p, A), (
FIGURE 6. Lemma 4: Joining two components of a scaffold with a flip. Clouds hide the rest of the components. Given a triangulated 2-manifold M, we convert it to a tree manifold T, apply Lemmas 3 and 4 to obtain a connected scaffold, and finally use an Euler walk through this scaffold to produce a facet path. This yields:6 Theorem 5. Every connected triangulated 2-manifold (possibly with boundary) has a facet path. Igor Pak [personal communication, Dec. 2001] found a different proof of this result for triangulated, genus-zero polyhedra (without boundary). After vertex truncation, a Hamiltonian cycle in the dual is found via Whitney's theorem on planar triangulations without separating triangles. This cycle is then converted to a facet path.

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3.3. Checkered Triangulations and Facet Cycles. We can strengthen this result slightly, by showing that most properly triangulated 2-manifolds actually have a facet cycle. This permits the strip layout of Lemma 2 to start on the left with any given triangle of the manifold. We call a polygon triangulation checkered if there is a 2-coloring of the triangles so that every white triangle has three (necessarily black) neighbors. See Figure 7(a). Lemma 6. A polygon triangulation with no interior vertices has a facet cycle if and only if it is not checkered. Proof. Let T be a triangulated polygon with no interior vertices. First we prove by induction that no checkered triangulation has a facet cycle. The base case is a single (black) triangle, which clearly has no facet cycle. In any other checkered triangulation T, we can always find a (Mickey Mouse) hat: two black ears adjacent to a common (white) triangle. If T has a facet cycle, it must contain a subcycle of six edges inside the Mickey Mouse hat and another subcycle through the rest of the triangulation. But the rest of the triangulation is checkered, so by the induction hypothesis, it has no facet cycle. See Figure 7(b).

FIGURE 7. Lemma 6: (a) A checkered polygon triangulation. (b) After removing three Mickey Mouse hats. Now suppose T is a noncheckered triangulation; in particular, T has at least two triangles. To prove the lemma, it suffices to show that T has an even scaffold. Assume without loss of generality that T has no hats; otherwise, we can remove them as described above. (This might actually eliminate every triangle in T, but then we've computed an even scaffold!) We cannot be left with a single triangle because T is not checkered. Thus, T has at least two "dunce caps": hats with only one adjacent ear. If we follow the algorithm in Lemma 3, removing hats whenever possible, the triangulation always contains at least one dunce cap, until either the algorithm removes

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every triangle or there are exactly three triangles left. Thus, the triangulation never consists of a single triangle, which implies that the algorithm constructs an even scaffold. D We say that a triangulated 2-manifold is simplicial if it is a simplicial complex, or equivalently, if its dual graph is simple (has no multi-edges or loops). Every manifold constructed from geometric triangles is simplicial. (A nonsimplicial manifold is shown below in Figure 8.) Lemma 7. For every connected simplicial 2-manifold (possibly with boundary) M., with the exception of a checkered polygon triangulation, there corresponds a noncheckered tree manifold T. Proof. Let A4 be a simplicial 2-manifold that either is multiply connected or has interior vertices. Assume without loss of generality that A4 has a checkered tree manifold T, because otherwise we have nothing to prove. This immediately implies that M. is not 2-colorable. Color the triangles of T black and white, so that adjacent triangles have opposite colors and every boundary edge of T lies on a black triangle. Because the dual 1-skeleton of M. is a simple graph, we can cut T into two simple polygons along some edge, and then reglue those pieces along some other pair of edges, to obtain another tree manifold T' of M.. Because M. is not 2-colorable, we must reverse the colors of one of those pieces to obtain a proper 2-coloring of T'. Because each piece has at least three edges, each piece has at least one edge that is on the boundary of both T and T'. It follows that T' has boundary edges adjacent to both black and white triangles, so T' is not checkered. D Combining the previous two lemmas, we conclude the following: Theorem 8. Every connected simplicial 1-manifold (possibly with boundary) has a facet cycle, except a checkered polygon triangulation. This theorem requires that we start with a simplicial complex. There are triangulated but nonsimplicial 2-manifolds that have no facet cycle, like the triangulation of the sphere shown in Figure 8. However, even improperly triangulated 2-manifolds have facet paths.

4. HIGHER DIMENSIONS In this section we generalize our results to higher dimensions. Lemma 2 generalizes in the obvious way, yielding from any facet path a vertexunfolding that places simplices in slabs bounded by parallel hyperplanes. We will show that any simplicial polyhedron has a facet cycle, and thus a

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FIGURE 8. An improper triangulation of the sphere with a facet path but no facet cycle. Edges with identically labeled endpoints are identified. vertex-unfolding. Like the results in the previous section, the proofs are almost purely topological, and thus actually apply to arbitrary triangulated manifolds, possibly with boundary, independent of any embedding.7 We follow the notation set in Section 3.1, and follow the same proof outline: from simplicial d-manifold M. to tree d-manifold T to even scaffold to connected even scaffold to vertex cycle. Lemma 9. For all d > 3, every tree d-manifold has an even scaffold, the exception of a single d-simplex.

with

Proof. Let T be a tree d-manifold with d > 3 and with more than one facet. An ear of T is a facet that is adjacent to only one other facet. A hat is a facet that is adjacent to at least one ear and at most one non-ear. Just as in the two-dimensional case, every tree d-manifold with more than one facet has at least two ears and at least one hat (and with only d — 1 exceptions, at least two hats). We prove the lemma by induction. If T is non-empty, we identify a small collection of simplices in T, find a cycle in the incidence graph of those simplices, and recursively construct an even scaffold for the remaining complex, which is still a tree manifold. It is fairly easy to construct an even scaffold for any subcomplex consisting of a hat and its ears, and to decompose any tree complex into a sequence of such complexes by removing a hat (and its ears) and recursing [8]. To keep things simple, however, we will consider only the four following cases. In fact, our results apply to pseudo-manifold A-complexes—sets of
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(1) If T is empty, there is nothing to do. (2) Suppose T consists of exactly three simplices: a hat H and two ears E and F. E and F share at least one vertex p; E and H share at least one vertex q ^ p; and H and F share at least one vertex r ^ Pi^- (In f ac t, we have d — 2 choices for each of these three vertices.) Then (p,E,q,H,r,F,p) is a cycle in the incidence graph of T. See Figure 9.

FIGURE 9. Case 2 of Lemma 9: hat H, ears E and F, T = {H, E, F}. (3) Suppose some hat H is adjacent to just one ear E. H and E share an edge pq. (In fact, they share an entire ridge.) We recursively construct an even scaffold for the subcomplex U \ {H,E}, and add the cycle (p, H, q, E, p). (4) Finally, suppose some hat H is adjacent to more than one ear and T has more than three facets. Let E and F be any two ears adjacent to H. E and F share an edge pq. (In fact, they share an entire face of dimension d — 2 > 0; this is the only step of the proof that requires d > 3.) We recursively construct an even scaffold for the subcomplex U \ {E, F} and add the cycle (p, E, g, F,p). The only tree manifold that does not fall into one of these four cases is a single simplex. D Once we have an even scaffold, we can make it connected using local flip operations as in the two-dimensional case; in fact the proof is slightly simpler because every component of an even scaffold is 2-connected. Let A and B be adjacent simplices that lie in different components of the even scaffold, and suppose the scaffold contains edges (p,-A), (q,A), (r, I?), and (s, B). The ridge A n B contains all but one vertex of A and all but one vertex of B, so

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without loss of generality, q and r are both in A n B. If we replace edges (q, A) and (r, B] with edges (q, B) and (r, A), we obtain a new even scaffold. Any node in the old component of A is still connected to p, then A, then q, and then to any node in the old component of B. Thus, the new even scaffold has one fewer component. Repeating this process for each adjacent pair of components, we obtain a connected even scaffold. Putting the pieces together yields the following: Theorem 10. For any d> 3, every connected simplicial d-manifold bly with boundary) has a facet cycle, except a single d-simplex.

(possi-

As we only need a path for the slab construction, we immediately obtain: Corollary 11. Every connected simplicial manifold (possibly with boundary) has a vertex-unfolding, which can be computed in linear time.

5. OPEN PROBLEMS (1) The obvious question left open by our work is whether the restriction to simplicial facets is necessary. Does every three-dimensional polyhedron with simply-connected facets have a non-overlapping vertex-unfolding? What if we require the facets to be convex? Our strip construction fails for polyhedra with nontriangular convex facets, because such polyhedra may not have facet paths. For example, the truncated cube has no facet path: no pair of its eight triangles can be adjacent in a path, but its six octagons are not enough to separate the triangles. If facets are permitted to have holes, then there are polyhedra that cannot be vertex-unfolded at all, for example, the box-on-top-of-a-box construction of Biedl et al. (Figure 7 of [5]). (2) A related reverse problem is: Given a collection of polygons connected in a chain at vertices, can they be glued along their edges to form a polyhedron? In other words, do they form a vertex-unfolding of some polyhedron? What about convex polyhedra? What is the complexity of these decision problems? (3) Finally, we mention the question which initiated our work, the generalization of edge-unfolding to higher dimensions. One can view this question as seeking a spanning tree of the dual 1-skeleton M* of a simplicial manifold M. that embeds geometrically in IR^"1 without overlap. It remains open for all d > 3 whether every d-polytope has such a ridge-unfolding.

ACKNOWLEDGMENTS We thank Anna Lubiw for a clarifying discussion, and Allison Baird, Dessislava Michaylova, and Amanda Toop for assisting with Figure 3.

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REFERENCES [I] E. M. Arkin, M. Held, J. S. B. Mitchell, and S. S. Skiena. Hamiltonian triangulations for fast rendering. Visual Comput., 12(9):429-444, 1996. [2] M. Bern, E. D. Demaine, D. Eppstein, and E. Kuo. Ununfoldable polyhedra. In Proc. llth Canad. Conf. Comput. Geom., pages 13-16, 1999. Full version: arXiv:cs.CG/9908003/. [3] M. Bern, E. D. Demaine, D. Eppstein, E. Kuo, A. Mantler, and J. Snoeyink. Ununfoldable polyhedra with convex faces. Comput. Geom. Theory Appl, to appear, 2002. [4] T. Biedl, M. Kaufmann, P. Mutzel. Drawing planar partitions II: HHdrawings. Proc. 24th Internal. Workshop Graph-Theoret. Concepts Comput. Sci. Vol. 1517, Lecture Notes in Comput. Sci., Springer-Verlag, pp. 124-136, 1998. [5] T. Biedl, E. Demaine, M. Demaine, A. Lubiw, J. O'Rourke, M. Overmars, S. Robbins, and S. Whitesides. Unfolding some classes of orthogonal polyhedra. In Proc. 10th Canad. Conf. Comput. Geom., pages 70-71, 1998. Elec. Proc. http://cgm.cs.mcgill.ca/cccg98/proceedings/. [6] E. D. Demaine, D. Eppstein, J. Erickson, G. W. Hart, and J. O'Rourke. Vertex-unfoldings of simplicial polyhedra. Technical Report 071, Dept. Comput. Sci., Smith College, Northampton, MA, USA, July 2001. arXiv:cs.CG/0110054/. [7] G. Frederickson. Dissections: Plane and Fancy. Cambridge University Press, 1997. [8] D. T. S. Geisel [Dr. Seussj. The Cat in the Hat Comes Back. Beginner Books. Random House, 1958. [9] A. Hatcher. Algebraic Topology. Cambridge University Press, 2001. [10] J. O'Rourke. Computational Geometry in C. Cambridge University Press, 2nd edition, 1998. [II] J. O'Rourke. Folding and unfolding in computational geometry. In Discrete Comput. Geom., volume 1763 of Lecture Notes Comput. Sci., pages 258-266. Springer-Verlag, 2000. Papers from the Japan Conf. Discrete Comput. Geom., Tokyo, Dec. 1998. E-mail E-mail E-mail E-mail E-mail

address: address: address: address: address:

edemaineQmit.edu ''Erik D. Demaine'' eppstein8ics.uci.edu ''David Eppstein'' [email protected] ' ' J e f f Erickson'' [email protected] ''George W. Hart'' orourkeScs.smith.edu ''Joseph O'Rourke''

VIEW-OBSTRUCTION THROUGH TRAJECTORIES OF CO-DIMENSION THREE

VISHWA C. DUMIR Centre for Advanced Study in Mathematics, Panjab University, Chandigarh 160 014, India RAJINDER J. HANS-GILL Centre for Advanced Study in Mathematics, Panjab University, Chandigarh 160 014, India

ABSTRACT. Here two generalizations of the view-obstruction problem are studied and the relevant constants are obtained for obstructing the view through (n — 3)-dimensional subspaces (flats) in the case of spheres. Together with the results known earlier, these suggest two conjectures which are stated in Section 2.

1. INTRODUCTION The view-obstruction problem was originally formulated by Cusick [2], though it had been studied in another formulation by Wills [16] and Betke and Wills [1] in the case of boxes. Its generalisations were formulated and studied by Dumir, Hans-Gill and Wilker [3,4] where rays were replaced by subspaces (flats). In [4], they observed that the problem of obstructing the view through lines is related to the billiard ball motion problem considered by Schoenberg [12,13,14] (see also Konig and Sziics [10].) Generalisations to higher dimensional trajectories were considered by Schoenberg [15] in the case of boxes. The problem is in some sense related to the problem regarding the covering minima considered by Kannan and Lovasz [9]. Dumir et al [3,4] solved the problem of obstructing the view through (n-l)-dimensional subspaces (flats) for convex bodies in Rn centred at the origin o. In the special case of spheres centred at o, the relevant constants for obstructing the view through subspaces (flats) of dimension n-2 were also 229

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determined, together with Markoff type chains for related extreme values. Here we shall determine the relevant constants for view-obstruction through (n-3)-dimensional subspaces (flats). These results together with earlier ones suggest conjectures for the cases when (n-k)-dimensional subspaces (flats) are considered (see conjectures I, II in Section 2).

2. DEFINITIONS AND STATEMENT OF RESULTS. Let Rn denote n-dimensional Euclidean space; Zn the integral lattice and A, the shifted lattice (^, ..., |) + Zn. Let C be a closed, centrally symmetric, convex body centered at the origin o and let dc*(K,K') be the C-norm distance between non-empty subsets K,K' C Rn. In [4] (see also [3]) we considered, for flats F C Rn, the quantity //(C, F) = d c (A, F) = inf {a > 0 : (aC + A) n F ^ 0}. For each dimension 1 < d < n, we define i>(C,d) = sup |z/(C,L) : L subspace of dimension d not contained in a co-ordinate hyperplane} F(C, -Z/) = sup{i/(C,F) :F translate of subspace L} and ]7(C, d) = sup{77(C, L) : L subspace of dimension d not lying in a co-ordinate hyperplane} =sup {^(C, F) : F flat of dimension d not contained in a hyperplane Xi = constant}. Clearly z/(C,n) = F(C,n) = z/(C,R n ) = 0. It is shown in [4] that to determine /'(C, d) and I^(C, d) we need consider only 'rational' subspaces (see Section 3 for definition). A method is developed for finding upper bounds for f(C,L) and z/(C,L) for rational subspaces L. Among other results, this method is applied in [4] and [6] to show that for the sphere Bn of diameter 1 and centre o in R n ,i/(B n ,?z — 1) = l/\/3 for n > 3, and z/(B n ,n — 1) = l/V^ for n > 2; also ^(B n ,n — 2) — w| and F(Bn, n — 2) = 1, for n > 4. Some related isolation results are also obtained. In [5,6,7], Dumir et al showed that z/(B 4 , 1) = WY|and ^(B 4 , 1) = J | and z/(Bs, 1) — y||. Here we shall use the isolation results for j/(B n ,n — 2) and I7(B n ,n — 2) and the method developed in [4] to prove the following results Theorem 1. z>(B n ,n - 3) < 1 for n > 4. Corollary 1. ^(B n ,n - 3) = 1 for n > 7. The corresponding results for v are Theorem 2. P(B , n - 3) < A/| for n > 4. \

71 i

/

—

W Zi *

-

Corollary 2. i>(B n ,n - 3) = J~ for n > 6.

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231

Corollaries 1 and 2 follow at once from Theorems 1 and 2 respectively on observing that for the subspaces LI : X\ + X-2 -f £3 = 0, #2 ~~X3 ~\~ X4 — 0} %5 ~f~ X6 H~ ^7 — 0

and

L2 : £1 + £2 — 0, #3 -f £4 = 0, £5 + XQ = 0

f(B n , LI) = 1 and I7(Bn, L 2 ) = v/f. These results suggest the following conjectures: Conjecture I: z/(B n ,ri — k) = w | for n > 2k if k is even and for n > 2k + 1 if k is odd. Conjecture II: Z>(B n ,n — fc) = v/| for n > 2k. The results mentioned above show that these conjectures hold for co-dimension k=0,l,2 and 3. Remark: Obvious generalizations of LI and L% to n-k dimensional subspaces provide equality cases for Conjectures I and II respectively. 3. IRRATIONAL SUBSPACES. A d-dimensional subspace L of Rn is called rational if it is {o} or has a basis consisting of vectors in Qn. If a subspace is not rational, it is called irrational. We observed in [3,4] that every irrational subspace L is contained in a unique rational subspace M(L) of minimum dimension. Also for p e Rn, we showed that z/(B n ,L + p) — f(B n ,M(L) + p) and hence Z7(B n ,L)=F(B n ,M(L)). Let n > 4 and let L be an (n-3)-dimensional irrational subspace of Rn, not lying in a co-ordinate hyperplane. Then dim M(L) > n — 2. In [3,4,6], it is shown that f(B n ,n — 1) = -^=, I7(B n ,n — 1) = -4=, f(B n ,n — 2) — A/| and I7(Bn, n — 2) = 1. These results yield Theorem 3.Let n > 4. IfL is an irrational subspace of dimension n-3 not lying in a co-ordinate hyperplane o/R n , then i/(B n ,L) < y| and I7(B n ,L)< 1. 4. RATIONAL SUBSPACES. In this section we shall drop the subscript n and denote the sphere Bn by B. 4-1 Reduction and known results. Now let L be a rational subspace of dimension n-3, not lying in a co-ordinate hyperplane. Then L-1 is a rational subspace of dimension 3 in Rn. As in [5,6] we choose a reduced basis GI , C2,03 of the 3-dimensional lattice T = L1 fl Zn

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by Minkowski's reduction process (see Sections 10.3 and v.2 of [8]). These points satisfy the following properties (i) | GI |= min{| p |: peF,p 7^ o}. (ii) | c2 |= min{| p |: c l 5 p can be extended to a basis of F). (iii) | €3 |= min{| p |: ci, c2, p is a basis of F}. Let A be the determinant of the two dimensional lattice generated by GI, c2. Then A 2 =| Cl | 2 | c 2 | 2 -( Cl .c 2 ) 2 . The following inequalities hold (1)

|ci < | c 2 | < | c 3 l

(2)

I c,-.Cj |< - | Ci |2 for i < j,

(3)

id iic 2 Hc 3 \
where A = det F. The inequality (3) is a classical result of Gauss (see Mahlerfll]). From (3) we get

m

=?' <-

A2

2

ksT'

The inequalities (1), (2) and (4) are used throughout the paper without reference to them. Let S be a rational (n-2) dimensional subspace of Rn containing L. Let A(S) be the determinant of the lattice S1 n Zn. By Theorem 4, Corollary 9 and Remark 1 of [4] we have 2fC\

A

A

If S-1 is the subspace generated by ci,C2, then using (4) we get (7)

^D^B.S C3

(8)

2

For the proof of Theorem 1 we shall use an isolation result which follows from Theorem 11 of [4]. Lemma 1. For S =< Ci,c 2 >""" we have t>(B,S) < -4= except when (i) | ci \2= 2,| c2 |2= 3,ci.c 2 — 1, in which case z/ 2 (B,S) = |,

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233

(ii) | GI | 2 =| c2 \2= 5,ci.C2 = 0, in which case z/ 2 (B,S) = |, (Hi) | GI \2 = 5, | c2 \2= 7,ci.c 2 = 0, in which case i/ 2 (B,S) = ^|, (iv) | ci |2= 3, A2 = -r2(mod 6),r = 1,2,3 in to^ic^ case z/ 2 (B,S) = - 4- r29 For the proof of Theorem 2, we need the analogous isolation result, which follows from Corollaries 2 and 3 in [6]; Lemma 2. For S = < Ci,c 2 >1 we have F(B,S) < 4= except when (i)\ d \2= 3,ci.c 2 = 0 and 3 <| c2 |2< 5, in which case Z/ 2 (B,S) = 3 + T^W, (ii) | GI | 2 =| c2 \2= 3,Ci.C2 = 1, in which case F 2 (B,S) = ^, (Hi) \c\ |2= 2, m which case

'-\

_

|(1 + ^) 2 ,7 A 2 = 1 (mo
Also we need a result, which follows from Theorem 5 of [6] by projecting onto the space of the first four coordinates. Lemma 3. IfL is defined by the equations xi + £2 — 0> X2 + £3 = 0,x 2 + a?4 = 0,then F(B,L) = ^ Proof of Theorem 1. We divide the discussion into cases depending on the value of z/(B,S). Case 1: ^ 2 (B,S) < \. Here (7) gives z/2(B,L)3. If | 03 |2= 2 then by (1), | GI |2 = | 02 |2= 2. It is easy to see that in this case L contains a point of A giving f(B,L) = 0. Case 2: | < ^ 2 (B,S) < f. Here (7) gives ^2(B,L)<|+ F f p < l i f |c3|2>4. It remains to consider the cases | 03 |2= 2,3. Using Lemma 1, it is easy to see that we need to consider the following case only: I C! |2- 2,| C2 |2 = | C3 |2- 3 , C i . C 2 - l,Ci.C 3 = 1.

Then c2.c3 = 0, 1 or -1 and A2 = 12 + 2(c2.c3) - 2(c2.c3)2 = 12,12,8, respectively. Here A2 = 5 and in case A =12, the ineguality (5) gives , —2

.

In case A = 8,c2.c3 = —1 and we can suppose without loss of generality, thatci = (1,1,0,0,0,0,...), c2 = (0, 1,1, 1,0,0,...) and c3 = (1,0, -1,0, 1,0, ...)

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or (0, !,-!,— 1,0,0,...). The second possiblity for C3 puts L into a coordinate hyperplane. So we need only consider the first possibility for c3. Orthogonalising these vectors we see that L is given by equations x\ + z2 — 0, -xi -f X2 + 2z3 + 2x4 — 0 and x\ — xi — 2x3 -f 8x4 + 5x5 — 0. Hence ,

1

--

10 -Taking x=(|, -\, |, |, -|,...) 6 A we obtain

r

- x2- 2x3 + 3z4 40

Case 3: ^ 2 (B,S) > f. This takes us into Case (iv) of Lemma 1 where | GI |2_ — 3, and v (B,S) = 3 + 3^75 with r — l,2,3,r 2 = -A2 (mod 6). Here the inequality (7) gives i/2(B,L)
(9)

2

2

ic 3 r

2

Therefore i/(B,L) < 1 if A > 2 r and | c3 | > 4. First we consider the case | c3 \2 — 3. Here | GI |2 = | c2 |2=: 3 and Ci.c 2 = 0 or 1. If Ci.c 2 = 1, then A2 = 8 giving r = 2 and A2 = 24 - 3( Cl .c 3 ) 2 - 3(c2.c3)2 + 2(ci.c 3 )(c 2 .c 3 ) > 16 since Ci.c3 and c2.c3 are 0 or ±1. Then (5) gives //(B,L) < 1. If ci.c 2 = 0 but ci.c3 ^ 0 or c2.c3 ^ 0 then renaming Ci,c 2 ,cs suitably will take us to the above case. So we can suppose that GI.CS = c 2 .C3 = 0. In this case it is easy to see (cf.[3], Lemma l)that i>(B,L) = 1. So we can now suppose that | c3 |2> 4. Now (9) gives z/(B,L) < 1 if A2 > 2r 2 . The condition A2 > 2r2 is always satisfied if r — 1 or 2. So let us suppose r = 3. Then it is easily seen that A2 > 2r2 = 18 and A2 = -9 (mod 6) are satisfied except when | c2 |2= 3,5 and Ci.c 2 = 0. If | c2 |2= 5 then A2 = 15, | c3 |2> 5 and (9) gives o o If | c2 \2= 3, then A2 = 9 and (9) gives

o

^ 2 (B,L) < i + i + -A- < 1, if | c3 |2> 6. o o I C3 I Since the case | c3 \2= 3 has already been discussed we are left with the case I Ci |2=:| 2

Here f (B,S) = |,A

2

C2 |2=

3 , C i . C 2 = 0, I C 3 | 2 =

4,5.

= 9 and

A2 - 9 c3

2

-3(Cl.c3)2 - 3(c2.c3)2 > 30.

VIEW-OBSTRUCTION THROUGH TRAJECTORIES

235

Therefore (5) gives

v

'

2 A2 2 ) — o ~*~ -r-2 — o

9 on

O

"A.

OU

This completes the proof of Theorem 1.

5. PROOF OF THEOREM 2. Here we divide the discussion into cases depending on the value of F(B, S). Case 1: F 2 (B,S) < \. Here (8) gives 1 2 3 F 2 (B,L) < - + i pr < -, because I c3 |2> 2. 2 I c3 \* 2 Case 2: F 2 (B,S) > \. Here we have to consider cases (i), (ii) and (iii) of Lemma 2. In cases (i) and (ii), F(B,S) < A/I and | c3 |2> 3, so that (8) gives

It now remains to consider | GI \2= 1. Here ^(B, S) is as given in Lemma 2(iii). In this case F 2 (B,S) < 1 and so
, C

if | c 3 | 2 > 4 .

I 3 I

If | c3 |2= 2, then | Ci | 2 =| 02 \2= 2. If further c;.Cj = 0 for each i < j, then I7(B,L) = w|. If c;.Cj ^ 0 for some i, j we can suppose that c\.c^ — 1. Then A2 = 3 and A2 - 6 - 2(Cl.c3)2 - 2(c2.c3)2 + 2(c1.c3)(c2.c3). _ 2

_ 2

Since A ^ 0,(ci.c 3 )(c2.c 3 ) = — 1 is not possible. Therefor A = 6 or 4. In case A = 6, F 2 (B,L) < | + | < |. In case A~ = 4, we can suppose that (ci.c 3 ,C2.c 3 ) = (1,0) or (1,1). Here we can suppose that GI = (1,1,0,0,...) and c2 = (0,1,1,0,0,...). There are now essentially four possibilities for c3. Two of these put L in a coordinate hyperplane. The other two are equivalent to taking c3 = (0, 1, 0, 1, 0, ...) so that L is given by the equation x\ + #2 — 0, x-2 + £3 = 0 and x% + £4 = 0. By Lemma 3, F(B, L) < */|. Now suppose that | c3 |2= 3. Then | c2 |2= 2 or 3. If | c2 \2= 3, then A2 = 6 or 5 and F 2 (B,S) = f or |f . Therefore F22 ( B , L ) < 6 9 /-~-^ 9

-^ \

c3

12

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Now let | c2 |2= 2. Here A2 = 4 or 3,1/ 2 (B,S) = 1 or f. When A2 = 4 we have Ci.c 2 = 0 and A = 12 — 2(ci.cs) 2 - 2(c2.c 3 ) 2 > 8. Therefore F(B,L)<1 + |^<|. When A2 = 3, we have Ci.c 2 — 1 and A 2 = 9 - 2( Cl .c 3 ) 2 - 2(c 2 .c 3 ) 2 + 2(c 1 .c 3 )(c 2 .c 3 ). Therefore */ 2 (B,L) <| + = ^ - < | i f A > 5. This inequality is satisfied except when (ci.c 3 )(c 2 .c 3 ) = —1. It is easy to see that this is not possible because L is not contained in a co-ordinate hyperplane. This completes the proof of Theorem 2. Acknowledgements: Considerable part of this paper was worked out jointly with Professor J.B. Wilker, who unfortunately expired in 1995 and our lively discussions on view-obstruction came to an end. The authors are grateful to the referee for pointing out some misprints.

REFERENCES [1] U. Betke and J.M. Wills, Untere Schranken fiir zwei diophantische Approximations-Functionen, Monatshefte fiir Math. 76 (1972), 214-217. [2] T.W. Cusick, View-obstruction problems, Aequationes Math. 9 (1973) 165-170. [3] V.C. Dumir, R.J. Hans-Gill and J.B. Wilker, Contributions to a general theory of view-obstruction problems, Can. J. Math. 20 (1993) 517-536. [4] _ Contributions to a general theory of view-obstruction problems II, J. of Number Theory 59 (1996) 352-373. [5] _ A Markoff type chain for the view-obstruction problem for spheres in R4, Monatshefte fiir Math. 118 (1994) 205-217. [6] _ View-obstruction and a conjecture of Schoenberg, Indian J. of Pure and Applied Math. 27 (1996) 323-342. [7] _ The view-obstruction problem for spheres in R5, Monatshefte fiir Math. 122 (1996) 21-34. [8] P.M. Gruber and C.G. Lekkerkerker, Geometry of Numbers, second edition, North-Holland, Amsterdam. [9] R. Kannan and L. Lovasz, Covering minima and lattice point free convex bodies, Annals of Math. 128 (1988) 577-602. [10] D. Konig and Sziics, Mouvement d'un point abandonne a 1'interieur d'un cube, Rendiconti del Circ. Mat. di Palermo 38 (1913) 79-90. [11] K. Mahler, On reduced positive definite ternary quadratic forms, J. of London Math. Soc. 15 (1940) 193-195. [12] I.J. Schoenberg, On the motion of a billiard ball in two dimensions, Delta, Madison, Wisconsin (1975) 1-18. [13] _ Extremurn problems for the motion of a billiard ball I. The Lp norm, 1 < P < oo, Indag. Math. 38 (1976) 66-75.

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[14] _ Extremum problems for the motion of a billiard ball II. The L^ norm, Indag. Math. 38 (1976) 263-279. [15] _ Extremum problems for the motion of a billiard ball III. The multidimensional case of Konig and Sziics, Studia Sci. Math. Hungarica 13 (1978) 53-78. [16] J.M. Wills, Zur simultanen homogenen diophantischen approximation I, Monatshefte fur Math. 72 (1968) 254-263. E-mail address: E-mail address:

vcdumirQpu.ac.in ' ' V . C. Dumir'' [email protected] ' ' R . J. Hans-Gill''

FAT 4-POLYTOPES AND FATTER 3-SPHERES

DAVID EPPSTEIN1 Department of Information and Computer Science, University of California, Irvine, CA 92697 GREG KUPERBERG2 Department of Mathematics, University of California, Davis, CA 95616 GUNTER M. ZIEGLER3 Institut fiir Mathematik, MA 6-2, Technische Universitat Berlin, D-10623 Berlin, Germany

ABSTRACT. We introduce the fatness parameter of a 4-dimensional polytope P, defined as 4>(P] = (/i + /2)/(/o + /a)- It arises in an important open problem in 4-dimensional combinatorial geometry: Is the fatness of convex 4-polytopes bounded? We describe and analyze a hyperbolic geometry construction that produces 4-polytopes with fatness (P) > 5.048, as well as the first infinite family of 2-simple, 2-simplicial 4-polytopes. Moreover, using a construction via finite covering spaces of surfaces, we show that fatness is not bounded for the more general class of strongly regular CW decompositions of the 3-sphere.

1. INTRODUCTION The characterization of the set ^-3 of /-vectors of convex 3-dimensional poly topes (from 1906, due to Steinitz [28]) is well-known and explicit, with a simple proof: An integer vector (/o, /i, /2) is the /-vector of a 3-polytope if and only if it satisfies • /i = /o + /2 — 2 (the Euler equation), Supported in part by NSF grant CCR #9912338 Supported by NSF grant DMS #0072342 3 Supported by a DFG Leibniz grant 239

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• /2 < 2/o — 4 (with equality for simplicial polytopes), and • /o < 2/2 — 4 (with equality for simple polytopes) . (Recall that by the definition of the /-vector, /^ is the number of fc-faces of the polytope.) This simple result is interesting for several reasons: • The set of /-vectors is the set of all the integer points in a closed 2-dimensional polyhedral cone (whose apex is the /-vector /(As) = (4,6,4) of a 3-dimensional simplex). In particular, it is convex in the sense that F$ = conv^) n Z3. • The same characterization holds for convex 3-polytopes (geometric objects), more generally for strongly regular CW 2-spheres (topological objects), and yet more generally for Eulerian lattices of length 4 (combinatorial objects [27]). In contrast to this explicit and complete description of jFs, our knowledge of the set ^-4 of /-vectors of (convex) 4-polytopes (see Bayer [2] and Hoppner and Ziegler [14]) is very incomplete. We know that the set J-± of all /-vectors of 4-dimensional polytopes has no similarly simple description. In particular, the convex hull of ^4 is not a cone, it is not a closed set, and not all integer points in the convex hull are /-vectors. Also, the 3-dimensional cone with apex /(A,$) spanned by ^-4 is not closed, and its closure may not be polyhedral. Only the two extreme cases of simplicial and of simple 4-polytopes (or 3-spheres) are well-understood. Their /-vectors correspond to faces of the convex hull of ^4, denned by the valid inequalities /2 > 2/3 and f\ > 2/o, and the (/-Theorem, proved for 4-polytopes by Barnette [1] and for 3-spheres by Walkup [34], provides complete characterizations of their /-vectors. (The g-Theorem for general simplicial polytopes was famously conjectured by McMullen [21] and proved by Billera and Lee [4] and Stanley [26]. See [35, §8.6] for a review.) But we have no similarly complete picture of other extremal types of 4-polytopes. In particular, we cannot currently answer the following key question: Is there a constant c such that all 4-dimensional convex polytopes P satisfy the inequality fi(P)+h(P)

<

c(f0(P)+h(P))l

To study this question, we introduce the fatness parameter def

fl(P)+h(P)

of a 4-polytope P. We would like to know whether fatness is bounded. For example, the 4-simplex has fatness 2, while the 4-cube and the 4-cross polytope have fatness || = |. More generally, if P is simple, then we can

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substitute the Dehn-Sommerville relations /2(P) = /i(P) + /3(P) - /o(P)

/i(P) = 2/0(P)

into the formula for fatness, yielding /i(P)+/ 2 (P) _ 3/ 0 (P)+/ 3 (P) ~ / o ( P ) + /3(P)~ /o(P) + /3(P) < ' Since every 4-polytope and its dual have the same fatness, the same upper bound holds for simplicial 4-polytopes. On the other hand, the "neighborly cubical" 4-polytopes of Joswig and Ziegler [16] have /-vectors 0(

}

(4,2n,3n-6,n-2)-2n-2, and thus fatness

5n - 6 6= — -> 5. ^ n +2 In particular, the construction of these polytopes disproved the conjectured flag-vector inequalities of Bayer [2, pp. 145, 149] and Billera and Ehrenborg

[14, p. 109]. The main results of this paper are two lower bounds on fatness: Theorem 1. There are convex ^-polytopes P with fatness 0(P) > 5.048. Theorem 2. The fatness of cellulated 3-spheres is not bounded. A 3-sphere S with N vertices may have fatness as high as (f)(S) = f2(-/V 1//12 ). We will prove Theorem 1 in Section 3 and Theorem 2 in Section 4, and present a number of related results along the way. 2. CONVENTIONS Let X be a finite CW complex. If X is identified with a manifold M, it is also called a cellulation of M. The complex X is regular if all closed cells are embedded [22, §38]. If X is regular, we define it to be strongly regular if in addition the intersection of any two closed cells is a cell. For example, every simplicial complex is a strongly regular CW complex. The complex X is perfect if the boundary maps of its chain complex vanish. (A non-zero-dimensional perfect complex is never regular.) The f-vector of a cellulation X, denoted f ( X ) = (/o, / i , . . . ) , counts the number of cells in each dimension: fo(X) is the number of vertices, f i ( X ) is the number of edges, etc. If X is 2-dimensional, we define its fatness as def

fl(X)

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If X is 3-dimensional, we define its fatness X as fl(X)+h(X)

If P is a convex d-polytope, then its /-vector /(P) is defined to be the /-vector of its boundary complex, which is a strongly regular (d— l)-sphere. If d = 4 we can thus consider (P), the fatness of P. The faces of P of dimension 0 and 1 are called vertices and edges while the faces of dimension d — I and d — 2 are called facets and ridges. We extend this terminology to general cellulations of (d — 1) -manifolds. The flag vector of a regular cellulation X, and likewise the flag vector of a polytope P, counts the number of nested sequences of cells with prespecified dimensions. For example, fois(X) is the number triples consisting of a 3-cell of X, an edge of the 3-cell, and a vertex of the edge; if P is a 4-cube, then /oi 3 (P) = 192. A convex polytope P is simplicial if each facet of P is a simplex. It is simple if its polar dual PA is simplicial, or equivalently if the cone of each vertex matches that of a simplex. 3. 4-POLYTOPES In this section we construct families of 4-polytopes with several interesting properties: • They are the first known infinite families of 2-simple, 2-simplicial 4:-polytopes, that is, polytopes in which all 2-faces and all dual 2-faces are triangles (all edges are "co-simple"). E.g., Bayer [2] says that it would be interesting to have an infinite family. As far as we know, there were only six such polytopes previously known: the simplex, the hypersimplex (the set of points in [0, 1]4 with coordinate sum between 1 and 2), the dual of the hypersimplex, the 24-cell, and a gluing of two hypersimplices (due to Braden [6]), and the dual of the gluing. (There are claims in Griinbaum [13, p. 82, resp. p. 170] that results of Shephard, resp. Perles and Shephard, imply the existence of infinitely many 2-simple 2-simplicial 4-polytopes. Both claims appear to be incorrect.) • They are the fattest known convex 4-polytopes. • They yield finite packings of (not necessarily congruent) spheres in R3 with slightly higher average kissing numbers than previously known examples [18]. Let Q C R4 be a 4-polytope that contains the origin in its interior. If an edge e of Q is tangent to the unit sphere S"3 C R4 at a point t, then the

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corresponding ridge (2-dimensional face) F = e^ of the polar dual P = QA is also tangent to S3 at t. Recall that the polar dual is defined as *

Furthermore, the affine hulls of e and F form orthogonal complements in the tangent space of S3, so the convex hull conv(e U F) is an orthogonal bipyramid tangent to S3 (cf., Schulte [24, Thm. 1]). We will construct polytopes E by what we call the E-construction. This means that they are convex hulls Ed=conv(guP), where Q is a simplicial 4-dimensional polytope whose edges are tangent to the unit 3-sphere S3, and P is the polar dual of Q. Thus P is simple and its ridges are tangent to 53. (We call Q edge-tangent and P ridge-tangent.) Proposition 3. If P is a simple, ridge-tangent ^-polytope, then the ^-polytope E = conv(P U Q) produced by the E-construction is 2,-simple and 2simplicial, with f-vector f ( E ) = (/2(P),6/o(P),6/ 0 (P),/ 2 (P)), and fatness ^)=6 / ° ( P )

Proof. Another way to view the ^-construction is that E is produced from P by adding the vertices of Q sequentially. At each step, we cap a facet of P with a pyramid whose apex is a vertex of Q. Thus the new facets consist of pyramids over the ridges of P, where two pyramids with the same base (appearing in different steps) lie in the same hyperplane (tangent to S3), and together form a bipyramid. The facets of the final polytope E are orthogonal bipyramids over the ridges of P and are tangent to S3. Since the 2-faces of E are pyramids over the edges of P, E is 2-simplicial. The polytope E is 2-simple if and only if each edge is co-simple, i.e., contained in exactly three facets of E. The iterative construction of E shows that it has two types of edges: (i) edges of P, which are co-simple in E if and only if they are co-simple in P, and (ii) edges formed by adding pyramids, which are co-simple if and only if the facets of P are simple. Since P is simple, its facets are simple and its edges are co-simple, so E is then 2-simple. This combinatorial description of E yields an expression for the /-vector of E in terms of the flag vector [3, 2, 14] of P. Since the facets of E are

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bipyramids over the ridges of P, the following identities hold: ME) = h(P)

ME) = /13(P)

ME) = /I(P) + /03(P)

ME) = /0(P) + /3(P).

Since P is simple, /03(P)=4/ 0 (P)

/13(P) = 3/i(P).

These identities together with Eider's equation and /i(P) = 2/o(P) imply the proposition. D The /-vector of P also satisfies /o(P) - /i(P) + / 2 (P) - /3(P) = 0

/i(P) = 2/ 0 (P),

in the second case because P is simple, so the fatness of E can also be written

Thus maximizing the fatness of E is equivalent to maximizing the ridge- facet ratio /2(P)//s(P), or the average degree of the graph of Q. It also shows that the jE-construction cannot achieve a fatness of 6 or more. In light of Proposition 3, we would like to construct edge-tangent simplicial 4-polytopes. Regular simplicial 4-polytopes (suitably scaled) provide three obvious examples: the 4-simplex A4, the cross poly tope 64A, and the 600-cell. From these, the E'-construction produces the dual of the hypersimplex, the 24-cell, and a new 2-simple, 2-simplicial polytope with /-vector (720, 3600, 3600, 720) and fatness 5, whose facets are bipyramids over pentagons. We will construct new edge-tangent simplicial 4-polytopes by gluing together (not necessarily simplicial) edge-tangent 4-polytopes, called atoms, to form compounds. We must position the polytopes so that their facets match, they remain edge-tangent, and the resulting compound is convex. It will be very useful to interpret the interior of the 4-dimensional unit ball as the Klein model of hyperbolic 4-space H4, with S3 the sphere at infinity. (See Iversen [15] and Thurston [30, Chap. 2] for introductions to hyperbolic geometry.) In particular, Euclidean lines are straight in the Klein model, Euclidean subspaces are flat, and hence any intersection of a convex polytope with HI 4 is a convex (hyperbolic) polyhedron. Even though the Klein model respects convexity, it does not respect angles. However, angles and convexity are preserved under hyperbolic isometries. There are enough isometries to favorably position certain 4-polytopes to produce convex compounds. If a polytope Q is edge-tangent to ,53, then it is hyperbolically hyperideal: Not only its vertices, but also its edges, lie beyond the sphere at infinity, except for the tangency point of each edge. Nonetheless portions of its facets

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and ridges lie in the finite hyperbolic realm. As a hyperbolic object the polytope Q (more precisely, QflH 4 ) is convex and has flat facets. The ridge r between any two adjacent facets has a well-defined hyperbolic dihedral angle, which is strictly between 0 and TT if (as in our situation) the ridge properly intersects H4. To compute this angle we can intersect r at any point t with any hyperplane R that contains the (hyperbolic) orthogonal complement to r at t. We let t be the tangent point of any edge e of the ridge, and let R be the hyperplane perpendicular to e.

FIGURE 1. A cone emanating from an ideal point t in H3 in the Poincare model, and a horosphere 5 incident to t. The link of t (here a right isosceles triangle) inherits Euclidean geometry from S. Within the hyperbolic geometry of .R = H3, every line emanating from the ideal point t is orthogonal to any horosphere incident to t. Thus the link of the edge e of Q is the intersection of Q fl R with a sufficiently small horosphere S at t. Since horospheres have flat Euclidean geometry [30, p. 61], the link Q D S is a Euclidean polygon. Its edges correspond to the facets of Q that contain e, and its vertices correspond to the ridges of Q that contain e. Thus the dihedral angle of a ridge r of Q equals to the Euclidean angle of the vertex r D S of the Euclidean polygon Q D S. This is easier to see in the Poincare model of hyperbolic space, because it respects angles, than in the Klein model. Figure 1 shows an example. To summarize:

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Lemma 4. A compound of two or more polytopes is convex if and only if each ridge has hyperbolic dihedral angle less than TT, or equivalently, iff each edge link is a convex Euclidean polygon. Compounds can also have interior ridges with total dihedral angle exactly 2?r. But since all atoms of a compound are edge-tangent, compounds do not have any interior edges or vertices. If Q is a regular polytope, then Q D S is a regular polygon. The following lemma is then immediate: Lemma 5. // Q is a regular, edge-tangent, simplicial ^-polytope, then in the hyperbolic metric of the Klein model, its dihedral angles are Tr/3 (for the simplex), ir/2 (for the cross polytope), and 3/T/5 (for the 600-cell). A hyperideal hyperbolic object, even if it is an edge-tangent convex polytope, can be unfavorably positioned so that it is unbounded as a Euclidean object (c/., Schulte [24, p. 508]). Fortunately there is always a bounded position as well: Lemma 6. Let be Q an edge-tangent, convex polytope in M whose points of tangency with Sd~l do not lie in a hyperplane. Then there is a hyperbolic isometry h (extended to all o / R d j such that h(Q) is bounded. Proof. Let p lie in the interior of the convex hull of the edge tangencies and let / be any hyperbolic motion that moves p to the Euclidean origin in IR rf . Since the convex hull K of the edge tangencies of h(Q) contains the origin, K^ is a bounded polytope that circumscribes Sd~l. Since K^ is facet-tangent where h(Q] is edge-tangent, h(Q] C K^. D In the following we discuss three classes of edge-tangent simplicial convex 4-polytopes that are obtained by gluing in the Klein model: Compounds of simplices, then simplices and cross polytopes, and finally compounds from cut 600-cells. There are yet other edge-tangent compounds involving cross polytopes cut in half (ie., pyramids over octahedra), 24-cells, and hypersimplices as atoms, but we will not discuss these here. 3.1. Compounds of simplices. In this section we classify compounds whose atoms are simplices. This includes all stacked polytopes, which are simplicial polytopes that decompose as a union of simplices without any interior faces other than facets. However compounds of simplices are a larger class, since they may have interior ridges. Lemma 7. Any edge-tangent d-simplex is hyperbolically regular. Proof. The proof is by induction on d, starting from the case d = 2, where the three tangency points define an ideal triangle in H2. All ideal triangles

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are congruent [30, p. 83]. Since edge-tangent triangles are the polar duals of ideal triangles, they are all equivalent as well. If d > 2, let B be a general edge-tangent d-simplex. On the one hand, there exists an edge-tangent simplex A which is regular both in hyperbolic geometry and Euclidean geometry. On the other hand, given the position of d of the vertices, there are at most two choices for the last vertex that produce an edge-tangent simplex, one on each side of the hyperplane spanned by the first d. By induction there exists an isometry that takes a face of B to a face of A and the remaining vertex to the same side. The edge-tangent constraint implies that this isometry takes the last vertex of B to the last vertex of A as well. D Proposition 8. There are only three possible edge-tangent compounds of ^-simplices: • the regular simplex, • the bipyramid (a compound of two simplices that share a facet), and • the join of a triangle and a hexagon (a compound of six simplices that share a ridge). Proof. Figure 2 shows all strictly convex polygons with unit-length edges tiled by unit equilateral triangles, or triangle jewels. Since the atoms of an edge-tangent compound of simplices are edge-tangent, they are hyperbolically regular by Lemma 7, and their edge links are equilateral triangles. Thus every edge link of a compound of simplices is a triangle jewel. Any three 4-simplices in a chain in such a compound share a ridge. In order to create an edge link matching Figure 2, they must extend to a ring of six simplices around the same ridge. Adding any further simplex to these six would create an edge link in the form of a triangle surrounded by three other triangles, which does not appear in Figure 2. D

FIGURE 2. The 3 possible edge links of edge-tangent compounds of 4-simplices. Two of the .E-polytopes produced by Proposition 8 were previously known. If Q is the simplex, then E is dual to the hypersimplex. If Q is the bipyramid, then E is dual to Braden's glued hypersimplex. However, if Q is the sixsimplex compound (dual to the product of a triangle and a hexagon), then E is a new 2-simple, 2-simplicial polytope with /-vector (27,108,108,27).

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Proposition 8 also implies an interesting impossibility result. Corollary 9. No stacked 4-polytope with more than 6 vertices is edge-tangent. See Schulte [24, Thm. 3] for the first examples of polytopes that have no edge-tangent realization. 3.2. Compounds of simplices and cross polytopes. Next, we consider compounds of simplices and regular cross polytopes. The edge link of any convex compound of these two types of polytopes must be one of the eleven strictly convex polygons tiled by unit triangles and squares, or squaretriangle jewels (Figure 3). See Malkevitch [20] and Waite [33] for work on convex compounds of these shapes relaxing the requirement of strict convexity.

FIGURE 3. The 11 possible edge links of edge-tangent compounds of 4-simplices and cross polytopes. If Q is a single cross polytope, then E is a 24-cell. We can also glue simplices onto subsets of the facets of the cross polytope. The new dihedral angles formed by such a gluing are 57T/6. The resulting compound is convex as long as no two glued cross polytope facets share a ridge. We used a computer program to list the combinatorially distinct ways of choosing a subset of nonadjacent facets of the cross polytope; the results may be summarized as follows. In addition to the 24-cell, this yields 20 new 2-simple, 2-simplicial polytopes. Proposition 10. There are exactly 21 distinct simplicial edge-tangent compounds composed of one regular 4-dimensional cross polytope and k > 0

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simplices, according to the following table: k 0 1 2 3 4 5 6 7 8 Total 1 1 3 3 6 3 2 1 1 21

#

We can also confirm that every square-triangle jewel arises as the edge link of an edge-tangent 4-dimensional compound. For every jewel other than the one in the center, we can form a convex edge bouquet consisting of simplices and cross polytopes that meet at an edge: we replace each triangle by a simplex and each square by a cross polytope. Since the central jewel has two adjacent squares, its edge bouquet is not convex. Instead we glue two cross polytopes along a facet so that the 4 ridges of that facet are flush, z'.e., their dihedral angle is TT. Thus we can "caulk" each such ridge with three simplices that share the ridge. The central jewel is the link of 6 of the edges of the resulting compound of 2 cross polytopes and 12 simplices. Simplices and regular cross polytopes combine to form many other edgetangent simplicial polytopes and hence 2-simple, 2-simplicial polytopes. In particular, these methods lead to the following theorem. Theorem 11. There are infinitely many combinatorially distinct 2-simple, 2-simplicial facet-tangent ^-polytopes. Proof. We glue n cross polytopes end-to-end. Each adjacent pair produces 4 flush ridges that we caulk with chains of three simplices. The facets to which these simplices are glued are not adjacent and so do not produce any further concavities. D The chain of n cross polytopes has /-vector (4n + 4,18n + 6,28n + 4,14n + 2). Filling a concavity adds (2,9,14,7), so after filling the 4(n — 1) concavities we get a simplicial polytope Q with /(Q) = (12n-4,54n-30,84n-54,42-26), which yields a 2-simple, 2-simplicial 4-polytope E with f ( E ) = (54ra - 30,252n - 156,252n - 156,54n - 30) by Proposition 3. Remark. Every 2-simple, 2-simplicial 4-polytopes that we know is combinatorially equivalent to one which circumscribes the sphere. Are there any that are not? We do know a few 2-simple, 2-simplicial 4-polytopes which are not .E-polytopes. Trivially there is the simplex. There are a few others that arise by the fact that the 24-cell is the .E-polytope of a cross polytope in 3

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different ways. Color the vertices of a 24-cell red, green, and blue, so that the vertices of each color span a cross polytope. If we cap one facet of a cross-polytope by a simplex and apply the ^-construction, the result is a 24-cell in which 6 facets that meet at 1 vertex are replaced by 10 facets and 4 vertices. If the replaced vertex is red, the replacement can be induced by capping either blue cross polytope or the green cross polytope and then applying the .E-construction; the position of the replacement differs between the two cases. If we replace two different red vertices, one by capping the green cross polytope and the other by capping the blue cross polytope, then the resulting polytope is 2-simple and 2-simplicial but not an E-polytope. This construction has several variations: for example, we can also replace three vertices, one of each color. 3.3. Compounds involving the 600-cell. If Q is the 600-cell, then E is a 2-simple, 2-simplicial polytope with /-vector (720, 3600,3600, 720) and fatness exactly 5. Again, we can glue simplices onto any subset of nonadjacent facets of the 600-cell, creating convex compounds with dihedral angle 147T/15. We did not count the (large) number of distinct ways of choosing such a subset, analogous to Proposition 10. It is not possible to glue a cross polytope to a 600-cell, because that would create an Il7r/10 angle (ie., a concave dihedral of 9?r/10) which cannot be filled by additional simplices or cross polytopes. The large dihedral angles of the 600-cell make it difficult to form compounds from it, but we can modify it as follows to create smaller dihedrals. Remove a vertex and form the convex hull of the remaining 119 vertices. The resulting convex polytope has 580 of the 600-cell's tetrahedral facets and one icosahedral facet. The pentagonal edge link (Figure 4(a)) of the edges bordering this new facet become modified in a similar way, by removing one vertex and forming the convex hull of the remaining four vertices (Figure 4(b)), which results in a trapezoid; thus, the hyperbolic dihedrals at the ridges around the new facet are 27T/5. This cut polytope is not simplicial, but we may glue two of these polytopes together along their icosahedral facets, forming a simplicial polytope with 4?r/5 dihedrals along the glued ridges. This compound's new edge links are hexagons formed by gluing pairs of trapezoids (Figure 5(a)). The same cutting and gluing process may be repeated to form a sequence or tree of 600-cells, connected along cuts that do not share a ridge. For such a chain or tree formed from n cut 600-cells, the /-vector may be computed as /(Q) = (106n + 14,666n + 54,1120n + 80,560n + 40), so the ^/-construction yields f ( E ) = (666n + 54,3360n + 240,3360n + 240,666n + 54)

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(a)

(b)

(c)

FIGURE 4. Edge figures of (a) a 600-cell, (b) a 600-cell with one vertex removed, (c) a 600-cell with two removed vertices, and (d) an icosahedral cap. and thus a fatness of 3360n + 240 666n + 54

560

TIT

5.045045.

Thus the fatness of the 2-simple, 2-simplicial polytopes formed by such compounds improves slightly on that formed from the 600-cell alone.

(a)

(b)

FIGURE 5. Edge links of compounds of cut 600-cells.

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EPPSTEIN, KUPERBERG AND ZIEGLER

It is also possible to form compounds involving 600-cells which have been cut by removing several vertices (as described above) so that two of the resulting icosahedral facets meet at a ridge. Each edge link at this ridge is an isosceles triangle formed by removing two vertices from a pentagon (Figure 4(c)). Thus the dihedral angle of the triangular ridge between the icosahedral facets is ?r/5. We can therefore form compounds in which ten of these doubly-cut 600-cells share a triangle, whose edges links are a regular decagon cut into ten isosceles triangles (Figure 5(b)). Yet other compounds of cut 600-cells and simplices are possible, although we do not need them here. The cut 600-cells also form more complicated compounds which require some group-theoretic terminology to explain. The vertices of a regular 600cell form a 120-element group under (rescaled) quaternionic multiplication, the binary icosahedral group. This group has a 24-element subgroup, the binary tetrahedral group, which also arises as the units of the Hurwitz integers (see Conway and Sloane [7, §2.2.6,8.2.1]). Let A be the convex hull of the other 96 vertices of the 600-cell; ie., A is formed by cutting 24 vertices from 600-cell in the above manner. The resulting polytope is the "snub{3,4,3}" (snub 24-cell) of Coxeter [8, §8.4,8.5]. Its /-vector is (96,432,480,144). Every icosahedral facet of A is adjacent to 8 other icosahedron facets, as well as to 12 tetrahedra. Thus A has 96 icosahedron-icosahedron ridges. We can build new hyperbolic, edge-tangent, simplicial polytopes by gluing copies of A along icosahedral faces and capping the remaining icosahedral facets with pyramidal caps of the type that we had cut off to form A. (The edge links of such a cap C are given by Figure 4(d).) The resulting polytope Q will be convex if at each icosahedral-icosahedral ridge of a copy of A, either 10 copies of A meet, or two caps and one or two copies of A do. Also at each icosahedral-tetrahedral ridge of a copy of A, either two copies of A or one each of A and a cap must meet. If two copies of A meet (in an icosahedral facet F), then they differ by a reflection through F. These reflections generate a discrete hyperbolic reflection group F since the supporting hyperplanes of the icosahedral facets (the facets of a hyperideal 24-cell, whose ridges are also ridges of ^1!) satisfy the Coxeter condition: When they meet, they meet at an angle of Tr/5, which divides TT. Thus the copies of A used in Q are a finite subset S of the orbit of A under F. The set S determines Q. Remark. That it suffices to consider the dihedral angles of adjacent facets follows from Poincare's covering-space argument: Let P be a spherical, Euclidean, or hyperbolic polytope whose dihedral angles divide TT. Let X denote the space in which it lives. Let Y be the disjoint union of all copies of P in X in every position, and let Z be the quotient of Y given by identifying two

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copies of P along a shared facet. The space Z is constructed abstractly so that P tiles it. We claim that Z is a covering space of X. Each p 6 Z lies in the interior of some face F of a copy of P. If F is a copy of P or a facet, this is elementary; if F is a ridge, it follows from the dihedral angle condition. Otherwise it follows by applying the covering-space argument inductively, replacing X by the link S of F and P by P n S. Since Z is a covering space, a connected component of Z is a tiling of X by P. See Vinberg [32] for a survey of hyperbolic reflection groups. There is no one best choice for S, only a supremal limit. One reasonable choice for S is the corona of a copy of A, i.e., A together with the set of all images under F that meet it, necessarily at a facet or a ridge. The corona of A is depicted by a simplified (and therefore erroneous) schematic in Figure 6; the reader should imagine the correct, more complicated version. The schematic uses a chemistry notation in which each copy of A is represented as an atom, each pair of copies that shares a facet is represented as a bond, and "7A" denotes a chain of 7 atoms. To extend the terminology, we call 10 copies of A that meet at a ridge a ring. The schematic is simplified in that the central copy actually has 24 bonds (not 6), each of the neighbors has 9 bonds (not 3), and there are 96 rings (not 6). 7A / \

A

A

\/ 7A FIGURE 6. An oversimplified schematic of a corona of A. Let Q be the union of these copies of A with the remaining icosahedral facets capped. To compute the /-vector of Q it is easier to view each atom as a copy of a 600-cell B, minus two caps for each bond. There are 1+24+96-7 = 697 atoms and 24 + 96 • 8 = 792 bonds in total. Thus Q has /3(Q) = 697/3(5) - 792 • 2/£(C) = 386520 facets, where f^(C) — 30 is the number of simplicial facets of a cap C.

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Counting vertices is more complicated. Let / be an icosahedron and let T be a triangle. Then /o(Q) = 697/0(£) - 792 - 2/0(C) + 792/0(J) + 96/0(T) = 72840, because after the vertices of the caps are subtracted, the vertices of each icosahedral facet at a bond are undercounted once, and after these are restored the vertices of each triangle at the center of a ring are undercounted once. The rest of the /-vector of Q follows from the Dehn-Sommerville equations: h(Q] = 2/3(Q) = 773040, /i (Q) = /o(Q)+/3(Q)= 459360. The polytope Q yields an E'-polytope with fatness

Note that since this bound arises from a specific choice of S rather than a supremal limit, this is not optimal as a lower bound of supremal fatness. 3.4. Kissing numbers. As mentioned above, another use of ridge-tangent polytopes P is the average kissing number problem [18]. Let X be a finite packing of (not necessarily congruent) spheres S3, which is equivalent to a finite sphere packing in M 3 by stereographic projection. The question is to maximize the average number of kissing points of the spheres in X . If P is ridge-tangent, its facets intersect the unit sphere S3 in a sphere packing X, in which the spheres kiss at the tangency points of P. Thus the ridge-facet ratio of P is exactly half the average kissing number of X. (Not all sphere packings come from ridge-tangent polytopes in this way.) The sphere packings due to Kuperberg and Schramm [18] can be viewed as coming from a compound consisting of a chain or tree of n cut 600-cells (i.e., atoms in the sense of Figure 6). Their average kissing numbers are 666n + 54 666 K = 2—> - w 12.56603. 106ra + 14 53 By contrast if Q is the compound formed from a corona of A as in Section 3.3, then the average kissing number of the corresponding sphere packing is

= Like the bound on fatness, it is not optimal as a lower bound on the supremal average kissing number. Here we offer no improvement on the upper bound K < 8 + 4\/3 « 14.92820

FAT 4-POLYTOPES AND FATTER 3-SPHERES

255

from [18], even though it cannot be optimal either. 4. 3-SPHERES

In this section we construct a family of strongly regular cellulations of the 3-sphere with unbounded fatness. Indeed, we provide an efficient version of the construction, in the sense that it requires only polynomially many cells to achieve a given fatness. (The construction is also a polynomially effective randomized algorithm.) Given TV, we find a strongly regular cellulation of S3 with O(N12) cells and fatness at least N. Note that there are also powerlaw upper bounds on fatness: An O(N1/3) upper bound for the fatness of a convex 4-polytope with N vertices follows from work by Edelsbrunner and Sharir [9]. The K6vari-S6s-Turan theorem [17] (see also [5, p. 1239] and [23, Thm. 9.6,p. 121]) implies an 0(TV2/3) upper bound on the fatness of strongly regular cellulations of 53, since the vertex-facet (atom-coatom) incidence graph has no K3;3-subgraph, and thus has at most O(/o(/o + /s)2/3) edges. Our construction provides an £l(N1/12) lower bound. The inefficient construction is a simpler version which we describe first. For every g > 0, S3 can be realized as HI U (Sg x /) U #2,

a thickened surface of genus g capped on both ends with handlebodies. (This is obtained from a neighborhood of the standard [unknotted] smooth embedding of Sg into S3.) If for some g we can find a fat cellulation of 55, we can realize S3 as a "fat sausage with lean ends," as shown in Figure 7. We cross the fat cellulation of Sg with an interval divided into TV segments to produce a fat cellulation of Sg x I. Then we fix arbitrary strongly regular cellulations of the handlebodies H\ and H^. If we make the sausage long enough, i.e., if we take N —> oo, the fatness of the sausage converges to the fatness of its middle regardless of the structure of its ends.

S9xl

FIGURE 7. S3 as a fat sausage with lean ends. It remains only to show that there are strongly regular fat cellulations of surfaces. The surface Sg has perfect cellulations with /-vector (l,2g, 1). Such a cellulation is obtained by gluing pairs of sides of a 4#-gon in such a

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EPPSTEIN, KUPERBERG AND ZIEGLER

way that all vertices are identified. It has fatness g, and it exists for arbitrarilyjarge #, but it is far from regular. However, its lift to the universal cover Sg is strongly regular, since any such cellulation can be represented by a tiling of the hyperbolic plane by convex polygons. (Indeed, if we take the regular 4^-gon with angles of 7r/(2g), which is certainly convex, then its edges and angles are compatible with any perfect cellulation.) Moreover, Mal'cev's theorem [19], states that finitely generated matrix groups are residually finite; this implies that every closed hyperbolic manifold admits intermediate finite covers with arbitrarily large injectivity radius. (See [11, §4] for a detailed exposition.) In particular Sg admits an intermediate cover Sg whose injectivity radius exceeds the diameter of a 2-cell. The cellulation of Sg is then strongly regular. Its genus is much larger than g, but its fatness is still 9-

The efficient construction is the same: It only requires careful choices for the finite cover Sg and for the handlebodies HI and H^- Among the perfect cellulations of Sg, a convenient one for us is a 4
X

i ~

X

2g+i

and (2)

(We interpret the indices as elements of Z/(4g).) Equation (1) expresses the identifications, while equation (2) expresses the boundary of the 2-cell.

FIGURE 8. Labelling edges of Sg.

FAT 4-POLYTOPES AND FATTER 3-SPHERES

257

We construct Sg as a tower of two abelian finite covers, which together form an irregular cover of Sg. (No abelian cover of Sg is strongly regular.) The surface Sg satisfies the usual requirements of covering-space theory (see Fulton [12, § 13b, 14a]): It is a connected, locally path-connected, and locally simply connected space. For any finite group A, the A-coverings of Sg are covering spaces of the form Y with Y/A = Sg , where A acts properly discontinuously on Y. These coverings, up to isomorphism, are classified by the set of group homomorphisms Hom(7Ti(S'5,s), A) [12, Thm. 14.a]. Furthermore, if A is abelian, then every such homomorphism maps the commutators in 7Ti(S5, x) to zero, so the .A-coverings are classified by Hom(Hi(Sg, Z ), .A). In other words, if A is any abelian group with n elements, then every homomorphism a : Hi(Sg) —> A defines an n-fold abelian covering of Sg. (If the homomorphism is not surjective, then the covering space is not connected [12, p. 193]. In this case we use a connected component of the covering space, which has the same fatness but smaller genus.) Now assume that q = 4g + 1 is a prime power and let a generate the cyclic group F*, where ¥q is the field with q elements. We can fulfill the assumption by changing g by a bounded factor. (Most simply we can let q = 5k. Or we can let q be prime, so that ¥ q — Z /g, by a form of Bertrand's postulate for primes in congruence classes. This result dates to the 19th century; see Erdos [10] for an elementary proof.) Define a homomorphism a : Hi(Sg] —> Fg by o"([a^i]) = at1, where [xi\ is the 1-cycle (or homology class) represented by the loop xi. Since o?9 = — 1, the definition of a is consistent with equation (1). Consistency with (2) is then automatic. Let S'g be the finite cover corresponding to a. To prepare for the subsequent analysis of Sg, we give an explicit combinatorial description of S'g. Let F° be a 2-cell of S'g and label its vertices

in cyclic order. See Figure 9. For each s 6 F9, let Fs and vf. be the images of F° and v® under the action of s. Since Sg has only one vertex, the vertices of S' may be identified with ¥ q. Thus, if we identify VQ with 0 6 ¥ q, then the action of ¥ q will identify VQ with s 6 F9. The structure of a further implies that c

_

9

If

vf. = s + l + a + cr + ...+ or *

t

= sH

Q-

— -I

a- 1

for all k e Z/(4p) and s <E Fg. See Figure 10. Using this explicit description, it is routine to verify the following (remarkable) properties of the surface S'g.

EPPSTEIN, KUPERBERG AND ZIEGLER

258

+a

FIGURE 9. Labelling the vertices of F°. l =

— s +

S +

1

VQ — S

a3-! a-1

FIGURE 10. Labelling the vertices of Fs. Lemma 12. The cellulation of the abelian cover S'g is regular and has f-vector

Every facet has q — l = 4^g vertices, while every vertex has degree q—l — 4g. The graph (or l-skeleton) of S'g is the complete graph on q + l vertices. The dual graph is also complete; any two facets share exactly one edge (as well as q — 4 other vertices).

259

FAT 4-POLYTOPES AND FATTER 3-SPHERES

Proof. In view of the combinatorial description above (Figure 10), all these facts follow from simple computations in the field Fg: • S' is regular — for each s 6 Fg, the vertex labels s + aa~i (0 < k < 4g) are distinct. • The 1-skeleton of S' is complete — for v,v' 6 ¥ q, v =£ v' there is a unique s 6 ¥q and k e Z/(4#) with

= s-\

ak - 1 — a —1

, v=s-\

a —1

The dual graph of S' is complete — for s,s' £ Fg, s ^ s', there are unique k,t 6 Z/(4g) such that ak , = s' + a —I a —I ak+i _ ! s+ = s' 4a —1 a —1

s+

a Theorem 13. Let n > I28g4, and let Z/n

fee a randomly chosen homomorphism, and let Sg be the finite cover of S'g corresponding to p. Then with probability more than ^, the cellulation of Sg is strongly regular. In order to prove Theorem 13, we need to more explicitly describe the condition of strong regularity as it applies to Sg. Let X be a regular cell complex and suppose that its universal cover X is strongly regular. Recall that the star st(v) of a vertex v in X is the subcomplex generated by the cells that contain v. The complex X is strongly regular if and only if the star of each vertex is. Suppose that v e X projects to v € X. Then the star st(tT), which is strongly regular, projects to the star st(v). The latter is strongly regular if and only if the projection is injective. In other words, X is strongly regular if and only if the stars of X embed in X. If X is not strongly regular, then there must be a path t connecting distinct vertices of st(v) which projects to a loop i in st(v). We say that such a loop obstructs strong regularity. We can assume that t is a pair of segments properly embedded in distinct cells in st(^), with only the end-points of the segments on the boundary of the cells, which implies that i is embedded if X is regular. Figure 11 gives an example of such a loop i in a regular cellulation of a torus.

EPPSTEIN, KUPERBERG AND ZIEGLER

260

3

1 F1

F3

0!

FO\

.-'" t 4

2

F4

1

FIGURE 11. A loop t that obstructs strong regularity in the torus S{. In our case, the surfaces S' are regular, but they have many obstructing loops. Theorem 13 asserts that, with non-zero probability, all such loops lengthen when lifted to Sg. Lemma 14. No loops in S' that obstruct strong regularity are null-homologous. Furthermore, all obstructing loops represent indivisible elements in Hi(S'g}. Proof. In brief, they are indivisible because they are embedded, and they are too short to be null-homologous. By Lemma 12, 5' is regular. By the discussion after the statement of Theorem 13, each obstructing loop I is embedded. If t separates S'g, then it is null-homologous. If t does not separate S'g, then it is indivisible in homology. (To show this, we can appeal to the classification of surfaces by cutting S'g along t. The classification implies that all non-separating positions for I are equivalent up to homeomorphism of S'g. It is easy to find a standard position for i in which it is indivisible in homology.) Thus it remains to show that no obstructing loop is null-homologous. First, we claim that any obstructing loop t can be supported on fewer than 4g edges of the 1-skeleton of S'g. We homotop the two segments of t to the boundaries of the 2-cell containing them, giving them each at most 2g edges. Thus I is represented by a sequence of at most 4g edges in S'g. The case of exactly 4g edges does not occur, since the endpoints of the loop coincide, and no two vertices v ^ v' of S' are opposite vertices in two different facets Fs. As in Lemma 12, this follows from the fact that for v,v' 6 F f f , v / v': there are unique s 6 ¥ q and k 6 Z/(4p) with ak - I , ak+29 - 1 v — s -\ —. a —1 a —I Second, we claim that any null-homologous loop in the 1-skeleton of S'g contains at least 4# edges. In other words, if / is a 2-chain on S'g and df •£ 0, then \df\ > 4g. Since S'g is orientable, we can regard / as a function on its 2-cells. Since / is non-constant, it attains some value t on k 2-cells with

v = sH

FAT 4-POLYTOPES AND FATTER 3-SPHERES

261

0 < k < q. Since any two 2-cells share an edge by Lemma 12, these 2-cells share k(q — k) > 4# edges with the complementary set of 2-cells, of which there are q — k. Since df is non-zero on these edges, \df\ > 4#, as desired. D Proof of Theorem 13. In brief, S'g has fewer than 64g4 obstructing loops t. For each one,

The expected number of obstructing loops that lift from S'g to Sg without lengthening is less than 64p4/n < ^. Thus there is a good chance that all obstructing loops lengthen. The homology group H\(S'g] is a finitely generated free abelian group: It is isomorphic to Z r f , for d = l + q(g — l). Thus it admits nd homomorphisms p to Z /n. Since this is a finite number, choosing one uniformly at random is well-defined. If c is any indivisible vector in Z d, then it is contained in a basis, and thus p(c) is equidistributed. In particular, if t is an obstructing loop, then [£] is indivisible by Lemma 14, so p([i]) is equidistributed in Z /n. It remains only to bound the number of obstructing loops in S'g. A star st(t> ) in Sg has 4#(4g— 2) points other than v itself. Without loss of generality, v projects to 0 in S'g. In this case the other vertices are equidistributed among the 4g non-zero elements of ¥ q. Therefore st(v) has 2

)

4 g (4(?-2)(4<7-3) 2

pairs of arcs connecting v to two vertices that are the same in S'g. These pairs represent all two-segment obstructing loops that pass through 0 and a nonzero vertex v1 (and some of these loops are homotopic). If we count such pairs of arcs for any pair of distinct vertices v, v' of S'g, then we find that the total number is not more than '-2\ (4<7 + l)4<7(4<7-2)(4(7-3) 2 ) 4 as desired. D Question 15. For each g > 1, what is the maximum fatness of a strongly regular cellulation of a surface of genus g ? Equivalently, how many edges are needed for a strongly regular cellulation of a surface of genus g ? Remark. One interesting alternative to the construction of 5' is to assume instead that q = 4g — 1 is a prime power, and to let a be an element of order 4g in F Q 2. The resulting g2-fold cover S" is almost strongly regular: the

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EPPSTEIN, KUPERBERG AND ZIEGLER

only obstructing loops are those that are null homologous in Sg. Another interesting surface is the modular curve X(1p], where p is a prime [25, §13]. The inclusion T(2p) C F(2) of modular groups induces a projection from X(2p) to the modular curve X ( 2 ] , which is a sphere with three cusp points. If we connect two of these points by an arc which avoids the third, it lifts to a cellulation of X(2p) with /-vector (p2 — 1, P'P2~'1\ P-^L). Like Sg, it has a few obstructing loops. Unfortunately we do not know a way to use either Sg or X(2p) to make fat surfaces of lower genus (or equivalently fewer cells) than Sg. Since Theorem 13 provides us with efficient fat surfaces Sg, the construction of fat cellulations of S3 only requires efficient cellulations of the handlebodies H\ and HI and an efficient way to attach them to Sg. In our construction the handlebody cellulations are a priori unrelated to the cellulation of Sg. Rather they are transverse after attachment, and each point of intersection will become a new vertex. Thus the question is to position the cellulations to minimize their intersection. We describe the cellulations in three stages: first, a dissection of H\ and H<2 individually into 3-cells; second, their relative position; and third, their position relative to the cellulation of Sg. Let y be the genus of Sg. A handlebody H of genus g can be formed by identifying g pairs of disks on the surface of a 3-cell. The result is a dissection of H into g 2-cells and one 3-cell, although it is not a cell complex because there are no 1-cells or 0-cells. We can still ask whether such a dissection is regular or strongly regular; this one is neither. However, if we replace each 2-cell by 3 parallel 2-cells, it becomes strongly regular. An example of the resulting dissection A is shown in Figure 12.

FIGURE 12. A strongly regular dissection A of a handlebody of genus 2. The surface 5^ (which for our choice of "g is isomorphic to Sg) has another standard perfect cellulation called a canonical schema in the computer science literature [31]. Using the labelling in Figure 8, we identify x4k

FAT 4-POLYTOPES AND FATTER 3-SPHERES

263

with %4k+2 (the even loops), and x
n=

=

Thus, by tripling the edges of the canonical schema in the Vegter- Yap construction, we can position A\ and A% so that the lean ends in the sausage have O(
and its fatness is Q(g}. This completes the efficient construction with unbounded fatness.

REFERENCES [1] David Barnette, Inequalities for f -vectors of 4-poly topes, Israel J. Math. 11 (1972), 284-291. [2] Margaret M. Bayer, The extended f -vectors of 4-poly topes, J. Combinatorial Theory Ser. A 44 (1987), no. 1, 141-151. [3] Margaret M. Bayer and Louis J. Billera, Generalized Dehn-Sommerville relations for polytopes, spheres and Eulerian partially ordered sets, Inventiones Math. 79 (1985), no. 1, 143-157. [4] Louis J. Billera and Carl W. Lee, A proof of the sufficiency of McMullen's conditions for f -vectors of simplicial convex polytopes, J. Combin. Theory Ser. A 31 (1981), no. 3, 237-255. [5] Bela Bollobas, Extremal graph theory, Handbook of Combinatorics (R. Graham, M. Grotschel, and L. Lovasz, eds.), NorthHolland/Elsevier, Amsterdam, 1995, pp. 1231-1292. [6] Tom Braden, A glued hypersimplex, 1997, Personal communication. [7] John H. Conway and Neil J. A. Sloane, Sphere packings, lattices and groups, 3rd ed., Grundlehren der mathematischen Wissenschaften, vol. 290, Springer- Verlag, New York, 1993.

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[8] Harold Scott MacDonald Coxeter, Regular polytopes, 2nd ed., Macmillan, New York, 1963, Corrected reprint, Dover, New York 1973. [9] Herbert Edelsbrunner and Micha Sharir, A hyperplane incidence problem with applications to counting distances, Applied geometry and discrete mathematics: The Victor Klee Festschrift (P. Gritzman and B. Sturmfels, eds.), DIM ACS Series in Discrete Mathematics and Theoretical Computer Science, vol. 4, Amer. Math. Soc., Providence, RI, 1991, pp. 253-263. [10] Paul Erdos, tfber die Primzahlen gewisser arithmetischer Reihen, Math. Z. 39 (1935), 473-491. [11] Gabor Fejes Toth, Greg Kuperberg, and Wlodzimierz Kuperberg, Highly saturated packings and reduced coverings, Monatsh. Math. 125 (1998), no. 2, 127-145. [12] William Fulton, Algebraic topology. A first course, Graduate Texts in Mathematics, vol. 153, Springer-Verlag, New York, 1995. [13] Branko Griinbaum, Convex polytopes, Interscience, London, 1967. [14] Andrea Hoppner and Giinter M. Ziegler, A census of flag-vectors of ^-polytopes, Polytopes — combinatorics and computation (G. Kalai and G.M. Ziegler, eds.), DMV Seminars, vol. 29, Birkhauser-Verlag, Basel, 2000, pp. 105-110. [15] Birger Iversen, Hyperbolic geometry, London Math. Soc. Student Texts, vol. 25, Cambridge University Press, Cambridge, 1992. [16] Michael Joswig and Giinter M. Ziegler, Neighborly cubical polytopes, Discrete Comput. Geom. 24 (2000), no. 2-3, 325-344, arXiv:math.CO/9812033. [17] Tamas Kovari, Vera T. Sos, and Pal Turan, On a problem of K. Zarankiewicz, Colloq. Math. 3 (1954), 50-57. [18] Greg Kuperberg and Oded Schramm, Average kissing numbers for noncongruent sphere packings, Math. Res. Lett. 1 (1994), no. 3, 339-344, arXiv:math.MG/9405218. [19] A. I. Mal'cev, On the faithful representation of infinite groups by matrices, Amer. Math. Soc. Transl. (2) 45 (1965), 1-18. [20] Joseph Malkevitch, Tiling convex polygons with equilateral triangles and squares, Discrete Geometry and Convexity (J. E. Goodman, E. Lutwak, J. Malkevitch, and R. Pollack, eds.), Ann. New York Acad. Sci., vol. 440, New York Academy of Sciences, 1985, pp. 299-303. [21] Peter McMullen, The numbers of faces of simplicial polytopes, Israel J. Math. 9 (1971), 559-570. [22] James R. Munkres, Elements of algebraic topology, Addison-Wesley, Menlo Park, CA, 1984. [23] Janos Pach and Pankaj K. Agarwal, Combinatorial geometry, J. Wiley and Sons, New York, 1995.

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[24] Egon Schulte, Analogues of Steinitz's theorem about non-inscribable polytopes, Intuitive geometry (Siofok 1985) (Amsterdam), Colloquia Soc. Janos Bolyai, vol. 48, North Holland, 1987, pp. 503-516. [25] Joseph H. Silverman, The arithmetic of elliptic curves, Graduate Texts in Mathematics, vol. 106, Springer-Verlag, New York, 1986. [26] Richard P. Stanley, The number of faces of a simplicial convex polytope, Adv. in Math. 35 (1980), no. 3, 236-238. [27] , A survey of Eulerian posets, Polytopes: abstract, convex and computational (Scarborough, ON, 1993), NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., vol. 440, Kluwer Acad. Publ., Dordrecht, 1994, pp. 301-333. [28] Ernst Steinitz, Uber die Eulerschen Polyederrelationen, Archiv fur Mathematik und Physik 11 (1906), 86-88. [29] John Stillwell, Classical topology and combinatorial group theory, 2nd ed., Graduate Texts in Mathematics, no. 72, Springer-Verlag, 1993. [30] William P. Thurston, Three-dimensional geometry and topology, vol. I, Princeton University Press, Princeton, NJ, 1997. [31] Gert Vegter and Chee K. Yap, Computational complexity of combinatorial surfaces, Proc. 6th ACM Symp. Comp. Geom., ACM Press, 1990, pp. 102-111. [32] Ernest B. Vinberg, Hyperbolic reflection groups, Russian Math. Surveys 40 (1985), 31-75. [33] William Waite, Combining squares and triangles, Cubism for Fun 53 (2000), 18-22. [34] David W. Walkup, The lower bound conjecture for 3- and 4-manifolds, Acta Math. 125 (1970), 75-107. [35] Giinter M. Ziegler, Lectures on polytopes, Graduate Texts in Mathematics, vol. 152, Springer-Verlag, New York, 1995, Revised edition, 1998. E-mail address: eppsteinQics.uci.edu ''David Eppstein'' E-mail address: gregQmath.ucdavis.edu ''Greg Kuperberg'' E-mail address: zieglerOmath.tu-berlin.de ''Giinter M. Ziegler''

ARBITRARILY LARGE NEIGHBORLY FAMILIES OF CONGRUENT SYMMETRIC CONVEX 3-POLYTOPES

JEFF ERICKSON1 Department of Computer Science, University of Illinois, Urbana, IL 61820 SCOTT KIM P.O. Box 2499, El Granada, CA 94018

ABSTRACT. We construct, for any positive integer n, a family of n congruent convex polyhedra in IR3, such that every pair intersects in a common facet. Our polyhedra are Voronoi regions of evenly distributed points on the helix (i,cost,sint). The largest previously published example of such a family contains only eight polytopes. With a simple modification, we can ensure that each polyhedron in the family has a point, a line, and a plane of symmetry. We also generalize our construction to higher dimensions and introduce a new family of cyclic polytopes.

1. INTRODUCTION AND HISTORY A family of
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a convex hull edge; the Schlegel diagram of the polar dual of any neighborly 4-polytope consists of a neighborly family of 3-polytopes. Neighborly polytopes were discovered by Caratheodory [7], who showed that the convex hull of any finite set of points on either the moment curve (£, £2, t 3 , . . . , td) 6 IRd or the trigonometric moment curve (cos £, sin i, cos 2t, sin 2 £ , . . . , cos kt, sin kt) 6 IR is a neighborly polytope. Caratheodory's proof was later simplified by Gale [13], who called these polytope families the cyclic polytopes and the Petrie polytopes, respectively, and showed that the two families are combinatorially equivalent. Cyclic polytopes were independently rediscovered by Motzkin [21, 16] and Saskin [23], among others. For further discussion of neighborly and cyclic polytopes, see Griinbaum [15] and Ziegler [32]. Dewdney and Vranch [10] showed that the Voronoi diagram of the integer points {(£,t 2 ,t 3 ) | t — 1,2, . . . , n } form a neighborly family of unbounded convex polyhedra. Klee [18] derived a similar result for any set of evenly distributed points on the trigonometric moment curve in even dimensions four and higher. Seidel [25] observed that for any d > 3, Descartes' rule of signs2 implies that any finite set of points on the positive branch of the ddimensional (polynomial) moment curve has a neighborly Voronoi diagram. More generally, the vertices of any neighborly polytope have a neighborly Voronoi diagram, since the endpoints of any polytope edge have neighboring Voronoi regions. Zaks [29] described a general procedure to modify any neighborly family of unbounded polyhedra of any dimension, where each polyhedron contains an unbounded circular cone, so that the resulting polytopes are symmetric about a flat of any prescribed dimension. Danzer, Griinbaum, and Klee [9] asked if there is a largest neighborly family of congruent polytopes. Zaks (with Linhart) [29] observed that Klee's Voronoi diagram of evenly distributed points on the trigonometric moment curve forms a neighborly family of congruent convex polyhedra in even dimensions four and higher, but left the three-dimensional case open. The largest previously published neighborly family of congruent 3-polytopes, discovered by Zaks [30], consists of eight triangular prisms. According to Croft, Falconer, and Guy [8, Problem E7], this was also the largest known collection of congruent convex bodies in IR3 with the property that every pair has even one (distinct) point of contact. Both Zaks [30] and Croft, Falconer, and Guy [8] conjectured that the largest neighborly family of congruent 3polytopes is finite (see also Moser and Pach [20, Problem 55]). In Section 2, we show that this conjecture is incorrect, by giving a constructive proof of the following theorem. 2 The number of real roots of a polynomial is no more than the number of sign changes in its degree-ordered sequence of non-zero coefficients.

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Main Theorem. For any positive integer n, there is a neighborly family of n congruent convex 3-poly topes. Like the earlier constructions of Dewdney and Vranch [10] and Zaks and Linhart [30], our construction is based on the Voronoi diagram of a set of points on a curve, namely the regular circular helix h(t) = (t, cos t, shit). An example of our construction is shown in Figure 1, and a single poly tope in our family is shown in Figure 3.

(a

(b)

FIGURE 1. (a) A neighborly family of sixteen congruent convex polytopes. (b) An exploded view of the same family. Our neighborly family (or a linear transformation thereof) was discovered in the late 1980s by the second author, who was inspired by the way playing cards overlap when they are fanned. However, except for a brief announcement by Gardner [14] (which was unnoticed by most of the mathematics community), the construction was never published. The same construction was independently discovered by the first author in 2001, as a result of his research on the complexity of three-dimesnional Voronoi diagrams [12]. In Section 3, we generalize our Main Theorem to higher dimensions, by constructing an arbitrarily large family of congruent convex polytopes in IRd, any ("rf/2~| of which share a unique common boundary face. We also introduce a new family of cyclic polytopes, generalizing both the classic cyclic polytopes and the Petrie polytopes.

2. THE MAIN THEOREM Our construction relies on the following observation, independently discovered by Bochis. and Santos [6, Lemma 4.2] (generalizing their earlier proof of a special case [5, Lemma 2.8]) and the author [12, Lemma 2.1]. We include the proof here for the sake of completeness.

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Lemma 1. Let /3(t) denote the unique sphere passing through h(t) and h(—t) and tangent to the helix at those two points. For any 0 < t < TT, the sphere j3(t) intersects the helix only at its two points of tangency. Proof. Since a 180° rotation about the y-axis maps h(t) to h(—t) and leaves the helix invariant, the bitangent sphere f3(i] must be centered on the yaxis. Thus, (3(t) is described by the equation x2 + (y — a)2 + z2 — r2 for some constants a and r. Let 7 denote the intersection curve of (3(t) and the cylinder y2 + z2 = 1. Every intersection point between /3(t) and the helix must lie on this curve. If we project the helix h and the intersection curve 7 to the xy-plane, we obtain the sinusoid y = cos x and a portion of the parabola y — 7(0;) = (x2 — r2 + a2 + l)/2a. These two curves meet tangentially at the points (t,cost) and (—t,cost). See Figure 2.

FIGURE 2. The intersection curve of the cylinder and a bitangent sphere projects to a parabola on the xy-plane. The mean value theorem implies that the equation 7(0;) = cosx has at most four solutions in the range —TT < x < TT. (Otherwise, the curves y" — — cosx and y" = j"(x) = I/a would intersect more than twice in that range.) Since the curves meet with even multiplicity at two points, those are the only intersection points in the range — TT < x < TT. Since 7(2;) is concave, we have 7(±7r) < COS±TT = —1, so there are no intersections with \x > TT. Thus, the curves meet only at their two points of tangency. D Lemma 1 immediately implies that the Voronoi diagram of any finite set of points on the helix (t, cos t, sint) in the range —IT < t < TT is a neighborly collection of unbounded convex polyhedra. To obtain a neighborly family of congruent polyhedra, we use the Voronoi diagram of evenly spaced points on the helix. For any positive integer n, let hn(t) = h(1iTt/n} and let l-in denote the infinite point set {hn(t} \ t G 2}. By Lemma 1, the Voronoi regions of any n +1 consecutive points in Hn form a neighborly family of convex bodies. Since the point set Hn is preserved by the rigid motion / 2yr 2?r ITT 2?r 2?r\ (x, y. z) f-» \ x -\ , y cos z sm —, y sm h z cos — , V n n n n nJ

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which maps each point hn(t) to its successor hn(t +1), these Voronoi regions are all congruent. The following more refined analysis of the Delaunay triangulation of "Hn, reminiscent of Gale's 'evenness condition' for cyclic polytopes [13, 24], implies that these Voronoi regions have only a finite number of facets, and thus are actually polyhedra. Lemma 2. For any integers a < b < c < d, the points hn(a), hn(b),hn(c), hn(d) are vertices of a simplex in the Delaunay triangulation of 1-Ln if and only if b — a = d — c = l and d — a < n. Proof. Call a tetrahedron with vertices hn(a), hn(b), hn(c), hn(d] local if 6 — a = d — c = I and d — a < n. Let a be the sphere passing through the vertices of an arbitrary local tetrahedron. Analysis similar to the proof of Lemma 1 implies that the only portions of the helix that lie inside a are the segments between hn(a) and hn(b) and between hn(c) and hn(d). Thus, all other points in Hn lie outside cr, so the four points form a Delaunay simplex. The local Delaunay simplices exactly fill the convex hull of Hn, and therefore comprise the entire Delaunay triangulation. Specifically, the only triangles that are facets of exactly one local tetrahedron have vertices hn(i), hn(i + l),hn(i + n) or hn(i — n + l),/i n (z'),/i n (i + 1) for some integer i. Thus, a tetrahedron is Delaunay if and only if it is local. D In light of the duality between Delaunay triangulations and Voronoi diagrams, Lemma 2 lets us exactly describe the combinatorial structure of the Voronoi regions of T-Ln. Let Vn(t) denote the Voronoi region of hn(t). This polyhedron has exactly 2n facets, in n symmetric pairs, as follows: • two unbounded (In — l)-gons shared with Vn(t ± 1), each bounded by 2n — 3 segments and two parallel rays; • two triangles shared with Vn(t ± 2); • 2n — 8 bounded quadrilaterals shared with Vn(t ± 3), Vn(t ± 4 ) , . . . , Vn(t±(n-1}}• two unbounded quadrilaterals shared with Vn(t±(n—1)), each bounded by two line segments and two parallel rays; • two wedges in parallel planes shared with Vn(t ± n), each bounded by a pair of rays. The two (2n — l)-gons are adjacent to all the other facets, including each other, and contain all the vertices of Vn(t); otherwise, the facets are adjacent in sequence. See Figure 3. The vertex of Vn(Q) furthest from the x-axis is the center of the sphere through the points /i ft (0), hn(l], hn(n — 1), and hn(n); this point has coordinates (TT, (2vr - 9)9/(2 - 2 cos 9), 0), where 9 = IK/H. Thus, all the Voronoi vertices of Hn lie in a cylinder of radius (2yr — 0)9/ (2 — 2 cos 0) w n — 1 around

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FIGURE 3. (a) One polytope in the neighborly family of sixteen shown in Figure 1. (b) An orthographic edge-on view of the same polytope. the x-axis. To transform our neighborly family of unbounded polyhedra into a neighborly family of polytopes, we intersect each Voronoi region Vn(t) with the halfspace ycos(2Trt/n) + zsin(2?rt/n) < n, which contains some positive area of every facet of Vn(t). This completes the proof of the Main Theorem. Zaks [31] describes an alternate proof, based entirely on the neighborliness of the Voronoi regions of l-Ln. For each integer 1 < t < n, place a triangle on the shared boundary facet between Vn(Q) and Vn(i). Now place congruent copies of these triangles on the boundary of every Voronoi region, so that the entire collection has the same screw symmetry as "Hn. Finally, for any integer t, let Cn(t) be the convex hull of the triangles on the boundary of Vn(t). The n + 1 congruent convex poly topes C n (0), Cn(l),..-, Cn(ri) form a neighborly family. To actually construct our neighborly family (or Zaks'), it suffices to compute the Voronoi diagram of the finite point set {hn(t} \ t = 0 , 1 , . . . , 3n} and then consider only the Voronoi regions of the middle n + 1 points hn(n), hn(n + 1 ) , . . . , hn(1n — 1), /i n (2n), since those Voronoi regions are the same as in the infinite point set %n. Figure 1 was computed using this method. Since a 180° rotation about the y-axis maps each point hn(t) to hn(—t), and thus preserves the point set Hn, the Voronoi region Vn(0) is rotationally symmetric about the y-axis. It immediately follows every Voronoi region of T-Ln has a line of 180° rotational symmetry. Clipping each Voronoi region by an additional halfspace as above retains this symmetry, since the clipping plane is normal to the symmetry axis. We can create a neighborly family of congruent polytopes with additional symmetries by taking the union of each clipped Voronoi region and its reflection across its clipping plane. Each

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resulting polytope clearly has bilateral symmetry about its clipping plane and 180° symmetry about the original Voronoi region's axis of symmetry, and therefore is centrally symmetric about the intersection point of the clipping plane and the symmetry axis. Theorem 3. For any integer positive integer n, there is a neighborly family of n congruent convex 3-polytopes, each with a plane of bilateral symmetry, a line of 180° rotational symmetry, and a point of central symmetry.

3. HIGHER DIMENSIONS A family of convex polyhedra in Rd is (strictly) k-neighborly if any subset of k polyhedra has a (d — k + l)-dimensional intersection, and no subset of k + l poly topes has a non-empty intersection.3 Arbitrarily large k- neighborly families of polyhedra are easy to construct in H2fc~1, for example, Schlegel diagrams of dual cyclic 2fc-polytopes [7, 13] or Voronoi diagrams of points on the moment curve [25]. However, arbitrarily large k-neighborly families of congruent polyhedra were previously only known in dimensions Ik and higher. The lowest-dimensional example is based on the Voronoi diagram of evenly distributed points on the trigonometric moment curve [18, 29] together with the origin (since otherwise the origin is on the boundary of every Voronoi polyhedron). In this section, we generalize our three dimensional results by considering regularly spaced points on the following generalized helix: h>k(t] = (t, cosi, sini, cos2i, sin2t, . . . , cosfct, sinfct) G JR2k+l. Theorem 4. Let P be any finite set of points on the curve hk(t) in the range 0 < t < 2-7T, for some non-negative integer k. The Voronoi diagram of P is a (k + 1) -neighborly family of convex polyhedra in H2fc+1. Proof. Consider the sphere a passing through k + l arbitrary points /ifc(ao), ^fc(oi), - . . , hk(a>k} € P and tangent to the generalized helix at those points, where 0 < OQ < ai < • • • < a& < 2?r. Any point hk(t) that lies on a satisfies the following (2A; + 3) x (2k + 3) matrix equation:

3

The second condition is necessary to rule out degenerate constructions such as the product of a (d — 2)-dimensional cube with n congruent planar wedges.

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1 ao cos ao sin ao 01— s i n a o c o sa o 1 0,1 cosai sinai 0 1 — sinai cosai

0 1

1 t

sna^ — sina/c cosa^ cos t sin t

cos ka,Q -ksinkdQ cos ka\

-ks'mkai

sin ka,Q k cos kaQ sin ka\ k cos ka\

cos kak -ks'mkttk coskt

sin kak k cos kak smkt

k + OQ 2a0 k + a\ 2ai

=0

£ + «it 2afc A; + t2

(Each of the even rows of this matrix is the derivative of the preceding row. To bound the number of zeros of F(t), consider its second derivative

F"(t) =

OQ 0 1 1 ai 0 1 ak 01— 00—

cosao —sinao cosai —sinai

sinao cosao sinai cosai

cos a/; sinak cos a^ c o s t —s i n t

cos kdQ —ks'mkdQ

—ks'mkai

sin kaQ k cos kdQ sinfcai kcoskai

cos kak —ksinkak —k^coskt

kcoskak —k^s'mkt

k + ag 2ao k + a\ 2ai

F"(t) is an affine combination of the functions cost, sin t, cos2t,sin2i, . . . , cosfc£,sinA;£, so it can be rewritten as as a polynomial of degree at most 2k in the variable elt. Thus, F"(t) has at most 2k zeros in the range 0 < t < 2?r. (This is essentially the argument used by Caratheodory to show that Petrie poly topes are neighborly [7].) Since ao, ai, . . . , a& are roots of F(t) of multiplicity two, they are the only roots in the range 0 < t < 2-Tr; otherwise, by the mean value theorem, F"(k] would have more than 2k roots in the range 0 < ao < t < a^ < 27r, which we have just shown to be impossible. Thus, the points hk(ao),hk(ai), . . . , hk(ak) h'e °n a sphere that excludes every other point in P and so have mutually neighboring Voronoi regions. D In fact, Theorem 4 is a special case of the following result, which follows from an easy generalization of the previous proof and Gale's evenness condition for cyclic polytopes [13, 24]. Define the mixed moment curve /J>d,k(t) as follows: ,k (t) = (t, t2 , . . . , td, cos t, sin t, cos 2t, sin 2*, . . . , cos Art, sin kt) G

For example, ^d,o(t) is the standard o'-dimensional moment curve, /io,jfc(£) is the 2fc-dimensional trigonometric moment curve, and p>\,k(t} is our generalized helix.

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Theorem 5. For any non-negative integers d and k, the convex hull of any finite set of points on the curve p>d,k(t) in the range 0 < t < In is a (d + 2k)dimensional cyclic polytope. Not surprisingly, we obtain arbitrarily large highly-neighborly families of congruent polytopes by considering the Voronoi diagram of the infinite evenly-spaced point set H^ = {hk(27rt/n) \ t 6 7L\. Theorem 6. For any non-negative integers n and k, any n + 1 consecutive Voronoi regions in the Voronoi diagram of 1-L^ form a (k + l)-neighborly family of congruent convex polyhedra. Proof. Fix an integer n. To simplify notation, let h(t) = hk(2-rrt/n), and let (ti, £2, • • • , tr) denote the convex hull of the points ft(£i), ft(<2), • • • > n(tr}Since the set 'H^ = {fi(t) \ t G 2} is preserved under a rigid motion mapping each point h(t) to its successor h(t + 1), the Voronoi regions of H^ are all congruent. Call a full-dimensional simplex with vertices in T-L^ local if all its vertices consist of k + 1 adjacent pairs within a single turn of the generalized helix; in other words, every local simplex has the form (a0, ao + 1, ai, ai + 1, • • • , afc, ak + 1}

for some integers ao, a\ , . . . , a& with a^ + l < ao+n and a^ + 1 < ai+i for all i. Analysis similar to Theorem 4 implies that every local simplex is Delaunay. The convex hull of %*, which we will call the Petrie cylinder, is the product of an 2fc-dimensional Petrie polytope with n vertices and a line orthogonal to that polytope's hyperplane. By Gale's evenness condition [13, 24], the facets of the Petrie polytope are formed by all sets of k adjacent pairs of points on the trigonometric moment curve. The faces of the Petrie cylinder are cylinders over the faces of the Petrie polytope. Call a facet of a local simplex that is not shared by another local simplex a boundary simplex. We easily observe that the boundary simplices are exactly the 2A:-simplices with one of the following two forms: (ak - n + 1, 01, ai + 1, a2, o2 + 1, . . . , ak, ak + 1}

(ai, ai + 1, 02, a2 + 1, . . . , ajt, a^ + 1, 01 + n)

The following infinite sequence of boundary simplices exactly covers one facet of the Petrie cylinder.

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- n + 1,01,01 + 1,02, 02 + 1, • • • ,ak,ak + 1) 1,02,02 + l,...,a/fc,0fc + l,ai + n) 1, 02, a.2 + 1, . . . ,flfc,Ofc + 1, ai + n, ai + n + 1) (a2, 02 + 1, • • • , Ofc, Ofc + 1, ai + n, ai + n02++1, n) Every facet of the Petrie cylinder is covered in this manner, and every boundary simplex lies on some facet of the Petrie cylinder. Thus, the union of the boundary facets is the boundary of the Petrie cylinder, so the local Delaunay simplices completely fill the Petrie cylinder and therefore comprise the entire Delaunay triangulation. It easily follows that each Voronoi region of T-L^ is a convex polyhedron with Q(nk) facets, and that any n + 1 consecutive Voronoi regions form a (A; 4- l)-neighborly family. As we already observed, these polyhedra are congruent. D We can easily modify our construction to obtain a (fc+l)-neighborly family of polytopes, by intersecting each Voronoi region with a halfspace strictly containing all its vertices. Each Voronoi region of H^ has a fc-flat of two-fold symmetry. As long as the boundary of the new halfspace is perpendicular to this central A>flat, the resulting polytope is also symmetric about this flat. Using a variant of Zaks' symmetrization procedure [29], we can ensure that each polytope is also symmetric about a flat of any specified dimension. Consider the Voronoi region V of h(0) in the Voronoi diagram of H^- Let p be the ray from the origin through /i(0), let 0+ and ~ denote the supporting hyperplanes of the only two parallel facets of V (shared with the Voronoi regions of h(n) and h(—n)), and let TT be a hyperplane normal to p at sufficient distance from the origin. Finally, let / be any flat that lies in TT, contains the point p D TT, and is either parallel or perpendicular to both + and (f>~ . The intersection of V and its reflection across / is a convex polytope that is obviously symmetric about / and whose boundary contains positive measure from every boundary facet of V. Applying this procedure to any n + 1 consecutive Voronoi regions of T-L^ we obtain our final result. Theorem 7. For any non-negative integers k, n, and r < 2k, there is a (k + l) -neighborly family ofn+l congruent convex polytopes mIR +1, each of which is symmetric about an r-flat. Acknowledgments. Thanks to Joseph Zaks for his insightful comments on an early draft of this paper, and to Victor Klee for sending a copy of his paper [18]. Figures 1 and 3 were produced with the help of the programs 'qhull' [2, 3] and 'geomview' [1, 19].

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[18] V. Klee. On the complexity of d-dimensional Voronoi diagrams. Archiv der Math., 34 (1980), 75-80. [19] S. Levy, T. Munzner, M. Phillips, C. Fowler, N. Thurston, D. Krech, S. Wisdom, D. Meyer, T. Rowley, and S. M. Robbins. Geomview, version 1.8.1, March 2001. (http://www.geomview.org). [20] W. Moser and J. Pach. Research problems in discrete geometry: Packing and covering. Tech. Rep. 93-32, DIMACS, 1993. [21] T. S. Motzkin. Comonotone curves and polyhedra. Bull. Amer. Math. Soc., 63 (1957), 35. [22] R. Rado. A sequence of polyhedra having intersections of specified dimensions. J. London Math. Soc., 22 (1947), 287-289. [23] K). A. IIIauiKHH [Yu. A. Saskin]. SaMenaHne o coce^rnux Bepiirawax na BLinyKJiOM MHororpanaHHHKe [A remark on adjacent vertices on a convex polyhedron]. Vcnexu MameM. HayK [Uspehi Mat. Nauk], 18 (1963), 209-211. [24] G. C. Shepherd. A theorem on cyclic polytopes. Israel J. Math., 6 (1968), 368-372. [25] R. Seidel. Exact upper bounds for the number of faces in d-dimensional Voronoi diagrams. Applied Geometry and Discrete Mathematics: The Victor Klee Festschrift, pp. 517-530. DIMACS Series Discrete Mathematics and Theoretical Computer Science, vol. 4, AMS Press, 1991. [26] P. Stackel. Z. Math. Phys., 42 (1897), 275. [27] H. Tietze. Uber das Problem der Nachbargite im Raum. Monatsh. Math. Phys., 16 (1905), 211-216. [28] H. Tietze. Famous Problems of Mathematics: Solved and Unsolved Mathematical Problems from Antiquity to Modern Times. Graylock Press, New York, 1965. Translation of Geloste und ungeloste mathematische Probleme aus alter und neuer Zeit, 2nd edition, Verlag C. H. Beck, Miinchen, 1959. [29] J. Zaks. Arbitrarily large neighborly families of symmetric convex polytopes. Geom. Dedicata, 20 (1986), 175-179. [30] J. Zaks. Neighborly families of congruent convex polytopes. Amer. Math. Monthly, 94 (1987), 151-155. [31] J. Zaks. Personal communication, April 2001. [32] G. M. Ziegler. Lectures on Polytopes. Graduate Texts in Mathematics, vol. 152. Springer-Verlag, Heidelberg, 1994. E-mail address: [email protected] E-mail address: [email protected]

ON THE NON-SOLIDITY OF SOME PACKINGS AND COVERINGS WITH CIRCLES

AUGUST FLORIAN Department of Mathematics, University of Salzburg, Hellbrunnerstrasse 34, Salzburg, Austria, A-5020 ALADAR HEPPES1

Vercse u 24/A, Budapest, Hungary, H-1124

ABSTRACT. An old conjecture of L. Fejes Toth [4] states that the packing (covering) consisting of the incircles (circumcircles) of the faces of an Archimedean tiling is solid if the number of regular polygons meeting at a vertex is 3, and non-solid if this number is greater than 3. The object of the present paper is to survey the state of the second part of this conjecture for the Euclidean and spherical tilings, and to prove it for all cases not settled before.

1. INTRODUCTION A set of open (closed) circles is said to form a packing (covering) on the sphere, or in the Euclidean or hyperbolic plane if every point of the sphere or plane belongs to at most (at least) one circle of the set. Following a concept introduced by L. Fejes Toth [4] we say that a packing (covering) is solid if no finite number of its members can be rearranged so as to form, together with the rest of the members, a packing (covering) not congruent to the original one. A solid packing of equal circles in the Euclidean plane is always a densest packing of equal circles, and similarly for coverings. A solid packing of n equal circles on the sphere is always a densest packing of n equal circles, and similarly for coverings. Note, however, that the converse is not true. Supported in paxt by OTKA T030012 and OTKA T037752 279

280

FLORIAN AND HEPPES

An edge-to-edge tiling consisting of regular polygons is said to be uniform if, for any two vertices, the symmetry group of the tiling contains an operation that maps one vertex onto the other. If all faces of a uniform tiling are congruent, the tiling is regular, otherwise it is called Archimedean. In the present paper, we shall refer to the incircles (circumcircles) of a tiling when we mean the incircles (circumcircles) of the faces of this tiling. L. Fejes Toth observed that the incircles of any trihedral regular tiling {p, 3}, with p > 2, form a solid packing, and the circumcircles of (p, 3}, with p > 2, form a solid covering (see also [14]). Furthermore, he stated the following

Conjectures: (i) The incircles of any trihedral Archimedean tiling form a solid packing. (ii) The incircles of any more-than-trihedral uniform tiling form a non-solid packing. (ni) The circumcircles of any trihedral Archimedean tiling form a solid covering. (iv) The circumcircles of any more-than-trihedral uniform tiling form a non-solid covering. These conjectures triggered intensive research. As the References show, they have been confirmed in many cases. Conjecture (i) has been proved for all trihedral Archimedean tilings in the Euclidean plane and on the sphere. In the case of covering (iii), however, progress is somewhat slow. For example, none of the coverings associated with the Euclidean tilings (3,12,12), (4,8,8) and (4,6,12) is known to be solid. In the present paper we shall be concerned with the non-solidity aspect, i.e., conjectures (ii) and (iv), for all Euclidean and spherical uniform tilings. We shall prove these conjectures in the cases that have remained open until now. The non-solidity of the spherical packings and coverings associated with the Archimedean tilings (3,3,3,3,4), (3,3,3,3,5) (see Figures 1 and 2) and (3,3,3,n), with n > 4 (see Figure 3 for n = 9) seems not to have been proved so far. For the sake of completeness, we will briefly discuss the other, simpler cases, too.

2. THE TILINGS (3,3,3,3,4) AND (3,3,3,3,5) Theorem 1. The incircles of the Archimedean tilings (3,3,3,3,4) and (3,3,3,3,5) form non-solid packings. Theorem 2. The circumcircles of the Archimedean tilings (3,3,3, 3,4) and (3, 3,3, 3,5) form non-solid coverings.

281

NON-SOLIDITY OF SOME CIRCLE PACKINGS AND COVERINGS

Fig. 1

Fig- 3

Fig. 2

These two tilings have similar structures. Both consist of two kinds of regular polygons, one of which is a triangle. The large faces (quadrangles and pentagons, respectively) are concentric with the faces of the spherical net (called the basic net) of a regular polyhedron, that is the cube for (3,3,3,3,4), and the dodecahedron for (3,3,3,3,5). The set of small faces (triangles) can be divided into two subsets. The members of the first subset have their centres at the vertices of the basic net. These will be called vertex triangles. The second subset consists of side-to-side pairs of triangles, each of them sharing a side with a large face. The midpoint of the common side of each pair coincides with the midpoint of an edge of the basic net. These triangles will be referred to as edge triangles. Proof of Theorem 1. The table below shows some parameters of the two associated packings. Let R and r denote the (spherical) radius of the large and the small circles (i.e., the incircles of the large and small faces), respectively. Let d be the (spherical) distance of the midpoints of two adjacent edges of the basic net. Observe that the midpoints of any two edges of the basic net have a distance > d.

R= r = d = 7T/3 =

(3,3,3,3,4)

(3,3,3,3,5)

0.41248 ... 0.23357 ... 1.04719 ...

0.33436 .. 0.13809 ... 0.62831 ..

7T/5

=

We shall show that the incircles of the edge triangles can be rearranged in a "non-congruent" way, which implies that the packing is not solid. Consider the incircles C\ and C
282

FLORIAN AND HEPPES

large circles and the set of incircles of the vertex triangles are symmetric with respect to the plane p, the reflected images of C\ and C<2, say C[ and C^, will not overlap any such circle. Next, we replace C\ and C<2 by C[ and C^, leaving all other circles of the original packing unchanged. To prove that the resulting system is a packing, we have to show that C( and C^ have no point in common with the incircle of any edge triangle belonging to a different edge. Consider the midpoints M' and M" of two edges of the basic net, and two circles with radius r containing M' and M", respectively. Since 4r < d, these two circles are indeed disjoint. It is easy to show that the new packing is not congruent to the original one. The original packing is invariant under rotation about the centre of any large circle of this packing, through the angle 27T/4 and 2?r/5, respectively. This is not true, however, for the circle C^ of the new packing, since the circles C[ and C'2 are different both from C\ and C^. This completes the proof of Theorem 1. D Proof of Theorem 2. In the tilings under consideration, any pair of adjacent edge triangles is enclosed by two large faces and two vertex triangles that have their centres at the endpoints of an edge of the basic net. The edge triangles are partly covered by the circumcircles of the four polygons. The remaining 'hole,' i.e., the uncovered region, is symmetrical about the central plane p containing the edge in question. This means that such a hole is covered not only by the pair of circumcircles of the two edge triangles, but also by their reflection in p. If one such pair is reflected and all other circles are left in their original position, a new covering is generated. This covering is not congruent to the original one, since the reflection destroys the rotatory symmetry of the original covering. D

3. THE TILINGS (3,3,3,n), FOR n > 4 Among Fejes Toth's conjectures (ii) and (iv), the last non-hyperbolic, as yet unverified cases concern the packings and coverings generated by the tilings (3,3,3,n), with n > 4. Our next goal is to prove that these are non-solid sets. A tiling of this sort consists of two large n-gonal faces centred at the "poles" and 2n triangular faces arranged along the "equator". Each triangle is sharing a side with one of the large faces and a single vertex with the other one (see Figure 3). Theorem 3. The packing on the sphere consisting of the incircles of ihr Archimedean tiling (3, 3,3, n) is non-solid for n > 4.

NON-SOLIDITY OF SOME CIRCLE PACKINGS AND COVERINGS

283

Proof of Theorem 3. On the unit sphere with centre O we choose an oriented great circle e, the "equator" , and a point E 6 e. Any point of the sphere will be determined by its "geographic" coordinates (A,/3). The longitude A is measured from the point E on the equator, and the latitude f3 measures the signed distance of the point from e (—-nfl < /3 < Tr/2). By distance we always understand spherical distance. Let R and r denote the radius of the incircles of the n-gons and that of the triangles of the tiling, respectively. We may assume that the large circles have their centres at the poles N (/? = 7T/2) and S (/5 = — Tr/2). The centres of the small circles are located at the vertices of two regular n-gons (of side > 2r) inscribed in two circles GI and C2 of radius R + r having their centres at the poles. In this packing every small circle touches exactly two others and a large one. If the point E is chosen properly, the set of the original centres is symmetrical about the plane A = vr/2 (37T/2), and none of the centres lying on ci is contained in the plane A = 0 (TT). To generate an alternative packing of the same circles, we move the centre of the "upper" large circle from AT to a point N'(Q, Tr/2 — e) and denote the circle with centre N' and radius R + r by c{ . The value of e > 0 is chosen so small that c'x is contained in the hemisphere (3 > 0, i.e., (1)

£<^-(R + r}.

We denote the centres of the small circles lying on c\ by AI (i — 1, . . . , n) such that (2)

0 < \(Ai) <

< X(Am}
We assign to each centre Ai a point A[ e c( with (3)

\(A'i) = \(Ai)

(i = l , . . . , n ) .

If e is sufficiently small, then for each A[ there are two points B[ and D\ on C2, both at distance 1r from A^, where X(B'i) < X(D'i). The region bounded by the arc B^D^ of 02 and the two geodesic arcs B^A'^ A^D^ will be called the quasi-triangle B^A^D^. We shall refer to the arc B^D^ as the base, and to the distance h^ of A't from the midpoint Mi of the base as the height of the quasi-triangle B'^D^. Furthermore, we shall denote the half-length of the base of a quasi-triangle with height h by s(h). It is clear that some of the quasi-triangles in the 'narrow' part of the ring enclosed by c( and 02 overlap. We shall use the following lemma to show that these overlaps can be removed: Lemma. Let r e (0, Tr/4) and (3o e (0, r) be given constants, and let I dc note the interval (00,2r). For every h 6 / there exists (up to congruence) exactly one quasi-triangle BAD with the base BD lying on a circle of radius

284

FLORIAN AND HEPPES

7T/2 — fa, AB = AD = 2r, and a height of length h. The half-length s of the base is a strictly decreasing and strictly concave function of h 6 /. Proof of the Lemma. Let B and D be two points on the circle c with centre S (ft = — 7T/2) and radius ?r/2 — fa. We denote the length of the geodesic arc joining B and D by 2a, where 0 < a < Tr/2. Let A be a point with 0 < /3(A) = f3 < 2r - fa such that AB = AD. We shall show, with an appropriate choice of a, that the quasi-triangle BAD satisfies the requirements of the lemma. Note that the height h of BAD is fa + 0, whence fa < h < 2r. We shall use the following notation: M' for the midpoint of the geodesic arc BD, M" for the midpoint of the straight line segment BD, F for the intersection of the geodesic arc AM' with equator e. The point M" has the normal distance sin fa from the segment OF. Since, furthermore, OM" = cos a and the three points O,M",M' are collinear, we find FM' = arcsin(sin/3o/cosa). Hence AM' = AF + FM' = /3 + arcsin

!^°. cos a

To satisfy the condition AB — 2r, we consider the right (spherical) triangle AM'B. From cosAB = cosBM' • cos AM' we obtain cos2r = cosacos/3\/l

^— — sin (3 sin fa cosz a

or

(4)

cos2r = cos /3\lcos2 a — sin2 fa — sin j3 sin fa.

For a = 7T/2 — fa: the right-hand side of (4) is — sin/? sin fa < cos2r, while for a = 0, the right-hand side of (4) is cos(/5 + fa) > cos2r. Thus (4) has a unique solution a — a((3) £ (0,7r/2 — fa). If we set y cos2 a - sin2 fa = /(/?), then (4) says that (5)

f((3] cos (3 — sin 0 sin fa = const.

Repeated differentiation of (5) yields (6)

-/sin/3 + /'cos/?-cos/3sin^ 0 = 0

and

(7)

-/cos/3-2/'sin/5 + /"cos/3 + sin^sin/5o =0.

NON-SOLIDITY OF SOME CIRCLE PACKINGS AND COVERINGS

285

^From (6) it follows that (8)

/'-sinA) = /tan/3,

which shows that /'(£) > 0

(9)

for 0 < /? < 2r - {fa. Combining (7) with (8) we obtain -/ cos p - f sin 0 - f sin j3 tan /3 + f" cos /3 = 0.

In view of (9), this implies that (10)

f"(fi) > 0.

Because the straight line segment BD has the length 2 sin a and the circle c has the (Euclidean) radius cos/^o, we find for the half-length of the base BD of BAD s

\(0] = cosp\)arcsin (

— I = cos/?o arccos

cos/V

or si COS po

=

f COS 0Q '

Hence

(12)

--

-- cos -V - «i' sin -= /"• cospo cospo cospo Combining (11) and (12) with (9) and (10), we see that

(13)

ai(0)<0,

s'M<0

for 0 < (3 < 2r - /3Q. Since h = (3 + /3Q and s(h) = si(/3), the proof of the lemma is complete. D Proceeding with the proof of Theorem 3, we consider the quasi-triangles B^A'iD't for i = 1,... ,m and recall that 0 < X(A'i) < TT (see (2) and (3)). We first give a lower bound to the sum of the heights h\ + • • • + h'm. If the subscripts i and j ^ i refer to a pair A^Aj symmetric about the half-plane A = 7T/2, then A^N + A'jN is equal to the length of the chord of c( through A'i and N. Since A(A^) is different from 0 and TT, this chord does not contain N'. Consequently, it has a length less than the diameter of c'1: so that

Thus we have

(14)

h'i + h'j = 27r - 1(R + r)- (A'iN + A'jN) > lit - 4(R + r) = 2h0.

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FLORIAN AND HEPPES

Here

h0 = TT - 2(R + r)

(15)

stands for the constant height of the quasi-triangles generated by the centres of the small circles of the original packing. We denote these quasi-triangles by BiAiDi (i = 1,... ,n) and remark that Dx = Bi+\, and Mi is the common midpoint of the arcs BiDi and B\D\. If A(^) = ?r/2, then A^N < R + r and (16)

tii

= Tr-(R + r)-A'iN>hQ.

Combining (14) and (16), we obtain (17)

til

+ ---+h'm> mhQ.

Next, we will show that the total length of the arcs B^D'^ for i — 1, , . . , m, is less than the total length of the arcs BiD{. To do so, we make use of the lemma by setting /?0 = Tr/2 - (R + r). Observe that r < 0.45769 . . . < ?r/4 for n > 4, and /3$ < r, or, equivalently,

This follows easily from the triangle inequality and the fact that the point of contact of two small circles lies on the equator. Since all points A\ have from N a distance between R + r — e and Tr/2, we have

for i — 1, . . . , m. If we take e < 2(R + 2r — ?r/2), then 7

(18)

^-(R + r)
z*

Thus all assumptions of the lemma are satisfied. By Jensen's inequality in conjunction with (17) we obtain /i r\\ (19)

n

\ / i f \ ^ n " I 2> s(/ij) < 2ms / —

'

'

m

r. fi \ — \ < 2ms(/io),

where 2ras (/IQ) is the total length of the bases of the m quasi-triangles BiA{Di. These bases fit together without gaps or overlaps to form the subarc BiDm of c2. Thus (19) states that m

(20)

In the following, we shall write b\ for the arc B[D'^ s^ for s(/i^) (that is, the half-length of 6'J, and SQ for a(/io)- From (21)

h'l
NON-SOLIDITY OF SOME CIRCLE PACKINGS AND COVERINGS

287

it follows that (22)

s(>S'2>--->s'm

and, by (19),

s'm < 80.

(23)

Let us now move the quasi-triangles B^A^D'^ for i = ra, ra — 1,..., along 02 towards Dm into new positions B"A"D" in such a way that D'^ = Dm and D" — B"+1. Obviously, the new quasi-triangles do not overlap. Setting B"D" — b'l and making use of (20) we see that the union of & " , • • • , b"m is a connected arc B'[Dm whose length is less than B\Dm. Thus (24)

X(B'{) > A(Bi).

If M" denotes the midpoint of 6", then, by (23), (25)

A(AC) > A(M m ).

Furthermore, we will show that (26)

A(Mf) > A(Mi),

for alH = 1, • • • , m — 1 as well. Suppose that (26) is true for i = k + 1, while (27)

A(Afi') < A(Af fc ).

^,From s'k+1 + s'k = M%M%+1 > MAMjf+1 if follows that (28)

^n^+1^0,

and, in conjunction with MkM'f!+l > M^M^+i = 2so, (29)

4+1+4>2s0.

Because s is a decreasing function of /i, inequality (29) implies (30)

s'i+l + s'i > 2s0

for alH < fc, whence

(31)

s( > s0.

^,From (28), (30) and (31) we conclude that the arc BiDm is covered by the union of 6' l 5 ... , 6'fc, &' fc ' +1 ,... , 6^, in contradiction to (20). Thus supposition (27) is wrong and, in view of (25), inequality (26) is proved for 1 < i < m. Note that, as a consequence of (25) and (26), all points -A'/,... ,A'^ lie between c[ and 02On the hemisphere TT < A < 2?r, a similar set of points can be generated by an analogous process. The points A", with m + 1 < i < n, are also contained in the ring between c[ and C2, and the points B" and D" lie on 02, where B'^+l = Bm+i = Dm. Note that, analogously to (24), A(£>") < X(Dn). After dropping one of the endpoints, B" or J9^, we end up with n points on

288

FLORIAN AND HEPPES

the upper hemisphere and a total of n points on the lower hemisphere such that no upper point is closer than 2r to a lower one. Observe that in the above construction the position of the points obtained depends continuously on e. If £ is sufficiently small, the new points are arbitrarily close to the original ones. Therefore, the distance between any pair of upper points is larger than 2r, and similarly for any pair of lower points. To summarize, we can state the following: The 2n circles of radius r with their centres at the points A", B", and the two circles of radius R centred at N' and S, form a packing not congruent to the original one. D Theorem 4. The covering of the sphere consisting of the circumcircles of the Archimedean tiling (3,3,3,n) is non-solid for n > 4. Proof of Theorem 4- Let a be the interior angle of the triangles. We first establish the relation between a and n. If / denotes the edge-length of the tiling, then (32)

I

TT

a

cos-= cos-sin-.

Since three triangles and a single n-gon meet at each vertex of the tiling, we have ,OON

v(33)

'

n

I .

3a

n

2

2

cos — = cos - s i n — .

Combining (32) and (33) we find the required relation sin(3a/2) ^ TT ——.—7-7- = 2 cos — sm(a/2) n or, equivalently. (34)

2 c o s a + l = 2 cos-. n

Let r and R be the circumradius of the triangles and the n-gons, respectively. Then trigonometric formulae yield , _. (35)

a TT 1 /I + cos a cos r = cot — cot — = 2 3 x/3 V 1 - cos a

and .

/

.

7T .

_

sin - = sin — sin R.

2 n Making use of (32) and (34), we finally obtain (36)

,/9

cos a)(3 + 2 cos a)

NON-SOLIDITY OF SOME CIRCLE PACKINGS AND COVERINGS

289

Let KI and K^ be the circles of radius R centred at the poles N and 5. Consider 2n circles of radius r whose centres are equidistantly arranged on the equator. Theorem 4 will be proved by showing that these 2n circles cover the ring between K\ and K^. Let C be one of the circles with centre A. Let 27 be the central angle of the arc on the boundary of K\ covered by C. Obviously, it is sufficient to show that

Let P be a point on the sphere with AP = r and NP = R. Since AN = Tr/2, we have cos r = sin R cos 7. Thus, by (35) and (36), inequality (37) is equivalent to

,orA (38)

/(I + cos a)(3 + 2 cos a) TT ^ >± i
Referring to (34), (38) can be written in the form

I (I + cos a) (3+ 2 cos a) V 6 But this is clearly true, as a > Tr/3. (<39}

<

\/3 + 2 cos a 2 '

D

4. SOME FURTHER UNIFORM TILINGS On the sphere there are six more-than-trihedral uniform tilings not discussed above. Two of them are the regular octahedral {3,4} and the regular icosahedral tiling {3,5}. The Archimedean tilings are (3,4,3,4), (3,5,3,5) (having faces concentric with the faces and vertices of {3,4} and {3, 5}, respectively), (3,4,4,4), and (3,4,5,4) (having faces concentric with the faces, vertices and midpoints of the edges of {3,4} and {3,5}, respectively). We shall show that the incircles and the circumcircles of these tilings form nonsolid sets. In each of the above tilings there is a proper subset of the edges forming the boundary of a regular n-gon II contained in a hemisphere, where n = 4,5,6,10,8 and 10, respectively. Let F\,... , FK be the faces of the tiling composing II. Furthermore, let c\,... , c^ be their incircles, and C\,... ,Ck their circumcircles. The incircle of II contains G I , . . . ,c&, and the circumcircle of II is covered by C i , . . . , C^. If ci,... , c^ (Ci,... , Ck) are rotated about the centre of II through an angle of an (a irrational) and all other circles are left in their original position, we obtain a packing (covering) not congruent to the original one. In the Euclidean plane there are seven more-than-trihedral uniform tilings. Two of them are the regular ones {3,6} and {4,4}. The systems of their incircles (circumcircles) are non-solid sets, because their densities are less than

290

FLORIAN AND HEPPES

(greater than) that of the incircles (circumcircles) of {6,3}. The Archimedean tilings are (3,6,3,6), (3,4,6,4), (3,3,3,4,4), (3,3,4,3,4), and (3,3,3,3,6). Their incircles form non-solid packings, and their circumcircles form nonsolid coverings. For the proof we refer to [6]. This completes the proofs of Fejes Toth's conjectures (ii) and (iv) (see Section 1) for the Euclidean plane and the sphere.

REFERENCES [I] Barany, I. and Dolbilin, N. P., A Stability Property of the Densest Circle Packing, Monatsh. Math., 106 (1988), 107-114. [2] Bezdek, A., Solid Packing of Circles in the Hyperbolic Plane, Studio, Sci. Math. Hungar., 14 (1979), 203-207. [3] Fejes Toth, G., Solid Sets of Circles. Studia Sci. Math. Hungar., 9 (1974), 101-109. [4] Fejes Toth, L., Solid Circle-Packings and Circle-Coverings, Studia Sci. Math. Hungar., 3 (1968), 401-409. [5] Fejes Toth, L., Solid Packing of Circles in the Hyperbolic Plane, Studia Sci. Math. Hungar., 15 (1980), 299-302. [6] Florian, A., Archimedean Tilings, and Solid and Non-Solid Sets of Circles, Rend. Sem. Mat. Messina, Ser. II, 6 (1999), 5-12. [7] Florian, A., Some Recent Results on Packing and Covering with Incongruent Circles, Suppl. Rend. Circ. Mat. Palermo, Ser. II, No. 65 (2000), 93-104. [8] Florian, A., An Infinite Set of Solid Packings on the Sphere, Sitzungsber. Osterr. Akad. Wiss. Math.-naturw. Kl., Abt. II, 209 (2000), 67-79. [9] Florian, A., Packing of Incongruent Circles on the Sphere, Monatsh. Math., 133 (2001), 111-129. [10] Florian, A. and Heppes, A., Solid Coverings of the Euclidean Plane with Incongruent Circles, Discrete Comput. Geom., 23 (2000), 225-245. [II] Heppes, A., Solid Circle-Packings in the Euclidean Plane, Discrete Comput. Geom., 7 (1992), 29-43. [12] Heppes, A., On the Solidity of the Hexagonal Tiling, Colloquia Math. Soc. J. Bolyai, 63, Intuitive Geometry 1991 (1994), 151-154. [13] Heppes, A. and Kertesz, G., Packing Circles of Two Different Sizes on the Sphere, Intuitive Geometry, Bolyai Soc. Mathem. Studies, 6 (1997), 357-365. [14] Imre, M., Kreislagerungen auf Flachen konstanter Kriimmung, Acta Math. Acad. Sci. Hungar., 15 (1964), 115-121. E-mail address: E-mail address:

august [email protected]. at ''August Florian'' hep9202Qella.hu ''Aladar Heppes''

ON THE mTH PETTY NUMBERS OF NORMED SPACES

KAROLY BEZDEK 1 Department of Geometry Eotvos University, Budapest, Hungary, H-1117 MARTON NASZODI 2 Department of Geometry Eotvos University, Budapest, Hungary, H-1117 BALAZS VISY3 Department of Geometry Eotvos University, Budapest, Hungary, H-1117

ABSTRACT. In this paper we study some old and new problems on the Petty numbers of finite dimensional normed spaces. Here the mth Petty number (m > 1) of the given finite dimensional normed space is the largest cardinality of a pointset having the property that among any m points always there are two lying at distance one. The main challenge is to find good upper bounds for the Petty numbers in terms of m and the dimension of the normed space.

1. INTRODUCTION A Minkowski space M = (R d , || ||) is just Rrf with distances measured using a norm || ||. A norm || || is completely determined by its unit ball

B = {xe Hd\ ||x|| < 1}, which is a compact convex set with nonempty interior (i.e. a convex body), centrally symmetric about the origin o in Rd. An elegant result of Petty [15] claims that the cardinality of any equilateral set in M.d is at most 2d i.e. the J

Part of this research was done while the author was visiting the Department of Mathematics at Cornell University in Ithaca, NY. Research was also supported by the Hung. Nat. Sci. Found. (OTKA), grant no. T029786. 2 Partially supported by the Hung. Nat. Sci. Found. (OTKA), grant no. T029786. 3 Partially supported by the Hung. Nat. Sci. Found. (OTKA), grant no. T029786. 291

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cardinality of any family of pairwise different translates of the unit ball B in Md such that any two translates are tangent is at most 2d . Motivated by this result and the results in [3] and [17] we introduce the following definitions. Definition 1. For the integer m > 1 and the d— dimensional Minkowski space Md let P(m,M r f ) denote the largest integer k such that there are k unit balls in Md (i.e. k pairwise different translates of the unit ball B in M d ) with the property that among any m of them there are two touching ones. We call P(m,M r f ) the mth Petty number of arrangements of unit balls in Md. Definition 2. For the integer m > 1 and the d— dimensional Minkowski space Md let P p a c fc(m,M d ) denote the largest integer k such that there are k non-overlapping unit balls in Md (i.e. k non-overlapping translates of the unit ball B in Md) with the property that among any m of them there are two touching ones. We call Ppack(m,~M.d) the mth Petty number of packings of unit balls in Md. Using our notation the above cited result of Petty can be phrased as follows: (A nice alternative proof of this inequality can be found in [10].) In 1983, P. Erdos raised the following question: What is the largest number g ( n ) with the property that every system of n non-overlapping unit disks in the Euclidean plane has an independent subsystem with at least g(n} members i.e. there are g ( n ) members no two of which are tangent. Recently, Csizmadia [8] showed that \^n\ < g ( n ) moreover, Pach and Toth [14] proved the inequality g ( n ) < f^w] for large n. These results naturally lead us to the following definitions. Definition 3. For the integer n > 1 and d— dimensional Minkowski space Md let Ipack(n:*M.d) denote the largest integer / such that every packing of n unit balls in Md (i.e. every arrangement of n non-overlapping translates of the unit ball B in Md) has an independent subsystem with at least / members i.e. there are / members no two of which are tangent. / p a c fc(n,M d ) is called the independence number of packings of n unit balls in M . Definition 4. For the integer n > 1 and d— dimensional Minkowski space M d let /Y/?, M r f ) denote the largest integer / such that every arrangement of n unit balls in Md (i.e. every arrangement of n pairwise different translates of the unit ball B in M r f ) has an independent subsystem with at least I

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members i.e. there are / members no two of which are tangent. I(n,~M.d) is called the independence number of arrangements of n unit balls in Md. In this paper we prove the following statements. The proofs are presented in separate sections. Obviously, the Petty numbers and the independence numbers are in inverse relation and satisfy some often used elementary inequalities that are summarized in the following claim. Proposition.

(1) I(n,Md) < I(n + l,M d ) and Ipack(n,Md) < Ipack(n + l,M d ) hold for all positive integers n > 1 and any d—dimensional Minkowski space M rf ; (2) P(m,M d ) + 2 < P(m + l,M d ) and Ppack(m,Md) + 2 < Ppack(m + 1, Md) hold for all positive integers m > 1 and any d—dimensional Minkowski space Md; (3) J(n,M d ) < Ipack(n^'M.d) holds for all positive integers n > 1 and any d—dimensional Minkowski space M rf ; (4) Ppack(m,Md) < P(m,Md} < (^"t^"1) hold for all positive integers m > 1 and any d—dimensional Minkowski space M rf ; (5) 7//(n,M r f ) > m, then P(m,Md) < n - 1; (6) If Ipack(n,Md) > m, then Ppack(m,Md) < n - 1; (7) //P(m,M d ) = 7 i - l , ^en /(n,M d ) = m; (8) If Ppack(m,Md) = n-l, then Ipack(n,Md} = m. Notice that (4) includes the Petty inequality Ppack(2,Md) = P(2,M r f ) < 2 . It seems a rather challenging question to find a better upper bound on P(m,M d ) than (2 t^"1) for m > 2. Our next claim does this for several values of m and d. d

Theorem 1. Let M d ,d > 2 be an arbitrary d—dimensional Minkowski space. Then (1) P(3,M d ) < 2-3d moreover, P(m,M d ) < ( m - l ) [ ( m - l ) 3 r f - ( m - 2 ) ] holds for all positive integers m > 3. Furthermore, (2) P(m,M d ) < (m - l)4 d holds for all positive integers m > 1 and (3) P(m,M 2 ) < 8(m — 1) holds for all positive integers m > 1. Corollary 1. 7 / n > 2 d + l , d > 2 are positive integers and M d is an arbitrary d—dimensional Minkowski space and I(n — l,M d ) < /( then ^j- < I(n,Md] and -^ < I(n,Md). 32 4

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It is easy to check that the method of the proof of Theorem 1 yields also the inequalities P padc (3,M rf ) < 3 r f ,P p a c / c (m,M d ) < (m-l)3 rf for all m > 3,d > 2 moreover, if n > 2d + l,d > 2 and Ipack(n - l , M d ) < Ipack(n,Md), then 3~dn < /pocitl^M^) holds for any d—dimensional Minkowski space M d . This we can improve as follows. Theorem 2. If n > l,d > 1 are positive integers and M d—dimensional Minkowski space, then

?s an arbitrary

Corollary 2. If m > l,d > 1 are positive integers and Md is an arbitrary d—dimensional Minkowski space, then Pr,nrk(m,Md}

<

Til • 2 • S^" 1 .

The question of determining P pa cfc( m 7 M r f ) and /'(mjM^) is a definitely hard looking problem for most norms of Hd including also the Euclidean norm. However, H d with the maximum norm can be treated in a rather short way as it is shown in the next statement.

Theorem 3. Ifm>I,d>l with the maximum norm, then Ppack(m,MdJ

are positive integers and M^ denotes Hd = P(ro,M?J = (m - l)2 d .

Moreover, we can prove the following partial results for some special Minkowski spaces. Theorem 4. (1) P(3,M 2 ) < 8. ( 2 ) P ( 3 , M ? ? ) = 7. (3) P(m,M^ ) < (m — l)3 rf holds for all positive integers m > 1. (4) P(m,Mf ) < (m - l)3 d /io/ds /or all m > 1 provided that dl/p < 2.

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Remark 1. In connection with Theorem 4 we mention the still wide open conjecture of Kusner [11] according to which P(2, M^ ) = d + 1 for all 1 < p < oo and d > 2. Finally, we call the attention to the norm /A whose unit ball is the difference body of a d—dimensional simplex in RA It is obvious that P(2, Mf ) > d + 1 for all d > 1 and that P(2, Mf A ) = 3. However, for d = 3 we have the following claim. Theorem 5. //M 3 cuboctahedra), then

denotes R3 with the /A norm (the balls of which are P(2,M?J = 5.

Remark 2. The highly interesting paper [12] shows that P(2,Mf ) > d + 2 holds for all d > 4. However, the exact value of P(2, Mf ) is not known to us for any d > 4. Finally, we cannot resist on raising the following question. Problem. Prove or disprove that if m > 2, d > 1 are positive integers and Mrf denotes an arbitrary d— dimensional Minkowski space, then P(m,M d ) <(m-

l)2 rf .

Remark 3. Recall that Petty [15] conjectures that each (/—dimensional Minkowski space contains d + 1 points at pairwise distance one. This challenging conjecture of Petty is still open for all d > 4. (For more details and partial results see the recent papers [4] and [9].) If this conjecture is true, then the inequality follows for any Minkowski space Mrf for all m > 1.

2. PROOF OF THE PROPOSITION The claims (1),(3),(5) and (6) follow from the definitions in a straightforward way. Proof of (2): If P(m,Md] = fc, then let Bi,...,Bk be a family of k pairwise different translates of the unit ball B in M such that among any

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m of them there are two touching ones. Then let P>A;+I, -^fc+2 be two touching unit balls in Md that are different from the unit balls P»i, . . . , B^. It is easy to see that the family 5i, . . . , 5^,5/ c+1 ,-B / t + 2 of k + 2 unit balls in Mrf possesses the property that among any m + 1 members always there are two touching ones. Thus, k + 2 < P(m + l,M r f ) finishing the proof of the first inequality of (2). The second inequality of (2) namely, P poc fc(m,M d ) + 2 < -Ppacfc(m + l,M d ) can be proved in the same way. Proof of (4): The inequality Ppack(m,Md) < P(ra,M d ) is trivial. Then recall the following fact from [1], p. 130. For any r, s > 2 let R(r,s] denote the minimum value of n such that every graph with n vertices contains either a complete subgraph of r vertices or an empty subgraph of 5 vertices. (A graph is called empty if it has no edges.) Then R(r,s) < ( r ^f.^ 2 ). Now, let P(m,M d ) = k and let J 9 i , . . . , _ 0 f c be a family of k pairwise different translates of the unit ball B in Md such that among any m of them there are two touching ones. Then let G be the graph whose vertices are the centers of B\ , . . . , Bk and whose two vertices are connected by an edge if and only if the corresponding two unit balls are tangent in Md. Thus, Petty's result [15] applied to the graph G easily implies that ^

finishing the proof of (4). Proof of (7): If P(ra, M rf ) = ra — 1, then every arrangement of n unit balls in Md has m members no two of which are tangent. Thus, m < J(n,M d ). Assume that m < /(n,M d ) say, m -f- m' = J(n,M d ) with some m' > 1. Recall that P(m,M d ) = n— 1 and so there are n— 1 (pairwise different) unit balls say, B\, . . . , Bn-\ in Md such that among any m of them there are two touching ones. Now, take an additional unit ball Bn different from the unit balls 5i, . . . , J9 n _i in Md. By assumption the arrangement B\, . . . , -B n _i, Bn of n unit balls must have m -f 1 independent members in M d and so the arrangement BI , . . . , Bn-\ possesses m members no two of which are tangent in Md, a contradiction. Thus, /(ri,M d ) = m indeed. Proof of (8}: This can be proved in the same way as (7).

3. PROOF OF (1) IN THEOREM 1 We prove our claim by induction on m > 3. Case of m = 3: Let P(3,M d ) = k and let Bi,...,Bk be the family of k pairwise different translates of the unit ball B in Md with the centers c i , . . . , Cfc such that among any three translates there are two touching ones.

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Then either there are two centers say, ct- and GJ with ||c; — Cj|| > 4 or the diameter diam|| ||{ci, . . . , cjj of the set {GI, . . . , c^} is at most 4 in Md i.e. diamy ||{ci, . . . , c^} — max{||cz- — Cj|| | 1 < i < j < k} < 4. First, assume that there are two centers say, c; and Cj with ||c2- — Cj|| > 4. By assumption any unit ball 5S, 1 < s < k different from the unit balls B{ and Bj must be tangent either to B{ or to Bj. Moreover, if BSl and 5S2, 1 < s\ < 52 < k are two unit balls tangent to the unit ball B{ (resp., B j } , then BSl must be tangent to BS2 because among the unit balls BSl,BS2,Bj (resp., BSl,BS2,Bi) there must be two touching ones. Hence, the unit ball Bi (resp., Bj} together with the unit balls _BS, 1 < s < k that are tangent to it form a family of unit balls in which any two members are tangent. Now, recall Petty's result [15] according to which the cardinality of any family of pairwise different translates of the unit ball B in Mrf such that any two translates are tangent is at most 1d. Thus, k < 2 • 2d < 2 • 3d finishing the proof of Theorem 1 . Second, assume that diani|| ||{ci, . . . ,c/t} < 4. Then it is easy to see that diani|| \\(Bi U • • • U Bk) = diam|| y (conv(#i U • • • U Bk)) < 6. As a result the isodiametric inequality (see for example [6], p. 93) implies that the d— dimensional volume Vol^(conv(.Bi U ••• U Bk)) of the convex hull conv(Bi U • • • U Bk) of the unit balls f?i, . . . , Bk is at most as large as the d— dimensional volume of a ball of diameter 6 in Mrf i.e. it is at most 3d • Vold(B). Finally, as among any three unit balls of the family BI , . . . , Bk there must be two touching ones we get in a straightforward way that any point of the convex hull conv(5i U • • • U Bk) belongs to the interior of at most two unit balls from the family BI, . . . , Bk- Thus, k • Vold(B) < 2 • Volrf(conv(51 U • • • U Bk)) < 2 • 3d • Vold(B). This completes the proof of Theorem 1 in the case m — 3. Case of m > 3; Let P(m, Md) = k and let Bi,...,Bk be the family of k pairwise different translates of the unit ball B in Md with the centers ci , . . . , cj. such that among any m translates there are two touching ones. If among any m — 1 members of the family of the unit balls B\, . . . , Bk always there are two touching ones, then by induction our claim follows. Thus, we may assume that there are m — 1 members of the family B\,...,Bk say, Bi,...,Bm-i such that no two of them are tangent. Now, for each i,l < i < m — 1 let F; be the subfamily of the unit balls of the family 5i, . . . , Bk that are tangent to Bi and add to F; also the unit ball Bi. As among any m members of the family B\,...,Bk there are two touching ones we get that

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Finally, for each z, 1 < i < m — I let Al - {xe Md I 1 < H x - C i U < 3). Obviously, for any Bj £ F,- different from Bi we have that Bj C A{. Moreover, as among any m members of the subfamily F? \ {Bi} there are two touching ones every point of A; belongs to the interior of at most m— I unit balls of F,\{Bi}. Thus, card(F t - \ {£,-}) Vol rf (fl) < (m - 1) • Vol d (A t -) = (m - I) • (3rf - l)Vold(5). As a result we get that m —l

k
finishing the proof of (1) in Theorem 1.

4. PROOF OF COROLLARY 1 According to the subclaim (5) of the Proposition if I(n,M r f ) = m, then P(m, Md) < n ~ 1. Assume that P(m, Md) - ri - I < n - 1. Then subclaim (7) of the Proposition implies that J(n',M d ) = m with n' < n. Finally, subclaim (1) of the Proposition and the assumption I(n— l,M d ) < /(n,M d ) imply that m - I(n',Md} < I(n - l,M d ) < /(n,M d ) = m, a contradiction. Hence, P(m,M r f ) = n - 1. Thus, subclaim (1) of Theorem 1 and Petty's result [15] imply that if m > 2, then n - 1 = P(m,M r f ) < (m - I)2 • 3d and so

n< 3d-m2 = 3d-I2(n,Md) d

provided that J(n,M ) = m > 2. Finally, applying again Petty's result [15] we get that I(n,Md) > 2 for all n > 2d + 1. In a very similar way, subclaim (2) of Theorem 1 implies the inequality ^ < I(n,M.d). This completes the proof of Corollary 1.

5. PROOF OF THEOREM 2 The following statement has been proved in [2]. Lemma 1. Let M d ,d > 2 be an arbitrary d—dimensional Minkowski space with the unit ball B centered at the origin o. // h(B) denotes the

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one-sided Hadwiger number of B i.e. h(B} denotes the maximum number of non-overlapping translates of B that are all (outer) tangent to B lying in a closed supporting half space to B, then

h(B} < 2 • 3d'1 - 1.

Based on Lemma 1 we prove Theorem 2 as follows. Let jBj, . . . , Bn be a packing of n,n > 1 unit balls with centers C i , . . . , c n in M^,c? > 1. Take a vertex say, c,- of the convex hull of the centers ci, . . . ,cn in Md. Then it is obvious that the number of the unit balls in the family J?i,...,.Z? n that are (outer) tangent to Bi is at most h(B). As the Lemma claims that h(B) < 2 • 3d-1 - 1 it follows by induction (on n) that

This completes the proof of Theorem 2.

6. PROOF OF COROLLARY 2 If Ppackfa, Md) = n — 1, then subclaim (8) of the Proposition implies that Ipack(n,Md) = m. Thus, Theorem 2 yields that 2.3d-i < m or equivalently, Ppack(™"> Md) + 1 = n < m • 2 • 3d"1 finishing the proof of Corollary 2.

7. PROOF OF THEOREM 3 Let P(m, M^ ) = k and let J?i, . . . , B& be an arrangement of k pairwise different translates of the unit ball B with centers GI , . . . , c^ in Md such that among any m translates there are two touching ones. Hence, among any m points of {GI, . . . , c^} always there are two lying at distance 2 in Md . Now, if c'- = |cz-, 1 < i < k, then among any m points of {c'1: . . . , c^.} always there are two lying at distance 1 in Mf . Let C = [0, 1) X • • • X [0, 1) be the "partially open" unit d— cube of Rd with 2d vertices of 0 — 1 coordinates. If Zd denotes the integral points of Rd, then the family {z + C \ z 6 Zd} is a tiling of Rd into pairwise disjoint "partially open" unit cubes. Thus, for each c'- there is a uniquely determined Z{ 6 Zd with c( G z; + C, where 1 < i < k. As the norm in Mf is the maximum norm it is easy to see that

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the multiset of the points Z i , . . . , z & (which is a pointset with points having multiplicity) has the property that among any m not necessarily different points always there are two lying at distance 1 in M^ . Finally, for each z; let u z be the uniquely determined 0—1 vector of R d whose coordinates are congruent to the corresponding coordinates of zt- mod 2, where 1 < i < k. It is obvious that the multiset of the points u i , . . . , u/- has the property that among any m not necessarily different points always there are two lying at distance 1 in M^ . As the points m , . . . , u^ are vertices of the unit d— cube C it is easy to see that the inequality k < (m — 1)2^ must hold. From this it follows that PpadbKM/J < P(m,Mfj

< (m-l)2d.

As the reversed inequality (m - l)2 d < P pacfc (m,MO < P(m,M?J is trivial, this completes the proof of Theorem 3.

8. PROOF OF (1) AND (2) IN THEOREM 4 Proof of (I) in Theorem 4. Theorem 3 implies that P(3,Mf oo ) = 8. Thus, we may assume that the norm is different from the maximum norm. We prove that the touching pairs graph G, constructed in the proof of subclaim (4) of the Proposition, has maximum degree at most 4. Taking any vertex say, c let G(c) denote the subgraph spanned by the vertices connected to c by an edge. Since among any 3 vertices there are always 2 connected by an edge, G(c) contains no empty triangle, and since the norm is not the maximum norm, G'(c) contains no complete subgraph on 3 vertices. Thus, we get from [1], p. 130, that |G(c)| < 5. It is easy to see that if |G(c)| = 5, then G(c) is a cycle on 5 vertices. This is however a contradiction, since if the norm is different from the maximum norm, then no unit disk can be touched by five unit disks in M2 such that any two consecutive unit disks in the cyclic ordering of the five unit disks are tangent. (This elementary fact one can prove in several different ways but, one can get an elegant and clean-cut approach using the angular measure introduced by Brass in [5].) Finally, it is obvious, that the number of vertices not connected to c by an edge is at most 3 thus, indeed G has at most 8 vertices. Proof of (2) in Theorem 4. Assume that P(3,Mf ) = 8. Then according to the subclaim in the proof of (1) of Theorem 4 every vertex of G must have degree 4. Taking an arbitrary vertex say, c let ci, 02, 03, 04 denote the vertices connected to c by an edge of G and let c5, ce, €7 denote the rest

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of the vertices of G. It is easy to see, that among ci, 02, 03, 04 there are always 2, or 3 edges. Having 2 edges between GI , . . . , 04 implies that the degree of these vertices in the subgraph G1 of G spanned by c, GI, ... ,04 is 2, so each vertex in the family {ci,... ,04} has 2 edges going to the vertex set {05, eg, CT}. Hence 8 edges connect G" to the vertices 05, eg, cy, a contradiction. If 3 edges occur between GI , . . . , 04, then c is in the convex hull of GI , . . . , 04 contradicting to the arbitrary choice of c.

9. PROOF OF THEOREM 5 We consider R3 with the cuboctahedron obtained from the unit cube [— |, |]3 by proper truncation (by taking the convex hull of the midpoints of the edges) as the unit ball for the norm || ||/ A . Then we take an arbitrary l&equilateral set. First, we prove that the angles of any triangle with vertices in the /^-equilateral set are non-obtuse. Let a, b and c denote the Euclidean lengths of the sides of a triangle of the /^-equilateral set. Since c2 = (a - 6)2 + 2a6(l - cos 7) > 2a6(l - 0037), 2

we have that cos 7 > 1 — ^ > 0. The later inequality is due to the fact that v/2 > a,6,c > 1. Hence 7 < 90°. If every triangle spanned by the points of the /^-equilateral set is acute, then Croft's result [7] yields, that we can have at most 5 points in the l&equilateral set, finishing the proof of Theorem 5. So, suppose that there is a right triangle spanned by three points of the /^-equilateral set in question. Then a — b = 1, c — \/2 for some proper labelling of the sidelengths moreover, each leg must be parallel to a coordinate axis of R3. Thus, we can assume that the vertices of the right triangle in question are a(0, 1,0), b(l,0,0) and c(0,0,0). By taking a closer look of the cuboctahedron we get that {p € R3 : ||p - a||,A - ||p - b||/A = ||p - c||/A - d(l, 1, 0) U El U

where E\ = {(x, y,z,) € R 3 : 0 < # = y < | , 2

= l} and

E2 = { ( x , y , z , ) € R3 : 0 < x = y < \,z = -1}. Since no 2 points of E\ (resp., E^) are at distance 1 from each other in the norm j| ||/A and the /^-distance between the sets EI and E% is at least 2, we get that at most d and one additional point from either EI or E<2 can belong to the //^-equilateral set. Thus, again we get at most five points for our /^-equilateral set. This finishes the proof of Theorem 5.

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BEZDEK, NASZODI AND VISY 10. PROOF OF (2) AND (3) IN THEOREM 1 AND (3) AND (4) IN THEOREM 4

We say that a distance h > 0 is realized within a given set of a metric space if there are two points of the given set whose distance is equal to h. We will need the following claim which is equivalent to Theorem 1 in [13]. Lemma 2. If a measurable set in a d-dimensional Minkowski space has upper density larger than m7l with the property that some distance is not realized in that set, then there is no configuration of k points having the property that there are two points lying at distance one among any m of the k points. Thus, in order to establish the inequality P(m,M d ) < (m — l)/c it is sufficient to find a set of upper density at least £ not realizing all distances. Proof o/(2) in Theorem 1. Take a saturated packing of unit balls in M and take the Voronoi cells V\, V?,... of the balls. As the Voronoi cell of each unit ball is the collection of points that are as near to the center of that ball as to the center of any other, it follows from the saturated property of the packing that each Voronoi cell is covered by the open ball of radius 2 concentric to the unit ball of the Voronoi cell. Let V? denote the contraction of Vi by factor ^ from the center of the unit ball contained in V;, i'. — 1, 2 , . . . . Finally, let V = |J£i V{. As Hd = (J^ Vl the upper density of V is at least ^j. Finally, as each cell V[ is contained in an open ball of radius \ in M rf , it easy to see that the distance 1 is not realized in V finishing the proof of (2) in Theorem 1. Proof of (3) in Theorem 4- Again we take a saturated packing of unit balls this time, in M^ . Also, just as before let Vi, V^ • • • denote the Voronoi cells of the unit balls in the given saturated packing. However, now we contract each Voronoi cell Vi by factor ^ from the center of the unit ball contained in Vi. Let V[ be the contracted cell obtained from Vt, i = 1 , 2 , . . . . As a result we get that each V{ is contained in an open ball of radius ^ in M^ . Consequently, the distance | is not realized in any V?. Moreover, if Vi and Vj are two Voronoi cells, then they can be separated by the perpendicular bisector hyperplane of the linesegment connecting the centers of the unit balls lying in the cells Vi and Vj. As a result it is easy to see that the distance between the contracted cells V-' and V-J is at least f.o Thus, in the * set V = Ui^i intV/ the distance | is not realized and the upper density of V is obviously at least ^. This completes the proof of (3) in Theorem 4 via Lemma 2. Proof of (3) in Theorem, 1. Using Lemma 2 it is sufficient to construct a set of upper density at least ^ not realizing all distances. Auerbach's lemma

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(for a proof see [16]) states that a proper choice of a basis of the Minkowski space ensures that the unit ball is inscribed in the unit ball of the l^ norm and is circumscribed the unit ball of the /i norm. That is Vx = (xi,x 2 ) €M2 = (R 2 ,|| ||) :max{|x 1 |,|x 2 |} < ||x|| < |zi|-Hz 2 |. For any 1 < p < oo denote by Bp(x,r) the open ball of center x and radius r according to the lp norm. Let A

'-= l'l=

oo

oo

U

U

1

*i((M2)>4)-

— 00 12 = — 00

Then the (upper) density of A is | while the distance ^ is not realized in A. Proof of (4) in Theorem 4- Now we want to find a set of upper density (\>}d not realizing a given distance, say 2. Let A:=

U

U •" U

This set is the desired one. Namely, the diameter of one open cell of A is i less than dp < 2, and the distance between any two cells of A is larger than 2.

REFERENCES [1] Agarwal, P.K. and Pach, J., Combinatorial Geometry, John Wiley and Sons. Inc., 1995. [2] Bezdek, K. and Brass, P., On k+— neighbour packings and one-sided Hadwiger configurations, Beitrdge Algebra Geom., (submitted), 1-4. [3] Bezdek, K. and Langi, Zs., Almost equidistant points on Sd~l, Periodica Math. Hung., 39 (1-3) (1999), 139-144. [4] Brass, P., On equilateral simplices in normed spaces, Beitrdge Algebra Geom., 40 (1999), 303-307. [5] Brass, P., Erdos distance problems in normed spaces, Computational Geometry - Theory and Applications, 6 (1996), 195-214. [6] Burago, Y.D. and Zalgaller, V.A., Geometric inequalities, SpringerVerlag, New York, 1988. [7] Croft, H.T., On 6— point configurations in 3— space, J. London Math. Soc., 36 (1961), 289-306. [8] Csizmadia, G., On the independence number of minimum distance graphs, Discrete Comput. Geom., 20 (1998), 179-187. [9] Dekster, B.V., Simplexes with prescribed edge lengths in Minkowski and Banach spaces, Ada Math. Hung., 86 (2000), 343-358. [10] Fiiredi, Z., Lagarias, J.C. and Morgan, F., Singularities of minimal surfaces and networks and related extremal problems in Minkowski space, DIM ACS Series in Discrete Mathematics and Theoretical Computer Science, 6 (1991), 95-109.

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[11] Guy, R.K. and Kusner, R.B., An olla podrida of open problems, often oddly posed, Amer. Math. Monthly, 90 (1983), 196-199. [12] Koolen, J., Laurent, M. and Schrijver, A., Equilateral dimension of the rectilinear space, Designs, Codes and Cryptography, 21 (2000), 149-164. [13] Larman, D.G. and Rogers, C.A., The realization of distances within sets in Euclidean space, Mathematika, 19 (1972), 1-24. [14] Pach, J. and Toth, G., On the independence number of coin graphs, Geombinatorics, 6 (1) (1996), 30-33. [15] Petty, C.M., Equilateral sets in Minkowski spaces, Proc. Amer. Math. Soc., 29 (1971), 369-374. [16] Pietsch, A., Operator Ideals, V. E. B. Deutscher Verlag Wiss., Berlin, 1978 and North-Holland, Amsterdam, 1980. [17] Rosenfeld, M., Almost orthogonal lines in Ed, DIM ACS Series in Discrete Mathematics and Theoretical Computer Science, 4 (1991), 489-493. E-mail address: E-mail address: E-mail address:

[email protected] ' ' K a r o l y B e z d e k ' ' [email protected] ''Marton Naszodi ' ' [email protected] ''Balazs V i s y ' 1

CUBIC POLYHEDRA

CHAIM GOODMAN-STRAUSS1 Dept. Mathematics, Univ. Arkansas, Fayetteville AR 72701, USA JOHN M. SULLIVAN2 Dept. Mathematics, Univ. Illinois, Urbana IL 61801, USA

ABSTRACT. A cubic polyhedron is a polyhedral surface whose edges are exactly all the edges of the cubic lattice. Every such polyhedron is a discrete minimal surface, and it appears that many (but not all) of them can be relaxed to smooth minimal surfaces (under an appropriate smoothing flow, keeping their symmetries). Here we give a complete classification of the cubic polyhedra. Among these are five new infinite uniform polyhedra and an uncountable collection of new infinite semi-regular polyhedra. We also consider the somewhat larger class of all discrete minimal surfaces in the cubic lattice.

1. INTRODUCTION We define a cubic polyhedron P to be any polyhedron whose vertices and edges are exactly the vertices and edges of the cubic lattice in E3, and which forms an embedded topological surface (or 2-manifold) in E3. (We pick fixed orthonormal coordinates on E3; and we view the cubic lattice as a cellcomplex with vertices at Z3.) It follows that cubic polyhedra are connected, non-compact and unbounded, and of course have faces that are among the square faces of the cubic lattice. Note that we allow adjacent coplanar squares, and still consider them to be distinct faces of the polyhedron. Research partially supported by NSF grant DMS-00-72573 and by the Consejo INacional de Ciencia y Tecnologia (CONACYT) of Mexico. 2 Research partially supported by NSF grant DMS-00-71520. 305

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A Hamiltonian path in a graph (1-complex) is a connected subcomplex which forms a 1-manifold and includes every vertex (0-cell) of the graph. By analogy, we could describe cubic polyhedra as the "Hamiltonian surfaces" in the 2-skeleton of the cubic lattice: they are the subcomplexes which form connected 2-manifolds including every edge of the lattice. (Banchoff [1] used such Hamiltonian surfaces in the 2-skeleton of the n-cube as examples of tight polyhedral surfaces; for n = 6 his example is a quotient of our cubic polyhedron PQ.) Note that we do not allow the configuration consisting of two touching cube corners (which would correspond to a pair of opposite triangular loops in the octahedron) because such a vertex would be a nonmanifold point. Given a cubic polyhedron P, every edge in the cubic lattice is incident to two faces of P, and we call it a crease or a flange depending on whether these faces are perpendicular or coplanar. Similarly, every vertex has valence six, and we find that (up to isometry) there are only two possibilities for the configuration of incident squares, as shown in figure 1. To see this, consider the vertex figure of the cubic lattice, a regular octahedron. The two vertex configurations correspond to the two possible Hamiltonian cycles in the 1skeleton of this octahedron. In the first possible configuration, a monkey-saddle (M) vertex, all six incident edges are creases, alternating up and down, making the vertex indeed a (polyhedral) monkey-saddle. The normal vector at an M vertex (meaning the average of the normals to the six incident faces) points along one of the four body-diagonals; the monkey-saddle configuration can occur in four possible orientations.

Figure 1: Up to rigid motion, there are three possible vertex configurations in a cubic polyhedron, a monkey-saddle M, a left-handed screw S, and a right-handed screw Z. The two flanges of a screw vertex lie along its axis line.

The second configuration, a screw vertex, has two incident flanges, in opposite directions along a common axis. This configuration comes in leftand right-handed versions, which we call S and Z vertices, respectively. The normal vector at a screw vertex points along one of the two face-diagonal

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lines perpendicular to the axis. There are six orientations for an S vertex or for a Z vertex, corresponding to the choice of axis and normal lines. Our problem of classifying cubic polyhedra comes down to figuring out all the ways that these vertex configurations can be fit together to fill up all of space. An uncountable number of polyhedra can be built, with surprising variety; but there is also certain rigidity which aids our classification, culminating in our Main Theorem 4.5. 1.1. Basic Constructions. Before giving some examples, we prove two useful lemmas which let us extend a finite configuration to a complete polyhedron. Lemma 1.1. Suppose we are given compatible configurations at the vertices within a rectangular box (which can be finite, infinite or bi-infinite in each of the coordinate directions). This can be extended to a complete cubic polyhedron, by repeated reflection in the sides of the box (which are planes at half-integer coordinate values). Proof. The edges meeting the boundary of the box do so perpendicularly (as do their incident faces) no matter whether they are creases or flanges. Thus the half of each such edge or face within the box reflects to the half outside; the fact that each vertex within the box has a legal configuration means the same is true at all reflected vertices. D Lemma 1.2. Given the configuration of faces along an edge e, the configuration at either endpoint v of that edge is determined uniquely by its type M, S or Z. In particular, if the vertex configurations at the ends of e are M and M; or S and Z; then these configurations are mirror images of each other. Proof. If e is a flange, then its vertices must be screws with axis along e, which (together with the flange normal) fixes their orientation. If e is a crease, then there are four possible orientations of that crease. If v is type M, it has four possible orientations; if it is type S (or Z) it has four possible orientations with axis perpendicular to e. In any case, the possibilities correspond bijectively to the four for the crease along e. D These simple lemmas immediately give us the two symmetric cubic polyhedra shown in figure 2: Proposition 1.3. Up to isometry, there is a unique cubic polyhedron PQ with all monkey-saddle vertices, and there is a unique cubic polyhedron PI with all screw vertices whose types S and Z alternate in checkerboard fashion. Proof. Existence follows from Lemma 1.1, starting with a single M or S vertex VQ in a 1 x 1 x 1 box. Rotating the configuration at VQ will just rotate

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Figure 2: The cubic polyhedra PQ (left) and PI (right) are generated by repeated reflection of a single monkey-saddle or screw vertex, respectively. The thin black lines in the picture of PI are its parallel axes.

the entire polyhedron. Uniqueness then follows from Lemma 1.2, working outward from VQ. D Of course, PQ is the infinite regular polyhedron {4,6} as described by Coxeter [3], with symmetry group transitive on flags. We can view PI as a new uniform polyhedron: its faces are all regular polygons and its symmetry group acts transitively on the vertices, but the faces or edges around a vertex fall into more than one transitivity class. Later we will see four more uniform examples among our cubic polyhedra. 1.2. Curvature and Topology of Cubic Polyhedra. Any cubic polyhedron P has, by definition, six squares meeting at every vertex. Thus it has an equal quantum —TT of total Gauss curvature at each vertex. The surface is an Alexandrov space with curvature bounded above by zero. This negative curvature spread uniformly throughout space suggests that P should have nontrivial topology everywhere, whether or not P is triply periodic. To examine this, consider an arbitrary loop in the edges of P (which are the edges of the lattice). It can be written (homologically) as a sum of square (four-edge) loops. Given a square loop 7, either 7 is spanned by a square in P and thus is trivial in 7Ti(P), or 7 is a closed geodesic in P and thus (because of the nonpositive curvature) is nontrivial. In fact, the square loop 7 is nontrivial in iri(P) only if it is nontrivial in Hi(P): Suppose there is a compact spanning surface K for 7 within P. If K is just the square convex hull of 7, then 7 is nontrivial. Otherwise K is not contained in the convex hull of 7, so it must have an extreme point away from 7. But neither kind of vertex in a cubic polyhedron can be an extreme point. Incident to any vertex v of the cubic lattice are twelve squares, four in each of the three directions. In a cubic polyhedron P, independent of whether v is a monkey-saddle or screw vertex, exactly two of the four squares in any

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direction are present in P. The missing squares exhibit nontrivial loops in Hi(P), which are thus equidistributed in space. (Note that these loops are not all independent.) We note also that every cubic polyhedron P is orientable. If not, there would necessarily be some orientation-reversing square loop. We merely need to check the nine possible 2x2 planar diagrams (denned below in Section 1.4) with a missing central square, to see that this is impossible. If a cubic polyhedron P has orientation-preserving translational symmetry with respect to some index k sublattice of the cubic lattice, then it projects to a compact orientable surface P in the quotient torus (which has volume k). This surface P has k vertices (with total Gauss curvature —fcTr),3k edges and 3k/2 faces, so it must have Euler number — k/1 and genus fc/4+1. (Note, however, that there are 3k/2 missing squares within this torus. They form loops that generate HI (P) but are clearly not independent since HI has rank only k/1 + 2.) For example, PQ and PI both have translational symmetry with respect to the even integer lattice 2Z3, with k = 8, so they have quotients of genus three. We can induce a smooth constant-curvature (hyperbolic) metric on a cubic polyhedron, by giving each square face the metric of a square in the hyperbolic plane sized to have internal angle Tr/3. 1.3. Minimality of cubic polyhedra. Our interest in cubic polyhedra arose from the fact that the first two examples mentioned above were reminiscent of certain classical triply-periodic minimal surfaces of Schwarz (see [5]). Indeed, we expect that PQ will relax to the P surface and PI to the CLP surface. (See figure 3).

Figure 3: The polyhedra PQ and PI relax to triply periodic minimal surfaces of Schwarz. the P surface (left) and CLP surface (right). In both esses, the reflection symmetries (across the dark lines) are preserved under relaxation.

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We have performed numerical simulations to confirm this, using Brakke's Evolver [2]. We subdivide the faces and let the geometry relax under a flow which decreases the Willmore bending energy [4]. (We use this rather than mean-curvature flow, since the minimal surfaces we flow towards are unstable.) Further experiments, starting from other cubic polyhedra, indicate that many, but not all, will relax to minimal surfaces in a similar way. (A typical example is shown in figure 4; we have used pictures like this on posters and in the first author's Ptolemy mathcard series.) We plan to report on these experiments in a future paper.

Figure 4: A typical example of a cubic polyhedron (left) and the minimal surface (right) to which it relaxes under the Willmore flow.

Our cubic polyhedra themselves are discrete minimal surfaces, in the sense of Pinkall and Polthier [6]. Just as a smooth surface is minimal if it is critical for surface area, a triangulated surface is defined to be discrete minimal if the first variation of its area is zero, under motion of any interior vertex. For a more general polyhedral surface, we introduce diagonals to triangulate it, and note that the condition of discrete minimality is independent of the choice of these diagonals. Given a surface P made of squares, it is not hard to check that P is discrete minimal at a vertex v exactly when v is the center of mass of the incident faces or edges. This happens whenever v has valence six, as in our cubic polyhedra, but also for exactly one other configuration at v: four coplanar squares. At the end of this paper, we briefly consider the calssification of this more general family of all discrete minimal surfaces within the cubic lattice.

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1.4. Examples with both types of vertices. We can combine monkeysaddle and screw vertices in an astounding variety of ways. To generate a cubic polyhedron from any of the complexes in figure 5, we can repeatedly reflect across the front and back bounding planes, and translate vertically and to the right and left. These examples are meant to suggest that the full class of cubic polyhedra is quite large and varied. We introduce some graphical notation to help discuss examples. Given a cubic polyhedron P, we use a planar diagram to illustrate a slice through P along some oriented plane TT in the cubic lattice. A planar diagram will have shaded squares corresponding to the faces of P in TT, black edges corresponding to the faces of P incident to TT from above, and grey edges corresponding to the faces of P incident to TT from below. So for example, a monkey-saddle vertex will always appear (up to congruence) as ™j-, and a screw vertex will appear (up to congruence) as either -[- or : |-, depending whether or not the axis of the vertex is normal to the plane of the diagram. Figure 5 also shows planar diagrams for the examples.

rv

Figure 5: Varied examples of cubic polyhedra, with their corresponding planar diagrams. These 1 x 4 x 4 boxes can be extended by translations in two directions to 1 x Z x Z boxes, and then by reflection to complete cubic polyhedra.

1.5. Configurations around a face in a cubic polyhedron. Let / be a (square) face, which we think of as horizontal, in a cubic polyhedron P. Each of the four edges of / is either a crease or a flange, but successive edges

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cannot both be flanges, because the vertex between them would then have two perpendicular flanges. Furthermore, if there are successive creases, the neighboring squares across them have to alternate up and down (from the plane of /). We deduce that a (partial) planar diagram in the plane of / must look like one of the four possibilities in figure 6.

ft

«-., •••••;•»•<•!

Figure 6: These are the four possible partial planar diagrams in the plane of a face /. At the left, a normal face has four creases, alternating up and down. Next, a face with one flange has three alternating creases. Finally, a face with two opposite flanges also has two opposite creases; there are then exactly two possibilities for the complete planar diagram around /, depending on whether the creases are to the same side or not.

We say that / is a normal face of P if no adjacent faces are coplanar with /, that is, if all its edges are creases. The four adjacent faces across these creases must alternate up and down. (This is the first case in figure 6.) The complete configuration of P in a neighborhood of / is then given (up to isometry) by one of the six planar diagrams in figure 7.

wmJ^^^x^ff^-\mmm

&r&'«'*w**>.; MMM

v* mi's/^'^* 1 •*•#•;

Figure 7: These six possible planar diagrams around a normal face, up to isometry and switching the colors of the edges, are the ways to complete the first partial diagram from figure 6, corresponding to the six dot diagrams shown in the text.

Figure 8: These three possible diagrams around a face with one flange are the completions of the second partial diagram from figure 6.

To see this, consider how monkey-saddle and screw vertices (whose axes must be vertical) occur in cyclic order around /. There are a priori six

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possibilities: ££§, £§, £^, ££, £J, and JJ, where these dot diagrams record the types of vertices in a slice through P by dots on the integer lattice, with unfilled dots o for M vertices and filled dots • for screw vertices. For each of these six possibilities, the fact that the central square / is normal means that there is a uniquely determined configuration as in figure 7. A face f of P which is not normal has, among its edges, either one flange or two opposite flanges. If there is one flange, it is the horizontal axis connecting two screw vertices. The other two vertices of / can be either monkey-saddles or screws (with vertical axis); the three resulting configurations are shown in figure 8. A face / with two (opposite) flanges among its edges must be in one of the last two configurations shown in figure 6. Note that the second one, where each flange connects two screws of the same handedness, is the unique configuration of a face / with two opposite creases bent to different sides: Lemma 1.4. Suppose a face f in a cubic polyhedron has two opposite edges which are creases, and the two adjacent faces across these edges lie on opposite sides of the plane containing f. Then the other two edges of f are flanges, and each of them is the common axis of two successive screw vertices of equal handedness. D We will not illustrate the nine possible diagrams around a missing square in a cubic polyhedron. Note, however, that if all four edges are creases, there are four configurations, corresponding to the dot diagrams with even numbers of black (or white) dots. In particular, the diagrams ^ and £J cannot occur around a missing square. The local configuration around a normal face / i s a 2 x 2 x l box in one of the six configurations shown in figure 7. We define a tower to be one of the (six) 2 x 2 x Z configurations obtained from these by reflections. Lemma 1.5. Given a configuration T in a vertical 2 x 2 x Z box, if all the central horizontal squares are present as normal faces in T, then T is a tower. Proof. Each layer of T is one of the six local configurations around a normal face. But each one of these can stack vertically only to its mirror image. So T must be generated by reflections from any of its layers. D

2. SCREW VERTICES IN CUBIC POLYHEDRA Because monkey-saddles do not have flanges, the flanges of any screw vertex in a cubic polyhedron must connect to further screws along the axis.

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Therefore, any screw vertex lies in a bi-infinite column of screws with a common axis line. Such a column Ca is specified by a sequence a : Z —)• {S, Z} specifying the handedness of each vertex. (Shifting or reversing the sequence results in a directly congruent column; interchanging S and Z results in a reflected column.) Lemma 2.1. There are uncountably many cubic polyhedra with all screw vertices. Proof. For any sequence a, the column Ca is a configuration in a 1 x 1 x Z box, which can be reflected to a complete cubic polyhedron Pa by Lemma 1.1. (See figure 9.) These polyhedra are congruent only when the corresponding columns are. D

bi-axis of twis

Figure 9: The column Cff of screw vertices corresponding the the sequence a = • • • SSZSS • • • (left); the slab Sa it generates; a planar diagram for the column (or for the slab); and the cubic polyhedron Pa it generates by reflection (right).

When two adjacent vertices in a column have the same handedness, we say the column has a twist along the flange joining them; the normal of a twist is the normal direction to that flange (and the bi-axis is the direction perpendicular to both the axis and the normal). Let o\ be the alternating sequence • • • SZSZ • • •. Then Cffl is the unique untwisted column, and Pffl is the polyhedron PI we saw in the introduction. Note that each Pa divides E3 into two congruent regions. To see this, shift any Pa by (1,1,0); this interchanges the components of the complement of Pa, but leaves Pa invariant. (The polyhedron PQ also divides space into two congruent regions, as seen by a body diagonal (1,1,1) translation.) If we (repeatedly) reflect a column Ca in one coordinate direction but not the other, we get a 1 x Z x Z box which we call the slab Sff; see figure 9.

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(We can also fill out a 1 x Z x Z box with reflected monkey-saddles; we call this configuration a sheet.) The axis of a slab is the direction of the axis line in any of its columns, and its normal is its direction of finite extent. (Again, the bi-axis is the direction perpendicular to both the axis and the normal.) Note that if two slabs in a cubic polyhedron intersect, they do so in a common column, and thus they have the same axis direction. Any slab is invariant under translation by two units along its bi-axis, because it was generated by reflections in planes a unit distance apart. We now construct a second uncountable family of cubic polyhedra, indexed by a bi-infinite ternary sequence r : Z —> {0, x,y}. The polyhedron PT is built from layers in horizontal planes. If r(n) = 0, then there is a sheet of monkey-saddles in the plane z = n; otherwise there is an untwisted slab Sffl oriented to have normal z and axis r(n). (Because Sai is untwisted, it fits against a copy of itself with or without a 90° rotation, or against a sheet, as illustrated in figure 10.)

Figure 10: The untwisted column Cffl (left). Any sequence of sheets of monkeysaddles (right) or untwisted slabs Sai (center), which can be rotated by a quarter-turn so that their normals agree but their axes are perpendicular, can fit together front-to-back to form a polyhedron PT.

Lemma 2.2. Two adjacent parallel columns must be reflections of each other. Proof. If two adjacent screw vertices have the same handedness, then either they share an axis in a common column, or they have perpendicular axes. In two adjacent parallel columns, corresponding vertices thus have opposite handedness, and are mirror images by Lemma 1.2. D Lemma 2.3. Three mutually perpendicular untwisted columns cannot be mutually adjacent.

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Proof. Two adjacent perpendicular columns touch in screws of the same handedness, but handedness alternates along any untwisted column. D Lemma 2.4. Let C be a column of screws in a cubic polyhedron P. If C has a twist, then C lies in a slab with the same normal as that twist. Moreover, if C has a pair of twists separated by an odd distance, P is congruent to some Pff, with all screw vertices. Proof. Suppose there are two successive screws of the same handedness in a column C. The edge connecting them is their common axis, a flange between two coplanar faces /i and /2- (See figure 11.) But each fi is then of the type described by Lemma 1.4, so its opposite edge is similarly a twist in a column parallel to C. If a column Ca has twists with odd separation distance, then their normals are perpendicular. Applying the argument above, adjacent to Ca in any direction there must be a mirror-image column. Repeating, we see that our polyhedron must be Pa. D

Figure 11: The faces on the flange of a twist in one column are in the configuration described by Lemma 1.4, and thus must connect to another twist in a mirror image column.

Consequently, if Sff is a slab in a polyhedron other than Pff, the sequence a must be formed by pairs SZ and ZS; since twists correspond to consecutive vertices of the same handedness, this ensures all twists lie an even distance apart. Lemma 2.5. Suppose P is a cubic polyhedron containing a twisted slab S — Sa. Unless P = Pa, every slab in P has the same normal as S. Proof. Any slab 5' with a different normal would intersect S along a twisted column C = Ca\ let n be the normal direction to some twist in C. Both slabs S and 5" have the same axis (along C} but distinct normals; let T denote the one whose normal is not equal to n. The consecutive columns in the slab T are reflections of one another, so they all are twisted and have normal n. By Lemma 2.4, each lies in a slab with normal n, so P = Pa. D

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Lemma 2.6. Let P be a cubic polyhedron with all screw vertices such that there is some column in P with a twist. Then all columns in P are parallel, and P is congruent to some Pa. Proof. Let C be a column with a twist between S vertices v\ and v-^. By Lemma 2.4, C lies in some slab with the same normal as that twist. Consider the vertices v[ adjacent to vi in that normal direction. If either v[ is Z, then it is a reflection of ^, with axis parallel to that of C, so we get a parallel column. Then this column is also in a slab, and propogating this argument, we have the desired result. But the v'i cannot both be S, for then each would have axis perpendicular to that of Vi. But then v( and v'2 would be adjacent S vertices with parallel axes, contradicting Lemma 2.2. D Theorem 2.7. Any polyhedron with all screw vertices is congruent to a Pa or a PT. Proof. Let P be a cubic polyhedron with all screw vertices. The vertices of the cubic lattice are partitioned into columns, which clearly can have axes in at most two directions. We may assume each column is untwisted, for otherwise the last lemma would apply. If all the axes are parallel, then P is congruent to some Pff by Lemma 2.2. Otherwise, we can at least partition the vertices into planes, within which all axes are parallel. Within each plane, the (untwisted) columns form a slab congruent to Sao. The family Pr includes all possible ways for these untwisted slabs to fit together, so we are done. D Note that any cubic polyhedron which is uniform in the sense of having regular faces and a symmetry acting transitively on the vertices must either be PQ or have all screw vertices and be some Pa or PT, where the sequences a and r must have transitive symmetry group. The possibilities (up to isometry) are exactly CTQ = • • • SSS • • • ,

TO = • • • xxx • • • ,

a\ = • • • SZSZ • • - ,

TI = • • • xyxy • • • ,

<72 = • • • SSZZSS • • • ,

TI = ••• xxyyxx • • • .

Of course, PTO = Pffl = PI is the uniform polyhedron already mentioned. The four others, Pao, Pff2, PT1 and PT2, are additional new uniform polyhedra, shown in figure 12. Of course, any all-screw polyhedron in the uncountable families Pa and PT is semi-regular in the weaker sense of having regular faces and congruent vertex figures.

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Figure 12: The four additional new uniform polyhedra PffQ, Pff2, PTl and PT2.

3. OPERATIONS ON CUBIC POLYHEDRA In order to understand cubic polyhedra that mix monkey-saddle and screw vertices, we next define two operations. Pushing a tower changes monkeysaddle vertices to screw vertices (and vice versa) within the four columns of the tower. Inserting a slab cuts a polyhedron along an appropriate plane, moves the two pieces apart, and adds a new slab of screws between them. Our Main Theorem 4.5 will say that applying these operations in turn, starting from PQ, suffices to create any generic cubic polyhedron. 3.1. Pushing towers. Remember that a tower is one of six possible configurations in a 2 x 2 x Z box obtained by mirror reflection from the local configuration around a normal face. Typical towers are shown in figure 13. Of course, a screw vertex in a tower is part of a column contained in the tower. Note also that the vertical faces within a tower are present or absent in checkerboard fashion. Given a tower in a cubic polyhedron P, we push the tower by moving each face within the tower one unit along the axis of the tower. Since all horizontal faces are present in the configuration, we need not worry about them. So we can equivalently describe the push as removing all the vertical faces within the tower and filling in vertical faces where before there were none. In figure 13 we show the result of pushing on each diagram from figure 7. The following lemma should be clear: Lemma 3.1. The result of pushing any tower is again a cubic polyhedron. After pushing, monkey-saddle vertices in the tower become screw vertices and vice versa. D As an exercise, we note that each polyhedron in figure 5 can be created by applying tower pushes to PQ, except the one in the upper right. (That one has no towers, and can be described instead by the techniques below.)

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Figure 13: At left, we see two typical towers (corresponding to the top pair of diagrams at right. They differ from each other (as do the other pairs) by a push operation, which moves all the vertical faces one unit up (or, equivalently, down).

The columns within a tower are untwisted. Conversely, the following lemma shows that it is not hard to find towers around untwisted columns. Lemma 3.2. Suppose P is a cubic polyhedron, and f is a normal face of P at least one of whose vertices is a screw v lying in an untwisted column. Then f is in a tower with axis along the normal to f. Proof. The local configuration around / looks like one of the six configurations in figure 7. The first of the six is ruled out, as it has only M vertices. Let v be a screw vertex of /, call its axis direction (normal to /) vertical, and assume without loss of generality that v is an S vertex. Above and below v are Z vertices u±, since this column has no twists. Thus there are parallel faces f± in P just above and below /. Since the edges of f± incident to v± are both creases, the configuration near f± looks like one of the diagrams in figure 7 or 8. But none of the possibilities with flanges (figure 8) can fit above or below the normal face /. Thus f± must also be normal, and in fact mirror images of /. Continuing in this way, we find the configuration to be part of a tower. D Lemma 3.3. Let P be any cubic polyhedron, and C be any column of screw vertices in P.IfC does not lie in a slab or in a tower, then C lies sandwiched between two slabs whose parallel axes are perpendicular to the axis of C. Proof. If the column C has a twist, it lies in a slab. So suppose C is untwisted, and call its axis direction vertical. If, among the horizontal faces incident to C, one is normal, the previous lemma applies, and C is part of a tower. Thus, we may assume that every horizontal slice through C is in

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the configuration "frf^r- But then the screw vertices to either side of C must lie in slabs, for they lie in columns with axes (the thin black lines) perpendicular to that of (7, and such columns must occur at every level of C. Therefore the column C lies between two slabs, as desired. (It is possible that C is also part of a slab itself.) D Our Main Theorem 4.5 will say that any cubic polyhedron either is a Poor is obtained from a PT by pushing some towers and then inserting slabs. Let us consider how to apply tower pushes to a PT. The polyhedron PQ has a checkerboard of towers in each coordinate direction. Any other PT has slabs with horizontal axis, and thus has no vertical towers. We find horizontal towers at any level except where r alternates from x to y. The set of horizontal towers is always nonempty, except in PT1 . If we are starting with PQ, we first push any (infinite, finite or empty) subset of the vertical towers. (A nontrivial push will create some flanges and thus destroy some horizontal towers, but it may also create new ones.) We now take the resulting polyhedron (or our starting Pr] and push any subset of its towers in the y-direction. (Note that if we push a tower at a level where r had a Oy or yy, we will destroy the slab at that level.) Finally, we push any subset of the remaining towers in the re-direction. We call any cubic polyhedron resulting from this procedure a pushed- PT. 3.2. Removing and inserting slabs. Given a cubic polyhedron P, we remove a slab by the following operation: delete the slab completely, and then join the two (half-space) components of the complement together by translating one of them by one unit along the slab normal (to bring them together) plus one unit along the slab bi-axis (to let them match). For example, in figure 14 we show the result of removing a slab from the cubic polyhedron partly indicated by the planar diagram. It is not hard, using such diagrams, to prove the following: Lemma 3.4. Let P be a cubic polyhedron containing a slab S; the result of removing S is again a (well-defined) cubic polyhedron. D We define inserting a slab to be the inverse of this operation. After the next lemma, we will explain exactly what we mean by inserting an infinite sequence of slabs, but we note now that any PT can be obtained from PQ by inserting (perhaps infintely many) untwisted slabs. We will now examine the conditions under which a slab can be inserted into a pushed- PT, and will find that the slab to be inserted is usually uniquely determined.

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m

Figure 14: To remove a slab, we delete it and join the two resulting half-spaces together by a translation in the direction of the slab normal plus the bi-axis. In this diagram (in the axis/normal plane) the bi-axis translation appears as swapping all edge colors in the right half-space.

Lemma 3.5. Let P be a pushed-Pr. If P has columns in all three directions, no slab can be inserted in P. If P has no vertical columns, then slabs can be inserted at any horizontal plane z = i + |7 i £ Z.

Figure 15: The sequence of tower pushes in a plane determine which slabs can be inserted. We start with the left figure, exhibiting the trivial pattern of squares on the front boundary plane of the image. We push towers with axes in this plane, producing the middle figure. We will then be able to attach the slab illustrated at right, since they both show the same pattern of squares.

Proof. If P has columns in all three directions, then any plane is cut perpendicularly by the axis of some column, and thus cannot be a candidate for slab insertion.

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If there are no vertical columns, consider a horizontal plane II given by z = i + ^. We will refer to the intersection of P with II as the pattern of P in II. This shows us the set of faces in P which are bisected by IT, and is always a union of squares. If we insert a slab along II, the boundary of the slab must have the same pattern. If P is a PT, the pattern we see is always the trivial pattern consisting of a square array of squares, as in figure 15 (left). (This is why the three possible layers in a PT can fit together in any order.) To determine the pattern of P in II, we need only look at which tower pushes in the plane H have been performed. (Note that horizontal tower pushes which occur one layer higher or lower in P will change the vertices just above and below II, but won't affect the faces cut by H or thus what slab we can insert. Also, depending on r, pushes in II may or may not be possible.) If any tower within II has been pushed, then all tower pushes at that level (or at levels just above and below II) must have parallel axes. The only exception is that if all x-towers within II are pushed, then we again see the trivial pattern in II (shifted by one unit), and any subset of the (new) y-towers in II can be pushed. If the pattern of P in II is trivial (meaning that either no towers or all towers in IT have been pushed), we can insert an untwisted slab with axis x or y. Indeed, we can insert any finite number of untwisted slabs, indexed by a finite sequence of x's and y's, or we can delete the half-space of P to one side of II and insert a half-space of some PT corresponding to an infinite sequence of x's and y's. Of course, the result of such an insertion is merely another pushed-P T . On the other hand, if the pattern is nontrivial, then all tower pushes have occurred with parallel axes, say along x, and are at an even distance apart from one another within IT. We now define a bi-infinite sequence w of letters u,p: reading along the y-direction in II, each tower with axis in the x-direction was either impushed or pushed. The substitution rules u M- ZS, p (->• SZ convert u; to a sequence a. Then the unique slab that can be inserted along II is SV, as illustrated in figure 15. Moreover, we can insert any finite number of 5 ff 's, each a reflection of the next, or we can delete the half-space to one side of II and insert a half-space of Pa. D If P is a pushed-PT with no vertical columns, we can now explain what we mean by insertion of a (possibly infinite) set of slabs. Beginning near the origin and working outwards, we examine the planes z — ±(i — 5), for ?' = 1,2 If a given plane has the trivial pattern, we can insert any finite or infinite sequence of untwisted slabs, indexed by choice of axis, x or y. Otherwise, there is a sequence a associated with the plane, as described in

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the proof of Lemma 3.5, and we can insert any finite or infinite number of copies of Sa. (To insert an infinite sequence of slabs at one level, we throw out the half-space bounded by the plane away from the origin, and then stop the procedure on that side.) In our Main Theorem 4.5, we will show that any cubic polyhedron can be obtained from a pushed-Pr in this way.

4. CLASSIFYING CUBIC POLYHEDRA We now show that, given any cubic polyhedron, we can apply the operations of the previous section in reverse to simplify it. First we consider cases where tower pushes alone can bring us back to PQ. Lemma 4.1. Let P be a cubic polyhedron in which all columns (if there are any) are untwisted with vertical axis. Then P can be obtained from PQ by pushing some set of vertical towers. Proof. The polyhedron P must be generated by reflection from any Z x Z x 1 box around a horizontal plane TT, since the screws lie in untwisted vertical columns and the monkey-saddles also lie in infinite vertical lines. It follows that every horizontal face in P is normal and part of a vertical tower. We will assign a parity modulo 2 to each tower as follows. Arbitrarily assign parity 0 to one tower, centered at some point (x,y). We may assume that the planar diagram in TT at this tower is (B. Any other tower is centered at some (x + a, y + b) with a = b (mod 2). To this tower we assign parity a if it has the same diagram H, and parity a +1 if it has the other diagram B. Note that two adjacent towers intersect in a column of screws if they have opposite parity, as in *^, while they intersect in a line of monkey-saddles if they have the same parity, as in iffl. Pushing a tower switches its parity and leaves all other towers unaffected. So we now simply push all towers of parity 1; this leaves all towers with the same parity, so we now have all monkey-saddle vertices, as desired. D Lemma 4.2. Let P be a cubic polyhedron with columns in all three coordinate directions. Then we can push some set of vertical towers to eliminate all vertical columns. All columns in the resulting polyhedron P' are untwisted and horizontal, and P' has at least one vertical tower (of monkey-saddles). Proof. There can be no slabs in P, and thus all columns are untwisted. We will first find pushes to remove all vertical columns. Choose any horizontal plane TT in the cubic lattice, and consider the set V of all screw vertices in P with horizontal axis. Since these lie in horizontal columns, the projection of V to TT consists of infinite lines of vertices, including at least one line in each direction. Its complement (within the set of

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all vertices in TT) is a (possibly infinite) union of rectangular regions. Each region is finite or singly infinite in each direction, and thus has connected boundary. In figure 16 we illustrate a schematic of the slice in TT. Somewhere above or below each occurence of the symbol * there is a screw vertex with horizontal axis. At the locations marked o there is either a monkey-saddle vertex o or a screw vertex with vertical axis •.

x O xOOO x *OQOQO * O * OOO * xOOOOO

x o xo O O x xOOOOO x O xO O O x xQOOOO x O xO O O x xOOOOO Figure 16: A schematic for a typical slice through a cubic polyhedron with columns in three directions.

Now consider some a x (3 rectangle R of vertices in the complement of the projection of K, and assume a > (3. The configuration around R is a a x /3 x 1 box which necessarily appears reflected in P to fill out a a x (3 x Z box. (Indeed, any screw vertex in R is part of an untwisted vertical column, and any monkey-saddle vertex lies above and below further monkey-saddles by the definition of R.) We will prove that we can push vertical towers within R to eliminate all screw vertices in R. Within R, horizontal faces appear in a checkerboard pattern, as seen in figure 17. Those along the boundaries of R (shown in lighter gray in the figure) are not in towers, since they are incident, at some level, to horizontal screws in V'. But all interior horizontal faces (in darker gray) lie in vertical towers, by Lemma 3.2 and the definition of R. Since P has no slabs, by Lemma 3.3 any column of screws must be in a tower. If (3 = 1 then there are no towers in /?, so there can be no screws in R, and we are done. We now assume j3 > 2; we will return later to the case (3 = 2. Note that interior vertices of R are incident to exactly two towers, while vertices along an edge are incident to exactly one. A corner vertex in R is incident to either one tower or zero towers, depending on whether or not there is a horizontal face in that corner of R. We assign a parity modulo 2 to each tower in R exactly as in the proof of Lemma 4.1, and then push all towers of parity 1. Again, this leaves all

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towers within R with the same parity, meaning that the interior of R has only monkey-saddle vertices. (Beginning with any polyhedron represented by the first configuration in figure 17, we have now obtained the second one.) X

X>yX

X-X

XfgX

bpb.p^x p'Q.O'O^ X

X-' : X

X-*-X

X

X--.-X

X,.,X

X?X

X

x,-p?^>p'"'b o?x x -< ; x%:p opb * x-< x-opob.ox xx << ; x

x o.do.OQX

X -X

X-"X

X -x

X

X- X

X~'X

X'-X

Figure 17: Four configurations illustrating the effect of pushing towers.

If all boundary vertices of R are also now monkey-saddles (as in the third configuration in figure 17), we have proved the claim, having eliminated all vertical columns from this portion of the cubic polyhedron. Otherwise, we claim that each screw vertex in R is incident to exactly one tower, while each monkey-saddle is incident to zero or two towers. (This is the case in the last configuration in figure 17.) Then pushing all towers in R will remove all vertical columns, as desired. (In other words, if we originally had pushed the towers of parity 0 instead of those of parity 1, we would have obtained all monkey-saddles.) The claim is equivalent to saying that there are screws along the edges of R, and screws in corners where a horizontal face is present, but monkey-saddles in the other corners. Suppose there is a screw in an inappropriate corner, that is, we see the configuration ''£,«. At some levels above or below TT, there are horizontal columns along one edge of R] at other levels there are perpendicular horizontal columns along the other edge of R. But these two possibilities cannot happen at successive levels, since then we would have three mutually perpendicular and adjacent untwisted columns, contradicting Lemma 2.3. This means that at some intermediate levels, we have monkey-saddles at all three positions marked *. This means we see the configuration F :=C^5 which is impossible as we noted in Section 1.5. Remember we have assumed there is at least one screw on the boundary of R. We cannot have the configuration i^, since, again, at some level we will have monkey-saddle vertices at the positions marked x and thus would have the forbidden configuration F. Similarly, we cannot have the configuration »o> for there are only monkey-saddles in the (non-empty) interior of /?, and again we find the forbidden configuration F. Thus, along an edge of R with one screw vertex, we continue to see screws until we reach a corner.

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Finally, we check that we cannot have a monkey-saddle at an inappropriate corner: we cannot see the configuration *'oi, for again at some level the positions marked * must be monkey-saddles, giving F yet again. So all corners must have the appropriate vertices, either i * or o*. In the latter case, we cannot have the configuration o o * , since we have only monkey-saddles in the interior of R, giving F once again. Therefore, since the boundary of R is connected and contains at least one screw, it consists entirely of screws (except monkey-saddles in the appropriate corners). Finally, suppose (3 — 1. At the end of R, we must see one of the configurations at the left in figure 18. Note that the configuration F' := J* is forbidden just like F, Since we cannot have either F or F', by induction the towers in the interior of R must consist of a sequence of the two configurations in the middle in figure 18, and the vertices of each tower are either all screws or all monkey-saddles. Moreover, each vertex in R belongs to just one tower. We simply push the towers that consist of screw vertices and leave the others alone. We are left with all monkey-saddles in R.

* O

* 0.

*, o• * •

x o

.o

°°

oo

• •"

•*

°°

• ••

••

oo

Figure 18: At left and middle, configurations arising when R is a x 2; at middle and right, configurations arising when R is a bi-infinite strip of width two, as in the proof of Lemma 4.3.

If we repeat the procedure described so far for each rectangle R, we can push towers to eliminate all screws with vertical axis, and obtain the polyhedron P'. It contains no vertical columns, but does contain the vertical towers we pushed in order to obtain it from P. D Lemma 4.3. Let P is a cubic polyhedron with no vertical columns and no slabs. Then P is a pushed-Pr. If P includes a vertical tower (of monkeysaddles), then it is actually a pushed-Po. Proof. If P has columns only in a single direction, we can apply Lemma 4.1. So suppose P has columns along both x and y. The proof will be very similar to the previous one. Let TT be the xz-plane, arid V the set of screws with axis in the x-direction. Projecting V to TT, we now find that the complement is a union of horizontal Z x {3 strips R. When (3 = 1, since P has no slabs, Lemma 3.3 again guarantees that the region R contains no screws. So we may assume j3 > 1.

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First, we will suppose that P does include a vertical tower. This tower gives a vertical strip S (of width 2) in which we see only monkey-saddles. The complement of S in each R is two semi-infinite strips R±. Each of these regions has corners and connected boundary, as in the previous proof. The argument then proceeds as before, with the monkey saddles along S playing the same role as the monkey-saddles we earlier found at some level along each boundary of R. We assign parities to the towers in R±, and push the towers of parity 1. Then if any screws remain along the boundary, we push all towers. The same arguments guarantee that this removes all columns from R±. Repeating for each region R±, we eventually remove all columns along y. We can then use Lemma 4.1 to remove the remaining columns (along x). If P contains no vertical tower, the argument is a bit more difficult. We work directly on the strips R of height /3 > 2. First assume /3 > 2. Assign parities to the towers in R, and push those of parity 1. Fix one boundary component of /?; the usual arguments imply that along this boundary now either all vertices are screws or all are monkey-saddles. If both boundary components have screws, push all towers in R to eliminate all the screws. However, if one boundary has screws and the other has monkey-saddles, there is no way to push towers to get rid of all screws; we are left with a slab (untwisted, with axis y and normal z] along this boundary of R. If (3 = 2, the situation is similar. Again, since we cannot have the configurations F or F', the towers in R either are a sequence of the two configurations in figure 18 (center) or are a sequence of the two configurations figure 18 (right). In the first case, we push the towers consisting of all screw vertices as before, removing all columns from R. In the second case, we push all copies of the far-right configuration. (In either case, the push we use is in fact the usual push of all towers of parity 1, although the accounting is more difficult here.) In summary, within in each region R, we push towers to eliminate all columns with axis y except, perhaps, for a single untwisted slab along one boundary of R. We arrange (by pushing all towers in R if necessary) that these slabs are always at the top of each region R. It follows that these slabs are separated by distance at least 3 from each other (since each R was at least 2 units high, and they are separated by at least one layer). Call the resulting cubic polyhedron P'. Because of these slabs, we cannot now apply Lemma 4.1 to P'. We could remove the slabs and then apply that lemma. This would show our original P could be obtained from PQ by a combination of tower pushes and slab insertions, but would not demonstrate that it is a pushed-Pr.

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Instead, we repeat the argument we have just given, but with the roles of x and y interchanged, starting from P'. We must be careful because P' does contain slabs. But we get (3 > 2 for each strip R directly, since the slabs (the only screws with axis y) are separated vertically by 3 units. We get rid of internal columns in R by pushing towers of parity 1 as usual. Along a boundary of J?, we again claim that either all vertices are screws or all are monkey-saddles. We can no longer find some height above TT at which there are monkey-saddles just outside R, since R is now bounded by slabs. But these slabs are untwisted so the configuration -io is still forbidden, and the argument goes through as before. We can thus find tower pushes in the x direction taking P' to a polyhedron with no columns except untwisted horizontal slabs. This is a Pr. So P', and then our original P, is a pushed-P r . D Corollary 4.4. Let P be a cubic polyhedron with no slabs. Then P is a pushed-PT. Proof. The columns of P must be untwisted. If they have parallel axes, apply Lemma 4.1. If they occur in two directions, apply Lemma 4.3. If they occur in all three directions, apply Lemma 4.2 and then Lemma 4.3. D Main Theorem 4.5. Any cubic polyhedron P is a Pa or is obtained by inserting slabs into some pushed-PT. Proof. If a cubic polyhedron P has all screw vertices, then it is a Pa or a PT, and we are done. Otherwise, the slabs in P do not fill out all of P. Suppose there are two slabs in P with different normals. They intersect along a column which we will call vertical. All columns must be vertical, since a horizontal column would cut one of the slabs. By Lemma 2.5 all columns are also untwisted (or we would have a Pa}. Thus Lemma 4.1 applies to show P is a pushed-PT. Thus we may assume all slabs in P have the same normal. Call this normal direction vertical and the slabs horizontal. We can remove all slabs; we are left with an a x Z x Z box P", with a being bi-infinite, infinite or finite depending on whether there were originally zero, one or two half-spaces of slabs. We reflect P" if necessary to get a complete cubic polyhedron P1 with no slabs, from which P can be obtained by inserting slabs. Now Corollary 4.4 applies to P', showing P' is a pushed-PT, as desired. D

5. DISCRETE MINIMAL SURFACES IN THE CUBIC LATTICE As we mentioned in the introduction, a complete surface built from faces of the cubic lattice is discrete minimal if and only if each vertex of the lattice

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has one of five possible configurations of faces. We call these M, S, Z, X and 0, where M, S and Z are the vertex configurations of cubic polyhedra, X represents a fiat vertex with four coplanar squares, and 0 is an empty vertex, not in the polyhedron. It would be interesting to extend our classification theorem to give a complete list of all such discrete minimal surfaces; we give a few partial results here. First, let us give some examples. A trivial family, indexed by subsets of Z, contains any collection of horizontal planes. These surfaces are not connected, and do not include every vertex in the cubic lattice. We can construct a more interesting example as follows: Attach infinite sequences of flat X vertices to the two flanges of a screw vertex, and then extend this 1 x 1 x Z box to a complete surface by reflection. This is clearly a discrete version of Scherk's doubly periodic minimal surface, and it does relax to that surface. (See figure 19.) In this polyhedron, there are two halfspaces containing stacked half-planes of flat vertices, separated by a layer of screw vertices which twist the half-planes 90 degrees. This polyhedron is connected, and does include every lattice vertex.

Figure 19: This discrete minimal surface (left) is not a cubic polyhedron, but does relax to the singly-periodic minimal surface of Scherk (right).

Lemma 5.1. Along the normal line of a flat vertex (or any coordinate line through an empty vertex) we find only empty vertices or flat vertices with that normal line. Proof. The empty and flat configurations are the only ones which omit edges, and they omit edges in colinear pairs. Thus all edges along any such line are omitted in the polyhedron. D Although this lemma constrains how flat vertices can appear, there is still more flexibility than for cubic polyhedra. For instance, a flat vertex can be

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surrounded on all four sides by screw vertices, as in figure 20. This 3 x 3 x 1 block can be extended by reflection.

Figure 20: This discrete minimal surface (left) is not a cubic polyhedron. Its screw vertices lie in intersecting horizontal columns. It has two-fold rotational symmetry around each of the eight lines shown through the central flat vertex, and thus four-fold rotational symmetry around the normal line there. It evidently relaxes to the minimal surface shown (right), with the same symmetries.

We will not attempt here to give a complete classification of all the discrete minimal surfaces in the cubic lattice.

REFERENCES [1] T.F. Banchoff, Tightly Embedded 2-Dimensional Polyhedral Manifolds, Amer. J. Math., 87 (1965), 462-472. [2] K.A. Brakke, The Surface Evolver, Experimental Math. 1:2 (1992), 141165. [3] H.S.M. Coxeter, Regular Polytopes, third edition, Dover (1973). [4] L. Hsu, R. Kusner and J.M. Sullivan, Minimizing the Squared Mean Curvature Integral for Surfaces in Space Forms, Experimental Math. 1:3 (1992), 191-207. [5] H. Karcher and K. Polthier, Construction of triply periodic minimal surfaces, Philos. Trans. Roy. Soc. London Ser. A 354 (1996), 2077-2104. [6] U. Pinkall and K. Polthier, Computing Discrete Minimal Surfaces and Their Conjugates, Experimental Math. 2:1 (1993), 15-36. E-mail address: E-mail address:

cgstraus
"NEW" UNIFORM POLYHEDRA

BRANKO GRUNBAUM 4350, USA

University of Washington, Seattle, WA 98195-

ABSTRACT. Definitions of polygons and polyhedra, more general than the ones traditionally accepted, allow the construction of "new" uniform polyhedra. Although regular polyhedra, as well as some other classes of polyhedra, have recently been discussed from this point of view, the uniform polyhedra seem not to have been considered till now.

1. INTRODUCTION By their nature, facts do not change. However, our interpretation of facts changes quite often, frequently due to changes in definitions. Uniform polyhedra (also called Archimedean by some - but we shall come back to this later), that is, polyhedra with regular polygons as faces, and with all vertices in a single orbit under symmetries of the polyhedron, have been studied for a long time. The fact that the family of convex uniform polyhedra consists - besides the regular polyhedra - of the infinite families of prisms and antiprisms together with thirteen individual polyhedra, has been established countless times. In contrast, the enumeration of all uniform polyhedra, convex and nonconvex, has been carried out only gradually, and much more recently. Only in 1953 was the complete list published [1], without a claim of completeness. In fact, that enumeration was proved to be complete in [15] and [14]; a different approach to the enumeration and a proof of completeness is reported to be contained in [16] - unfortunately, I have not had the opportunity to see this work, and probably would not have been able to overcome the language barrier in any case. Illustrations and data can be found in [1], [10] and [17]. However, these "facts" should be replaced by new 331

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ones as soon as more inclusive definitions are accepted for regular polygons, for polyhedra and for their symmetries. There are many reasons for generalizing the traditional definitions. The main motivations are the wish to avoid needless restrictions, and to introduce consistency among the concepts and their applications - which is lacking in the traditional literature. Detailed critiques of the traditions can be found in [6], [7], [8] and [9], among others. Here I shall not repeat these arguments; instead, I shall only briefly describe what I believe are much more appropriate concepts. After that I shall present an account, as complete as I could make it within bounds for this article and with available time and energy, of the "new" uniform polyhedra.

2. POLYGONS AND POLYHEDRA An n-gon, for some n > 3, is a cyclically ordered sequence of arbitrary points labeled Vi, V^-, • • •, Vn, (called vertices), together with the segments EI determined by pairs of vertices Vi, Vi+i adjacent in the cyclic order (the edges}] each edge is incident with, and only with, the two vertices that determine it. If the value of n (the size of the n-gon) is not relevant, we speak of a polygon. Here we restrict attention to planar polygons, that is polygons all vertices of which are coplanar. A polygon is regular if each of its flags can be mapped onto any other flag by a symmetry. (A flag is a pair consisting of a vertex and an edge incident with it.) All this should sound familiar - countless publications give these definitions or equivalent ones. However, most of them - in particular, almost all those devoted to the study of polyhedra - interpret (tacitly or explicitly) the definition of polygons as including the qualifier "distinct" when referring to "arbitrary points". Despite appearances, this is not a minor difference. To begin with, the unrestricted definition includes the possibility of two or more vertices, adjacent or not, to be situated at the same point of the plane. These vertices are still distinguished by their labels, and each is incident with just the edges specified in the definition. Also, edges can be of zero length, collinear, overlapping, or coinciding in pairs or larger sets, passing through vertices not incident with them, and intersecting at triple or multiple points. In contrast, each (labeled) vertex of a polygon appears only once in the cyclic sequence describing the polygon; in other words, the polygon does not "revisit" any vertex, although it may return to the point representing a vertex. These possibilities require that "symmetry" be defined in a way that meaningfully accounts for them. To achieve this, by symmetry we understand a pair consisting of a permutation of the vertices that preserves incidences and adjacencies, and a compatible isometric map of the polygon onto itself. An illustration is given in Figure 1.

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1,8

Figure 1. An octagon, with two edges of zero length, and three coinciding edges. The only symmetry (besides the identity) is the reflection in a vertical mirror, paired with the permutation (16) (25) (34) (78) of the vertices. No proper rotation is a symmetry of this octagon. Here, and in all diagrams, to avoid clutter vertices are labeled only by numerals. For regular polygons this understanding of symmetries means that for every n and d with 0 < d < n there is a regular polygon that can be denoted {n/d} and obtained by the following construction. Start with a fixed circle, and a point chosen as vertex V\. Locate vertex V<2 on the circle, at arc distance lifd/n from V\ in the positive orientation of the circle, and continue analogously for n steps. Clearly, the n vertices will be represented by distinct points if and only if n and d are relatively prime; otherwise some points will represent several distinct vertices. It is also clear that {n/d} and {n/e} with d + e = n differ only by the orientation; since this is not important in the present context, we may restrict attention to 0 < d < n/2. In fact, since {n/0} is the trivial regular polygon, with all vertices at the same point, and since {n/d} with d — n/2 is a polygon with two sets of n/2 vertices, each set represented by the one point, hence not suitable for the production of polyhedra interesting in the present context, we may assume that 0 < d < n/2. A polyhedron is best described as being a geometric realization of an underlying combinatorial object which we call an "abstract polyhedron". An abstract polyhedron is a family of objects called vertices, edges, and faces, some pairs of which are incident, subject to the conditions which we state here informally (formal statements appear in [9]). Each edge is incident with two distinct vertices and two distinct faces. If two edges are incident with the same two vertices [faces], then these edges are incident with four distinct faces [vertices]. For each flag (triplet of mutually incident vertex, edge, face) there is precisely one other flag with the same vertex and face. Each face is a cyclically ordered sequence of vertices and edges, and analogously the faces and edges incident with any vertex form a cyclically ordered sequence (the vertex star of that vertex). Finally, any two faces [vertices] are connected by a chain of mutually incident faces [vertices] and edges.

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A (geometric) polyhedron is obtained by a mapping of an abstract polyhedron into 3-space in such a way that vertices are mapped to points, edges to segments (possibly of zero length) and faces to polygons. Two polyhedra are combinatorially equivalent, or of the same combinatorial type, if they have the same underlying abstract polyhedron. As in the case of polygons, a symmetry of a polyhedron is a pair consisting of an incidence-preserving automorphism of the underlying abstract polyhedron together with a compatible isometric map of the polyhedron onto itself. With this understanding of symmetries, the definition of uniform polyhedra given above remains valid. An illustration of these concepts is provided by the two combinatorially equivalent uniform polyhedra in Figure 2; one is a traditional prism, the other is "new".

9,12

7,10 11

(a)

(b)

Figure 2. Two combinatorially equivalent uniform polyhedra. Both are prisms; the prism in (a) has basis {6/1} while the one in (b) has basis {6/2}. The prism in (b) can be interpreted as arising by vertex-doubling (see Section 3) of the prism with triangular basis.

It is customary to designate uniform polyhedra by a symbol of type (p. q . r . . . . ) , which specifies the cyclic sequence of the sizes of faces surrounding one (hence every) vertex of the polyhedron. Among the possible choices, the lexicographically first is usually selected. A refinement of this notation takes into account the orientation of the faces incident to a vertex, with respect to the centroid O of the polyhedron. All faces of such a cycle have one side selected as the "outer" one, the selection being such that the "outer" sides of adjacent faces agree. If the "outer" side of a face is not visible from O its size receives a + sign in the symbol (but this sign is usually omitted); if the "outer" side is visible from O, its size receives a sign —; if the plane of the face contains O, the sign is ±.

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3. GENERATING "NEW" UNIFORM POLYHEDRA We turn now to a description of the various methods for the construction of "new" uniform polyhedra that have been found so far. 1. Vertex-doubling replaces each vertex by two, one red and one green. For each face, we follow around it, but connect by edges only vertices that differ in color. Hence vertex-doubling doubles the number of vertices and edges; it also doubles the size of odd-sized faces, and doubles the number of even-sized faces without change in their size. As is easily seen, a new polyhedron results if and only if at least one face of the starting polyhedron has odd size; if the starting polyhedron is uniform, so is the one obtained by vertex-doubling. For example, the polyhedron (6/2.4.4) in Figure 2(b) arises by vertex-doubling of the prism (3.4.4). Similarly, (3.6.6) leads to (6/2.6.6), as illustrated in Figure 3. If all faces of a polyhedron P have even sizes then vertex-doubling produces two separate polyhedra, each congruent with P.

h,H

(a)

(b)

Figure 3. (a) The uniform truncated tetrahedron (3.6.6), with the obvious faces such as [a b c a], [a i h I j a], Here and in the sequel, when listing the vertices of a face we repeat the first vertex in order to stress the cyclic nature of the symbol, (b) The uniform polyhedron (6/2 .6.6), obtained from (3 . 6 . 6) by vertex-doubling. Instead of using colors, we distinguish doubled-up vertices by upper and lower case characters. Note that all the faces of the polyhedron in (b) are hexagons - four of type {6/2}, and eight of type {6} = {6/1}. The faces incident with the vertex A are: [A b C a B c A], [A c D f G i A] and [A i H I J b A}.

2. Face-doubling replaces each face by one red and one green face, with edges joining only faces of different, colors; hence the number of edges is also doubled. Face-doubling doubles the valence of odd-valent vertices, replaces even-valent vertices by two vertices each. Face-doubling results in

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a polyhedron if and only if the starting polyhedron has at least one oddvalent vertex. Since all vertices of a uniform polyhedron have the same valence, face-doubling is applicable only to uniform polyhedra of odd valence; in such cases, it produces another uniform polyhedron with the same number of vertices. For example, the 3-valent ( 3 . 6 . 6 ) leads to the 6-valent (3.6.6.3.6.6); here the different fonts are substituting for different colors. Notice that in the latter polyhedron the cycle of faces incident with a vertex winds twice around the vertex. Hence it is reasonable to say that its vertex rotation is 2. Similar explanations hold for the vertex rotation data for other polyhedra. It should be observed that if a uniform polyhedron is odd-valent and has some faces of odd size, then both vertex-doubling and face-doubling are applicable; the result of carrying out both constructions is independent of order, and produces an even-valent uniform polyhedron with all faces even-sized. In the example ( 3 . 6 . 6 ) , the resulting polyhedron is (6/2.6.6.6/2.6.6), with 24 six-valent vertices, 72 edges, eight faces of type {6/2} and 16 faces of type {6}. 3. Deleting one transitivity class of faces of a uniform polyhedron, accompanied by the replacement of each remaining face by one red and one green copies, leads in many cases to one or two new uniform polyhedra. To achieve this, we join the different-colored faces at the edges that were incident to the deleted faces, and we join the faces at their other edges while insisting that all the joining edges either be incident with faces of the same color, or else that all be incident with faces of different colors. It may happen that these two procedures yield congruent polyhedra - for example, if the triangles are deleted from ( 3 . 4 . 4 . 4 ) - but for (3.8.8) with the triangles deleted the two resulting uniform polyhedra (8.8.8.8) and (8.8. 8.8) are distinct. The precise conditions under which the two choices result in isomorphic polyhedra remain to be determined. 4. Slitting along one transitivity class of edges (that is, deleting some or all edges of this class), accompanied by duplication of all faces, with no new vertices. Call one face of each pair red, the other green. In place of each edge that was slit, two edges are introduced. Each is incident with one red and one green face, but the incidences can happen in one of two ways: the faces either overlap, or else they are on opposite sides of the slit. This construction is applicable whenever there is a perfect matching in the graph of vertices and edges of the starting uniform polyhedron; such that the subgroup of the symmetry group which maps the matching onto itself leaves all vertices in one transitivity class.

"NEW" UNIFORM POLYHEDRA

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Figure 4. Deleting the triangles of the uniform truncated cube (3.8.8), and replacing each octagon by a pair of differently colored octagons, leads to two distinct uniform polyhedra, as explained in the text. Both have symbol (8.8.8.8), but are not of the same combinatorial type: in one there are two cycles of three faces around each "hole", while in the other there is only one cycle of length six.

e,E

d,D

a,A

c,C

b,B

d,D

b,B

a,A

f,F

(a)

(b)

Figure 5. The two kinds of slits possible on a cube. The two edges at each slit are shown slightly curved, to simplify their visualization. The pairs of faces obtained by duplication are distinguished by lower and upper case characters. If we indicate each edge by the labels of the faces incident with it and put in parentheses, one of the "new" uniform polyhedra obtained in (a) can be described by the cycle of faces incident with vertex V as (a(aA)A(Ae}e(eB)B(Bb)b(bE)E(Ea)a), the other as (a(aB)B(Be)e(eA)A(Ab)b(bE)E(Ea)a). According to our conventions, the latter has the symbol ( 4 . 4 . 4 . 4 . 4 . 4 ) and vertex rotation 2, while the former has symbol (4. —4. —4. — 4 . 4 . 4 ) and vertex rotation 0. The alternatives in case (b) are the same, and so the resulting uniform polyhedra have the same symbols and the same vertex rotations. However, the polyhedra resulting in case (b) are distinct from the ones in case (a), since - among other distinctions - their symmetry groups are different.

For example (see Figure 5), on a cube make two parallel slits on the front face, and two parallel slits on the back face. One possibility is that all four

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slits are parallel (Figure 5a), another that not all are parallel (Figure 5b). Both variants of incidences at slits work in each case, for a total of four distinct polyhedra. 5. Double covers of nonorientable uniform polyhedra. The orientable double cover of each nonorientable uniform polyhedron is itself a uniform polyhedron. An easy way to visualize this is to begin with the old procedure for deciding that a surface is nonorientable: start painting and continue till all reachable parts of the surface have been painted; if both sides of the surface are covered, the surface is nonorientable. Now, assume you use latex or oil paint, let it dry, and then make the original surface disappear. The double cover consisting of the paint is left over. This construction clearly doubles the numbers of faces, edges and vertices of the polyhedron, but leaves its (unsigned) symbol unchanged. The simplest example is the nonorientable heptahedron (3 . ±4 . — 3 . ±4), also known [17] as the tetrahemihexahedron. Its orientable double cover is combinatorially equivalent to the cuboctahedron ( 3 . 4 . 3 . 4 ) . 6. Including additional polygons as faces. The vertices of many traditional uniform polyhedra determine regular polygons that are not faces of the polyhedron. In some instances, one can include such polygons, doubled up if necessary, and redefine adjacencies so as to obtain new uniform polyhedra. For example, introducing in the regular octahedron ( 3 . 3 . 3 . 3 ) pairs of equatorial squares leads to a uniform polyhedron (3. ± 4 . 3 . ± 4 . 3 . ± 4 . 3 . ±4). Similarly, introducing the twelve regular pentagons determined by the neighbors of each vertex of the regular icosahedron ( 3 . 3 . 3 . 3 . 3 ) , we obtain two uniform polyhedra, one with symbol ( 3 . 5 . 3 . 5 . 3 . 5 . 3 . 5 . 3 . 5 ) and the other with symbol (3. — 5 . 3 . — 5 . 3 . — 5 . 3 . — 5 . 3 . — 5). Additional examples of this kind are given in Table 1. 7. Special constructions. There are a few additional uniform polyhedra I have found, which appear not to fit any systematic construction method. Naturally, it may well be that I just do not see the possibly larger scope involved. In any case, here they are. (i) The sixty diagonals of the faces of the regular dodecahedron can be organized into thirty squares, and also into twelve regular pentagons, see Figure 6. Double-up the pentagons, and color one of each pair red, the other green. One pair of opposite sides of each square, parallel to the edge of the dodecahedron which is "between" them, is colored red, the other pair green. Pentagons are adjacent only to squares, along edges of squares of their own

"NEW" UNIFORM POLYHEDRA

339

color. This yields a uniform polyhedron with symbol (4.5.4.-5.4.5.4.-5.4.5.4.-5). It is orientable, with "density" 5 and is topologically of genus 24.

Figure 6. The construction of the polyhedron ( 4 . 5 . 4 . — 5 . 4 . 5 . 4 . — 5 . 4 . 5 . 4 . — 5). On top is shown the regular dodecahedron that serves as the scaffold for the construction together with one of the pentagonal faces of the new uniform polyhedron, and one of its square faces. The pentagon represents a pair of coinciding faces, one red, the other green. The square has red edges [ac] and [ij], and green edges [ci] and \ja\. The lower diagram shows the cycle of faces that are incident with the vertex a. In this modification of the "vertex figure" used in [1] and [17], the view is along the line from vertex a towards the center of the polyhedron, and each directed line segment represents one of the faces.

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(a)

(b)

(c)

Figure 7. Different levels of truncation of the regular polyhedron {5/2,5}. In (a) and (b) one of the pentagonal faces and one of the decagonal faces are emphasized. In the full truncation (shown in (c)) the emphasized pentagon represents both faces; the resulting uniform polyhedron has symbol (5 .10/2 .10/2).

Figure 8. An approximation to the uniform polyhedron with symbol (4 .10 .10/2). One face of each kind is emphasized. The uniform polyhedron has vertices at the vertices of the convex uniform polyhedron ( 3 . 4 . 5 . 4 ) . (ii) Starting with the regular Keplerian polyhedron (5/2,5), we truncate its "points". The results of different levels of truncation are illustrated in parts (a) and (b) of Figure 7. If the truncation proceeds to the end, so that only the dodecahedral "core" is left (diagram (c) in Figure 7), a uniform

"NEW" UNIFORM POLYHEDRA

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polyhedron (5.10/2.10/2) is obtained. Notice that it has sixty vertices, three of which are at each vertex of the dodecahedron. Each pentagon of the dodecahedron represents both a pentagonal face of this uniform polyhedron and a decagonal face of type {10/2}. Similarly, from the other Keplerian polyhedron {5/2,3}, we may obtain another uniform polyhedron with sixty vertices and with symbol (3 .10/2.10/2). (iii) Figure 8 shows an analogous approximation to a uniform polyhedron (4.10.10/2), obtained by replacing the pentagons in (3.4. 5 .4) by decagons of type {10/2}, deleting the triangles, and introducing the decagons that surround the pentagons in (3 .4. 5 .4). Another "modification and replacement" produces the uniform polyhedron (6/2 .10/2 . —10) from the convex uniform polyhedron (3.5.3.5). There probably exist other such possibilities.

5/2 X \

/\

s^\

I / 5/2

Figure 9. The vertex figure of Skilling's "great disnub dirhombidodecahedron", sketched from [14]. Since the edges leading to vertices C,D,E,F are incident with four faces, to make a uniform polyhedron each of these edges has to be split into two; this is possible, by having two pairs of faces each determining an edge. Skilling proposed to have the cycle of faces proceed as AEFBDFHDCGECA. However, five other cycles seem as effective: AEFDBFHDCEGCA, AECGEFHDFBDCA, AECDBFHDFEGCA, AEGCEFHDBFDCA, and AEGCDFHDBFECA, for a total of six uniform polyhedra. Additional uniform polyhedra can be obtained from these by applying the different constructions described in Section 3. (iv) Skilling [14] described a "polyhedron" that is uniform according to the criteria imposed in [1] and accepted in his paper, with one exception: some

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edges belong to four faces. However, as Skilling points out on p. 123, this object is a polyhedron if the exceptional edges are interpreted as two distinct edges which happen to be represented by the same segment although they are determined by different pairs of faces; in other words, it is a polyhedron in the sense adopted here. The vertex figure - as shown in [14] - appears in Figure 9; the caption gives the resolution of the "double edges" as given by Skilling, together with five other possible resolutions.

4. THE MAIN TABLE

In Table 2 we give a detailed survey of the traditional uniform polyhedra, slightly modifying the presentation in [1] and [10], as well as a summary of results of the methods of construction of new polyhedra, following their general exposition given is Section 3. From the constructions in Section 3, and from the table, it is clear that there are hundreds of "new" uniform polyhedra (besides the several infinite families). It would be very nice if it were possible to determine all such polyhedra, and to explicitly describe the parameters for each of them. I do not have the time, nor the energy, to undertake such an effort. Moreover, finding a venue for the publication of a detailed accounting of all uniform polyhedra may be quite challenging. In any case, the data presented here are the best I could do. While acknowledging its shortcomings (and even that a few errors may have crept in), I do hope that the presentation here will provoke some investigators to devote part of their energies to the study of "new" uniform polyhedra, and of other special classes of general polyhedra. Here are some explanations of the entries in Table 2: ± in the symbol indicates that the plane of the face passes through the center. WS is the Wythoff symbol, which describes a method of generating the polyhedron; see Coxeter et al. [C]. VW stands for "vertex winding"; it is the winding number of the vertex figure. d is the "density" as given by Coxeter et al., that is, the winding number (of the surface) with respect to the center. NO indicates that the polyhedron is nonorientable. X is the Euler characteristic, and g is the genus of the (orientable) map which is isomorphic to the polyhedron. V, E, F denote the numbers of vertices, edges and faces, respectively. T = equilateral triangle; S = square; P = regular pentagon; H = regular hexagon; O — regular octagon; D = regular decagon; Pg = pentagram {5/2}; Og =octagrarn {8/3}; Dg = decagram {10/3}.

"NEW" UNIFORM POLYHEDRA

343

is the number in the Coxeter et al. paper [1]; W# is the number in Wenninger's book [17]. B# is the number of the in-text figure, or, if preceded by a Roman numeral, in the appropriate plate, in [B]. Note that some polyhedra shown in [B] are only isogonal, not uniform. H# is the number in [10]. The "Notes" in the last column refer to the traditional uniform polyhedra. CP stands for "coplanar" and means that there are pairs of coplanar faces. NEP stands for "no edge pairs" and means that there are pairs of vertices incident with pairs of faces but not defining an edge. OR means that the polyhedron is orientable even though there is no density defined. O? means that I do not know whether the polyhedron is orientable or not. Nl to N5 denote the various methods for obtaining "new" uniform polyhedra, as described in Section 3. The numbers shown in these columns indicates how many distinct uniform polyhedra are possible in each method of construction. 1* means that there are either one or more possibilities, depending on the parity of n\ an exponent + means that there may be additional possibilities which I did not investigate, while ? means that I do not know the answer.

5. REMARKS (i) It seems wasteful (and inappropriate) to use the terms "uniform" and "Archimedean" interchangeably. While Archimedes may well have discovered (at least some of) the various regular- faced polyhedra in which all vertices are surrounded by the same cycle of polygons - these Archimedean polyhedra are both conceptually and effectively not the same as the uniform polyhedra. In the latter, symmetries have to act transitively on the vertices and this is not a consequence of the characteristic property of Archimedean polyhedra, nor is it reasonable to assume that Archimedes had any grouptheoretic ideas. The distinction between "uniform" and "Archimedean" is clearly demonstrated by the "pseudo-rhombicuboctahedron" of J. C. P. Miller and V. G. Ashkinuze. An analogous distinction is necessary between the uniform quasirhombicuboctahedron (17, p. 132] and the nonuniform pseudo- quasirhombicuboctahedron discovered by Jones [11] in 1994. The existence of two concepts and two words makes it desirable and possible to pair them in a logical way. (ii) A few additional uniform polyhedra are described in [9]. Their descriptions have not been repeated here in order to avoid excessive length. (iii) Two other kinds of generalization are possible (see [51). First, one may admit infinite polyhedra, provided they are discrete (that is, every compact set meets only a finite number of vertices, edges and faces). Such polyhedra

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have been studied by various authors, but except in very restricted situations the results are fragmentary. The most detailed account for a special class is that of Jones [12], but even this is incomplete. Second, the definition of "polygon" may be extended to include nonplanar ones, which may then be used as building blocks of polyhedra. However, the relevant literature is restricted to regular polyhedra of this kind (see [2], [3], [13]), with no study of the possibilities of uniform polyhedra with such "skew" faces. The work of Farris [4] was a start in this direction, but the intended development seems not to have been published.

Figure 10. The matching in the graph of the regular icosahedron, used in the construction of the uniform but nonregular polyhedron having all faces congruent, described in Section 5. (iv) Any traditional uniform polyhedron with all faces congruent regular polygons is a regular polyhedron. This is not the case for the "new" uniform polyhedra. For example, let a "slitting" operation be applied to the regular icosahedron along the matching indicated in Figure 10. Let the faces be doubled-up, to yield one red and one green face. Across the edges of the matching, let faces of different colors be adjacent, while along the other edges faces of the same color are adjacent. It is clear that this determines a uniform polyhedron of type ( 3 . 3 . 3 . 3 . 3 . 3 . 3 . 3 . 3 . 3 ) with all faces congruent equilateral triangles - but the twelve faces adjacent to edges of the matching are not in the same orbit under symmetries of the polyhedron, as the other eight faces. (v) From the constructions described above, one may get the impression that all "new" uniform polyhedra (except the regular ones) have pairs of

"NEW" UNIFORM POLYHEDRA

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coinciding faces. However, this is not always the case. For example, the polyhedron (4.10.10/2) described above and in Figure 8 has no such faces; other examples can be found as well. Table 1. The following table lists the uniform polyhedra obtainable by introducing some of the additional regular polygons determined by the vertices and edges of certain traditional uniform polyhedra. In each case, pairs of coinciding (but distinct) polygons are introduced. Such polygons are indicated in the table by bold-faced numerals in italics. The list is certainly incomplete since it does not deal with possibilities of additional faces in "new" uniform polyhedra; probably there are omissions even with respect to traditional uniform polyhedra. Starting uniform polyhedron "new" uniform polyhedron (3.4.3.4) (3.±4.-3.±4) (3.4.4.4) (-3.4.4.4) (3.4.5.4) (3A-5/2.4) (3.5.3.5) or (5.10.-5.-10) (3.±6.-3.±6) (-3.6.5.6)

(3.6A.6.3.6A.6).

(3.6.5/2.6)

(3.10.6.10.5/2.10.6.10)

(-3.8.4.8)

(-3.4.8.4A.4-8.4). (3.5.±10.5.-3.5.±10.5). (-3.4.10.4.5.4.10.4) 3.6.10.6.-5/2.6.W.6). (3.4.5/2.4.3.4.5/2.4.3.4.5/2.4)

(3.±10.-3.±10) (-3.10.5.10) (3.10.-5/2.10) (3.5/2.3.5/2.3.5/2) (3.8/3.4.8/3) (4.5.4.5/2) or (4.G.-4.6) (4.±6.-4.±6) (5.G.-5.-6)

(3.8A.8A.8A.8) (-3.8/3A.8/3A.8/3A.8/3). (3. 10 A. 10.5. 10 A. 10) (3.10/3A.10/3.-5.10/3A.10/3). (3.10.5.10.3.10.5.10) (3.4.±6.4.-3.4.±6.4). (-3.10/3.6.10/3.5.10/3.6.10/3).

(3.4.8/3.4.4.4.8/3.4) (4.6.5.6A.-6.5/2.-6) (4.3.±6.-3.-4.-3.±6.3) (5.5/2.6.5/2.-5.5/2.-6.5/S)

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REFERENCES [I] H.S.M. Coxeter, M.S. Longuet-Higgins and J. C.P. Miller, Uniform polyhedra, Philos. Trans. Roy. Soc. London (A), 246 (1953/54), 401-450. [2] A. W. M. Dress, A combinatorial theory of Grunbaum's new regular polyhedra. I. Griinbaum's new regular polyhedra and their automorphism group, Aequationes Math., 23 (1981), 252-265. [3] A. W. M. Dress, A combinatoriai theory of Grunbaum's new regular polyhedra. II. Complete enumeration, Aequationes Math., 29 (1985), 222-243. [4] S. L. Farris, Completely classifying all vertex-transitive and edgetransitive polyhedra. I. Necessary class conditions, Geom. Dedicata, 26 (1988), 111-124. [5] B. Griinbaum, Regular polyhedra - old and new, Aequationes Math., 16 (1977), 1-20. [6] B. Griinbaum, Regular polyhedra, In Companion Encyclopaedia of the History and Philosophy of the Mathematical Sciences, I. GrattanGuinness, ed., Routledge, London, 1994. Vol. 2, pp. 866-876. [7] B. Griinbaum, Polyhedra with hollow faces, in: POLYTOPES: Abstract, Convex and Computational, Proc. NATO-ASI Conference, Toronto 1993. T. Bisztriczky, P. McMullen, R. Schneider and A. Ivic' Weiss, eds., pp. 43-70, Kluwer Acad. Publ., Dordrecht, 1994. [8] B. Griinbaum, Metamorphoses of polygons, in: The Lighter Side of Mathematics, Proc. Eugene Strens Memorial Conference, R. K. Guy and R. E. Woodrow, eds., pp. 35-48, Math. Assoc. of America, Washington, D.C., 1994. [9] B. Griinbaum, Are your polyhedra the same as my polyhedra? (to appear). [10] Z. Har'El, Uniform solutions for uniform polyhedra, Geometriae Dedicata, 47 (1993), 57-110. [II] R. H. Jones, The pseudo-great rhombicuboctahedron, The Mathematical Scientist, 19 (1994), 60-63. [12] R. H. Jones, Enumerating uniform polyhedral surfaces with triangular faces, Discrete Math., 138 (1995), 281-292. [13] P. McMullen and E. Schulte, Regular polytopes in ordinary space, Discrete Comput. Geom., 17 (1997), 449-478. [14] J. Skilling, The complete set of uniform polyhedra, Philos. Trans. Roy. Soc. London (A), 278 (1975), 111-135. [15] S. P. Sopov, Proof of the completeness of the enumeration of uniform polyhedra, Ukrain. Geom. Sbornik, 8 (1970), 139-156 [in Russian]. [16] I. Szepesvari, On the number of uniform polyhedra. I, II, Mat. Lapok, 29 (1977/81), 273-328 [In Hungarian]. [17] M. J. Wenninger, Polyhedron Models, Cambridge University Press, 1971. E-mail address: grunbaumQmath.washington.edu ' ' B . Griinbaum''

ON THE EXISTENCE OF A CONVEX POLYGON WITH A SPECIFIED NUMBER OF INTERIOR POINTS

KIYOSHI HOSONO 1 Department of Mathematics Tokai University 3-20-1 Orido, Shimizu Shizuoka, 424-8610 Japan GYULA KAROLYI 2 Department of Algebra and Number Theory Eotvos University Kecskemeti u. 10-12 Budapest, H-1053 Hungary MASATSUGU URABE 3 Department of Mathematics Tokai University 320-1 Orido, Shimizu Shizuoka, 424-8610 Japan

ABSTRACT. For any integer k > 0, let g(k] be the smallest integer such that every set of points in general position in the plane with at least g(k) interior points contains the vertex set of a convex polygon with exactly k interior points. Let g%(k} denote the corresponding number if the point sets are restricted to those whose convex hull is a triangle. We prove that g(k) is finite if <7s(A;) is so.

1. INTRODUCTION In 1935, Erdos and Szekeres [3] proved that for every integer t > 3 there is a smallest number f ( t } such that every set P of at least /(/) points, no three collinear, contains a subset of points whose convex hull has precisely t vertices. An extremal case of this problem was studied by Horton [4] who showed that it was not possible in general to specify both the number of hull vertices Supported in part by OTKA Grants #26358 and #30012. Visiting the Theoretical Computer Science Institute of the ETH Zurich. Research partially supported by Hungarian Scientific Research Grant OTKA F030822 3 Research partially supported by Grant-in-Aid for Scientific Research of the Ministry of Education, Culture, Sports, Science and Technology of Japan, 13640137 2

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and the number of interior points: there exist arbitrarily large point sets P which do not contain at least 7 vertices of any convex polygon whose interior is disjoint from P. This result has very recently been extended by Nyklova [6], see also [2] for a survey of related problems. Let P be a fixed set of points, in general position in the plane. Assume that C C P is the vertex set of a convex n-gon. Such a set C will be referred to as a convex n-gon, or n points is convex position. For any subset 5* C P, those points of P that lie in the interior of CH(S] will be called the interior points of 5*. Here CH(S) stands for the convex hull of S. A convex polygon C C P is called empty if it has no interior points. In [1], Avis et al. initiated the following problem. Find conditions which guarantee that P contains a subset of points with precisely k interior points. For any integer k > 0 let g(k) denote the smallest integer such that every set of points in general position in the plane containing at least g(k) interior points has a subset with exactly k interior points. Similarly, let, for any ra > 3, gm(k) denote the smallest integer such that every set of points in general position in the plane whose convex hull has m vertices and which has at least gm(k) interior points also contains a subset with exactly k interior points. Obviously, 3. It is observed in [1] that gm(k] < (m - 2)(g5(k) — 1) + 1. Thus, if g%(k} is finite, then so is gm(k) for every m > 3. This, however, has no implication about the finiteness of g(k). In this note we study the relationship between gs(k) and g(k). Although we cannot decide if g(k) is finite for any specific integer k > 3, we strongly believe that the following result is an essential step in that direction. Theorem Let k denote a nonnegative integer. If gs(k) is finite, then g(k) is also finite. To prove the theorem, we will need an extension of a result in [5] found recently by Valtr. Let K denote a fixed nonnegative integer. If every triple in P determines a triangle with at most K points in its interior, then P is said to be K -convex. Theorem A [7] For any K > 0 and N > 3, there is a smallest integer f ( K , N ) such that every K-convex set of at least f ( K , N ) points in general position in the plane determines an empty convex N-gon.

ON THE EXISTENCE OF A CONVEX POLYGON

353

Since the only known upper bound on /(A', TV) is quite enormous, we will make no attempt to calculate a reasonable upper bound for g(k) in terms of

2. PROOF OF THE THEOREM Fix k > 1, assume that gs(k) is finite, and write K = gz(k} — 1. Suppose that NQ is a large enough but finite number that we will specify later, and put g'(k} = f(K,No). Let P denote any set with at least g'(k) interior points, in general position in the plane. We are to prove that P contains a convex polygon with exactly A; interior points. If a, 6, c are consecutive vertices of CH(P) such that Aa&c is empty, then P\{b} has also at least g'(k) interior points. If, moreover, P\{b} contains a convex polygon with exactly k interior points, then so does P, too. Therefore we may assume Assumption 1 No three consecutive vertices ofCH(P) triangle.

determine an empty

If there are three points a, 6, c G P such that Aa&c has at least gs(k) interior points, then we are done. Thus we may make Assumption 2 For every a, 6, c G P, Aa&c has at most K interior points. In other words, we assume that P is A'-convex. Note that |P| > N0. Before we can continue with the proof we have to introduce some terminology. First, let x be an arbitrary point in P \ C . Denote by lx and rx the points at which the left and right tangents from x to CH(C] respectively touch CH(C}. The arc corresponding to x, denoted by A(x) is the set of those points in C which lie in the interior of £\xlxrx. The extended arc corresponding to x is A'(x) = A(x] U {lx,rx}. We say that x,y € P \ C are separated if their extended arcs are disjoint, that is A'(x) fl A'(y} = 0. The cardinality of the arc of x will be called the type of x and denoted by t ( x ) = |A(a;)|. Note that t ( x ) > 0, by Assumption 3. Next, the region corresponding to x, denoted by R(x] is the interior of the region obtained by removing the points of CH(C] from Axlxrx. We define a partial order X on P \ C by putting x X y iff R(x] C R(y}.

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Finally, the height of x £ P \ C is defined as follows. It is 0 if and only if a: is a minimal element in the partial order X, that is, if R(x) n P — 0. In this case we also say that x is a closest point to C', and write h(x) = 0. In general, h(x] is the length h of the longest chain x0 -< xi -< ... -< x^ = x. Note that h is not necessarily a rank function on P \ C. Figure 1 below shows two separated points x and y with t(x) = 3 , t ( y ) — 2.

Figure 1 Lemma 1 For every x 6 P \ C we have h(x] < K. Proof Assume on the contrary that h(x] — K for some x £ P\C. Consider the triangle T — Axlxrx. There is a chain XQ -< x\ -< ... -< XK = x with h(x{) = i for 0 < i < K. Then XQ,XI, ... , XA"_I are all interior points of T. Moreover, as t(xo) > 0, there is an other point y £ C in A(XQ) C A(x^) C T. Thus, T contains at least K -f 1 points of P in its interior, contradicting Assumption 2. Suppose that h(x) — 0 and t ( x ) > k for a point x £ P \ C. Write A(x) = {7/1,1/2,... ,yt(x)}, wnere Vo = ^,yi,2/2, • • • ,yt(x),yt(x)+i = rx are consecutive vertices of C. Then Ax/^t/^+i has exactly k interior points, and we are done. Thus, we may make Assumption 4 Every point x which is closest to C satisfies 1 < t(x) < k—l.

First we consider points whose type is 1. Suppose that x i , X 2 , . . . ,x^ are pairwise separated points closest to C, each of type 1. Then the convex polygon whose vertex set is Cu{x!,X2,... ,£;„•} has exactly k interior points. Hence, we may assume that there are at most k — l pairwise separated points with h(x) = 0 and t(x) = 1.

ON THE EXISTENCE OF A CONVEX POLYGON

355

Next, fix any integer 2 < t < k — 1. In this case we can write k = qt + r, where the integers g,r satisfy 1 < q < k and 0 < r < t — 1. Suppose that XQ, x\ , . . . , Xkq are pairwise separated points closest to C, each of type /, such that 1XQ , rxo , 1X1 , rxi , . . . , lXkq , rXkq follow each other in this order along the boundary oiCH(C] (they are not consecutive vertices, though). Since t < A;, there is an 0 < i < k — 1 such that AlXiq+1rXiq+qxo contains no point of A(XQ}. This is because any two such triangles has only one common point, namely x0. Write A(x0) - {yl,y2,... ,yt}, where y0 = lXQ,y\,yi, . . . ,yt,yt+i = rxo are consecutive vertices of C. There is an index j, 0 < j < t — r, such that r a 7 i . g + g ,t/j,xo,yj+r+i,^xi 9 + i are vertices of a convex pentagon. Let z\ = r xiq+i ? z?-> • • • izs — lxiq+q be consecutive vertices of C. Define Q as the vertex set of

Figure 2 depicts the case k = l,t — 3.

Figure 2 The only interior points of Q are the points of A(xiq+i), A( A(xiq+q), and the points j/j+i, yj+2, • • • •> 1/j+r- That is, Q has exactly qt + r = k interior points, and we are done. Hence, we may assume in this case that there are at most kq < k(k — 1) pairwise separated points with h(x) — 0 and t(x] - t. Given that (A; - 1) + (k — 2)k(k — 1) = (k — I)3, based on Assumption 4 we may summarize these assumptions in

HOSONO, KAROLYI AND URABE

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Assumption 5 The maximum number of pairwise separated points x with h(x) — 0 is at most (k - I) 3 .

P\C

Suppose that yo,2/i, • • • 5 2//+2 are consecutive vertices of C, we call such a set of points an arc of length I + 2. The interior of the region bounded by the two rays yo2/i and yi+zyi+i and the / line segments yiyi+i (1 < i < I) is called a domain of length I, it is an unbounded region if the two rays do not intersect. Figure 3 shows two domains, each of length 3. We write l(D) — I for the length of such a domain D, and we say that it is the domain supported by the arc A — {yi,yi,... ,T// + I}. For this we also use the shorthand notations A(D) — A and D(A) — D. Note that if a domain D contains a point x £ -P, then R(x) C D. In the sequel we suppose that

Figure 3

Lemma 2 There are domains D\,D NQ, it follows from Assumption 5 and the choice of NQ that there is a point yo £ C which is not contained in any extended arc belonging to some point closest to C. Suppose we have already denned yj. Go along the boundary of CH(C) in counter-clockwise direction, starting at yj, and let Xj denote a point closest to C whose extended arc is first encountered. Let Xj denote all the points closest to C which are not separated from Xj, they are all contained in a domain Dj whose length is at most 2k. Put / = rx^ and repeat. Then every point closest to C belongs to some Dj. Note that X Q . X \ * X I , - . . are pairwise separated, therefore it follows from Assumption 5 that the number of different domains obtained this way is at most (k — I) 3 . This proves the assertion.

ON THE EXISTENCE OF A CONVEX POLYGON

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Suppose that for some 0 < h < K — 1 we have already defined domains D[h\ D%\ . . . , D$h such that Mh < (k - I)3, every point x with h(x) < h is in one of the Mh domains D\ ' and Mh

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, that is, A'(x] <£. Aj

'. In this case there are

consecutive vertices ZQ, zi,... , 2S of A'(x) (s > k + 1) such that ZQ ^ Aj

,

^i is an endpoint of yr, and zs € >i'(y)- We claim that A^o^fc+i does not contain any point of P \ C. Assume on the contrary that Azo^-^jt+i contains a point x' G P \ C. Then x' G -R(x), thus x' -< x and /i(V) < /i. Since the segments ZiZ{+i (0 < i < k} are inside Az0xzk+i and are visible from x, at least one of them must be visible from x' as well. This implies that A'(x') contains at least one point from { ^ i , ^ ? - - - -,Zk\- Since z\ is an endpoint of Aj , it follows from the construction of the arcs A\ +l' that z\,,.. ,Zk # U^A] '. Consequently, x' £ Uj_i-DJ , contradicting the assumption that every point w with h(w) < h is in one of the Mh domains D\ . Thus we conclude that in this case Q = {ZQ,X, 2fc+i} has exactly k interior points, namely 21,^2?... , 2jfc. This proves the lemma.

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Choose now N0 = max{(& - l) (3fc + l),2kMK + (k - I)3} + 1, this also determines the value of g ' ( k } . Based on the above inductive procedure and Lemmas 1 and 3, we find that either P \ C is contained in the union of at most (k — I)3 domains whose total length is at most 2kMK, or there is a Q C P with exactly k interior points. Assume the first case. By the choice of 7V"o, there is a point y € C which is not contained in any arc that support one of these domains. Let x, ?/, z be consecutive vertices of C. It follows that these are also consecutive vertices of CH(P), contradicting Assumption 1. Consequently, it must be the second case, proving that g'(k) is an upper bound for g ( k ) . Thus, the proof is complete.

REFERENCES [1] D. Avis, K. HOSONO, AND M. URABE, On the existence of a point subset with a specified number of interior points, to appear in a special issue of Discrete Math., in honour of Helge Tverberg (2001) [2] I. BARANY AND GY. KAROLYI, Problems and results around the ErdosSzekeres convex polygon theorem, in: Discrete and Computational Geometry, Japanese Conference, JCDCG 2000(3. Akiyama, M. Kano, and M. Urabe, eds.), Lecture Notes in Computer Science 2098, Springer, 2001, pp. 91-105 [3] P. ERDOS AND G. SZEKERES, A combinatorial problem in geometry, Compositio Math. 2 (1935) 463-470 [4] J. MORTON, Sets with no empty 7-gons, Canadian Math. Bull. 26 (1983) 482-484 [5] GY. KAROLYI, J. PACK, AND G. TOTH, A modular version of the Erdos-Szekeres theorem, Studia Sci. Math. Hungar. 38 (2001) 245-259 [6] H. NYKLOVA, Almost empty polygons (preprint), KAM-DIMATIA Series (2000) No. 498 [7] P. VALTR, A sufficient condition for the existence of large empty convex polygons, to appear in Discrete Comput. Geom. (2001) E-mail address: E-mail address: E-mail address:

[email protected] ''Kiyoshi Hosono' 1 [email protected] ' ' G y u l a K a r o l y i ' ' [email protected] ''Masatsugu Urabe''

ON-LINE 2-ADIC COVERING OF THE UNIT SQUARE BY

BOXES

JANUSZ JANUSZEWSKI Instytut Matematyki i Fiziky ATR, Kaliskiego 7, 85-796 Bydgoszcz, Poland MAREK LASSAK Instytut Matematyki i Fiziky ATR, Kaliskiego 7, 85-796 Bydgoszcz, Poland

ABSTRACT. We present a method of 2-adic on-line covering of the unit square by arbitrary sequence of boxes of side lengths 2~k for k e {0,1,2,...} whose total area is at least 8.25. 1. INTRODUCTION We say that a sequence Ci,^,... of sets of d-dimensional Euclidean space Ed permits a covering of a set C C Ed if there are rigid motions £7i,CT2,... such that C C USi aiCi- Results on covering by sequences of convex bodies are given by many authors. In particular, see [1], [11] and the survey paper [2]. On-line covering is a covering in which every d with i > 1 is presented only after the motion <j;_i has been provided. The online covering is considered in [3-8], and the survey of results concerning the on-line covering is given in [9]. 2.

ON-LINE 2-ADic COVERING BY SEGMENTS

Kuperberg [8] proposed the so-called 2-adic on-line covering of the interval [0,1] by sequences of segments of lengths 2~ fc , where k € (0,1,2,... }, with the requirement that each segment is put in such a way that both its endpoints become multiples of its length. The question is finding an efficient method for this kind of covering. Such on-line covering methods are proposed 359

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in [5], [6] and [10]. Let us add that Kuperberg formulated this question as a problem about a good strategy in a game. As in [6], we may imagine a player called dealer who cuts pieces of length 2~k, where k G {0,1,2,... }, from a rope of fixed length V(l). The second player called placer uses them for the 2-adic covering of the interval [0,1]. The placer wins if he is able to cover the interval. The papers [5] and [6] propose the so-called method of the third interval for the 2-adic covering of the interval [0,1]. For the purpose of the application in Section 3, we present it below in a slightly more general form. Namely, we consider the covering of the interval [0,2 -i ] by segments of lengths from the set {2~*,2~ i ~ 1 ,... }. The case t — 0 gives the original covering method of the third interval. Let us recall the description of the 2-adic covering method of the third interval. At every moment of the covering process we consider the greatest number b £ [0,2 -t ] such that the whole interval [0,&] is covered. When a segment 5, say of length 2~ r , is given to us, we look for the greatest integer a such that a1~T < b. If the interval [(a -f h — l)2~ r ,(a + /i)2"~ r ], where h € {1,2,...}, is a subset of [0,2~*], then we call it the h-th interval. We look for such an interval of length 2~r which is not totally covered: first we check the third interval (if it exists), then the second interval (if it exists), and the first interval, until we find an interval not totally covered. The segment S is put on this first not totally covered interval counted in the above presented order. From [5], as well as from [6], it follows that the method of the third interval allows us to cover the interval [0,2 -t ] by every sequence of segments of lengths from the set (2~ i ,2~ t ~~ 1 , • • • } whose total length is at least 2 •1~t. In the proof of our Theorem we also need the following modification of this estimate. We can ask what is the maximum of the total length of segments "used for the covering" (instead of the question what total length of segments "permits the covering"). Now the worst situation for the placer is when he obtains the last segment of length 2~*. This is possible because the dealer is not limited with a fixed length of the rope. From the considerations in [5] or [6] we conclude the following estimate. Lemma 1. The total length of segments used for the covering of interval [0,2"*] by the method of third interval is not greater than 3 • 2~~*. We omit an easy proof of the following Lemma 2. Lemma 2. Assume that we execute the covering process by the method of the third interval. Imagine that at a stage of the covering process we remove from the interval [0,2~*] a number of partially covered intervals of the form [a2~ m ,(a-|- l)2~m}. Denote by P the union of those removed intervals. The

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ratio between the total length of segments from the sequence used for the covering of [0,2~*j \ P to the total length of the covered part of [0,2~ f ] \ P is not greater than 3. The paper [10] presents a somewhat more efficient methods of the 2-adic on-line covering of an interval. On the other hand, they are not as simple as the above method of the third interval. This is why in our basic algorithm presented below, we prefer to apply the method of the third interval and the related method of the fifth square as tools. At the end of the paper we comment on how to apply the methods from [10] for the 2-adic on-line covering by 2-standard boxes. 3.

THE PROBLEM OF 2-ADIC ON-LINE COVERING BY BOXES

Let an orthogonal coordinate system be given in Ed. Each set of the form {(#1,... ,x
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for the placer which always permits him a covering with possibly small fixed value of V(d).

4.

A TWO-DIMENSIONAL COVERING METHOD

For every non-negative integer s the unit square /2 is the union of 4s squares of side length 2~ s . We label these squares by numbers 1,...,4 S and we denote them by Kh(2~s), where h = 1 , . . . , 4 S , Simultaneously we require Kh(2~s) C Kg(2-s+l) for h = 4(0 - 1) + 1,.. - ,40 and for every 5 € {1,2,...}. Denote by T the family of all the squares of the form A'/ l (2~ s ) for alls G {0,1,2,...} and he {!,...,4 s }. By the bottom of the box {(x,|/); x\ < x < #2, y\ < y < 2/2} we mean the set {(x,y); x\ < x < x^ and y = y ^ } . It is convenient to assume that w\ < W2 for each 2-standard box from the sequence. Then each box is put into a 2-adic position by a translation. Of course, such an assumption w\ < w^ does not limit our consideration; just if this inequality does not hold, we first rotate the box by 90°. By the height of the box we mean w^. Below we present the covering method. When a successive box Si from the sequence is given to us, first (see (I) below) we assign a square K(Si) in which it is put, and then (see (II) below) we show a specific place in this square, where S{ is put. (I) Let us assign the square K(Si] into which our box Si is put. For this purpose first we look at the height of Si. Of course, the height has the form 2~ n . We find the not yet totally covered square Kz+i(2~n) with the smallest possible value z G {0,1,2,...}. Next we look for squares Kz+5(2~n), A' z +4(2~ n ), A'2 + 3(2~ n ), A' 2+ 2(2~ n ), if they exist, and we choose the first of them (in the above presented order) which is not totally covered yet. If none of those four squares exists or if they are totally covered, we choose the square A' z _j_i(2~ n ). Denote the chosen square Kz+m(2~n) by K(Si). Into this square Si is put. Let a(Si) be the number of values from amongst z + 1,2 + 2,2 + 3,2 + 4,2 + 5 which are at most 4n. Obviously m G {1,... ,a(Si)}. We say that a box Si is put into a far square K(Si) if this square is A' z+m (2~ n ) with m G {3,4,5}. (II) Now we describe the place in K(S\} where Si is put. Let G be the set of points p = (xi,#2) of the bottom of K(Si) such that the segment pp*, where p* = ( x i , # 2 4- 2~ n ), is contained in this subset of K(Si) which is covered either by boxes preceding Si of the height equal to the height of 5,-, or by boxes preceding Si which are higher than S.; and which are put into far boxes. The rectangle Si is used for the covering of K(Si) in such a way that the bottom of 5,- is used for the covering of the bottom of K(Si) according

ON-LINE 2-ADIC COVERING

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to the method of the third interval, where G is counted as the part already covered in the bottom. The above method will be called the algorithm of the fifth square and the third interval. Observe that if Si is a square, it covers the whole square K(Si). Thus in particular, when all the 2-standard boxes 5i, 6*2,... are 2-standard squares, then the described method is nothing else but the method of covering by 2-standard boxes as given in [5].

Theorem 1. Each sequence of 2-standard boxes of total area greater than or equal to 8.25 permits a 2-adic covering of the unit square. Proof. We apply the algorithm of the fifth square and the third interval described above. By the part of Si used for covering of a square A'/ l (2~ r ) we mean the set: 1) Si provided K(Si) = Kh(1-T], 2) Si n A'/ l (2~ r ) provided Si is put in a far square and provided the height of Si is greater than 2~ r , 3) 0 otherwise. Below we define the families P,P\,P2,1Z of squares from T. A square from T is considered to belong to P if during the covering process it has the form K(Si) = Kz+m(1~n\ where m £ {1,2}, for a box Si such that a(Si) = 5. We include a square from F into P\ if it does not belong to P and if during the covering process it has the form K(Si) — K2+i(2,~n), for a box Si such that a(Si) < 5. A square from F is counted to be in P^ if it does not belong to P U P\ and if during the covering process it has the form K(Si) = Kz+2(2~n), for a box Si with a(Si) < 5. We include a square from F into P, if during the covering process it has the form K(Si) = Kz+m(2~n), where m 6 {3,4,5}, for a box Si such that after putting 5*;, the square K(Si) is totally covered. Denote by p, p\, p2, r the total area of squares from the families P,Pi,P2,K, respectively. Denote by Q the union of those parts of the boxes Si which are not used for the covering of squares from P U PI U P? U Tl. By the part of the box Si used for the covering of Q we mean the set S i f } Q . Let q denote the area of

Q. Observe, that each box used for the covering can be presented as a. finite number of parts, such that each of them is used for the covering of squares from 7£ U P U Pi U P% or for the covering of Q.

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If 0 < k < /, then each five successive squares of the form /i'/l(2~') are subsets of at most two squares of the form Kg(2~k). Of course, if we have here two squares, they are successive. Thus if an interior point of S{ covers again a point covered earlier, then Si is put in K(Si) = Kz+m(2~n), where m G {1,2}, i.e. Si is used for the covering of a square from PUPiUP?. At this moment Ji' 2 + 3(2~ n ),..., A' a+a (5.)(2~ n ) are totally covered. Consequently, squares from 71 have disjoint interiors, r -f q < 1 and p < |(r + < / ) < § • If a square from 'R U P U P\ U P? is covered only by boxes of the height equal to the height of this square, then by Lemma 1 we see that the total area of boxes used for the covering of this square is at most 3 times larger than its area. It can happen that in a square A'/ l (2~ n ), for n > 0, there are contained parts of boxes of heights greater than 1~n. Consider a square of the form Kh(1~n} and denote by D the subset of Ji'/ l (2~ n ) which is covered by thode boxes of heights greater than 2~n which are put into far squares. We can treat D as covered only by boxes of height 2~n in this meaning that it is possible to cover D only by boxes of height 2~n according to our covering method, and by putting boxes of exactly the same area as the area of the parts of higher boxes used for the covering of D. This means that the total area of the parts of the boxes used for covering of any square from Jl U V U PI U P<2 is not greater than 3 times the area of this square. Let us notice that A"/l(2~71) can be partially covered by higher boxes which are not put into far squares. Those boxes cover points of squares from P U P\ U PI and therefore they are taken into consideration in our evaluation. Points from Q can be covered only by parts of boxes which are put into far squares (by those parts which are not used for covering of squares from 71 U P U PI U PI}. Thus, by Lemma 2, the total area of the parts of boxes used for covering of Q is not greater than 3. Obviously, if a square from P\ is totally covered, then the unit square is totally covered as well. Assume that a square of side length 2""1 belongs to PI. Then there is no square of side length of the form ( 2 ~ 2 , 2 ~ 3 , . . . } which belongs to PI UP?. In this case at most one square of side length 1 and at most one square of side length 2"1 belongs to P\, and at most one square of side length 2~~l belongs to P?. By the considerations presented before Lemma 1 we conclude that each sequence of 2-standard boxes of fixed height 2~n and of total area not smaller than 2 • (2~ n ) 2 permits a covering of the square of the height 2~ n . Thus each sequence of 2-standard boxes of total area not smaller than 3(q + r + p) + 2pi + 3p2 < 3(1 + f) + 2(1 + \) + 3 • \ = 8.25 permits a covering of the unit square by our method. Assume now that no square of side length 2"1 belongs to PI. Observe that there is at most one square of the form 7^(2""), for each n > 0, which

ON-LINE 2-ADIC COVERING

365

belongs to PI- Consequently, p? < (|) + (^) + • • • = |- Moreover, there is at most one square of the form /i.'/l(2~n), for each non-negative integer n 7^ 1, which belongs to PI. Thus p\ < I -f ^. In this case our sequence permits a covering of the unit square provided its total area is not smaller than 3(q + r + p)) + 2pl + 3p2 < 3(1 + f ) + 2(1 + ^) + 3 • i < 8.25. Observe that if a square from P\ is totally covered, then J2 is totally covered too. Thus in the above proof the total area of boxes used for the covering of the unit square is not greater than 8.25 + 1. This gives the following Corollary. Corollary 1. When we apply the algorithm of the fifth square and the third interval, the total area of boxes used for the covering of the unit square is not greater than 9.25.

5. REMARKS Imagine that the covering method from [10] replaces the method of the third interval in our covering method considered in the preceding Section. ^From [10] we see that each sequence of segments of total length not smaller than ^ permits a 2-adic covering of the unit interval. Thus the factor 3 in Lemmas 1 and 2 can be changed into ~. As a consequence, after an analogous consideration like in the proof of the Theorem, we obtain the following lower estimate : f (q + r + p)) + ±j-pl + f p2 < f (1 + §) + ^(1 + ?) ~^~ T ' \ ~ ^H' Thus, each sequence of 1-standard boxes of total area greater than or equal to 7|| permits a 2-adic covering of the unit square. The covering method of (n + l)-st strip is described in [4]. This method can be adapted for a description of a 2-adic covering method of Id by 2standard boxes. It is sufficient to take (instead of the segment- to- segment method) the method of the third interval, and to take n = 2 in the inductive description for d > 2. Considerations like the one in Lemma from [4] lead to the conclusion that each sequence of d-dimensional 2-standard boxes of the total volume greater than or equal to 3 • 5d~l permits a 2-adic covering of Id . So this kind of algorithm gives V(d) of the order of magnitude 5d. The authors wish to announce a new algorithm which gives an estimate of the order of magnitude 2d. The algorithm will be presented and discussed in a forthcoming paper.

REFERENCES [1] H. Groemer, Covering and packing properties of bounded sequences of convex sets, Mathematika 29 (1982), 18-31. [2] H. Groemer, Covering and packing by sequences of convex sets, in Discrete Geometry and Convexity, Annals of the New York Academy of Science, vol. 440, 1985, pp. 262-278.

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[3] J. Januszewski, M. Lassak, On-line covering by boxes and by convex bodies, Bull. Pol. Ac.: Math., 42 (1994), 69-76. [4] J. Januszewski, M. Lassak, Efficient on-line covering of large cubes by convex bodies of at most unit diameters. Bull. Pol. Ac.: Math., 43 No. 4 (1995), 305-315. [5] J. Januszewski, M. Lassak, G. Rote, G. Woeginger, On-line q-adic covering by the method of the n-th interval and its application to on-line covering by cubes, Beitr. Alg. Geom. 37 (1996), 51-65. [6] J. Januszewski, M. Lassak, G. Rote, G. Woeginger, Solution of Problem 74, Math. Semesterber., 43 (1996), 94-100. [7] W. Kuperberg, On-line covering a cube by a sequence of cubes, Discrete Comput. Geom., 42 (1994), 83-90. [8] W. Kuperberg, Problem 74-' Ein Intervaliiberdeckungsspiel, Math. Semesterber. 41 (1994), 207-210. [9] M. Lassak, A survey of algorithms for on-line packing and covering by sequences of convex bodies, Bolyai Society Mathematical Studies 6 (1997), pp. 129-157. [10] M. Lassak, On-line algorithms for q-adic covering of the unit interval and for the covering a cube by cubes, Beitr. Alg. Geom., to appear. [11] J. W. Moon, L. Moser, Some packing and covering theorems, Colloq. Math. 17 (1967), 103-110. E-mail address: E-mail address:

[email protected] ''Janusz Januszewski'' [email protected] ''Marek Lassak''

AN EXAMPLE OF A STABLE, EVEN ORDER QUADRANGLE WHICH IS DETERMINED BY ITS ANGLE FUNCTION

JANOS KINCSES1 Bolyai Institute, University of Szeged, Aradi vertanuk tere 1., Szeged, Hungary, H-6720

ABSTRACT. In connection with the problem of determining a convex polygon from its angle function we prove, that the midpoint square 5 of the inscribed square of the unit circle is distinguishable and, moreover, it is stable in the sense that each quadrangle close enough to S is distinguishable as well.

1. INTRODUCTION Geometric tomography, roughly speaking, is the subject where the aim is to determine a set if we know some partial information on it, such as the Xray pictures, the sections or the projections of the set. The first monograph [1] of this relatively new theory was published in 1995. Some years ago (see [4]) the author of the present paper proposed to investigate the following tomography type question. For a convex disk K (a compact convex set with nonempty interior) inside the unit circle C we define the angle function of K as a(K, x} — the measure of the visual angle of K at the point x e C. The problem is to determine the set from its angle function. The situation is not as good as in the case of X-ray pictures because there is no uniqueness. The angle function does not determine the set even in the case when it is a circle concentric to C, that is when the angle function is constant. We say that a convex disk is distinguishable if there is no other convex disk with Research partially supported by Hungarian NFS Research grants no. T032542 and T029786. 367

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the same angle function. With this notation Green [2] proved the following result. Theorem of Green. A circle K concentric with C is distinguishable if and only if a(K,.) is irrational mo( modulo IT or TT — a(K,.) = ^TT where p and q are relatively primes and q is odd. However, in [4] we proved that if we restrict our investigations to the class of convex polygons then there is uniqueness. Theorem 1. If P\ and P^ are convex polygons and a(Pi,x] = a(P2,x) for any x £ C then P\ = P<2. In [3] we discussed the question whether each convex polygon is distinguishable or not. The best we could prove is that if a convex polygon is not distinguishable then its tangent homeomorphism must have a periodic point with even length period. Specially, we derived that each triangle is distinguishable. However, the method of [3] breaks down for general periodic polygons with even period. The first example where it does not work is the midpoint square S of the inscribed square of C. In this paper we prove, by investigating the differentiability of the angle function, that S is distinguishable and, moreover, it is stable in the sense that it has a neighborhood such that any quadrangle of this neighborhood is distinguishable.

2. THE DIFFERENTIABILITY OF THE TANGENT MAP AND THE ANGLE FUNCTION

In [3] we introduced the tangent homeomorphism induced by a convex set. Fix an orientation on the unit circle C (let it be the counterclockwise orientation). For a convex disk K C C and for x 6 C let T(K,x) be the point of intersection of C with the right (with respect to the orientation of C) supporting line of K passing through the point x. The mapping T(Kj .) : C —>• C is an orientation preserving homeomorphism of the unit circle into itself which we call as the tangent homeomorphism induced by K. We introduce the arc length parameter on C and we will use the covering map ofT(K,x) which is denoted by f(K,x). It is well known that /(-K", .) : R -» R is a strictly monotone increasing, continuous function such that ",0) =T(K,0) <2?r 2ir) = f(K,x) + '2-n

T(K, x mod 27r) = f ( K , x) mod 2?r.

A STABLE, EVEN ORDER AND DISTINGUISHABLE QUADRANGLE

369

The connection between the tangent homeomorphism and the support function p(K, x) of K can be expressed by the following equations: ^ '

f(K,x] x

= = ip(x) — a.rccosp(K,(p(x)),

where ^p(x) is the angle of the outer normal of K to the line xT(K, x) supporting the set K (obviously, f(K,x) — (p(x) =
2a(K,x) = IK + f~l(K,x) -

f(K,x).

It is well known [5] that both the right and the left derivative of the support function of K exist. Using formulas (1) and some elementary calculus, we have that both the right and the left derivative of f(K,x) exist and we can calculate these derivatives.

and

From this we have that

f K r\ — _V Jt ri(J\,X) \ Similar reasoning gives that L - ^ Jlf,'(K \J^-i-x} )—

Let t+(x) (t~(x)) be the largest (smallest) point of K on the supporting line xT(K,x) and m(x) be the midpoint of the segment xT(K, x). It is well known that p(((p(x}} (p[(ip(x))) is the signed distance of the points m(:r) and t+(x) (m(x) and t~(x}). Using this, we have that }\

and,

r,ITjr

N

\t-(x)T(K,x)\

This implies that if t+ (x) ^ t~ (x} then (3)

0

and /(K, x) is differentiate if and only if the line xT(K, x) supports the set K in exactly one point. Now we can state the main result of this section on the differentiability of the angle function.

KINCSES

370

Theorem 2. The right derivative and the left derivative of the angle function exist. The angle function is differentiate if and only if each supporting line of K through x supports K in exactly one point. PROOF. The existence of the right and the left derivatives is clear from the formulas (la) and (2), and moreover we have that

and

1

2a[(K,x) =

f{(K,x). 27T))

If one of the supporting lines of K passing through x supports A" in a segment with positive length then, by formula (3), we obtain that a'r(K,x)>a[(K,x), that is, a(K, x) is not differentiate.

D

3. THE EXAMPLE Consider the square abed inscribed in the unit circle (7, and let 5* = pqrs be the midpoint square of abed (see Figure la). Theorem 3. The square S is distinguishable. PROOF. The points a,6,c, and d are periodic points of T(S, a;) with period 4, that is, T^(S,x) — x for x = a,b,c,d. We show that these are the only periodic points of T(5, x). For this, consider an arbitrary point x 6 C. Because of symmetry, we may suppose that x is in the arc ab.

T(S,x)

FIGURE IB If x is in the arc ap\ (p\ is the intersection of the line pq and the arc ab (see Figure la)) then elementary calculation gives that \pT(S,x)\ > \pb\ — \pa\ > \xp\. By this and the similarity of the triangles xap and bT(S, x)p,

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we have that the length of the arc 6T(5, x) is greater than the length of the arc ax. If x is in the arc p\b (see Figure Ib) then similar reasoning gives that the length of the arc T(5, x)c is smaller than the length of the arc xb which implies that the length of the arc ax is smaller than the length of the arc bT(S,x). Iterating this, we have that the arc aT^(S, x) is longer than the arc ax, that is, x cannot be a periodic point. Suppose, on the contrary, that there is a compact convex set K ^ S such that a(K, x) = a(S, x) for any x e C. Neither K nor S can contain the other, so there exists a common supporting line of the two sets, that is, there is an XQ 6 C such that T(S,XQ) = T(K,XQ). By the equality of the angle functions, we have that T(n\S,x0) = T^(K:x0) for any n > 0. The subsequence T^k> (S,XQ) is convergent and the limit is a or & or c or d (these are the only periodic points). This gives that the square abed is circumscribed around K as well. Theorem 2 implies that a(K,x) — a(S,x) is differentiate at the points a, 6, c, d and that each side of the square abed contains exactly one point of K (and at least one of these points does not belong to the set {p, q, r, s } ) . We may suppose that there is a point k\ ^ p of K on the segment pb. In this case the point k2 = K D be cannot be on the segment qc: because otherwise 0 = 2a'(S,&) = 2a<(*,6) = l f y - M > 0 . \kib\ \bk2\ On the same way, we have that k$ = K fl cd € rd and £4 = K n da e ds.

Consider now the point p\ = Cr\pq (see Figure 2). Because of the position of the points {ki}, the point T(K,p\] must be on the arc bq2 and the point T^1^^,^!) must lie on the arc dT(~l\S,pi). By Theorem 2, a(S,x} is not differentiate at pi, so at least one of the lines piT(~l\K,pi) or piT(K.pi] must intersect K in a segment. But a(S, x) is differentiable at each point of the arc dT(~l)(S,pi), so the line piT(~l\K,p\) supports K in exactly one

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point. This means that the line p\T(K,pi) must support K in a segment and it follows that a(S,x) is not different iable at T(K,p\). But qi is the unique point on the open arc 692 where a(S, x) is not differentiate. This gives that T(K,p\] = q\. By the equality of the angle functions, the length of the arc q\qi must equal to the length of the arc T Elementary calculations gives that

4

which contradicts to the fact that the point T^~l\K,p\] must stay on the D

Corollary 4. There is a neighborhood of S (in the sense of the Hausdorff metric) such that any quadrangle from this neighborhood is distinguishable. PROOF. We know from [3] that if a polygon has no periodic point then it is distinguishable. If a quadrangle Q is close enough to S and it has at least one periodic point then each periodic point of Q is close enough to one of the points a, 6, c, d and for each periodic orbit the midpoint quadrangle is close enough to Q. If we choose a neighborhood of S such that for any quadrangle Q = p*q*r*s* from this neighborhood and for any periodic orbit a*b*c*d* of it we have that the corresponding arc q\q^ is longer than the arc d*T(~l\Q,p^) then we can repeat the argument of the proof of Theorem 3 starting from the second paragraph. D Remark. We believe that more careful analysis of the above method may give the proof that each quadrangle which has periodic point with period four is distinguishable.

REFERENCES [1] Gardner, R.J., Geometric tomography, Encyclopedia of Mathematics and its Applications, 58, Cambridge University Press, Cambridge, 1995. [2] Green, J.W., Sets subtending a constant angle on a circle, Duke Math. J., 17, 1950, 263-267. [3] Kineses, J., On the determination of a convex set from its angle function, (submitted). [4] Kineses, J. and Kurusa, A., Can you recognize the shape of a figure from its shadows?. Contributions to Algebra and Geometry, 36 (1995), No. 1, 25-35. [5] Schneider, R., Convex bodies: The Brunn-Minkowski theory, Encyclopedia of Mathematics and its Applications, 44, Cambridge University Press, Cambridge, 1993. E-mail address:

[email protected] ''Kineses Janos''

SETS WITH A UNIQUE EXTENSION TO A SET OF CONSTANT WIDTH

MARTON NASZODI Department of Geometry, Eotvos University, Budapest, Hungary BALAZS VISY Department of Geometry, Eotvos University, Budapest, Hungary

ABSTRACT. Three topics are dealt with in this paper. We show that the sum of diametrically maximal sets in a Minkowski space is not necessarily diametrically maximal. A sufficient condition is given to guarantee the nonexistence of non-trivial set of constant width in a Minkowski space with a polytope as a unit ball. Then we give a topological proof of a theorem that enables us to obtain another set of constant width from a given one in Euclidean space.

1. INTRODUCTION A Minkowski space is a finite dimensional real Banach space, that is Rn equipped with a norm - not necessarily the Euclidean one. Any Minkowski space is determined by it's unit ball which is an o-symmetric convex set which intersects any line through o in a non-void closed segment, and vice versa any such body considered as a unit ball determines a Minkowski space. It is well known that there is only one vector topology on Rn so topology is independent of the metric. Notations: From now on W1 denotes an arbitrary Minkowski space, the metric is denoted d(.,.), a ball is: J5(x, r) :— {y € Mn : d(x,y) < r}, the sphere is 5(x,r) := {y G Mn : rf(x,y) = r}. Without the radii -B(x) := Key words and phrases. Set of constant width, Minkowski geometry, diametrically maximal set. 373

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5(x, 1), S(x.) := S(x, 1). The boundary of H is bd(H}. For a set ^ C Mn denote

B(H] := f| Definition 1.1. For a A" C W1 bounded set its diameter is diam(K) := sup{d(x,y) : x,y G A'}. A' C Mn is diametrically maximal if for any K' D A' diam(K') > diam(K). The following proposition is due Eggleston [1]. Proposition 1.1 (Eggleston). A' C Mn convex, diam(K) — 1. Then K is diametrically maximal if and only if B(K) — K. This notion has a twin: constant width. Definition 1.2. The distance of two parallel hyperplanes in a Minkowski space is twice the radius of the largest ball that fits in the strip between them. Another definition would be: Choose any point on one hyperplane and take the radius of the ball which is supported by the other hyperplane. This is the same as the previous definition. A K C K.71 convex set is of constant width w if the distance of any two parallel supporting hyperplanes is w. In the Euclidean space the two notions coincide that is a set is of constant width if and only if it is diametrically maximal. Eggleston showed that any set of constant width is diametrically maximal in any Minkowski space, but he presented a three-dimensional Minkowski space with a smooth and strictly convex unit ball in which he constructed a diametrically maximal set that was not of constant width [1]. In section 2 we answer the question of whether diametric maximality is invariant under Minkowski additon of sets as is constant width. In section 3 a condition on the unit ball is given to block the existence of a non-trivial (i.e. other than the ball) set of constant width. Section 4 presents a way of altering sets of constant width in Euclidean space. 2. NON-ADDITIVITY OF DIAMETRICALLY MAXIMAL SETS

Looking at the definition of constant width one easily verifies that this is an additive property, that is if A', L C Mn are sets of constant width then the Minkowski sum K + L is also of constant width. This observation motivates the question weather diametric maximality is also additive. Another result motivated Groemer [3] in formulating this question. It is obvious that any bounded set can be covered by a diametrically complete set of the same diameter. Groemer proved that any set K of diameter 1 has either 1 or

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continuum many diametrically maximal sets containing K of diameter 1. The similar result for sets of constant width (that is: If a bounded set K has more than one covering sets of constant with diam(K) then K has continuum many.) is trivial using the additivity. Theorem 1. There is a Minkowski space in which there is a diametrically maximal set K and 3s > 0 for which K + 5(o,£) is not diametrically maximal. This means diametric maximality is not additive since the ball is obviously diametrically maximal in any Minkowski space. The following proposition is needed: Proposition 2.1. In any Minkowski space a smooth set K is of constant width if and only if K is diametrically maximal. Proof. We only have to prove that a smooth diametrically maximal set is of constant width. Suppose diam(K) = 1. Take a supporting hyperplane H that supports K in a point on the boundary p. Using proposition 1.1 it is trivial that a point p is on the boundary of K if and only if it has a diametrically opposite point q that is c?mm(p,q) = 1 and q G K. Since diam(K) — 1 we have K C -B(p) H B(q). Then any support hyperplane of -B(q) at p is a support hyperplane of K at p. Since K is smooth the only support hyperplane of K at p is H. But -B(p) D B(q) is symmetric to ^^ so the reflection of H to ^^ *s a support hyperplane of -B(p) at q, let this be H'. Then again since K C -B(q),p € B(q) H' is a support hyperplane of K. So the parallel pair of H is H' and according to the second definition of the distance of two parallel hyperplanes the distance of H and H' is 1. D Proof of the theorem. Take the Minkowski space constructed by Eggleston, that is a Minkowski space where the unit ball is smooth, and there is a set K that is diametrically maximal but not of constant width. Suppose diam(K} = 1. For every e > 0 take K + 5(o, s). Since j?(o,£) is smooth K -f f?(o, e) is also smooth. If this was diametrically maximal for all £ > 0 then this would be a set of constant width (1 + Is}. Width in a direction1 is a continuous function from the set of compact convex sets equipped with the Hausdorff-metric to 9?, so K would be a set of constant width 1 what K is not. So for some £ > 0 the set K + B(O,E) is not diametrically maximal. D

1 Given H a hyperplane. The width of K in the direction H is the distance of the two parallel supporting hyperplanes of K parallel to H.

376

NASZODI AND VISY 3. A MlNKOWSKI SPACE WITHOUT NONTRIVIAL SETS OF CONSTANT WIDTH

A ball of a Minkowski space is the trivial example for a set of constant width. Are there other sets at all? In Euclidean space there are, the Reuleaux triangle is an example in the plane. Here we show that there are Minkowski spaces in which there is no other set of constant width but the ball. Theorem 2. If the unit ball of the Minkowski space is a polytope with a vertex p for which every facet of the ball contains p or —p then there is no other set of constant width in the Minkowski space than the balls. Corollary 3.1. In any Minkowski space with a double cone as the unit ball there are no nontrivial sets of constant width. n

For example: In l\ (EJ1 with the ||x||i := ^ \Xi\ norm) the ball is the only i=i set of constant width. Proof. We need an equivalent form of constant width. A set X has the support intersection property if for any support hyperplane H of the unit ball B(o) with p £ B(o) n // there are two points pi G X and p2 G X one on each supporting hyperplane of X parallel to H such that the line Pip2 is parallel to the line op. In [1] Eggleston showed that this property is equivalent to constant width. So suppose X C Mn is a compact convex set having the above property in a Minkowski space described by the theorem. We might assume that diam(X) = 1. Take a hyperplane H supporting the unit ball B(o, 1) only in p, that is in the given vertex. Take the two supporting hyperplanes of X parallel to H. There must be two points pi £ X and p2 G X one on each hyperplane such that pi — p2 — p because of the support intersection property and the fact that diam(X} — 1. Since X is diametrically maximal B(X] = X. Then X - B(X] C J5( Pl ) n B(p2). On the other hand 5(pi) n J3(p 2 ) = B(PI~^P2,1/2) since these points are in the same position as ^ and ^ and one easily checks that the fact that any facet of the unit ball contain p or -p implies that J5(f) n B(=f) = B(o, 1/2). So X C f?(P!+P2 ; 1/2), which is already a set of constant width 1. Two sets of constant width 1 cannot properly contain each other so our set is a ball. D

4. UNIQUE EXTENSION IN EUCLIDEAN SPACE ^From now on all of our work takes place in the d-dimensional Euclidean space Ed.

UNIQUE SETS OF CONSTANT WIDTH

377

Some trivial observations we will use without refering to them: Claim 4.1. Let J,K C Ed be arbitrary sets. Then the following hold • JCK => B(J] D B(K] • diam(K) < 1 & K C B(K] • if K is compact and convex then B(K) = B(bd(K)}, where bd(K) is the boundary of K. The following proposition that claims the equivalence of the spherical intersection property and constant width in the Euclidean space is also known: Proposition 4.1. Let W C Ed of diameter 1. Then W is of constant width if and only ifW = B(W) The following lemma is due to Sallee and will be useful. Lemma 4.1. Suppose x,y and z lie on a circular arc of radius 1 centered at c with |y — z| < 1 and x lies between y and z. Then 5(y) n 5(z) C 5(x) and

Proof. For the first containment suppose the contrary. Choose a point r € (.8(y) n B(z))\J3(x). Then since |r - y| < 1 < |r - xj, r lies in the open half space containing y which is determined by -ff^^y], the perpendicular bisector of the line segment [x,y]. Likewise r lies in the open half space containing z determined by //^[x^z]. A look at the 2-dimensional projection of these half spaces onto the plane spanned by x, y and z shows that the intersection of these half spaces is at distance > 1 from both y and z. Hence r does not exist. For the equation suppose q 6 B(y) n B(z) D 6*(x). Then the same argument shows that q lies in the intersection / of the two closed half spaces which are determined by TT^fxjy] and jff-^x, z], and which contain y and z respectively. But the only point of this closed wedge at distance 1 from x is the point c. D Definition 4.1. Let W C Ed. Then for a t € W t' € W is said to be diametrically opposite iffd(t,t') = diam(W). Remark 4.1. Let W be a smooth set of constant width. Let ' denote the diametrically-opposite-point-mapping: ,. bd(W) -» bd(W) ''

t

H+

t'

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where for t G bd(W) t' 6 bd(W} is the unique point on bd(W) that is at a distance diam(W) from t. This is a homeomorphism of the boundary of W onto itself. The following theorem was stated by G.T. Sallee in [2] but his proof is incomplete and - in my view unreparable. Theorem 3. Let W C Ed be a smooth set of constant width 1 and let p E Ed not in W but so close to W that the following are true: B(p) PI W ^ 0 and For any t € W n S(p) let t' be in B(p). Then there is a unique set Z, which contains (B(p) fl W) U {p} and of constant width I. This set is Z = B((B(p) n W) U {p}) Proof. First we prove unicity using that the given set Z is a set of constant width 1 and contains (W fl B ( p ) ) U {p} i.e. good. Then we prove that Z is indeed good. Suppose Z\ is also a good set. Then Z\ D (B(p) H W) U {p}. Using proposition 4.1 we get Zi = B(Zi) C B((B(p) n W) U {p}) = Z so Z\ is a subset of Z. Both being sets of constant width 1 it implies Z = Z\. For the proof of Z being good we have to show first that Z D (J9(p)n W)U {p}, which is obvious because diam((B(p) fl W) U {p}) < 1 and second that Z is of constant width 1. According to proposition 4.1 we need to show that B(Z) = Z. Z D (B(p) n W) U {p} so B(Z) C B ( ( B ( p ) KW}\J {p}) = Z. Thus we need to prove B(Z) D Z that is Vs 6 Z : B(s) 2 Z.

(1)

It is of course enough to show that B(s) D Z for every s £ bd(Z). For a point t E 6rf(VF) n 5'(p) let arc(t) be the shorter circular arc with center t from t' to p. The condition of the theorem ensures that in this case

t' e B(p). F := {x 6 bd(W) n B(p) : x; 6 B(p)} Three claims will show 1. Claim 4.2. 6d(Z) D (5(p) n W) U F U |J{arc(t) : t e 5(p) n bd(W}} Proof. Z obviously contains the first and the second part and because of lemma 4.1 it also does the third part. For a point in Z to be in bd(Z) there must be a point in (-B(p) n W) U {p} that is at a distance 1 from it. This

UNIQUE SETS OF CONSTANT WIDTH

379

FIGURE 1. In the picture F is the union of the two arcs from ti to t'2 and from t? to t^. opposite point is p for the points of the first set, t' for t for the points of the second set and t for the points of arc(t). D Claim 4.3. Vs 6 (S(p) n W) U F U (J{arc(t) : t £ 5(p) D bd(W}} : B(s} D Z that is formula (1) is true for these points. Proof. The containment holds for s £ (S(p) D W) U F by the definition of Z. The lemma proves it for any point on the arcs. D Claim 4.4. (5(p) n W) U F U \J{arc(t) : t £ S(p) n bd(W)} is a compact d — I dimensional manifold. Proof. B(p) n W is homeomorphic to a rf-dimensional disk since it is a compact convex body with nonempty interior. S(p) and bd(W] are two transversal hypersurfaces i.e. their support hyperplanes are different at any common point. This is pretty clear using the fact that for an x point on the boundary its opposite point is the one on the line perpendicular to the unique support hyperplane at x at a distance 1 from x, so if the two surfaces had the same support hyperplane at a common point then p would be in W. So S(p) H bd(W}, their intersection is a closed (d — 2)-dimensional manifold in bd(W). We also know that S(p) D bd(W) divides bd(W] into two parts, one outside -B(p) and one inside. Both parts are closed manifolds with boundary S(p)C\bd(W). Hence its image by the diametrically-opposite-point-mapping is also an d — 2 manifold V", which divides bd(W} into two parts. One manifold in which the points have their opposite points in the ball and the other one in which the points have their opposite points out of the ball. Using the closeness condition of the theorem on p we get that V C B(p}. So we can say that F is a submanifold of bd(W n B ( p ) ) , and divides it into two

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parts. Take the one containing S(p) n W. This is (S(p) fl W) U F because F = (B(p) n bd(W}} n (5(p) n 6d(W))'. So this part of bd(W n 5(p)) is an d — 1 dimensional manifold with boundary V. Now attach the union of arcs to it. The things to show are as follows: No.l: this union is also a manifold with boundary V No.2: this union does not intersect the selected part of bd(W) anywhere else than in V. The second claim is of course true since these arcs have no common point with bd(W] other than the corresponding t'. This follows from the fact that for any t € bd(W) there is a unique opposite point on bd(W). The first claim is also true since we can give an exact parametrisation of the open arcs with V x (0,1) and we have points of V at one end of these arcs and p at the other. The parametrisation is injective since for a point s G arc(t) the center of the arc is completely determined by t',p and s by the second part of the lemma. So s can not lie on another arc. D Now combining the three claims we have an d — 1 dimensional submanifold in bd(Z) for which B(s) 3 Z for any s point of it. But there cannot be an d — 1 dimensional manifold a proper submanifold of another so they are equal. So we get B(s) 3 Z for all s G bd(Z) which completes the proof. D As an immediate corollary we get: Corollary 4.1. IfW is a smooth set of constant width one in the Euclidean space then there is an £ > 0 such that any for point p in the e — neighborhood ofW ( B ( p ) r \ W ) ( J p has a unique extension to a set of constant width one. This property does not hold for non-smooth bodies as demonstrated by points near the vertex of a Reuleaux triangle.

REFERENCES [1] E.G. Eggleston, Sets of Constant Width in Finite Dimensional Banach Spaces, Israel J. Math., 3 (1968), 163-172 [2] G.T. Sallee, Reuleaux polytopes, Mathematika 17, (1970), 315-328 [3] H. Groemer, On Complete Convex Bodies, Geometriae Dedicata 20, (1986), 319-334 E-mail address: E-mail address:

runarciQcs.elte.hu ''Marton Naszodi'' [email protected] ''Balazs V i s y ' 1

THE NUMBER OF SIMPLICES EMBRACING THE ORIGIN

JANOS PACK1 Courant Institute, NY Academy of Sciences

and Renyi Institute, Hungarian

MARIO SZEGEDY Department of Computer Science, Rutgers University, New Brunswick

ABSTRACT. Using the Upper Bound Theorem for polytopes and Gale transforms, Uli Wagner and Emo Welzl have recently proved the following remarkable theorem. For any absolutely continuous probability distribution in c?-space, the probability that the convex hull of d + 1 randomly and independently selected points contains the origin is at most l/2 d , and this bound is tight. We present two very short proofs for the planar version of this result, and discuss some related questions.

1. INTRODUCTION Pick three points on the perimeter of the unit circle around the origin O, independently with uniform distribution. What is the probability that their convex hull contains 01 There is a short and sweet argument that goes back at least to the sixties (see Wendel [3]), which shows that the answer is 1/4. For any point x on the circle, let — x denote the point diametrically opposite to x. For any distinct points x\, #2, and x% on the circle, consider the unordered triples T = {£1X1,62X2,£3X3}, where each £,- = +1 or —1. Observe that out of these 8 triples precisely 2 induce a triangle which contains the origin in its interior, and the claim readily follows. In exactly the same way one can argue that the probability that the simplex induced by d -f 1 randomly, independently, and uniformly selected Supported by NSF grant CR-00-86013, PSC-CUNY Research Award 63382-00-32 and OTKA-T-032452. 381

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points on the surface of the unit sphere in rf-space contains the origin in its interior is l/2 d . Moreover, this statement remains true for any absolutely continuous distribution symmetric about the origin, i.e., when the measure of any half-space bounded by a hyperplane through the origin is precisely 1/2. By introducing an ingenious continuous analogue of the Upper Bound Theorem (cf. [4]), Uli Wagner and Emo Welzl have proved that for every absolutely continuous distribution in (/-space, not necessarily symmetric to the origin, the probability defined above cannot exceed 1/2 . They raised the question whether their theorem can be established by a simpler and "more illuminating" direct argument, at least for d = 2. In the following two sections, we describe two such arguments. Both methods solve some discrete variants of the problem, from where the Wagner- Welzl result follows by passing to the limit. 2. DISCRETE DISTRIBUTIONS ON A REGULAR 71-GON Let n > 3 be an odd integer, and let V — (t>i, 1*2, • - • , vnj be the vertex set of a regular ri-gon in the plane, centered at 0, where the indices are taken modulo n. Assume that the elements of V are numbered in such a way that the angle ViOvi+i is equal to (1 — l/n)?r for every 1 < i < n. Let P be a discrete probability distribution on F, for which P[vi] = pi > 0 and ^j- pi = 1. The set of those indices i for which pi ^ 0 is called the support of P and is denoted by supp(P). Theorem 1. The probability that the triangle determined by three randomly and independently selected points of V contains O in its interior is at most i (l — ^j). The maximum is attained if and only if P is the uniform distribution PQ. Proof: The probability we have to maximize is 6 times S(P) = Y^ where the sum is taken over all triples (z,j, k} such that the triangle contains O in its interior. Fix a distribution P for which S(P) attains its maximum, and assume that supp(P) is also maximal under this condition. We are going to show that S(P) does not exceed S(Po). Observe that there are no two consecutive indices i + 1,2 i + 2 (£ supp(P). Indeed, if i G supp(P) and z'-f l,z'+2 ^ supp(.P), then choose a small constant £ > 0. For the distribution P' defined b (1)

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383

we have S(P') — S(P), and the support of P' is larger than that of P. This contradicts the maximality of P. Since n is odd, it follows that there are two consecutive indices i:i -f 1 6 supp(P). We separate two cases depending on whether the support of P is full or not. CASE A: supp(P) ^ {1,2,... , n}. There exist (not necessarily disjoint) indices i and k such that {i} U {i + 1, i + 3, i + 5,... , k} U {k + 1} C supp(P), and {i + 2, i + 4,... , k - 1} £ supp(P). For the distribution P' defined in (1) we have S(P) - S(P') = pl+le[(pk+l + pk+3 + ... + #_!)(Pi+3 + Pi+5 + Pi+7 + • • • + Pi ~ £)] > 0,

whenever £ > 0 is sufficiently small. This yields Pk+i + Pk+3 + • • • + Pi-i > Pi+3 + Pi+s + Pi+r + ... + PiExchanging the roles of i and k + 1, we obtain, by symmetry, that Pk+2 + Pk+4 + • • • + Pi > Pk+l + Pk+3 + Pk+5 + • • • + Pk-2-

Comparing the last two inequalities, it follows that Pi+i = Pi+3 = Pi+5 = • • • - Pk - 0, contradicting our assumption that e.g. i + 1 6 supp(P). CASE B: supp(P) = {1,2,..., n}. Using the same argument as before, now we obtain that for every z, Pi+2 + Pi+4 + • • • + Pi-l

= Pi+3 + Pi+5 + • • • + Pi-

Substituting i with i -f 1, we obtain Pi+3 + Pi+5 + • • • + Pi = Pi+4 + Pi+6 + • • • + Pi+i • Therefore, pi+i = pi+2 holds for every i. The discussion of the case of equality is left to the reader.

D

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PACK AND SZEGEDY 3. COUNTING ALTERNATING SUBSEQUENCES OF LENGTH 3

It was just a matter of convenience that we assumed that V = {^i,... , vn} is the vertex set of a regular n-gon centered at 0. The proof of Theorem 1 applies to every set V with the property that any line connecting 0 with an element Vi G V has precisely (n — l)/2 points of V on both of its sides. In other words, V has the antipodality property with respect to 0: if we draw n rays from O through all elements of V, for any two consecutive rays there will be a third one lying in the cone induced by their reflections about 0. Theorem 2. Let n be an odd positive integer, and let V be a set ofn points in the plane such that V ( J { O } is in general position and the number of triangles induced by V that contain O in their interiors, is as large as possible. Then V has the antipodality property with respect to 0. Conversely, every set V which has the antipodality property with respect to 0, maximizes the number of triangles containing 0 in their interiors. Proof: Fix an x-y coordinate system in the plane with O as the origin, and assume without loss of generality that no point of V = {^i, 1^2? • • • ? vn} lies on the x-axis. Let 0 < at- < ir be the counter-clockwise angle from the xaxis to the line Ov{. Suppose without loss of generality a.\ < 0.1 < ... < an. For every i'. = 1, 2 , . . . , w, let sign(i) = -f if the y-coordinate of V[ is positive, and let sign(i) = — otherwise. Notice that, for every i < j < k, the triangle ViVjV^ contains 0 in its interior if and only if (sign(z'),sign(j'),sign(fc)) is an alternating sequence, i.e., ( + , - , + ) or (-,+,-). If we rotate our coordinate system until after the x-axis passes through the first point of V, the sequence (sign(l),sign(2),... , sign(n)) changes according to the following rule: the first or the last element will change its sign and move to the other end of sequence. We call this operation shifting. Obviously, the number of alternating subsequences of length 3 does not change during shifting. If S = (sign(l),sign(2),... ,sign(n)) itself is an alternating sequence and n is odd, then every other sequence obtained from S by shifting is also alternating. Note that S is an alternating sequence if and only if V has the antipodality property with respect to O. Thus, it is sufficient to verify the following Lemma. Let n be a (not necessarily odd) positive integer^ and lei S = (51,52,... ,5 n ) be a sequence of + and — signs, for which the number of alternating subsequences of length 3 is as large as possible.

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Then the maximum is f

/(«) / •"v

=

i

/

9

- \

n.\n~ — ii

if n is odd, n• {•n is even.

For odd n, this maximum is attained if and only if S is an alternating sequence. For even n, the maximum is attained if and only if in every string of consecutive members of S the number of plus signs and the number of minus signs differ by at most 2. It remains to prove the Lemma. Let n > 3, and assume that we have already established the assertion for all sequences whose length is smaller than n. By shifting, if necessary, we can achieve that s\ = sn = +. Deleting Si and sn from 5, we are left with a sequence S' consisting of p plus and m minus signs, p + m = n — 2. Clearly, /(51), the number of alternating subsequences of length 3 in 6*, satisfies

f(S) = f(S') + m + pm < f(n - 2) + (p + l)m < f(n - 2) + L^J

• r^i]

- /(n),

as required. If n is odd, then equality can hold only if S' is an alternating subsequence of length n — 2, and p + 1 = m. Therefore, S' must start and end with minus signs, and S must be alternating, too. One can also check that the cases when equality holds for even n are precisely those characterized in the theorem. D 4. OPEN PROBLEMS Both Theorem 1 and Theorem 2 immediately imply the planar case of the result of Wagner and Welzl mentioned in the abstract. Corollary. [1] For any absolutely continuous probability distribution in the plane, the probability that a triangle induced by 3 randomly and independently selected points contains 0 in its interior is at most 1/4. Equality holds here if the measure of any half-plane bounded by a line passing through O is 1/2. Problem 1. (Unicity) Is it possible to argue, based on the discrete variants of the result, that all distributions for which the bound 1/4 is attained in the Corollary satisfy the condition that the measure of every half-plane bounded by a line passing through O is 1/2? Problem 2. Can one extend the above arguments to higher dimensions? In order to generalize our proofs to 3-space, one should solve the following planar problem. Given n points in general position in the plane, colored red and blue. We want to maximize the number of multi-colored 4-tuples with

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the property that the convex hull of its red elements and the convex hull of its blue elements have at least one point in common. In particular, we want to show that when the maximum is attained, the number of red and blue elements are roughly the same.

REFERENCES [I] U. Wagner and E. Welzl: A continuous analogue of the Upper Bound Theorem, Discrete and Computational Geometry 26 (2001), 205-219. [2] E. Welzl: Entering and leaving j-facets, Discrete and Computational Geometry 25 (2001), 351-364. [3] J. G. Wendel: A problem in geometric probability, Math. Scand. 11 (1962), 109-111. [4] G. M. Ziegler. Lectures on Polytopes, Springer-Verlag, New York, 1995. E-mail address: E-mail address:

[email protected] ''Janos Pach'' [email protected] ''Mario Szegedy''

HELLY-TYPE THEOREMS ON DEFINITE SUPPORTING LINES FOR fc-DISJOINT FAMILIES OF CONVEX BODIES IN THE PLANE

SORIN REVENKO Institute of Mathematics and Informatics, Academy of Sciences of Moldova, Chi§ina, Moldova VALERIU SOLTAN Department of Mathematical Sciences, George Mason University, Fairfax, Virginia 22030, USA

ABSTRACT. A line / definitely supports a family T of convex bodies in the plane if it supports every member of T and all the bodies are in the same closed half-plane determined by /. T is called k-disjoint, where k (> 2) is a given positive integer, if no k elements of F have a common point. The paper contains new Helly-type results on the existence of a line definitely supporting a fc-disjoint family of convex bodies in the plane.

1. INTRODUCTION Dawson and Edelstein (see [2], [3]) proved several Helly-type theorems on the existence of a supporting hyperplane for a separated family of convex bodies. We need some definitions to describe their results for the planar case. By a planar convex body we mean a compact convex set with non-empty interior in the plane. In what follows, T denotes a family, maybe infinite, of convex bodies in the plane. We will say that a family T is k-disjoint, where k (> 2) is a given positive integer, if no k of the bodies from T have a common point. In particular, 2-disjointness means usual disjointness. We will also need a weaker variant of disjointness: a family T is called non-overlapping if no two of its members have common interior points. 387

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A family T is said to have the property D if there is a line / definitely supporting J7, i.e., a line supporting every member of T such that all the bodies are in the same closed half-plane determined by /. Similarly, JF has the property D(n) if |JF| > n and every n-membered subfamily of T has the property D. Trivially, D =^> D(n) for any positive integer n < |JF|; similarly, D(n) =^ D(m) for any positive integers m,n with m < n < \F\. The results from [2] can be summarized for the planar case as follows. (1) D(4) => D for a disjoint family F of convex bodies in the plane, with 1^1 > 4(2) D(3) =^ D for a disjoint family T of convex bodies in the plane, with | T\ > 5. Dawson and Edelstein [2] mentioned that similar results could be obtained for the case of collections of convex bodies in the plane which have restrictions on their positions less stringent than disjointness. This was realized in [5] for the case of a non-overlapping family T , as follows. (3) D(4) => D for a non-overlapping family J- of convex bodies in the plane, with \F\ > 4. In contrast to the statement (2) above, one can construct a non-overlapping family T of an arbitrary large cardinality with the property D(3) but not the property D (see [5]). Our purpose here is to further extend the results by Dawson and Edelstein [2] on definite supporting lines for the case of fc-disjoint families T of convex bodies in the plane. We mention here the only known results from [2] regarding fc-disjoint families. (4) There is a 3-disjoint family J7, \T\ — 10, with the property D(4) but not the property D. (5) There is a 3-disjoint family T of an arbitrary large cardinality with the property D(3] but not the property D.

2. MAIN RESULTS Theorem 1. D(6) => D for a k-disjoint family T of convex bodies in the plane, with k>2 and \J-\ > max {k,6}. In connection with the statements (1) and (2) above, we pose the following problem.

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Problem. For a given integer k > 3, does there exist a positive integer 715 (A;) such that D(5) => D for any fc-disjoint family T of convex bodies in the plane with at least n$(lK} members? The following two theorems give a positive answer to this problem for the cases k = 3 and k = 4, respectively. Theorem 2. D(5) =£• D for any 3-disjoint family T of convex bodies in the plane, with \F\ > 5. Theorem 3. D(5] => D for any 4~disjoint family T of convex bodies in the plane, with \F\ > 7.

\ Figure 1 Six triangles in Figure 1 give an example of a 4-disjoint family with the property -D(5) but not the property D. In other words, ns(4) = 7. A similar problem regarding the property D(4) has a negative answer for k > 4. An obvious modification of the example shown on Figure 4 from [2] gives an arbitrary large 4-disjoint family of convex bodies in the plane, with the property -D(4) but not the property D. The only positive case here is k = 3, as stated in Theorem 4. Theorem 4. £>(4) bodies in the plane.

D for any 3-disjoint family T of at least 77 convex

Based on Theorem 4, one can pose the following problem.

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Problem. Determine the minimum positive integer 71,4(3) such that -D(4) =£> D for any 3-disjoint family T of at least 77,4(8) convex bodies in the plane. From (4) and Theorem 4 it follows that 11 < n 4 (3) < 77. 3. AUXILIARY LEMMAS In what follows, we need some auxiliary lemmas, the first of them being obvious. Lemma 1. Any two disjoint convex bodies in the plane have exactly two definite supporting lines. D Lemma 2. ([2]) Any three convex bodies in the plane, with empty intersection, have at most three definite supporting lines. D We will say that a family F of convex bodies in the plane has the (p, q)property, provided it contains a subfamily of p members with at most q definite supporting lines. Lemma 3. D(p + q) => D for any family T, \F\ > p + q, of convex bodies in the plane, with the (p, q) -property. Proof. Assume, for contradiction, the existence of a family T, \F\ > p + q, for which D(p + q) does not imply D, and choose a subfamily H = {Ai,... , Ap} C F with at most q definite supporting lines. Let / i , . . . , /;, i < g, be all of these lines. Since any line /j, 1 < j; < i, does not support T definitely, there is a convex body Bj G T which either is not supported by lj or, being supported by lj, does not lie together with H in the same closed half-plane determined by lj. By the hypothesis, there is a definite supporting line / for the family {A\,... , Ap, J5i,... , J5Z-}. Since / is distinct from any of / i , . . . , /j, we get a contradiction. D We will say that n distinct lines / i , . . . , ln determine a convex polygonal domain jR, maybe unbounded, if every intersection Sj = /j PI R, i = 1,... , n, is a nontrivial line segment or a half line and s i , . . . , sn describe all the sides ofR. Lemma 4. ([!]) If n (> 6) lines in the plane are such that any five of them determine a convex polygonal domain, then all of them determine a convex polygonal domain. D

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4. PROOFS OF MAIN RESULTS Proof of Theorem 1. By Kelly's theorem, T contains three members with empty intersection (otherwise, all the members of T would have a common point, in contradiction with the &-disjointness assumption). By Lemma 2, T has the (3,3)-property, which gives -D(6) =£• D according to Lemma 3. D Proof of Theorem 2. Due to Theorem 1, it suffices to prove that -D(5) =>• -D(6) if \F\ > 6. So, let Q = {Ai,... ,A$} be a subfamily of six members of a 3-disjoint family T. Denote by \i a line definitely supporting the family Q \ {Ai}, i = 1,... ,6. Assume, for contradiction, that no line definitely supports Q. Then the lines / i , . . . ,/e are pairwise distinct. (Indeed, if, for example, li = h, then li would support Q.) Under this assumption, we are going to prove the following two claims.

Figure 2 Claim 1. No three of the lines / i , . . . , /e have a common point. Suppose, for instance, that /i,/2?^3 have a common point, a. By a symmetry argument, we may assume that H\ n H^ C H?, where Hi means the closed half-plane determined by ^ and containing Q \ {Ai}. Since all of ^4,^5,^6 lie in HI n HI, and /s supports H\ n H% only at the point a

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the bodies A^A^^A^ are supported by £3 at a, and hence they have a as a common point. The last is in contradiction with 3-disjointness of T. Hence /i,/2^3 have no common point, as claimed. For any i = 1,... , 6, each line of the family {li,... , IQ} \ {li} supports AI. By Claim 1, these lines form a convex pentagonal domain. From Lemma 4 it follows that the lines / i , . . . ,/e determine a convex hexagonal domain jR.

Claim 2. Any body A{ € Q, i = 1,... , 6, has a point in common with some other two bodies from Q. To prove Claim 2, we consider separately the following two cases. Case 1. The domain R is bounded. Without loss of generality, we may assume that the position of lines is as shown in Figure 2. Let a i , . . . ,OG be the vertices of R, and 61,62 > &3 be the points of intersection of its diagonals [01,04], [02,05], and [03,06]. We are going to show that

(6)

&i e A4 n A5 n A6, 62 e AI n A2 n A&, 6 3 e A 2 n A 3 n A 4 .

Trivially, the inclusions (6) immediately imply Claim 2. We prove here only the first inclusion from (6); the other cases are similar. Denote by

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z

\?zij%35^5) ZG some points of contact of A$ with the lines ^i,/2^3^5)^6? respectively. Clearly, [22,^3] intersects [02,61] and [25, £5] intersects [05,63]. Since A± is convex, it contains the segment [61,63]. Similarly, [61,62] C AQ. Finally, since A$ is supported by each of the lines lines £i,/2>^3^4,^6) the whole triangle A(6i, 62,63) belongs to A5. Thus, 61 € A± n A5 n AQ. Case 2. The domain R is unbounded. In this case we may assume, without loss of generality, that the lines / i , . . . , /s are as it is shown in Figure 3. Let 01,... , 05 be the vertices of R, and o~ic, a^d be the half-lines that bound R. Choose in R any half-line 036. Let 61,62,63 be points of intersection of diagonals [01,04], [02,05], and the half-line 03^, as shown in Figure 3. Similarly to Case 1, one can check the inclusions (6), that lead to the proof of Claim 2. As Claim 2 contradicts the assumed 3-disjointness of .7-", the family Q is definitely supported by a line. D Proof of Theorem 3. Due to Theorem 1, it is sufficient to prove that .D(5) =>• -D(6) for a 4-disjoint family T with \T\ > 7. We are going to prove a stronger statement: D(5) => D(7) for a 4-disjoint family T with \T\ > 7. Assume, for contradiction, the existence of a 4-disjoint family T of seven convex bodies in the plane, that has the property -D(5) but not D. By Theorem 1, one can find a subfamily Q = {Ai,... , AQ} C T with no definite supporting line. Due to -D(5), this means the existence of six distinct lines / i , . . . ,/6 such that li definitely supports the family Q \ {Ai} for all i = 1,... ,6. Let A? be the remained body in T \ Q. Claim 3. Any four bodies from Q, together with Ay, are definitely supported by at least one of the lines l\,... , IQ. Indeed, assume for a moment that all of A\, A-z, A$, A±, A? are not supported by any of the lines l\,... , / e - Due to -D(5), there is a new line lj definitely supporting {^1,^2, ^3, A±, A-?}. As the family T is 4-disjoint, one can find among Ai,A2,A^,A4 some three bodies with empty intersection (otherwise AI n A% n A3 n A± is nonempty). If they are AI,^2, and AS, say, then four distinct lines /4,/5,/e^r support {Ai,A2,A3J, contradicting Lemma 2. Due to Claim 3, for any pair of indexes i and j, with 1 < i < j < 6, the family T \ {Aj, Aj} is definitely supported by at least one of the lines /i, lj. One of li,lj also supports Ay. This easily implies that at least five of the lines / i , . . . ,/e support Ay. Assume, without loss of generality, that Ay is supported by / i , . . . , 1$.

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By repeating the considerations from Claim 1, we obtain that no three of the lines / i , . . . , 1$ have a common point. Hence / i , . . . , £5 form a convex polygonal domain, and IQ either supports this domain passing through one of its vertices, or cuts off a convex hexagonal domain. In both cases, by applying the same arguments as in the proof of Claim 2, we get that there is a point common to AQ,A^ and some two other bodies from Q, contradicting 4-disjointness of T. Proof of Theorem 4. Assume the existence of a 3-disjoint family T of at least 77 convex bodies in the plane, with the property D(&) but not the property D. We will show that this contradicts Lemma 2. Claim 4. Any four bodies from T are definitely supported by at least two lines. Let A,B,C,K be any four bodies in T. Due to -D(4), there is a line / definitely supporting these bodies. Assume for a moment that / is the unique line definitely supporting (A, B, C, K}. Denote by £ the maximal subfamily of T that includes {A, B, C, K} and is definitely supported by I. In what follows, we will consider two cases: |£| > 45 and |£| < 44. Case 1: \£\ > 45. Since the family JF does not possess the property D, one can find a body M ^ C. By Lemma 2, there exist at most three lines that definitely support A, B, M and any other body from £ \ {A, B}. Since |£ \ {A, B}\ > 43, we can find a line l\ which is a definite support for A, B, M and for at least 15 other bodies from £. Denote by L\ the maximum subfamily of £ that includes {A, B} and is supported by /i- Obviously, \C\\ > 17. Without loss of generality, we may assume that C belongs to L\. Since I is the unique line definitely supporting {A, B, C, K}, we have K ^ L\. As M ^ £, we conclude that L\ is a proper part of £, and there is a body N e £ \ LI. Again, at most three lines definitely support A, C, N and any other body from £1, and, since \C\ \ {A}\ > 16, one can find a line 1% definitely supporting all of A,C,N and at least six other bodies from L\. Denote by £2 the maximum subfamily of C\ that includes {A, C} and is supported by /2- Obviously, (£2) > 8 and K £ £2^From those three lines that definitely support all of A, K, N and any other body from £2 \ {A} one can find a line £3 such that at least three bodies from £2 \ {A} are definitely supported by £3. Since these three bodies are also definitely supported by any of /, /i, and l^-, we get a contradiction with Lemma 2. Case 2: |£| < 44. Then |^"\£| > 33. By Lemma 2, there can be at most two more lines (besides /) definitely supporting {A, B,C} and any other body

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from F\C. Hence one can find a line l\ definitely supporting the family {,4, B,C} UJ\fi, where A/"i is a subfamily of T\£ with |M| > 17. Since K is not supported by /i, among two lines that remain to be definite supports for A, B, K and any other body from A/i one can find a line 1-2 supporting the family {A, B,K} U A/2, where A/2 is a subfamily of A/i with |A/2| > 9. Since C is not supported by /25 one can similarly find a line 1% definitely supporting the family {A, (7, K} UA/s, where A/3 is a subfamily of A/2 with |A/s| > 5. Finally, among two more lines for B,C,K and any other body from A/3 there exists a line £4 definitely supporting the family {B, C, K} U A/4, where A/4 is a subfamily of A/3, with |A/4| > 3. Hence all the members of A/4 are definitely supported by each of the lines /i, /2, fa, ^4> again contradicting Lemma 2 (and completing the proof of Claim 4). ^From Claim 4 we conclude that any body in F\ {A, B, C} shares at least two definite supporting lines with A, 5, C. Since there are at most three such lines and since \F \ {A, B, C}\ = 74, one can find a pair of lines, li and 1-2 say, such that l\ definitely supports a family £1, 1-2 definitely supports a family £2 (both including {A, B, C}) such that \d n £2| > 25 + 3 = 28. Let K and M be elements of T \ {yl, 5, C} not supported by /i and /2, respectively. By Lemma 2, there are at most three definite supporting lines for the family {A,K,M}. Due to £>(4), every member of (Ci D £2) \ {-4} shares one of these lines as a definite common support with {A, K, M}. Since |(£i fl £2) \ {-4} I > 28 — 1 = 27, one of these lines supports at least 10 members of C\ H £2 (including A). Denote this line by /s, and let £3 be the family of all the bodies from £ (including {A, K, M}) definitely supported by /3. From the above we conclude that \C\ n £2 fl £3] > 10. Chose any body N £ £3. At most three lines, say /4,/s,/6, have to be definite supports for {K, M, N}, and any other body from C\ D £2 D £3. Hence, at least three bodies from £1 D £2 D £3 share four definite supporting lines, namely, /i,/2> fa and one of /4,fa,/6- The last is in contradiction with Lemma 2. D

5. ACKNOWLEDGMENT The authors wish to thank a referee for many helpful comments on an earlier version of this paper.

REFERENCES [1] I. Barany, H. Bunting, D. Larman, J. Pach, Rich cells in an arragement of hyperplanes, Linear Algebra Appl. 226/228 (1995), 567-575. [2] R. Dawson, M. Edelstein, Families of bodies with definite common supports, Geom. Dedicata 33 (1990), 195-204.

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[3] R. J. M. Dawson, Helly-type theorems for bodies in the plane with common supports, Geom. Dedicata 45 (1993), 289-299. [4] S. Revenko, V. Soltan, Helly-type theorems on transversality for setsystems, Studia Sci. Math. Hungar. 32 (1996), 395-406. [5] S. Revenko, V. Soltan, Helly-type theorems on common supporting lines for non-overlapping families of convex bodies in the plane, Geom. Dedicata 77 (1999), 225-237. E-mail address: E-mail address:

sorinrevenko9hotmail.com ''Sorin Revenko'' vsoltan9gmu.edu ''Valeriu Soltan''

COMBINATORIAL APERIODICITY OF POLYHEDRAL PROTOTILES

EGON SCHULTE Department of Mathematics, Northeastern University, Boston, Massachusetts 02115, USA

ABSTRACT. The paper studies combinatorial prototiles of locally finite faceto-face tilings of euclidean d-space Ed by convex d-polytopes. A finite set P of prototiles is called combinatorially aperiodic if P admits a locally finite face-to-face tiling by combinatorially equivalent copies of the prototiles in P, but no such tiling with a combinatorial automorphism of infinite order. The paper describes some properties of combinatorially aperiodic protosets and their tilings, and also discusses some open problems and conjectures.

1. INTRODUCTION Aperiodicity is a fascinating phenomenon. Traditionally, an aperiodic protoset is a set of tiles in euclidean d-space Ed that admit a tiling of Erf by congruent copies, but no such tiling with a non-trivial translational symmetry. In the light of the discovery of the Schmitt-Conway-Danzer tile and its tiling properties ([4]), the notion of aperiodicity has been revised to require the stronger condition that no tiling by the protoset have a euclidean symmetry of infinite order. The purpose of this short note is to introduce a combinatorial analogue of this stronger notion, called combinatorial aperiodicity. Congruence of the tiles is here replaced by combinatorial equivalence of the tiles, and the interest is in locally finite face-to-face tilings of Erf by convex polytopes each combinatorially equivalent to a polytope from a finite protoset of convex polytopes. At this point it is still open whether or not combinatorially aperiodic protosets actually exist in any dimension. We describe some results about combinatorially aperiodic protosets and their tilings, and also discuss some open problems and conjectures. 397

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2. BACKGROUND A tiling T of Ed is a countable family of closed subsets of Ed, the tiles of T, which cover Ed without gaps and overlaps ([8]). We shall assume that the tiles are convex d-polytopes. A tiling T by convex d-polytopes is said to be face-to-face if the intersection of any two tiles is a face of each tile, possibly the empty face. All tilings are taken to be locally finite, meaning that each point of Ed has a neighborhood that meets only finitely many tiles. A protoset of a tiling T of Ed is a minimal subset of tiles of T such that each tile of T is congruent to one of those in the subset. The tiles in the subset are the prototiles of T, and the protoset is said to admit the tiling T. We also use this terminology for shapes and sets of shapes that are under consideration for being prototiles or protosets of tilings, respectively. A tiling T in Erf is called periodic if its symmetry group is a crystallographic group and thus contains translations in d linearly independent directions. A non-periodic tiling has no (non-trivial) translational symmetry at all. A protoset is called weakly aperiodic if it admits a tiling of E but if all such tilings are non-periodic (see, for example, [7]). Moreover, a protoset is (isometrically] aperiodic if it admits a tiling of Ed but if no such tiling has a symmetry of infinite order. In the euclidean plane, each symmetry of infinite order is a translation or a glide reflection; thus aperiodicity and weak aperiodicity are equivalent concepts for plane tilings. There exists a considerable body of literature about aperiodicity of protosets, especially planar protosets. The first appearance of an aperiodic protoset was Berger's (1966) discovery of a set of 20426 square tiles with colored edges, so-called Wang-tiles; in particular, this proved that the socalled "Tiling Problem" is undecidable (that is, no algorithm exists that, upon being fed a protoset, decides whether it admits a tiling or not). Many (colored or otherwise) decorated or non-decorated aperiodic protosets have been discovered since then, but all have at least two prototiles. Some famous examples found by Robinson (1971), Penrose (1974, 1978) and Ammann (1977) are surveyed in Griinbaum & Shephard [8] and Senechal [17]; others occur in, for example, [2, 3, 10]. There are aperiodic protosets with only two prototiles in any dimension d > 2 (Goodman-Strauss [6]). The tile of [4], known as the Schmitt-Conway-Danzer tile, comes remarkably close to being a single aperiodic prototile in dimension 3. A tiling by directly congruent copies of it (that is, mirror images are not allowed) cannot have translational symmetry but must have screw rotational symmetry of infinite order in one direction; if mirror images are allowed, then periodic tilings do exist. The existence of a, single aperiodic prototile, an aperiodic monotile, is an important open problem about tilings in euclidean spaces.

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Aperiodic protosets and their non-periodic tilings have truly amazing properties ([8, 17]). Three basic construction techniques for the protosets and their tilings have emerged over time. One type of construction involves decorated tiles and matching rules. For example, the two famous Penrose rhombs have their vertices colored black or white, and have orientations given to some of their edges; considered are only those tilings for which the colors at the vertices and the orientations of the sides of adjacent tiles match. In another type of construction, the aperiodicity is based on the existence of a hierarchical structure on the tilings. These tilings are denned by a substitution rule and an inflation process. The substitution rule tells us how each prototile is decomposed into tiles at a smaller scale, each a homothetic copy of some prototile by a fixed factor A"1; the tilings are then produced by an iterative process of subdividing tiles and expanding by the factor A. Finally, the projection method produces the tiles of a non-periodic tiling of Ed by projecting onto Ed certain d-faces of the Delone tiling that is associated with a lattice in a higher-dimensional space (superspace) En; the d-faces that are projected down are those whose projection onto the orthogonal complement of Ed in En lies within a certain "acceptance region" (window).

3. COMBINATORIAL APERIODICITY We now investigate a combinatorial version of aperiodicity in tilings by convex polytopes. In this, congruence of the tiles is replaced by combinatorial equivalence of the tiles. We are interested in locally finite tilings T of Erf by (non-decorated) convex polytopes each combinatorially equivalent to one polytope from a given finite set P of convex polytopes, the combinatorial protoset of combinatorial prototiles of T. Again, P is said to admit the tiling T. As before, we use similar terminology also for polytopes and sets of polytopes that are under consideration for being combinatorial prototiles or combinatorial protosets. The tilings T with a single combinatorial prototile are called monotypic. Throughout we insist on convexity of the tiles. Combinatorial prototiles have been studied in detail in [14, 15]. For each d > 3 there are polytopes, called nontiles, which are not (single) combinatorial prototiles of monotypic tilings of E^ that are face-to-face. On the other hand, each convex 3-polytope does admit a monotypic tiling of 3-space; however, this tiling will not be face-to-face in general. A finite protoset P of convex d-polytopes is called combinatorially aperiodic if P admits a locally finite face-to-face tiling T by combinatorially equivalent convex copies of the prototiles, but if no such tiling has a. combinatorial automorphism of infinite order. Thus an automorphism of a locally finite face-to-face tiling with protoset P must have finite order.

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Note that we are not requiring the stronger condition that each tiling T must have a finite combinatorial automorphism group F(T). This condition would lead to a stronger notion of combinatorial aperiodicity than is intended here. Very little seems to be known about general properties of combinatorial automorphisms of face-to-face tilings T that are not induced by isometries of the underlying space. For example, it does not seem to be obvious that an automorphism of T of finite order must leave one face of T invariant. If a combinatorially aperiodic protoset admits a face-to-face tiling by congruent copies of the prototiles, then such a tiling cannot have a symmetry of infinite order (because this would yield a combinatorial automorphism of infinite order); that is, in the context of face-to-face tilings, a combinatorially aperiodic protoset is also isometrically aperiodic provided it does admit a tiling by congruent copies. Combinatorial aperiodicity requires a property to hold for a larger class of tilings (and in this sense it is stronger than isometrical aperiodicity), but it does not require the existence of a tiling by congruent copies (and in this sense it is weaker). Every tiling of the real line by closed intervals has a combinatorial automorphism of infinite order, namely the "shift" in one direction. Thus there is no combinatorially aperiodic protoset on the line. The same is also true for the plane but is not quite as obvious.

Theorem 3.1. There is no combinatorially aperiodic protoset in the plane.

Proof. Let p, q be positive integers such that - + - < |, and let {p,q} denote the regular tiling of the hyperbolic plane by regular p-gons, q meeting at each vertex. Then there exists a tiling T of the euclidean plane which is combinatorially equivalent to {p,q} ([9, 15]). To construct T we begin with a convex p-gon which contains the origin o as an interior point. We now add new convex p-gons such that, after each step, the resulting finite patch of tiles is star-shaped with respect to o (every ray emanating from o intersects the boundary of the patch in exactly one point), and every vertex on the boundary is in at most two tiles. More precisely, if x is a vertex on the boundary of an existing patch which is contained in two tiles, we add q — 2 new p-gonal tiles with vertex x, which, together with the two existing tiles, fully surround x; with some care, this new patch is again star-shaped, and each new vertex is in at most two tiles and sufficiently far away from the origin. The latter condition guarantees that the iteration of the process will yield a tiling T' of the entire plane. The tiling is indeed combinatorially equivalent to the hyperbolic tiling (p, q} because it is "freely generated"

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from p-gons where q meet at each vertex. Note that the combinatorial automorphism group of T' is isomorphic to the symmetry group G of {p, q} and thus contains elements of infinite order. Now suppose that P := {-Pi,... , Pn} is a combinatorially aperiodic protoset in the euclidean plane, with Pi a pi-gon for each i. The above arguments show that we cannot have n = 1; in fact, if p := p\ and q is chosen properly, we obtain a tiling T := T' by pi-gons which has an automorphism of infinite order. We can also rule out the case n > 2. It is not difficult to see that, if q is even, each p-gonal 2-face of the hyperbolic tiling {p, q} is a fundamental region for the subgroup H of G that is generated by the reflections in the sides of the 2-face (see, for example, [12, p.269,275]). In fact, G is the semidirect product of H and the symmetry group of the 2-face. Clearly, this subgroup H contains elements of infinite order. Now, if we dissect some 2-face P in an edge-to-edge manner into smaller polygons, then the images under H yield an H -invariant tiling in which every old 2-face is dissected in exactly the same way. If each smaller polygon is a p;-gon for some z, with each i actually occurring, we then have an edge-to-edge tiling of the hyperbolic plane, in which each tile is a pi-gon for some i, again with each i occurring, and which admits H as a group of symmetries. We now apply this observation with p :— (p\ + ... -\-pn} — 2n, using the following dissection of the p-gon: split the boundary of the p-gon into n edge-disjoint paths, with the ith path consisting of pi — 2 edges, and then join their endpoints to the center of the p-gon. In our final step we employ the isomorphism between {p, q} and T7, and pass from the resulting tiling of the hyperbolic plane to a corresponding tiling T of the euclidean plane by dissecting each tile of T' into convex p;-gons as dictated by the isomorphism. Then T is a locally finite face-to-face (edge-to-edge) tiling of E2 with protoset P. Moreover, by construction, T has combinatorial automorphisms of infinite order, including those that correspond to elements of infinite order in H. D We now discuss higher dimensions. Problem 3.2. Are there combinatorially aperiodicprotosets in dim. d>3? Call a set {Pi,... , Pn} of convex rf-polytopes facet-forming if there exists a convex (d -f l)-polytope Q such that each facet of Q is combinatorially equivalent to some P;, and each P4 is combinatorially equivalent to some facet of Q. A vertex of a convex cf-polytope is simple if it is contained in exactly d edges. Theorem 3.3. Let P := {Pi,... ,Pn} be a facet-forming set of convex dpolytopes. If PI, ... ,Pn can be realized as the types of facets of a (d + 1)polytope Q which has a simple vertex, then {Pi,... ,Pn} admits a locally

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finite face-to-face tiling of Ed which is periodic. Thus P is not combinatorially aperiodic. Proof. We use the same idea as in [14, Thm.3]. Let x be a simple vertex of Q, and let T be the d-simplex spanned by the d + 1 neighbors of x in the boundary complex of Q; then T is the vertex-figure of Q at x. We now project the facets of Q which do not contain x, centrally from x into T. This yields a face-to-face "tiling" of T into poly topes each combinatorially equivalent to some PI. We then map T affinely onto the standard fundamental region (d-simplex) for the symmetry group of the cubical tessellation in Ed; this results in a corresponding tiling of the fundamental d-simplex. In the final step we apply all the symmetries of the cubical tessellation to generate a tiling T of the entire space. Clearly, T is periodic. Each tile in T is combinatorially equivalent to a facet of Q which does not contain x. If each prototile Pz is represented by one of these facets, then T is indeed a locally finite face-to-face tiling with protoset P. However, if there exists a prototile Pj which only is represented by a facet which contains x, then Pi will not occur as a tile of T. In this case we replace Q by a new polytope Q', for which each PI is represented by a facet not containing the simple vertex, and then apply the above construction to Q' instead of Q. To obtain Q' we adjoin a small projective copy of Q along a facet of Q which does not contain x (as described in, for example, [13, p.121]), and repeat this construction once more if necessary until the desired property holds. D The basic idea of the proof works in more general circumstances. If the (d + l)-polytope Q has a vertex-figure T that admits a suitable tiling T' of E rf , then we can map the "tiling" in T obtained by projection, into the tiles of T' to create a new tiling T with tiles comb, equivalent to PI, ... , Pn. We now describe a construction of tilings which shares some of the features of the construction of non-periodic tilings by inflation. Recall that a ddiagram V := {D} U C in Ed consists of a convex rf-polytope D and a finite (face-to-face) rf-complex C of convex polytopes, such that D is the union of the members in C, each proper face of D is a member of C, and each member of C intersects the boundary of D in a (possibly empty) member of C ([18]); we call D the support of T>, and the d-polytopes in C the facets of V. The Schlegel diagrams of convex (d-\- l)-polytopes yield examples of d-diagrams, but not every diagram is a Schlegel diagram. Lemma 3.4. Let V := {Di} U C\ be a d-diagram in Ed with support D\, and let DQ be a facet of V contained in the interior of D\. Let : Ed H^ Ed be a similarity transformation that maps DQ onto D\. Define T := (U n > 0 Vn(Ci \ (A)})) U {A)},

T' := U neS (pn(Ci \ (A)}).

COMBINATORIAL APERIODICITY

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Then, a) T is a tiling o/E d , in which every tile is similar to a facet of C\, with every facet of C\ actually occurring; b) T' is a tiling o/Erf minus a point in the interior of DQ, in which every tile is similar to a facet of C\ \ {Do}, with every facet of C\ \ {-Do} actually occurring. Proof. For each integer n define Dn := (pn(D0) and Cn := (pn~l(C\); then {Dn} U Cn is a d-diagram, the image of T> under (pn~l. Then T and T' are obtained as the union of the complexes Cn \ {Dn-\}, in the case of T with n > 1 and with DQ added as the initial tile. It helps to think of the tiles in Cn \ {Dn-i} as the tiles of level n (they lie between the boundaries of Dn-\ and Dn}, and, in the case of T, of the initial tile DQ as an additional tile of level 1. Note that the poly topes Dn are not tiles, except for DQ in the case of T. By construction, each tile distinct from DO is similar to a c/-polytope in C\ \ {-Do}? this will generally not be true for the initial tile DQ of T, so that we must include -Do and take C\ instead of C\ \ {-Do}. It is immediate that T and T are actually tilings as stated. In fact, (p maps DQ onto the strictly larger set D\, and hence (p must be a similarity transformation that expands with a factor c > 1. If TO and RQ, respectively, are the inradius and circumradius of the polytope DQ, then the inradius and circumradius of Dn are rn = cnro and Rn = cnRo for each n. But rn —>• oo as n —» oo, and Rn —* 0 as n —> — oo, and therefore we have the desired tiling property. The point of Ed that lies in each set Dn is the singular point of T' (that is, the point not covered by T'}. D Let T be a locally finite face-to-face tiling of Ed (or a subset of Ed) by convex d-polytopes. A homomorphism of T (or of a subcomplex of T, respectively) is a mapping of the set of faces of T (or of the subcomplex) onto itself that is incidence-preserving. An automorphism of T then is a bijective mapping <£> of T for which both (f> and (f>~1 are homomorphisms. The set of all homomorphisms of T forms a semigroup. This is an analogue of the semigroup of self-similarities arising in the context of quasicrystals ([!]). The tiling T of Ed produced in Lemma 3.4 will generally not have any automorphisms; however, if we remove the initial (open) tile -Do, then we obtain an injective homomorphism of infinite order, namely the mapping induced by (f>. For the tiling T of Ed minus a point, this mapping is a genuine automorphism. Theorem 3.5. Let P :— {Pi,... ,Pn} be a facet-forming set of convex dpolytopes. Then P admits a locally finite face-to-face tiling of Ed, which has only finitely many similarity classes of convex tiles, and which has an injective homomorphism of infinite order if one (open) tile is removed.

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Proof. The result is a generalization and a stronger version of [14, Thm.2]. We apply the construction of Lemma 3.4 with a suitable initial diagram V := {£>!} U Ci and initial facet DQ of C\. We begin by constructing a (d + l)-polytope Q with a pair of facets that are translates of each other. Let Q' be any convex (d + l)-polytope such that each facet of Q' is combinatorially equivalent to some Pi, and each P{ is combinatorially equivalent to some facet of Q'. We now alter Q' as follows. First, by adjoining small projective copies of Q' along facets of Q' if need be (again see, [13, p.121]), we can assume that Q' has a pair of disjoint facets, and that every Pi (still) occurs as the type of a facet of Q'. Second, by applying a suitable projective transformation which maps the intersection subspace of the corresponding pair of supporting hyperplanes to infinity, we can also assume that these two disjoint facets are parallel. Third, with yet another projective transformation we can move a hyperplane that is parallel and very close to one of these facets, to infinity; the effect is that the resulting polytope Q" now has a pair of parallel facets, where one facet, F (say), is so huge that the orthogonal projection onto the affine hull of F maps the set Q" \ F into the interior of F. Finally, to obtain the desired polytope Q we reflect Q" in the affine hull of F, and adjoin Q" and its mirror image along F. Then Q has a pair of facets (parallel to F) that are translates of each other. Moreover, each facet of Q is combinatorially equivalent to some P4, and each Pi is combinatorially equivalent to some facet of Q. We now take a suitable Schlegel diagram of Q. If FQ and FI are the two distinguished facets of (Q, we project from a suitable point beyond FI, onto the supporting plane of FQ. The result is a rf-diagram D whose support D\ is the image of F\ under the projection; the initial facet is given by DQ := FQ and is homothetic to D\. Now Lemma 3.4 applies. D The last two theorems support the following conjecture. Conjecture 3.6. A facet-forming set of convex d-polytopes is not combinatorially aperiodic. The second part of Lemma 3.4 can sometimes be used to construct arbitrarily large (in a combinatorial sense) patches of tiles in situations when a global tiling of space is not known to exist. For example, it is not known if there is a monotypic face-to-face tiling of Erf all of whose tiles are d-cubes with one vertex cut off. On the other hand, there is such a tiling of Erf minus a point. In fact, if we cut off one pair of antipodal vertices of the (d + 1)cube by hyperplanes orthogonal to the corresponding connecting diagonal, we obtain a convex (d -f l)-polytope Q which has as facets two simplices and Id -f 2 cubes with one vertex cut off. Then, if we project from a point beyond one of these two simplices, we arrive at a Schlegel diagram V of Q

COMBINATORIAL APERIODICITY

405

to which Lemma 3.4 applies (with simplices DQ and D\). Moreover, if the projection point is on the diagonal of the original cube, the facets of V distinct from DQ and D\ form only two congruence classes, and hence the tiles of the corresponding tiling form only two similarity classes. In dimension 3, there is a lot of freedom to choose the metrical shape of the tiles within a combinatorial equivalence class of tiles. This is generally not true in dimensions d > 4. The answers to Problem 3.2 may thus be different in the two cases d = 3 and d > 4. We conjecture that, if d > 4, there are indeed protosets of convex c?-polytopes which admit locally finite face-to-face tilings, but only tilings with automorphisms of finite order. In other words, we have Conjecture 3.7. There are combinatorially aperiodic protosets in dimensions d > 4. In fact, there might even exist a combinatorially aperiodic protoset for which every tiling has a trivial combinatorial automorphism group. It would also be interesting to settle the existence of a combinatorially aperiodic monotile (single combinatorially aperiodic prototile). As we mentioned before, there are examples of convex polytopes in dimension 3 which are not (single) prototiles of monotypic face-to-face tilings ([14]). On the other hand, every convex 3-polytope that is simplicial (has triangular 2-faces) does admit such a tiling ([9]); hence, for the class of simplical 3-polytopes, no obstruction arises from the condition to admit a tiling. It would already be interesting to decide the question for simplicial 3-polytopes. Problem 3.8. Does every simplicial convex 3-polytope admit a locally finite face-to-face tiling o/E3 which has a combinatorial automorphism of infinite order? A similar question may of course also be asked for finite protosets consisting of simplicial convex 3-polytopes. Note that there are simplicial 3polytopes which are not facet-forming ([13]); therefore we cannot expect to obtain a complete answer to Problem 3.8 by solving Conjecture 3.6. We have not addressed normality in the above. (Recall that a tiling by convex polytopes is normal if its tiles are uniformly bounded.) If we require the face-to-face tilings T to be normal, the picture changes completely and we arrive at a different notion of combinatorial aperiodicity. For example, there are simplicial 3-polytopes that do not tile face-to-face and normally ([5]), and hence these are excluded from the onset because they do not admit a tiling. It is also not known if every facet-forming set of convex rf-polytopes admits any normal tiling of Ed at all, let alone a normal face-to-face tiling.

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Another variant of the problem may allow tiles which are topologically more complicated, like general solid handlebodies, knotted or unknotted. Examples of such tilings have been described in, for example, Kuperberg [11].

REFERENCES [I] Baake, M. and Moody, R.V., Similarity submodules and semigroups, in Quasicrystals and Discrete Geometry, Patera, J. (ed.), Fields Institute Monographs, Vol. 10, Amer. Math. Soc. (1998), 1-13. [2] Culik, K., II and Kari, J., An aperiodic set of Wang cubes, J. Universalis Comp. Sci. 1 (1995), 675-686. [3] Danzer, L., Three-dimensional analogues of the planar Penrose tilings and quasicrystals, Discrete Math. 76 (1989), 1-7. [4] Danzer, L., A family of 3D-spacefillers not permitting any periodic tiling, in Aperiodic '94, World Scientific (1995), 11-17. [5] Danzer, L., Griinbaum, B. and Shephard, G.C., Does every type of polyhedron tile three-space, Structural Topology 8 (1983), 3-14. [6] Goodman-Strauss, C., An aperiodic pair of tiles in En for all n > 3, Europ. J. Combinatorics 20 (1999), 385-395. [7] Goodman-Strauss, C., Open questions in tiling, Preprint (January 2000). [8] Griinbaum, B. and Shephard, G.C., Tilings and Patterns, Freeman & Co. (1986). [9] Griinbaum, B., Mani-Levitska, P. and Shephard, G.C., Tiling threedimensional space with polyhedral tiles of a given isomorphism type, J. London Math. Soc. (2) 29 (1984), 181-191. [10] Kari, J., A small aperiodic set of Wang tiles, Discrete Math. 160 (1996), 259-264. [II] Kuperberg, W., Knotted lattice-like space fillers, Discrete Comput. Geom. 13 (1995), 561-567. [12] McMullen, P. and Schulte, E., Flat Regular Polytopes, Annals of Combinatorics 1 (1997), 261-278. [13] Perles, M.A. and Shephard, G.C., Facets and nonfacets of convex polytopes, Acta Mathematica 119 (1967), 113-145. [14] Schulte, E., Nontiles and nonfacets for euclidean space, spherical complexes and convex polytopes, J. Reine Angewandte Mathematik 352 (1984), 161-183. [15] Schulte, E., Tiling three-space by combinatorially equivalent convex polytopes, Proc. London Math. Soc. (3) 49 (1984), 128-140. [16] Schmitt, P., An aperiodic prototile in space, Informal notes (1988). [17] Senechal, M., Quasicrystals and Geometry, Cambridge Univ. Pr. (1995). [18] Ziegler, G., Lectures on Polytopes, Springer-Verlag (1994). E-mail address:

[email protected] ''Egon Schulte''

SEQUENCES OF SMOOTHED POLYGONS

G. C. SHEPHARD 17 Mill Green, Stoke Holy Cross, Norwich NR14 8PB, England, U.K.

ABSTRACT. Starting from a given polygon, consider the sequence obtained by taking the midpoints of the edges of each polygon as the vertices of the next polygon in the sequence. The mapping from one polygon to the next is usually known as the "midpoint map". Such sequences have been studied for over a century. Here we consider sequences in which each vertex of a polygon is any prescribed linear combination of the vertices of the polygon which precedes it. A general method is given for determining the behaviour of such sequences of polygons, and all possibilities are described and illustrated. This behaviour is more complicated, and therefore more interesting, than that of, for example, sequences of numbers. The method depends on the expression of a polygon as a vertex sum of regular components.

1. INTRODUCTION We begin by presenting the basic concepts, notation and terminology which we require. These differ slightly from those used by previous authors on this subject. By a rooted n-gon P we mean a column vector Pv = bo,^i,...,t; n -i] T of complex numbers V{. We interpret these complex numbers as points in the complex plane, and refer to them as the vertices of P. Clearly, any permutation of the vertices changes P into a. different rooted n-gon, so that not only is P oriented, but one of its vertices (the first) is distinguished, and we call it the root of P. In diagrams, the root vertex of P is denoted by a 407

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SHEPHARD

small black disc, and the orientation of P is indicated by an arrowhead on one of its edges, usually that joining the first (root) and second vertices. The n vectors obtained by cyclic permutation of the elements of the vector pv denote what we shall call associated rooted n-gons; each set of such n-gons corresponds to an oriented (unrooted) n-gon in the usual sense of this word. With the standard definitions of vector addition and scalar multiplication, the set of all rooted n-gons forms an n-dimensional vector space over the complex numbers. We denote this by Pn. Scalar multiplication has a simple geometrical interpretation: for any complex number x, xpv is the rooted n-gon obtained from P by multiplying it by the factor |x| and rotating it counterclockwise through the angle arg x. Addition of the vectors pv and qv of two rooted w-gons P and Q corresponds to an operation on the polygons which we call vertex addition (to distinguish it from other methods of adding polygons described in the literature) and denote by P=^Q. Let u = exp(27ri/n) be an n-th root of unity. Then we write R^n/l\ #( n / 2 ), . . . , J R( n /( n ~ 1 )) for the regular rooted n-gons and regular rooted star n-gons with vectors

(i) These will be called standard (rooted) regular polygons] they are centred at the origin and the root of each is the point 1. We include also the point n-gon R(n/Q} consisting of the point 1 repeated n times with vector For each rf, let 1^n'd' represent the one-dimensional subspace of Pn generated by the vector r^n'd'. By the above remark on scalar multiplication we see that the vectors of 1?(nld) (d = 1,2, . . . , n — 1) are all the regular (n/cf)-gons centred at the origin and of any circumradius. Similarly, 7£(n/°) is the one-dimensional subspace generated by r^71/0); its vectors are the point n-gons each consisting of a single point repeated n times. Clearly, any two distinct subspaces Ti(nld^ intersect in the zero vector, and as the sum of the dimensions of all these subspaces is n, which is the dimension of Pn so we deduce the existence of the direct sum

pn =

(2)

It follows that for every rooted polygon P, the vector pv can be written uniquely in the form (3)

Pv

= cor<»/°> + cir("M + c 2 r<»/ 2 ) + • • • + c^r^1)),

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409

for suitable complex numbers Q, or, in geometrical terms, Theorem 1. Every rooted n-gon can be expressed uniquely in the form (4)

P = C-o=H=C f 1 =H=...=tt=C n _i,

where Cd (d = 1,2,... , n — 1) is the regular (n/d)-gon centred at the origin with vector c^r^^^ and CQ is the point polygon with vector CQr^/Q\ that is, the point CQ (the vertex centroid of P) repeated n times. 60, Ci,... , C n _i are called the regular components of P, and, in particular, Cd (a regular (n/cQ-gon) is called the dth component of P. As P is completely specified by the coefficients CQ,CI, ... ,c n _i, we write PC - ( c o , c i , . . . , c n _ i ) T where the subscript C means that elements of the vector are the coefficients in (3) in contrast to the subscript V in pv in which the polygon is specified by its vertices. If, for some d, Cd = 0 then we say that the cf-th component is absent or missing from this representation of P. In Figure 1 we show an example of a pentagon and its regular components. In Figures 2, 3, 4 and 7 the polygon (labelled the Initial Polygon) has just two regular components, namely the pentagon C^5/1) oriented counterclockwise, and the pentagram C(5/3) oriented clockwise. The polygons of Figures 5, 6, 8 and 10 have four regular components as indicated. From now on, it is convenient to consider polygons for which the 0-component is absent. The presence of such a component simply corresponds to a translation of the polygon. As will be seen from Figure 1, the point polygon Co consists of the vertex centroid of P repeated n times. It might be thought that Theorem 1 is artificial in that it relates to rooted polygons (in contrast to unrooted polygons). This is not so, for if two rooted polygons are associated, then each regular component of one is associated with the corresponding regular component of the other. On the other hand, given a set of regular (unrooted) regular n-gons, choosing their roots in all possible ways and taking their vertex sum can lead to as many as nn different rooted n-gons. Though as we shall see (Proposition B), all these will have the same signed area. Write fi for the n X n matrix whose (z,j)-th element is u/J where u = exp(27rz'/n). Here the integers i,j, denoting the rows and columns of 0, run independently from 0 to n — 1 (and not, as is more usual, from 1 to n). fl consists of the column vectors (n/O) ' V

(n/l) 1 V

r (n/(n-l))

' ' ' ' ' V

stacked side by side. Clearly pv = fip c , which leads immediately to a computationally simple way of determining the regular components of any

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SHEPHARD

given polygon P, namely pc = $l~lpv. Here, the (i,j)-th element of O"1 is w-*J'. The following results are of interest, but as these are not used in the sequel, we omit proofs. That stated in Proposition B is both surprising and unexpected, see [10]. Proposition A. Pn is an inner product space and the n subspaces 'R,(n>d> are mutually orthogonal. Here the inner product is defined in the usual way by (x | y) - ^Xiyi

where x - (x 0 , • • • ,-^n-i) and y = (y 0 , • • • ,2/n-i).

Proposition B. If P is expressed in terms of its regular components as in (4), then A(P] = A(C0) + A(Ci) + • • • + A(Cn-i) where A(P) means the signed area of the polygon P. 2. LINEAR TRANSFORMATIONS If A is an n X n matrix then a transformation qv — Apv transforms a rooted n-gon P into another such polygon Q. However, without restrictions, this definition is not satisfactory since it depends critically on the choice of root of P. It is clearly desirable that associated polygons should map into associated polygons. Consider the n X n matrix Z defined by

{

1 if j = i + l(modn) t 0 otherwise.

Then for any rooted n-gon P, Zpv is an associated polygon. The corresponding change of root for Q leads to a vector Zqv, so for associated polygons to map into associated polygons, Zqv = AZpv, or (Z~lAZ)pv = qv ~ Apv, that is, A commutes with Z. Hence, from now on, we shall restrict attention to linear transformations represented by matrices which commute with Z. Such matrices are well-known; they are called circulant matrices [4] and may be written (5)

B = b0I + 6jZ + b2Z2 + • • • + bn^Zn~l

6 n _l

(6)

6r,_9 /t— Z

&o bn /i

?2

1 1

61

..

"U

6n

•

03

••

&n-i •

&n /t —' I.

00 /

SEQUENCES OF SMOOTHED POLYGONS

411

The matrix B is specified by its first row, namely the vector bp = (60,61,... ,& n -i) which is traditionally known as a smoothing vector. (The subscript F is to remind us that the elements of the vector are the "first row" of the matrix B.) The corresponding linear transformation is called a smoothing operation, and is denoted by S(bp}One smoothing operation which has been extensively studied, and we shall discuss later, is the mid-point map with smoothing vector (|, ^ , 0 , 0 , . . . ,0). This transforms a polygon P into one whose vertices are the mid-points of the edges of P. The sum of two smoothing operations S(bp] and S(cp) can be defined as S(bF + cp) and their product as the product of the corresponding matrices B and C. That this is a smoothing operation, and moreover is commutative, follows from the representation (5). In other words the set of smoothing operations corresponds to polynomials in Z over the complex numbers with the relation Zn = I. We deduce: Theorem 2. With the above addition and multiplication, smoothing operations on n-gons form a linear associative algebra of degree n over the complex numbers. If the vertex centroid of P is at the origin, that is, P has no 0-component, then for all vectors bp the smoothed polygon S(bp)P also has vertex centroid at the origin. In particular, if P is a regular (n/oQ-gon, then every smoothing of P is also a regular (n/d)-gon with the same value of d. If P is represented by one of the vectors in (1) or (I'), then using the smoothing operation represented by the matrix (6) we see

t-0

or S(b}R(n/d"> = \dR(n/d"> where n-l

(7)

, ,dt

Ad = t-Q

This is equivalent to the statement that the vectors in (1) and (!') are eigenvectors of every smoothing matrix, and the corresponding eigenvalues are given by (7) (see [4]). It is sometimes convenient to specify a smoothing operation S(bp) by its eigenvalues, and we write

where the subscript E stands for "eigenvalues". From (7) we see that &£ = £lbp where fi is the matrix defined in the previous section. Hence

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SHEPHARD

the smoothing vector bp can be recovered from its list of eigenvalues by bp = Q~lbE- There is thus an analogy between smoothing vectors bp, bp and the vectors pv, pc which specify polygons. However, whereas multiplication is defined for smoothing vectors, it is hard to see any geometrical significance in the corresponding operation on rooted polygons. Using the eigenvalues A^, it is simple to exhibit the result of smoothing a given polygon P. If P is expressed in terms of its regular components as in (4) (see Theorem 1), and bp is any smoothing vector then S(bp}P = A0C0 + Aid

+ ' ' • + \n-lCn-i.

So, if we apply this smoothing operation t times we arrive at (8)

S'(6F)P = AjCo + Aid + • • • + A^Cn-i

(9)

5 4 (6 F )P - Aicofl< n/0 > + XfaRWV + • • • + A^cn-i^"-1".

This last equation will be fundamental in our treatment of the iteration of smoothing operations. We conclude with a simple result. Proposition C. // the components of the smoothing vector bp are all real, then for d ^ 0, and, if n is even, d ^ n/2, the eigenvalues occur in complex conjugate pairs: Aj= A n _; fori = l , 2 , . . . , [ n / 2 ] . Further, AQ and, if n is even, A n / 2are

rea

l-

3. ITERATION OF SMOOTHING OPERATIONS We now consider the sequence of polygons {St(bp)P}^.0 that arises by repeated smoothing of a polygon P using a given smoothing operation S(bp}. Unlike the theory of sequences of real or complex numbers, the behaviour of such sequences is more complicated (and interesting). To describe this we need two concepts. Say that two sequences of polygons {Pt}^.Q and {Qt}^.0 are asymptotic if p(Pt,Qt) —> 0 as t -» oo. Here p is any metric on the space of rooted polygons, for example, if P = ( V Q , V I , . . .,^ n -i) and Q = (WQ,WI,. .. , w n _ i ) then we may write p(P,Q) — sup \V{ — W{\. The second concept we require is that of a standard sequence. This is a sequence of the form (9) in which, for each value of d = 0 , . . . , n — 1, either Cd = 0 or | Ad | = 1. As stated above, we shall assume that every polygon has no 0-component. For a given P and smoothing vector bp with eigenvalues A^ = mj(exp(o^?'), let M(P, 6) = sup*(md) where the * signifies that the supremum of the absolute values rrid is to be taken over all values of d for which the components

SEQUENCES OF SMOOTHED POLYGONS

413

are present in P (that is, c& ^ 0 in relation (4)). The set of d for which md = M(P, 6) is denoted by D and the corresponding A^ are called dominant eigenvalues of the smoothing operation S(b). We begin with the following simple result: Theorem 3. (a) // M(P,bF) > 1 then the sequence {S^bp^P}^ diverges, that is, the polygons in the sequence become larger and larger without bound. (b) If M(P,bp) < 1 then the sequence {St(bp)P}^:Q converges to (is asymptotic to) a point polygon in which all the vertices coincide with the origin. Proof. For part (a) we see that if M(P, bp) > 1 then there exists a value of d such that md > 1 and Q ^ 0. Hence the coefficient A*Q increases without bound. So does the
in the form (9), that is, Pt = £ ^dCdR^n/d\ and let {Qt} be the standard d=o sequence with

(10)

Qt

then Pi and Qt differ only in terms for which the eigenvalues A^ satisfy \\d\ < 1- Hence A^ —> 0 as t -» oo, and we deduce that p(Pt,Qt) -*• 0. Thus {Pi} is asymptotic to the standard sequence {Qt}D Notice that Qt is constructed by deleting the terms from Pt for which Xd< 1. In view of the above, and to avoid the trivialities of Theorem 3, from now on we shall employ a process called normalisation of a smoothing vector

414

SHEPHARD

when applied to a given polygon. This simply means that we multiply the smoothing vector by a factor 1/ra where m = supm^, thus ensuring that d£D

the dominant terms have eigenvalues with modulus 1. We now describe the possible types of standard sequences {Qt}- Write Qt in the form (10), where A^ = m ( fexp(a t ;z), and say that a^ (in radians) is rational if it is a rational multiple of TT, or equivalently, a^ is a rational number of degrees. There are four possibilities: (a) All the a^ (d £ D] are equal to 0, so that the eigenvalues A^ = 1 for d £ D. Then the polygons Qt of the standard sequence {Qt}^0 are all congruent to a fixed polygon Q. We call this a constant sequence. (b) The angles ad (d £ D) are all equal but non-zero. Then the polygons Qt are rotations of a fixed polygon Q about its vertex centroid. There are two subcases: (case 6j) If the angle of rotation ad is rational with denominator 5, then the sequence repeats after s steps and we call it rotationally periodic. If, on the other hand, (case 62) the angle of rotation a^ is irrational, then no repetition occurs and we say the sequence is rotationally non-periodic. (c) The angles ad (d £ D) are rational but not equal, then the polygons in the sequence repeat after a certain number of steps, but they are not rotations of a fixed polygon. This is case c\ and we say that the sequence is periodic. The same applies if the ad (d £ D] are irrational, but all their differences are rational (case 02). In this case we say the sequence is nonperiodic. (d) At least two of the angles ad (d £ D) differ by an irrational angle. Then the polygons Qt are all different; there is no repetition or periodicity of any kind. In this case we say the sequence is divergent. All these possibilities are illustrated by examples in Section 5. Using the above we can now state the procedure for determining the behaviour of any sequence {Pt} of polygons obtained by repeatedly applying a normalised smoothing vector to a given polygon P: From bp determine b^ and apply a normalising factor. Hence, from the coefficients Cd of the regular components of P, find the set D of dominant eigenvalues. Construct the standard sequence {Qt}^-0 as above. The behaviour of {Pt} as t —^ oc is described by the type of the standard sequence

{Qt}£o-

Since smoothing operations are represented by circulant matrices, it follows that every vertex of polygon in the sequence {Pt} is exactly the same linear combination of the corresponding vertices in the preceding polygon. Hence sequences of unrooted polygons can be classified into the same four types (a), ... , (d) above. However the periods may differ. For example, in Figure 3, the rooted polygons of the standard sequence rotate with period 5; the corresponding unrooted polygons are constant. In Figure 5 the

SEQUENCES OF SMOOTHED POLYGONS

415

rooted polygons have period 10, but the unrooted polygons have period 2. Generally, where both rooted and unrooted polygons are periodic, then their periods will differ by a factor equal to some submultipole of n.

4. AFFINE REGULARITY A polygon is called an affine regular (n/d]-gon if it is the image of a regular (n/fl?)-gon under an affine transformation (a projective transformation that preserves the line at infinity, or, equivalently, a linear transformation plus a translation). Examples of affine regular polygons appear in Figures 8, 9 and 10. A regular polygon P is inscribed in a circle S. An affine transformation T maps P into an affine regular polygon PI inscribed in an ellipse Si, which is the affine image of S under T. We require the following. Theorem 5. An n-gon with two regular components of the form P = n-d (for d — 1,2, . . . , [(n — l)/2}) is affine regular. Proof. If d ^ 0, d ^ n/2 the r-th vertex of Cd^Cn-d is of the form vr = cdudr + cn-duj~~dr . Write cd = m^exp(^0 and cn_d = ran With u> = exp(m) where a = IK /n, we obtain vr = md exp(0d + rda)i + mn-d exp( n _d - rda)i = md exp(V> + X + rda)i + mn^d exp(V> — X ~ where 2^ = 4>d + (j)n-d and 2% = 4>d ~ n-d- Hence (11) vr = exp(^z')(radexp(x + rda)i + mn-dox$(— (x + rda)i) = expd>i}((md + mn_d} cos(x + rda] + i(md - m n _d) sin(x + rda)}. Since, for r = 0 , . . . , n — 1, the points cos(x + rda) -f- isin(x + rda] are the vertices of a regular (n/cf)-gon inscribed in a unit circle, the points represented by the second factor in (11), namely (md + mn_d) cos(x + rda} + i(md - m n _ r f ) sin(x + rda), are the vertices of an affine regular (w/cQ-gon Q inscribed in an ellipse with semiaxes (m^ + <mn_d) and (rad — m n _ n _^)/2. D The next result is the analogue, for affine regularity, of Theorem 1. Corollary 1. Every polygon can be written uniquely in the form

where each Ad is an affine regular (n/d)-gon.

416

SHEPHARD

Proof. If d = 0 (a point polygon) or, if n is even and d = n/2, let Ad = Tl(n/d) which are also affine regular. For other values of d let Ad =• K(n/d)_y_'fr(n/(n-d))f Then the theorem i mp ii es t h a t Ad is the space of affine regular (w/cf)-gons. Pairing off the terms in (2) in this way leads to

from which we obtain the unique representation, analogous to (4),

of P as a vertex sum of [(n + 2)/2] affine regular n-gons.

D

Corollary 2. // A^ and Xn-d are complex conjugates of modulus I, and P = Cd^Cn-d is an affine regular n-gon, then so is PI — X^Cd^ Xn-dCn-dMoreover, P and PI, have the same circumscribed ellipse. Proof. Let A^ = exp(#i), \n-d — exp( — #z), then a calculation similar to that in the proof of the theorem yields, instead of (11), vr - exp(i/)i)((md

+ m n _ d )cos(x + rda + 9}

+ i(md - m n _ d ) sin(x + rda + 0)) from which the statement follows immediately.

D

Affine regular (n/d}-gons like P and P1? which are inscribed in the same ellipse, are said to arise, one from the other, by an affine rotation. More precisely, we may say that PI is obtained from P by an affine rotation through angle 9. The theorem and its corollaries have immediate consequences for standard sequences, and hence for the behaviour of sequences of smoothed polygons: // there are two dominant eigenvalues \d and \n-d, then the polygons in the standard sequence are all affine regular (n/d]-gons, and the sequence {Pn} of smoothed n-gons is asymptotic to such a sequence. If the two dominant eigenvalues \d and \n-d are complex conjugates, then the polygons in the standard sequence are all affine regular (n/d)-gons which are affine rotations of the initial polygon. We distinguish two cases, that in which the affine rotation is rational (case e\) and that in which it is irrational (case e
SEQUENCES OF SMOOTHED POLYGONS

417

Examples illustrating these statements appear in Figures 8, 9 and 10. The remark about unrooted polygons at the end of the previous section applies to these sequences also.

5. EXAMPLES We now give nine examples of sequences of smoothed polygons which illustrate all the possibilities listed in the previous two sections. These are adequately illustrated by using pentagons (labelled Initial polygons in the diagrams) namely Pi,P 2 ,P3 and P4 defined by their regular components as follows: Pi: Pc = (0,10,0, -5 + 4t,0), P2 : pc = ( 0 , 9 + 9i,-10-4«,-10,6i), P3 : pc = (0,1 + 3i, 3 + 6«, 2 + i, -6 - 2t), P4 : pc = (0,7, -3 - 6i, 1 + 2i, 0). These are shown, with their regular components, in the Figures 2 to 10. Each example uses a different smoothing vector. In each case we give the bp vector, the set £), the normalisation factor n, and the arguments a:; of the dominant eigenvalues. These arguments are given in degrees (rather than radians) both for computational convenience, and also because it enables one to see immediately which angles are rational and which are not. The irrational angles are given to six places of decimals, and although, in some cases we cannot give a proof, it seems reasonable to assume irrationality in all such cases. Figure 2: Polygon PI, smoothing vector bp = (2,1,0,0,1), D = {!}, n — 1/2.618034, OL\ — 0°. The sequence converges very rapidly to a standard (constant) sequence all of whose terms are congruent to a fixed regular pentagon. (Case (a) of Section 3.) Figure 3: Polygon PI, smoothing vector bF = (1 + z, 1 + z, 0,0,0), D = {!}, n = 1/2.288246, a\ = 81°. The sequence is asymptotic to a standard sequence of regular pentagons of which successive terms are rotations through angle «i, and hence the sequence of smoothed polygons is rotationally periodic with period 40. (Case (b\] of Section 3.) The corresponding sequence of unrooted polygons has period 8. Figure 4: Polygon PI, smoothing vector bp = (2,1,0,0,0), D = {!}, n = 1/2.497212, <*i = 22.386178°. The sequence is asymptotic to a standard sequence of regular pentagons of which successive terms are rotations through angle ai, and hence is rotationally non-periodic. (Case (6 2 ) of Section 3.) The same applies to the corresponding sequence of unrooted polygons.

418

SHEPHARD

Figure 5: Polygon P2, smoothing vector bF = (2,1,2,0,0), D = {1,2,3,4}, n = 1/2.236068, Oi\ — ~a4 — 72°, a3 = — a2 =• 36°. The sequence is asymptotic to a standard sequence of period 10, though as unrooted polygons, the period is 2. (Case (GI) of Section 3.) Figure 6: Polygon P3, smoothing vector bp = (2 + 4i, 1 + 2z,2 + 4i,0,0), D = {1,2,3,4}, n = 1/5, <*i = -44.565051°, a2 = 27.434949°, a3 = -80.565051°, a4 = -8.565051°. These angles are irrational, but all their differences are rational. Hence the sequence is asymptotic to a standard sequence which is non-periodic. (Case (c 2 ) of Section 3.) The corresponding sequence of unrooted polygons consists of two distinct polygons which rotate through an irrational angle. Figure 7: Polygon PI, smoothing vector bp — (2cos36°, 2cos72°, 1,0,0), D = {1,3}, n = 1/1.543362, 01 = 49.613822°, «3 = 22.386177°. As the difference between these angles is irrational, the standard sequence has no periodicity or repetitions of any kind. The sequence of smoothed polygons is therefore divergent as is the corresponding sequence of unrooted polygons. (Case (d) of Section 3.) Figure 8: Polygon P2, smoothing vector bp — (1,1,0,0,0), D — {1,4}, n = 1/1.618034, a\ — —04 = 36°. This is the mid-point map. The standard sequence, to which the sequence of smoothed polygons is asymptotic, consists of regular affine pentagons, successive terms of which are obtained by an affine rotation of 36° and hence the standard sequence has period 10. It will be observed that iterates 13 and 14 are almost identical with iterates 3 and 4. The corresponding sequence of unrooted affine regular polygons has period 2. (Case (EI) of Section 4.) Figure 9: Polygon P4, smoothing vector bF = (0,1,0,0,2), D = {2,3}, n = 1/2.497212, a2 = -a3 = 13.613822°. The standard sequence, to which the sequence of smoothed polygons is asymptotic, consists of affine regular pentagrams, successive terms of which are obtained by an irrational affine rotation. (Case (e 2 ) of Section 4.) The same is true of the corresponding sequence of unrooted polygons. Figure 10: Polygon P3, smoothing vector bp = (2 + 2i,0,1 + z,0,0), D = {2,3}, n = 1/3.531591, a2 = 22.613822°, a3 = 67.386178°. The standard sequence, to which the sequence of smoothed polygons is asymptotic, consists of affine regular pentagrams, but these are not obtained by an affine rotation of a fixed polygon. (Case (/) of Section 4.) The same is true of the corresponding sequence of unrooted polygons.

SEQUENCES OF SMOOTHED POLYGONS

419

6. COMMENTS AND REFERENCES The earliest paper concerning sequences of smoothed polygons seems to be that of Darboux [5] in 1878. He investigated the midpoint map bp — (1,1,0,0,..., 0). Independently the same material has appeared many times, for example, Roseman [9], Huston [6], Cadwell [2], [3] and Berlekamp et al. [1]. The latter paper discusses the shapes (convexity or self-intersections) of the polygons in the sequence. Cadwell also considers sequences in which the vertices of a polygon are obtained be dividing the sides of a polygon in a prescribed ratio bp = (p, 1 — p,0,0,... ,0). He also considers the case in which the vertices are the bisectors of a diagonal of a polygon, bp = (1,0,1,0,... ,0). Cadwell also gives an equation equivalent to our (9). Another special case can be found in Kasner [7] and Kasner & Comenetz [8]. Here the vertices of each polygon is the centroid of a specified number of vertices of the previous polygon in the sequence. Schoenberg [10] investigates not only the midpoint construction, but sequences constructed using a general cyclic transformation. He shows that a polygon can be expressed as a vertex sum of regular polygons, a result also to be found in Berlekamp et al. [1]. The analogous problem for skew polygons has also been investigated, see [1], [3], [5], [10] and [11].

420

SHEPHARD

Figure 1

SEQUENCES OF SMOOTHED POLYGONS

Initial polygon

421

Iterate 1

Iterate 4

Figure 2

422

SHEPHARD

Iterate 2

Iterate 4

Figure 3

SEQUENCES OF SMOOTHED POLYGONS

423

Initial polygon

Iterate 1

Iterate 2

Iterate 3

Iterate 4

Figure 4

SHEPHARD

424

Initial polygon

Iterate 1

Iterate 2

Iterate 4

Iterate 5

Figure 5

SEQUENCES OF SMOOTHED POLYGONS

425

Initial polygon

Iterate 1

Iterate 2

Iterate 3

Iterate 4

Figure 6

\

426

SHEPHARD

Iterate 2

Iterate 6

SEQUENCES OF SMOOTHED POLYGONS

427

Iterate 4

7 Iterate 13

Iterate 14

Figure 8

SHEPHARD

428

Iterate 6

429

SEQUENCES OF SMOOTHED POLYGONS

Iterate 6

Iterate 7

Figure 10

430

SHEPHARD

REFERENCES [I] E. R. Berlekamp, E. N. Gilbert and F. W. Sinden, A polygon problem, American Math. Monthly, 72 (1965), 233-241. Reprinted in Selected Papers on Algebra, Mathematical Association of America, Washington, B.C., 1977. [2] J. H. Cadwell, A property of linear cyclic transformations, Math. Gazette, 37 (1953), 85-89. [3] J. H. Cadwell, Chapter 3 in Topics in Recreational Mathematics, Cambridge, 1966. [4] P. J. Davis, Circulant Matrices, Wiley-Interscience, New York, 1979. [5] G. Darboux, Sur un probleme de geometrie elementaire, Bull. Sci. Math., (2) 2 (1878), 298-304. [6] R. E. Huston, Solution to Problem 3547, American Math. Monthly, 184185. [7] E. Kasner, Centroidal polygons and groups, Scripta Math., 2 (1934), 131-138. [8] E. Kasner and G. Comenetz, Groups of multipoint transformations with application to polygons, Scripta Math., 4 (1936), 37-49. [9] M. Rosenman, Problem 3547, American Math. Monthly, 39 (1932), 239. [10] I. J. Schoenberg, The finite Fourier series and elementary geometry, American Math. Monthly, 57 (1950), 390-404. [II] I. J. Schoenberg, The harmonic analysis of skew polygons as a source of outdoor sculptures, in The Geometric Vein: the Coxeter Festschrift, pp. 165-176, Springer, New York, 1981. [12] I. J. Schoenberg, The harmonic analysis of skew polygons as source of outdoor sculptures, Chapter 9 in Mathematical Time Exposures, Math. Association of America, Washington D.C., 1982. E-mail address:

[email protected]

(<

G . C. Shephard''

ON A PACKING INEQUALITY BY GRAHAM, WITSENHAUSEN AND ZASSENHAUS

JORG M. WILLS Department of Mathematics, University of Siegen D-57068 Siegen, Germany

ABSTRACT. In 1972 Graham, Witsenhausen and Zassenhaus proved a tight packing inequality for 0-symmetric convex discs. Although this inequality is rather special, it is crucial for finite packings in the plane and for infinite nonlattice packings. Their proof is rather elaborate and uses a lot of angle classification and calculation in Minkowski geometry. We give an angle-free proof only based on the distance function.

In 1972 Graham, Witsenhausen and Zassenhaus [5] proved a tight packing inequality for 0-symmetric convex discs. Although this inequality is rather special, is is the crucial step in the proof of a fundamental inequality for finite packings, which was first proved by N. Oler [6]. Special cases and weaker results were proved before by L. Fejes Toth (cf. [2]), by Rogers [7] and Groemer [3]. For details cf. [2], [4] or [5]. The general proof in [5] is very elegant, but the proof of the special case is rather elaborate and uses a lot of angle classification and calculation in Minkowski geometry. We give an angle-free proof only based on the distance function. For this let K be an 0-symmetric convex body in the plane and let \\x\\ be the norm induced by K, i.e. with 1) ||Ax|| = A ||x|| for A > 0, 2) \\x + y\\ < \\x\\ + \\y\\ and 3) ||ar|| = 0 <S> x = 0. So "length" is meant with respect to this norm or the distance function of K. Further A(A') denotes the critical determinant (for lattice packings of A'). 431

432

WILLS

Theorem (Graham, Witsenhausen, Zassenhaus 1972) Let Q be a convex quadrangle, such that any diagonal is at least as long as any side. Then

where s,t are the lengths of an appropriate pair of adjacent sides and s',t' the lengths of the two other sides. Proof. We first solve the obvious case, when Q is a parallelogram: Lemma 1. Let P — conu(0,x,y,x + y) be a parallelogram with s = \\x\\, t = ||y||, 0 < s < t < \\x - y\\ and t < \x + y\\. Then (1)

V(P) > &(K)st

Proof. Let P be the given parallelogram and P' the parallelogram with edges parallel to those of P and with edgelengths 2. By assumption the diagonals of P' are > 2. Hence P' is an elementary cell of a packing lattice of A', so V(P') > A(A'). Let A be the linear map with P = A(P'), which preserves orientation. As A maps each ray from 0 into itself, one has V(P) = V(P'}st and so V(P) > A(A>i. For the general case we need three lemmas and the following construction: If Q — conv(zi,£2,£3,£4) is the given quadrangle, then its diagonal £2^4 cuts Q into two triangles X",X" with x\ G X", £3 G X%, and x\x^ cuts Q into X2,X4 with x2 G X2, x4 6 X'4. Reflection of X'2 at x2 and X'4 at x4 yields X'2 and X'4. Now P = \j]'4 Xt with Xi = X-' — Xi is a parallelogram with vertices y;, i = 1,... ,4. The segments 0?/2- are translates of the edges of Q, hence 2 < ||y;||, i = 1, ... ,4 and the edges of P are translates of the diagonals of Q, hence (2)

max | |y; 1 1 < min||y t - - y,-+i||. i i Here and in the following the indices are considered mod 4, i.e. y$ = y\. Now let Zi = Oyi n bdK, i = 1,. . . ,4. Then ||^-|| = 1, i = 1,. . . ,4. Lemma 2.

\\zt - Zi+i\\ > 1,

z= l,...,4

Proof. Indirect. Let j^j — 2j+i|| = a < 1 for some j G [1,4]. Without restriction let \\yj\\ < H ^ + i H . With t/j+i = /'^;+i we have t/j = (1 — A)//ZJ with some A G [0,1). Hence yj - y]+l = X(-yj+l) + (1 - \}(nzj - yj+i) and \\yj - yj+i|| < A/z + (1 — A)^a < ^ = ||yj+i|| which contradicts (2).

ON A PACKING INEQUALITY

433

Lemma 3. Let x,y,z £ bdK, \\x-y\\ < 1 and z on the shorter arc (x,y) of bdK . Then \\x-z\\ < 1. Remark. As y ^ —x, the shorter arc (z,y) is well defined. Proof. We consider the 0-symmetric hexagon H = conv(±x, ±y, ±(x — y ) ) . Then ± z , ± y £ bdK and ±(x - y) £ mi A', i.e. H C K. This implies that the shorter arc (x, y) lies in the triangle T = conv(x, y, x + y). So 2 £ mi T. Hence z — x £ int H C int K and \\z - x\\ = \\x — z\\ < I. Lemma 4. // \\Zj + £j+i|| < 1, for some j £ [1,4], then \\Zj + 2j-i|| > 1 and \\zj+l + Zj +2 || > IProof. Indirect. We assume that for some j £ [1,4] : \\zi +zi-i\\ < 1

and

\\zi + zi+i\\ < 1

For z = — Zj we then have \\z- Zj-i\\ < 1 and ||^ Now either Zj+% = z or Zj+i lies in one of the shorter arcs of ( z , Z j - i ) or (z,Zj+i) of bdK. From lemma 3 follows z

3+2 ~

Z

J-1

<

Or

which contradicts lemma 2. This proves lemma 4.

Now the proof of the theorem is obvious: Lemma 2 and 4 show that lemma 1 can be applied to the parallelogram P or, equivalently, to the triangles Tj = conv(0,2zj,22j + i) and Tj+z = conv(0,2^j+2,2zj+3). Hence and

For the corresponding triangles Xj, Xj+2 of P we get V(Xj)

>

£

&(K)st

and

V(Xj+2) >

£

With V(Q] = V(Xj) + V(Xj+2) this proves the theorem

434

WILLS

REFERENCES [1] J.H. Folkman and R.L. Graham, A packing inequality for compact convex subsets of the plane, Canad. Math. Bull., 12 (1969) 745-752. [2] G. Fejes Toth and W. Kuperberg, Packing and Covering with convex sets, 799-860 in: Handbook of Convex Geometry, Vol. B, North Holland, Amsterdam, 1993. [3] H. Groemer, Uber die Einlagerung von Kreisen in einem konvexen Bereich, Math. Z. 73 (1960) 285-294. [4] P. Gritzmann and J.M. Wills, Finite Packing and Covering, p. 861-897 in: Handbook of Convex Geometry, Vol. B, North Holland, Amsterdam, 1993. [5] R.L. Graham, H.S. Witsenhausen and H.J. Zassenhaus, On tightest packings in the Minkowski plane, Pacific J. Math. 41 (1972) 699-715. [6] N. Oler, An inequality in the Geometry of Numbers, Acta Math. 105 (1961) 19-48. [7] C.A. Rogers, The closest packing of convex two-dimensional domains, Acta Math. 86 (1951) 309-321. E-mail address:

[email protected] ''Jorg M. Wills''

COVERING A TRIANGLE WITH HOMOTHETIC COPIES

ZOLTAN FUREDI1 Department of Mathematics, University of Illinois at Urbana-Champaign 1409 W. Green Street, Urbana, IL61801, USA and Renyi Institute of Mathematics of the Hungarian Academy of Sciences Budapest, P. 0. Box 127, Hungary-1364.

ABSTRACT. Let A0 be a triangle and let H = {Ai,..., An} be a family of positive homothetic copies of AQ. We prove a conjecture of A. Bezdek and K. Bezdek (1984) that if the total area of triangles from H is at least twice the area of AQ, then there exist translates of AI, . . . , An that cover AQ.

1. TRANSLATION COVERINGS Let C be a disk, i.e., a convex, compact set on the Euclidean plane with interior points. Let H = {Ci,..., C,-,... } be a (finite or infinite) sequence of disks. We say that H permits a translation covering of C if there exist translations r,- such that C C U t -r,-(C;). Moser and Moon [11] showed that if Q is the unit square and H is a family of squares of sizes x\, x^, •.. with total area ^ x\ > 3 and with sides parallel to the sides of Q, then H permits a translation covering of Q. This is the best possible bound as one can see from the case when x\ = X? = x% = 1 — £, £4 = • • • = 0. L. Fejes Toth proposed the following question. Suppose that each d is a positive homothetic copy of C. How large the sum of areas of the sets Ci must be, so that C can be covered by translates of these sets? Denote by f ( C ) the ratio of the minimum (infimum) of ]T Area(C;) to Area(C). The above cited theorem of Moon and Moser states f ( Q ) = 3. This easily implies f ( C ) < 12 for every disk (A. Bezdek and K. Bezdek [1]) and it was recently improved to f ( C ) < 22/3 by Januszewski [8]. One can observe that for every Supported in part by NSF grant DMS 0140692 and OTKA grant T 032452. 435

436

FUREDI

C one has f ( C ) > 2 (two copies of sizes 1 — e can't cover C). A. Bezdek and K. Bezdek [1] conjectured that this is achievable for every triangle A. Here we prove this conjecture and show that /(A) = 2. Theorem 1. Let AQ be a triangle and let H be a family of positive homothetic copies of AQ. Suppose that ^ A e H Area(A) > 2 Area(Ao). Then there exist translates of the members of H that cover AQ. The minimum density of a covering a (large) convex region with small equal sized positive homothetic triangles is conjectured to be 3/2 and this can be obtained by a hexagonal lattice arrangement. (See, e.g., in the excellent monograph of Pach and Agarwal [12]). So the factor 2 (probably) cannot be decreased below this value, even if the triangles are all small. In contrast, in the case of the unit square, Q, Moon and Moser showed that a total area (1)

>l + 2zi

ensures a translation covering. However, it seems to be impossible to imitate such hexagonal arrangements if the triangles have different sizes. So we use the method of Moon and Moser, the triangles of H are ordered in layers with minimum overlaps. This way we are able to keep track the area of the covered region. However, in these arrangements, we only use a small part of most of the triangles A € H, namely, an inscribed parallelogram, and waste the rest of the A. As every inscribed parallelogram has area at most | Area (A), we waste a lot and apparently loose a factor of 2 immediately. So proving /(Ao) < 2 seems impossible by denning layers and the placements of the triangles in this way, but as we can see in the next sections, still can be done with extreme care. (The real improvements comes by adding the triangle A s +i, see below in Section 2). One can generalize Moon and Moser's theorem for every convex disk C. Denote by ^(C) the infimum of the densities of coverings the plane by translated congruent copies of C. Let H be a family of positive homothetic copies of C. Theorem 2. For every e > 0 there exists a 6 = 8(e,C) > 0 such that the following holds. If each member of H is smaller then 6C and for the total area we have Y^tfeH Area(#) > (^(C) + £)Area(C), then there exist translates of the members o/H that cover C. The proof of this Theorem is straightforward and it is postponed to Section 5. 2. CONSTRUCTING THE COVERING OF A 0 The aim of this section is to define a translation of each member of H. In the next section we will show that they really cover AO.

COVERING A TRIANGLE WITH TRIANGLES

437

As translation coverings are affine invariant, we may suppose that AQ (and all members of H) are isosceles, right angled triangles, with angles 90°, 45° and 45°. We may also suppose that the coordinates of the vertices of AQ are (0,0), (1,0), and (0,1). Let a(A) denote the length of the sides of the triangle A. We have Area (A) = |a(A)2 and for the total area (2)

Area (A) > 2 Area(A 0 ) = 1.

We order the triangles in decreasing order by their sizes and form disjoint subgroups Hi,H2 ... of H, at the same time we define right angled trapezoids Ti,T2,... such that some translations of the members of H^ cover Tk. The base lengths of Tk are denoted by b^-\ and &&, its heights is hk, hk = bk — bk-i and side lengths h^ and \/2/^, with vertices (0,1 — bk-\), (0, 1 - 6 fc ), (&*, 1 - 6jb), and (bk-i,l - &*-i). The first group, HI, consists of a single triangle of the largest size from H, let us denote its side length by 61. Place this largest triangle such that it covers the upper part of AQ, i.e., define the (now degenerate) trapezoid T\ by its three vertices (0, 1), (0, 1 — 61), and (61, 1 — b\). Whenever k > 2 and HI , . . . , H/t-i had already been defined together with TI, . . . , Tk_i then we proceed as follows. Consider the rest of the triangles R/j := H \ (Hi U • • • U Hfc_j ), and denote by a\ > a^ > . • . the lengths of the sides of the members of R^, with at- = a(A;). If R^ is empty the procedure stops. Also if

a(A) < 6^_! := b,

(3)

then our procedure decomposing H stops. Otherwise let s be the maximum integer, such that (4)

£

(at- - ia.) < b.

l
Here s > 2, because a\ < b. We have that (5)

b<

(If |Rfc| = 5, then here we introduce as+i = 0, and take A s +i as a triangle of side length 0). Define H^ as the family of these s + 1 largest triangles of

R;.Now we are ready to define the trapezoid Tk and its translation covering by Hfc. As b — bk-i is already known, the only thing is needed to define Tk

FUREDI

438

FIGURE 1. The trapezoid Tk is covered by the members of its heights hk := h. Define h by the following equation (6)

P (a; - h)+ °s+1

=b

'

Adding (6) and (5) we obtain (after simplification) that (7)

ias+1 < h.

Similarly, comparing (6) to (4) one obtains that (8)

h < -as + —as+i 2 2s

For 1 < i < s we place A; such that its apex vertex (the vertex with the right angle) lies on the line y = l-b — h(=l — 6^). Then the triangle A; meets the line y = 1-6 in an interval /; of length a;-/i (> 0 by (8)). Place the triangles such a way that /i, / 2 , . . . ,/s-i form a continuous segment starting at the y axis, i.e., the apex vertex of At- is at (]Ci<j<; aj — (i—l)h, 1 — 6 — h ) for 1 < i < s — 1. These triangles cover T^ except a parallelogram of horizontal side length as — h -+- |a s _)_i. Translate As such that /s covers the other end of the upper base [(0,1 — 6), (6,1-6)] of the trapezoid Tk, i.e., the apex vertex of As is at (6+/i-a s ,lb — h). Then the uncovered region, Tk \ Ui<;< s A z -, is a homothetic copy of A s+ i (with afactor —1/2), a triangle with vertices (6 — (as — h) — |as+i, 1 — 6), (6- (as - h), 1 -6), and (6- (as -h),l — b— ^a s+ i). (These points are in T^, by (7)). There is a unique translate of A s +i covering this region, its apex must go to (6 — (as — h) — ^a s+ i, 1 — 6 — ^as+\}. We obtained that

COVERING A TRIANGLE WITH TRIANGLES

439

3. AREA ESTIMATES Suppose that the above procedure stops after I steps, i.e., H = HI U • • • U H^ U R^+i and we have obtained the trapezoids 7\, . . . , T(, with heights / & ! , . . . , /ty and bases bk = Si<» 1, i.e., our procedure results in a translation covering of AQ. Let Uk (vki respectively) denote the size of a largest (a smallest) triangle in Hfc, (1 < k < £), and u^+i is the size of the largest triangle in R^+I. If R^+ 1 = 0, then we take Uf+i = 0. Using the notations of the previous section Uk = «i and Vk = as+iElementary calculation shows that (4)-(6) imply that oa'? ^

(9)

aib +

^

+ h(b +k

^~

The proof of this inequality is purely algebraic, it does not use any geometry, therefore it is is postponed to the next section as Lemma 1. The sum of the middle two terms in (9) is, obviously, twice the area of Tk- So reformulating (9) we obtain the following upper estimate for the total area of the members of Hfc (2 < k < I). ]T Area (A) < ~ukbk-i + 2 Area(T fc ) - -vkbk. A6H fc

Since ui = vi > u% > V2 > • • - > Uk > v^ > • • • > vi > ui+i , we have that (10)

Area (A) < -ukbk-i + 2 Area(r fc ) - -uk+ibk.

The above inequality holds for k = 1, too, because 61 = u\ and we can define

]P Area(A) - -b\ < -&i(2&i - «2) = ~uib0 + 2 Area(Ti) - -u2bi. Adding up (10) for all k one obtains that Area (A) < 2 I V

(11)

= 6? -

Area(T fc ) )

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Moreover (3) implies that Area (A) < -ue+i X ]T

a(A) < -ue+lbe.

Adding this to (11) we obtain that Area (A) < &|. This and (2) imply 6^ > 1, so we obtained that HI U • • • U H^ really forms a n translation covering of AQ.

4. PROOF OF AN INEQUALITY Lemma 1. Suppose that b > a\ > a? > • • • > as > 0 and as > as+i > 0 are real numbers and s > 2 is an integer satisfying (4) and (5), i.e., .

.

\

•>

,

_L

,

Define h as in (6), z'.e., 6y Z^i
+ hb + hb + ^ "

Proof. It is high school algebra. First, observe that the upper bound part of (12) implies that h > |a s +i- Similarly, the lower bound part of (12) implies that (13)

s ( i a 5 - / i ) + i a s + 1 > 0,

thus as > |as -f ^a s +i > h. We distinguish two cases. If \as > h, then (|a; + /i)(az- - h) = \a\ + /i(|az- — h) > \a\ because |a; > ^as > /i. We get the following lower bound for the sum of the first two terms of R:

\

>

As the sum of the last two terms of R, i.e., (/i - |as+i)(6 + ft) is positive we obtain the desired lower bound for R.

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441

Suppose that h > \as. R - h(h - -a s+ i) = -aib + 2hb - -as+ib > b(-ai +1h- -as) (^,

I

},

Vi<.-<-

(ai ~

LX 1a

^

(^

1 A

) ^— I —a\ + 2hL —2—aJs I 2 s+i I x V2

/

X ( -(ai + a,) + 2(h - -a,) , , , X -(oi + a,)

!, + a,) X -(ai

[•^as -h}-}- -a i ] X 2 ( h - -a, z s+ / V ^ Here the last four lines are positive. The first line is at least Xli as, in the third Z)i 0 and in the fourth line the first factor is positive by (13). We obtain that ^ 1? / 1 _ 1 \ 1 2(h--

-at-

2

D

5. THE CASE OF SMALL COPIES Here we sketch the proof of Theorem 2. First, we fix a direction on the plane, say the line L is parallel to the x axis. We may suppose that Area(C) = 1 and its perimeter is bounded (less than 8, say), otherwise we use an affine transformation. We also know that 1 < $ < 3/2. (This upper bound is due to Besicovitch [2] and usually attributed to Fary [4], see again, e.g., [12]. The idea is that every disk contains a large symmetric hexagon). Next, (a slight generalization of Moon and Moser's proof gives that) there exists a ^i > 0 such that the following holds. If Qi,$2? • • • are squares with

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sides parallel to L, each of them has size at most 6\ and total area exceeding (14)

then these squares permit a translation covering of C. The definition of i? implies that there exists an integer no such that, if Ci, . . . , Cn are congruent copies of C, n > no = no(<s,C) and Q is a square with (15)

n Area(Ci) > (tf + if)Area(Q), 8 then {Ci, . . . ,Cn} permits a translation covering of Q. Define 8 =: (I/no) X Consider H the family of positive homothetic copies of C with members smaller than 8C. If the size of H 6 H is between (1 + ^e)~kC and (1 + |)~ fc+1 C, then replace it with a homothetic copy of C of size exactly (1 + ^e)~kC. Do this with every member of H. Obviously, it is sufficient to make a translation covering of C using the members of the obtained new family, H'. We have that

If H' contains ariQ + 6 copies of the same size, Ci, . . . ,Cano+b, where a and b are integers with no > b > 0, then leave out 6 of those and replace the rest with a squares Qi, . . . , Qa of the same sizes and with sides parallel to L such that equality holds in (15). Then these squares can be covered by the copies of C they replace so we are done if we show that the obtained new family H" (consisting only of squares) can cover C. ^From each size in H' we have deleted at most UQ — 1, so their area altogether is at most

(n0 - 1) x ( max Area(#') )x V ^

'

\H>€W

'/

(l + -£)~k < -£. 8

8

Thus we get for the area of the squares in H ^ Area(#") > — Area(H') #»eH" + 8£ V Then (14) finishes the proof.

e > Area(C) + £. / n

Naturally, we can extend this proof for every dimension. 6. CONCLUSION, REMARKS Note that our Theorem holds for infinite family of triangles satisfying (2). Indeed, almost all of our inequalities are strict (most notable Lemma 1), so

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443

after a few steps in the procedure defining the trapezoids, one does not need total area 2, a slightly less will do. This can make H finite. One can get a slightly better (i.e., thinner) covering of AQ if in the definition of Hjt not only As+1, but A s + 2,..., A2S-i too are placed between At and A^+i along the upper base line of Tk.. However, the density of the obtained covering will be very very close to 2, even if the triangles are small. As every planar convex set C can be sandwiched between two homothetic triangles AI C C C A2 with ratio at most 9/4 (see [5]), Theorem 1 implies f(C) < 2(A2/Ai) 2 < 10.125. However, we can easily get a better bound from inscribed and circumscribed parallelograms. Let Ci,C2,... be a family of positive homothetic copies of C with Ci = X{C and £),- x\ > 8. One can find parallelograms P' C C C P" such that P' = XP" and A > 1/2. Further, Pi Q Ci C P/'. Then 1 + 2xi\ < X2(Y,ix*}- Tnus Moon-Moser's theorem (1) implies that P" can be covered by some translates of the parallelograms P/ giving

f(C) <

+

< 8.

This is rather close to the best known general upper bound, f ( C ) < 22/3 (see [8]). If H can contain positive and negative homothetic copies of AQ then to ensure a translation covering one needs total area at least 4 Area(Ao). This was conjectured by K. Boroczky and proved by Januszewski [7] and for the more special case of H being a finite sequence purely of homothetic copies of —AO was proved earlier by Vasarhelyi [13]. She also considered translation coverings of the triangle AQ when H consist of homothetic copies each of them rotated by a certain angle (f>, see [14]. Moon and Moser's result was rediscovered by Groemer [6] and A. Bezdek and K. Bezdek [1]. They both extended it to higher dimensions proving that f(Q^) = 2d — 1, where now Q^ is the d-dimensional cube. An algorithm for packing or covering a given set K with a family of sets {Ci} is an on-line method if the sets Ci are given in sequence, and Ci+i is presented only after Ci has been put in place, without the option of changing the placement afterwards. Januszewski, Lassak, Rote, and Woeginger [9] proved that in Euclidean d-space, every sequence of cubes of total volume greater than or equal to 2d + 3 can cover the unit cube in the on-line manner. A slightly better estimate is proved by Lassak in [10]. These volume bounds are astoundingly good, considering the best possible bound of 1d — 1 for the analogous off-line problem.

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7. THE CASE OF CUBES Finally, the author cannot resists to add here a lovely short proof for f(Q^) < 2d,

(16)

where Q is the d- dimensional unit cube, a slightly weaker result than the optimal f(Q^) = 2d - 1 proved in [1] and [6]. The idea of this proof appears even in the paper of Moon and Moser [11], but not in this explicit form. Proof. Let H be a (finite or infinite) family of cubes homothetic to Q^d> with total volume at least 2d. We define the translation covering in two phases, first trimming and then gluing. If the side length a of H £ H is between 2a < a < 2a+1 where a is a (usually negative) integer, then replace H with a cube, #', with side length 2a. We obtain the family H' of sizes powers of 2. We have that Vol(ff') > //'6H'

£

Vol(H) > 1.

W6H

Now, whenever we find in our family 1d cubes of the same size we glue them together to get a cube of double of that size. Finally, we arrive at a family of cubes, H", all sizes are powers of 2, where every size is represented at most 2d — 1 times. If the largest size is at most 2~ 1 , then for the total volume we get vol(#") < (2d -

£Vol(#') = £ #'€H'

#"eH"

However, this total volume exceeds 1. So the largest size in H" is at least 2° = 1, so the cubes of H' can perfectly cover Q^ and we are done. n

8. ACKNOWLEDGMENTS The author is greatly thankful to A. Bezdek for thoroughful discussions about the problem of coverings. The author is also indebted to I. Barany, J. Pach and I. David Berg for helpful conversations, to P. Brass and M. Lassak sending a copy of [3] about triangle approximations, to K. Boroczky, Jr. for pointing out an error in the references, and to the referee for a number of helpful comments.

REFERENCES [1] A. BEZDEK AND K. BEZDEK: Eine hinreichende Bedingung fiir die Uberdeckung des Einheitswiirfels durch homothetische Exemplare im ndimensionalen euklidischen Raum. (German) [A sufficient condition for the covering of the unit cube by homothetic copies in the n-dimensional Euclidean space] Beitrdge Algebra Geom. 17 (1984), 5-21.

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[2] A. S. BESICOVITCH: Measure of asymmetry of convex curves. J. London Math. Soc. 23 (1948), 237-240. [3] P. BRASS AND M. LASSAK: Problems on approximation by triangles, (manuscript). [4] I. FARY: Sur la densite des reseaux de domaines convexes. (French) Bull Soc. Math. France 78 (1950), 152-161. [5] R. FLEISCHER, K. MEHLHORN, G. ROTE, E. WELZL, AND C. YAP: Simultaneous inner and outer approximation of shapes. Algorithmica 8 (1992), 365-389. [6] H. GROEMER: Covering and packing properties of bounded sequences of convex sets. Mathematika 29 (1982), 18-31. [7] J. JANUSZEWSKI: Covering a triangle with sequences of its homothetic copies. Period. Math. Hungar. 36 (1998), 183-189. [8] J. JANUSZEWSKI: Translative covering of a convex body with its positive homothetic copies. International Scientific Conference on Mathematics. Proceedings (Zilina, 1998), 29-34, Univ. Zilina, Zilina, 1998. [9] J. JANUSZEWSKI, M. LASSAK, G. ROTE, AND G. WOEGINGER: Online
OPEN PROBLEMS

ANDRAS BEZDEK 1 Department of Mathematics, Auburn University, AL 36849-5310, U.S.A and Renyi Institute of Mathematics, Hungarian Academy of Science Budapest, Hungary, H-1053

Authors of the volume Discrete Geometry: In Honor of W. Kuperberg's 60th Birthday and participants of the Discrete Geometry Special Session, organized by Andras Bezdek and held within the Annual Joint Mathematics Meetings of the American Mathematical Society in January, 2001 in New Orleans, Louisiana were asked to contribute to an open problem collection. This resulted in the following list of problems.

1. PROBLEMS Problem 1. ANDRAS BEZDEK, Auburn University, Auburn, AL and Renyi Institute of the Hungarian Academy of Sciences, Budapest, Hungary The well-known result of Tarski on covering a circular disk with planks (a plank or a strip is a plane region between two parallel lines) states that if the disk is covered by planks, then the sum of their width's is equal to or greater than the disk's diameter, and if equality occurs, then the strips can cover the disk only in parallel fashion. Tarski conjectured, and T. Bang [B] proved the following generalization: should a closed convex region K be covered by a collection of strips, then the sum of the widths of the strips must be at least the minimal width of K. Several further generalizations in various directions of this statement have been obtained, and we propose to investigate the following one, not yet considered: lr The author was partially supported by the Hungarian National Science Foundation, grant numbers T016131, T16387 and T38397

447

448

BEZDEK

It seems that in case of disks the same conclusion holds true even if the disk to be covered by the planks is reduced to an annulus by cutting a small hole in its center. In other words, if the planks can cover the annulus, then they can cover the whole disk as well. If this turns out to be true, then the natural question of the maximum size of the hole arises. Also, generalizations are possible to other (centrally-symmetric) convex regions, punctured in the center, and to higher dimensions. Only the case, when the convex region is a square is settled (see Bezdek [Be]). Problem 2. ANDRAS BEZDEK, Auburn University, Auburn, AL and Renyi Institute of the Hungarian Academy of Sciences, Budapest, Hungary According to standard terminology we say a family K of convex regions in the plane has property T(k) if any subfamily containing at most k members has a transversal. We also say that the family of sets has property T — k if there is a common transversal for all but at most k members of K. One of the most difficult open question is due to Katchalski and it is to show that if K is a family of disjoint translates of a given convex set, then T(3) implies T - 2. It was long believed that in a family of at least 6 unit disks T(4) implies T. As a result, in case of disks, that variation of the above problem of Katchalski, when one assumes T(4) was not even considered. Recently E. Goodman in a joint paper with R. Pollack, B. Aronov and R. Wenger points out that in contrary to the belief: even for arbitrary large families of unit disks T(5) is necessary to get T. A. Bezdek in 1991 constructed an arrangement to show that T(3) not necessarily implies T — 1. In view of this observation we suugest to revisit Katchalsky's problem in case of unit disks. Based on the above remarks we expect that T(4) implies T — 2 or even T — I. Problem 3. TIBOR BISZTRICZKY, University of Calgary, Calgary, Canada A d-multiplex M(d,n) is a convex rf-polytope with an ordered set of vertices XQXI, ... , xn, and the facets FQ, jFi, . . . , Fn given by

under the convention that Xj = 0 for j < 0 and Xj = xn for j > n. M(d, n) was introduced in [Bl] by Bisztriczky, and studied in [BBS] by Bayer, Bruening and Stewart, and in [D] by Dinh. Among other interesting properties, M(d,n) is a generalization of a d-simplex. Each face and each quotient polytope of M(d,n) is a multiplex, and M(d,n) is self-dual. Let P denote a convex rf-polytope with an ordered set of vertices. We say that P is multiplicial if each facet of P is (d — l)-multiplex with respect to the ordering. Clearly, M(d,n) and any simplicial polytope is multiplicial.

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At present, the only known examples of non-trivial multiplicial poly topes are the so-called ordinary (2m + l)-polytopes (see Bisztriczky [B2]). Do there exist non-trivial multiplicial cf-polytopes when d = 2m > 4? Problem 4. PETER BRASS, City College, New York, NY What is the maximum length of a simple polygon with n vertices, contained in a unit radius disc, for n odd? The maximum should probably be a supremum, the probable extremal polygons use the same edge many times, so they are not simple, but they are limits of simple polygons. When n is an even integer, the maximum length is In by the simple reason that each edge has length at most 2, and we can have each edgelength almost equal to 2 by going back and forth near a diameter of the circle. When n is an odd integer the extremal polygon probably is an "isosceles triangle" inscribed in the unit circle, where the longest side is the base and that is the limit of n — 2 sides of the n sided polygons (these sides go "back and forth"). The length of this longest side will be a solution of a numerical extreme value problem. Problem 5. ROBERT J. MACG. DAWSON, Saint Mary's University, Halifax, Nova Scotia, Canada Given two vectors vi = (x\,yi) and v% — (#2,2/2) in the plane, let L = {avi + bv-2 : a,6 € Z} be the lattice that they generate. There does not in general exist a rectangle which, together with its ^-translates, tiles the plane; but there always exist pairs of rectangles that do so. For instance, let R\ have one vertex at the origin and the other at (x\,y<2\ while #2 has one vertex at the origin and the other at (#2, —y\}- Counting degrees of freedom provides an easy argument against the existence of a one-rectangle solution that varies continuously as a function of the period vectors. (If xi/x% is rational, there does exist a single rectangle whose translates tile the plane; for instance, take the height to be the greatest common divisor of x\ and X-2-]

Conjecture: given a lattice in Rd generated by d arbitrary vectors, it requires dl "bricks" (presumably corresponding in some way to the terms of the determinant) to yield such a continuous tiling by translation. (In contrast, note that if the coordinates in at least d — 1 directions are commeasurable, a single brick still suffices!)

450

BEZDEK

Problem 6. FERENC FODOR, Bolyai Institute, Szeged University, Szeged, Hungary Let Pn be the convex n-gon of unit diameter whose perimeter is a maximum. Show that Pn always has axial symmetry. In particular, try to prove this when n =• 1k, where k > 3. Pn is known for those n which have an odd divisor. Also, P$ is known. The first unsolved case is n = 8. Analogous problems concerning maximal area are disscussed in [G] by Graham. Problem 7. ZOLTAN FiJREDl, University of Illinois at Urbana-Champaign, Urbana, IL Let n > k,d > 0 be integers. Determine S(n,d,k) the minimum number s of fc-dimensional affine planes AI , . . . , As C Rn such that they meet in an inner point every rf-dimensional face of the n-dimensional hypercube. The case k = n — I, d = I (hyperplanes cuts the edges of the cube) this problem was proposed by O'Neil [0]. He conjectured S(n, n— 1,1) = n which was established for small values. For n < 4 see the papers of Emamy-K. [E] and most recently the case n — 5 was established independently in [EUT] by Emamy-K., Uribe and Tomassini and in [Z] by M. Ziegler. However, it is known that 5(6,5,1) = 5 (see again [Z]), where the construction is due to Peterson (see in [Z, S]). Additivity implies S(n,n — 1,1) < [5n/6]. Impagliazzo, Paturi ands Saks [IPS] investigated S(n,n — l,d) (in inverse form). See the survey [S] of Saks. The case d — 1 belongs to the so-called stabbing number problems. For further references see, e.g., [AM] by Aronov and Matousek. Problem 8. RAJINDER J. HANSGILL, Centre for Advanced Study in Mathematics, Panjab University, Chandigarh, India Consider the line Ln^ through the origin and the point (1,..., 1,2,3, ...,rz) of Rd+k, where the number of 1's is k-}-1, k > I. Find those points which are closest to this line among those points whose coordinates are odd integers. For k = 0, it is known that (1,..., 1) is one such point and all others can be obtained from this in an obvious manner. For k — 1, it is seen that (!,...,!) is not a nearest point for some small values of n, but it appears to be so for all large n. Is it true that there exists N = n ( k ) such that for all n •> N, (1,..., 1) is in the set of points with odd integral coordinates which a r e nearest t o t h e line L 1

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451

Problem 9. DAN ISMAILESCU, Hofstra University, Hempstead, NY The following problem was posed by L. Fejes Toth in [F]: prove that if n great circles on a sphere are in general position, then they cannot divide the sphere too evenly - that is, the ratio between the areas of the largest cell and smallest cell is unbounded as n goes to infinity. In connection of this problem we would like to ask the following question (see also [I]): Consider n continuous functions /i,/2, • • •/n; each from [0,?r] into [0,1] and satisfying the following four properties: - each function has at most one extremum on (0,7r), - no three graphs pass through the same point, - /i(0) ^ /,-(0) and /.-(TT) ? /XTT) for i ^ j, - each pair of graphs cross exactly once over the interval (0,?r). Notice that the graphs partition the rectangle [0,7r] X [0,1] into 1 + ^2— cells. Let R(fi,..., fn] be the ratio of the area of the biggest cell and the area of the smallest cell. Is it true that liminf -R(/i, •.. ,/ n ) = +00? If this question has an affirmative answer, then it will immediately imply L. Fejes Toth's conjecture via Archimedean projection. Problem 10. JANOS KINCSES, Szeged University, Szeged, Hungary For a convex disk K (a compact convex set with nonempty interior) inside the unit circle C we define the angle function of K as ax(x} = the measure of the visual angle of K at the point x 6 C. We say that K is distinguishable, or shortly a P-set, if any other convex disk has a different angle function. Question 1. 7s it true that each convex polygon is a V-set? Question 2. 7s it true that most convex sets (in the sense of Baire category) are V-sets? For related results see Kineses [Kl] [K2] and Kurusa [KK]. Problem 11. GREG KUPERBERG, University of California, Davis, CA Question 1. The fatness of a 4-dimensional convex polytope is denned as 0(-P) = / ? p S + f ?p{? where fn(P) is the number of n-dimensional faces of P. We conjecture that the fatness of 4-dim. convex polytopes is bounded. Question 2. The fatness of a strongly regular, polygonal tiling T of a surface is denned as (T} — Jo(-L , / >)-rj2(Jf \ i / / T \) , where fn(T] is the number of n-cells of the tiling. Here a tiling of a surface is strongly regular if the intersection of two tiles is either empty, or a single edge or a vertex. What is the maximum fatness of f

/
452

BEZDEK

a strongly regular tiling of a closed surface of genus g? Or at least, what is its rate of growth as a function of g? Problem 12. KRYSTYNA KUPERBERG, Auburn University, Auburn, AL The following problem was stated by Kirby in [Ki] as Problem 5.17 by Freedman: Given a finite set of points X in dB3, let T be a tree in B3 of minimal length with dT = X. IsT unknotted, that is, is there a PL imbedded 2-ball in B3 containing T? Here the minimal tree means the Steiner minimal tree, i.e., nodes may be added. A knotted arc in such a tree would be a arc that together with an outside unknotted arc give a knot K in R3. In such case we say that the tree contains K. There is an example of a finite set of points X in dB3 whose minimal tree contains the trefoil knot answering Freedman's question (see [Ku]). The paper of K. Kuperberg [Ku] contains the following questions: 1. (M. Freedman) What does the set of fc-tuples in S2 whose minimal tree is unknotted look like in the &-fold product S2 X • • • x S2? In particular, what is the measure of this set? 2. (W. Kuperberg) What is the minimum number of vertices of order 3 in a knotted minimal tree for a finite subset of 52? The knotted tree described in [Ku] has 6 vertices of order 3. 3. (G. Kuperberg) There is a finite set whose minimal tree is knotted on the surface of an ellipsoid with one of the axes much shorter than the other two axes. What are the strictly convex closed surfaces in R3 containing a finite set whose minimal tree is knotted? In addition, interesting natural questions are: 1. Which knots can be realized in a minimal tree of a finite set on the unit sphere? It is known that any 2-bridge torus knot can be realized in such a minimal tree. 2. Is there a finite set on the unit sphere whose minimal tree is knotted but does not contain a knotted arc? Problem 13. WLODZIMIERZ KUPERBERG, Auburn University, Auburn, AL If a right prism (of finite height) whose base is a convex polygon tiles R3 by its congruent replicas, must the polygon tile R2? (Conjecture: yes.) Comment: It is necessary to assume that the prism is right. Allowing that the latteral edges of the prism are inclined to the base plane by an angle that differs from 90° by any £ > 0 makes it possible to produce a counterexample. The requirement that the base polygon be convex is necessary as well.

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Problem 14. WLODZIMIERZ KUPERBERG, Auburn University, Auburn, AL Classify all convex polyhedra K that can tile R3 by translates of K and translates of — K. Must each such K admit a double-lattice tiling? (A tiling is double-lattice if it is the union of a lattice arrangement A + K and its symmetric image A — K for some lattice A.) Comment: For the analogous problem in R2, it is known that a convex polygon K tiles the plane by translates of K and of —K if and only if K is a jo-hexagon, that is, a convex hexagon with a pair of parallel opposite sides of equal length (degeneracies allowed), and therefore it also admits a double-lattice tiling of the plane. Problem 15. WLODZIMIERZ KUPERBERG, Auburn University, Auburn, AL Determine the maximum number of unit circular cylinders of infinite height that can be brought in contact with a unit ball in R3. Comment: The number is conjectured to be six. Aladar Heppes and Laszlo Szabo [HSZ] proved that the number is at most eight, and Peter Brass and Corola Wenk [BW] improved it to: at most seven. There are several non-congruent configurations of six cylinders around the ball, but the topological space of all possible configurations of six cylinders has not been yet identified. Problem 16. WLODZIMIERZ KUPERBERG, Auburn University, Auburn, AL What is the smallest number of cubes of unit edge-length in Rd that can cover a larger cube? It is known that d + 1 unit cubes suffice, and it is conjectured that d + 1 are needed, too. Problem 17. WLODZIMIERZ KUPERBERG, Auburn University, Auburn, AL A packing of unit circles in the Euclidean plane is saturated if it cannot be improved by placing an additional unit circle anywhere, and it is completely saturated if it cannot be improved by removing any finite number of circles and replacing them with a greater number of unit circles. A saturated packing of unit circles is barely saturated if there is room for an addtional circle of radius 1 — e for every £ > 0. Question: Is there a circle packing that is completely saturated and barely saturated at the same time? Comment. This problem is related to a problem posed by Gabor Fejes Toth: Is there a completely saturated non-lattice circle packing? If there were no such packing, then the densest lattice circle packing would be uniquely determined by complete saturation. But observe that there is a non-lattice, completely saturated packing of squares...

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Problem 18. JOSEPH MALKEVITCH, York College, Jamaica, NY Let T be a triangulation of the plane with at least 4 vertices (i.e. T is a plane graph with every face a triangle). The following is conjectured: T can be realized by a convex 3-dimensional polytope P so all of the faces of P are isosceles triangles. The author proved that one can not hope for P to have all its faces congruent isosceles triangles. For more details on the subject see [M] by Malkevitch. Problem 19. JANOS PACK, GABOR FEJES TOTH, GEZA TOTH, Renyi Institute of the Hungarian Academy of Sciences, Budapest, Hungary A family T — {A\,A-2,... ,An} of convex bodies (compact convex sets) in rf-space is said to be in general position if no rf+ 1 of them have a common supporting hyperplane. T is said in con-vex position if none of its members is contained in the convex hull of the others, i. e., if bd conv^JF), the boundary of the convex hull of the union of all members of JF, contains a piece of the boundary of each A{. Evidently, any two members of T is in convex position. If any k members of J- are in convex position for some fixed k > 3, then we say that T has property P^. Generalizing the Erdos-Szekeres theorem, Bisztriczky and G. Fejes Toth [BF1] proved that there exists a function f ( n ) tending to infinity (as n —> oo) such that any family of pairwise disjoint convex bodies in general position in the plane, which satisfies property PS, has at least f ( n ) members in convex position. The lower bound on f ( n ) has been improved subsequently in [BF2] by Bisztriczky and G. Fejes Toth and in [PT1] by Pach and G. Toth. Much better lower bounds are proved under the stronger condition that T has property P^ for some fixed k > 3 in [TG] by G. Toth. Similar results are true for families of not necessarily disjoint plane convex bodies, provided that any two of them have at most two boundary points in common [PT2]. If the boundaries of the sets are allowed to cross four times, then no such theorem holds for families with property P^. Question 1. Is it true that for every j > 2 there exists a constant k — kj and a function f ( n ) = fj(n) tending to infinity satisfying the following condition: Let T be any family of n plane convex bodies in general position, any two of which have at most j boundary points in common. If J- has property P&, then it has at least f ( n ) members in convex position. In 3-space, for any k there exist arbitrarily many pairwise disjoint convex bodies with property P& such that only a bounded number of them are in convex position. On the other hand, Bisztriczky and G. Fejes Toth (unpublished) have some positive results under stronger conditions. A family of convex bodies in d-space is called separable if any two of its members can be

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separated from every third one by a hyperplane. In particular, Bisztriczky and G. Fejes Toth showed that there is a function g(n) = #3(71) tending to infinity such that every separable family of n convex bodies in general position in 3-space with property P-j has at least g(n] members in convex position. Question 2. For every k > 1 and j or every n construct a separable family of n convex bodies in general position in 3-space with property P^ such that not all of them are in convex position. Question 3. Is it true that for every d > 3 there exists a constant k = kj and a function g(n] — gd(n) tending to infinity satisfying the following condition: Any separable family of n convex bodies in general position in d-space with property P^ has at least g(n) members in convex position. Problem 20. EGON SCHULTE, Northeastern University, Boston, MA (1) An abstract polytope P is called neighborly if any two vertices of P are incident with an edge of P. Enumerate the regular or chiral abstract polytopes that are neighborly. (2) A pair of distinct vertices of an abstract polytope P is called a diagonal of P. The diagonals fall into diagonal classes, consisting of diagonals that are equivalent under the automorphism group of P. Enumerate the regular or chiral abstract polytopes with exactly two diagonal classes. For references see [MS] by McMullen and Schulte. Problem 21. VALERIU SOLTAN, George Mason University, Fairfax, VA It is well-known that any convex body K in the plane can be touched by six non-overlapping translates of K. Along this line we ask: Let A'i,... ,KQ be a family of six convex bodies in the plane. Can one of these bodies, say Ki, be touched by five non- overlapping translates of the j + K j t J 6 {!,... ,6}\{i}f If the answer is "yes", can we replace number six by seven? Problem 22. ISTVAN TALATA, Eotvos University, Budapest, Hungary The Hadwiger number H(K) of a d-dimensional convex body K C Rd is the maximum number of neighbours (^touching translates) of K occuring in packings with translates of K. The lattice kissing number HL(K] of K is defined analogously with the restriction considering only lattice packings with translates of K.

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Let K be a convex cylinder, obtained as the direct product of a (d— ^-dimensional convex body KQ and a segment s. It is well-known that H(K) = 3H(K0) + 2. It is also not difficult to show that HL(K) > 3HL(K0) + 2, just one has to consider the lattice packing of K whose translation vectors are direct products of translation vectors of those two lattice packings with translates of KQ and s, respectively, that realize the maximum number of neighbours. It is interesting that for lattice kissing numbers of cylinders we have only an inequality above and there seems to be no analogous way to get equality like in the proof of the mentioned formula for Hadwiger numbers of cylinders when HL(KQ) ^ H(Ko), which very well can be the case for a (d — l)-dimensional convex body KQ, when d > 4 (see [T] by Talata), Question. Is there a d-dimensional convex cylinder K = KQ x s for some d>4 such that HL(K) > 3# L (/i 0 ) + 2? Problem 23. JORG M. WILLS, Universitdt Siegen, Siegen, Germany Let Cn -f Bd, Cn = ( c i , . . . , cn} be a finite packing of unit balls in Euclidean d-space Ed. Let V denote the volume and p > 0 a (boundary controlling) parameter. A packing Cn + Bd, for which y(convCn + pBd] is minimal (for given d, n and p) is called dense. If Cn is a subset of a packing lattice, Cn + Bd is called a lattice packing; if the lattice is critical, Cn + Bd is called critical. Further dimconvC n is called the dimension of the packing. The following 3 problems are due to several authors (Betke, Gritzmann, Henk, Schnell, Schiirmann, Wills), who also gave partial solutions (e.g. for d = 2 in Schuermann's thesis 2000). Question 1: Is it true that dense sphere packings have extreme dimension, i. e. 1 or d for n > d; and 1 or n — 1 for n < d?. Question 2: Is it true that dense lattice packings of spheres are critical? Question 3: Is it true that dense sphere packings are nonlattice, if p and n are sufficiently large? Problem 24. JOSEPH ZAKS, University of Haifa, Haifa, Israel We say that a dimension d > 2 is special if every unit distance preserving mapping / : Qd —> Qd is an isometry. What are the special dimensions? This is the rational analogue of the Beckman Quarles Theorem (On isometries of Euclidean spaces, Proc. Amer. Math. Soc. 4 (1953), 810-815). Dimensions 2, 3 and 4 are special, d is also known to be special if d > 6 and even , and if d is odd and has the form d — 2k2 — 1, for k > 3. For which subfields F of R is it true that every unit distance preserving mapping / : Fd —> Fd is an isometry, for all d > 2. Does the field of all the algebraic numbers have this property?

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Problem 25. JOSEPH ZAKS, University of Haifa, Haifa, Israel Can every n mutually disjoint segments in the plane be illuminted from some | points? I have an example of 9k + 2 mutually disjoint segments in the plane that need at least 4k light sources for illumintion. Thus, about (|)n light sources might be needed.

REFERENCES [AM] B. Aronov and J. Matousek, On stabbing triangles by lines in 3-space Comment. Math. Univ. Carolin. 36 (1995), 109-113. [B] T. Bang, A solution of the plank problem, Proc. Amer. Math. Soc. (1951), 990-993. [Be] A. Bezdek, Covering an annulus by strips, J. of Discrete and Computatonal Geometry (2002), 1-4 (accepted). [BBS] M.A. Bayer, A.M. Bruening, J. Stewart, A combinatorial study of multiplexes and ordinary polytopes, Discrete and Computational Geometry, 27 (2002), 49-64. [Bl] T. Bisztriczky, On a class of generalized simplices, Mathematika, 43 (1996), 274-285. [B2] T. Bisztriczky, Ordinary (2m + 1) -polytopes, Israel J. Math, 102 (1997), 101-123. [BF1] T. Bisztriczky and G. Fejes Toth, A generalization of the ErdosSzekeres convex n-gon theorem, Journal fur die reine und angewandte Mathematik 395 (1989), 167-170. [BF2] T. Bisztriczky and G. Fejes Toth, Convexly independent sets, Combinatorica 10 (1990), 195-202. [BW] P. Brass and C. Wenk, On the number of cylinders touching a ball, geom. Dedicata 81 (2000), no. 1-3, 281-284. [D] T.N. Dinh, Ordinary polytopes, Ph.D. thesis, The University of Calgary, 1999. [E] M.R. Emamy-K., On the cut-complexes of the 5-cube, Discrete Mathematics 78 (1989), 239-256. [EUT] M.R. Emamy-K., L. Uribe and I.C. Tomassini, Cut number of the 5-cube and a geometric approach, Lecture in University Puerto Rico, March 2002. [F] L. Fejes Toth, On spherical tilings generated by great circles, Geom. Dedicata 23 (1987), no. 1, 67-71. [G] R.L. Graham, The Largest Small Hexagon, J. Combin Theory Series A: 18 (1975), 165-170. [HSZ] A. Heppes and L.E. Szabo, On the number of cylinders touching a ball, Geom. Dedicata. 40 (1991), no. 1, 111-116. [IPS] R. Impagliazzo, R. Paturi, and M. Saks, Size-depth tradeoffs for threshold circuits, SIAM J. Comput. 26 (1997), 693-707.

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D. Ismailescu, Slicing the Pie, manuscript 2001, to appear in Discrete and Computational Geometry. [Kl] J. Kineses, On the determination of a convex set from its angle function, (submitted). [K2] J. Kineses, An example of a stable even order quadrangle determined by its angle function, in Discrete Geometry, In Honor of W. Kuperberg's 60th Birthday (Ed. Andras Bezdek), Pure and Applied Mathematics, A series of Monographs and Textbooks, Marcel Dekker (2002) 367-372. [KK] J. Kineses and A. Kurusa, Can you recognize the shape of a figure from its shadows?, Contributions to Algebra and Geometry, 36 (1995), No. 1, 25-35. [Ki] R. Kirby, Problems in low-dimensional topology, 35-473, in: Geometric Topology by William H. Kazez, Studies in Advanced Mathematics, AMS/IP; Proceedings of the 1993 Georgia International Topology Conference, Athens, Georgia. [Ku] K. Kuperberg, A knotted minimal tree, Communications in Contemporary Mathematics 1 (1999), 71-86. [M] J. Malkevitch, Convex isosceles triangle polyhedra, Geombinatorics 10 (2001), no. 3, 122-132. [MS] P. McMullen and E. Schulte, Abstract Regular Polytopes, Cambridge University Presss, to appear in 2002. [O] P.E. O'Neil, Hyperplane cuts of an n-cube, Discrete Math. 1 (1971), 193-195. [PT1] J. Pach and G. Toth, A generalization of the Erdos-Szekeres theorem to disjoint convex sets, Discrete and Computational Geometry 19 (1998), 437-445. [PT2] J. Pach and G. Toth, Erdos-Szekeres-type theorems for segments and noncrossing convex sets, Geometriae Dedicata 81 (2000), 1-12. [S] M. Saks, Slicing the hypercube, in Surveys in combinatorics, 1993 (Keele), 211-255, London Math. Soc. Lecture Note Ser., 187, Cambridge Univ. Press, Cambridge, 1993. [T] I. Talata, On a lemma of Minkowski, Period. Math. Hungar. 36 (1998), 199-207. [TG] G. Toth, Finding convex sets in convex position, Combinatorica 20 (2000), 589-596. [Z] M. Ziegler, New ways of slicing the 6-cube, preprint, Padeborn University, Germany, March 2002. See the "Cut number home page" http: //www .uni-pad.erborn.de/cs/cubecuts E-mail address:

bezdeanQauburn.edu ''Andras Bezdek''

INDEX

D affine regularity 415 algorithm 215, 361 angle function 367 antipodal points!94, 201 B bipyramid 69 C cellulation 241 chessboard Ramsey number 79 circle arrangement 27 covering 19 packing 279 supporting 104 spherical 103 supporting 104 collinear points 185 coloring 79 combinatorial aperiodicity 399 type 334 constant width 374 convex body 387 hull 50, 60, 201, 352, 381 position 49, 352 covering on-line 359, 443 triangles 435 2-adic 360 CW complex 241

Delaunay triangulation 271 Delone cell 105, 120 density of packing 279 diameter 374 distance preserving map 193 double lattice 453 duality 20 E edge unfolding 216 ellipsoid 202 Erdos problem 186, 292 Erdos-Szekeres problem 49, 185, 351 Euler equation 239 F f-vector 239 facet 71 cycle 216 fatness of a polytope 239, 451 of a tiling 451 Fejes Toth G. problem 19 Fejes Toth L. problem 88, 111, 279 fixing system 85 fixing system, primitive 85 flag vector 242

G Gale transform 381 general position 49 generalized helix 273 geometric tomography 367 459

INDEX

graph 79, 178, 194, 300 Griinbaum's problem 87 H Hadwiger number 455 Hadwiger's theorem 1 Hamiltonian path 306 Kelly's theorem 1, 387 horocycle 27 hyperbolic geometry 239 hyperboloid 2 hyperplane 34, 230

illumination 88, 457 independence number 292 integer hull 33 interval 13 isometry 193, 244 K Katchalski's problem 448 k-disjoint family 387 Keplerian polyhedron 340 kissing number 254 Klein model 244 knotted arc 452 Kuperberg 254, 256, 359, 431

lattice 20, 33, 231 kissing number 455 packing 456 critical 456 cubic 305 M Melchior's inequality 187 mid-pint map 411 minimal tree 452 Minkowski metric 203 space 373, 291

theorem 37 mixed moment curve 274 monkey-saddle 306 monotile 398 N Newton-Gregory problem 103, 112 O orchard problem 185 P packing barely saturated 453 completely saturated 453 saturated 453 Penrose rhombus 399 perimeter 450 Petrie cylinder 275 polytope 275 Petty number 292 Pick's theorem 39 plank 447 Poincare model 245 polygon 227, 267 convex 367 empty 52 regular 332 smoothed 407 polyhedron 215, 267, 331 cubic 305 uniform 331 abstract 333 polytope cyclic 267 neighborly 59, 267 4-polytope 59, 239 universal 59 totally-sewn 60 d-polytope 69, 201, 242, 267 universal 59 primitive vectors 34 protoset 398

INDEX

weekly aperiodic 398 prototile 397 R Ramsey like number 79 Reuleaux triangle 376 rich edge 39 facet 34 right prism 452 rooted n-gon 407

saturated packing 302 separable family of solids 455 shortest path problem 19 simplex 198, 225, 242, 381 solid covering 280 solidity 279 sphere 207, 229 packing 254 spherical Law of Sin 104 polygon 104 square 359 stabbing number problem 450 Steiner-system 190 strip 447 (r,q)-structure 28 support hyperplane 375, 387 Sylvester-Gallai problem 27 symmetric body 431

tangent map 368 Tarski's problem 447 tile, aperiodic 397 tiling 20, 207 Archimedian 280, 343 locally finite 398 non-periodic 398 spherical 280 uniform 280 topological pyramid 71

461

translation covering 435 transversal line 2 triangle 208 triangulation 207, 216 U unit distances 193 universal 4-polytope 60 Upper Bound Theorem 381 V vertex doubling 335 figure 71 star 333 unfolding 216 view obstruction 229 Voronoi diagram 268, 302