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ARBELOS

PRODUCED FOR

PRECOLLEGE

PHILOMATHS

~.

I

1985 - 1986

EDITED BY

PROFESSOR SAMUEL L. GREITZER

RUTGERS UNIVERSITY

Copyright © 1985

Committee on the American Mathematics Competitions

Mathematical Association of America.

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arbelos PRODUCED FOR PRECOLLEGE PHILOMATHS

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Orthocenter H of

~BC

is the center of

a circle. Lines DD', EE', FF' are on the midpoints A' ,B' ,C t of the aides. Prove: AD = BE = CF.

1985-1986:

No. 1

September, 1985

Copyright ~ The Mathematical Association of America, 1986

i PREFACE

We are surprised and grati£ied to start this £ourth year o£ the Arbelos. It £eels like a new beginning. First, acknowledgements £or solutions go to C. Kenneth Fan, Mark Kantrowitz, Ambati J. Krishna, Sherman Lau and David Morin. The solutions were to some extent usual, but there were some which showed ingenuity. Please continue to send in your solutions as soon as you can. Perhaps it is worth-while to restate the purpose o£ this publication. Part o£ it has the purpose o£ presenting material that may be new or un£amiliar to the reader. Now we do not know just what is new or un£amiliar to everyone. We can just hope that there will be enough to make the Arbelos worthy o£ readin~. Thus, Pro£essor Klamkin, writing about the cover problem £or Arbelos 4 - March, 1985, provided us with many re£erences. It was £irst apparently stated by Aleksandrov in 1882, and was not original with him. It was apparently rediscovered by Morley in 1888. There was a generalization by Rosenbaum in 192), etc. Note that Morley thought the theorem was new and prepared his own proo£. In Arbelos, we hope the reader will find a problem new to him and prepare his own - perhaps ingenious - proof. Also, Dimitrios Vathis, writing from Chalcis, Greece, pointed out that there were problems that he found in previous literature. We agree, but hope they were new to most readers. Any text on projective geometry, for example, contains the theorem that the midpoints o£ the diagonals of a complete quadrilateral are collinear. We hoped. that this would be new to enough readers to make it worth including. So = let us know what you would like to see in Arbelos, give us sug~stions, and help us make Arbelos the publication you want it to be. Samuel L. Greitzer Editor

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1

CONTINUED FRACTIONS CONTINUED We first developed some of the theory of contin­ ued fractions some time ago - in Arbelos. No.4 - of March 1983 (see p.13). There we used them to solve linear Diophantine equations. Some problems we have encountered make it useful to continue our development. Let us start with some review.

A continued fraction has the form N

b1

b

= a O + a- + ..2 1

a2 +

b

b

.J + • • • +.-ll an

a3

Each new fraction is added to the denominator of

the previous one.

We shall make some changes in this definition

for our special purposes. First. let us have each

b = 1. The resultinp; continued fractions are then

cilled Simple continued fractions. Next, for the

sake of ease in typing, we shall write the contin­ ued fraction in the form N = (a O' a 1 , a 2 , • • • , an) Finally, let us agree to have all the a

r

positive.

If we drop all fractions after a given one. we can do the arithmetic required to get a result, which we will call a partial convergent. For example, consider the fraction 39/17. We divide to get

12=2+-2=2+~=2+ 1

17

17

2 +

1

17/5

1

3 + 5/2

That is, 39/17

=2 + 1

3

+

2/5

=

1

3 + '2 + 1 2

= (2,3,2,2).

The partial converp;ents

are po/qo= 211 , P1 /q 1 = 7/3 , P2 /q 2 = 16/7. Of course, the final conver~nt will be 39/17.

2

For the moment, let PO = a O' qo

= 1.

Then

P1 = a Oa 1 + 1 and Q1 = a 1 (work it out !), and P2 = a Oa 1a 2 + a O + a 2 ' Q2 = a Oa 1 + 1. At this juncture, imagination takes over. We note that it is possible to rewrite P and Q2 as 2 Q Q2 = a 2 1 + QO· P2 = a 2P1 + Po By induction, we easily find that, in general, Pn = a nPn_1 + Pn- 2 Qn = a nQn_1 + Qn-2. These relations hold for all values of n. Notice that these last two expressions look a lot like simultaneous equations. Let us e1ininate the tems containing an from these. We get (Pnq n-1 - qnPn-1)

= -(Pn-1qn-2

- qn-1 Pn-2)

=

(-1)2(Pn_2q n_3 = qn-2Pn-3) = ••• =(-1)n(P1q O - Q1PO)· If we substitute the values for P1 ,PO,Q1,QO' we get just ~ 1. That is, (A)

If this last expression is divided by QnQn­ l' we get 1

~n-1 Frau this, we may conc1\de that successive convergents are alternately smaller and larger than the final convergent (if any) and that the set of convergents approaches a ~1mit as n increases. Moreover, we can now justify claiming that Pn , Qn are relatively prime. Any common factor of Pn and Q would have to divide 1. n

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3 Let us apply what we have to a problem. We take a problem that we have encountered so often that we can give the answer before the proposer finishes giving the problem. It is: A man cashes a check at a local lank. He spends 68¢ and then finds that he has twice the amount of the original check left. What was the amount of the original check? The original check being for 100x + y cents, the condition canes to 100y + x - 68 = 2(10Ox + y). or 98y - 199x = 68. Were we to change 199/98 to a continued fraction, we would get 199/98 = (2,32,1,2). Note: The procedure we have described for changing to a continued fraction is just the Euclidean Algorithm. Thus, 2

98)199

!2£1£ 3)98

.22)1

2 3 2 2

1)2

and we take the partial quotients. We have already remarked. that the final converg­ ent is, in this case, 199/98. From formula (A), then 199(qn_l) - 98(Pn_l) = : 1. Thus, the next-to­ last, or penultimate convergents yield a solution to 199Y - 98x = : 1. We fiddle with signs to get +1. Thus,

98(-67) + 199(33)

= 1.

Therefore

98(-67)(68) + 199(33)(68) Also

98y

- 199x

= 68

•

= 68

4 Subtracting and rearranging, we get 98 _ 113~f68~ + x 199 - Tb7 68 + y. x = 98t - 2244

Then,

y = 199t - 4556 and, for t = 23, we get x = 10 and y original check was for $10.21.

= 21.

The

Now there is nothing in the formulas we have derived. so far to indicate that they must be used. for finite continued fractions only. We therefore extend our use of continued fractions to square roots. This is a bit more difficult than before, and we will not prove all our assertions. Let us apply our results to find ~ as a continued. fraction. To do this, we repeatedly separate expressions, as they occur, into an integr~l part and a fractional part less than unity. Thus, ,/I9 = 4 + (,/19 - 4) and we let a = 4. O invert the fractional part and rationalize, thus __....1""--- =

.Jt9 -4 3

;/I9 -

2

,/19

+ 4

3

=2

..Ji9 -

+

2

3 . = -/19 + 2 = 1 + .,If9 -3

5

and a

1

= 2.

5

5~ = -J19 + 3 = 3 + .Ji9 -3

---,=:o....

:;i9 -

3

_......;;;;.2_

;/19 - 3 _-",,-5_

,/19-2 _--",,-3_ -Ji9 - 4

2

= ,fi9

+ 3

2

=1

+

5 =

.J19

.y'i9 - 4

3 + 4

-2

5

+ 2 = 2 +

= v'19

.Ji9

3

=8

+

09 -4

and a 4 = 1. and a

5

= 2.

and a6 = 8.

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5 We have found,J19 = (4,2,1,3,1,2,8). There are several points we must comment on at this point. First, note that we got back to at a6. Therefore, the whole process repeats from this point on. The line above the numbers on line 1 show which numbers will repeat. In general, when we reach the last number of the

-09

repeating part, an' we will fim it equal to 2a O• Next, note that, except for the last number of the repeating part, the numbers are "palindromic" ­ they read the same from left or from right. This is always true. We omit the proofs - they can be foum in texts on number theory (some of them). Assume now that we have developed a continued fraction for, say, -./N. We have reached the last repetend. Then, by virtue of our fundamental rules

involving three successive partial convergents (see

Page 2 of this article), we may write:

(If:N + a O)Pn-l

...jN = (::yrN

+ Pn-2

+ a)q + q

o n-l n-2

Now for a little algebra. We clear fractions:

q Nqn_l + (a O n-l +

qn-2)~ = (aOPn-l

Recall that, if a a = c and b = d. Nq n-l

+ ~ = c + d~ , then

For us, this means that

-ap +p

- 0 n-l n-2

&Oqn-l + qn-2

= Pp-l

+ Pn-2) +

Pn-l~

6

Again, since these look like simultaneous equations, we eliminate the terms involving a ' O 2 2 Pn-1 - N qn-2 = ~ 1. That is, to and get solve the equation x2 _ N y2 = 1 try the penultimate convergents of the continued fraction expansion of ~. We shall call a quadratic equation of this form a Fermat equation, since it was he who first used such equations. As an example, let us solve First, expand

=)

~

1

-v'f4

5

_.Ji4+)

5

so a

=

~

O =)

=1+.J14-2

5

_.J!4+2=2+.J!4-2 2 2

--=~2_ = -y'14 + ./i4 - 2 5

;Ji4'- 3

as a continued fraction.

(,/14 - ))

+

-~j14-1-4=---) -;/I4~1=4~---2

x 2 - 14 y2 = 1.

2 = 1 +

+ ) = 6 +

so a 1 = 1. so a 2 =

2

.J14 -5 )

so a)

~

so a4 = 6.

- )

= 1.

Therefore, ,jf4 = (),1,2,1,6) Before we complete the solution of this problem, suppose we try to approximate the square root of 14 from the data we have. The formulas on pa~ 2 of this article make all this mechanical. We prepare the follOWing tables

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i. ,.<.• ;.•• i.••

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,.

7

a

3 1

3 4

2 11 1 15 6 101

1 1 3 4

3.000 4.000 3.667

3.750 3.741 27 We hope the entries are self-evident. Note how the values of p/q approach each other alternately from above and below. The value of ~ is 3.742. Now the penultimate terms are 15, 4, and, if you check, you will find that (15)2 - 14(4)2 = 1. Perhaps the reader would like to solve this problem, modeled after one in "Canterbury Puzzles", by H. Dudeney. To fulfill their obligation to provide same military service for their shire, each of thirteen villages contributed the same number of villeins, that number happening to be a perfect square. They arranged themselves into thirteen squares. At last, the shire reeve (sheriff to you) arriVed, whereupon he and the villeins rearranged themselves into a perfect square. How many men did each Village supply, and how large was the ultimate square ? (It is assumed that all squares are solid). To wax a bit personal, we fim it is fun to work out the continued fraction eqUivalents of square roots, as in the table above, and check for the accuracy of the approximation, as in the table above. In so doing, we have noticed the following:

8

-.f5

= (1,1,2) ;f8 = (2,1,4) -vTI = (),g) -y'j

,fi4

= (4,I:8)

,fE

= (5,1,10)

= (2,4)

-/iO = (),6) .,ff7 = (4,8) -J26 = (5,10) . . m = (6,12)

Does this situation hold in general ? Find, if it exists, some general rule. We have noted that we have restricted our work to simple continued fractions. There remain many areas to explore. Far example, consider the equation:

2 x - lOx - 1

= O.

We can write this (by division by x) in the form: x

= 10

+

.1x = (10),

root. How about

and we can approximate a positive

solvin~ x 2 - ax - b

=0

?

Can we use continued fractions to find cube roots ? What about special functions and special numbers? We

thi~

that e

= (2,1,2,1,1,4,1,1,6,1,1,8,1,1,10, ••• )

2h

at 2 Is it? We bel~eve tan x =.!...2£- 2S...... 2S...... 1 -) - 5 - ?

- .

•• True or false

References: The reader can find good materials in the usual excellent texts: Higher Algebra, by Hall and Knight, Algebra, by Barnard & Childs, Algebra, by Chrystal. "Continued Fractions", by C.D. Olds, is highly recommended. We have learned a lot from "Continued Fractions", by A.N. Khovanski1. There is always the classic text by Wall. Finally, there are texts on Number Theory that have sections on Con­ tinued Fractions.

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1.

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9

TRIG PRACTICE The following problems involve trigonanetry in sane fom or other. We would like to see solutions in any form. 4 4 4 2 2 2 E.l If, in AABC, a + b + c = 2c (a + b ) then LC

= 45

u

or 135°.

E.2

In AABC, with area K, show that sin A + sin B + sin C = K/(Rr) where R, r have the usual meaning.

E.)

Prove that (sin2x + csc 2x)2 + (cos 2x + sec 2x)2 ~ 25/2.

E.4

If P is a point inside AABC such that LPAB = LPBC = LPCA = ro, prove that cot ill = cot A + cot B + cot C

(point P has a special name in geometry).

E.5

Show that the

SlDl1

of the three fractions

a-b b-c c-a 1 + ab ' 1 + be ' 1 + ca

is equal to their product.

E.6

Prove that n 2n 1n 4'TT . 'in 6~ ?Tl' cosrs·cosrs·c0st5.cos15·c0st5.cos15·c°stS equals (2)-7.

Note: These problems are arranged in order of increasing difficulty - as we found them to be for us. Send in your solutions!

10 THE FERMAT POINT

About the year 1610, Fermat sent the following problem to Torricelli (inventor of the barometer): Given a triangle with no angle greater than 1200, find a point P within the triangle such that the eum of the distances from this point to the vertices is a minimum. Much has been written on the problem, and we shall include some references at the end of the article. Of the many proofs, we show one that seems to be simplest. We start with a Lemma: Let triangle ABC, with AB = BC = CA, be inscribed in a circle, and let P be any point on the arc BC. Then PB + PC = PA.

A

Proof: We use Ptolemy's Theorem (see, for example, Arbelos, January 198), p.5). Let the side of the equilateral triangle be denoted by s. Then s.BP + s.CP = s.PA, or PB + PC = PA. Now let ABC be a triangle with none of its angles greater than 120°. On side BC of this triangle, we construct equilateral triangle BCQ (externally), as shown in the diagram on the next page. Select any point P on the minor arc BC. From the lemma, we have PB + PC = PQ. Therefore, PA + PB + PC = PA + PQ. This will be a minimum when points A, P, Q are collinear.

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Q.

Now, an~les BPQ and CPQ are each 60~, so that = 120 • Furthermore, when APQ is a straight line, LBPA = LCPQ = 1200 , since both are the supplements of the equal angles BPQ and CPQ. Therefore, the desired point is such that LAPB = LBPe = LCPA = 120°.

LBPC

In the three plus centuries that it has been around, this problem has been worked on by many mathematicians. In addition to solutions by Steiner and others by geometric means, there have been others usinQ; analytic geometry and even physics. It is interesting to compare such methods of solution. Incidentally, we state without proof the added condition that, if the angle at A equals or exceeds 120°, then the Fermat Joint will be at A. Some of the references to the problem are to be found in the followings What is Mathematics - Courant & Robbins - p.)54 Modern Geometry - Johnson - p.221 Mathematical Gems - Honsberger - p.24 Maxima and Minima without Calculuc - Niven - p.15? Geometry Revisited - Coxeter & Greitzer - p.S)

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12 HOM<X;ENEOUS COORDINATES

We beKin this article with a problem that is well-known and quite old.

A

"In MBC, lines AA', BB'. CC' divide the sides Be, CA, AB in the ratio 1/2. Prove that the ratio of' the area of' ~N to that of' MBC is 1/7. The problem seems to date back to 1891. We f'irst met with it at the end of' the World War (No.1. that is). We solved it by means of' Menelaus' Theorem then. We hope to present at least three other solutions. We did try to solve the problem umer the assump­ tion that the sides were divided in the ratios 1;1, m;l, n;l. It was tough! Many years later, we mentioned this problem am our dif'f'iculty to a colleague, who happened to have been a graduate of' Cambridge University. It took him one day to cane up with a solution which, he said, he had worked out by using areal coordinates. It became necessary f'or us to f'im out about these coordinates, am we have used what we discovered to solve a number of' hard problems by usinp; them. We assume now that all of' us know rectangular coordinates. Many of' us may know polar coorQinates. However, f'ew of' us have used oblique coordinates, y in which the axes make an angle other than 90 with each other. Yet they can be very usef'ul. There are also occasions when, instead of' using two numbers, (x,y) to locate a point in a plane, we may f'ind it more usef'ul to use three or more numbers. For example, suppose we start with the

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linear equation ax + by + C = O. Replace the

x by x/z and. the y by y/z and clear fractions in

the resultin~ equality, and we have

ax + by + cz = 0 For points on the (usual) X-axis, y = O. For points on the (usual) Y-axis, x = O. Now, however, we seem to have a Z-axis to explain. Usually, this is referred to as the ideal axis, or even infinity. We will not be using this fonn, so we leave it. Instead, let us start with the triangle CAB, where OA is our usual X-axis am OB·our usual Y-axis. We will represent the area of a triangle, say ABC, by (ABC). Then we shall define the coordinates of the point P as the A ratios of the areas of of the triangles with P as vertex to that of (OAB). Then the three coordinates of P will be Pt = fOPB~ ABC ' P2 = fOPAj ABC ' P3 = fAPBj ABC .One can see why these would be called areal coord inates. There are two remarks we might make at this juncture. First, note that the sum of the areal coordinates of a point is unity. This enables us to transfonn any equation into a hmogeneous one. Next, any point in the plane can be represented by three coordinates, am we can see that the point (tx, ty, tz) and the point (x, y, z) are the same. One may also locate the point P by using the distances PO, PA, PB, in which case we have trilinear coordinates.

14

By using the definition, we find the coordinates of the vertices of our coordinate triangle. We find. that for 0, we get (0,0,1); A, we get (1,0,0); o B, we ~t (0,1,0). Next, for a point r! on axis OA, we find the coordinates (1,0,1). For the point lit on OB, we have coordinates (O,1,m). And for the point N' on AB, we find (1,n,O). Note that we should divide these by (1 + i), (m + 1) and (n + 1) respectively, if we are to have the sum of the coordinates equal to unity. z Now consider the determinant x y First, this is linear in x,y,z. PJ = 0 Next, it equals zero when QJ (x,y,z) = (P1,P2,PJ) and when (x,y,z) = (Q1,q2,qJ). Hence the determinant is the linear equation determined by the points (P1,P2,PJ) and (Q1,Q2,QJ). We now find. the equations of the lines on AM', BL', ON'. The line on B, J1is z x y 0 1 0 = 0 , or x = 1z. 0 1 1 I For the line on A, M, we have x y 1 0 = 0 0 1 or my = z.

.

~j

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I.!.•

•

j.i.•••

!.I.•

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,.

I·

15 For the line on 0, N~ we have

x

y

z

o

0 1 100

= 0,

or y = nz. We proceed next to find the points of intersec­ tion of these lines (the L, M, N or our original diap;ram). Again we use determinants. I I The intersection of lines BL, AM is x y or(lm, 1, m). I , The intersection of lines AM, ON is or (1, m, mn). I

I

1

0

o

m

x

y

z

o

-m

1

n

-1

o

~

Y -1

z 0 or (1, ln, 1). 1 0-1 Note again that we should divide by (lm + m + 1), (mn + n + 1), (In + 1 + 1) respectively, if we want to have the sums of the coordinates equal to uiDity. Now the area of triangle (ABC) is 1 0 .!. 0 1 2 which equals 1/2. 0 0 The area of the triangle IJo1N of figure I will be lm 1 m 1 In 1 = lm + mI n 1 mn which simplifies to (lmn - 1)2. Hence the ratio

The intersection of lines ON, BL is

~

-

I~~' r~ m~1

~ 1~1

•

16 of the area of (LMN) to that of (ABC) will be

~lmn

- 1)2

(1m + m + 1 (mn + n + 1)(nl + I + 1)

•

For the reason far the occurrence of the denaninator see the note on the previous page. In the special case of the problem which we started, we have

= m = n = 1/2

I

and when we substitute in our fonnula, we find

= 1/?

that (LMN)/(ABC)

We do have an additional "plus' in the fonnula. A necessary and sufficient condition for the three I

I

,

lines AM, ON, BL to be concurrent is

that the

nwnerator of the fonnula be zero, or lmn

= 1.

This

is Ceva's Theorem! (We apologize far the careless use of L, M, N. In the formula, .,e are referring to the diagram that is at the start of the article. Now we shall I , I . use the third diagram, with L, M, N on sides OA, OB, AB respectively.) Let us now find the ratio of the area of (til!) to that of (ABC). For the area of (:DfN), we find

I

1

1

0 n

o

1

m

0

=

1mn + 1

Now refer to the first note to see why the ratio is

11

lmn + 1 + 1)(m + 1)(n + 1) •

If If, M', N' do not form a triangle, and are col­ linear, we must have lmn = -1. But this is Menelaus' Thearem.

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17

e e

e

e

•e •e e e e e e

• • ••

The following solution for the original problem canes from "Mathematical Snapshots", by H. Steinhaus. The sides of trian~le LMN have been extended and the resulting figure has many congruent triangles. Congruent triangles have been given the same label. The reader should be able to match the various pieces of trian~les to show that the seven equal triangles (1 + a), (x-2 + 2), (b + 3), (x - 4 + 4), (c + 5), (x - 6 + 6) and x fill the triangle ABC like a jigsaw puzzle, making its area 7x.

tit

e e e e e e e e

• e e

e e e e e e

(; ..-.----------~ C A second solution canes fran "Bypasses", by Z.A. Melzak. It suggests using an orthogonal projection to make, say, triangle LMN equilateral. Now both the general formula and the special case can be obtained using the angles of «o(;J where they occur.

18

o

A o

o

o

o

c

o The prettiest solution for the original problem canes fran ''Geanetry'', by R.S.M. Coxeter. Imbed the triangle in a lattice (which can be detennined from the fi~e itself}. Now we can apply Pick's Theorem, which we all should know. The theorem states: The area of any simple polygon whose vertices are lattice points is given by the fonnula 1

Area = 2 b + C - 1. A simple polygon has no intersecting sides; b stands for the number of lattice points on the boundary of the polygon; c stands for the number of lattice points inside the polygon. In the case of triangle ABC, we get (ABC)

=12

3 + 3 - 1

=1. 2

For triangle (I1'1N), we have Then

!3+0-

1 =

!.

~~~~ = ~}~ = ~.

Two final points. First, Pick's Theorem should be familiar to everyone. Second - a problem. Let L'M'N' be a triangle whose vertices are on OA, OB,

AB. Prove that, among triangles OL 'M " AL' N' and BM'N', and L'M'N', the last cannot have the small­ est area except when L', M', N' are midpoints.

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19

MANY CHEERFUL FACTS

We have previously mentioned various averages or means (see Arbelos #2 - 1984 and #3 - 1983). For two nlD1lbers, we can have all sorts of means. x +

y

There is the arithmetic mean ( AM ) = 2 ' the geanetric mean (GM) = .,;xy, and the harmonic mean

(HM)

= 2/(1 +

x

1). y

We also have the Root-mean­

•

square, whiCh~ll call the quadratic mean (QM). It equals ~ ;. Y for two numbers. We have

•

e

also found that QM ;> AM ~ GM ~ HM.

We would like to see where the QM fits in the

continued set of inequalities given above. The easiest way is to check squares of these terms to'\avoid inconvenient square roots. So we write 2 22+ 2 + 2 (QM) - (AM) = x 2 Y - (x 2 Y) • Expanding, we 22222 2 x Z x Y ~ x y ~ (x - y) 2. get "2 + 2 ­ 4: - 2 .. 4 = 4 - 2 - 4 = 2

. > O.

•e e •e ••

e e e

e e e e

e e

e e e e

­

e e

e e

This tells us that, of the four means. the QM is greatest. That is, (QM) ~ (AM) ~ (GM) ~ (HM).

There is equality when x = y. The QM is often useful in same problems involving inequalities. Keep on the lookout in case such a situation arises. We never did get answers to the problem on the cover page of Arbelos #3. January 1983. It should be easy to guess the answernow. However, it should be easier to work out the lengths of the lines in the dla~am as well. Send solutions to Dr. S. Greitzer Mathematics Dep't Rutgers University New Brunswick, NJ 08903

20

•

OLYMPIAD PROBLEMS 1) Prove that for any natural number n, the number (2~) divides the least common multiple of the numbers 1,2, ••• ,n. (Bulgaria) 8 5

2) Consider all sums 0 f the f orm 1 ~L., e k k 5 -±1 5 t. 2 5 + • • • ±. 19855 , where e = '±1. k

What is the smallest non-negative value assumed by a sum of this type ?

A strictly increasing function [O,lJ satisfies

f(O)

= 0,

(Canada) f

defined on

f(l) = 1, and

< f~X+Y) - f(xj < 2

2 - f x) - f(x-y ­ for all x and y such that 0 < x

1

Show that f(l/J) :So 76/135.

y

< 1.

(Finland)

4) Note: We represent the area of A4BC by (ABC). Let ABC be a triangle with angle bisectors AA , 1 BB , CC and incenter I. If 1 1 (IA B) + (IB 1C) + (IC A) = ~ (ABC), show that 1 1 ABC is an isosceles triangle.

5) The divisors of the natural number arranged in the increasing series

(Rumania) n are

Xl < x2 < ••• < ~. Find all numbers n for which x 2 + X 62 ­ 5 (USSR)

1 = n.

Please - when sendin~ in solutions, refer to the problems as 01 2 , etc.

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21

THE GAMBLER'S RUIN

The problem I A gambler am his opponent agree to play a game in which the gambler wins one dollar if a tossed coin comes up heads and the opponent wins one dollar if the coin comes up tails. The gambler starts with ~ dollars, the opponent with J2 dollars. What is the probability that the gamb­ ler will lose all (his ruin) and what is the expected number of games it will take to ruin him? This problem has been around for quite a while in various forms. It may be considered a random walk problem with absorbing barriers at Q and at a+b, for example. Also, it has been solved in a number of ways. At any rate, it pops up fairly often, and it has interesting eventualities. In its simplest farm, we have the ~mbler with a dollars to start. His opponent has b dollars. After the first toss of the coin, the-gambler has either a-1 dollars or a+1 dollars. In general, when the ~mbler is at situation x, he has reached this from x-1 or x+1 with equal probability 1/2 for each. Since weare talking about "ruin", let q(x) be the pro1:Bbility of ruin for the gambler startin~ when he has x dollars. Then

q(x)

=~ q(x=l)

+ ~ q(x+l)

or

q(x-1) - 2 q(x) + q(x+1) = O.

Now this is a difference equation or a recursion equation. There was an article on such equations as far back as our first issue, in September 1982. Since such equations happen to occur in many situations ( like Olympiads ) , we will take this occasion to outline briefly methods for solving some of the simpler difference equations, am then apply these to the problem at hand.

We define E(f(x»

= f(x+l),

E2(~(x»

=f(x+2),

and, in general Ek(f(x)) = f(x+k). For example,

22 if f(x) = aX, then E(f(x)) = a x+1 = a.ax , E2 (f(x)) = a x+2 = a 2 .ax , etc. Let us now restrict ourselves to a situation with E2f(x)) as the m ax1mum "power" of E. That is, we are givena E2 (f(x)) + a E(f(x)) + b f(x) = O. If we let f(x) = k X, and substitute in the above, we get (k 2 + a k + b).k x = O. This expression will be zero only when k is a root of the quadratic equation in parentheses. If these roots are r , r , 2 1 the solution sought will take the form f(x) = A.(r )X + B.(r ? 1 2 where A and B must be found by suitable "boundary" conditions. If the difference equation has two equal roots, say r

1

=r 2,

then the associated solution will be f(x) = (A + Bx).(r1 )x.

In the case of our problem of the "gambler's ruin", we do not have to consider equations above the quadratic. The equation in this case has two equal roots (= 1), and therefore the solution has the form q(x) = (A + BX).(1)x = (A + Bx). As for boundary conditions, ruin is complete when X

= 0,

or q(O)

= 1,

and also q(a+b)

= O.

These

1

give us A = 1 and B = - a+b • Finally, when we simplify and use a, b, the probability of the gambler's ruin comes out t~ be q(a) chance of winning all is a+b •

b = a+b.

His

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23

Now for the expected number of plays necessary to achieve "ruin". For a start at x , we label this N(x). If the gambler wins the first game, he will be at N{a+l), and the game will continue just as at N{a). Thus, in general, the expected number of games, given a win at the first trial, is

N(x) = ~ N(x-l) + ~ N(x+l) + 1. As for boundary conditiona, we obviously have N{O)

=0

and

N{a+b)

= O.

Again, we have a difference equation, which we write in the form N{x+l) - 2 N{x) + N(x-l)

= -2.

. When the left side of this equation is set equal to zero, we know that, as before, the solution of the resulting equation is

N{x)

= {A +

Bx).{l)x

I

=A + Bx.

(This is called the complementary solution.) To take care of the -2, we assume a particular solution with the form N{x)

=P +

Qx + Rx 2

am substitute in the difference equation. This yields

P + Q(x+l) + R(x+l)2 - 2P - 2Qx - ZRx 2 + P + Q(x-l)+ R(x-l)2 = -2. On simplifying all this, we arrive

at R = -1., am the full solution of the difference equation is of the form

N(x)

=A +

Bx - x2 •

Using the boundary conditions, we have

A

=0

and

B = (a+b), so, at last, N(x) = (a+b)x - x2 • If we start with a dollars for the gambler and b dollars for the opponent, we find that ­ N{a) = a.b •

24 We note that, for the probability that the ~mbler will be rUined, the result is simple and easy to underatant am accept. It is different with the expected number of games. Take the case, for example, where the gambler has one dollar and the opponent has nIne dollars. We see that the first game will, if the gambler loses, finish the process. Yet the formula says that the expected number of games in this case is 1 .x 9 = nine1 To some of us, this appears to be a paradox. We did assume that the gambler would win the first game. Now the expected number of plays would equal 2 x 8 = 16. Next, it may be that the gambler alternates between, say, ~ and a +1, so that the game continues forever. There are quite a number of cases where a result goes against our intuition in probability. Usually, a careful examination of the initial comitions helps resolve the paradox. For example, in tossing a coin, we usually assume that it will fall heads or tails with equal probability 1/2. Actually, no two coins are identical and the center of mass of a coin is not necessarily at the center of the coin. In assuming a probability of 1/2 for the fall, we are not precise. We just hope that our assumptions will not affect our results too much. Finally, let us assume that the gambler and his opponent each start the game with $100. Then each has a fifty-fifty chance of Winning, which seems reasonable, but the expected number of tosses until the game is over am the gambler (or his opponent) ruined comes out to 10,000. We hope the reader finds this surprising. We do. We leave it to the reader to determine, assuming that there are many such ~mes, how ofter the gambler will be ruined, or more reasonable, if he starts with ~ dollars, how often he will again have a dollars.

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25 ..

I

KURSCHAK KORNER

This column features those problems of the famous Hungarian Kiirschak Contest which have not yet appeared in English trans­ lation. For the problems of 1894-1928, the reader should consult Hungarian Problem Books!. & !!. in the MAA's New Mathematical Library series (available from the MAA hdqrs.); the problems (with "brief summaries of their solutions) covering the years 1966-1981 were translated into English by Prof. Csirmaz of Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers.) Student readers are also invited to set aside an uninterrupted 4-hour period to compose complete, well-written solutions to the problems below, and to submit their work to the address given below for critical evaluation. 1/1958. Of six points given in the plane no three are collinear. Show that one can choose three of the six points so that if a triangle is constructed with them as vertices, then one of the angles of the triangle will measure at least 120°. 2/1'958. Prove that if u and v are integers such that u 2 +uv + v 2 is divisible by 9, then both u and v are multiples of 3. 3/1958. In convex hexagon ABCDEF the opposite are parallel (i.e., AB DE, BC EF and CD FA). that the areas of ~ACE and ~BDF are equal.

II

II

II

sides Prove

Dr. George Berzsenyi 2040 Chevy Chase Beaumont, TX 77706

CONTENTS COVER: i 1 9 10 12 19 20 21 25

Geometry Problem Preface Continued Fractions Continued Trig Practice Fermat Point Homogeneous Coordinates Many Cheerful Facts Olympiad Problems The Gambler's Ruin Kurschak Korner

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arbelos

PRODUCED FOR

PRECOLLEGE

PHILOMATHS

A

,

i·

,

&I'-.....x...------~C FQR is the AX, BY, CZ QR, HP, PQ AX, BY, CZ

orthic triangle of ABC.

are perpendicular to

respectively. Prove that

are concurrent.

1985-1986& No. 2

November, 1985

Copyright C> The Mathematical Association of America, 1985

i

PREFACE The number of readers who send in answers is never very large, unfortunately. Having lots of time, we welcome more such. First, a reader asks for solutions to the problems. This seems a fair request, and we are adding solutions to some problems in this issue. However, we prefer to give solutions sent in by readers if these show some ingenuity. The number of these have been few. We have always been asking for original articles, and have published just two so far. Please send in your efforts. We may publish them. However, they must be suitable for the nature of the publication and for the preparation of the readers. Ambati Jaya Krishna has sent in some materials but these are unfortunately beyond the supposed preparation of our readers and go beyond what we have been presenting so far. Nevertheless, there is no total loss. There are two problems that one might like. I)

Given

a + b + C + d + e = Tf2/16 a 2 + b2 + c 2 + d 2 + e 2 = Tf4/1024

What are the values of a, b, c, d, e attain its maximum ? II)

if

e

is to

Given the continued fraction Evaluate.

Readers might find hints or help by consulting "Theory of Numbers" by Hardy and Wright, or "Higher Algebra" by Hall am Knight. Let's have some letters! Dr. S.L. Greitzer Rutgers University

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1

THOUGHTS ON PROBLEM SOLVING We are in receipt of a set of solutions to some of our problems in the March, 1985, issue of ARBELOS, and they have set us to thinki~~ about problem solvin~ and one purpose of ARBELOS. Let us share our thou~ts. On Pa~ 8, there appears the followinp; problemz Factor a(b2 _ c 2) + b(c 2 _ a 2 ) + c(a2 _ b2 ).

Our correspondent, Ken Burton of Bellaire High School, Texas, expanded this, rearranged terms, added or subtracted where useful, and proved the assertion.

However, let the expression to be S = 0 when a = b. Hence (a - b) is a 5 = 0 when b= c. Hence (b - c) is a S = 0 when c = a. Hence (c - a) is a

factored be S. factor of S; factor of S; factor of S.

Hence S = A(a - b)(b - c)(c - a) with A constant, since, except for A, both expressions ahve degree J.

Now let a = 1, b we are done.

,. I.

= 2,

c

= J,

say. We find A = 1, and

On Page 22, we have the following problema

The sum of all the face angles about all of the vertices except ~ of a given polyhedron is equal to 5160 degrees. Find the sum of all of the face angles of the polyhedro•• Our correspondent assumed that, just as a ~oly~on can be trian~la ted, a polyhedron can be dissected into tetrahedra. For all we know, this may not be true. We would like a reference to show that it is. We went to Euler's Formula, V - E + F = 2. Let f i be the number of faces with i edges. Then f) + f 4 + f + • • • + f n = F 5 Also, Jf) + 4f4 + 5f + + nf = 2E. n

5

Now 2f) + 2f4 + 2f 5 + • + 2fn = 2E. Hence f) + 2f + )f + • • • + (n-2)f n = 2(E-F)= 2(V-2). 4 5

•

2

But the sum of all the face of the polyhedron equals L

= 'rTf3

+ 2'rTf4 +

L

= 2'Tr(V

3'rTf 5 + •

an~les

of all the faces

+ (n-2)'rTf • Therefore, n

- 2).

Dividing 5160 by 36€f, we fim that V = 17, am the sum of all the face angles equals 5400 degrees. As a space filler on Page 8 t we had the following: Let at b am c be positive numbers. Prove that abc ~ (a + b - c)(b + c -a)(c + a - b). Our correspoment checklld out each factor to fim when it would be positive or negative, am developed a good. proof. What we did was the following, with obvious results:

2 a ~ a 2 _ (b-c)2 = (a + b - c)(a - b + c), b2 > b2 _ (c_a)2 = (b + c - a)(b - c + a), c 2 ~ c 2 _ (a_b)2 = (c + a - b)(c - a + b). Multiply, am a 2b2c 2 ~ (a + b - c)2(b + c - a)2(c + a _ b)2. Now take the square root am we are done. Finally, on Page 22, we have the following problem, where we have altered the l' notation somewhat. 0 Fim the point 0 inside the ~~r • tria~le at the right for which 0­ alp + b/q + c/r is a minimum, where p, q, rare perPendicular to sides a, b, c respectively. Here our correspoment used rectangular coordinates am partial derivatives to derive a solution. It was correct but took four pages to develop. We would like to present another solution and would welcome your comments on it.

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3

Note that (ap + bq + cr) = 2A (the area of the • triangle) and is therefore a constant. Then we seek

I •

I

(a b + -c)( ap + bq + cr ) = a minim\.U1l, say m. - + -

•

•

p

q

r

2 Multiply, and a + b2+ c 2 + ab(.l4S) + bc(~) + ac(!+E) • qp rq pr 222 • is to be a minimum. We discard the a + b + c • But

I •

• , • • • • •

what is left will be a minimum when each of the expres­ sions in parentheses equals 2, by a we11-nown inequa1­ ity. This occurs when p = q = r, so that the point 0 desired is the incenter of the triangle. The solutions by Ken Burton were correct, neat am a pleasure to read. We would welcome more like them.

• • • •

We have just two mild comments to make. First, we present novel theorems am proofs in the hope that the reader will remember them and use them. Thus, on Page 7, there is an example of factoring as we have , done it. Naturs11y, we hope the reader will use what • has been presented in an issue of ARBELOS to solve • problems in the manner presented. • Secom , we prefer not to use a sledgehammer to • crack a nut. We have thus far avoided problems using, • for example, calculus, am certainly not partial • derivatives. The Olympiad problems thus far have not • need calculus. However, if it becomes advisable to use mathematics • other than what we have thus far provided, am if our • readers want such materials, just let us know, and we • will present such materials.

!. • , •

the

Incidentally, Ken Burton did an excellent job on Bu1~rian problem on page 22 of the same issue.

; . Keep workinp; on the problems and sent in your • solutions. They will be duly acknowledged. •

•

•• •• • •

Dr. Samuel L. Greitzer Mathematics Department

Rutgers University

New BrunSWick, N. J. 08903

4 RELEVANCE IN MATHEMATlr-S We all know that geometry has had a hard time of it in our secondary school curricula. However, one of its strongest critics, has confessed that perhaps he has been too harsh as well as mistaken. This article is provoked by a remark in a book we were readinf.!;, stating that, " ••• the nine-point circle now has a distinct flavor of beautiful irrelevance ••• ". To add to the situation, the same book then went on to discuss the Schwarz Function. Frankly, we question which is more irrelevant - the nine-point circle or the Schwarz function. There is no question that mat~ematics changes with the times. However, nothing is actually discarded. For example, figurate numbers have not disappeared. They exist now as Ferrers Graphs. To wax historical, there was a time (until the end of the ninete~nth century, when the theorem stating that base angles of an isosceles triangle are equal was taut#lt in what we would now agree is archaic.

I'

Q

Given the triangle ABC with AB equal to AC, one first extended AB to P and AC to Q, with BP = CQ, drew lines BQ and CP, and Then proved that triangles APC and AQB were cotlgruent. One then proved triangles BPC and CQB congruent. Now, angles ABC and ACB, as supplements of equal angles, were proved equal.

Why the theorem was proved this way we do not kno""f. There were students who found this proof difficult, and who were dissuaded from further study of mathe­ matics. Also, there is a fanciful similarity of the diagram to a bridge. Therefore, this theorem with its diagram, was called the Pons Asinorum, or the Bridge of Fools.

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5 In the early days of this century, the nine-point circle seemed to take the place of the Pons Asinorum. The would-be mathematician had to have a proof of the theorem. For the (veryfew) who do not have such a proof, we dissect the theorem and indicate a proof.

I.

A·

~.

Ha..

I.

I.

A

A

(H\ We have separated the diagram for the theorem, which will be shown later, into three diagrams. We have the same triangle, ABC. At the right, in Figure (M), we have the midpoints of the sides forming the Medial Triangle, MaMhMc • It is similar to triangle ABC and the ratio 1:2 wi~h the original. The two are homothetic. Its circumradius is 1/2 that to the triangle ABC. The diagram in the middle (E) joins the midpoints of lines AH, BH and CH. It is called the Euler tri­ angle, and points E , E , E are called Euler points. b a c Note that this triangle is also similar to triangle ABC, the ratio of corresponding lengths is also 1/2, and, in fact it is also similar to triangle ABC. Thus the medial triangle and the Euler triangle are con­ gruent. The diagram at the left (H) joins the feet of the altitudes of triangle ABC. It is called the orthic triangle and is not necessarily similar to any of the other triangles. We shall now show that, with all the triangles in one diagram, the three triangles have the same circumcircle, which explains the term nine­ point.

•

Since E

am M are midpoints of BH am BA, line b c McEbis parallel to line AH am equal to half of it. Similarly, line EcM

is parallel to line AH am equal b to half of it. Since EbE is parallel to Be which is c perPendicular to AH (or AH) , EbE MbM is a rectangle a c c and is inscriptible in a circle. This rectangle has diagonals EbM

c

am EcM , which are diameters of the b

circumcircle. In the same manner ( you may draw in the lines ), you will fim that M E E M ia a rectangle with c a c a diagonals E M am EM. Having a common diap;ona.l, a a c c it follows that we already have six points on the same circle. Finally, we should note that EaMJIaEb is also a rectanp;le, and that the three rectangles have three diagonals among them. Now the diagram looks like this:

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We have the points E M EbMbE M all in the same a a

c c

circle. We have the three diameters EaM , EbM , EcM ' b a c Because EaHaMa is a right angle, Ha lies on the circle. Because EbHbM b is a right angle, Hb lies on the c1.rele.

Because E H M is a right angle, H lies on the circle. c c c c And thus we have arrived at the nine points which lie on the same circle. The center of the circle is at N, the intersection of the diagonals ME, MbE , EM. a a b c c We merely mention that this circle, the nine-point circle is tangent to the incircle and to the three ex-circles. A proof of this would take us too far afield. One last remark about this circle. In the diagram below, I ~i A

C

r""'--..,.-~~----...,

8 we have drawn B'C' on A parallel to BC, C'A' on B parallel to CA, and A'B' on C parallel to AB. Tr.is triangle is similar to triangle ABC with homothetic ratio 2;1. Now if we draw the circumcircle of ABC, we note that it apsses through the midpoints of the sides of triangle A'B'C', hence the circumcircle of ABC is the nine-point circle of A'B;C'. Also, AH cor­ responds to line OM , so AH = 2 'C+1 • a

a

From the two diagrams on the previous page, we can, using our imagination, derive many interesting properties of the various lines in the figures. We may, in a future article, do so. Right now, however, we have decided to concentrate on the orthic triangle.

"'-------lIl'---..;:aL'c

Remember that we have lots of right angles in this figure. Thus, LABH and LACH are equal - both are complements of LA. Next, BH HH and CH HH are cyclic. a

c

a

c

Now LH BH and LH H H are equal, as are LHbCH and c

c a

LHbHa H. Thus, angles HcHa Hand HbHa H are eq\'~a.l. That . is - each altitude bisects the angle of the orthic triangle to which it is drawn. Incidentally, if one labels the angles, one finds that MHbH "- lillH H·....· ACH H (but not &1 HbH ). This c c a a b -- a c is a sort of "trivia" to remember.

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9

Let us find the perimeter p of the orthic triangle. First, AH = b cos A and AH = c cos A, from the b c

diagram. Using the Law of Cosines,

222 222

(HbH) = b cos A + C cos A - 2bc cos Acos A 22 2 2 c = cos 2 A(b + c - 2bc cos A) = a cos A, whence HbH HaH b

=c

c

=a

cos C.

triangle is p

cos A. Similarly, H H

c a

Addin~,

=a

=b

cos Band

the perimeter of the orthic

cos A + b cos B +

C

cos C.

Using the Law of Cosines a~in, we substitute and 2+ 2 2 2+ 2 b2 2+ 2 2 ( b c - a) b(c a - ) + c(a b ~c) ~t P = a 2bc + 2ca 2ab which simDlifies to 2a2b~ + 2b~c2 + 2 c 2a 2 _ a 4 _ b4 _ c 4 p = ?abc This is a nice exercise in factoring. When we let c2

= (a

+ b)2, the numerator becomes zero, and we

have a factor. (The reader should try to substitute). Similarly, we find that, in addition to (a+b)2_ c 2, (b+c)2_ a 2 and ~c+a)2_b2 are factors. There is a problem here. The expression we wish to factor has degree four, but the product of the terms we have found to be factors has degree six. However, (a+b)2 - c 2 = (a + b + c)(A + b - c) (b+c)2 a 2 = (b + c + a)(b + c - a) (c+a)2 b2 = (c + a + b)( c + a - b) and we conclude that the actual relatively prime factors are just four in number. We test a bit, and finn p

= (a+b+c)(a+b-c)(b+c-a)(c+a-b)/(2abc.

10 This should remim us of Heron's formula for the area of a triangle. In fact, if the area of

~BC

is

A, we just multiply and divide by 16, and we get p

= &:::,2/(abc).

This formula is short and neat. We just make one more adjustment. Recall that we have shown in a previous issue that

A

= (abc)/4R, p

=

and we have, finally,

?,6./B.

If we let K be the area of the orthic triangle, and

\I

we find

the radius of the incircle of this trian9le, K = v A/R. Of course, we can keep on and

invent more formulas. Space keeps us from doing much more. However, let us do something about a value forv. Referring to our diagrams, we see that v

= HHa cos

Again labeling the parts of the diagram, we get (HHa)/a cos B : b cos C/(CH c )' which gives us HH = ab cos B cos C/(CH ) am when we substitite, a c v = (ab cos A cos B cos C)/CH • Remembering that A

= (CH c )c/2 = abc/4R, we v = 2R cos A cos

c

substitute and find that B cos C.

Now we can find the area of the orthic triangle, compare this area with that of the original triangle, and, by letting the imagination roam, invent many new and curious properties of this triangle.

A.

•• •• •• •• ••• •• •• • • •• •• ••• •• •• •• •

i.••

• •• •• •• •• •• •• ••• •• ••• ••• •• •• ••• •• •• •• ••

11

We believe we have shown enough to make it at reasonable that there is fun as well as profit from the investigation into a geometric situation. We present here some problems that the reader may like to tackle. a) What is the area of the orthic triangle ? b) What is the ratio between the area K of the orthic triangle and the area of triangle ABC ? c) Prove that, in acute triangle ABC, the orthic triangle has the least perimeter of all triangles that are inscribed in triangle ABC. (We recommend that the reader look at the proof in ''What is Mathematics ?", by Courant and Robbins, page 348. One might take a look at pages 349 and 350 as well.) d) In the diagram at the right,

perpendiculars Ha P and Ha Q

are constructed, and line

PQ drawn. Prove that the

length of PQ equals half

f3 the perimeter of the orthic triangle of ABC. e) In the diagram at the right, ABC am A' BC are inscribed

in the same circle on the

same l:ase Be (as shown).

H and H' are the ortho-

centers of the triangles.

Prove that HH' and AA'

are equal and parallel.

c.. 1-lG...

AI

"

The reader who thinks all this irrelevant can now go back to the area of mathematics considered more relevant - say catastrophe theory, normed semi-groups, etc.

•

12 PRCXttISCUOUS EXAMPLES P1)

When the polynomial

f(x)

is divided by

(x - a)(x - b), what is the remainder? P2)

Solve for

XI

eX - a)ex - b) _ ex - c)ex - d) X -

a - b

-

X -

C - d

P3)

Show that, given a,b,c > 0, then 2 2 2 2 2 a (1 + b ) + b (1 + c 2 ) + c (1 + a ) > 6abc.

P4)

Solve

(x + y)2

P5) Factor a(b4 _ c 4 ) p6)

= (x'+ + b(c4

1)(y - 1).

_ a 4) + c(a4 _ b4 )

Given p an odd prime, solve x

2

- 1

= 17p.

Prove that the product of four successive integers is never a perfect "quare. P8)

Show that three times the sum of the squares of three natural numbers is also the sum of four perfect squares.

Note I These examples can, of course, be solved by brute force, which may take some time. They can also be solved by looking at them in a "different" way. Try to find that different way.

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,e

••• • •• •• •

••• • •• •• ••• •

13 THOUGHTS ON AN OLYMPIAD PROBLEM Problem 1 on the 1985 International Mathematical Olympiad went as follows: A ~ircle has centre on the side AB of the cyclic quadrilateral ABrn. The other three sides are tangent to the circle. Prove that AD + B" = AB. Many contestants found this problem simple, and most of the solvers used trigonometry to solve it. In addition, students and adults - leaders and deputies included had their own solutions.We found the solution by Greg Patruno, USA deputy, so pretty that we wish to show it here.

D

With the dia~am as shown, let the an~le at C be 2a. Since the quadrilateral is cyclic, the an~le at A equals 180" - 2a. Since R, Sand T are tangent points angles OrR an~ oes equal a each, and the angle at "OS equals 90 - a. Also, tangentrR equals tangent rs. _Now rotate triangle oes until it takes the position arc. By addition of angles, we find that the triangle AOC is isosceles, with OA = AC = AT + TC = AT + CR. Similarly, we see that OB we finally arrive at AB = AD + Be.

= BR

+ TD. Adding these,

OA + DB = AT + CR + BR + TD, or

14

While we are at it, let us look at the fifth problem of the 1985 IMO. It goes as follows I A circle with centre 0 passes through the vertices A and C of triangle ABC, and intersects the ~eg­ ments AB and Be again at distinct points K.~~ N. respectively. The circumscribed circles of the triangles ABC and KBN intersect at exactly two distinct points B and M. Prove that angle CJt1B

is a right angle.

This appeared to be perhaps the most difficult problem of the entire IMO. This reminded us of a "theorem" we had stated in the Arbelos of March. 1985 - page 13. and we wondered i f it might be possible to use this "theorem" to solve the problem. It was !

•• •• •• •• •• ••• •• •• •• •• •• •• •• •• •• •• •• •• •• •

e e

15

e

First, line AC is the common chord of two circles. Therefore, line CAP is a radical axis. Similarly, NIl{ and BM are radical axes of their circles. These three lines must intersect at a point, P, the radical center of the circles. This gives us a large triangle PBC.

e e

Now OX is perpendicular to AC (and bisects it), and OY is perpendicular to BC ( and bisects CN.) If we can show that

e e e e e

•

•

(BM)2 _ (MP)2

+

(PX)2 _ (Xc)2

+

(Cy)2 _ (YB)2 = o.

e e e

then, from the perpend icularity of OX to CP and OY to ('!B, it will follow that OM is perpend.icular to BP. Notice first that (PM) (PB) = (PA) (PC) •

•

We now operate on each of the three pairs of tenns in the above.

e

• (BM)2 - (MP)2 = (BM + MP)(BM - MP) =

:: (BP)(BM - MP) = (BP)(BM) - (BP)(MP) = (BP)(BM) - (PA)(PC).

• (PX)2 - (Xc)2 = (PX + XC)(PX - XC) = (pc) (PA}.

:: (Cy)2 - (YA)2 • (CY + YA)(CY - YA) = -(CB)(BN}.

•

­­

e e e e

• e e

e It

e

e e e

­• e

Now add the three expressions that are underlined, and have arrived at the question - does (PB)(BM) that is, are

= (CB)(BN)

?

triangle BMN and triangle BCP similar ?

Well, since quadrilateral ACNK is cyclic, the angle Cl

at C equals angle BKN which equals angle BMN, and

angle MBN is common to both triangles. The triangles are similar, am (BM)(BP) does equal (BN)(B('!). Therefore, OM is perperrlicular to PB, am angle :sMO is a

ri~ht

angle.

This is an example of the use of simple properties of the triangle to prove a complicated result. We hope you like it.

• •• •• .

16 MAXIMA ANn MINIMA

Just as analytic p;eometry has made it possible to solve problems easily that are quite difficult to solve by synthetic geometry, calculus has made it possible to solve problems involving maxima or minima that are difficult otherwise. However, just as there .. are problems that may best be done by synthetic p;eometry. there are problems involving maxima or minima that can .. be done without calculus. •

.

In this connection, we recommend that the reader try to refer to two sources that are very interesting. The first is an article that appeared in Scripta Mathematica, Volume 18, entitled "No Calculus Please", by Butchart and Moser. We were amazed at what could be done by the two authors. The second is No. 6 in the Dolciani Series, published by the Mathematical Association, entitled "Maxima and Minima Without r.alculus", by Professor I. Niven. We ourselves have used methods of our own on occasions where they were simpler than formal calculus methods. We present here some of these methods, with a few problems. The point is that these methods offer the opportunity to readers to use their ingenuity. First, a simple observation: from the identity (x + y)2 _ (x _ y)2 = 4xy, we can see that a) if xy is a constant, (x + y) will be a minimum when (x - y)

= 0;

that is, when x

= y:

b) if (x + y) is a constant, xy will be a maximum when (x - y)

= 0;

that is, when x

= y.

Both results are easily extended to take care of the eventualities that, if the product of n terms is a constant, the sum of the terms will be smallest when all terms are equal, and if the sum of E terms is a constant, their product will be greatest when all terms are equal.

•• •• •• ••• •• •• ••• •• •• •• •• •• •• ••

e

•e •• e e •e

e

e

•e

e

e

17

Thus, consider a rectangle with terimeter 4p. Then we may consider its sides to be (p + a) and (p - a), and its area will be (p + a)(p - a) = K. Since the sum p + a + p - a = 2p is constant, the rectangle has maximum area when p + a = p - a, a = 0, and the rectangle is a square. Of course, it is even easier to note that K = p2 - a 2 which makes K a maximum when a is zero.

r - ------.-; I

,

I t

NIIl

:

e

e

e

e e

e

e e

e

e

•e

e e

-e

e

•e e e e

e e

e

e

Incidentally, this solves a popular calculus problem - that of constructing an enclosure of three sides, the f·onrth being a wall, using a given amount of fencin~. Clearly, if we reflect the enclosure as shown at the left, the resulting rectan~le must be a square. Hence y = 2x.

Next, if a triangle has constant perimeter, what shape will make its area a maximum? Here we use Heron's Formula in the form 2 K = s(s - a)(s - b)(s - c). Dropping the factor ~, which is a constant, K t,;~ill be a maximum when s - a = s - b = s - c. or a = b = c • The triangle is then equilateral. ¥ ~

yo'

Here is another popular problem: An oJ)en-top box is to be made from a square piece of cardboard whose side is ~ by cutting equal squares at the corners and foldin~ to form the box. What would be the maximum volume ?

):

We cut out the squares with side x and fold, The volume of the desired box will then equal 2 V = (a - 2x) .x •

~ a..~'

We will use a smidgeon of ingenuity here. Multiply

18 by 4 to get

4V

= (a

of the three factors

- 2x) 2 .4x. Now we have the sum

= (a

- 2x) + (a

2x) + 4x,

which is a constant. The volume will be a maximum or when x = a/6. The maximum volume will then equal 2aJ /27. Sometimes, one has to when

a - 2x

= 4x,

manipulate a bit to get a constant sum or product, and therein lies the fun and challenge. Thus, let us find when (x - J)2(x - 2) is a maximum. We multiply by

-2

to get

(x - J)2(4 - 2x). The

sum of the factors is (x - J) + (x - J) + (4 - 2x), which is a constant. Then

x - J

=4

- 2x, x

= 7/3,

and the maximum is 4/27. (a) Try your hand at maximizing where x less than a.

(a + x)2(a - x)3

Now what positive number added to its reciprocal gives the least possible sum ? Here, we write x + -1 = min. Since the product x ~ -1 = constant, x x the minimum occurs when x = 1/x, or x = 1. The minimum value is 2. To find when -a + -b + xy is a minimum, we note that x y the product of the three terms is a constant, abe a b \ We get a minimum when - =- = xy. It is easy to get x

y

y = bx/a, and, after a bit of algebra, the value of the minimum. (b) Find this minimum, from the result derived above, and then - by calculus.

(c) Find the least value of y = aekx + be-kx •

•• •• ••• •• ••• •• •• ••• •• ••• •• •• •• •• •• •• •• •••

•e e •e e

e

e e e e e e e e e

• e e

e e e e e e e e e e

•

­

e e

e e e

e

e

­• e

e

19 Sometimes, a very simple property will be enough to solve a proble~. For example, recall that the square of a real number is never negative. Now, when is

25 + 24x - 16x2

a maximum? Here, we will

"complete the square", thus: 2 25 + 24x - 16x 2 = 34 - 9 + 24x - 16x = 34 - (3-4x)2. This is obviously a maximum when () - 4x) = 0, or x = 3/4. The maximum .value is then )4. This is quite easy by calculus, but there can be more difficult ones. (d) When is 5 + 12 sin A - 9 sin2A a maximum, and what is that maximum value? Again, consider f(x) = ax 2 + bx + C (a positive). Multiply by 4a, and 4a2x 2 + 4abx + 4ac = 4f(x) .a. Completing the square we just add b2 and get

(2ax + b)2 + (4ac - b2 ) = 4af(x).

This will be· a minimum when when

(2ax + b) = 0, or

x = -b/2a.

It may be that an inequality relation can be used to solve a problem. For exami:le, we know that sin A + sin B = 2 sin A+B cos A-B 2 2 and since a cosine cannot exceed unity, we find ~

~

sin A + sin B < sin(A + B) o < A ,B < 180 • 2 2 First, this inequality can be extended to three or more an~les. That is, sin A + sin B + sin C < i (A + B + c) .3 _sn) and so on. Equality will occur when A = B = C. Similar results exist for other trigonometric functions,

20 so that

cos A + cos B

~---~-

2

< cos(

f!:

A

t'

tan A + tan B < tan(A

and

C'

-90::: A,B :: 90

o~

2

t

A,B, 90 •

These are all examples of Jensen inequalities, namely f(x)

T fey) > f(x + Y) for convex functions. The 2 2 inequalities are reversed for concave functions.

Problem: what is the minimum value for the sum of the sines of the angles of a triangle ? We may sin A + sin B + sin C = ( A + B + C) = sin 60° sin 3" 3 or sin A + sin B + sin C = 3 i/3 / 2. This will occur when A = B = C = 60' • We have already discussed the Arithmetic- mean ~eometric-

mean inequality.

ations, we have

AM

~

Usin~

the usual abbrevi­

GM , with equality

occurrin~

when all the elements are equal. We illustrate with a problem of a type communicated to us by Professor Eugene Levene, of Adelphi University. the problem is: 4 2 f(x) = 4x - ax 3 + bx - cx + 5 = 0 has four distinct roots, r , r , r , r 4 • We are given 1 2 3 I) r /2 + r /4 + r /5 + r 4 /8 = 1. 1 2 3 Find the roots and the values of a, b, c. The arithmetic mean of

terms in I is 1/4. The geometric mean of the terms in I is r r r r 1 2 3 4 320

~he

= .!.42 320 =

1 256

=

(l)~ 4;

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• •• •• •• •• •• ••

•• •• •• •• ••

••• •• •• •• •• •• •• •• ••

21 Since the AM and the GM are equal, the terms must all be equal to each other (and to 1/4). Hence r 1/2 r1

= r 2/4 = r)/5 = r 4 /8 = 1/4, and, solving, = 1/2, r 2 = 1, r) = 5/4 and r4 = 2.

The rest of the problem is now trivial. We now offer some problems for the reader to do. e) For what dimensions will a box without a top and with a fixed volume have dimensions a minimum ? f) A roofer wishes to make an open gutter of maximum capacity whose bottom and sides are each 4" wide and whose sides have the same slope. What should be the width across the top? g)

Given the function

u - (x - l)(x - 6)

(x - 10)

examine for maxima and/or minima.

h)

Given the function

u

=a

sin x + b cos x,

where a and b are positive reals, examine for maxima and/or minima. i)

If

a + b +

C

= 1,

for l/x + l/y + l/z

show that the minimum value is

9.

j) Prove that the quadrilateral of maximum area inscribed in a circle of fixed radius is a square. (Hint: Use the formula for a quadrilateral with sides a, b, c, d inscribed in a circle, namely, 2 K = (s - a)(s - b)(s - c)(s - d).

22 APrERTHOUGHTS In our development of the properties of the orthic triangle, we wished to exhibit some such areas as factoring, properties of trigonometric theorems, and the like. The reader knowa that we have a special interest on reviving geometry. With this in mind, let us look once more at the orthic triangle. Examine the diagr~ at the left. Looking at it, we remember

A

that there are right angles at Ha , Hb , Hc • Therefore AHcHH b is cyclic, the circle having diameter AH. Using the Law of Sines, we get

(HbH c ) = (AH)sin A. However, 2R = a(sin A). Eliminate the sine term, we have 2R(Ha Hb )

= a(AH).

Now we have shown that, if we construct the line OM a , from the circumcenter of the circumcircle of triangle ABC to the midpoint M of side BC, then a

(AH) = 2(OM a ). Substitution gives us R(HbH c ) = aCOMa)' But, if the area of triangle (ABC) is 6, then, first R(HbH c ) = 2(OBr). Similarly, R(HcHa ) = 2(or.A) and R(HaH b ) = 2(OAB). Adding,

we get R(HbH + H H + H Hb ) = 26. Finally, the c c a a perimeter of the orthic triangle equals 26/R. We find this shorter and easier than the proof we gave previously. Moreover, it is all geometry!

••• ••• •• •• •• •• •• •• •• ••• •• •• ••• •• •• •

,.!.•• ,.••

23

••• •• ~ •• ••• •• ••• •• •• •• •• •• •• •• •• ••

· •

let H P am. H Q be a a perpendiculars from H to a the sides am. draw PQ.

A~in.

I'

Again. APHaQ is cyclic. By the Law of Sines. L..-~::"-

~

c

(PQ) = (AHa )sin A. Once more. 2R(sin A) = a. Eliminate sin A. am. we arrive at

= a(AHa ) = 26. =AIR. Comparing

2R(PQ)

from which follow

(PQ) with the first result. we see that (PQ) equals half the perimeter of the orthic triangle. Again - all geometric! Now it might be fun to fim. such additional data

as the radius of the incircle of the orthic triangle. the area of the orthic triangle. am. much more. Avoirdupois vs. Metric

Some steam seems to have gone out of the drive for metrization. but we still meet with either (or both) in problems in science and mathe~atics. Hence it is useful to have some way of changing fran Avoirdupois to Metric (am. vice verse). We present the followin~ approximationsl Kilometers 1 2 3 5 8 13 Miles 1 1 2 3 5 8 etc. As for temperatures try the followingl to change fran

Fahrenheit to Centigrade - subtract 30 and take half; Centigrade to Fahrenheit - double am. then add 30. You will fim. the results close enough for most purposes.

24 INDETERMINATE FORMS We have most probably encounteted indeterminate forms. These usually take the form % (or perhaps 00/(0) for some given value of the variable. One such form, which we will find handy to examine is f(x)

= sin

x/ x, which

is indeterminate for x = O. To accustom ourselves to some useful notation, we will be

O....c.........:;;;----~T

.5

lookin~

for

Lim sin x • Substitution does result in -00 •

x=O

x

In the early days of tri~onometry, the various functions were actually I~presented by lines in a unit circle. In the diaP,Tam above, which shows part of a unit circle, the line AT is the tan~ent of the angle x at 0, the arc TN is equal to the an~le x, and the line NS is equal to the sine of the angle x. By inspection, we have sin x < x < tan x • Dividing each of the terms by sin x, we get 1 < x < 1 sin x cos x

o

Now, if we let x Thus L1m

x=o

sin x/ x

= 1.

We conclude that we can sometimes determine the limit of an expression by usin~ a dia~m. When the diagram is a ~ph, the point of indeterminacy is called a sinRular point. If we can give the point a value so as to make the ~ph continuous, we can call the point a removable sin~larity. Our aim is to present various ways of findin~ the value of an indeterminate form, and, for awhile, we will use sin x/x for our examples.

• •• •• ••• •• •• ••• •• ••• •• ••• ••• ••• •• •• •• •• •

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,.

•• •• •• ••• •

•

25 Just as Gauss used to take pleasure in finding a "chiliad" or so of prime numbers, we used to take time to find expansions for various functions. We used either Maclaurin's series or Arbogast's method. for further use, we ~ive you some of our results. sin x = x - x 3/3! + x 5/5! all x cos x tan x eX In(l+x)

=1 =x =1 =x

Now, to find

4 2 - x /2! + x /4! - • • • + x 3/3 + 2x 5/15 +

all x

-~2 < x < ~2

+ x + x2/2! + x 3/3! + - x 2/2 + x3/3 _ x 2/4 + Lim sin x

x=o

x

, suppose we substitute the

series above. We may divide the denominator x into 4 2 the numerator, getting 1 - x /3! + x /5! ­ and substitute x = 0, to get Lim sin x = 1. Somehow x=O x we believe this is easier than using the diagram.

"

Most of us have seen I'Hopital's rule, which we just sketch for our readers. (There is a story that I'Hopital paid Leibnitz for the formula, but we do not listen to libelous stories - as a rule.) Suppose we have functions f(x) and ~(x) such that f(a) = 0 and g(a) = O. Then f(a)/g(a) = 0/0 which is indeterminate. Let us rewrite the form above

~s f(a+h)h- f(a)/ g(a+h)h- g(a) • But in the limit, each expression becomes a derivative, and we have

.

Lim fL!l = Lim £:~ There is a hint here - if x=a SIX) x=a g (XJ the ratio of the d~rived expressions is indeterminate, we can continue to use second derivatives, etc.

•

26 A Lim sin x , we find that Using l'Hopita1's rule on x=O x Lim cos x when we differentiate, we get = 1.

x=O

1

We now have three ways of operating on an indeter­ minate form. The skill lies in selecting the one that is easiest. C id th Lim -Ja + x - -.,fa - x (= -0°) • ons er e x=O x We wouldn't think of a diagram for this. We might expand each by using the binomial theorem, but we think this might be too complicated. Of course, we could use l'H~pita1's rule, but this looks quite complicated, too. So - we will multiply numerator and denominator by ~a + x + -Ja - x. This gives 2x

xC,fa + x + ~a _ x)

• Cancelling and letting

x = 0, we find the value to be

1/"';;:.

It is a fact that many students, once they are aware of a general method for doing anything, stick to that method through thick and thin. I~enuity consists of selecting the shortest, fastest and simplest method. For example, examine Lim x - sin x

x=o

x

3

Using l'Hipita1's rule reqUires three differentia­ tions. Substitution is shortest and easiest. Question: What is the value of this form ? Suppose we try Lim x(1/1-x). Think first of x=1 what method or methods we might use. Then, take logarithms. This gives us log x

1 - x •

Now we may use 1 'H~pi tal's rule, which.

will give us the value -1, whence the limit is e- 1 •

•• ••• •• •• •• ••• ••• •• •• ••• •• •• •• •• •• ••• •

I.i.I. •• ,.•• ••

•• •• •• •• • ••

•• •• •• •• •• •• •• •• •• ••

I.

27 Here is another problem, with some hints for its solution: Find -2 x Lim(tan x) x=O x Here we took logarithms first and then substituted f or the tangent term. Question: what is the value of this indeterminate ? Since 00 is not a number, an expression that takes the form 00/00 is indeterminate, and we can use our methods on such forms •Anyway , we can always change such forms jo the 2form 0/0, if need be. Thus for Li m 2x - 23x + 43 x=o 5x - x - 7x we merely divide allover by x3 and we easily find

the desired limit.

Question: what is the limiting value for this?

We conclude with some problems for which you should

try to find the best way to proceed. Find Lim (ex _ e- x _ 2x)/(e x _ 1)3

x=o

Find Lim (sin x)sin x •

x=O

Find

Lim x + tan x - tan 2x x=O 2x + tan x - tan 3x

Find

Lim x loge1 + x) x=o 1 - cos x

28

OLYMPIAD PROBLEMS 6}

A polyhedron has 12 faces and is such that: (i) all faces are isosceles triangles, (ii) all edges have length either x or y, (iii) at each vertex e. ither 3 or 6 edges meet, and (iv) all dihedral angles are equal. Find the ratio x/yo

(BraZil)

7) Let A be a set of positive integers such that for any two elements x,y of A, x - Y ~ xy/25. Prove that A contains at most nine elements. Give an example of such a set of nine elements. (Czechoslovakia)

8) Prove that a triangle with angles a, ~, y, circumradius R and area A satisfies tan ,g + tan ~ + tan.x < 9R 2/4A. 2 2 2 ­ (German Democratic Rep) The weight w(p) of a polynomial P, where p(x) = Eaix i , with integer coefficients a i is defined as the number of its odd coefficients. For i

= 0,1,2, ••• ,

let qi(x)

= (1

+ x)i. Prove

that for any finite sequence 0 ~ i 1 < i ••• < in 2

The inequality

w(q + q + • • • + q ) > w(q ) i i i i n 1 2 1 (Netherlands) 10)

Decompose the number 51985 into a product of three integers, each of which is larger than

100

5 • (USSR)

Please send your solutions to your editor,

Dr. Samuel L. Greitzer Mathematics Department Rutgers University New Brunswick, NJ 08903

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29

.. , KURSCHAK KORNER This column features those problems of the famous Hungarian Ktirschak Contest which have not yet appeared in English trans­ lation. For the problems of 1894-1928, the reader should consult Hungarian Problem Books I & II in the MAA's New Mathematical Library series (available from the MAA hdqrs.); the problems (with brief summaries of their solutions) covering the years 1966-i981 were translated into English by Prof. Csirma~ of Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers.) Student readers are also invited to set aside an uninterrupted 4-hoUf period to compose complete, well-written solutiom to the problems below, and to submit their work to the address given below for critical evaluation.

1/1962. Let n be a natural number. Prove that the number of ordered pairs of natural numbers, (u, v) , for which the least common multiple of u and v is n, is equal to the number of positive divisors of n 2 • 2/1962. Prove that it is impossible to choose more than n diagonals in a convex polygon of n sides se that each pair of diagonals should have a common point. 3/1962. Let ABCD be a tetrahedron and assume that P is a point, distinct from D, in its interior or on its boundary. Prove that among the segments PA, PB and PC there is one which is shorter than one of the segments DA, DB and DC. Dr. George Berzsenyi 2040 Chevy Chase Beaumont, TX 77706

CONTENTS COVER: Geometry Problem i

Preface

1 Thoughts on Problem Solving 4 Relevance in Mathematics 12 Promiscuous Examples 1) Thoughts on an Olympiad Problem

16 ~ Maxima and Minima 22 Afterthoughts 2) Avoirdupois vs. Metric 24 Indeterminate Forms 28

Olympiad Problems

29 Ktirschak Korner

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e e e e e e

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arbelos

PRODUCED F'OR PRECOLLE~E

P}TIIJOMA THS

e

e e e e e

ABeD is inscribed in semicircle AE with unit radius GA. (

e

e e e e e e e e e e

-------4l-------4

1Io-=....

f=­

cn = c, DE = d. 2 2 Prove: a + b + c 2 + d 2 - a'be - 'bed is less than 4.

AB

= a, Be = b,

e

e

e e tit tit

e e

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e

e

e

1985-1986: Copyright~

No. 3

January, 1986

The Mathematical Association of America, 1986

1

COMMENTS ·We are happy to report that our mail has oocome much heavier and interesting during the Fall. We will do our oost to reply to some of it. But - keep your letters coming. We enjoy hearing from you all. Ms. Regina Grieco, of New Windsor, N.Y., was quite complementary. We do include solutions to problems occasionally, but prefer that readers supply them. Professor p,. Levine, of Adelphi University, has sent us some ooautiful problems, requiring ingenuity to solve. 'ile have included one in this issue, ann hope to present others in the future. Thomas Chung, of Apple Valley, Minnesota, sent in a fine solution to the "Lewis Carroll" Diophantine equation problem, and equally good solutions to our trigonometry exercises. Congratulations are due. Ambati Jaya Krishna has sent us several problems thus far. 'I'l e fear that they are not qui te "right "for Arbelos. Remember that it is "Produced for Precollege Polymaths". We have simplified one problem and it will 00 found in this issue. We hope to receive more problems which may be appropriate. Credit is here given to David Morin, of South Paris, f1aine, for his solutions to our "term-enders". S. Nam, of Dayton, Ohio, has sent in an excellent solution to our Cover problem of September, 1985. Incidentally, I used Stewart's Theorem to get a short proof. Stephen Smith, of Thornhill, Ontario, Canada wrote in suggesting that there be more references to some of the mathematical material. We do try to make what we write as self-contained as possible, but we have occasionally provided references. We are dubious about his suggstion that some articles on computers be included in future issues.

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E.L. Wilmer, of Stuyvesant High School, in New York City, solved one of our cover pag-e problems. He asks for permission to publish some Arbelos problems in the school mathematics publication. We are not sure this can be legally done. We suggest that he write Professor Mientka about this. The address will be found on the back cover. Through the y8ars, there have been repeated requests fer problems involving calculus. We may have, to accede to these requests, not to solve problems usually found in texts, but, as is our purpose, to involve calculus in special problems, provided such solutions are better than solutions not involving calculus.

We have a special interest in Stuyvesant High

School, since we were a student there at the beginning of the century. In fact, that is where we encountered the first problem of our "Many ~heerful Facts", way back in 1920. There are some schools that are quite ~ood in New York City, like the Bronx High School of Science and Hunter ~ollege Hi~h School. When we tau~ht at Bronx Science, that school was per-eminent in mathematics. It is now starting to re~in some of its excellence, and we eX])ect this improvement to continu8. '?ight now, we are trying to proo.uce an index of subject matter covered in our Arbc:os issues to date. This is surprisingly difficult to do, since we had no idea of how much we had presented. We hope to have something available before -the summe~ of 1986. At any rate, many thanks for the letters and for the complementary r~marks contained in them. Send in solutions, ideas for subjects you would like us to cover, suggestions for improvement, etc. We read them with pleasure and they make the task of preparing futur~ issues leas of a chore. Write! Dr. Samuel L. Greitzer Mathematics Department Rutgers Universi~y New Brunswick, NJ

08903

3 THE ELLIPSE Several problems and articles have come to our attention in recent weeks ~volvin~ the ellipse, so we thought it might be useful to consider this curve. We hope that the reader will find something useful and new in this article. First, "ellipsis" is defined in our dictionary as " omission of a word or words in a sentence". :;:xample, "He is taller than his brother (is tall)". Peaders should look up "parabole" and Hyperbole". I early r.reece, ronic sections were considered to be sections of a ri~ht circular cone (with one nappe) by a plane perpendicular to one element. acute right obtuse

(a)

(b)

• •• ••• •• •• •• •• •• ••

/L ••• (c)

When the angle at the vertex was acute, the section was called an ellip~~; when the angle at the vertex was a ri~ht an~le, the section was called a parabola; when the angle at the vertex was obtuse, the section was called an hyperbola. Note two peculiarities of this definition: no way a section be a circle, and only half of any hyperbola was obtained. In later Greek mathematics, Appolonius and Archimedes, as well as Euclid, used a definition that corrected these omissions • co~li

There are numerous ways of defining the ellipse.

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4

The most popular is to think of one as the locus of points such that the sum of the distances any point to two distinct points, called foci (sin~lar;focus) is a constant.

Although Dandelin's method of describing a conic is well-known, we will sketch it for the ellipse. The diagram above shows a cone and a section. We insert two balloons, one above the inteTse~ting plane and one below, so that each is tangent to the intersecting plane and to the surface of the cone. The points of tangency are F1 and F • For any point P on the curve 2 of intersectioi'!., PF 1 will equal PM, because tangents to a sphere from an external point are equal, and PF 2 will equal PN. Since PM + PN = MN is constant, we have P? 1 + PF2 = MN = constant. This, then, is a second way of-definin~ an ellipse, and it is the most common. Precisely the same procedure shows that the section of a circular cylinder by a plane not perpen­ dicular to an element is also an ellipse. This gives us a third definition of an u~lipse. What is more, we can now ~et the circle as the locus of points on a section whose plane is perpendic­ ular to an element.

5

Y

•

If we use algebra, let us place F

at (-c,a) and 1 at (c,a) on a rectangular coordinate system as

F

2 shown above. The definition ,tells us that FP

equals a constant - say 2a. That is, + c)2 + ~2 + c)2 + y2

V«X

.,i'cx _

1

+ PF

2

= 2a.

When we rationalize tbe above ( a pedestrian bit of alge bra), we end with 2 _ = 1.

y_ L 2 + _.oL. 222

a

Let 2 L 2 a

a

2

_ c2

+

x:b2

a

- c

= b2 , and we end with the familiar

=

•• •• •• •• •• •• •

1, or b2x 2 + a 2y2

= a 2 b2

We will call either one of these the "canonical" form for the equation of an ellipse. From it, we see that the curve is s~~etric with respect to both axes. Where it cuts the X-axis, we have x = a and x = -a. Where it cuts the Y-axis, we have y = band y = -b. The curve can be inscribed in a rectangle whose base is 2a and whose altitude is 2b. (Not) incidentally, the ratio cia = e is called the " excentric ity" of the ellipse, and it is less than one - another case of ellipsis. Bemember c = ae. Next, consider the diagram at the top of the next page:

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6 \/

We have two concentric circles. The smaller one has radius = OM = b. The larger one has radius = ON = a. Then, from the diagram, OV = a.cos e and PV = MU = b.sin e. Eliminating the trigonometric functions, we have once more, x 2/a 2 + y2/b2 = 1, the canonical form for the ellipse - still another definition

= ~(x + c)2 + y2. If we solve in 1 2 . the =a b 2 for 2 y and substitute 1n first expression (letting b2 = a 2 _ c 2 ), or b2 = a 2 (1 _ e 2 ), we find (more destrian algebra) Now, let 2 2 b x + a 2 y2

PF

1

=

(x +

PF

~e)2

2

+ (1 - e )(a

2

- x

2

=

a + ex.

e(x + ~). Now (x +~) is the e e equation of a line perpendicular to the X-axis, and

Write this last as

we have found that the distance from a point on the ellipse from the focus (-c,a) is equal to

e

times

its distance from the line x = -a/e. We call this line a directrix. Hence we have the last definition for an ellipse - it is the locus of points whose distance from a given focus F equals e times its 1

distance from a fixed line - the directrix. There are two directrices - one involving F 1 and one inv~l"'ing F2. Also, we have at five ways of defining - and constructing an ellipse.

7

v

Suppose (see diagram above) that we have tangent

UV a tangent to the ellipse b2x 2 + a 2 y2 -_ a.2~2 ,_. Th e easiest way to find the slope of the tangent is by calculus. We ~et b2x.dx + a 2y.dy = 0, from which the

*

2

2

slope is = - b x/a y. Draw PN perpendicular to the

tan~nt

at P and

the X=ax-s at N. Such a line is called a normaL Because of perpendicularity, the slope of 2 2 PN = + a Y/b x. The equation of the line on P and N would then be intersectin~

(y - Yt)/(x - Xt) = a2Yt/b2Xt· To find the coordinates of N, we let y = 0, and get (after some algebraic manipulation)

x = xte 2 •

We now remind the reader that a line bisecting an angle of a triangle divides the side it cuts into segments proportional to the sides ad.:jacent to those segments. Were PN and angle bisector, therefore, we would have (F P)/(F N) = (F P)/(F N). Let us see. 2 2 t 1 The length of F N is (c + e 2x) and the length of 1 F2N is (c - e 2 x). Remember that c = ae, and we can

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8

substitute, thus: for FiN, c + e 2x

= ae

2

+ e x

=

e(a + ex). Similarly, F N equals e(a - ex). However, 2 we have already shown that FiP = (a + ex). Also, F P = (a - ex). Therefore, we really have 2 (F P)/(F N) = (F P)/(F N). That is, normal PN is i i 2 2 really the bisector of the angle F PF • Finally, the i 2 angles marked a. are equal, so that the tangent makes equal angles with the radii vectores at point P. If there were such a thing as an elliptical pool table, a ball placed at F1 and struck would bounce off the cushion and pass through F2' regardless of

the direction in which it was struck. A noise made at one focus of an ellipse would travel outward and finally pass through F • This is the principle of

2 whispering galleries. One speaks in a normal voice at one focus, and the only place one can hear this voice would be at the other focus. Books have been written examining only the ellipse,

and we will not take the space to provide more of

its properties. We add only one further property:

2 2 2 2 2 2 the area of the ellipse b x + a y = a b is equal to nab. We can justify this by resorting to an orthogonal projection of a circle onto a plane not parallel to the plane of the circle. Or, one could use calculus. We will provide the reader with some problems based on ideas that have already been covered in previous issues of Arbelos. Concepts of projective geometry will be found very useful. Try them, for fun.

9

1)

Determine the envelope of all ellipses whose axes lie on the coordinate axes, whose centers are at the origin, and which have equal areas.

2)

An ellipse is inscribed in triangle ABC. From a focus F. lines are drawn to vertices A, B, C and to tangent point D on side AB. Prove that LAFD + LBFC = 1800 •

3)

Two pairs of paralle I tangents are drawn to an ellipse. Parallels to these are drawn through a focus and intersect these tangents in four points. Show that the four points lie on a circle.

4)

All escribed parallelograms of an ellipse have the same area.

5)

F P and F P are constructed perpendicular to 1 1 2 2 a line tangent to an ellipse at point T. Prove

6)

7) 8)

that (F P )(F P ) is a constant. 2 2 1 1 PQRS is a trapezoid inscribed in an ellipse, and PR and QS intersect at X. Tangents at Q and R intersect at T. Show that XT and PQ are parallel. The major axis of an ellipse is greater than any other line passing through the center of the ellipse and terminating on the ellipse. If F P and F P form a right angle, what is a 1 2

special relation between a and b ?

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10 Arithmetic Practice Sometimes, one has to relax from the ri~ors of difficult mathematical problems. One might read, listen to music, play ball, or else. We think it mi,q;ht be fun and relaxation for our readers to try the problems below. They are taken from ''Dubbs' Arithmetical Problems", written for students in grades 6, 7 and 8. The book was published in 1893. It cost 24 cents. (See what inflation does !) We have selected five eighth grade problems. Because of the age of the problems, we will use Roman numerals. solvin~

I.

Two circles, each 20 feet in diameter, are placed in such a way that the circumference of each just touches the center of the other. What is the area of the space between the centers?

II.

A, B, C, and D, whose ages are respective ly 19, 17, 15, and 13 years, inherited $13750, which was ~o divided that their respective shares at 1()'t: simple interest, amounted to equal sums when they ~rrived' at the age of 21 years. What was the share of each?

III. A concert hall is and 16 feet high. that a spider can a northwest lower corner ?

72 feet long, 38 feet wide What is the shortest distance crawl by floor a~d wall from corner to southeast upper

IV.

Four apples are worth as much as 5 plums; three pears are worth as much as 7 apples; eight apricots are worth as much as 15 pears. If five apples sell for two cents, what is the smallest whole number of cents that will buy an equal number of each of the four kinds of fruit?

V.

What is the greatest number that will divide 1166, 1558, and 2244, and leave the same remainder in each case ?

Send in your answers on a postcard, with your name and address, to Dr. Greitzer.

11

Rlementary Arithmetic? '·Te are always on the lookout for topics that may be interestinp-; and useful to our faithful readers. Some topics have been rather difficult (the Gamma Function), and some complicated ( continued fractions) but these have their uses in problem solvin~. We have just returned from a meeting of teachers in Manchester, N.H., where the main subject for discussion was "problem solving". We found the problems the speakers ~resented as examples to be actually childish, and hold out little hope for the success of the teaching of problem solving by the participants. At any rate, when we returned, we took a look at an arithmetic book we bought at a stall at a very low price. The book is titled '1)uboo' Arithmetical Problems", was published in 1893, and cost, when new, 24 ¢ .• It was intended for students in grades six throup-;h eight. It contained some problems of such difficulty that we thought it might be fun and relaxing for our readers to see some. First, as to the nature of the contents. There were problems on all sorts of business situations ­ taxes, insurance, interest, discounts, investments, commis~ions - what seemed to be a complete course in business matters. The student also had to kno~ about fractions, including complex fractions. There were problems involvinp.: lengths, areas and volumes of such figures as pri::=-.ms, cylinders, pyramids, cones, the circle and sphere - a good course in plane and solid geometry. Students had to know about arithmetic and geometric progressions, the Pythago~an Theorem, and how to compute square roots and cube roots. We take leave to present some problems for your delectation. We will solve some.

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12

(a)

What is the greatest number that will divide 1166, 1558, and 2244, and leave the same remainder in each case ?

Division by this number will result in a quotient, different for each division, plus the rame remainder. Thus, 1166

= Q1D +

r, 1558

= Q2D +

r, 2244

= Q3D +

r.

Subtract the first equality from the second, and we have 392

= XD.

Subtracting the second equality from

the third yields 686 = YD. Therefore, all we want is hi~hest

the

~uclidean

common factor of 392 and 686.

Usin~

the

alp-orithm, we find that this HCF is 98.

The remainder in each case will be 88. It would seem as if elementary school students were expected to know the Euclidean algorithm. (b)

We don't like this subject, so we will just present this problem. If 12 terriers can kill 600 rats in 15 minutes, how many rats can 30 terriers kill in 10 minutes? (1,000).

Continuing to geometry, we present this problem: (c)

The sides of a field are respectively 216, 63, 135, and 180 rods, of which the first two form a right angle. What is the area of the field?

Using the Pythagorean Theorem, the student must have found that one diagonal of the quadrilateral was 225 rods. Now we are not sure they had met with Heron's Formula, so we checl: and find that 225, 135 and 180 are also sides of a right angle. The rest of the problem is routine. However, consider how much the eighth p;rade student had to know to be able to solve.

13

The next problem is reminiscent of a popular (d)

pU$~le:

A watchmaker sold two watches at the same price each, gaining 2~ on one, losing 2~ on the other. If he lost $8 by the transaction, what was the cost of each watch ?

We leave the solution of this problem to the reader. (Can you get the answers - $80 and $120 ?) We remember eX:l.r.lples of the next type from our own experiences in elementary schooL (e)

A jockey sold a horse· for $131.25 and thereby gained as many percent as the number of dollars he paid for it. What did the horse cost?

Again, we leave it to the reader to decipher this problem and find the cost. ($75). Now we come to problems that would be difficult for many students (and some teachers). This time, we will not provide solutions. (remember that our book has answers, so we know these. However, we did solve each problem on our own, just for fun.) (f)

A concert-hall is 72 feet long, 38 feet wide, and 16 feet high. What is the shortest distance that a spider can crawl by floor and wall from a northwest lower corner to a southeast upper corner? (Caution: check all possibilities !)

There is some confusion about the statement of the next problem. We don't know whether each side track consists of two rails side by side, or just one rail. We shall assume that each track has two rails. Now (g)

A railroad company built three side tracks, 6699, 8671, and 9367 feet in length. What is the greatest length of rail that could be used in their construction ?

We assume that the company wants to use just one length of rail for the whole construction.

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14

We now corne to some problems that are worthy of being includes in a contest for gifted students. They are, we think, far from simple. Here goes: (h)

A, B, C, and D, whose ages were respectively 19, 17, 15, and 13 years, inherited $13750, which was so divided that their respective shares, at 10% simple interest, amounted to equal sums when they arrived at the age of 21 years. What was the share of each ?

(i)

A man, dying, left $23500 to be divided among his widow, son, and daughter, on condition that if the daughter dies and the son survived, he should have 5/8 of the money, and the widow 3/8 of it. But if the son died and the daughter lived, she should have 5/9 of the money and the widow 4/9 of it. Both son and daugh~~ survived. What did each of the heirs revlve ?

(j)

Four apples are worth as much as five plums; three pears are worth as much as seven apples; eight apricots are worth as much as fifteen pears. If five app~s sell for two cents, what is the smallest wh61e number of cents that will buy an equal number of each of the four kinds of fruit?

If you solve the last four problems, perhaps you would send us the solutions, on a post card. and we will acknowledge them. There are three final observations: first, the numbers in the examples and problems were sometlmes huge. The students were expected to work without any calculators. There were none. Next, most of the topics taught then are now found in upper classes, if at all. Third, the language is impeccable. The text consists of promiscuous examples and miscellaneous problems. How many of now would recognize the difference ?

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15 MANY CHEERFUL FACTS We assume that our readers have read and remembered most or all of the algebraic theorems we have presented in Arbelos, or have encountered them in their reading, or even met them in class. Therefore, if one remembers Cardan's Method for solving a cubic, one can then solve any other cubic having the same form. Or, if one knows Heron's Formula, one can find the area of a triangle whose sides are known. We can consider solving problems like these as mere "pedestrianism". The skilled solver looks at such rules and formulas in a different way, and can therefore solve difficult or even seemingly impossible problems by means of a different outlook. For example, we have already met the Factor Theorem and the Remainder Theorem. For the sake of the three who do not recall these, we restate them. When a polynomial f(x) is divided by (x - h), we get a quotient Q and a remainder R. Symbollically, f(x)

= Q(x

- h) + ~.

From this; we easily derive two results: (a) if R

= 0,

(b) f(h)

=R

(x - h) is a factor of f(x)

Now consider f(x)

(Factor Thm) (Remainder Thm)

= (x-r1 )(x-r2 )(x-r3 ) ..• (x-rn ).

We can also write this in the expanded form as f(x) = ax n + bx n- 1 + cx n- 2 + • • • + k. Since (x

r ) is a factor of f(x), R = 0 for this case. 1 + k = O. However, R = ar n + br n-1 + cr n-2 + 1n 1 1 n-2 + br n-1 + + k = 0 Also, R = ar + cr 2 2 2 and so on for each of the r • i Now suppose we abbreviate a bit, and let si equal the sum of the i-th powers of the roots. When we add all n remainders, we get

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16 as

n

+ bs

n-

1 + CS

n-

2 + • • • + nk

= 0.

We first met up with this when asked to solve x + y + z =6 x2 + y2 + z 2 = 14 x 3 + ? + z3 = 36 As stated in our first paragraph, we suddenly got the idea of lettin~ x, y, z be roots of the cubic ap3 + bp2 + cP + d

= 0.

+ b = 0,

1 aS + bs + 2c = 0, 2 1 aS + bs + CS + 3d 2 1 3

Then we could write

Since we knew that

aS

= 6,

s1

s2

= 14,

s3

= 0.

= 36,

we easily

found a, b, c and were able to write the cubic as p3 _ 6p2 + l1P - 6

= 0.

This factored easily, and we found that x, y, z be the various arrangements of

could

1, 2, 3, six in number.

The following problem occurred on the USA Math Olympiad, for 1972. It was difficult then. Try to solve it now. x + Y + z =3 222 x + y + Z =3

x5 And just for fun,

+

r

+

~iven

z5

=3

the cubic equation

x 3 - x - 1 = 0, find the value of s6. Sometimes, looking at a familiar situation in a different way may lead to something unique in the way of a solution.

17

It may be that all this is not too startling. Many students know thp recursion formulas for powers of roots before. In the literature, they are referred to as Newton's Formulas. We think the next example is more surprising. It arrived thanks to two of our correspondents ­ Professor E. Levine, of Adelphi University, and Dimitrios Vathis, from Chalcis, Greece. Let's take the problem a la Levine first. It goes ­ 4 You are given the equation x - ax3 + bx 2 - cx + 30

= O.

The roots are r , r , r , r and are real. If 1 2 3 4 r1 r2 ~ r 4

~ + ~ + 6 + S- = 2, solve for the roots

(and find the value of

b).

Why we thought of inequalities in this case we cannot explain. However, we did, thus: We have already presented the Arithmetic Mean - Geo­ metric Mean inequality before. That is, we know that AM > GM. Somehow, one never thinks of the possibility that the two may be equaL We have four fractions whose sum equals 2. Hence the AM of the four fractions is 1/2. The GM equals _ 1

'. 4jr1r 2r 3r 4 "\ 480 ' and,(since r 1r 2r r 4 - 3 0), equals 2"'

3 The AM and GM are therefore equal. This means that r

1

r

2 . -.'"

~

>:'4 _ 1

1, r 2 =22' 1 = r 4 = 4. Now you can find a, b, c and d. We think this is an unusual way to use inequalities.

2= 5" r = 3 , anc' 3

= "8 - 2"'

Hence

r

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e

18

• It

e

We take another problem. The idea comes from • Ambati JayaKrishna but we have simplified it quite a It bit. Suppose we are given the biquadratic equation 4 2 • f(x) = ax + bx 3 + cx + dx + e = 0

It which has roots, r , r , r , r • We also know that 1 2 3 4 It sl = 12 and s2 = 36. Can we solve the equation?

e

e

First, we cannot use Newton's recursion formulas here, because just trying them will not give us d and e. However, we just happen to notice that (r + r + r + r ) 2 = 144 = (r 2 + r 2 + r2 r )x4 . +2 4 4

1 2 1

2

3 3 and we suddenly remember ~auchy's inequality, which says (La b)2 < (La 2)(Lb 2) i i i i· We have equality if and only if the sequences a and b are proportional.

•

e e e

e e e e Rewriting the expressions for s1 and s2 thus, (r xl + r x1 + r x1 + r 4x1)2 = e 1 2 2 3 2

2 2 2 2

2 2 e

(r + r + r + r4 )(1 + 1 + 1 + 1 ) 2 1 3

e

2 e or 12 = 36 x 4, we see that the condition for equality e in Cauchy's expression holds, so that we must have

•

­

e

•e e

Thus f(x)

= (x

4 - 3) •

We hope this solution is sufficiently unique to rouse the interest of the inquiring student.

Naturally, we cannot guarantee that every e has a unique solution of this type or other, e

hope to have shown that, sometimes at least,

e

e e

•e

e e e

e e

problem but we do one can invent an ingenious solution for what would otherwise seem to be an impossible problem. You have to look for it.

19 The Pappus-Guldin Theorems We present here a pair of theorems that will be found useful to mathematicians and physicists al1k€ . They were originally stated by Pappus (300 AD) but were rediscovered by Guldin about 1200 years later. We state these theorems after Leibnitz, who gave a generalization about 1690 AD. The statements ares (a) If a plane curve moves so that the path of its center of gravity is always perpendicular to the plane of the curve, then the area of the surface generated is equal to the product of the length of the curve and the len~th of the path described by the center of gravity of 'thp curve. (b) If a plane surface moves so that the path of its center of gravity is always perpend icular to the plane of the surface, then the volume of the solid generated is equal to the area of the surface multiplied by the length of the path described by the center of gravity

of the surface.

We will restrict ourselves to the cases where the path of the center of gravity is a circle, but we can always arrange for the case where such paths consist of pieces of circular arcs. First, let us try to justify these rules. In the first case, let s be the length of the curve, y its distance from th~ axis about which the curve is to be rotated, and y the distance from the center of gravity of the curve. Since the curve behaves as i f all its "mass" were concentrated at the center of gravity, we find, on summing, that, for each piece ds, Jyds = Ys. Since Jyds equals area, A = Ys. (Please note that we are only "justifying" «ur result and not presenting an ironbound proo~. In the same way, let A be the area of the curve in statement (b), and dA an element of area. Then

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e

•e • •e e e

e e e e e

e

e

tit

e

e e

­

• •

e e

e e e

e e e e e e

e e

•ee e tit e

e

20

2nyA :': D:2Try i,\A

= 2rrJJydydx = rrJy2dx = Volume

V.

We shall restrict ourselves to the case where the curve or the surface lies in a plane, rotates about an axis lying in that plane, this axis not intersecting the curve or surface. The path of the center of gravity will be a circle. The ingenious reader will be able to extend these results. First, the area of a circle. Let the circle be generated by a rotating radius of length R with one end on point O. The center of gravity of the radius is at the midpoint. Therefore, the area of the circle will be (R)x(nR) = nR 2 • Next, for the volume of a ~ight prism with base B and altitude h. When the base B rises (lika ~n elevator), its center of gravity travels a distance h to the upper base. Therefore the volume produced is V = Bh. (Note that here we did not have the center of gravity traveling in a circle). To find the volume of a circular cylinder with rad1~.~s of the base R and height h, consider a rectangle with altitude h and base R rotating about altitude h. The center of gravity of the rectangle is halfway up the rectangle at a distance of R/2 from the axis of rotation. The area of the

is Rh. Therefore, the volume is (Rh)x(nR)

rec~ngle

= nR

h.

A right circular cone results when a right triangle is revolved about one leg. In the figure at the left, triangle VAB has altitude VB = h and bas AB = R. The center of gravity of the triangle is at G, where the medians intersect • Hence the volume 11'

V = (Rh/2)(rrR/3) '= ~2h.

21 Of course, we all knew all this already. Now consider the torus. First, let the surface rotated be a circle with radius R. Let the center of this circle be at a distance r > R from the axis about which rotation will take place, this axis lying in the plane of the c irc Ie. The center of gravity 2f the circle is at its center and its area is nR • Hence the volume of the

2

2

torus is (nR )(2nr) = 2n2R r. Incidentally, what is the formula for the surface area of this torus ? Now assume we have an equilateral triangle with side s. If this triangle be revolved about one edge as axis, what is the volume of the solid thus formed ? And what if the triangle were to be revolved about an axis passing through one vertex and parallel to the side opposite? Physicists can use Pappus' theorems to locate centers of gravity. Where would the center of gravity of a semt~trcular arc be located? Well, i f the arc were to be revolved about the semidiameter, we would end up with a spherical surface with area 4nR 2 • The length of the arc is nR. Hence 4rrR 2= (nR) (2ny), or y = 2R/n • Where would the center of gravity of a hemisphere be located ? Here we know the volume of the hemi­ sphere, ~3. It is formed when the semicircle with (

radius R is revolved 180 • Therefore, we have 2 3 ~

= (nR 2 /2)(21TY)

We end with one problem as physicists might find quadrilateral, how would gravity of such a figure

2R

, or y ~ • that mathematicians as ~ell interesting; Given a one locate the center of ?

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22

Pythagorean Diversions

We are all aware that, if a = m2 - n2 , b = 2mn, and c = m2 + n2 , then a, b, c can be sides of a right triangle. If m and n are relatively prime,

It

e

e e e

e

e

with one odd and the other even, then a, b, c may form a primitive

pytha~orean

to present some curiosities

e

Diophantos has the following problem: Find. a pythagorean triangle in which the bisector of one of the acute angles is rational.

­

e

A

fit

e

•e

f t

\3

CL

e e

­­ -­e e e

e

e e

e

e

e e

--e

involvin~ p~~hagorean

triples.

e

e

triple. We would like

We know that an angle bisector divides the side to which it is drawn in the same ratio as the adjacent sides. Thus AT = xc and ~T = xa. Since xa + xc = b, we finally find that ab CT = a + c • Now we use the

theorem on trian~le BeT to find BT. We leave the details of the computations to the reader

pytha~orean

and give our result only- t 2

= a 2c/m2 •

If we let

c be a perfect square, we can find a rational t. For example, let m a

=4

2

- ]2

= 7,

b

=4

= 24,

and

and c

n

=4

= 3. 2

+

Then

32

= 25.

The bisector of the acute angle that cuts the side whose measure is an even number (namely b), should then equal 8 ]/4. We note that the bisector must end at the side that has even number measure. Can you prove this? Next, what are the dimensions of another right triangle, not congruent to the above, which has a rational bisector ?

•

23

We remember that the Pythagoreans held numbers in reverence. They had lucky and unlucky numbers, abundant numbers and deficient numbers, perfect numbers, and so on. They noticed that the numbers 222 1 ,5, 7 or 1, 25, 49 formed an arithmetic progression. The common difference, 24

was called

a congruum (not to be confused with congruence). Naturally, they wondered whether there were any other similar triples. a congruum is difficult, but we can do in a special case. Consider the following: 2 2 a 2 + b = c, and c 2+ k = x2 ,c 2 - k = y2 Findin~

somethin~

If we allow the congruum k to equal 2ab, we ~t (a? + b2 ) + 2ab = (a + b)2, (a2 + b2 ) - 2ab = (a - b)~ and we have three squares in arithmetic progression. 2 2 When we let m = 4, n = 3, we get (4 + 3 ) = 25, The congruum is 2x4x3 example.

=

24, which explains our

For a triangle with sides 5 - 12 - 13, we have 52 + 122 = 13 2 • We take as congruum k = 2x5x12 = 120. And we have

169 + 120

= 289 = 49.

120 169 Obviously, there are an infinity of such relations available. We have noticed that every congruum, for the special case we have accepted, is a multiple of 24. Can any reader prove this? And, finally, are there any cases where four perfect squares are in arithmetic progression? We have not (yet) found a solution to this problem.

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24

We read, with interest, a note from Lewis Carroll in which he said he had worked on a problem for hours without solving it. The problem - find three pythagorean triangles with e~(-lal areas.

e

We can get two triangles easily. The triangles with sides 20 - 21 - 29 and 12 - 35 - 37 both have areas of 210. But a third ?

•

Thanks to Henry Dudeney's "Canterbury Puzzles", we have a formula that will p;ive us three such trian~les. It is the following: Let m = u 2 + v 2 + uv

e e e e e e

e e

It

e

­

•

e

e e e e e e e

e e e e e e e e e

­

e e

n = u P

2

- v

= 2,.uv

2

+ v

2

If you form three pythago~an triangles with generators m and n, m and p, m and n+p, says Lewis Carroll, you will get three triangles with equal areas •

Starting with u m = 37,

n

= 7,

= 4, p

v

= J,

he got, first,

= 33.

The generators of the three triangles were then

37 and 7 , 37 and 33 , 37 and 40. The right triangles had dimensions (a) (b) (c)

1418

280

1320 2442

2)1

2960

2969

518

2458

\-le feel that there are smaller triangles than the ones above. A few minutes on a computer should ~ive us better results.

Finally, Montucla says that, if one accepts fractional results, one can get an infinite number of such triangles. He missed the possibility of multiplying by the common denominator of these fractional sides. Hence it is possible to find, say, four right triangles with equal areas.

•

25 Olympiad Problems 11)

Find the maximum value of 2 2 2 sin e + sin e + • • • + sin en 1 2 subject to the restrictions

e1 12)

T

e2

+ ••• +

0 <

en = n.

ei

~

nand

(Canada)

Prove that the product of two sides of a trian~le is always ~eater than the product of the diameter of the inscribed circle and the diameter of the circumscribed circle. (Israel)

1'3)

Determine all functions fIR ~R the followin~ two conditions: a) f(x + y) + f(x - y) x,y in R, b) lim f(x)

satisfyin~

= 2.f(x).f(y)

= O.

for all

(Viet Nam)

x=><>

14)

E , E and E) are three mutually intersecting 1 2 ellipses, all in the same plane. Their foci are respectively F2 ,F) ; F),F 1 ; and F 1 ,F 2 • The three foci are not on a straight line. Prove that the common chords of each pait of ellipses are concurrent.

15)

(Gt. Britain)

Determine the ran~e of w(w + x)(w + y)(w + z) where x, y, z and ware real numbers such that x + y + Z + W = x 7 + y7 + z7 + w7 = O. (USA)

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KURSCHAK KORNER

This column features those problems of the famous Hungarian Kiirschak Contest which have not yet appeared in English trans­ lation. For the problems of 1894-1928, the reader should consult Hungarian Problem Books 1 &'.!!. in the MAA's New Mathematical Library sertes (available from the MAA hdqrs.); the problems (with brief summaries of their solutions) covering the years 1966-1981 were translated into English by Prof. Csirmaz of Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers.) Student readers are also invited to set aside an uninterrupted 4-hour period to compose complete, well-written solutions to the problems below, and to submit their work to the address given below for critical evaluation. 1/1959. Prove that if x, y and z are distinct integers and if n is a non-negative integer, then the sum

x"

(x-y) (x-z)

+

y" (y-x)(y-z)

+

z" (z-x)(z-y)

•

is also an integer.

• • •

2/1959. Let a, l3 and y denote the angles of elevation of a tower measured at distances of 100, 200 and 300 meters from the foot of the tower, respectively. Find the height of the tower if a+l3+y=90°.

•

• 3/1959. Three brothers visited a patient in the hospital • on the same day. Their wives were also visiting the patient on that day. None of them were there more • than once, and yet each of the brothers met both of , • his sisters-in-law at the patient's bed. Prove that • one of the brothers must have also met with his wife

•

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there. Dr. George Berzsenyi 2040 Chevy Chase Beaumont, TX 77706

CONTBNTS COVER: r.eometric Inequality 1

~omments

3

The Ellipse Arithmetic Practice Elementary Arithmetic ? Many Cheerful Facts Pappus-Guldin Theorems Pythagorean Diversions Olympiad Problems Kurschak Korner

10 11

15

19 22 25 26

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arbelos

PRODUCED FOR

PRECOLLEGE

PHILOMATHS

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Given: Inscribed triangle ABC with angle bisector AM. MP is perpendicular to radius AO extended if necessary. P lies on side AC. Prove: AP = AB.

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1985-1986~

No.4

March, 1986

Copyright ~ The ~1athematica1 Association of America, 1986

i PREFACE

AmonB; the very few readers who write us, we now have Ambiti Jaya Krishna, who writes on stationery from Johns Hopkins University. We are including in this issue an article of which he is a co-author, which we hope will be instructive. In the Preface of our November issue, we did present two problems bised on ideas by Mr. Krishna. We have received solutions to one of these from two readers - Richard Chai, of Livin~ston, N.J., and Ed Mullins, of Hurricane, W.V. Both used the Cauchy Inequality formula to solve it - and very ~ood they were. In addition, we received a solution to Olympiad Problem No.7, of the same issue, due to Rahul Pandharipande, of Urbina, Ill. The problem, from Czechoslovakia, was quite complicated, and the solution very clear. We were not aware of it, but now know that Arbelos is read in other countries - in Europe and South America. We have gone international at last. Once a~in, as we have for 10, these many issues, we are asking for letters. These may include solutions to problems, problems you would like to see in an issue, suggestions for articles you would like to have presented, and, most appreciated, articles you have written and wish to have included. Finally, perhaps you would find it useful to let others know about Arbelos. We can always use the subscriptions. Your Editor (etc.) Dr. Samuel L. Greitzar Mathematics Department Rutgers University New Brunswick, NJ 08903

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An Introduction to Number Theory

In this article we present a trick with which you can amaze your non-mathematical friends ! What is the cube root of 262144 ? Why, 64, of course ! This may not be obvious now; however, this article's purpose is to make this fact obvious. Without further delay, let us proceed. We first

construct a tablet 3

-n

o

••

•• •• •

•• •• •• •• •• •• ••

,.

1

-n 0

1

1

2

8

3

27

4

64

5 6

125

216

7

343

8

512

9

729

We notice that the last digit of every number in the

n3 column is distinct. Therefore we know the last digit of any perfect cubet e.g., since 262144 ends with a 4, we see that the cube root ends with a 4. We probose the follOWing strategy for determin­

ing the cube root of any perfect cubet First divide

the cube N into two parts - namely, the three right­

hand digits and the other N - 3 digits. Again, in

the case of 262144, we have 262 and 144

. Now we determine the other digits of N113. We note that (x + 1)3- > x3 • We claim that the

•

2 1 cube root of N is lOx + the last digit of N / 3 (previously explained). Since

73 > 262 > 63 , (262144)1/3

= 10(6)

+ 4

or 64. The proof is straightforward. All that is required is the memorization of a few cubes. So ­ what is the cube root of 1815848 ? why, 122, of course. This small example of number theory will, we hope, persuade more students to venture into the realm of mathematics. Am bati J aya Krishna Mrs. Gomathi S. Rao Prof. A. Murali Mohan Rao

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Editorial Notes I 1) My copy of "Mathematical Recreations am. Essays" by W. W. Rouse Ball, Edition 13, Chapter 1, has some very nice methods for discovering numbers

from what would appear to be insufficient

information. 2)

"Amusements in Mathematics", by Dudeny, page 13, has some information on digital numbers that the reader will fim. useful. For example, the digital number (what you get when you add the digits of a number until just one is left) for a cube is 1, 8, 9, 1, 8, 9, ••• • Now 1dl0968, the number in the article subnitted to us, has a digital sum equal to 6, so it isn't a perfect cube • I took the liberty of substituting 1815968, which is a perfect cube • S. Greitzer

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Tetrahedra In view of the present status of geometry in the usual classroom, it is not surprising that very little, if any, time is devoted to solid geometry. Those of us who are acquainted with rnathernatics contests held in other countries know, however, that solid geometry is still considered important enough to be studied. Many such problems involve the tetra­ hedron, so we shall present SODle data about this figure. First, there are four points in )-space, called vertices. These points are not co-planar. Next, we join these points to one another, which is done by using six line segments. These are called edges. Finally, each triad of edges forms a flat surface, called a face. There are four of these. Note that for this figure, if we let V be the number of vertices, E the number of edges, aM F the number of faces, then V - E + F = 2.

e.­ Any two faces form an angle, called a diJledral angle. For example, one such angle is A-BC-D. We can measure such angles by means of associated plane angles. Select a point P on the "vertex" of the angle and construct UP and VP lying in the faces of the angle and perpendicular to Be. The number of degrees in LJJPV is the number of degrees in the dihedral angle. Angles like ABC, BCD, etc., are called face angles. .

4

We use right triangle AHD to find that the altitude AH is equal to of the regular

s-/6/3, and we can now find the volume

te~edron to be equal to s3_/2/12.

Per.haps we might mention that, on the cover of the very first of our issues, in September, 1982, Me had a formula for the volume of any tetrahedron in terms of the six edges. In most cases, we can find the base and altitude of a te~hedron and use the rule. V

= Bh/3,

which we have previously derived elsewhere.

It is easy to show that the "medians" of the tetrahedron, lines from the vertices to the centroids of the opposite faces, do intersect. Assume that they intersect at a point O. Remember also that for the regular tetrahedron, these medians are also altitudes. Drawing in all the medians divides the tetrahedron into four pyramids which, in this case, happen to be equal in volume. Also, recall that the volumes of two pyramids with equal bases have a ratio equal to the ratio between their altitudes (ond medians). This means that the altitude of, say, O-BCD is 1/4 that of A-BCD, and we have the Thma the median of a tetrahedron divide each other in the ratio of 3 to 1.

D

(;.-

There is a theorem due to Varignon that says, Thm: The midpoints of the sides of any quadrilateral form a Paralle logram lying in a plane.

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5 An altitude of a tetrahedron is a line on a vertex perpendicular to the opposite face. There is no guarantee that two altitudes will intersect, but, Thml If two altitudes of a tetrahedron intersect, so do the other two altitudes. It is possible to construct a sphere passing through all four vertices. We leave the method of doing this to the imagination of the reader. In order to construct a sphere inscribed in a tetrahedron, one must bisect the dihedral angles - at least three of these. Naturally, the bisector of a dihedral angle is a plane. All points in this plane are equidistant from the faces forming the dihedral angle - a fact ueeful to know. Three such angle bisectors meet at a point, which is the center of the inscribed sphere. Also, Thml

If the centers of the circumscribed sphere and the inscribed sphere coincide, the faces of the tetrahedron are con~ent.

We have encountered a number of problems involving either a regular tetrahedron or an isosceles tetra­ hedron. Let us examine the regular tetrahedron first. For a start, all four faces are equilateral triangles. If we let each edge equal s, then the area of each f~ce will equal 2 s 0/4 and. the whole surface area equals s -(3.

To find the volume, we need the alti tude AH. Now H is the intersection of the medians of ABeD, and we know that DH is 2/3 of DM. Therefore, DH = s-fj/3.

6

For a regular tetrahedron, the 11nes joining the midpoints of opposite edges form three squares with edges equal to s/2. This gives us the Thml the lines joining opposite edges have length s-/2/2. We have noticed that a plane bisecting dihedral angle A-BC-D of a regular tetrahedron will also bisec t the edge AD. This leads us to c onjec ture I Thml In any tetrahedron A-BCD, the bisector of the

dihedral angle A-BC-D divides the edge AD in the same ratl0 as the areas of faces ABD to ACD. This is a sPaCe analog of a well-known theorem in plane geanetry Since we have already referred to the general tetrahedron, we go on to discuss the isosceles tetrahedron. This is defined as a tetrahedron such that opposite edges are equal.

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A~--~---~A

If we cut the isosceles tetrahedron along edges AB, AC am AD, am spread the figure as above, we see that the four faces are congruent a,d that the sum of the angles at each vertex equals 180 • Query I If the sum of the angles at each vertex of a tetrahedron is 180 , is the tetrahedron isosceles?

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If we apply Varignon's theorem to an isosceles tetrahedron, we will not usually find. that the lines joining the midpoints of four edges form a square. However, they will form a rhoabus, and we may says Thms In an isisceles tetrahedron, the lines joining the midpoints of opposite edges intersect at right angles and are perpendicular to the

edges. Incidentally, the above was stated as a problem on the national mathematical olympiad of one country. Possibly we should mention trihedral angles. These are formed by the intersections of three planes. They are to be found at the vertices of every tetrahedron. It might be useful to remember sone properties of these angles for tetrahedrons. Since there are three face angles at each trihedral angle, the sum of any two of these face angles is greater than the third face angle. There is a pretty theorem that is worth recalling.

Thms If two trihedral angles have the three face

angles of one equal respectively to the three

face angles of the other. their corresponding

dihedral angles are equal.

We end with a question - or a problem, if you likes Querys Is it possible for a quadrilateral that is not planar to have all four angles right angles? We hope that if, in the future, the reader c OIIles

up against a problem involving a tetrahedron or more,

it will not be with absolutely no previous knowledge.

8 MANY CHEERFUL FACTS ••• Even bef'ore the study of' geometry went into its present decline. there had been changes in the content - usually without any replacements f'or what had been removed. Nevertheless. it is usef'u1 f'or the would-be problem solver to be aware of' some of' these. because they may make the solution of other problems easier. We shall start by presenting two of' these simple results without proof's. because the proof's will be obvious to all.

A

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f.J

(I)

(II)

In diagram (I). the points p. Q. R divide the lines on which they lie into proportional segments. Hence p. Q. R lie on a line parallel to line A:&:. As a special case. if' p. Q. R are midpoints of' segments XA. XB. XC. then p. Q. R are collinear. We cannot f'ind this. even as an example. in our more recent geometry texts. In diagram (II). A:&:D is a parallelogram with AC one diagonal. If'. through a point on this diagonal. lines are drawn parallel to the sides. six areas are f'ormed. The areas through which the diagonal does not pass are equal. All one has to remember is that a diagonal of a parallelogram divides the f'igure into equal parts. We have f'ound this in a text twenty years old, as a problem. The converse is also true, although we could f'ind no proof' of' this. The reader should be able to prove this.

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•• •• •• •• •• •• •• •• •• •• ••• •

I. I. I.

9

Euclid used this theorem to prove the Pythagorean Theorem. Yet both theorems, (I) and (I) have sunk into innocuous desuetitude. We will use them to prove a famous problem in projective geometry. ~--~--"-----I5

fB (III) In diagram (III). ABCD is a complete quadrilateral wi th edges ABE, BCF, ECD, FDA. Note that there are four edges, six vertices and three diagonals.These diagonals are AC, BD and EF. We wish to prove that the midpoints of these three diagonals lie on a line. First, to explain the various lines we have added to the quadrilateral. We first drew FS and ES to produce a parallelogram. Then we drew lines through point C paralle1 to AE and AF to make more paralle1­ ograms.Finally, we added lines through Band D parallel to AE andAF, as shown, to make more parallelograms. (We made no assumption about point R). In parallelogram A~, C lies on diagonal BF. Therefore area (AC) equals area (CQ). We hope our shorthand notation is not confusing. Next, in parallelogram AEPD, C lies on diagonal DE. Hence area (AC) equals area CP). This means that area (CQ) and area (Cp) are equal. By the converse of (II), therefore, C, R, and S are collinear and lie on the diagonal CS of a parallelogram.

10

Now AC, AR, aM AS are lines from point A to the three collinear points C, R, S. Call the midpoint of AC (x), the midpoint or AR (y), and the midpoint or AS (z). Then, from (I), (x), (y), (z) are collinear.

•

Now (x) is the midpoint of diagonal AC. However, • (y) is the midpoint or diagonal AR of parallelogram ABRD, or which the other diagonal is BD. Since the • diagonals of a parallelogram bisect each other, (y) • must be the midpoint or diagonal BD. Finally, the • midpoint of AS lies midway between E and F of diagonal. EF. This proves the theorem. • We did give this theorem as a problem in the Arbelos for September 1983, but got no solutions. Here is one - without projective geometry - based on very elementary theorems.

• • .. •

To go on to another subject - we have received some canments on the solution or Problem I or the 1985 IMO by Greg Patruno. This gives us a chance to present another solution. Before doing so, let us note that solutions are or two tyPes. On seeing one the reader is wont to say, "How did he ever think of that?". On seeing the other type, the reader might say, ''Why did 'nt I think of that ?". Which type of solution is better, we leave it to the reader to decide. Obviously, Greg's solution belongs to the first type.

•

The more ordinary problem solver would probably consider using trigonometry in some form. Unfortun­ ately, the student normally gets very little trig, mainly confined to using sines and cosines, and sometimes even tangents. He rarely gets to use a cotangent, a secant or a cosecant, and practically never a versed sine. (Haversines are out I) And now to the problema It saysa A circle has centre on the side AB or the cyclic quadrilateral ABeD. The other three sides are tangent to the circle. Prove that AD + Be = AB.

•• •• • •• •• •• ••• •• ••

•• •• •• •• •• •• ••• •• ••• • • •• • •• •• •• •• •• •• •• •• •

11

A.

_ _...::I~

~----J.....:::::I~:::::::::"

o

In the diagram, let angle A = 2a and B = 213. The labeling of the other angles is shown in the diagram, and. should cause no difficulty for the reader to comprehend.. Now suppose the radius of the circle to be unity. Then AT = cot 2a

TD = tan 13

BR = cot 213 RC = tan a

Be = cot 213 + tan a 2 2 tan S AD+:ocl=1-tana+ 1 2- tan 13 + tan a + tan S. 2 tan a 2 2 = 1 + tan a + 1 + tan S 2 tan a

2 tan 13 2

2 = seo a + sec @ 2 tan a 2 tan 13

AD = cot 2a + tan 13

=2

1 sin a cos

= csc 2 a

+

+ (1

1 2 sin ~ cos 13

csc 2 S.

But, referring to the diagram, we find that AO = csc 2 a , BO = csc 2 13, A0 + OB = AB = AD + :ocl.

so that QED

Any good. student should be able to solve the problem in this way, subject only to errors in computations or lack of knowledge about secants and cosecants. (Note I We did not believe this problem should have been on the Olympiad, but we were outvoted.)

12 Finally. a colleague. Professor Eugene Levine of Adelphi College. sent us this problem. He did not include a solution. The problema Given v 4 c S = sin42 + sin4" 7 + sin412t + + sin 87 •

T

4"j) = sin)

+ sin4,..p 0

find the value of

+ sin413"

+

4

0

+ sin 88 •

S + T.

Our first solution made use of De Moivre's Theorem.

as well as a formula for the sum of a series of cosines in arithmetic progression. We felt this to be a pedestrian method, but it

took time. The sum sought is

work'ed~

although it

27/2.

When we were done. we had a "flash" and were able to solve the problem in very short order. We didn't

•• •• •• •• •• ••

need De Moivre at all. and we didn't even use all of the formula for the sum of cosines. In fact. we found an almost mental solution. However. we still feel the second solution may be pedestrian. Hence we offer the above for your consideration. Can you find a short. simp}.e solution that does not use esoteric knowledge ? Send your solutions to your editor as soon as you can. We may use this problem in some mathematical contest, and a snappy clever solution would be appreciated.

•• • •• • ••

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•

•• •• •• •• •• •• •• •• <. <. •• •• ••• •• •• •• •• ••• •• •• •

I.

13 SEEING THINGS DIFFERENTLY

The difference between the average and the better student often lies in the ability to recognize and use information in novel ways. We assume that all our readers know about the Binomial Theorem. There are some students, however, who are not aware that the theorem holds for any exponent, integral, fractional, negative, etc. In many cases, one cannot use the well-known Pascal Triangle, and must rely on the general development, n-1 b + n(n-1) n-2 b2 + ( a + b)n = a n + n 1 a lx2 a • '0 • We have referred to the value of the Binomial Theorem in solving problems in prolabillty and in Combinatorics. But - keep an eye open for other uses. For example I Find the sum of the infinite series 1 + } + 3x5 + 3x5xZ + 3x5xZx9 + 4 4x8 4x8x12 4x8x12x16 ••• When we factor the products in the denominators, we

~t

1 +

22'21

+ ~(4-2) + 3x5x Z(4- 3 ) +

lx2

lx2x3

•••

and we

take note of the factorials in each denominator. To make the numerators more amenable, we distribute some of the 2-s and get to 1 + (3/2)(1) + (3/2)(5/2) (1)2 + (3/2)(5/2)(Z/2)(1)3 2 lx2 2 1x2x3 2 and so on. To make the numerators follow the rule ­ n, n(n - 1), n(n - l)(n - 2), we must introduce negative numbers. With a little care, we find that our series is a development of

(1 -

~)-3/2,

easily find the value of this to be

and we can

2-/2.

14

Now we invite the reader to try to evaluate ­ 1 +

1 )

+ 2.x 5 6 x 12

+

-? x 5 x 8 6 x 12 x 18

+

•••

Now we go to another subject where imagination can be useful in problem solving - Inequal1ties. We are aware of the usual inequal1ties, the Arithmetlc Mean (AM), the GeOllletrlc Mean (GM) and the Harmonic Mean (HM). We should also know the Cauchy-Schwartz- Bun­ iakowsky Inequality. We will stick to the name Cauchy. Hoping that the reader will understand our notation, 2 we write this (ra2 )(I:b2 ) ~ (fab) • It is easy to forget that these inequal1ties hold under certain conditions, and there are situations when it becOllles useful to consider when equality occurs. We will illustrate this with a simple examples Given f(x) = x) - ax 2 + bx - ) = 0 , with roots (r /2) + (r /)) + (r)/4) 2 1 Solve for the three roots. r , r , r), such that 1 2

=~ •

We have hinted at a method of proceeding. The AM (r /2) + (r /)) + (r)/4) is 1/2. The GM of the 2 1 same terms is (r r r)/24)1/) = (1/8)1/) = 1/2. 1 2 Now we have equal1ty, which means that we have of

(r /2) 1

= (r2/)) = (r)/4),

Finally, r

=

=

so each is equal to

1/2.

=

1, r )/2 , r) 2. 2 1

Morals keep an eye open for a unique or original

way of using otherwise innocent-looking rules. This may be the difference between success and failure on a contest.

• • •• •• •• •• •• • •• •• •• •• •• •• ••• •• ••

••• • •

15 \

EXERCISES

. • The following problems are taken from various Polish • Olympiads. Try to find the method for each that will • produce the simplest solution. •

• •

1) AABC and 6DEF are inscribed in the same circle.

•

Prove that the intersections of chords AB and DE,

•

Be and EF. and CD and FA are collinear.

•

2)

.• •

••

•

J)

Prove that for positive integers a and b,

6 6

2 2 a + b x a + b x a J + bJ < a + b 2 2 2 2 Show that if a, b, c are positive numbers such

that

= 1,

abc

then

a+b+c

> J.

y3 - x J

:

4)

Find all integer solutions of

:

5)

Show that for any natural number

•• • •• •• •• ••

•• •• •• •

•

6)

7)

1 + 2 + J + ••• + n digits

2,

n,

cannot terminate with the

1a+b 1 +c1 =

_1_

_1_ n c

n

a

1 then a+b+c'

1 = -----:==---for all odd n.

an + bn + c n

Show that for natural numbers x, y, x n

n

= Zn ,

such that

x

x

must equal or exceed

y

the sum

4, 7 or 9.

Show that if

and

= 91.

+ y

and n,

the minimum value for n.

16 Wilson '5 Theorem - etc. It has occurred to us that we have been presenting various topics in mathematics on a one-time basis, without using them later. Therefore we will develop Wilson's theorem, but in a roundabout way, using some of what we have already presented in previous issues, together with some new materials and some new results. First, we state Wilson's theorems A necessary and sufficient condition for a natural number p to be a prime number is to have (p-1)! + 1 divisible by p. Or, (p-1)!; -1 (mod p). First, note that this condition is both necessary and sufficient. Second, note that it is not of much use in determining when a number is prime once the number gets large. Even showing that 17 is prime takes a lot of arithmetic. There are lots of ways of proving the theorem, some very short. We will amble our way to it. In our very first issue - of September 1982 ­ we presented some information on the calculus of differences. Let us recall some of this information. We write f(x) in the form u ' Then f(x+1) = ux + ' 1 x We then define Aux = ux +1 - ux • Then, writing A(AU ) = A(Ux + - tlx ) = AUx +1 - Aux = (ux +2 - u x +1 ) 1 x - (ux +1 - uX )2= ux +2 - 2ux +1 + ux ' We wrcte this in the form Au. It is not difficult to show that x Anu = u - (n)u + (n)u + u x x+n 1 x+n-1 2 x+n-2 - , , , - x' We also wrote the rules in terms of the A. Thus,

• • • •

• •

• • • • • • • .. .. • •

•• •• ••• ••• • • •

• u +1 = u + Au = (1 + A)u , where we separated the x x x x operational symbol from the u • Continuing, we find • 2 x 2 u +2 = u + 2Au +A u = (1 +A) u ,and, generally,. x x x x x •

•• •

;.• I

••• •• ••• •• •• •• ••• •• •• •• •• •• ••• •• •• •• •• ••

17

When we first introduced this, we used it to find the n-th term of a sequence and the sum of n terms of a series.

=

If we define ux + Eux ' and, in general, 1 n u +n = E u , we have another way of writing u + • x x x n Separating the operational symbols as before, we have b.n = (E _ l)n, or better still, En = (1 + b.)n. We used this last form and notation to solve difference equations. The reader might consult any text that does work of this sort or perhaps the issue of Arbe10s for September, 1982. n Now for something new. Let u = x • Then x b.u = (x + 1)n _ x n = nx n- 1 + • • • x 2 b. u = n(x + 1)n-1 _ ftX n- 1 + • • • = n(n_1)x n- 2 + ••• x and continuing in the same way, we finally arrive at n n

b. x

= n!.

Note that when we reach this point, there

=

are no terms containing x. In fact, if we let x 0, n we get b.nO = n! • There are some useful results of deve1op:1Ag the various differences of zero, which we will leave for another time to examine Now we note that b.n(x n ) = (E - l)n{x n ). Expanding, _ (n)E n- 1 + (n)E n- 2 _ • • • Jxn

n!

= (En

n!

= (x+n)n

1

2

- (~)(x+n-1)n + (~)(x+n_2)n _ • • •

Let x = 0, and we get a curious formula - namely, n! nn _ (~)(n_1)n + (~)(n_2)n _ (~)(n_3)n + • • • •

=

•

18

We could continue fran this fonnula, but we go n on to do the same operations for 11 x - 1. Now nl

= (x

x

=

+ n)n - 1} - (~)(x + n - l)n - 1} + ••••

Select a prime number p, and let n = p - 1. Then (p-l)1 (x+p-l)p-l- 1) - (Pl1)(x+P-2)P-l ­ 1} + ••• Next, let x = 1 in the above, and we get

=

(p - 1)1

=(pp-l

_ 1} _ (~l)(P_l)P-l - 1} +

(~t)(P-2)P-l _ 1) - •••

(A)

Now we need Fermat's Little Theorem. We derived it before, but will do so again, differently. Let

••• ••• •• •••

p equal a prime number, and!: a number relatively prime to p ( though that isn' t really necessary. Then (a)p

= (1

+ 1 + ••• + l)P

=l P +

l P + ••• + l P + pX,

where X includes multinomial coefficients and pX is divisible by p. Then a P = a + pX, or finally,

=

aP a (mod p). We go back to formula (A) above. From the second term above, each tenn in brackets equals zero, from Fermat's theorem. Also, the first term is a power of p, so it is divisible by p. The only thing left at the right in (A) is the -1. Hence we can deduce that (p - 1)1 :. -1 (mod p) and we have Wilson's Theorem. We leave the question of the converse theorem to the reader, who will find proofs in most texts on Number Theory. We recODlJ1lend ''The Theory of Numbers", by Niven and Zuckerman, or "Introduction to Number Theory", by J .E. Shockley.

• •• ••• •• •• •• •• •• •

j.•

,.

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19 From this point on, there are a number of tracks one can take. For example, the separation of symbols of operation from thin~ operated on looks interest­ ing. How far can one go from there? For example, if the reader refers to any text on calculus, he or she will find Taylor's Theorem, which we choose to write as followss 2 2 3 f(x+h) = f(x) + ~!nf + ~! n f + ~! n 3f + ••• If we write then Eux

U

= (1

x for f(x) and u +9 = Eux for f(x+h), hD

h2n

2

h~

+ -1' +• -2' + -3' • • + ••• )ux

= ehD u. x

and we seem to get E = ehD ,and also 1 + l:i. = e hD Can we find derivatives by using differences ? And are there other cases involving other symbols that can be similarly separated? Consult any good text on numerical analysis for suggestions ! However, we proceded in another direction. If p is an odd prime, then p-1 is even, and (p-1)! has an even number of numbers from unity to (p-1) in it. We applied what we like to call the Gauss method to this product, and we found that 1 x (p - 1) ! -1 (mod p) 2 x (p - 2) ; -4 (mod p)

3 x (p - 3)

=-9

(mod p)

and, on multiplying, (p-1)! ~ (_1)r(1 22232 •••r 2 ), where we have let r = (p-1)/2. We notice that when r is even, or p = 4t + 1, then (p - 1)! (r!)2, and when r is odd, then (p - 1)! ~ _(r!)2. We conclude that, when p = 4t + 1,

=

20

= (r!)2

(p - 1)! + 1 If we let (r!)

+ 1

is divisible by

be equal to say, s, then we have,

a new theorem that says that, for a prime. p 4t + 1, there is a number

form

p. of the

s , for which

1).

p 1(s2 +

When we multiply (on the right) by the square of a number relatively prime to p, we naturally get p (s2 u2 + u 2 ) or p (A 2 + B2 )

I

I

which brings up the question - for what primes p is

p

representable as the sum of two squares? Notice that 5 = 12 + 22 and 13 = 22 + 32 , but we cannot write either?

nor

11 as sums of squares.

This will explain the "etc." in the title of this article. We started with Wilson's the01'il')M and have gotten, somehow, to the start of Warin~'s theorem. In the Arbelos for March, 1983, we presented an article on the Farey sequence. We will recall some of this. Consider the table below:

Q 1 1

Q

1

1 1

121

Q 1 1 1

3

~

1

231

Q 1 11~ 1 1 4 323 4

1

1 1

~11

~

1 4 1

543

525

3

451

Q _

1

1

and so on.

• •• •• ••• ••• •• ••• •• •• ••• •• •• •• •• •• •••

I.••

•• •• •• ••

•• ••

•• ••

•

•

•

•

•••

•• •

••

••

•• •• •• •• •

21

These are Fa:rey sequences. They have the following properties: the terms in any sequence are listed in ascending order; the denominators in the n-th sequence do not exceed n; given any two successive terms,

~I ~

ad - be = -1. We used this last condition

,

to find solutions to linear Diophantive equations.

In order to continue, we define the mediant of two successive terms of a Farey sequence of order N to

be equal to the sum of the numerators divided by the

sum of the denominators. That is to say, the mediant a c a + c of band d is b + d. Of course, the mediant will not belong to the sequence of order N, since b + d

will exceed N, but it will be long to a: sequence of

higher order.

Again we will amble on toward a solution to a

problem, by attacking another pro.1em first - to

whit - how closely can we approximate an irrational

number A by a rational number h/k with denominator k ?

In the Farey sequence of order N, we can find two successive terms, alb and c/d, such that ~<\<.£

bAd·

Take the mediant a

b

<

~.

<

a + c b + d. Then we must have either

a + c

b+d

or

Now we recall that the mediant is not in the Fa:rey sequence of order N, so b + d ~ N + 1.

22

In the inequality at the bott~ of the previous page and at the left, subtract alb all around, and O< A

a < a + c a _ 1 < 1 b + d - b - b(b + d) "'T:(N~+-l~)-:-b

- b

In the inequal1ty at the bottom of the previous page and at the right, subtract (a + c)/(b + d), and O
d-

A
d-

a+c_ 1 <.....-....:1=---r­ b + d - deb + d) (N + l)d'

We have shown that, given an irrational number h

it

there is a fraction

IA - ~I <

with denominator

A,

k < N, for

(N +1 1 )k •

We are well on the way to deriving Hurwitz's Theorem, but we won It - for now.

•• • •• •• •• ••• •• •

Now suppose we don It have an irrational number, but the rational number 11m reduced to lowest terms am with m > N. It may happen that 11m is equal to a + c the mediant b + d • Then we may not have a strict inequal1ty, and we can only write ­ If 11m is a reduced fraction with denominator

m < N, there exists a fraction h/k with

denominator k < N for which

1m1 - ith\ ~

1

(N + l)k

Let us now assume the existence of positive integers A and n such that

n

there exist integers

2'

divides s

and

+ 1. We show that

A

t

Note that we do not assume that

such that n n

is prime.

= s2

+ t2•

•• • ••• •• •• •• •• ••

;.•• I

••• <. •• • • ••• •• <. <. ••• •• • •• •• •• •• • ••• •• •• ••

:. ;.

,.

23

In our last formula, take

i'

N

= b-viii] <

n. Then there

is a fraction rls in lowest terms such that

I~ - ~ Therefore

(0 <

eN ; l)s

lAs - rnl

~

N

~

1

S

= [-JiiJ

< N) + 1 <

-In

Let As - rn = t, and square the expression above, and we have t 2 + s2 = (As _ rn)2 + s2 = s2(A 2+1) _ ZArsn + r 2n2 , and since each term on the right is divisible by n 2 2 t + 8 0 (mod n). 2 But t + s2 < (~)2 + «(~])2 ~ 2n. Hence t 2 + s2

=

is a positive multiple of n less than 2n. This means that we must have t 2 + s2 = n, and we are done. Note that, i f

n divides

a 2 + b2 , and a and b

are relatively prime, then there exist integers s, t such that n = s2 + t 2 • We have the identity (a2 + b2 )(c 2 + d 2 ) (ac + bd)2 + (ad _ be)2.

=

But (Farey sequences again) we can always select a c am a d so that (ad - be) = 1. Then we will have (a2 + b2 )(c 2 + d 2 ) = (ac + bd)2 + 1. This has the form n = t 2 + s2. and we have the converse. We have done qUite a bit of &IIlbllng, and have prol:ab1y taxed the patience of the reader, so we will rest for awhile. Note I Have you noticed that if you can yite2 n = (s + it)(s - it), then n = s + t ? For instance, 13 = (3 + 2i)(3 - 2i) = 9 + 4 • What kinds of n can we write in this form ?

24 OLYMPIAD PROBLEMS

16)

1Qt a and b be integers and n a positive intep;er. Prove that bn- 1a(a+b)(a+2b) ••• (a+9n-1)b)/n! is an integer. (Bulgaria)

17)

A convex quadrilateral is inscribed in a circlE" of radius 1. Prove that the difference between its perimeter and the sum of the lengths of its diagonals is greater than zero and less than 2. (BraZil)

18) The sphere inscribed in the tetrahedron ABCD

• •• •• •• •• •• •••

touches the side ABO and DBC at points K and M respectively. Prove that angle AKB

=angle

DMC.

(USSR) 19) Let k be a positive integer. Define and u to or

1

= 1. Let u ~ater

o = 0

1 - un­ 2 for n equal

than 2. Show that for each integer n

= ku

U

n-

u 3 + u 3 + • • • + u 3 is a n

2 1 multiple of u + u + + un·

2 1 (Canada) n, the number

20) Let n 2: 2 and n , n2 , • • • , ~ ~ 1 be 1 natural numbers having the property that

n

n n2 (2 • - 1), ~ (2 n2 - 1) ..... f\rJ(2 '<-1 - 1), and fi \ (2 n -:. - 1)' Show that n1 = n2 = 1. 1 (Romania)

I

= •••

• •• •• • •• •• •• •• •• ••

•

•

•

•

•

•

•

•

• •

• • •

25 KURSCHJ{)C KORNER

This column features those problems of the famous Hungarian Kurschak Contest which have not yet appeared in English trans­ lation. For the problems of 1894-1928, the reader should consult Hungarian Problem Books! & .!!. in the MAA's New Mathematical Library series (avatlable from the MAA hdqrs.); the problems (with brief summaries of their solutions) covering the years 1966-1981 were translated into English by Prof. Cslrmaz of Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers.)

•• ••• •

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•

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•

•

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•

•

•

•

•

•

•

•

•

•

•

•

•

•

Student readers are also invited to set aside an uninterrupted 4-hour period to compose complete, well-written solutions to the problems below, and to submit their work to the address given below for critical evaluation.

1·3·5····· (2n-1) , where n is a positive 1/1934. Let A = 2·4·6·····2n integer. Prove that there is at least one knteger among the members of the sequence A,2A,4A,8A, •.• ,2 .A, .••. 2/1934. Consider all non-self-intersecting polygons inscribed in a given circle. Which one of them will yield the largest value for the sum of the squares of its sides?

3/1934. Assume that an inf ini te number of rectangles are given in the first quadrant of the co-ordinate plane with vertices (0,0), (O,m), (n,O), (n,m),

where nand m are integers. two rectangles in the given is contained in the other.

Prove that one can find set so that one of them Dr. George Berzsenyi 2040 Chevy Chase Beaumont, TX 77706

• CONTENTS COVER I Geanetry Problem i

1

3 8

13 15

16 24

25

Preface

An Introduction to Number Theory

Tetrahedra

Many Cheerful Facts

Seeing Things Differently Exercises Wilson's Theorem - etc. Olympiad Problems Kurschak Korner

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••

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arbelos

PRODUCED FOR

PRECOLIEGE

HfILOMATHS

,..

A

,

,. '.

~ Givenl Equilateral triangle ABC with vertices on parallel lines as shown. The distances between the lines are as shown.

Finil The area of triangle ABC. Hintl Construct the triangle first.

1985-1986:

No.5

May,

1986

Copyright ~The Mathematical Association of America, 1986

i

PREFACE

With this issue, we close the most encouraging year since we started p:r:oducing Arbelos. We have received several articles, including the excellent one from Professor Stanley Rabinowitz. The number of lettGrs from subscribers has increased, and some of these have come from as far off as Greece. It seems we have become "international" ! Moreover, not all letters and comments have come from students only. We consider it an honor to have received mail from some mathematicians who have taken the trouble to read the articles, solve the problems, and offer useful suggestions. A letter from Japheth Wood, from Princeton, NJ, contained solutions to several problems, together with useful suggestions. We got solutions, for the first time, to our "c over Problem" from two readers. One was from S. Nam, from Dayton, Ohio (sorry but he/She did not include a first name). The other was from Rahul Pandharipande, from Urbana, Illinois. Both used the same combination of geometry, trigonometry and inequalities. In view of the present status of plane geometry in our schools, we recommend that readers try the cover problems. We also recommend reading a lot. Japheth Wood called our attention to a problem in Coxeter's "Geometry" that we had missed. Obviously, he has done a lot of careful reading. A final note about Problem 24 of the Olympiad Problems page. We attended the First Ibero-american Mathematical Olympiad in Colombia, where this was one problem. We were proud to have been involved in a small way with the events leading to this Olympiad, and hope to attend the Second Olympiad in Urup;ua.y next December. There is no lack of interest in mathematics and in problem solving. We are certain that it will continue to expand. Meanwhile, until next September, Have a goej year !

•• •• •• ••• •• ••• •

• •• ••• •• • •• •• •

4-.JtA+ :

•

•• ••• ••• •• •• •• •• ••• •• ••

,.••

••• •• •• ••• •• •• ••

1

A Useful Trigonometric Substitution Stanley Rabinowitz Digital Equipment Corporation Nashua, NH In this article, we will show that the substitution x

t = tan­ 2

is very useful in solving certain types of trigonometry problems. First, let us investigate formulas for the tangent of a half angle. One formula that is frequently taught is

x tan - = ± 2

1 - cos x 1 + cos x

This formula is frequently useful when you are solving for x/2 and know which quadrant the angle x falls in. Unfortunately, the formula is not very useful in proving trigonometric identi­ ties or inequalities because of the ambiguity of the ± sign. In fact, depending on the angle x, sometimes the positive square root should be taken, and sometimes the negative square root. For theoretical (as opposed to computational) problems, an exact formula is required, one in which there is no ambiguity In

sign. For this purpose, a good formula to know is

x sinx tan- = - - - ­ 2 1 + cosx This formula can be derived from the familiar formulas for the

sine and cosine of a sum and difference: sin(x

+ y)

= sinxcosy

+ cosxsiny

(1)

•

2

sin(x - y) = sinxcosy - cosxsiny cos(x

+ y)

(2)

= cosxcosy - sinxsiny

(3)

+ sinxsiny

(4)

cos(x - y) = cos x cos y

Adding equations (1) and (2) and also (3) and (4) gives sin(x

+ y) + sin(x - y)

= 2sinxcosy

cos(x

+ y) + cos(x - y)

= 2 cos x cos y

or letting x + y = 2A and x - y = 2B, we find sin 2A

+ sin 2B

= 2 sin(A

+ B) cos(A -

B)

cos 2A

+ cos 2B

= 2cos(A

+ B) cos(A -

B)

Letting x = 2A, y = 2B and dividing these two equations gives x +y sin x + sin y tan - - = - - - - ­ 2 cos x + cos Y

This, by itself, is a very useful formula to know. Furthermore, letting y = 0 gives us our half-angle formula x smx tan- = - - 2 1 + cos x

Also, the familiar identity, sin 2 x

+ cos 2 X

in the form smx 1 + cosx

1- cos x

smx

(5) = 1 may be written

•• ••• •• •• •• •• •• ••• ••• •• •• •• •• •• •• •

•e e •e e

•

e

J

which shows that an equivalent half-angle formula is x 1 - cos x tan - == - - - ­

tit

e e e e

e e e

e

e e e e

e

•• e e

e e e e

e e e e e e e e e

•e

e

2

(6)

Sinx

Letting t == tan~, we can solve equations (5) and (6) for sinx and cos x in terms of t. We get . SIn

2t

(7)

x == 1 + t 2 1 - t2

cos X = 1 + t 2 •

Now we see why the substitution, t == tan

(8) ~

is so useful. It

lets us express both sin x and cos x as rational functions of the single variable, t. Let us now see how this substitution can be useful in solv­ ing various problems.

Example 1 (Equations). J. T. Groenman asked in [1] to find all x in (O,21r) satisfying 176cos x + 64sinx == 75 cos 2x + 80sin 2x + 101. We can solve any problem like this with the aid of the substitution t == tan(x/2). If the equation we are trying to solve contains any trigonometric terms other than sin and cos (such as tan, cot, sec, or csc), we first express these in terms of sin and cos. Then, if there are any expressions involving multiple angles (or sums or differences), we expand these out using the appropriate formulas. In this example, we use the formulas for the sin and cos of 2x to get 75(cos 2 x-sin 2 x)+160sinxcosx-64sinx-176cosx+l01 ==

o.

4

Now we apply our substitution, using relations (7) and (8) to transform sin x and cos x into expressions in terms of t. We get

+ ----.,----::-:-:,-------° (1 + t )2 - . 352t 4

-

448t 3

248t 2

-

192t

2

Since the numerator must be equal to 0, and dividing out by 8t, we find that 44t 3

-

56t 2

-

3lt

+ 24 = 0.

Any rational solution to this must have a denominator that divides 44. Trying a denominator of 2, we find that t = 1/2 is a solution. Dividing out the factor (2t -1) gives us a quadratic that we can easily factor, and so we find that the complete factorization is t(2t - 1)(2t - 3)(llt

+ 8)

= 0.

Thus, the complete solution is

o 2 arctan(1/2) x- { 2 arctan(3/2} 2 arctan( -8/11) where arctan z denotes the angle in [0,71") whose tangent is z. Example 2 (Inequalities). Find the range of x in the interval [0,271") for which 8sinx

+ cosx

> 4.

•• •• •• •• •• ••• •• •• •• •• • •• •• •• •• •• •• ••• ••

e

e e

•e •e fit

e e e e e e e

• e

•e

­ e

e e e e e e e «It e e e e e e e

•

e e e

e

e

5

Applying the substitutions in (7) and (8) show that this problem is equivalent to

(t - 3)(5t - 1) 2 < o. l+t The quantity on the left is 0 when t == 3 or when t == 1/5. Since 1 + t 2 is always positive, the sign of the left-hand side depends only on the sign of (t - 3)(t - 1/5). Between 1/5 and 3 this expression is negative; elsewhere it is positive. Thus

t lies in the interval (1/5,3) and hence x lies in the interval (2 arctan k, 2 arctan 3). Example 3 (Integration). For those of you who know some calculus, the substitution t == tan

~

will let you integrate any

rational function of sin x and cos x by changing it to a rational function of t. This latter expression can then be integrated by expanding into partial fractions. For details, consult a calculus textbook such as [3]. As we have seen, the substitution t == tan

~

is a very effec­

tive method of solving certain types of problems. However, it should only be used as a last resort because its use can involve very complicated expressions. Its effectiveness makes it ideal for computer applications. For hand calculation, it is advis­ able to first look for other tricks or

substitutions~

Ingenuity

will normally lead to a simpler solution than brute force. The following example illustrates this point.

Example 4. Murray Klamkin asked in [2] to find all x in

(0, 27r) satisfying ·

81s1n

10

81 x+cos 10 x==-. 256

••

6

In this case, the simpler substitution, t

= sin 2 x,

1- t

= cos 2 x

can be used. This substitution may be used any time that all the exponents of sin x and cos x are even. In this case we get

8It 5

+ (1 -

t)5 = ­

81 256

which expands out to

4096t 5

+ 256t 4

-

512t 3 + 512t - 256t

+ 35 =

O.

The powers of 2 appearing as coefficients suggests that we make the further substitution t = T / 4 which yields

4T 5

+T4

-

8T 3 + 32T 2 - 64T

+ 35

•• •• •• •• •• ••

= O.

In this form, T = 1 is seen to be a solution (twice) and the polynomial factors as

Since T = 4 sin 2 x is always non-negative, 4T3 can never be 0 and so T

t =

i- and sin x = ± t.

+ 9T2 + 6T + 35

= 1 is the only solution. This yields

Thus, x

= 1f /6,

51f /6, 71f /6, or ll1f /6.

References [1] J. T. Groenman, Problem 1020, CRUX Mathematicorum. 11(1985)51.

[21 Murray S. Klamkin, Problem D-3, AMATYC Review. 4(1983)63. [3] George B. Thomas, Jr., Calculus and Analytic Geome­

try (third edition). Addison-Wesley Publishing Company, Inc. Reading: 1960.

•• ••• •• •• ••• ••• •• •

•• •• ••• •• •• ••• •• • ••• •• ••• •• •• •• ••• •• •• •• •

j.

7

THE PENTAGON Recently, we have come upon several articles about the regular polygon of seventeen sides, and how to construct it. This was certainly one of the great achievements of Gauss, being the construction of a new polygon unknown to the Greek geometers. We couldn't help noting that, of late, there has been nothing in our geometry relating to the construction of, say, a re~lar

pentap;on or decap;on. We had found. such con­

structions fun. To find out about such constructions, we went lack to our old geometry text, "Plane Geometry", by Schultze and Sevenoak, printed in 1925. First, we came again on the golden mean, or golden section, or golden ratio.

I' b _-', A__a.._--.ol--_--..:.._

13

If the line AB is divided at point P into pieces a, b a b so that b = a + b ' we say that AB has been divided in "extreme and mean ratio'" The rectangle below with

h

b

a.~lr

dimensions according to the golden mean was considered by the ancient Greeks to have the most pleasing shape.

•

8 This may well be so - our flag has about the proper dimensions.

ab = a+b b = ba + 1,

Let us write Then

-r

=1

+ 1, or

-r 2 - -r - 1

= O.

quadratic ;or -r, we easily find that

and let

ab = -r.

Solving this

-r

= ~2+

1.

This, then, is the golden mean. The number

-r

is well-known. We have referred to it

in previous articles. As a continued fraction, examp I e, ~• -- 1 + I1 + I1+1 I + • • ., so tha t ~• and we are back to the equation for the golden

for 1 -- 1 + :r­ mean.

Note also that since we can construct a segment equal to

~ (in several ways), we can construct a length

equal to -r. We can use the golden mean to construct a regular pentagon or a regular decagon as did the Greeks, but we prefer to do a little extra work because it may be helpful for the future. We will work with a circle with unit radius. The extension to a circle of any radius is obvious. A

• •• •• •• •• •• •• •• • •• • •• •• ••• •• •

•

•e •e e •e e

e

e e e

e e

• •e e e e

e

e

e

e

e

e e e e

e e e

e e e

e e tit e e

--

9

Let AB be a side of a regular pentagon inscribed in the unit circle. Call this side

~.

c Then LADE = 72 ,

c am, in right triangle MOB, LMOB = 36 ,and we have

~ = sin 36°. We must fim sin 36°. (We have done some of this before, but it doesn't hurt to repeat it.) First, note that sin 36° = cos 54". Let am we may write

x = 1ff ,

sin 2x = cos 3x. From De Moivre 's

Theorem, (cos x + i sin x)3

= cos

3x + i sin 3x.

We expand on the left and separate real fran complex tenns. We then equate am find that 2 cos 3x = cos3x - 3 sin x.cos x 2 sin 3x = 3 cos x.sin x - sin3x •

Using the first of these equalities, (and line 7 above)

we have

sin 2x = cosJx - 3 sin2x.cos x 2 sin x.cos x = cos3x -

3 sin2x.cos x

and dividing by cos x (which is not zero) we get 2 sin x

= cos 2x

2 - 3 sin x

= 1 - 4

2

sin x. This is a

quadratic in sin x, namely 4 sin2x + 2 sin x - 1

= O.

Solving, we have fourd that

o

sin x = sin 18

= -J5-1 4

o Now it is easy to construct ~, then ~ -1, 18 ,

o

and eventually, sin )6. then

se~ent S,

Now we construct segment s/2,

and we have constrmted a side of our

regular pentagon (in a unit circle). Admittedly, this is complicated and messy, but it will

work~

10

The reader has certainly noticed that if, in the diagram above, PQ is a side of a regular decagon, then LPOQ is

36' ,

then PQ/2

and, if M is the midpoint of side PQ,

= t = sin

c

1S ,

•

Perhaps it is easier, if one

wants a regular pentagon, to first construct a regular decagon and then connect alternate vertices. This 1s certainly easier but still messy. ,€

Let us find cos 36 • We start with cos 2x

=1 -

2 sin2x.

and we have found 2 0 cos 36

=1 -

2 sin 36

=

= 1 - 2(~15 -

cos 36

2 ., sin 36

()

=

1)2/16

= ,}54+

1

-J5+1 4 • Now

= 6 +16G"'-/5

,so that

16 '15 . and a side of the regular pentagon

10 - 2

in a unit circle is

...J.10

-22..}5. Constructing this

would indeed be a canp11cated procedure, but we can

•• •• •• •• •• •• •• •• •• •• ••

construct our regular pentagon, at least. With a 11t tIe labor (and a lot of trigonanetry) , we have managed to derive ways to construct our regular pentagon and our regular decagon. Where does the golden mean cane in ? Let us tie things together.

•• ••• •• •• ••

e e

11

•e

e e e e e e e ee e

e e e e e e e e e e e e e e e

e e e e e e e

e e

e

tit

e

e

The Greeks did their construction as follows. Let OA be a radius of a circle. Divide OA in extreme and mean ratio at D. Mark off AB = a. Then the angle at center 0 equals J6e , and AB is a side of a regular decagon. We leave it to the reader to prove that triangles ABD and ABO are similar isosceles triangles, that the angle at A is 72v and the angle at 0 is J 6tJ • This should raise the roader's estimation of the Greek geometers.

D

.14.' ~-~-----f------------I

A

12

We now present our own way of constructing a regular pentagon. In the diagram, we have a unit circle with perpemicular diameters. We work as follows. (a) Bisect radius OA at M; then OM

(b) with radius MB, draw arc

= 1/2.

Be meeting radius OA'

at C. (c) since OM

oc =

= 1/2

(-/5 -

and OB

= 1,

MB

= -/5/2.

Hence

1) /2 •

(d) With radius Be, mark off chord BD. Since Be

2

= B02

Be = BD =

+ OC

2

=1

+ (6 - 2...[5)/4

= (10

-

2-J"5)/4.

--aha - 20 2

Notice that, in line with all our

~vious

computations,

OC equals a side of the regular inscribed dec.agon, and BD equals a side of the regular inscribed pentagon. We happen to think this

t~

As an added curiosity"

easiest method. since OB would be a side

of a regular inscribed hexagon, we have proved that "When the square of a side of a regular hexagon is added to the square of a side of a regular decagon in the same c ire Ie, the sum equals the square of a side of the regular inscribed pentagon in the same circle. " We invite the reader to get out the proper tools am perform the construction - just for fun.

•• •• •• ••• •• •• • ••• •• •• •• •• •• •• •• •• •• •

•• •• •••

••e •

e e • e •

e

13 Constructions afford a way of seeing some of the

relations one derives by other methods. Once one has solved a construction problem, it remains in the

memory. Moreover, constructions provide a way of using sane of the things one has learned elsewhere. For example, how would one construct the canmon

tangents to two circles1 We add some problems (with

hints). Given H , Hb , H the feet of the altitudes of a a c triangle in position, reconstruct the triangle.

e e e

•e •e

e e e e e e e e e e e e

e

e e e e

e

(Hint: Draw a triangle with Ha , Hb , Hc the feet of the altitudes, then join them to make the orthic triangle. What relation is there between an angle of this triangle and the altitude at the vertex ?) If P, Q, R are midpoints of three sides of a parallelogram, construct the parallelogram. (Hint: you should be able to draw the diagonals.

Construct an eqUilateral triangle with one vertex on

each of three parallel lines.

(Hint: See the problem on the cover page.)

From a point outside a e1rcle, draw a line cutting

the circle so that the chord inside the circle is

half the lellgth of the whole line.

14

MANY CHEERFUL FACTS

We have already presented a method. of fiming the sum of a series by means of finite differences (see our issue of September, 1982). Sometimes this method. is too complicated am may even not be applicable in some given case. We would like to present another method. that may be easier. Given a series S = u

+ u + • • • + un. If we can 2 1 write u l = sl - s2' u 2 = s2 - s), ~ = s) - s4' am so on, the series takes the form. S = (sl- s 2) + (s2- s )) + (s)~s4) + ••• + (sn_l- s n), then the terms other than the first am last cancel each other, am we have

S = s1 - sn. Such a series

is called a "telescoping" series.

•• •• •• •• •• ••

The simplest example we can think of is the series + _1_ + _1_ + + 1 W ite it S = _1_ lx2 2x) )x4 ••• n(n+l)· r as _ ( 1) (1 1) (1 S - 1 -"2 + "2 - J + J am the sum is obviously 1 _ n 1 - n+l - n + 1 •

-"41)

+ ••• +

n-

(1

1) n+1

Usually, a bit of work is needed to put the series in the desired form. For example, let us consider S =_1_+_1_+_1_+ lx4 4x7 7xl0 •

. .

We cannot write this simply in the form

_ ( 1 -"41)

S -

+

(1"4 -"71)

+. •• +

( )n-2 1

1)

- )n+l •

because the expression in any parenthesis does not equal the correspoming term of the original series.

••• •• •• •• •• •• •• •

••• ••• •• • •• •• •• •• •• •• •• •• •• •• •• ••

,.

•• •• •• ••

15 In this case, we make use of partial fractions. We 1 write the general term, (3t-2)(3t+1) in the form

=

1

-rO::-:t~-~2)r'i('"='3~t+~1~)

A B + Ot+l) Ot-2) , am remelll be r t hat

the numerator 3(A + B)t + (2A - B) must equal 1. Then (A + B) A = 1/3

=0 am

Solving these, we have

B = -1/3. Now we may write the series

= (ill _ ill) 1 4

S

= 1.

(2A ­ B)

and

+

(ill _ ill) 4 ?

+

...

+

(..1L1 _ !L.L) 3n-2 3n+1'

telescope, simplify the result, am fim that S -

n

- 3n + 1·

Exercise. Fim an expression for the sum of the seriest T

= 1X~X3

+ 2x5x4 + • • • + n(n+ltCn+2) •

The philomath should. add partial fractions am the method of telescoping series to his/her armamentarium. As another instance of telescoping series am to present another rule that students will fim useful, let us examine the seriest S = sin(a) + sin(a+d) + sin(a+2d) + ••• + sin(a+ n-l d). Here the angles are in arithmetic progression, but the sines are not. Here we will need sane trigonaaetry. First,

cos(x - y) cos(x + y)

so

= cos = cos

x.cos y + sin x.sin y x.cos y - sin x.sin y.

cos(x - y) - cos(x + y)

=2

sin x.sin y.

Next, we multiply each term in our sine series by 2 sin

d '2.

The

~neral

2 sin ~ sin(a + td)

=

term then becomes 0 ~ t ~ n-1.

=

Now let x a + td am y d/2 in our formula for the difference of two cosines (see above) , and we have

16 the general term in the form cos(a + t - 1!2 d) - cos(a + t + 1!2 d). Substituting successive values for t, from 0 to n-1, we fUrl cos(a - 1/2.d) - cos(a + 1/2.d) cos(a + 1/2.d) - cos(a + 3/2.d) cos(a + 3/2.d) - cos(a + 5/2.d)

. . . . . .

cos(a + n-3!2.~) - cos(a + n-l!2.d) which telescopes to 2. sin % • S

= cos(a

- 1/2.d) - cos(a + n-l!2 .d).

•• •• •• •• •• •

Using the formula for the difference of cosines again, we let

x - y

=a

and get 2.sin %.S

- 1/2.d

x + y

= 2.sin(a

=a

+ n-l!2.d, and

+ n-l!2.d).sin(nd/2),

We divide by 2.sin d/2, am finally arrive at _ sin nd 2( ( / ~ S - sin d 2 sin a + n-1 2.d//. This formula is worth adding to your list of cheerful facts. We can make it easier to remember by noting the multiplier

s;~n~ ~ and that the term inside

the other sine term is the average of the first and last terms of the angles in the original series. nd 2 ( i first angle + last angle) Th a t i s, S = sin sin d 2 s n 2 The reader who is willing to go through the same process for the sum of a series of cosines of angles in arithmetic progression will find that theformula is

••

•• •• •• •• •• •• •• •• •

•• •• ••• •• •• •• ••• •• •• ••• •• •• •• ••• •• •• •• •• •

'.

17

c = sin

m. 2 ( cos first ahgle + last angle) sin d 2 2 •

Exercises & Show that

2

sin x + sin 3x + sin 5x + ••• + sin(2n-1)x sin 2x + sin 4x + sin ox + ••• + sin 2nx

nx = sin sin x·

=

sin nx. sin (n+1)x sin x cos 2x + cos 4x + cos 6x +

... + cos 2nx =

sin nx. cos(n+1)x sin x F ind

1'1' + J.!! + cos 2!! + 171'1' cos 19 cos 19 19 ••• + cos 19 •

If we are faced with the problem of summing a set of squares of sines or a square of. cosines, we can always proceed by first using the formulas 2 2 2 sin a = 1 - cos 2a or 2 cos a = 1 + cos 2a am. then using our formulas already derived. Note I We have tried without success to derive the formulas for the sum of sines am. the sum of cosines by other methods, but have not had much success. We tried using De Moivre's Theorem, Euler's formulas and other devices. This is because we consider the multiplication by 2 sin

'2d

a "trick". We welcome

other methods that are more "natural" from our readers. Please write if you have any other method. S. Greitzer

18

LIGHT FARE

1)

Piranhas and perch are fighting. A perch can kill one or two piranhas, is he ld at bay by three piranhas, am can be killed by four piranhas in three minutes. How long will a fight between four perch am thirteen piranhas last ?

2)

Three arabs have 17 camels, which they are to divide in the ratios 1/2, 1/3 am 1/9. The usual solution has another arab donate one camel, whereupon the three take 1/2 of 18 = 9, 1/3 of 18 = 6, and 1/9 of 18 = 2. This would leave one camel, which the fourth arab takes back. This solution is wrong. How can the camels be shared correctly ?

3)

On one side of a bad scale, a cheese weighed 16 poums. On the other side, it weighed 9 poums. What is the true weight of the cheese?

4)

It is said that the mathe~tician De Morgan was x years old in the year x • However, Jenkins was a

2 + b2 years old in the year a 4 + b 4 ,

2 2m years old in the year 2m , and. 4 3n years old in the year 3n •

How old was Jenkins each time ?

5)

We have two three-digit numbers of which the first is three times the sec om. If we reverse the digits of the secom number, that number will be four times the number obtained by reversing the first number. What are the numbers?

6)

A triangular field has a square field on each of its sides. One field has an area of 74 acres, another has an area of 116 acres, am the third field has an area of 370 acres. Fim the area of the triangular field.

You may send in your solutions, preferably by the end. of July. Label your answers "LF" followed by the number.

••• •• •• •• ••• •

•• •• ••• •• •• •• •

••• •

•e •• e

19

MORE ON DIOPHANTINE EQUATIONS

In the Arbelos for March, 198) (pp 10-17), we first investigated the linear Diophantine Equation. Among the methods we presented for solving these, we included (a) guessing, (b) Euler's method, (c) using the Farey e sequence, (d) using continued fractions, and (e) using tit con~uences. Our own favorate is guessinp;.

• •

e

e

To refresh the memory, suppose we are given the linear Diophantine Equation ax + by = c. We restrict ourselves to positive values for a, b, c, x, y and assume that (a,b,c) = 1.

e e

e • :

We first guess x o' YO such that ax O + byO = 1. Then acx O + bcyO = c.

Subtract ax + by = c , and we have

• •

a(cx O - x) + b(cyO - y) = O. We transform this into a proportion, as follows I

tt

ee

cy - y _0_ _

b and we may write

x - cx o cyO - y

a

•

e e •

-=

= at

and

Finally, we solve these for x, y,

e :

x = cx

O + bt

e

e

e e

e

= bt.

and arrive at

t

will produce

to positive values for

cyo -

at >

o.

Let us apply all this to a problem. We wish to

: find. positive x, y to satisfy

e

O

Y = cyO - at.

Substituting any integer value for

e solutions x and y. e If we restrict ourselves e x and y, we must have e cx + bt > 0 a.nd o e

x - cx

7x + 9Y = 400.

20

First, guess

7(4) + 9(-3)

=1

7(1600) + 9(-1200)

So Since

+

7x

9Y

7(1600 - x) + 9(-1200 - y)

1 = 1200

Therefore,

+ y 1600 - x'

9 am. finally,

= 1600

x

•

= 400 • = 400, = o.

7t

= 1200

+ Y

9t

= 1600

- x

- 9t

J = 7t - 1200

Substitute any integer value for

t , am. we get

solutions of the original equation.

There are some interesting problems that may arise. First, given

= c,

ax + by

with a, b, c

= 1,

positive, and with (a,b,c)

given, all

how many solutions

• ••• •• ••• •• • • ..

are there where x and y are positive? Second, given

..

ax + by, what values of

• •

c

will produce

n

solutions?

In the problem we have just solved, we would like to have

1600 - 9t > 0

and

7t - 1200 > O.

The first inequality gtves us t >

sec ond gives us

172

171~. ~

t

~

t < 17?§ • The

In integer values for t, 177 •

There will be positive solutions for x, y t = 172, 173, 174, 175, 176, 177. For example, let that

x

= 25

am.

t

= 175, y = 25.

when

•

..

•

• • •

..

am. we fim., on substituting, .. •

•• •.

•

••• • •• • •

21

Going back to the inequafities

cx

.cyO - at > 0, and solving these for

O

t

+ bt > 0

and

, we find

cyo

< -a­

.Since we know a, b, c, x

. • (integer) limits for

o am Yo' we can detennine the

t. The number of solutions in

.positive integers for x, y

will equal the number of

•

.integers between these limits. •

When we do not know the value of

c , the problem

.becanes more canpl1cated. At this point, we decide to \ : use the properties of the Greatest Integer function • • This can be defined in three ways, as followsl

.(a)

[ x ]

~ x < [ x]

.(b)

x - 1

< [ x ]

:(c)

0

~

~

+ 1

x

x - [ x ] < 1

.We agree to use whichever definition is handiest.

•

Assuming that

.let us replace :(d)

x

is not an integer, and using (b), by

-x. This gives us

-x - 1 < [-x] < -x

~~

x-1<[x]<x •

• Adding, we have .when

•

x

x

-2 < [ x ] + [-x] <

is not an integer,

o.

This means that

[ x ] + [-x]

= -1.

.We leave it to the reader to show that when

.integer,

•• •• ••

- L-x]

is an

[ x ] + [-x] = O•

• Again, fran (d), we have •

x

x + 1 > - [-x] > x, so that

is the least integer greater than

x.

22

Let us return to the inequality at the top of the previous page. Using the properties of the Greatest Integer function, we findl the smallest value for

t

_ [cx OJ b eyo -J. a

is

r

the greatest value for t is CX cyO O The difference is [ aJ + [ T J. This should tell us the number of positive values for a given It is easy to show that, where integers,

Lu

u

and.

v

c•

are not

J + [ v J < [u + v J < [ u J + [ v J + 1­

Substitution of the expressions above in this, we cyO cx O cyO CX cyO CX O O have [ a J + [b J < [ a + b J < [ a J + [b J + 1. However, the middle term in this inequality equals c ( ) _ c ab axo + byo - ab Therefore, if

•

N is the number of

are N positive solutions, then

t's for which there

N~· [c/abJ

or

N .~ [c/ab J + 1. In the illustrative example, c/ab

= 400/63,

and

(400/63J is 6, so we expect to have six solutions. Can you find a value of

c

for which

7x + 11y = c

will have four solutions in positive integers ? Because of our restrictions on a, b, c, x, y, the solutions for N usually boil down to N ~ [c/abJ.

••• •• ••• •• •• •• • ••

•

•• •• •• •• •• •• •••

• • • •

23

• The same issue of Arbelos had the following problem • to which nobody subnitted an answers • • • • • • •

On a certain planet, there are two inimical foms

of life. The Septicapita have seven heads but only

two legs each, while the Pentapoda have only two heads but do have five legs each. One day an odd lot of Septicapita encountered an odd lot of Pentapoda am a wild me lee ensued. Heads and legs were flying allover - one observer counted 180 of both. How many of each type were involved in the fracas ?

• Solutions The Septicapita contributed 7 heads am 2 . • legs each. The Pentapoda contributed 2 heads am 5 , legs each. Let x = the number of heads and y = the • number of legs. Then • 9x + 7y = 180 .Now 9(4) + 7(-5) = 1 , .so 9(720) + 7(-900) = 180 • • Subtracting, .or •

•

=0

.2_y+900 7 - 720 - x

• am

•

9(x - 720) + 7(y + 900)

= 720

x If

x

and

- 7t y

Y

= 9t

- 900.

are to be positive, we must have

• 900/9 < t < 720/7

.There are two possible solutions - when t = 101 or

.102. For t = 101, we find 13 Septicapita am 9 Penta­

.poda. For t = 102, there are 6 of one and 18 of the

• other.

• However, we must discard the second solution, since • we did say that there were an odd lot at each animal. • Our apologies for not presenting a complete set of

.rules for solving this type of problem, but we do

• believe there is enough here to enable a student to

.solve similar problems more easily.

•• •• ••

24 OLYMPIAD PROBLEMS 21)

Let p be a prime. For which k can the set (1,2,3, ••• ,k) be partitioned into p subsets with equal sum of elements?

22)

(Poland)

A collection of 2n letters contains two each of n different letters. The collection is partitioned into n pairs, each pair

containin~

two letters

which may be the same or different. Denote the number of distinct partitions by u • (Partitions n

differing in the order of the pairs in the partition or in the order of the two letters in the pairs are not considered distinct.) Prove that u = (n + l)un _ n(n2- 1) un_2 •

n+l (Great Britain)

23)

For each P inside the triangle ABC, let A(P), B(P) and C(P) be the points of intersection of the lines AP, BP and CP with the sides opposite to A, Band C respectively. Determine P in such a way that the area of the triangle A(P)B(P)C(P) is as large as possible. (Monp;ol1a)

24)

A

non-ne~tive inte~r

fen) is assip;ned to each

•

•• ••• •• ••• •• • • •• •• •• •• ••• •

positive integer n in such a way that the followi~ • conditions are satisfied: (a) f(mn) = f(m) + fen) for all positive integers m, n;

(b)

(c)

= 0 whenever f(10) = O. fen)

the units digit of n is 3;

Find the value of f(1985).

(Colanbia)

•• •• •• • •

••

25

<.•

PROBLEM SOLVING

• A good. deal of time and effort are being devoted • to the concept of teaching problem solving. Having .been connected with preparing superior students for { .mathematics contests for at least fifty years, I have • cane to the conclusion that one cannot teach problem • solving. It is a skill that is innate, like the ability of visualizing objects in space. However, it can be • improved in those who do possess problem solving skill. .We would like to show this by referring to problems

• fran various sources •

• Our first problem canes from a Polish Olympiad. I

•

• Problem a A conV6lX pentagon has the property that every

diagonal cuts off a triangle with unit area.

• Find the area of the pentagon.

I

::Solutiona 'We will sketch our solution, since the

f;.k----_~ "

.fine details can be supplied by .the reader. First, &3CD am

.e:.ECD have unit area and share the

• base CD. Therefore the altitudes D to base CD fran B and E must be

.equal am parallel.They fom opposite sides of a

.parallelogram whose bases, CD and BE are parallel. In

.fact, every diagonal must be parallel to a side of the

.pentagon. • But then, PBAE is a parallelogram, because PB am AE eare parallel, as are FE am AB. Therefore, the area of ~BE is unity. • Next, trianKles PBE am. PDC are similar - equal a~les eam that sort of thing, so 6PBE/6PDC = (BrIPD)~. Also, eAJ3PC/APDC :: (BP/PD) = (1 - a)/a. Equating, we arrive at •

1/a = (1 -a) 2/a 2 , which is easily solved to give

.a = 3 -2 -J5

We can now fim the area of each piece of

::the pentagon, and find

•• •• •

K = 5 +2-./5 •

26

There is nothing esoteric about the solution. Any p;ood. student knows all the data used. Yet this was a difficult problem. The difficulty apparently lay in the recop;nition ,.am use of the data. There are students who may read the solution am be unable to repeat it. The next problem comes from the Hungarian magazine K8Mal. In this, as in all other problems, we suggest that the reader try to solve before reading the solution.

= f(x)

Problema If f(x - y2)

+ (y2 - 2x).f(y) find the

value of f(1986). Solutiona Let y

= 1.

Then f(x - 1)

= 1,

f(O)

= f(l)

- f(l)

for x

= 2, = J,

f(l)

= f(2) = f(J)

- Jf(l)

f(2)

f(n-l)

- 5f(1)

= fen)

= f(x)

x (1 - 2x).

• • • • • • • •

•

and so on to

~ries

We add,

2 1 + J + 5 + ••• + (2n - 1) = n , am finally arrive at f(O) = 0 = fen) - n2 .f(1), or 2 fen) = n .f(1). note that the series

f(1986)

•

tt

- (2n - l).f(l).

We have arrived at a telescoping

Let f(t) equal A, am

•

•

for x for x

••

= (1986)2xA •

How would a teacher of :Jroblem solvinp; teach that first step - letting y = 1 ? It is possible that a student with a good. memory could eventually arrive at a solution of the previous problem. We challenge the teacher to present the secom problem to his class (am/or to other teachers). The number of correct solutions as listed in K~al was not bad. Hungary has a strong mathematics program, looates its good. prospects, am provides a lot of practice by expert teachers.

• •• •• •• • • •

• • • • • • •

••• •

•• e •• e

e e e

e e

e e e e

27

Our third problem also comes from K~al. Examine it first am see how you might attack it ! Problem: Solve for

XI

Lx] + [2x] + L4x] + L8x] + [16x] + [J2x]

~

Solutions This 1s not nearly as difficult as were the first two problems. A talented student could solve i"tr, he used suffic ient care. First,

x

=[

x ] + f, where

nx = n [ x ] + nf

Then

O~ f

< 1. 0 < nf <: n.

At this point, one uses judgment, and writes

n [ x ]

~

For all six terms, theretQre, we may write

e

63[ x ] ~ [ x

e

57.

:

That is,

e

J+

[me]

~

e• e

n [ x ] + n - 1.

[2x] + [4x] + ••• + [32x] ~ 6)[ x ] +

(Questions why the

57 ?).

63 [ x ] ~ 12345 ~ 63 [ x ] + 57. Now

12288 ~ 63 [ x ] ~ 12345.

••

Thus,

•

There can 't be any integer between 195 2i and 20 195 21. Doesn't look like there is a solution!

e •

e e e e

e e

e e e e

195 211 ~ [x] ~ 19? 20 21 •

The next problem comes from a Polish Olympiad. We include it because it can be done rather mechanically, which would take lots of time and paper. The Polish solution starts with something "out of a hat", Tf 1Tf t;Tf ?n em _ Problems Evaluate sinra.sint8.sint8.sint8.sint8.~N. Solutionl The diligent student has seen this kind of problem before, and. will proceed to combine

--

e e

12J45.

28

terms at opposite ends. In thifll c68e, we doubt that this method would work. If it did, it would take a lot of work. (We tried it at first). The Polish "hat trick" started with sin might try to see where to go from this.

=!

We first removed sin ~ and sin n sin .21! sin .f1! _ had sin 18. 18. 18 - 2N.

~~. You

tIT =1.

Then we

Now we used a method usee elsewhere in this issue ­ we multiplied both sides of this last equation by rr

2 cos 18 • This gave us .2! .2! .21! l!! - J., .2! 2 cos 18. sin 18. sin 18. sin 18 -,~N cos 18 •

•• •• •• •• •• •••

2n .21! 1!! ­ 4N cos 18·

n sin 18. sin 18. sin 18 ­ However, sin l!! 18 = cos 2rr 18. Make the change and

multiply by two. Now we haw 2rr

or

We can now use resulting in That is,

2rr

.21! -

n

8

2 cos 18. sin 18. sin 18 - N cos 18 • 4n 2!! _ n sin 18. sin 18 - 8N cos 18 • cos(x-y) - cos(x+y)

=2

sin x.sin y,

rr - cos .2!! rr • cos 18 18 -- 16N cos 18

n cos 18

= 16N cos

rr ,and 18

N

= 1/6 1 •

We don't think many students in a "problem solving" class could find this solution. The lesson - read a lot of solutions by expert problem solvers and hope that some of what you read may remain with you and be of use when you are faced with a tough mathematics problem.

• • •• •• ••

•• •• •

•

•e

29

e e e e

..

,

KURSCHAK KORNER

This column features those problems of the famous Hungarian • Kiirschak Contest which have not yet appeared in English trans­ elation. For the problems of 1894-1928, the reader should consult e Hungarian Problem Books! & !!. in the MAA's New Mathematical Library series (available from the MAA hdqrs.)j the problems (with brief summaries of their solutions) covering the years 1966-1981 were translated into English by Prof. Csirmaz of • Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers.) .

e e

e

­

e e e e e e e e e e e

e e e

e e e e e e e e e e e e

Student readers are also invited to set aside an uninterrupted 4-hour period to compose complete, well-written solutions to the problems below, and to submit their work to the address given below for critical evaluation. 1/1958. Of six points given in the plane no three are collinear. Show that one can choose three of the six points so that if a triangle is constructed with them as vertices, then one of the angles of the triangle will measure at least 120°. 2/1958.

Prove that if u and v are integers such that is divisible by 9, then both u and v are multiples of 3. U

Z

+ UV + v 2

3/1958. In convex hexagon ABCDEF the opposite are parallel (i.e., AB DE, BC EF and CD FA). that the areas of ~ACE and ~BDF are equal.

-•

II

II

II

sides Prove

Dr. George Berzsenyi 2040 Chevy Chase Beaumont, TX 77706

CONTENTS COVER, Our Geometry Problem. i 1

7 14 18 19 24 25 29

Preface A Useful Trigonometric Substitution The Pentagon Many Cheerful Facts Light Fare More on Diophantine Equations Olympiad Problems Problem Solving Kurschak Korner

••• •• •• •• ••• •• •• ••

•

•

•• • •

•• •• •• •

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j­

le Ie je ~e

le ~e

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le ~e

je

Ie je

1e

Ie je

le e Ie

je

je je

,ee

je

je je

e Ie

l

je

!e je

\e

•• •• •

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e

,

;

I

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,

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