No title

481

SKOLIAD

No. 121

Lily Yen and Mogens Hansen

Please send your solutions to problems in this Skoliad by 1 June, 2010. A

opy of Crux will be sent to one pre-university reader who sends in solutions before the deadline. The de ision of the editors is nal. Our ontest this month is the Niels Henrik Abel Mathemati s Contest, 2008{2009, Se ond Round. Our thanks go to yvind Bakke of the Norwegian University of S ien e and Te hnology, Trondheim, Norway, for providing us with this ontest and for permission to publish it. La reda tion  souhaite remer ier Rolland Gaudet, de College  universitaire de Saint-Bonifa e, Winnipeg, MB, d'avoir traduit e on ours. Con ours mathematique  Niels Henrik Abel, 2008{2009 e 2 ronde

Duree  de 100 minutes

. On dispose les entiers positifs impairs dans un tableau triangulaire tel qu'illustre.  Quelle est la somme des entiers dans les sept premieres  rangees  ? 1

13

7 15

1 3 5 9 11 ... ...

. Un grand panier ontient beau oup d'oeufs. Si on enleve  les oeufs deux a la fois, a la toute n il ne restera qu'un seul oeuf. La m^eme hose se produit si on enleve  les oeufs trois a la fois, ou quatre ou inq ou six a la fois, mais si on enleve  les oeufs sept a la fois, on vide le panier ompletement.  Le panier

ontenait au moins quel nombre d'oeufs ?

2

. Deux des angles d'une etoile  a inq pointes sont 28◦ et 37◦ , tel qu'illustre.  Tous les sommets se trouvent sur un er le, ou le point indique est le entre du er le. Determiner  la mesure de l'angle x. 3

................................................. .......... ........ ........ ......... ...... ........ .. .... ......................... . ........ ..... ....... . . ... .... ........ . ... ...... ................ . . . . . . . . ... .... ................. ... ... .. . . . . ... .... ....... ........ .. ......... ... x◦.... ............ ......... .. . . . . . . . . . . . ... . ......... .... .... .... ..... ..... . . . . . . . . . . . . . ... . . .... .......... .. . . ........ ◦ ... ........... ..... .... .. .......... . . ..... . .................................28 .............................................................................................................................. ... .... ... .... ... ... ... .. .... .. ... .. .... .. .. .. ....37◦..... .. ... . .... . ... .. .... ..... ... .... .. ..... .... .... ... .... .... ............ ..... . . . . ....... ........ ....... ........... ........ ...............................................

r

4. Lan elot et inq autres hevaliers sont assis autour d'une table ronde. Or ha un des six hevaliers trouve moyen de se faire ennemi ave ses deux voisins. Combien de manieres  y a-t-il d'asseoir les six hevaliers de fa on a

e que Lan elot garde son siege  et qu'au un des six hevaliers soit assis voisin d'un de ses deux ennemis ?

482 . La moyenne arithmetique  A, de deux nombres reels,  x et y , est 12 (x + y) √ y et la moyenne geom  etrique  G, est xy . Determiner  si 3A = 5G. x 5

6

. Une fon tion f est telle que pour tout entier positif n, on a f (n + 1) =

f (n) 1 + af (n)

ou a est un nombre reel,  f (1) = 1, et f (9) =

,

1 . 2009

Determiner  a.

. Le grand er le a un rayon egal  a √30 . Les π

er les moyens sont tangents l'un a l'autre, au entre du grand er le. De plus, les er les moyens sont tangents aux petits er les, et le grand er le est tangent a tous les er les qu'il

ontient. Determiner  la surfa e de la region  ombragee.  7

. Determiner  la somme de tous les entiers positifs n, tels que 2009 + n2 est le arre d'un entier positif. 8

9. Les points (23, 32), (8, 41) et (17, 45) sont les points milieux des ot ^ es  d'un triangle. Determiner  la plus grosse valeur possible de x + y, ou (x, y) est un sommet du triangle.

. Karine et Mathieu lan ent une pie e  de monnaie. Chaque fois que la pie e  de monnaie montre fa e, Karine gagne un point et haque fois qu'elle montre pile, Mathieu gagne un point. La personne gagnante est la premiere  personne a atteindre six points ou aussi atteindre au moins quatre points ave une avan e d'au moins trois points. Combien de jeux dierents  sont-ils possibles ; 'est-a-dire  quel est le nombre de suites dierentes  de lan ers de pie e  de monnaie, du debut  jusqu'a la de laration  du gagnant ? 10

Niels Henrik Abel Mathemati s Contest, 2008{2009 2nd Round

100 minutes allowed

. Arrange the positive odd numbers in a triangular diagram as shown. What is the sum of the numbers in the rst seven rows? 1

13

7 15

1 3 5 9 11 ... ...

. A large basket ontains many eggs. If you remove the eggs two at a time, a single egg remains in the basket. The same happens if you remove the eggs three at a time, or four or ve or six at a time, but if you remove the eggs seven at a time, you empty the basket ompletely. At least how many eggs were there in the basket? 2

483 .............................................. ........... ........ ........ ...... ....... ........ ........ ....................... . ........ .... ........ . . ........ ... ... . ... ..... ............... . . . . . . . . ... ... .... ......... .......... ... ... . . . ... . . . .... ... .................. .................. ... x◦... . . . . .. . . . . . . . . . . . . . . ......... ..... .. .... ...... ...... . . . . . .. . . . . . . . . .... . .......... .. .... ... .. . ............... ........ ◦ .... ... .......... ... . ........... .... .... ........ ................................28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... ... ... .. .... .. . ... ... .. .... .. ... .. .. .... . . .. ◦ . . ....37 .. . .. . .... .. ... .. .... ..... ... .... ... ..... ... .... .. .... .... . . ...... ... .... ....... ..... ....... ....... ........ ............ ......... ...........................................

. Two of the angles in a ve-pointed star are and 37◦ , as shown. All the verti es lie on a ir le, and the indi ated point is the entre of the ir le. Find the measure of angle x. 3

28◦

r

4. Sir Lan elot and ve other knights sit at a round table. All six knights manage to make enemies of both their neighbours. In how many ways an the six knights sit around the table if Sir Lan elot is to keep his seat and no one sits next to either of his new enemies?

. The arithmeti mean, A, of two real numbers, x and y, is their geometri mean, G, is √xy . Find xy if 3A = 5G.

5

1 (x + y) 2

and

. A fun tion, f , is su h that for ea h positive integer n,

6

f (n + 1) =

f (n) 1 + af (n)

where a is a real number, f (1) = 1, and f (9) =

, 1 . 2009

Find a.

. The large ir le has radius √30 . The π medium ir les are tangent to one another at the entre of the large ir le. Moreover, the medium ir les are tangent to the small ir les, and the large ir le is tangent to all the

ir les it ontains. Find the area of the shaded region.

7

8. Find the sum of all positive integers, n, su h that 2009 + n2 is the square of a positive integer.

. The points (23, 32), (8, 41), and (17, 45) are the midpoints of the sides of a triangle. Find the largest possible value of x + y where (x, y) is a vertex of the triangle. 9

10. Kari and Mons are tossing a oin. Ea h time they toss heads, Kari earns a point, and ea h time they toss tails, Mons earns a point. The person who rst rea hes six points or who has at least four points and leads by at least three points wins. How many dierent games are possible; that is, how many dierent sequen es of oin tosses end in a win?

484 Next follow the solutions to the British Columbia Se ondary S hool Mathemati s Contest 2007, Final Round, Part B [2009 : 65{68℄. 1. Joan has a olle tion of ni kels, dimes, and quarters worth $2.00. If the ni kels were dimes and the dimes were ni kels, the value of the oins would be $1.70. Determine all of the possibilities for the number of ni kels, dimes, and quarters that Joan ould have.

Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Suppose Joan has n ni kels, d dimes, and q quarters. Then the total value of the oins is 5n + 10d + 25q = 200, and with the ni kels and dimes swit hed 5d + 10n + 25q = 170. Subtra ting the latter equation from the former yields that 5d − 5n = 30, so d = n + 6. Substituting d = n + 6 into the equation 5n + 10d + 25q = 200 yields (28 − 5q) 5n + 10(n + 6) + 25q = 200, so 15n + 60 + 25q = 200, so n = . 3 Sin e n ≥ 0, it follows that q ≤ 5. You may now try all the possible values for q in turn: q

0

1

(28 − 5q) n= 3

28 3

23 3

2

3

4

5

6

13 3

8 3

1

The only integer solutions for (n, d, q) are (6, 12, 2) and (1, 7, 5). Also solved by OSCAR XIA, student, St. George's S hool, Van ouver, BC.

. A 3 × 3 × 3 ube is formed by sta king 1 × 1 × 1 ubes. Determine the total number of ubes with sides of integral length that are ontained in the 3 × 3 × 3 ube.

2

Solution by Gesine Geupel, student, Max Ernst Gymnasium, Bruhl,  NRW, Germany. The ube ontains one 2×2×2 ube for ea h of the 8 verti es in addition to 27 small 1 × 1 × 1 ubes and the 3 × 3 × 3 ube itself. In all 36 ubes.

Also solved by JEREMY TSE, student, Burnaby North Se ondary S hool, Burnaby, BC; JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; and OSCAR XIA, student, St. George's S hool, Van ouver, BC.

. The lengths of the sides of a triangle are 13, 13, and 10. The ir ums ribed ir le of a triangle is the ir le that goes through ea h of the three verti es of the triangle and here has its entre inside the triangle (see the diagram at right). Find the radius of the ir ums ribed

ir le. 3

........... ................... ................ . . . . . . .... . ... .. ... .... ... ... ... .... ... .. .... ... . ... ... .. . ... ... ... .. ... .. .... .. ..................................................... ............ ............ ...........

485 Solution by Os ar Xia, student, St. George's S hool, Van ouver, BC. Let r be the radius of the ir le through the verti es A, B , C , and let M be the midpoint of AB . Sin e △ABC is isos eles, ∠AM C = 90◦ . Therefore, the Pythagorean Theorem √ applies and CM = 132 − 52 = 12. It follows that M O = 12 − r. Using the Pythagorean Theorem again, AM 2 + M O 2 = AO 2 , so we have 52 +(12−r)2 = r 2 . Solving the equation yields that r = 169 . 24

C ............................................................... .......... ........ . .. .. ....... . ..... . . ... .. .. . ..... .. ..... ..... .... . . .... . . . . . .... ... . ... . . . . . ... . . . ... ... .... .... ... .. . ... . ... . . . . ... ... .... .... .... ... ... . . . . .... ... 13 .. . 13 . . . ... ...O . . .... . . . ... ... .............. .. . . . ... . . . . . ... . ... ..... ... . . . . . . . ... .. .... ... ....... ... ... ... .. .... ... ..... ... ... .. ........ ... ..... .. ... . . . . ..... ... ... ... .. ..... ... ....... ... ... ........ .. .... ...... ........................................................................................................... ...... .. 5 ......... B A ..............5 ................ M ......................... ................

Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.

. The game of End View onsists of a tableau with a four by four grid, one additional row at the top and at the bottom, and one additional olumn on the right and on the left. The letters A, B, and C are pla ed in the four by four grid in su h a way that every letter appears exa tly on e in ea h row and ea h olumn. This means that there will be exa tly one empty square in ea h row and ea h olumn. Letters are pla ed in the additional rows and olumns as hints, at the end of some rows and olumns of the four by four grid, to indi ate the nearest letter that an be found by reading that row or olumn of the grid. The diagram below shows the starting tableau and the resulting solution tableau for a game of End View. 4

................................... .................B......................C................. ... ... ... ... ... C ... ... ... ... ... ... B ... ... A ... ... ... ... ... ................................................... .....A...............B.............. ... ... ... .... .... .... .. ... ... .... .... ..... .... .... ... .... .... .... .... ... ... .... .. .... ...................................................................................................................................................... .... .... .... .... .... .... .. .. .. ...................................................................................................................................................... ..... ..... ..... .... .... .... .. .. .. ...................................................................................................................................................... ..... ..... ..... .... .... .... ... ... ... .... .... .... .... .... .... ... ... ... ... ... ... . . .

rrr rrrrrrrrrrrrrrrrrrrrrrrrrr rrr

Start

The diagram for another game of End View is shown at right. Fill in this tableau with the omplete solution. Give a justi ation of the steps that you used to nd the solution.

................................... .................B.......................C................ ... ... B A C .. .. ... C ... C B A ... ... ... ... . . ... A ... A C B .... B .... ... ... A C B ... ... ................................................. .....A...............B.............. ... ... ... .... .... .... .. ... ... .... .... ..... .... .... ... .... .... .... .... ... ... .... .. .... ...................................................................................................................................................... .... .... .... .... .... .... .. .. .. ...................................................................................................................................................... ..... ..... ..... .... .... .... .. .. .. ...................................................................................................................................................... ..... ..... ..... .... .... .... ... ... ... .... .... .... .... .... .... ... ... ... ... ... ... . . .

Solution

................................. ........................................................... ... ... ... ... ... A ... ... ... ... .. ... B ... ... ... ... ... .. ... .................................................. .............C................C...... .. .. .. .... .... .... ... .... .... .... .... .... .... .... .... . ... . ... .... ..... .... .... ... .. ........................................................................................................................................................ .... .... .... ... ... ... .. ... ... .......................................................................................................................................................... .... .... .... ... ... ... .. ... ... .......................................................................................................................................................... ... ... ... .... .... .... .... .... .... ... ... ... .... .... .... .... .... .... .... .... .... . . .

486 Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON.

............................... ................................................ ... ... ... .. ... ... ... ... .. ... ... ... ... ... ............................................... .............................. .... .

.... .

............................... ................................................ ... ... ... .. ... ... ... ... .. ... ... ... ... ... ............................................... ..............................

............................... ................................................ ... ... ... .. ... ... ... ... .. ... ... ... ... ... ............................................... ..............................

.... .

.... .

.. .. .. .... .... .... 6 .... .... .... ... .. ... .... .... 5 .......................................................................................................................................... ... ... ... .... .... .... .... .... .... 4 .......A . . . . ....................................................................................................................... ... .. ... ... ... 3 ...........................................................................................................................B .............. ... ... ... .... .... .... ... ... ... 2 ... ... ... .... .... .... ... .... .... 1 .... C .... C ..... 1 2 3 4 5 6

.... .

.... .

.... .

.... .

.... .

.. .. .. .... .... .... 6 .... .... .... ... .. ... .... .... 5 ........................................................................................................A ... ... .................................. ... .... .... .... .... .... .... 4 .......A . . . . ....................................................................................................................... ... ... ... ... ... ... 3 ....................................................C B ... ... ... ................................................................................. .... .... .... . . . . . . 2 C .... .... .... .... .... .... ... .... .... 1 .... C .... C ..... 1 2 3 4 5 6

.. .. .. .... .... .... 6 .... .... .... ... .. ... .... .... 5 .......................................................................................................................................... ... ... ... .... .... .... .... .... .... 4 .......A . . . . ....................................................................................................................... ... ... ... ... ... ... 3 ....................................................C B ... ... ... ................................................................................. .... .... .... . . . . . . 2 C .... .... .... .... .... .... ... .... .... 1 .... C .... C ..... 1 2 3 4 5 6

Introdu e a oordinate system as in the gures above. Then onsider

olumns 3 and 5. Sin e ea h must have ea h of the three letters, and C must be the losest one to the bottom, only one of the bottom two spa es of these

olumns an be o

upied with C, and the other spa e not o

upied. Now, in olumn 5, if C o

upied (5, 3) then B would not be the losest letter to the right extra olumn. Therefore (5, 2) must be C. Then in row 2, there annot be another C, so in olumn 3, C must not be at the bottom square, so it must be at (3, 3). We have arrived at the middle gure above. In row 4, A must be the farthest to the left. Therefore the A in olumn 5

annot be at (5, 4). It also annot be at (5, 3), sin e B must be the farthest right in row 3. Therefore an A is at (5, 5). We have arrived at the rightmost gure above. Now in olumn 3, A annot be at (3, 5), so it must be at (3, 4). Then B must be at (3, 5). Then in row 4, B and C must be in the rightmost two squares, and sin e olumn 5 has a C, the C goes in (4, 4), and the B in (5, 4). See the leftmost gure below.

............................... ............................................... ... ... ... .. ... ... ... ... .. ... ... ... ... ... ............................................... ............................... ... ..

... ..

... ..

... ... ... ... ... ... 6 ... ... ... ... ... ... .... .... .... 5 ....................................................B ... .. A .. ................................................................................ .... .... .... . . . . . 4 .......A . A . C .. B ................................................................................................................................. ... ... ... ... ... ... 3 ....................................................C B .. .. ... ................................................................................... ... ... ... . . . . . . 2 . . C . ..... ..... ..... ... ... ... ... .... .... 1 ... C ... C .... . . . 1 2 3 4 5 6

............................... ............................................... ... ... ... .. ... ... ... ... .. ... ... ... ... ... ............................................... ............................... ... ..

... ..

... ..

... ... ... ... ... ... 6 ... ... ... ... ... ... .... .... .... 5 .............................C .. A ... B .... ..................................................................................................... ..... ..... ..... . . . . . 4 .......A . A . C .. B .................................................................................................................................. ... ... ... ... ... .. 3 ............................A B .. C .. B ... ............................................................................................................ .... .... .... . . . . . . 2 B ...... ...... A ...... C ... ... ... ... .... .... 1 ... C ... C .... . . . 1 2 3 4 5 6

From here, it is easy to omplete the tableau. The last C has to go in Finally, row 2 has A at in the rightmost gure immediately above.

(2, 5). In row 3, B must be at (4, 3), and A at (2, 3). (4, 2) and B at (2, 2). The nished tableau is shown

Also solved by OSCAR XIA, student, St. George's S hool, Van ouver, BC.

. Determine all of the positive integer solutions, x and y, to the equation

5

1 1 1 − = x y 12

.

487 Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. 1 Sin e both x and y are positive, x1 > 12 , so x < 12. Solving the 1 equation x1 − y1 = 12 for y yields y = values for x in turn:

y=

12x . 12 − x

You may now try the possible

x

1

2

3

4

5

6

7

8

9

10

11

12x 12 − x

12 11

12 5

4

6

60 7

12

84 5

24

36

60

132

Thus the only positive integer solutions for (x, y) are (3, 4), (4, 6), (6, 12), (8, 24), (9, 36), (10, 60), and (11, 132). Also solved by OSCAR XIA, student, St. George's S hool, Van ouver, BC.

Slope m 2

√1

Here is a proof that if perpendi ular lines have slopes m and M , then mM = −1:

+m

m

1

−M

2



1 + m2 + 1 + M 2 = m2 − 2mM + M 2 ; 2 = −2mM ; mM = −1 .

√ 1+M

2 p 2 p 1 + m2 + 1 + M2 = (m − M )2 ;

Slope M √

√

This on ludes another Skoliad. This issue's prize for the best solutions goes to Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. We are looking forward to the results of our readers' eorts.

488

MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by High S hool and University Students. It ontinues, with the same emphasis, as an integral part of Crux Mathemati orum with Mathemati al Mayhem. The Mayhem Editor is Ian VanderBurgh (University of Waterloo). The other sta members are Monika Khbeis (Our Lady of Mt. Carmel Se ondary S hool, Mississauga, ON), and Eri Robert (Leo Hayes High S hool, Frederi ton, NB). Mayhem Problems

Veuillez nous transmettre vos solutions aux problemes du present numero avant le 1er mars 2010. Les solutions re ues apres ette date ne seront prises en

ompte que s'il nous reste du temps avant la publi ation des solutions. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes.  . Propose par l'Equipe de Mayhem. Trouver le nombre d'entiers positifs formes  de trois hires dont le produit donne 36.

M413

 . Propose par l'Equipe de Mayhem. On onsidere  la liste des entiers positifs, ranges  en ordre roissant, pouvant e^ tre exprimes  omme la somme de 21 entiers (non ne essairement  positifs). Determiner  le 21e entier de ette liste. M414

 . Propose par Ne ulai Stan iu, E ole se ondaire George Emil Palade, Buzau, Roumanie. Les ot ^ es  AB et CD d'un trapeze  ABCD sont paralleles.  Si AB = 15, CD = 30, AD = 9 et BC = 12, trouver l'aire du trapeze  ABCD. M415

M416. Propose  par Bru e Shawyer, Universite Memorial de Terre-Neuve, St. John's, NL.
Montrer que 9 est un diviseur de 10n +3 4n+2 +5 pour tous les entiers n non negatifs. 

. Propose par Mihaly Ben ze, Brasov, Roumanie. Soit M = {x2 + 4xy + y2 : x, y ∈ Z}. Montrer que le nombre 2022 appartient a M , mais pas le nombre 11. M417

489 . Propose par Georey  A. Kandall, Hamden, CT, E-U. Dans la gure, F est sur GE et Q sur le prolongement de GE . De plus, A et H sont sur P G de sorte que QA

oupe P F en B et P E en C , et que QH oupe P E en D . Montrer que M418

AB

·

BC M419

CD DE

·

EF FG

·

GH HA

= 1.

P . ....... ... ..... ... .. .. ... .. .. ... ... .... . . ... .. .. ... ... .... ... .. ... ... .. ... ............ .. . . ... ... ............... .... ... .......... ... . ... . ......... ... . . . . . . . . .. .......... .... .. . . . . . ........ ... .. .. ......... . . . ........ . .. ... .. .............. .................................. ... ............ ................. .... ... . ........ . ... .................. ........ ... .. .................. . . . ........ ................. .... .. .. ........ . . .................. ..... ... .. . . . . . . . . ... ............................... ................ ... .. . ........... ..... . . . . ..................................................................................................................................................................................................................................

A

B

C

H

R

G

D

F

E

Q

 . Propose par Joe Howard, Portales, NM, E-U.

Soit a, b et c les longueurs des ot ^ es  d'un triangle. Montrer que a(b + c) a2

+ bc

+

b(c + a) b2

+ ca

+

c(a + b) c2 + ab

≤ 3.

................................................................. . Proposed by the Mayhem Sta.

M413

Determine the number of three-digit positive integers whose digits have a produ t of 36. . Proposed by the Mayhem Sta.

M414

The positive integers that an be expressed as the sum of 21 onse utive (not ne essarily positive) integers are listed in in reasing order. Determine the 21st integer in this list. . Proposed by Ne ulai Stan iu, George Emil Palade Se ondary S hool, Buzau, Romania.

M415

In trapezoid ABCD, AB and CD are parallel. If AB = 15, CD = 30, and BC = 12, determine the area of the trapezoid.

AD = 9,

. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL.

M416

Prove that 9 divides 10n + 3


4n+2 + 5 for all nonnegative integers n.

. Proposed by Mihaly Ben ze, Brasov, Romania.

M417

Let M = {x2 + 4xy + y2 : x, y ∈ Z}. Prove that the number 2022 is in M but that the number 11 is not in M .

490 . Proposed by Georey A. Kandall, Hamden, CT, USA. In the diagram, F lies on GE and Q lies on GE extended. Also, A and H are on P G so that QA interse ts P F at B , QA interse ts P E at C , and QH interse ts P E at D . Prove that M418

AB CD EF GH · · · = 1. BC DE F G HA

P . ....... ... ..... ... .. .. ... .. .. ... ... .... . . ... ... .. ... .. ... ... .. ... ... .. ... ............ . . . ... .. ............ .... . ... . ........ ... ... . . ... . . . ......... .. . . . . . . . . .. .......... .... .. . . . . . ........ ... .. .. ......... . . . ........ . .. ... .. .............. .................................. ... ........... ................. .... ... ........ . . ... ................. ........ ... .. ................... . ........ . . ................. .... .. ...... .. . . .................. ... .. ................. ................ . . . . ................. ........ ... ... .. . ............ ..... . . . . ...................................................................................................................................................................................................................................

A

B

C

H

R

G

F

D

E

Q

. Proposed by Joe Howard, Portales, NM, USA. Let a, b, and c be the side lengths of a triangle. Prove that

M419

a(b + c) b(c + a) c(a + b) + 2 + 2 ≤ 3. 2 a + bc b + ca c + ab

Mayhem Solutions

. Proposed by the Mayhem Sta. Determine all pairs (x, y) of integers for whi h 4x2 − y2 = 480.

M382

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. First, we note that if (x, y) is a solution of the given equation with x and y integers, then so are (x, −y), (−x, y), and (−x, −y). Hen e, it suÆ es to nd all solutions (x, y) in whi h x ≥ 0 and y ≥ 0. Sin e 4x2 and 480 are both even, then y2 is even, so y is even. We thus set y = 2z for some nonnegative integer z . This yields 4x2 − (2z)2 = 480, or 4x2 − 4z 2 = 480, or x2 − z 2 = 120, or (x − z)(x + z) = 24 · 3 · 5. Next we note that (x − z) + (x + z) = 2x, whi h is even, so x − z and x + z must both be even integers or both odd integers. Sin e their produ t is 120 (whi h is even), then ea h is even. Also, x − z ≤ x + z sin e z ≥ 0. We make a hart to summarize the possible values for x − z and x + z , knowing that they are even positive integers whose produ t is 120. We obtain 2x by adding x − z and x + z , and we re over z by subtra ting x from x + z : x−z 2 4 6 10

x+z 60 30 20 12

2x 62 34 26 22

x 31 17 13 11

z 29 13 7 1

y 58 26 14 2

491 Therefore, the nonnegative solutions are (31, 58), (17, 26), (13, 14), and (11, 2). Thus, the omplete integer solution of 4x2 − y 2 = 480 onsists of the 16 pairs (±31, ±58), (±17, ±26), (±13, ±14), and (±11, ±2), where all possible ombinations of signs are taken.

Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; MATTHEW BABBITT, student, Albany Area Math Cir le, Fort Edward, NY, USA; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; JOSE  HERNANDEZ SANTIAGO, student, Universidad Te nologi a  de la Mixte a, Oaxa a, Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA. There were four in orre t and one in omplete solutions submitted.

. Proposed by the Mayhem Sta. In re tangle ABCD, P is on side BC and Q is on side DC so that = 1, AP = P Q = 2 and ∠AP Q = 90◦ . Determine the length of QD .

M383

BP

Solution by Ja lyn Chang, student, Western Canada High S hool, Calgary, AB. Using the given information, we draw the diagram at right. In the diagram, A B ∠AP Q = ∠ABP = ∠P CQ = 90 AB = DC , and AD = BC .

,

Sin e △AP B has a right angle at B , AP = 2, and BP = 1, then △AP √B is a 30◦ { 60◦ { 90◦ triangle, so AB = 3 and ∠AP B = 60◦ . Next, we see that

∠QP C = 180◦ − ∠AP Q − ∠AP B = 180◦ − 90◦ − 60◦ = 30◦ .

....................................................................................................................... ... ..... ........... .. ... ........ ... .. .. ........ ... ... ... ........ .. ........ 2 ... .... ........ .... .. ... ........ .. .. .. ........ ... ... ... ........ ........ .. ... ..... ........ .. .. .. ........ .. ... ........ .... ... ........ .. ... ... ... ... ..... ... .. ... .... .. ... . . ... ... . . ... .. .. ... .. .... ... .. .. ... .... .. ... . ... ... .. ... ... ... . ... ... . .. . ... ... .. .. 2 . . ... ... ... . . ... ... . ... . ... ... . .. . ... ... .. .. ... ... ... ... . ... ... . ... . ... ... .. ... ... ... . .. ... ... ... ... . ... .. ... ... .. .. . .. ...................................................................................................................

Sin e △QP C has a 30◦ angle and a 90 angle, then it is also a 30◦ { 60◦ { 90◦ triangle. Therefore, ∠P QC = 60◦ and D Q QC = 12 QP = 1. √ Lastly, QD = DC − QC = AB − QC = 3 − 1. ◦

.. ... ....... ... ..... .......... ...... .........

◦

1 P

C

Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; MATTHEW BABBITT, student, Albany Area Math Cir le, Fort Edward, NY, USA; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; G.C. GREUBEL, Newport News, VA, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA;  IES \Abastos", WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; RICARD PEIRO, Valen ia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; CHRISTOPHER WIRIAWAN, student, Surya Institute, BSD City, Indonesia; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA.

492 . Proposed by Kunal Singh, student, Kendriya Vidyalaya S hool, Shillong, India. B In the diagram at right, the point E is on AB and the point D is on AC 1 su h that AE = EB = DC = 1 E and AD = 2. Determine the ratio 1 of the area of quadrilateral BCDE to the area of triangle ABC . 2 A C D 1 M384

..... .... ........ ..... .... ..... .... . . ..... .. ..... .... . ..... . . ..... .. . . . . ..... ... .............. ..... . . ........ . ..... . . . . . ........ ..... .. . . . . . ..... . ........ .. . . ..... . . . . . ........ .. ..... . . . . . . . ... . ..............................................................................................................................................

Solution by S ott Brown, Auburn University, Montgomery, AL, USA. We use the notation [△ABC] for the area of △ABC and [BCDE] for the area of quadrilateral BCDE . First, we note that [△ABC] = [△AED] + [BCDE]. We will determine the ratio of [△ABC] to [△AED] and use this to determine the required ratio. To do this, we use the property that triangles with equal altitudes have their areas in the same ratio as the lengths of their bases. Therefore, [△ABC] [△ADB] AC AB 3 2 3 [△ABC] = · = · = · = [△EAD] [△ADB] [△EAD] AD AE 2 1 1

.

Thus, [△EAD] is 13 of [△ABC]. This implies that [BCDE] is 23 of [△ABC]. In summary,

[BCDE] 2 = . [△ABC] 3

Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; MATTHEW BABBITT, student, Albany Area Math Cir le, Fort Edward, NY, USA; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JACLYN  GYUSZI, Dimitrie Leonida CHANG, student, Western Canada High S hool, Calgary, AB; SZEP  IES \Abastos", Valen ia, Te hnologi al High S hool, Petrosani, Romania; RICARD PEIRO, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; CHRISTOPHER WIRIAWAN, student, Surya Institute, BSD City, Indonesia; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA. There was one in omplete solution submitted.

. Proposed by Mihaly Ben ze, Brasov, Romania. The base 10 integer N = 1 · · · 114 · · · 44 starts o with 2009 onse utive digits 1 followed by 4018 onse utive digits 4. Prove that N is not a perfe t square. M385

Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Considering N modulo 16, we see that N ≡ 12 (mod 16), as follows: 1 · · · 114 · · · 44 ≡ 4444 (mod 16) ≡ 44 (mod 16) ≡ 12 (mod 16)

.

(sin e 10 000 is a multiple of 16) (sin e 400 is a multiple of 16)

493 Any integer x is ongruent modulo 16 to one of 0, ±1, ±2, ±3, ±4, or 8; so x2 is ongruent to one of 0, 1, 4, 9, 0, 9, 4, 1, or 0. Therefore, every perfe t square is ongruent to 0, 1, 4, or 9 modulo 16, and we on lude that N annot be a perfe t square. ±5, ±6, ±7,

Also solved by MATTHEW BABBITT, student, Albany Area Math Cir le, Fort Edward,  GYUSZI, Dimitrie Leonida Te hnologi al High S hool, Petrosani, Romania; NY, USA; SZEP  JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a  de la Mixte a, Oaxa a,  IES \Abastos", Valen ia, Spain; BRUNO SALGUEIRO FANEGO, Mexi o; RICARD PEIRO, Viveiro, Spain; CHRISTA SOESANTO, student, Surya Institute, BSD City, Indonesia; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA. There were two in omplete solutions submitted.

. Proposed by Ne ulai Stan iu, George Emil Palade Se ondary S hool, Buzau, Romania. Determine all real numbers x for whi h

M386

p p 2 + 4x − 2x2 + 6 + 6x − 3x2 = x2 − 2x + 6 .

Solution by Bruno Salgueiro Fanego, Viveiro, Spain. We have

p p 2 + 4x − 2x2 + 6 + 6x − 3x2 q
q
= 4 − 2x2 − 4x + 2 + 9 − 3x2 − 6x + 3 p p = 4 − 2(x − 1)2 + 9 − 3(x − 1)2 p p ≤ 4 − 2(0) + 9 − 3(0) = 2 + 3 = 5 = 0 + 5 ≤ (x − 1)2 + 5 = x2 − 2x + 6 .

For the rst expression to a tually equal the nal expression, it must be that both inequalities are a tually equalities, and so (x − 1)2 = 0 or x = 1. Thus, the only possible solution to the given equation is x = 1. We an verify by substitution that x = 1 is indeed a solution. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; PAUL BRACKEN, University of Texas, Edinburg, TX, USA; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; G.C. GREUBEL, Newport News,  IES \Abastos", Valen ia, Spain; FRANCISCA SUSAN, student, Surya VA, USA; RICARD PEIRO, Institute, BSD City, Indonesia; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA. There were two in orre t and one in omplete solutions submitted.

. Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. Temperature an be measured in degrees Fahrenheit (F ) or in degrees Celsius (C ). The two s ales are related by the formula F = 1.8C + 32. When a two-digit integer degree temperature in Celsius is onverted to Fahrenheit and rounded to the nearest integer degree, it turns out the ones and tens digits of the original Celsius temperature C sometimes swit h pla es to give the rounded Fahrenheit equivalent. Find all two-digit integer values of C for whi h this o

urs. M387

494 Solution by the Mayhem Sta. Consider a two-digit temperature C = 10a + b in degrees Celsius, where a and b are integers with 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9. The equivalent temperature in degrees Fahrenheit is F =

9 9 9 C + 32 = (10a + b) + 32 = 18a + b + 32 . 5 5 5

We want the rounded version of this real number to equal 10b+a. Therefore, 1

10b + a − 2 100b + 10a − 5 −325 315

Sin e b

9

≤ 18a + b + 32 5 ≤ 180a + 18b + 320 ≤ 170a − 82b < 82b − 170a

< < < ≤

1

10b + a + ; 2 100b + 10a + 5 ; −315 ; 325 .

≤ 9, then 82b ≤ 738. Sin e 82b − 170a > 315, then 170a < 423 83 = 2 . Sin e a is an 82b − 315 < 738 − 315 = 423, when e a < 170 170 integer, then a ≤ 2. Therefore, we only need to try a = 1 and a = 2. If a = 1, the inequalities be ome 315 + 170(1) < 82b ≤ 325 + 170(1) 3 or 485 < 82b ≤ 495 or 5 75 < b ≤ 6 ; sin e b is an integer, then b = 6. 82 82 Similarly, if a = 2, then b = 8. Hen e, the two possibilities are C = 16 (giving F = 60.8, whi h rounds to F ≈ 61) and C = 28 (giving F = 82.4, whi h rounds to F ≈ 82).

Also solved by MATTHEW BABBITT, student, Albany Area Math Cir le, Fort Edward, NY, USA; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; G.C. GREUBEL,  Newport News, VA, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, IES \Abastos", Valen ia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; There was one in omplete solution submitted. Most submitted solutions involved an expli it or impli it omplete enumeration of ases from C = 10 to C = 39 after some examination of bounds.

Problem of the Month Ian VanderBurgh

What's in a de nition? Mathemati s is littered with them. Often, we pay attention to them; sometimes we treat them a bit avalierly. Here are two problems involving geometri sequen es. In the se ond of these problems, the pre ision of our de nition turns out to ae t the answer. Problem 1 (2009 Ameri an Invitational Mathemati s Examination) Call a 3-digit number geometri if it has 3 distin t digits whi h, when read from left to right, form a geometri sequen e. Find the dieren e between the largest and smallest geometri numbers.

495 Problem 2 (2009 Eu lid Contest) If log2 x, (1 + log4 x), and log8 4x are onse utive terms of a geometri sequen e, determine the possible values of x.

So what's a geometri sequen e? Many of you will know this already, but by way of reminder, here is our rst attempt at a de nition: De nition 1: A geometri sequen e is a sequen e of numbers in

whi h ea h term after the rst is obtained from the previous term by multiplying by a onstant.

Often, we would all the rst term in the sequen e a and the multiplying fa tor r, whi h gives the sequen e a, ar, ar2 , ar3 , . . . . (You may noti e that I've deliberately avoided the phrase \ ommon ratio" { stay tuned!) Let's use this version of the de nition to solve the rst problem. Solution to Problem 1 The smallest 3-digit integers have hundreds digit 1. Let's see if any of these integers is geometri . Call the tens digit of our

andidate number r (note that r is an integer). Sin e the hundreds digit is 1, the tens digit is r, and the digits form a geometri sequen e, then the units digit is r2 . The andidate 3-digit integer is as small as possible when r is as small as possible. Sin e the digits are distin t, then r 6= 1 (otherwise r = 1 would give 111) and r 6= 0 (otherwise r = 0 would give 100). So the smallest

andidate o

urs when r = 2, whi h yields the integer 124, whi h must be the smallest 3-digit integer that is geometri . The largest 3-digit integers have hundreds digit 9. Let's see if any of these integers are geometri . Consider a andidate integer and suppose that the multiplying fa tor between onse utive digits is R. Then the tens digit is 9R and the units digit is 9R2 . Sin e we want this integer to be as large as possible, we try the dierent possibilities for 9R. If 9R = 9, then R = 1, whi h would give the integer 999, whi h violates the ondition of distin t digits. If 9R = 8, then R = 89 ; in this ase, 9R2 = 64 , whi h is not an 9 7 2 , whi h is not integer. If 9R = 7, then R = 9 ; in this ase, 9R = 49 9 2 an integer. If 9R = 6, then R = 3 , when e 9R2 = (9R)R = 6R = 4, whi h yields the integer 964, whi h is thus the largest 3-digit integer that is geometri . Thus, the dieren e between the largest and smallest 3-digit integers that are geometri is 964 − 124 = 840. At this point, you're probably wondering about the preamble { the definition doesn't seem to be ae ting anything so far. Here's another ra k at the de nition of a geometri sequen e: De nition 2: A geometri sequen e is a sequen e of numbers with the property that if a, b, c are onse utive terms, then b2 = ac.

496 And another one: De nition 3: A geometri sequen e is a sequen e of numbers with the property that if a, b, c are onse utive terms, then ab = bc .

Again, you may wonder what the big deal is all about. So I have a question for you: is 1, 0, 0 a geometri sequen e? What do the dierent versions of the de nition tell you? Solution to Problem 2 First, we express the logarithms in the three terms using a ommon base, namely the base 2. We obtain: 1 + log4 x = log8 4x =

log2 x

1

log2 x ; log2 4 2 log2 4x log2 4 + log2 x 2 1 = = + log2 x . log2 8 3 3 3

1+

= 1+

Next, we make the substitution u = log2 x to make the next al ulations less

umbersome. In terms of u our sequen e is thus u, 1 + 12 u, 23 + 13 u. Sin e this sequen e is geometri , then 2  1 1+ u 2

3(2 + u)2 12 + 12u + 3u2 0 0



2 1 + u 3 3



;

=

u

= = = =

4u(2 + u) (multiplying 2 4u + 8u ; u2 − 4u − 12 ; (u − 6)(u + 2) ;

by 12) ;

and so u = log2 x = 6 or u = log2 x = −2, hen e x = 64 or x = 41 .

So what's the big deal? Let's look at what the sequen es are for the two possible values of x. If x = 64 (or u = 6), the sequen e is 6, 4, 83 , whi h is geometri and seems pretty inno uous. If x = 14 (or u = −2), the sequen e is −2, 0, 0. Oh dear! Why is this a problem? Whi h de nition are you using? This sequen e is geometri by De nition 1 and De nition 2, but a

ording to De nition 3 it is NOT geometri . So the hoi e of de nition (that is, one's parti ular onvention)

hanges the answer to Problem 2. Using De nition 1 or De nition 2, the answer is x = 64 or x = 41 ; but using De nition 3, the answer is x = 64 only. There is a happy ending to this saga, though. Lu kily, as the 2009 Eu lid Contest was being pre-marked, the markers were alerted to this dilemma of dieren es of de nitions and both versions, with proper justi ation, were a

epted as orre t. So pay attention to de nitions { are they ompletely pre ise? And think

riti ally about seemingly equivalent de nitions { are they really equivalent?

497

THE OLYMPIAD CORNER No. 282 R.E. Woodrow

We begin this number with the problems of the Austrian Mathemati al Olympiad 2007 National Competition, Final Round, Part 1, written May 17th 2007. Thanks go to Bill Sands, Canadian Team Leader to the 2007 IMO in Vietnam, for olle ting them for our use. AUSTRIAN MATHEMATICAL OLYMPIAD 2007 National Competition { Final Round { Part 1

May 17, 2007

. We are given a 2007 × 2007 grid. An odd integer is written in ea h of its ells. Let Zi be the sum of the numbers in the ith row and Sj the sum of the numbers in the j th olumn for 1 ≤ i, j ≤ 2007. Furthermore, let 2007 2007 Q Q A= Zi and B = Sj . Show that A + B annot be equal to zero. 1

i=1

j=1

2. Determine the largest possible value of C(n) for all positive integers n, su h that  2

(n + 1)

n X

j=1

a2j − 

n X

j=1

aj 

≥ C(n) ,

holds for all n-tuples (a1 , a2 , . . . , an ) of pairwise distin t integers. . Let M (n) = {−1, −2, . . . , −n}. For ea h nonempty subset of M (n) we form the produ t of the elements. What is the sum of all su h produ ts?

3

4. Let n > 4 be an integer. The n-gon A0 A1 . . . An−1 An (with An = A0 ), is ins ribed in a ir le, is onvex, and is su h that the lengths of the sides are Ai−1 Ai = i for 1 ≤ i ≤ n. Let φi be the angle between the line Ai Ai+1 and the tangent to the ir um ir le of the n-gon at Ai . (Note that the angle between any two lines is at most 90◦ .) Determine the value of

Φ =

n−1 X

φi .

i=0

Next we ontinue with the problems of the two days of Part 2 of the National Competition Final Round Austrian Mathemati al Olympiad 2007. Again we thank Bill Sands, Canadian Team Leader to the 2007 IMO in Vietnam, for olle ting them for us.

498 AUSTRIAN MATHEMATICAL OLYMPIAD 2007 National Competition { Final Round { Part 2

June 5-6, 2007

. Determine all nonnegative integers a < 2007, for whi h the ongruen e x2 + a ≡ 0 (mod 2007) has exa tly two distin t nonnegative integer solutions smaller than 2007. (In other words, there exist two nonnegative integers u and v ea h less than 2007, su h that u2 +a and v2 +a are divisible by 2007.) 1

. Determine all sextuples (x1 , x2 , x3 , x4 , x5 , x6 ) of nonnegative integers satisfying the following system of equations:

2

x1 x2 (1 − x3 ) = x4 x5 , x2 x3 (1 − x4 ) = x5 x6 , x3 x4 (1 − x5 ) = x6 x1 ,

x4 x5 (1 − x6 ) = x1 x2 , x5 x6 (1 − x1 ) = x2 x3 , x6 x1 (1 − x2 ) = x3 x4 .

. Determine all rhombuses with sides of length 2a, for whi h a ir le exists with the property that ea h of the four sides of the rhombus interse ts the

ir le produ ing a hord of length a.

3

4. Let M be the set of all polynomials P (x) with pairwise distin t integer roots and integer oeÆ ients whose absolute values are all less than 2007. What is the highest degree among all polynomials in M ?

. We are given a onvex n-gon with a triangulation, that is, a division into triangles by noninterse ting diagonals. Prove that the n orners of the n-gon

an ea h be labelled by the digits of 2007 su h that any quadrilateral omposed of two triangles in the triangulation with a ommon side has orners labelled by digits that sum up to 9. 5

. We are given a triangle ABC with ir um entre U . A point P is hosen on the extension of U A beyond A. Let g denote the line symmetri to P B with respe t to BA and h the line symmetri to P C with respe t to AC . Let the lines g and h interse t at the point Q. Determine the set of all points Q as P varies on the ray U A beyond A. 6

The next problems we give are those of the XXI Olimpiadi Italiane della Matemati a written at Cesenati o, 11 May 2007. Thanks go to Bill Sands, Canadian Team Leader to the 2007 IMO in Vietnam, for olle ting them for our use.

499 XXI OLIMPIADI ITALIANE DELLA MATEMATICA Cesenati o, 11 May 2007

. A regular hexagon is given in the plane. For ea h point P of the plane, let ℓ(P ) be the sum of the six distan es of P from ea h line determined by a side of the hexagon, and let v(P ) be the sum of the six distan es of P from the verti es of the hexagon. (a) For whi h points P of the plane does ℓ(P ) take its least value? (b) For whi h points P of the plane does v(P ) take its least value? 1

. Polynomials with integer oeÆ ients, p(x) and q(x), are similar if they have the same degree and the same oeÆ ients (possibly in dierent order). (a) If p(x) and q(x) are similar, prove that p(2007) − q(2007) is even. (b) Is there an integer k > 2 su h that p(2007) − q(2007) is divisible by k whenever p(x) and q(x) are similar? 2

. Triangle ABC has entroid G, D 6= A is a point on the line AG su h that AG = GD , and E 6= B is a point on the line GB su h that GB = GE . The midpoint of AB is M . Prove that the quadrilateral BM CD an be ins ribed in a ir le if and only if BA = BE .

3

4. On Barbara's birthday Alberto invites her to play a game. Given the numbers 0, 1, 2, . . . , 1024, Barbara removes 29 of them. Then Alberto removes 28 numbers from the ones that remain. Next, Barbara removes 27 numbers, and so on, until only two numbers a and b remain. Alberto then gives |a − b| euros to Barbara. Find the largest amount of euros that Barbara is ertain to win, regardless of how Alberto plays.

. Let x1 , x2 , x3 , . . . , be the sequen e of integers de ned by x1 = 2 and xn+1 = 2x2n − 1 for n ≥ 1. Prove that, for every positive integer n, the numbers n and xn are oprime.

5

. For ea h integer n ≥ 2, nd (a) the greatest real number cn su h that

6

1 1 1 + + ··· + ≥ cn 1 + a1 1 + a2 1 + an

for any positive real n-tuple (a1 , a2 , . . . , an ) with a1 a2 · · · an = 1; (b) the greatest real number dn su h that 1 1 + 2a1

+

1 1 + 2a2

+ ··· +

1 1 + 2an

≥ dn

for any positive real n-tuple (a1 , a2 , . . . , an ) with a1 a2 · · · an = 1.

500 As a further set of problems for your puzzling pleasure we give the Final Round of the 56th Cze h and Slovak Mathemati al Olympiad, Mar h 18-21, 2007, edited by Karol Horak  and translated by Miroslav Englis. Thanks go to Bill Sands, Canadian Team Leader to the 2007 IMO in Vietnam for olle ting them for us. th 56 CZECH AND SLOVAK MATHEMATICAL

OLYMPIAD Final Round

Mar h 18-21, 2007 1. A hess pie e is arbitrarily pla ed on a square of a n × n (n ≥ 2) square hessboard. It then alternately makes straight and diagonal moves. A straight move is from one square to another one with a ommon side. A diagonal move is from one square to another one with exa tly one point in

ommon. Find all n for whi h there is a sequen e of moves, starting with a diagonal move from the original square, su h that the pie e passes through all the squares of the hessboard and through ea h square exa tly on e.

. In a y li quadrangle ABCD let L and M be the in entres of triangles respe tively. Let R be the interse tion of the perpendi ulars from the points L and M onto the lines AC and BD, respe tively. Show that the triangle LM R is isos eles. 2

BCA and BCD ,

3. Denote by N the set of all positive f : N → N su h that for any x, y ∈ N,

integers and onsider all fun tions


f xf (y) = yf (x) .

Find the least possible value of f (2007).

. The set M ontains all of the integers from 1 to 2007 (in lusive) and if then M ontains the arithmeti progression with rst member n and dieren e n + 1. De ide whether there must exist a number m su h that M ontains all integers greater than m. 4

n ∈ M,

5. Triangle ABC is a ute with |AC| 6= |BC|. The points D and E lie on the interiors of the sides BC and AC (respe tively) su h that ABDE is a y li quadrangle, and the diagonals AD and BE interse t at P . If the lines CP and AB are perpendi ular, show that P is the ortho entre of triangle ABC .

. Find all ordered triples (x, y, z) of mutually distin t real numbers whi h satisfy the set equation

6

{x, y, z} =



x−y y−z z−x , , y−z z−x x−y



.

501 A nal set of problems for your pleasure over our winter break is the Sele ted Problems of the 2007 Taiwanese Mathemati al Olympiad. Thanks on e again are due to Bill Sands, Canadian Team Leader to the 2007 IMO in Vietnam, for olle ting them for our use. 2007 TAIWANESE MATHEMATICAL OLYMPIAD Sele ted Problems

1

. Prove the following statements:

(a) If 0 < a, b ≤ 1, then 1 1 2 + √ ≤ √ √ 2 2 1 + ab a +1 b +1

;

1 1 2 + √ ≥ √ √ 2 2 1 + ab a +1 b +1

.

(b) If ab ≥ 3, then

2

. Find all positive integers a, b, c, and d su h that 2a = 3b 5c + 7d .

3. Given △ABC and its ir um ir le, prove that the Simson lines of two diametri ally opposite points are perpendi ular and interse t on the ninepoint ir le of the triangle.

. Let ABCD be a onvex quadrilateral. Prove or disprove that there exists a point E in the plane of ABCD su h that △ABE is similar to △CDE . 4

5

. Find all fun tions f

: R → R,

su h that for all real numbers x and y,


f (x)f yf (x) − 1 = x2 f (y) − f (x) .

. Consider the following variation of the game of Nim. A position onsists of k piles of stones, with ni ≥ 1 stones in pile i. Two players alternately move by hoosing one of the piles, permanently removing one or more stones from that pile, and, optionally, redistributing some (or all) of the remaining stones in that pile to one or more of the other remaining piles. On e a pile is gone, no stones an be added to it, and the player who takes the last stone wins. Determine whi h ve tors of positive integers (n1 , n2 , . . . , nk ) represent a winning position for the rst player and whi h ve tors represent a winning position for the se ond player. 6

502 We now return to the 54th Cze h Mathemati al Olympiad 2004/2005, Category B, 10th Class [2008 : 342{344℄ and a solution that ould not be t into the last number of the Corner. . Let ABC be an a ute triangle. Let K and L be the feet of the altitudes from A and B , respe tively. Let M be the midpoint of AB and let H be the ortho entre of triangle ABC . Prove that the bise tor of ∠KM L bise ts the line segment HC .

K3

Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; and Titu Zvonaru, Comane  sti, Romania. We give the solution of Amengual Covas. Let N be the midpoint of CH . Equivalently, we will prove C that M N bise ts ∠KM L. Triangle ALB has a right angle at L and a median LM of N length 12 AB . Triangle AKB has K a right angle at K and a median KM of length 12 AB . Hen e, L LM = KM . Triangle CLH has a right H angle at L and a median LN of qq length 12 CH . Triangle CKH has a right angle at K and a meA B M dian KN of length 12 CH . Hen e, LN = N K . Sin e LM = KM and LN = N K , it follows that M N is the perpendi ular bise tor of segment LK . Sin e △KM L is isos eles, this perpendi ular bise tor is in fa t the bise tor of ∠KM L, as desired. ... ........ ... .... .. ... ... .. ..... ...... . .... . .. ... .... .. .. .... .. ... .... .. .... .... .. .. . . .... . .... ..... ... . . . . . ............................. ....... .. . . . ............... . ... .... ......... ........................... . . . . . .. . .. .. . . . . . . . . .. ..... .. . ..... .......................... ......... .... ...... .. ... . . .. .. .... .. ............ .. ..... .. .. .... .... ............. .... .... ........ . .... ........................ .. ....... . . . . .... . ....... . .. ................... . . .... . . . .. . .. . .... ... ... ..... ............................... .... .... . . . . . . . . . . . . . . . .. .... ............. .... . .. ............ ...... .... . . . .......... .. . .. . . .... . . . . ... .. ............. ... .... ... . . . . . . . . . . . ... .. .... . ............... ... .. . . . . . . . .......... .... ... .. ... .. . . .. . . . . .... . . . . . . . . . .......... ... .. . .... .. ........ . . . . . . . . . . ........... . ... ... . .. ...... . . . . . ........... ...... .. ... . ......... ... ........ ...... . ...........................................................................................................................................................................................................................

... ........ ........ ......

.......... .............

We also return to one more solution to a problem from the rst Round of the 23rd Iranian Mathemati al Olympiad [2008 : 345℄. . Find all fun tions f

6

: R+ → R+ , su h that for all x, y ∈ R+

(x + y)f f (x)y = x2 f f (x) + f (y) ,

we have

where R+ denotes the set of positive real numbers.

Solution by Mi hel Bataille, Rouen, Fran e. There is no su h fun tion. To prove this, assume that f is a solution. Taking y = x in the fun tional equation, we obtain the identity
f xf (x) =


x · f 2f (x) . 2

(1)

503 Let a, b ∈ R+ be su h that f (a) = f (b). Then
(a + b)f bf (b)

= =


(a + b)f f (a)b



a2 f f (a) + f (b) = a2 f 2f (b)

and by using (1) it follows that (a + b) 2b · f 2f (b) = a2 f (2f (b) . Hen e (a + b)b = 2a2 , or (a − b)(2a + b) = 0, and we on lude that f is inje tive. √ Now, let c = 1 +2 5 so that c + 1 = c2 > 0. Taking x = c and y = 1

2 in the
fun tional equation
yields (c + 1)f f (c) = c f f (c) + f (1) , or f f (c) = f f (c) + f (1) . Sin e f is inje tive, we have f (c) = f (c) + f (1), whi h ontradi ts f (1) ∈ R+ . This ontradi tion ompletes the proof.





Next we give solutions from our readers to problems of the Se ond Round of the 23rd Iranian Mathemati al Olympiad [2008 : 345℄. . Let a, b, and c be nonnegative real numbers. If

3

a2

1 1 1 + 2 + 2 = 2, +1 b +1 c +1

then show that ab + bc + ca ≤ 32 . Solved by Mi hel Bataille, Rouen, Fran e; Oliver Geupel, Bruhl,  NRW, Germany; and Titu Zvonaru, Comane  sti, Romania. We give Bataille's solution. Let a, b, and c satisfy the onstraint. Then a2 b2 c2 + + = 1 a2 + 1 b2 + 1 c2 + 1

(sin e

x2 1 =1− 2 ) +1 x +1

x2

(a+b+c)2 ≤

or



a2 a2

+1

+

and the Cau hy-S hwarz Inequality yields b2

b2

+1

+

c2 c2

+1






 a2 +1 + b2 +1 + c2 +1 ,

(a + b + c)2 ≤ a2 + b2 + c2 + 3 .

Finally, 2(ab + bc + ca) inequality follows.

= (a + b + c)2 − a2 + b2 + c2




≤ 3,

and the

504 We look next at solutions from our readers for the Third Round of the 23rd Iranian Mathemati al Olympiad given at [2008 : 346℄. . Let ABC be a triangle whose ir umradius equals the radius of the ex ir le whi h is tangent to the side BC . Let this ex ir le tou h the side BC and the lines AC and AB at M , N , and L, respe tively. Show that the

ir um entre of triangle ABC is the ortho entre of triangle M N L.

1

Solved by Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Comane  sti, Romania. We give the solution of Bataille. We use the usual notation for the elements of △ABC , in parti ular, O is the ir um entre of ABC and the A-ex ir le has entre Ia and radius ra . First, we show that LN 2 − LM 2 = ON 2 − OM 2 , whi h implies that LO ⊥ M N . Observing that Ia , L, B , and M are on the ir le with diameter BIa , we see that ∠LIa M = B . Sin e Ia L = Ia M = ra = R, it follows that B LM = 2R sin . Similarly, sin e ∠LIa N = 180◦ − A, it follows that 2 LN = 2R cos

A . 2

2

LN − LM

2

Now, the area of abc 2R2 = . b+c−a 2

As a result,

  2 B 2 A = 4R cos − sin = 2R2 (cos A + cos B) . 2 2 2

△ABC

is

ra (s − a) = R(s − a)

as well as

Using the Law of Cosines, this leads to

LN − LM

2



a b2 + c2 − a2 + b c2 + a2 − b2 = 2(b + c − a)

abc , 4R

.

hen e

(1)

BM = s − c and CM = s − b, we have that −−→ −−→ −−→ −→ −→ → − (s − b)M B + (s − c)M C = 0 , and so aOM = (s − b)OB + (s − c)OC .

On the other hand, sin e

It follows that a2 OM 2

= (s − b)2 R2 + (s − c)2 R2 + 2(s − b)(s − c)R2 cos 2A
= R2 (s − b)2 + (s − c)2 + 2(s − b)(s − c) − 4R2 (s − b)(s − c) sin2 A = R2 a2 − (s − b)(s − c)a2 ,

and so OM 2 = R2 − (s − b)(s − c). −− → −→ −− → −→ −→ Similarly, from AN = sb AC we obtain bON = sOC − (s − b)OA and a similar al ulation yields ON 2 = R2 + s(s − b). As a result, ON 2 − OM 2 = (s − b)(s + s − c) =

(a + b)(c + a − b) 2

.

(2)

505 A omparison of the righthand sides of (1) and (2) yields the desired equality by ex hanging the

LN 2 − LM 2 = ON 2 − OM 2 . We obtain N O ⊥ LM roles of L and N , and the result follows.

5. Let n > 1 be an integer, and let the entries of (a1 , a2 , . . . , an ) be pairwise distin t positive integers whi h are oprime in pairs.
Find all su h n-tuples for whi h (a1 + a2 + · · · + an ) | ai1 + ai2 + · · · + ain for 1 ≤ i ≤ n. Solution by Oliver Geupel, Bruhl,  NRW, Germany, modi ed by the editor. We prove that there are no solutions. First we will show by indu tion that for ea h positive integer m n X

i=1

! ak

n X

am k

k=1

!

.

(1)

This holds for m = 1, 2, . . . , n by hypothesis. Now assume that (1) holds for m = M − n, M − n + 1, . . . , M − 1, and onsider the polynomial MP −1 n Q P (x) = xM −n (x−ak ). It an be written as P (x) = xM + bm xm , m=M −n

k=1

where the oeÆ ients bm are integers. We obtain 0 =

n X

P (ak ) =

k=1

n X

aM k

n X

ak

k=1

pℓ

be a prime power dividing

n X

bm

m=M −n

k=1

hen e, by the indu tion hypothesis, the indu tion. Next we will prove that

Let

+

M −1 X



!

k=1

!

  n  n P M P ak ak , whi h ompletes

i=1

k=1


2 n −n .

n P

am k

. If

ak k=1 ϕ(pℓ ) ak

p

(2)

divides any

ak ,

say

p | a1 ,

then by Euler's Theorem we obtain ≡ 1 mod p for 2 ≤ k ≤ n; n
P ϕ(p ) ak ≡ n − 1 mod pℓ . therefore, applying (1), we on lude that 0 ≡ k=1 Otherwise, if p does not divide any pk , we obtain from (1) and Euler's Theon
P ϕ(p ) rem that 0 ≡ ak ≡ n mod pℓ . Consequently, pℓ | n(n − 1), whi h k=1 proves (2). n P Sin e ak ≥ 1 + 2 + · · · + n = 12 n(n + 1) > 12 n(n − 1), it follows ℓ


ℓ

ℓ

k=1 n P

from (2) that

k=1


ak = n2 − n .

Let f (1) = 1, and for ea h integer k ≥ 2 let

506 f (k)

be the highest prime fa tor of k. If n ≥ 8, then

n P

k=1 2

ak ≥

n P

k=1

f (ak ) ≥

1+2+3+5+7+11+13+· · ·+(2n−1) = n2 −7 > n −n, a ontradi tion. Sin e f (a1 ) + f (a2 ) + · · · + f (an ) ≤ n(n − 1), the remaining ases are easily

eliminated: n 3 4 5 6 7

n(n − 1) 6 12 20 30 42

{f (ai )}n i=1 1, 2, 3 1, 2, 3, 5 1, 2, 3, 5, 7 1, 2, 3, 5, 7, 11 1, 2, 3, 5, 7, 11, 13

{ai }n i=1

the same none 1, 22 , 3, 5, 7 none the same

(1) fails for m=2

|

m=4

|

m=6

Next we turn to solutions of a problem of the Romanian Mathemati al Olympiad 2006, Final Round, 9th Grade, given at [2008 : 346{347℄. . (Dan S hwarz) Find the maximum value of real numbers su h that x + y = 1. 1



x3 + 1 y 3 + 1 if x and y

are

Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Comane  sti, Romania. We give Bataille's solution. The maximum value is 4. Set p = xy, then we have

x3 + 1 y 3 + 1 =

=

Now,

(x + y)2

(xy)3 + (x + y)3 − 3xy(x + y) + 1

p3 − 3p + 2 = 4 + (p + 1)2 (p − 2) . 1

= , hen e p − 2 < 0. It then follows that xy ≤ 4 4
3
3 x + 1 y + 1 ≤ 4 with equality if and only if p = xy = −1 (in addition

to x + y = 1); that is, if and only if x, y are the numbers

√ 1+ 5 2

and

√ 1− 5 . 2

We next look at readers' solutions to problems of the Final Round of the Romanian Mathemati al Olympiad 2006, 10th Grade, given at [2008 : 347℄. . (Vasile Pop) Let M be a set with n elements and let P(M ) denote the set of all subsets of M . Find all fun tions f : P(M ) → {0, 1, 2, . . . , n}, with the following two properties:

1

(a) (b)

f (A) 6= 0 for

any A 6= ∅, and

f (A ∪ B) = f (A ∩ B) + f (A△B), A△B = (A ∪ B)\(A ∩ B).

for all

A, B ∈ P(M ),

where

507 Solution by Mi hel Bataille, Rouen, Fran e. The fun tion whi h maps ea h subset of M to its ardinality satis es the requirements, and we will show that there are no other solutions. Let f satisfy (a) and (b). If A, B ∈ P(M ) and A ( B , then f (B) = f (A ∪ B) = f (A ∩ B) + f (A△B) = f (A) + f (B\A) ,

hen e f (B\A) > 0 by (a) be ause B\A 6= ∅. It follows that f (B) > f (A). Now, let M = {m1 , m2 , . . . , mn } and observe that sin e f is \stri tly in reasing", we have

0 ≤ f (∅) < f {m1 } < f {m1 , m2 } < · · · < f (M ) ≤ n .

The n + 1 images under f are distin t
integers in {0, 1, 2, . . . , n}; hen e
f (∅) = 0, f {m1 } = 1, f {m1 , m2 } = 2, . . . , f (M ) = n. It follows (sin e subsets may be relabelled) that the image f (A) of any subset A of M equals the ardinality of A. π
2. (Iurie Borei o) Prove that for all integers n > 0 and all a, b ∈ 0, we 4

have

n

n

sin a + sin b

(sin a +

sin b)n

n

≥

n

sin 2a + sin 2b

(sin 2a + sin 2b)n

.

Solved by Mi hel Bataille, Rouen, Fran e; and Oliver Geupel, Bruhl,  NRW, Germany. We give Bataille's version. [Ed.: The notations Sx = sin x and Cx = cos x will be employed for typographi al reasons.℄ Without loss of generality, we assume n ≥ 2 and a > b (equality holds for n = 1 or a = b and a, b play symmetri roles). We want to prove where


n
n San + Sbn S2a + S2b + T2 ≥ T1 T2


n
n n S2a + S2b Sa + Sbn + T1 ,

n−1 X

 n−1 X n n−k n n−k k = Sa Sb = Sb Sak ; k k k=1 k=1 n−1 n−1 X n n−k X n n−k k k = S2a S2b = S2b S2a . k k k=1 k=1

Inequality (1) is equivalent to La + Lb ≥ Ra + Rb where La Ra

n−1 X

 n n−k k n = S S2b Sa ; k 2a k=1 n−1 X n n = San−k Sbk S2a ; k k=1

n−1 X

 n n−k k n Lb = S S2a Sb ; k 2b k=1 n−1 X n n−k n Rb = Sb Sak S2b . k k=1

(1)

508 Now, using the formula sin 2x = 2 sin x cos x, (or S2x = 2Sx Cx ) we have La − Ra Lb − Rb

It follows that

n

= 2

n−1 X k=1

n

= 2

 n

n−1 X k=1

k

 n k

Sa2n−k Sbk Can−k




Cbk


Sb2n−k Sak Cbn−k

Cak

−

Cak

−

Cbk

!


!


, .


1 La + Lb − (Ra + Rb ) n 2 n−1 X n
Sak Sbk 
n−k
n−k  = Cbk − Cak n−k . Sa S2a − Sb S2b k 2 k=1

Sin e a > b, we have Cbk − Cak > 0 and Sa S2a n−k ea h k, and so La + Lb − (Ra + Rb ) > 0, as desired.


>

Sb S2b


n−k

for

Next we move to the November 2008 number of the Corner and solutions from our readers to problems of the 2005/6 British Mathemati al Olympiad, Round 1, given at [2008 : 408℄. than 6. Prove that if n − 1 and n + 1 are both 1. Let n be an integer greater
prime, then n2 n2 + 16 is divisible by 720. Is the onverse true?

Solution by Titu Zvonaru, Comane  sti, Romania. Sin e n − 1 and n + 1 are both prime, n is even. If n = 6k + 2, then n + 1 = 3(2k + 1) is not prime. If n = 6k + 4,
then n − 1 = 3(2k +
1) 2 2 2 2 is not prime. Therefore, n = 6k and n n + 16 = 36k 36k + 16 =
144k2 9k2 + 4 is divisible by 144. If n = 5k + 1, then n − 1 = 5k is not prime. If n = 5k + 4, then n + 1 = 5(k + 1) is not prime. If n = 5k, then n2 is divisible by 5. If n = 5k + 2 or n = 5k + 3, then n2 = 5j + 4 where j = 5k2 + 6k + 1 is an integer, hen e n2 + 16 is divisible by 5.
By the pre eding two results, n2 n2 + 16 is divisible by 5 · 144 = 720
. The onverse is not true. For example, if n = 48, then n2 n2 + 16 is divisible by 720, but n + 1 = 49 is not prime. 2. Adrian tea hes a lass of six pairs of twins. He wishes to set up teams for a quiz, but wants to avoid putting any pair of twins into the same team. Subje t to this ondition:

(i) In how many ways an he split them into two teams of six? (ii) In how many ways an he split them into three teams of four?

509 Solution by Oliver Geupel, Bruhl,  NRW, Germany. (i) To set up the rst team, one member of ea h pair has to be sele ted. There are two possible hoi es for ea h of the six pairs, thus 26 hoi es in total. Finally, the order of the two teams is irrelevant; hen e a fa tor of 12 applies. We on lude that the teams an be arranged in 25 = 32 ways. (ii) To set up the rst team, Adrian an rst hoose four of the six
pairs and then pi k one person from ea h sele ted pair. This an be done in 64 ·24 ways. Let a, b, c, d be the four remaining members of the sele ted four pairs, and let S , T be the two pairs that were not hosen for the rst team. To build the se ond team, Adrian has to hoose one member from ea h of the two pairs
S and T and two extra persons from a, b, c, d. This an be done in 22 · 42 1 ways. Finally, the order of the three teams is irrelevant; hen e a fa tor of 3! 1 6
4
6 applies. Therefore, the teams an be arranged in 3! 2 = 960 ways. 2 2 3. In the y li quadrilateral ABCD , the diagonal AC bise ts the angle DAB . The side AD is extended beyond D to a point E . Show that CE = CA if and only if DE = AB .

Solved by Ri ardo Barroso Campos, University of Seville, Seville, Spain; and Titu Zvonaru, Comane  sti, Romania. We give the solution of Barroso Campos. If CE = CA, then △AEC is A isos eles, hen e ............... ∠CED = ∠CAD = ∠CAB .

Also

∠CDA = 180◦ − ∠CBA so ∠DEC and △BAC are ongruent and DE = AB . Sin e ∠DAC = ∠CAB we have CD = CB . If DE = AB , then we have DE = AB , CD = CB , and ∠EDC = ∠ABC , hen e CE = CA.

........... ................. ........ ...................... ............. ..... . .... . . . . . . . . .. . . . .... ... ....... .... .... ... ... .. .............. . . . . . ... D............... ... ... ... . . . . . .. . . . . . . . . . .. .. .. ....... ........ . . . . . . . . . . . . ... ... . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . ... ... .. . . . . . . . . . . . .... .. . .. ....... .. ... E ......................................................................................... .. .... ............ . . . . . . . .. C ....................................... ..... .................... ..................

B

Now we ontinue with solutions to problems of Round 2, 2005/6 British Mathemati al Olympiad given at [2008 : 409℄. . Find the minimum possible value of x2 + y2 given that x and y are real numbers satisfying

1

xy x2 − y 2




= x2 + y 2

and

x 6= 0 .

Solved by Mi hel Bataille, Rouen, Fran e; and Oliver Geupel, Bruhl,  NRW, Germany. We give the write-up of Bataille. We show that the required minimum is 4.

510 Let x, y satisfy the onstraint. Then x2 + y2  su h that (cos θ, sin θ) = p 2x 2 , p 2y x +y

6= 0,

and there is an angle θ

.

y2
x + 2 2 2 The given equation xy x − y = x + y2 an then
1 sin θ cos θ cos2 θ − sin2 θ = 2 , that is, sin 4θ = 2

be rewritten as 4 . Sin e x +y x2 + y 2 2 2 sin 4θ ≤ 1, we  obtain x + y ≥ 4. To omplete the proof, we observe that for (x, y) = 2 cos π8 , 2 sin π8 the onstraint is satis ed and x2 + y2 = 4.

3. Let ABC be a triangle with AC > AB . The point X lies on the side BA extended through A, and the point Y lies on the side CA in su h a way that BX = CA and CY = BA. The line XY meets the perpendi ular bise tor of side BC at P . Show that ∠BP C + ∠BAC = 180◦ .

Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Mi hel Bataille, Rouen, Fran e; Oliver Geupel, Bruhl,  NRW, Germany; and Titu Zvonaru, Comane  sti, Romania. We give the solution by Amengual Covas. Let the bise tor of ∠A meet BC at D and extend XY to meet BC at E . P Triangle XAY is isos eles sin e AX = BX − BA = CA − CY = AY , hen e ∠AXY = ∠AY X . The exterior angle of △XAY at A is ∠A, so ∠A = ∠AXY + ∠AY X = 2∠AXY . Thus, ∠AXY = 12 ∠A = ∠BAD. X The angles ∠AXY and ∠BAD q are equal, hen e ADkXY and DE AX = BD AB

Sin e

.

bise ts ∠A, we have from whi h we obtain

DC − BD BD

=

CA − AB AB

A

(1)

AD

DC AC = , BD AB

.. ............... ............. .. ................. . . . ... .. .... ... ... ... ... ... ... ... ... ... .. ... .. .. ... .... .... ..... . . ... ... .... ... ... ... .... .... .... .... ... ... .. .. .. .. ... .. ..... .... . . . ... .. ... ..... .... ... ... ... . . . . ... ... ... . .. . . . . ... ... ..... .. .. . . ... ... . .. .. . . . ... ... .... .. .. . . . ... ... .. .. .... . . ... .. ... .. .. ... . . . . .... ................ .. . ... . . ... ........ .. .. .. ... . . ... ...... . . .. ... . . . . . . . . . . ... . ........... .... ... .. . . . ... ... .. .. ........... .... . . . ... ... ........ ... .. . . . .. ... ... ..... . . .. . . . . . . . . ... ... ... .... .. . . .. . . . . . . . . ... . ... ............................ . . . .. . . . . . . . ... ... ..... .... ... ............................. . .. . . . . . ... ... .. ... .................... ... .. ........... . ... . . . .............. .. ... .. . .. .......... . . . . . . . .. . . . . . . . . .. ................... ... ..... . ........ . . . . . . . . . . ........ ... ... . . . ... ................................................................................................................................................................................................................

=

qq

B

CA − CY AB

qY q

D

=

AY AB

M

=

C

E

AX AB

.

(2)

From (1) and (2) we have DE = DC − BD. Also DE = DC − EC , so BD = EC . Let M be the midpoint of BC . Then M is also the midpoint of DE and △P DE is isos eles, sin e DM = BM − BD = M C − EC = M E . Be ause ∠P ED = ∠ECY + ∠CY E = ∠BCA + ∠AY X = ∠C + 12 ∠A, ea h of the base angles at D and E is ∠C + 12 ∠A. Consequently, ∠DP E = ∠B − ∠C

and

∠DP M =

1 (∠B − ∠C) . 2

511 Then, PM

1

= DM · cot ∠DP M = (BM − BD) · cot (∠B − ∠C) 2   a ac 1 = − · cot (∠B − ∠C) 2 2 b+c a(b − c) 1 = · cot (∠B − ∠C) 2 2(b + c)

where a, b, and c are the sides of △ABC in the usual order. tan 12 (∠B − ∠C) c We make the substitution bb − = to obtain 1 +c tan 2 (∠B + ∠C)

PM =

or

1 2

a 2 tan 12 (∠B + ∠C)

tan (∠B + ∠C) =

Also tan


BM ∠BP M = ; PM

tan

a/2 PM

=

,

BM PM

.

so




1 1 ∠B + ∠C = tan ∠BP M = tan ∠BP C . 2 2

From this it follows that ∠BP C = ∠B + ∠C , and therefore
∠BP C + ∠BAC = ∠B + ∠C + ∠A = 180◦ , as desired.

we have

We now nish with readers' solutions to problems of the Bulgarian National Olympiad 2006 given at [2008 : 409{410℄. 1. (Aleksandar Ivanov) Consider the set A = {1, 2, 3, . . . , 2n }, n ≥ 2. Find the number of subsets B of A, su h that if the sum of two elements of A is a power of 2 then exa tly one of them belongs to B .

Solution by Oliver Geupel, Bruhl,  NRW, Germany. Let  Bn be the olle tion of all admissible subsets B where n ≥ 2. Then B2 = {1}, {1, 2}, {1, 2, 4}, {1, 4}, {2, 3}, {2, 3, 4}, {3}, {3, 4} , hen e, |B2 | = 8. Consider the bipartite graph with the two sets of nodes Bn and Bn+1 , where a node B ∈ Bn is adja ent to a node B ′ ∈ Bn+1 if and only if B ⊂ B ′ . For 2n + 1 ≤ k ≤ 2n+1 − 1 we have k ∈ B ′ if and only if k − 2n ∈ B . Thus, the elements of B ′ ex ept 2n+1 are uniquely determined from B , whereas 2n+1 may or may not be a member of B ′ . Therefore, ea h node in Bn has degree 2, while ea h node in Bn+1 has degree 1, hen e, |Bn+1 | = 2|Bn |. From |B2 | = 8 and the re ursion |Bn+1 | = 2|Bn | we on lude that there are exa tly |Bn | = 2n+1 subsets with the desired property.

512 . (Oleg Mushkarov, Nikolai Nikolov) Let R+ be the set of all positive real numbers and f : R+ → R+ be a fun tion su h that 2

for all x > y > 0.

p f (x + y) − f (x − y) = 4 f (x)f (y)

(a) Prove that f (2x) = 4f (x) for all x ∈ R+ .

(b) Find all su h fun tions.

Solution by Mi hel Bataille, Rouen, Fran e, modi ed by the editor. (a) If a > b > 0, then f (a) − f (b)

= f



a+b a−b + 2 2



− f



a+b a−b − 2 2

r     a+b a−b f > 0, = 4 f 2



2

hen e f is in reasing. Sin e f is also bounded below by 0, f has a ( nite) limit as x approa hes 0 from the right. Let l0 = lim f (x). Note that l0 ≥ 0. x→0 Sin e f is in reasing, f has a limit from ea h side at ea h a ∈ R+ , and +

lim f (x) = la ≤ f (a) ≤ ra =

x→a−

Setting px

lim f (x) .

x→a+

in the given relation and letting y → 0+ , we obtain + l0 − l0 = 4 , hen e l0 = 0. Now p setting x = a and letting y → 0 in the same relation yields ra − la = 4 f (a)l p 0 = 0, thus f is ontinuous at a. From f (a + x) − f (a − x) = 4 f (a)f (x), we dedu e f (2a) = 4f (a) (by ontinuity at a and l = 0). Sin e a is arbitrary, the result follows. (b) The only solutions are the fun tions x 7→ kx2 , where k > 0. Let f be any solution. Then f (nx) = n2 f (x) holds for n = 1, 2. Assume that it holds m where m ≥ 2 is an integer. Then,
for n = 1, 2
, . . . ,p from f (m +
1)x − f (m − 1)x = 4 f (mx)f (x), we obtain the relation f (m + 1)x = (m − 1)2 f (x) + 4mf (x) = (m + 1)2 f (x). 2 Thus, by indu tion, f(nx) = for all positive integers n and all  n f (x) p p2 · x = 2 f (x) for all positive integers p, q . x > 0. It follows that f q q Lastly, if r ∈ (0, ∞), then there is a rational sequen e {rn }∞ n=1 onverging to r, and by the ontinuity of f we dedu e f (rx) = r2 x. In parti ular, f (r) = r 2 f (1) for all r ∈ (0, ∞). This ompletes the proof. = 2y

l02

The last Corner for this Volume of CRUX with MAYHEM is now omplete. Send me your ni e solutions and generalizations in the New Year!

513

BOOK REVIEWS Amar Sodhi

A Certain Ambiguity: A Mathemati al Novel By Gaurav Suri and Hartosh Singh Bal, Prin eton University Press, 2007 ISBN-13: 978-069112-709-5, hard over, 292 pages, US$27.95 Reviewed by , Halifax, NS A Certain Ambiguity is subtitled A Mathemati al Novel. Mathemati al it ertainly is, and it is novel, but a mathemati al novel... . The book opens with Ravi Kapoor re alling the day his mathemati ian grandfather (bauji) gave him a al ulator. The gift was to initiate bauji's plan to \get Ravi passionate about mathemati s" and together with the gift ame an arithmeti al teaser that set him along the way. Unfortunately, the day after the plan was set in motion, bauji died and young Ravi was abandoned to a s hool system dedi ated to rote learning and the a

umulation of fa ts. Although Ravi's grades were ex ellent his s hooling was, in his words, a joyless endeavour. However, the high grades and a bequest from bauji (eventually) enabled Ravi to enter Stanford. The young Kapoor's undergraduate years were initially those of a dilettante; he dabbled in this and that but no subje t had lasting interest for him. His eventual major, E onomi s, was hosen to satisfy his father who felt it would make Ravi attra tive to a wide range of orporate re ruiters. Just as it seems that our boy is destined to be ome an a olyte in the servi e of Mammon, he meets Dr. Ni o Aliprantis, mathemati ian, jazz saxophonist manque and tea her extraordinaire. Ravi is invited to join Ni o's Math 208 lass \Thinking About In nity", and this is where the story really begins. The authors use Aliprantis to draw us into a mathemati al feast. We are fed tasty morsel after tasty morsel that serve to addi t the neophyte and bring a smile to the lips of the ognos ente. From Zeno to in nite sums,

ounting to Cantor, the in nity of prime numbers and the irrationality of the square root 2. The table is well set and the servers, Ni o, Ravi, and other members of the Math 208 lass lay out the dishes and anti ipate our needs like all good wait sta. However, if the food does not suit the reader's palate there is little nourishment in the hara ters, that is until the authors introdu e a new literary and mathemati al thread. Ravi dis overs that as a young man his grandfather, formally known as Vijay Sahnis (VS), had spent some time in a New Jersey prison. The in ar eration ame as news to Vijay Sahnis' surviving family and Ravi set about to unravel the mystery. He tra ks down a trans ript of dis ussions between VS and a respe ted New Jersey judge, John Taylor. Using trans ripts and newspaper arti les together with some augmentation, the authors develop a situation and hara ters that hold the reader's attention

Mark Taylor

514 while a new mathemati al line is developed. The new line introdu es geometry and the idea of formal axioms. Very soon the Judge and VS are dis ussing Eu lid's fth postulate and this leads them to non-Eu lidean geometries and eventually into the very nature of mathemati s. The dialogue between VS and the Judge is interwoven with observations from members of the Math 208 lass and this allows the introdu tion of the Continuum Hypothesis and mention of the works of Godel  and Cohen. Suri and Bal su

eed admirably in des ribing and explaining some beautiful mathemati al results in su h a way that they are a

essible to people with little formal training in the dis ipline. A personal quibble: The book ontains over a dozen \Journal Entries" of one sort or another as ribed to various mathemati ians with the authors' a knowledgement that the ontents of many are either apo ryphal or titious (there is no attempt to de eive the reader { ea h entrant is explained in the notes at the end of the book). For my taste, the time and eort spent in onstru ting most of the journal entries would have been better employed in providing the reader with some good re ipes. In one part in the book Ni o invites a group of his students to a simple dinner onsisting of a Greek salad, lamb marinated in a garli sau e, and homemade pita bread. The preparation of the meal is des ribed but not in detail. Re ipes would have been most appropriate. Ravi might have re ipro ated Ni o's generosity with an Indian meal; perhaps pakoras, dhall, ri e, happati, a good hi ken dish or rogan gosht, brinjal, bindi, and kheer. A good kheer re ipe is hard to ome by. Finally the question: To read or not to read? Imagine you are in a do tor's or dentist's waiting room sifting through dog eared opies of Readers Digest, Ma Lean's, Time, Field and Stream, E um Se um Hog Breeders Quarterly, et ., when you ome a ross a opy of Crux with Mayhem. If you pi k it up and s an through it and nd anything of interest in it, I daresay you will nd more than a little to enjoy in A Certain Ambiguity.

515

PROBLEMS

Toutes solutions aux problemes dans e numero doivent nous parvenir au plus tard le 1er juin 2010. Une e toile (⋆) apres le numero indique que le probleme a e te soumis sans solution. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. Dans la se tion des solutions, le probleme sera publie dans la langue de la prin ipale solution presentee. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes. . Propose par Pham Huu Du , Ballajura, Australie. Soit a, b et c trois nombres reels  positifs. Montrer que

3488

a b c + + ≤ 2 2 2 2a + bc 2b + ca 2c + ab

s

a−1 + b−1 + c−1 a+b+c

.

. Propose par Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Soit n un entier non negatif.  Montrer que 3489

1 2n−1

  n X √ 2n k ≤ k k=0

s    2n n 22n + . n

. Propose par Mi hael Rozenberg, Tel-Aviv, Israel. Soit a, b et c trois nombres reels  non negatifs  tels que a + b + c = 1. Montrer que √ √ √ (a) 9 − 32ab + 9 − 32ac + 9 − 32bc ≥ 7 ; √ √ √ √ (b) 1 − 3ab + 1 − 3ac + 1 − 3bc ≥ 6. 3490

. Propose par Dorin Marghidanu, Colegiul National \A.I. Cuza", Corabia, Roumanie. Soit a1 , a2 , . . . , an+1 des nombres reels  positifs ou an+1 = a1 . Montrer que 3491

n X

i=1

n a4i 1X ≥ ai , (ai + ai+1 )(a2i + a2i+1 ) 4 i=1

⋆. Propose par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

3492

Soit P un point a l'interieur  d'un tetra  edre  ABCD deqsorte que ∠P AB , ∠P BC , ∠P CD et ∠P DA soient tous egaux  a arccos 32 . Montrer que ABCD est un tetra  edre  regulier  et que P est son entre de gravite. 

516  . Propose par Va lav Kone ny, Big Rapids, MI, E-U. Soit AB0 C0 un triangle re tangle d'hypotenuse  c = AB0 et de ot ^ es  a = B0 C0 et b = C0 A. On de nit  les points Bi sur AB0 et Ci sur AC0 de sorte que Bi Ci soit perpendi ulaire au ot ^ e AC0 et tangent au er le ∞ P ins rit du triangle re tangle pre  edent  ABi−1 Ci−1 . Trouver S = ACi en i=1 fon tion de a, b et c. 3493

. Propose par Mi hel Bataille, Rouen, Fran e. Soit n un entier, n > 1 et, pour haque k = 1, 2, . . . , n, soit

3494

σ(n, k) =

X

1≤i1 <···
i1 i2 · · · ik .

Montrer que n X

k=1

n X n+1−k ln n · σ(n, k) ∼ (n + 1)! ∼ · σ(n, k) , n+1−k ln n k=1

ou f (n) ∼ g(n) signi e

f (n) →1 g(n)

si n → ∞.

. Propose par Cosmin Pohoata , College  National Tudor Vianu, Bu arest, Roumanie. Soit a, b, c trois nombres reels  positifs ave a + b + c = 2. Montrer que 3495

X 1 a + ≤ 2 b+c

y lique

X

y lique


X 1 a2 a2 + bc ≤ + 2 b+c 2 b + c2

y lique

.

. Propose par Elias C. Buissant des Amorie, Castri um, Pays-Bas. Montrer la validite des equations  suivantes : ◦ ◦ ◦ (a) tan 72 = tan 66 + tan 36 + tan 6◦ . (b) tan 84◦ = tan 78◦ + tan 72◦ + tan 60◦ ;

3496

⋆

4 P

[Ed. : Le proposeur a ajoute six autres identites  de la forme f (θ) = tan ki θ i=1 = 0 pour ki ∈ Z et θ = 2π/n ave n|360, i i non in luses par manque d'espa e.℄ . Propose par Salem Maliki , etudiant,  College  de Sarajevo, Sarajevo, Bosnie et Herzegovine.  Soit P un point a l'interieur  d'un triangle ABC et designons  par r le rayon du er le ins rit de ABC . Montrer que max{AP , BP , CP } ≥ 2r. 3497

517 . Propose par Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Soit Fn le ne nombre de Fibona

i, 'est-a-dire  F0 = 0, F1 = 1, et Fn = Fn−1 + Fn−2 pour n ≥ 2. Montrer que, pour tout entier positif n, on a 3498

s

Fn+3 + Fn

s

Fn + Fn+2 > 1+2 Fn+1

s

Fn + Fn+3

s

Fn+1 Fn + Fn+2

!

.

⋆

. Propose par Bernardo Re aman,  Institut Alberto Merani, Bogota, Colombie. Dans un b^atiment de n etages,  numerot  es  de 1 a n, on a un ertain 1 et n, et peut-^etre a d'autres nombre d'as enseurs qui s'arr^etent aux etages  etages.  Pour haque n, trouver le nombre minimal d'as enseurs requis dans le b^atiment orrespondant pour que deux etages  quel onques soient relies  par au moins un as enseur express. Par exemple, si n = 6, neuf as enseurs suÆsent : (1, 6), (1, 3, 4, 6), (1, 5, 6), (1, 4, 6), (1, 2, 4, 5, 6), (1, 2, 5, 6), (1, 2, 6), (1, 3, 5, 6) et (1, 2, 3, 6). 3499

3500

 . Propose par Paul Bra ken, Universite du Texas, Edinburg, TX, E-U.

On de nit  la fon tion f (a) pose β = −f (1) + 21 f


1 4

− 12 f

1 ∞ Y (2k − 1) 2k

k=1

(2k)

1 2k−1

=

∞ P

k=1
− 41

ln k k(k + a)

pour a

. Montrer que

∈ (−1, ∞),

et on

= 2− 2 ln(2)+1−γ · eβ , 3

ou γ est la onstante d'Euler. ................................................................. . Proposed by Pham Huu Du , Ballajura, Australia. Let a, b, and c be positive real numbers. Prove that

3488

a 2a2 + bc

+

b 2b2 + ca

+

c 2c2 + ab

≤

s

a−1 + b−1 + c−1 a+b+c

.

. Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a  de Catalunya, Bar elona, Spain. Let n be a nonnegative integer. Prove that 3489

1 2n−1

  n X √ 2n k ≤ k k=0

s    2n 2n n 2 + . n

518 . Proposed by Mi hael Rozenberg, Tel-Aviv, Israel. Let a, b, and c be nonnegative real numbers su h that a + b + c = 1. Prove that √ √ √ (a) 9 − 32ab + 9 − 32ac + 9 − 32bc ≥ 7; √ √ √ √ (b) 1 − 3ab + 1 − 3ac + 1 − 3bc ≥ 6. 3490

. Proposed by Dorin Marghidanu, Colegiul National \A.I. Cuza", Corabia, Romania. Let a1 , a2 , . . . , an+1 be positive real numbers where an+1 = a1 . Prove that n n X a4i 1X ≥ ai , 2 2 3491

i=1

⋆.

(ai + ai+1 )(ai + ai+1 )

4

i=1

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Let P be a point in the interior of tetrahedron ABCDqsu h that ea h of ∠P AB , ∠P BC , ∠P CD, and ∠P DA is equal to arccos 23 . Prove that ABCD is a regular tetrahedron and that P is its entroid. 3492

. Proposed by Va lav Kone ny, Big Rapids, MI, USA. Let AB0 C0 be a right triangle with hypotenuse c = AB0 and legs a = B0 C0 and b = C0 A. De ne the points Bi on AB0 and Ci on AC0 so that Bi Ci is perpendi ular to the leg AC0 and tangent to the in ir le of the ∞ P previous right triangle ABi−1 Ci−1 . Find S = ACi in terms of a, b, and c. 3493

i=1

. Proposed by Mi hel Bataille, Rouen, Fran e. Let n > 1 be an integer and for ea h k = 1, 2, . . . , n let

3494

X

σ(n, k) =

1≤i1 <···
Prove that n X

k=1

i1 i2 · · · ik .

n X ln n n+1−k · σ(n, k) ∼ (n + 1)! ∼ · σ(n, k) , n+1−k ln n k=1

where f (n) ∼ g(n) means that

f (n) →1 g(n)

as n → ∞.

. Proposed by Cosmin Pohoata , Tudor Vianu National College, Bu harest, Romania. Let a, b, c be positive real numbers with a + b + c = 2. Prove that 3495

1 2

+

X

y li

a b+c

≤


X a2 + bc

y li

b+c

≤

1 2

+

X

y li

a2 b2 + c2

.

519 . Proposed by Elias C. Buissant des Amorie, Castri um, the Netherlands. Prove the following equations: (a) tan 72◦ = tan 66◦ + tan 36◦ + tan 6◦ . 3496

(b)

⋆ tan 84

◦

= tan 78◦ + tan 72◦ + tan 60◦ ; 4 P

[Ed.: The proposer gave six more relations of the form f (θ) = tan ki θ = 0 i=1 for ki ∈ Z and θ = 2π/n with n|360, not in luded here for la k of spa e.℄ . Proposed by Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina. Let P be a point in the interior of triangle ABC , and let r be the inradius of ABC . Prove that max{AP , BP , CP } ≥ 2r. 3497

. Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a  de Catalunya, Bar elona, Spain. Let Fn be the nth Fibona

i number, that is, F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. For ea h positive integer n, prove that 3498

s

Fn+3 + Fn

s

Fn + Fn+2 > 1+2 Fn+1

s

Fn + Fn+3

s

Fn+1 Fn + Fn+2

!

.

⋆

. Proposed by Bernardo Re aman,  Instituto Alberto Merani, Bogota,  Colombia. A building has n oors numbered 1 to n and a number of elevators all of whi h stop at both oors 1 and n, and possibly other oors. For ea h n, nd the least number of elevators needed in this building if between any two

oors there is at least one elevator that onne ts them non-stop. For example, if n = 6, nine elevators suÆ e: (1, 6), (1, 5, 6), (1, 4, 6), (1, 3, 4, 6), (1, 2, 4, 5, 6), (1, 2, 5, 6), (1, 2, 6), (1, 3, 5, 6), and (1, 2, 3, 6). 3499

3500

. Proposed by Paul Bra ken, University of Texas, Edinburg, TX, USA.

∞ P ln k f (a) = k(k + a) k=1

β = −f (1) + 12 f 14 − 12 f − 41 . Prove that

De ne the fun tion

1 ∞ Y (2k − 1) 2k

k=1

(2k)

where γ is Euler's onstant.

1 2k−1

for

a ∈ (−1, ∞),

= 2− 2 ln(2)+1−γ · eβ , 3

and set

520

SOLUTIONS No problem is ever permanently losed. The editor is always pleased to onsider for publi ation new solutions or new insights on past problems. We belatedly a knowledge a orre t solution to #3340 by \Solver X", dedi ated to the memory of Jim Totten, whi h we had previously lassi ed as in orre t. Our apologies. . [2008 : 483, 486℄ Proposed by Mihaly Ben ze, Brasov, Romania. For a ∈ R de ne a sequen e (xn ) by x0 = a and xn+1 = 4xn − x2n for all n ≥ 0. Prove that there exist in nitely many a ∈ R su h that the sequen e (xn ) is periodi .

3389

Similar solutions by George Apostolopoulos, Messolonghi, Gree e and Mi hel Bataille, Rouen, Fran e. and ap = 2(1 − cos θp ). For ea h positive integer p, let θp = 2p2π −1 Clearly, aj 6= ak for j 6= k, so it suÆ es to show (xn )  that the nsequen e
 is periodi for a = ap . If x0 = ap , then x n = 2 1 − cos 2 θ holds for p
 n n = 0. Moreover, if we assume xn = 2 1 − cos 2 θp to be true, then using the formula 2 cos2 y = 1 + cos 2y, we have 
 
2 = 8 1 − cos 2n θp − 4 1 − cos 2n θp

 = 4 − 4 cos2 2n θp = 2 1 − cos 2n+1 θp . 
 Thus, for all nonnegative integers n, we have xn = 2 1 − cos 2n θp , and so 

 xn+p − xn = 2 cos 2n θp − cos 2n+p θp  

 = 4 sin 2n−1 θp 2p − 1 sin 2n−1 θp 2p + 1 = 0, xn+1

where the latter equality follows from sin This shows that (xn ) is p-periodi .




2n−1 θp 2p − 1 = sin 2n π = 0.

Also solved by ARKADY ALT, San Jose, CA, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; CHARLES R. DIMINNIE and ROGER ZARNOWSKI, Angelo State University, San Angelo, TX, USA; OLIVER GEUPEL, Bruhl,  NRW, Germany; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; ALBERT STADLER, Herrliberg, Switzerland; and the proposer. Peter Y. Woo, Biola University, La Mirada, CA, USA sket hed the iterates of the fun tion f(x) = 4x − x2 and indi ated that periodi points are found by interse ting the graph of the line y = x with the graphs of these iterates, the nth iterate having 2n−1 \bumps" on it. The fun tion f(x) = 4x − x2 is known as the logisti fun tion, and its dynami s have been extensively studied. For example, see hapter 10 of the book by Heinz-Otto Peitgen, Harmut Jurgens,  and Deitmar Saupe, Chaos and Fra tals: New Frontiers of S ien e, 2nd ed., Springer.

521 . [2008 : 483, 486℄ Proposed by Mihaly Ben ze, Brasov, Romania. Prove that if A, B , C , and D are the solutions of

3390

X2 =



3 −5 5 8



,

then A2007 + B 2007 + C 2007 + D2007 = O, where O is the 2 × 2 zero matrix. Solution by Oliver Geupel, Bruhl,  NRW, Germany, modi ed by the editor. We generalize as follows: Let M ∈ M2 (C) have two distin t nonzero eigenvalues in C. Then the equation X 2 = M has exa tly four distin t solutions A, B , C , D ∈ M2 (C) and moreover, if m is any odd positive integer, then Am + B m + C m + Dm = 0. Proof. First we show that if M is a diagonal matrix with distin t diagonal entries, then any solution to X 2 = M is also a diagonal matrix. Indeed, let 

with λ

6 µ. = b = c = 0.

λ 0

0 µ



=



a b c d

2

=

a2 + bc (a + d)c



It follows immediately that a2

Se ondly, if



λ 0

0 µ



=



a 0 0 d

2

6= d2 ,

(a + d)b d2 + bc



hen e a + d

6= 0.

Thus

, then a2 = λ and d2 = µ. Thus, if 

λ 0 0 µ



ω and σ are hoi es of square roots of λ and µ, the equation X =   ±ω 0 has exa tly four distin t roots : . [Ed: Re all that λ and µ are 0 ±σ 2

nonzero.℄ Now let M be any 2 × 2 matrix with distin t nonzero eigenvalues λ,  µ.  λ 0 Then, there exists an invertible matrix V su h that V −1 M V = 0 µ . 2 It is easy to see that X is a solution = M if and only if Y = V −1 XV  to X  λ 0 is a solution to Y 2 = V −1 M V = 0 µ . By the rst part it follows that X 2 = M has exa tly four solutions, namely V

Then



±ω 0

0 ±σ

Am + B m + C m + D m  2ω m − 2ω m = V 0



V −1 .

2σ m

0 − 2σ m



V −1 = 0 .

Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; DIONNE BAILEY, ELSIE CAMPBELL, CHARLES DIMINNIE and ROGER

522 ZARNOWSKI, Angelo State University, San Angelo, TX, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; REBECCA EVERDING, student, Southeast Missouri State University, Cape Girardeau, MO, USA; CODY GUINAN, student, Southeast Missouri State University, Cape Girardeau, MO, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; WILLIAM M NEARY, Charleston, MO, USA; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; CRISTINEL MORTICI, Valahia University of T^argoviste, Romania; NANCY MUELLER and SETH STAHLHEBER, Southeast Missouri State University, Cape Girardeau, MO, USA; JENNIFER PAJDA, student, Southeast Missouri State University, Cape Girardeau, MO, USA; JOEL SCHLOSBERG, Bayside, NY, USA; SKIDMORE COLLEGE PROBLEM SOLVING GROUP, Skidmore College, Saratoga Springs, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; TITU ZVONARU, Comane  sti, Romania; and the proposer. There was one in omplete solution submitted. Some solvers observed that the problem is solved on e it is known that X2 = M has exa tly four roots, for then the roots an be grouped in pairs ±X. Barbara remarked that in general if R is a unitary ring where 2 is not a zero P divisor, N is a positive odd integer, and S = {x ∈ R : x2 = θ} is nite for some θ ∈ R, then xN = 0. x∈S

. [2008 : 483, 486℄ Proposed by Mi hel Bataille, Rouen, Fran e. Let ABCD be a onvex quadrilateral su h that AC and BD interse t in right angles at P , and let I , J , K , and L be the midpoints of AB , BC , CD , and DA, respe tively. Show that the ir les (P IJ ), (P J K), (P KL), and (P LI) are ongruent if and only if ABCD is y li . 3391

I. Solution by Va lav Kone ny, Big Rapids, MI, USA.

The midpoint quadrilateral IJ KL has sides parallel to the diagonals of ABCD, when e it is a re tangle. Be ause ABCD is onvex, the point P lies inside the re tangle. Moreover, be ause I and J are the midpoints of BA and BC , the line IJ is the perpendi ular bise tor of BP and, therefore, ∠P IJ = ∠J IB . Sin e IJ kAC , we have ∠J IB = ∠CAB , when e ∠P IJ = ∠CAB .

Similarly, J K is the perpendi ular bise tor of CP and ∠P KJ = ∠CDB ,

be ause they both equal ∠CKJ . Furthermore, sin e P is inside the re tangle (so that ∠P IJ < ∠LIJ and ∠P KJ < ∠LKJ ), both ∠P IJ and ∠P KJ are a ute. Be ause two triangles with a ommon side that subtend a ute angles have equal ir umradii if and only if those angles are equal, we dedu e that ∠P IJ = ∠P KJ ⇐⇒ (P IJ ) and (P J K) are

ongruent.

Be ause ABCD is onvex, the verti es A and D lie on the same side of the line BC , when e ∠CAB = ∠CDB ⇐⇒ ABCD

is y li .

523 We on lude that if just the two ir les (P IJ ) and (P J K) are ongruent, then ABCD must be y li . Be ause there was nothing spe ial about the ir um ir les (P IJ ) and (P J K), we have as a onverse, if ABCD is

y li , then any two onse utive ir um ir les of the hain (P IJ ), (P J K), (P KL), (P LI) are ongruent and, onsequently, all four are ongruent. II. Solution by D.J. Smeenk, Zaltbommel, the Netherlands, expanded by the

editor. The ir les (P IJ ), (P J K), (P KL), and (P LI) are the nine-point (or Feuerba h) ir les of triangles ABC , BCD, CDA, and DAB , respe tively (be ause ea h ontains the midpoints of two sides and the foot of an altitude); thus if A, B , C , D are four points in any order on a ir le for whi h AC and BD interse t orthogonally at P , the radius of ea h of the ninepoint ir les equals half the radius of the large ir le. The onverse is not so obvious|without using the onvexity of ABCD, all we an on lude from the ongruen e of the ir les (P IJ ), (P J K), (P KL), and (P LI) is that the ir les (ABC), (BCD), (CDA), and (DAB) are themselves ongruent. Indeed, if two from the latter set of four ir les are distin t, then these four ir les are related as in TiTei a's theorem: If three ongruent ir les pass through a ommon point, then their other three interse tion points lie on a fourth ir le of the same radius and, moreover, the four interse tion points form an ortho entri quadrilateral (meaning ea h point is the ortho entre of the triangle formed by the other three). (See problem 3337 [2009 : 191-192℄, where that theorem is dis ussed and referen es are provided.) Thus, either the points A, B , C , D form an ortho entri quadrilateral, or they must lie on a ir le. Sin e an ortho entri quadrilateral an never be onvex (exa tly one of the four triangles formed by three of the four points must be a ute with the fourth point|the ortho entre|inside), we see that onvexity for es ABCD to be y li . Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e; OLIVER GEUPEL, Bruhl,  NRW, Germany; JOHN G. HEUVER, Grande Prairie, AB; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; JOEL SCHLOSBERG, Bayside, NY, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane  sti, Romania; and the proposer.

. [2008 : 484, 486℄ Proposed by Mi hel Bataille, Rouen, Fran e. Let A, B , C , D, and E be on y li with V and W on the lines AB and AD , respe tively. Show that if the line CE , the parallel to CB through V , and the parallel to CD through W are on urrent, then triangles EV B and EW D are similar. Does the onverse hold? 3392

524 Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA, modi ed by the editor. [Ed.: Note that the problem has been arefully worded to allow the points A through E to be any ve distin t points on a ir le (in any order). Be ause of this, we must use dire ted angles to avoid numerous spe ial

ases.℄ Let P be the point where the parallel to CB through V interse ts the parallel to CD through W . We shall prove a modi ed version of the result: P lies on CE if and only if the triangles EV B and EW D are dire tly similar. Be ause V ∈ BA, W ∈ DA, and A, B , D, and E lie on a ir le, we have ∠EBV = ∠EBA = ∠EDA = ∠EDW. (1) Furthermore, be ause C lies on that same ir le, ∠CBA = ∠CDA. However, be ause P V kCB and P W kCD, we have ∠P V A = ∠CBA and ∠P W A = ∠CDA, when e ∠P V A = ∠P W A and, therefore, the points A, V , P , and W also lie on a ir le. Note that \P lies on CE " means that E is on the transversal CP of the parallel lines W P and CD . Thus, P ∈ CE ⇐⇒ ∠DCE = ∠W P E .

But, ∠DCE = ∠DAE = ∠W AE , when e P ∈ CE ⇐⇒ E

lies on the ir le ontaining A, V , P , W.

This, in turn, is equivalent to ∠AW E = ∠AV E . Sin e ∠DW E and ∠BV E = ∠AV E , we dedu e nally that P ∈ CE ⇐⇒ ∠DW E = ∠BV E .

= ∠AW E

(2)

From (1) and (2) we now have two pairs of equal orresponding angles, whi h

ompletes the proof that P lies on CE if and only if the triangles EV B and EW D are dire tly similar. Also solved by APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e;   Y,  Big Rapids, MI, USA; JOEL OLIVER GEUPEL, Bruhl,  NRW, Germany; V ACLAV KONECN SCHLOSBERG, Bayside, NY, USA; and the proposer, who provided two solutions. Note that in the statement of the problem the triangles are only required to be similar, not dire tly similar as all but one of the solvers assumed. The ex eption was Kone ny, who joked that it was obvious that the onverse does not hold be ause of the way the question was worded. He and Bataille both provided ounterexamples that require a pi ture. For an easier ounterexample, start by hoosing the points A and E on opposite ends of a diameter, and omplete the on guration with P ∈ CE. Then, from the featured solution, we know that the triangles EV B and EW D are dire tly similar; moreover they have right angles at the

orresponding verti es B and D. Re e t V in the line EB to the point V ′ . Then EV ′ B and EV B are oppositely oriented ongruent triangles, so that the triangles EV ′ B and EW D are, indeed, similar triangles; however, the parallel to CB through V ′ is dierent from the parallel to CB through V so that it ould not interse t the parallel to CD through W at a point of the line CE.

525 . Corre tion. [2008 : 484, 486; 2009 : 108, 110℄ Proposed by Dragoljub Milosevi , Gornji Milanova , Serbia. Let ABC be a triangle with a = BC , b = AC , c = AB , and semiperimeter s. Prove that 3393

A z+x B x+y C 9π y+z · + · + · ≥ 2 x a(s − a) y b(s − b) z c(s − c) s

,

where the angles A, B , and C are measured in radians and x, y, and z are any positive real numbers.  Solution by Sefket Arslanagi , University of Sarajevo, Sarajevo, Bosnia and Herzegovina. √ √ z z+x By the AM{GM Inequality we have y + ≥ yz , ≥ xz , and 2 2 √ x+y ≥ xy ; hen e (y + z)(z + x)(x + y) ≥ 8xyz . Applying the AM{GM 2 Inequality on e more and then using the pre eding inequality, we have y+z A z+x B x+y C · + · + · x a(s − a) y b(s − b) z c(s − c)

≥ 3

r

(y + z)(z + x)(x + y) ABC · xyz abc(s − a)(s − b)(s − c)

≥ 6

r

ABC abc(s − a)(s − b)(s − c)

3

3

.

(1)

2r 3A ≥ Now, using the inequality (see 6.59, p. 188 of [1℄) and π R 2s2 ≥ 27Rr (see 5.12, p. 52 of [2℄); and also using the well-known relations abc = 4RF , F = rs, and F 2 = s(s − a)(s − b)(s − c), where F is the area of triangle ABC , we have

Q

ABC abc(s − a)(s − b)(s − c)



≥

« 2rπ 3 27R π3 „ 2 2« = 54R 2 r 2 s2 r s 4Rrs s

≥

π3 272 27π 3 · = 54s2 4s4 8s6

„

.

(2)

The desired inequality now follows dire tly from (1) and (2). Equality holds if and only if A + B + C = π3 and x = y = z . Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; and the proposer.

Referen es

[1℄ O. Bottema, R.Z . Djordjevi , R.R. Jani , D.S. Mitrinovi , and P.M. Vasi , Geometri Inequalities, Wolters{Noordho, Groningen, 1969. [2℄ D.S. Mitrinovi , J.E. Pe ari , and V. Volene , Re ent Advan es in Geometri Inequalities, Kluwer A ademi Publishers, Dordre ht/Boston/London, 1989.

526 . [2008 : 484, 486℄ Proposed by Dragoljub Milosevi , Gornji Milanova , Serbia. Let ABCD be a tetrahedron with hA and mA the lengths of the altitude and the median from vertex A to the opposite fa e BCD, respe tively. If V is the volume of the tetrahedron, prove that 3394

(hA + hB + hC + hD ) m2A + m2B + m2C + m2D




128 3

≥ √ V

.

Solution by Oliver Geupel, Bruhl,  NRW, Germany. First we onsider the sum of altitudes. Let [XY Z] denote the area of triangle XY Z . By the AM{HM Inequality we obtain hA + hB + hC + hD

1



=

3V

≥

3V ·

[BCD]

+

1

+

1

[CDA] [DAB] 16

+

1 [ABC]

[BCA] + [CDA] + [DAB] + [ABC]

Next we ompute the √ median. BC 2 + CD 2 + DB 2 ≥ 4 3[BCD],

three fa es, hen e,



.

By Weitzenbo k's  inequality we have with similar inequalities for the other

AB 2 + AC 2 + AD 2 + BC 2 + BD 2 + CD 2 √
≥ 2 3 [ABC] + [BCD] + [CDA] + [DAB] .

We also have 9m2A

→ → − → − → − 2 − = B + C + D − 3A   − → − 2 → − → − 2 → − → − 2 → = 3 B − A + C − A + D − A

→ → − 2 → − → − 2 → − → − 2 − − C − B − D − C − B − D
= 3 AB 2 + AC 2 + AD 2 − BC 2 − CD 2 − DB 2 ,

with y li variants holding for the other three medians. Therefore, 9 m2A + m2B + m2C + m2D




= 4 AB 2 + AC 2 + AD 2 + BC 2 + BD 2 + CD 2 √
≥ 8 3 [ABC] + [BCD] + [CDA] + [DAB] .




Finally we put everything together: (hA + hB + hC + hD )

m2A

+

m2B

+

m2C

+

m2D




√ 8 3 128 ≥ 48V · = √ V 9 3

.

527 This ompletes the proof. Equality holds if and only if the tetrahedron is regular, as an be seen from the ondition for equality in Weitzenbo k's  inequality.

  University of Also solved by ARKADY ALT, San Jose, CA, USA; SEFKET ARSLANAGIC,  Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE, Rouen, Fran e; MIHALY BENCZE, Brasov, Romania (two solutions); CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; OLEH FAYNSHTEYN, Leipzig, Germany; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; TITU ZVONARU, Comane  sti, Romania; and the proposer. One in omplete solution was submitted. By employing the Hadwiger{Finsler Inequality, Ben ze proved that 2 2 2 (hA + hB + hC + hD ) m2 A + mB + mC + mD




128 32r X ≥ √ V + (AB − AC)2 , 9 3

where r is the inradius of the tetrahedron and the sum is over the 12 pairs of edges of the tetrahedron that share a ommon vertex.
Janous proved that (hA hB hC hD )1/4 m2A + m2B + m2C + m2D ≥ √323 V , from whi h the proposed inequality follows on a

ount of the AM{GM Inequality. 3395. [2008 :484, 486℄ Proposed by Tai hi Maekawa, Takatsuki City, Osaka, Japan. Let triangle ABC have ortho entre H and ir umradius R. Prove that 4R3 − (l2 + m2 + n2 )R − lmn = 0, where AH = l, BH = m, and CH = n.

Solution by Mi hel Bataille, Rouen, Fran e.
The relation 4R3 − l2 + m2 + n2 R − lmn = 0 holds only for a uteangled or right-angled triangles. We show, more generally, that
4R3 − l2 + m2 + n2 R − ǫlmn = 0 ,

(1)

where ǫ = −1 if one of the angles of △ABC is obtuse and ǫ = +1 otherwise. Let O be the ir um entre of △ABC and A′ , B ′ , C ′ the midpoints of the sides BC , CA, AB , respe tively. It is well known and easy to prove that l = 2OA′ , m = 2OB ′ , and n = 2OC ′ . If A ≤ 90◦ , then ∠BOC = 2A and OA′ = R cos A; but if A > 90◦ , then we have ∠BOC = 360◦ −2A and OA′ = R cos(180◦ −A) = −R cos A. Thus, ǫlmn = 8R3 cos A cos B cos C . Using the well-known trigonometri identity for the angles of a triangle, we obtain

cos2 A + cos2 B + cos2 C = 1 − 2 cos A cos B cos C ,
l2 + m2 + n2 R

The relation (1) follows.

= =

4R3 cos2 A + cos2 B + cos2 C
4R3 1 − 2 cos A cos B cos C .




528 Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS,   University of Sarajevo, Sarajevo, Bosnia and Messolonghi, Gree e; SEFKET ARSLANAGIC,  BENCZE, Brasov, Romania; CAO MINH QUANG, Nguyen Herzegovina (two solutions); MIHALY Binh Khiem High S hool, Vinh Long, Vietnam; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLEH FAYNSHTEYN, Leipzig, Germany; OLIVER GEUPEL, Bruhl,  NRW, Germany; JOHN G. HEUVER, Grande Prairie, AB; JOE HOWARD, Portales, NM,  C,  USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; DRAGOLJUB MILOSEVI Gornji Milanova , Serbia; JOEL SCHLOSBERG, Bayside, NY, USA; D.J. SMEENK, Zaltbommel, the Netherlands; ALBERT STADLER, Herrliberg, Switzerland; TITU ZVONARU, Comane  sti, Romania; and the proposer. Alt, Bataille, Geupel, Janous, S hlosberg, and Zvonaru ea h noted that the extra ondition is required and either suggested adding a hypothesis or provided a ounterexample for obtuseangled triangles. The other solvers assumed the triangle is a ute-angled.

. [2008 : 484, 487℄ Proposed by Neven Juri , Zagreb, Croatia. Let n be a positive integer, and for i, j , and k in {1, 2, . . . , n} let

3396

aijk

=

1 + mod(k − i + j − 1, n) + n mod(i − j + k − 1, n) + n2 mod(i + j + k − 2, n) ,

where mod(a, n) is the residue of a modulo n in the range 0, 1, . . . , n − 1. For whi h n is the ube with entries aijk a magi ube? (Here "magi " means that the sum of aijk is onstant if two indi es are xed and the third index varies, and also the sums along the great diagonals of the ube are equal to this onstant.) Solution by Oliver Geupel, Bruhl,  NRW, Germany, modi ed by the editor. We will prove that the ube is magi for all n and that the magi sum is n(n3 + 1) Sn = . Let Ln = {0, 1, 2, . . . , n − 1}, and for onvenien e write 2 bijk = k − i + j − 1, cijk = i − j + k − 1, and dijk = i + j + k − 2. Then aijk = 1 + mod(bijk , n) + n mod (cijk , n) + n2 mod (dijk , n) .

For xed j and k, the n numbers bijk , 1 ≤ i ≤ n, are pairwise in ongruent modulo n, sin e k − i1 + j − 1 ≡ k − i2 + j − 1 (mod n) implies i1 = i2 . Hen e, {mod(bijk , n) : 1 ≤ i ≤ n} = Ln . Similarly, {mod(cijk , n) : 1 ≤ i ≤ n} = {mod(dijk , n) : 1 ≤ i ≤ n} = Ln . It follows that n X

aijk

=

n +

1 + n + n2

i=1

X
n−1

l

l=0

= =

n +

1 + n + n2


n(n − 1)

2

n n3 − 1 n n3 + 1 = = Sn . n + 2 2

529 Analogously,

n P

aijk = Sn

j=1

and

n P

k=1

aijk = Sn .

Now we ompute the four diagonal sums. Sin e biii = ciii = i − 1, we have {mod(biii , n) : 1 ≤ i ≤ n} = {mod(ciii , n) : 1 ≤ i ≤ n} = Ln . n P

n−1 P

n P

n(n − 1) Hen e, mod(biii , n) = mod(ciii , n) = l = . Next, we 2 i=1 i=1 l=0 have diii = 3i − 2. If 3 ∤ n, then {mod(diii , n) : 1 ≤ i ≤ n} = Ln so n P n(n − 1) mod(diii , n) = . 2 i=1 If 3|n, then n = 3m for some positive integer m. Hen e, n X

mod(diii , n)

=

i=1

=

n X

mod(3i − 2, n) = 3

i=1 m X

3

(3i − 2) = 3

i=1

=

3m(3m − 1) 2

=



m X

i=1

mod(3i − 2, 3m)

3m(m + 1) 2

n(n − 1) 2

 − 2m

,

as before. Therefore, the sum along the main diagonal is n X

aiii

n X

=

i=1

i=1


1 + mod(biii , n) + n mod (ciii , n) + n2 mod (diii , n) 2

=

n+ 1+n+n


n(n − 1) 2

=


n n3 + 1 2

Next we onsider the diagonal {(n + 1 − i, i, i) : notational onvenien e, let i∗ = n + 1 − i. Then

= Sn .

1 ≤ i ≤ n}.

For

bi∗ ii = i − (n + 1 − i) + i − 1 = 3i − n − 2 , ci∗ ii = (n + 1 − i) − i + i − 1 = − i + n ,

di∗ ii = (n + 1 − i) + i + i − 2 =

Hen e, n X

i=1

ai∗ ii

=

n X

i=1

i + n − 1.

1 + mod(3i − n − 2, n)


+ n mod (−i + n, n) + n2 mod (i + n − 1, n)
X
n−1 n n3 + 1 2 = n + 1+n+n l = = Sn , 2 l=0

by the same argument as above. The sums of the entries in the ells of {(i, n + 1 − i, i)|1 ≤ i ≤ n} and {i, i, n + 1 − i)|1 ≤ i ≤ n} an be al ulated in a similar manner to yield the same value Sn .

530 Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; and the proposer. There was one partly in orre t solution. The proposer and the editor had ta itly assumed that the entries of the ube must be 1, 2, . . . , n3 in whi h ase the ube would be magi if and only if n is odd. This was proved by the proposer. Apparently, neither Curtis nor Geupel made this assumption in their solutions.

. [2008 : 485, 487℄ Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a  de Catalunya, Bar elona, Spain. Evaluate Z n√ 2 1 n − x2 lim 2 dx . −x n→∞ 3397

n

0

2+x

Solution by Oliver Geupel, Bruhl,  NRW, Germany. Let I denote the above integral. Making the substitution x have 1 I n2

=

=

= nt,

√ 1 − t2 dt −nt 0 2 + (nt) √ √ Z 1 Z √1 n 1 − t2 1 − t2 dt + dt . 2 + (nt)−nt 2 + (nt)−nt √1 0 n Z

we

1

(1)

Observe that for 0 < t < 1, we have √ 1 − t2 1 0 < < −nt 2 + (nt) 2

and for

1 √
we have

√ n < nt

,

(2)

and tt < 1, so that

1 1 1 √ (−√n) < 2 + (nt)−nt < 2 + n−n 2+ n

.

(3)

From (2) we obtain 0 <

Z

0

√1 n

√ 1 − t2 1 dt < √ −nt 2 + (nt) 2 n

(4)

and from (3) we obtain 1 √ (−√n) 2+ n

Z

1 √1 n

p 1 − t2 dt

< <

Z

1

√ 1 − t2

dt 2 + (nt)−nt Z 1 p 1 1 − t2 dt . 2 + n−n √1n 1 √ n

(5)

531 Letting nR tend to in nity in (5), we see that the integral in the middle R1√ √ 1 1 2 approa hes 2 0 1 − t dt. Sin e 0 1 − t2 dt = π4 , we obtain from (1), (4), and (5) that lim

1

n→∞ n2

Z

n

0

√ n2 − x2 2+

x−x

dx = 0 +

1 2

·

π 4

=

π 8

.

Also solved by MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU, Hong Kong, China; ALBERT STADLER, Herrliberg, Switzerland; and the proposer. One solution was submitted with an in omplete justi ation of the al ulations. 3398. [2008 : 485, 487℄ Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Given the equation

j

n 10

k

+

 j k 2n n n − 10 · 10⌊log10 n⌋ = 10

3

,

(a) show that n = 5294117647058823 is a solution,

(b) ⋆ nd all other positive integer solutions of the equation. Solution by Oliver Geupel, Bruhl,  NRW, Germany. We will prove that all solutions are given by
3b 1016k − 1 n = , 17 where k is a positive integer and b is any integer with 1 ≤ b ≤ 5. The solution in (a) is obtained by setting k = 1 and b = 3. Ea h positive integer n has an unique representation n = 10a + b, where a and b are nonnegative integers and b ≤ 9. Let l + 1 be the number of de imal digits of n. It is easy to he k that there is no solution with n ≤ 9, hen e we an assume that l ≥ 1. Then, the number n is a solution if and

only if

a + 10l b =

or equivalently a =

where

2 (10a + b) , 3


3 · 10l − 2 b 17

,

and 0 ≤ b ≤ 9 . For l = 1 we get 28b = 17a, whi h is impossible be ause 17 is not a divisor of 28b unless b is zero. Therefore, we have l ≥ 2. The inequality involving a will now be satis ed if and only if 1 ≤ b ≤ 5. 10l−1 ≤ a < 10l

532 (3 · 10l − 2)b

[Ed:

10l−1 ≤ a < 10l ⇐⇒ 10l−1 ≤ ≤ 10l − 1 ⇐⇒ 17

17·10l−1 ≤ 3·10l −2 b ≤ 17· 10l −1 . Now 17·10l−1 < 30·10l−1 −2 be ause l ≥ 2, and sin e b is a nonnegative integer the rst inequality is equivalent to b ≥ 1.   ` ´ 17 · 10l − 1

The se ond inequality be omes b ≤ 3 · 10l − 2 = 5 + Sin e b is an integer, this is equivalent to b ≤ 5.℄ Thus, n is a solution to the given equation if and only if

2 · 10l − 7 3 · 10l − 2

.

(1)

3 · 10l ≡ 2 (mod 17)

and 1 ≤ b ≤ 5. By Fermat's Little Theorem, 1016+l ≡ 1016 (mod 17), and an easy

omputation shows that the only solution l of (1) with 1 ≤ l ≤ 16 is l = 15. Thus, (1) is equivalent to l ≡ 1 (mod 16), or l + 1 = 16k. Putting everything together, we have that n is a solution if and only if n

= 10a + b = 10 · =


3 · 10l+1 − 3

where k ≥ 1 and 1 ≤ b ≤ 5.

17

3 · 10l − 2 17

b =

!

b+b


3b 1016k − 1 17

,

Also solved by RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; JOEL SCHLOSBERG, Bayside, NY, USA; and ALBERT STADLER, Herrliberg, Switzerland.  Part (a) was also solved by MICHEL BATAILLE, Rouen, Fran e; MIHALY BENCZE, Brasov, Romania; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; KEITH EKBLAW, Walla Walla, WA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was one in omplete solution submitted. The proposer remarked that 13 · 5294117647058823 = 1764705882352941, that is, the de imal digits of the rst solution to the equation are rotated ve pla es to the right upon multipli ation by 13

3399. [2008 : 485, 487℄ Proposed by Vin entiu R adules u, University of Craiova, Craiova, Romania. Prove that there does not exist a positive, twi e dierentiable fun tion f : [0, ∞) → R su h that f (x)f ′′ (x) + 1 ≤ 0 for all x ≥ 0.

Comments by Mi hel Bataille, Rouen, Fran e; and Walther Janous, Ursulinengymnasium, Innsbru k, Austria. Bataille indi ated this problem was posed in the College Mathemati s Journal by the same proposer in January, 2008. A solution and a generalization appeared in the January, 2009 issue of that journal (problem 869, pp. 60-61).

533 Janous indi ated that this problem was also posed in the Swiss journal Elemente der Mathemati k in the problem se tion of Heft 4, 2008, by the same proposer. Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; ROY BARBARA, Lebanese University, Fanar, Lebanon; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was one in orre t submission. Barbara showed that if f is negative and satis es the given inequality, then −f is positive and also satis es the inequality. Hen e, the ondition "f positive" an be omitted. On the other hand, by re-s aling, the number 1 an be repla ed by ǫ. The result an be reformulated as: Let ǫ > 0. There does not exist a twi e dierentiable fun tion f : [0, ∞) → R su h that f(x)f ′′ (x) + ǫ ≤ 0 for all x ≥ 0. Barbara further observed that ǫ annot be omitted. The fun tion f(x) = log(x + 2) satis es f(x)f ′′ (x) < 0 for all x ≥ 0. This leads to another reformulation of this result: let f : [0, ∞) → R be a twi e dierentiable fun tion su h that f(x)f ′′ (x) < 0 (or ≤ 0) for all x ≥ 0. Then, f(x)f ′′ (x) tends to zero as x tends to in nity.

. [2008 : 485, 487℄ Proposed by Yakub N. Aliyev, Qafqaz University, Khyrdalan, Azerbaijan. For positive integers m and k let (m)k = m(1+10+102 +· · ·+10k−1 ), for example, (1)2 = 11 and (3)4 = 3333. Find all real numbers α su h that 3400

j

10n

k k j p 5 − 9α (1)2n + α = (3)2n − 6

holds for ea h positive integer n, where ⌊x⌋ is the greatest integer not ex eeding x. Solution by the proposer. The answer for α is the union of 16 losed and disjoint intervals: α∈

 2 7 [ m − 66m − 11

m=−8

100

,

5 − 6m 9



.

(1)

To prove this, we rst let n = 1. Then   k j √ 5 − 9α 10 11 + α = 33 − 6

Set m =

j

5 − 9α 6

k

.

(2)

. Then m ≤

5 − 9α 6

< m + 1,

whi h is equivalent to −

6m + 1 5 − 6m < α ≤ 9 9

.

(3)

534 √

From (2) we obtain 33 − m ≤ 10 m2 − 66m − 11 100

11 + α < 34 − m.

≤ α <

Solving for α yields

m2 − 68m + 56 100

.

(4)

It is easy to he k that 5 − 6m 9

and −

6m + 1 9

<

m2 − 68m + 56

<

m2 − 66m − 11

100

100

Then, from (3) and (4), it follows that m2 − 66m − 11 100

≤ α ≤

.

5 − 6m 9

.

(5)

From (5) we obtain su

essively m2 − 66m − 11

≤

5 − 6m

,

(6)

100 9 (3m + 1)2 ≤ 600 , j √ k |3m + 1| ≤ 10 6 = 24 ,

from whi h we see that inequality (6) is stri t for m = −8, −7, . . . , 7. Thus, the equality (2) may only hold for values of α satisfying the inequality (5) for m = −8, −7, . . . , 7; these are pre isely the values of α overed in (1). We shall now prove that the given equality does indeed hold for these values of α. We note that the given equality is equivalent to ea h of the following double inequalities: p (1)2n + α < (3)2n − m + 1 , p −m ≤ 10n (1)2n + α − (3)2n < −m + 1 , s  2n  102n − 1 10 − 1 −m ≤ 10n +α − < −m + 1 , 9 3 p −1 − 3m ≤ 10n 102n + 9α − 1 − 102n < 2 − 3m ,

(3)2n − m ≤ 10n

10n (9α − 1) −1 − 3m ≤ √ < 2 − 3m , 102n + 9α − 1 + 10n     5 − 9α 9α − 1 5 − 9α < 2−3 −1 − 3 ≤ r 6 6 9α − 1 1+ + 1 2n 10

.

(7)

535 Note that the double inequality (7) holds for α = 19 , and that 9α − 1 9α − 1 9α − 1 r ≤ r < 2 9α − 1 9α − 1 1+ + 1 1+ + 1 2 2n 10

10

holds ( onsider separately the ases α > to prove (7), it suÆ es to show that −1 − 3

and



5 − 9α 6

9α − 1

and α < 19 ). Therefore, in order

9α − 1 ≤ r 9α − 1 1+ + 1 2



(8)

10

≤ 2−3

2

1 9

Inequality (8) an be written as



5 − 9α 6



.

√ 11 + α m ≥ − , √ 10 + 3 11 + α √ simpli es to m ≥ 33 − 10 11 + α. This last inequality

(9)

30α +

whi h to the left inequality of (5). Inequality (9) simpli es to 

5 − 9α 6



≤

5 − 9α 6

is equivalent

,

whi h is obvious. Remark. The given equality holds asymptoti ally for any value of α. This

an be easily proved by xing α and letting n → ∞ in inequality (7).

Also solved by OLIVER GEUPEL, Bruhl,  NRW, Germany. One in omplete solution and two in orre t solutions were submitted. The proposer mentioned that some problems on equalities involving the integer part fun tion and a parameter had appeared earlier [1℄, [2℄.

Referen es

[1℄ I. Bluskov, Problem 1650, Crux Mathemati orum Vol. 17 (1991) p. 141 [2℄ I. Bluskov, Problem 3216, Mathemati s in S hool (Russian) Vol. 3 (1988) p. 60

536

YEAR END FINALE As I enter my nal year as Editor-in-Chief of CRUX with MAYHEM, the all has been put out by the Canadian Math So iety for a new Editor-in-Chief, with nominations being taken by the CMS Publi ations Committee. Any potential nominee should know that (if needed) I am willing to serve as a Co-Editor for all of 2011 and to serve as an Editor-at-Large beyond that. This past year JOHAN RUDNICK took over the reins from GRAHAM WRIGHT as Exe utive Dire tor of the CMS and also as Managing Editor of CRUX. I appre iate very mu h Johan's interest in CRUX and our dis ussions, and also Graham's ontinuing support in his role as Consultant towards the end of the year. The last Publi ations Committee meeting I attended marked the end of Graham's term, so it is now nally safe to say, \Graham, all the best for you in your retirement!" DENISE CHARRON and LAURA ALYEA have done a wonderful job this past year administering CRUX at the CMS headquarters in Ottawa. Our past CRUX editor BRUCE SHAWYER ontinues to ontribute to CRUX and is an inspiration for me. I also thank our past CRUX editor BILL SANDS for his sound advi e and fantasti proof reading (it is my theory that in a previous life Bill was an eagle that learned to read). My olleague and past CMO Chair TERRY VISENTIN has also lent his experien e to proo ng the opy. In the New Year the CRUX board warmly wel omes two new Problems Editors: DZUNG MINH HA of Ryerson University and JONATAN ARONSSON of the University of Manitoba. I hope they nd the problems moderating every bit as fas inating as I do. This past year LILY YEN and MOGENS LEMVIG HANSEN ame on board as Skoliad Editors. Thank you Lily and Mogens for your wonderful ontributions, and for generously agreeing to ontinue editing Skoliad beyond your rst year. ILIYA BLUSKOV and MARIA TORRES are stepping down at the end of 2009 as Problems Editors, and I thank them for their servi e. Iliya, it has been a pleasure and fun as well, may you nd every design that you sear h for (ex ept those that do not exist). Maria, good lu k with your future proje ts and I am grateful for your generous oer to help with Spanish translations in the future. Gra ias! JOHN GRANT M LOUGHLIN is stepping down as Member-at-Large to take up editing the Edu ation se tion of the CMS Notes. John, best of lu k in the future and thank you for your in redible eort for the JIM TOTTEN spe ial issue this past year, whi h made it that mu h more spe ial. JEFF HOOPER, our Asso iate Editor, has pulled my ba on out of the re more than on e this past year, and I am most grateful for it! IAN VANDERBURGH, our Mayhem Editor, ontinues to outpa e this Editor-in-Chief as he hurns out yet another Problem of the Month olumn. Ian, let's see if I an at h up this year! A thank you is due to ROBERT WOODROW for managing the Her ulean task of assembling the Olympiad Corner this past year. This was also AMAR SODHI's rst year as Book Reviews Editor, and a proli one judging from the long list of reviews in the index that follows. I am ever grateful for the work of my olleague and Arti les Editor, JAMES CURRIE, who has kept CRUX with a healthy supply of arti les whi h makes it that mu h easier to omplete an issue. I thank CHRIS FISHER for his myriad ontributions to CRUX beyond his duties as Problems Editor. Chris' sense of humour is infe tious and brings some mu h needed levity into the pro ess of editing. I thank EDWARD WANG for ontinuing as Problems Editors this past year despite a very high workload, and for his perseveran e during a reallo ation of support sta within his department.

537 I thank MONIKA KHBEIS and ERIC ROBERT for their ontinuing servi e on the

Mayhem sta. I thank JEAN-MARC TERRIER for translating the problems that appear in CRUX with MAYHEM, and ROLLAND GAUDET for translations of the Skoliad

ontests. So great was Jean-Mar 's zeal for translating this past year that he razily requested that opy be sent over the holiday, to whi h this editor instantly responded with the required erti ate of insanity! I thank JILL AINSWORTH and JOANNE CANAPE at the University of Calgary; TAO GONG, JUNE ALEONG, and LOUIS MOUSTERAKIS at Wilfrid Laurier University, for their support in produ ing LATEX les. A big thank you goes to Mr. MATHIAS PIELAHN, the CRUX journal assistant, for his help with managing the mountains of information we re eive as input. Thanks go to JUDI BORWEIN, MICHAEL DOOB, CRAIG PLATT, and STEVE LA ROCQUE for te hni al support and putting CRUX up on the net. TAMI EHRLICH and the people at Thistle Printing ontinue to print ne quality issues of CRUX, whi h are always satisfying to re eive in the mail. The Department of Mathemati s and Statisti s at the University of Winnipeg

ontinues to provide support, and I thank the Dean of S ien e, ROD HANLEY, for keeping up the ommitment of the University of Winnipeg. I espe ially thank LYNNE TOTTEN, for her kind words about our eorts on the JIM TOTTEN spe ial issue, and for providing a sense of perspe tive when it was most helpful. I thank my wife CHARLENE for her support during this past year, and spe ial friends PETER ARPIN and RANDALL PYKE for their spe ial help with CRUX. Though this last year has seen a global re ession, our subs riptions have remained almost steady. This speaks to how tremendously CRUX is valued by you, the readers. I thank all of you for your support of CRUX in these diÆ ult nan ial times, and I thank you even more for all your wonderful mathemati al ontributions, whi h is the stu that CRUX is made of. It is an honour to partake of your ontributions almost ea h and every day, and a wonderful sour e of enri hment for me. I wish everyone happiness, pea e, and ful llment in the New Year. Va lav  (Vazz) Linek

Crux Mathemati orum

with Mathemati al Mayhem Former Editors / An iens Reda teurs:  Bru e L.R. Shawyer, James E. Totten

Crux Mathemati orum

Founding Editors / Reda teurs-fondateurs:  Leopold  Sauve & Frederi k G.B. Maskell Former Editors / An iens Reda teurs:  G.W. Sands, R.E. Woodrow, Bru e L.R. Shawyer

Mathemati al Mayhem Founding Editors / Reda teurs-fondateurs:  Patri k Surry & Ravi Vakil Former Editors / An iens Reda teurs:  Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je Hooper

538 INDEX TO VOLUME 35, 2009

Skoliad Lily Yen and Mogens Lemvig Hansen (∗Editor for February)

February Mar h April May September O tober November De ember

No. 114 ................................................ 1 No. 115 ............................................... 65 No. 116 .............................................. 129 No. 117 .............................................. 194 No. 118 .............................................. 263 No. 119 .............................................. 354 No. 120 .............................................. 417 No. 121 .............................................. 481 Mathemati al Mayhem Ian VanderBurgh February ........................................................ 11 Mar h ........................................................ 71 April ....................................................... 134 May ....................................................... 200 September ....................................................... 270 O tober ....................................................... 363 November ....................................................... 425 De ember ....................................................... 488 ∗

Mayhem Problems

February Mar h April May September O tober November De ember

M376{M381 .......................................... 11 M382{M387 .......................................... 71 M388{M393 ......................................... 134 M394{M400 ......................................... 200 Totten M1{Totten M10 ............................. 270 M381, M401{M406 ................................. 363 M407{M412 ......................................... 425 M413{M419 ......................................... 488

Mayhem Solutions

February Mar h April May September O tober November De ember

M338{M343 .......................................... 13 M344{M349 .......................................... 73 M350{M356 ......................................... 136 M357{M362 ......................................... 202 M363{M368 ......................................... 275 M369{M375 ......................................... 366 M376{M380 ......................................... 427 M382{M387 ......................................... 490 Problem of the Month Ian VanderBurgh February ........................................................ 18 Mar h ........................................................ 78 April ....................................................... 142 May ....................................................... 207 September ....................................................... 281 O tober ....................................................... 372 November ....................................................... 432 De ember ....................................................... 494

539 Mayhem Arti les

The Tanker Problem Ross Honsberger ................................................. 283 The Olympiad Corner R.E. Woodrow February No. 275 ............................................... 20 Mar h No. 276 ............................................... 80 April No. 277 .............................................. 144 May No. 278 .............................................. 211 September No. 279 .............................................. 290 O tober No. 280 .............................................. 375 November No. 281 .............................................. 435 De ember No. 282 .............................................. 497 Book Reviews Amar Sodhi Putnam and Beyond, by Razvan Gel a and Titu Adrees u Reviewed by Jeff Hooper .......................................... 36 Mathemati al Conne tions: A Companion for Tea hers and Others, by Al Cuo o Reviewed by Peter S. Brouwer .................................... 37 Benjamin Franklin's Numbers: An Unsung Mathemati al Odyssey, by Paul C. Pasles Reviewed by Jeff Hooper ......................................... 102 Polynomia and Related Realms: Un ommon Mathemati al Ex ursions, by Dan Kalman Reviewed by Edward J. Barbeau .................................. 160 The Shape of Content: Creative Writings in Mathemati s and S ien e, Edited by Chandler Davis, Marjorie Wikler Sene hal, and Jan Zwi ky Reviewed by Georg Gunther ...................................... 162 The Mobius  Strip: Dr. August Mobius's  Marvelous Band in Mathemati s, Games, Literature, Art, Te hnology, and Cosmology,

by Cliord Pi kover Reviewed by Edward J. Barbeau .................................. 230 The Contest Problem Book IX: Ameri an Mathemati s Competitions (AMC 12) 2001-2007,

Compiled and edited by David Wells and J. Douglas Faires Reviewed by John Grant M Loughlin ............................. 231 A Taste of Mathemati s, Volume IX, The CAUT Problems, by Edward J. Barbeau Reviewed by Amar Sodhi .......................................... 232 All-Star Mathlete Puzzles, by Di k Hess Reviewed by Andy Liu ............................................ 304

A Mathemati al Mosai : Patterns & Problem Solving (New Expanded Edition), by Ravi Vakil

Reviewed by John Grant M Loughlin ............................. 305

A Mathemati al Mosai , by Ravi Vakil

Reviewed by Jim Totten ........................................... 305

Sink or Float? Thought Problems in Math and Physi s, by Keith Kendig

Reviewed by Nora Franzova ...................................... 390

540 Number Theory Through Inquiry,

by David C. Marshall, Edward Odell, and Mi hael Starbird Reviewed by Jeff Hooper ......................................... 391 Geometri Origami, by Robert Gerets hlager Reviewed by Georg Gunther ...................................... 393 Problems from Murray Klamkin: The Canadian Colle tion

Edited by Andy Liu and Bru e Shawyer Reviewed by Bill Sands ........................................... 452 Cro heting Adventures with Hyperboli Planes, by Daina Taimina  Reviewed by J. Chris Fisher ....................................... 453 Mathemati al Mindbenders, by Peter Winkler Reviewed by Amar Sodhi .......................................... 455 A Certain Ambiguity: A Mathemati al Novel, by Gaurav Suri and Hartosh Singh Bal Reviewed by Mark Taylor ........................................ 513 Crux Arti les James Currie

Velo ity Analysis: An Approa h to Solving Geometry Problems Peng YuChen ............................................................... 38 Geometri Constru tions of Mixtilinear In ir les Cosmin Pohoata .......................................................... 103 A Useful Inequality Revisited Pha.m Van Thua.^ n and L^e V~ .............................................. 164 Jim Totten's Rea h John Grant M Loughlin ................................................... 260 The British Columbia Se ondary S hool Mathemati s Contest Clint Lee .................................................................. 307 A Duality for Bi entri Quadrilaterals Mi hel Bataille ........................................................... 310 A Study of Knight's Tours on the Surfa e of a Cube Awani Kumar ............................................................. 313 Several Symmetri Inequalities of Exponential Kind Arkady Alt ................................................................ 457 Problems February 3401{3413 ..................................................... 42 Mar h 3393, 3414{3425 ............................................. 108 April 3426{3438 .................................................... 172 May 3419, 3424, 3439{3450 ....................................... 233 September TOTTEN 01{TOTTEN 12 ..................................... 320 3451{3462 ................................................... 325 O tober 3439, 3463{3474 ............................................. 395 November 3464, 3475{3487 ............................................. 463 De ember 3488{3500 .................................................... 515 Solutions February 3301{3312 ..................................................... 47 Mar h 3313{3325 .................................................... 113 April 3326{3329, 3331{3337 ....................................... 177 May 3330, 3338{3350 ............................................. 238 September 2557, 3351{3362 ............................................. 330 O tober 3363{3375 .................................................... 400 November 3376{3388 .................................................... 468 De ember 3389{3400 .................................................... 520

541 Mis ellaneous Editorial .................................................................... 193 A photo of Jim Totten in his oÆ e .......................................... 257 In Memoriam: James Edward Totten, 1947-2008 .......................... 258 Totten Commemorative Issue ...............................................259 Editorial .................................................................... 353 Year End Finale ............................................................. 536

Proposers and solvers appearing in the SOLUTIONS se tion in 2009: Proposers 3400

Yakub N. Aliyev Arkady Alt

3384, 3385, 3391, 3392

Pantelimon George Popes u

3347, 3348, 3349, 3353, 3378, 3379, 3380, 3389, 3390 3325

3305, 3306, 3315

Stanley Rabinowitz

Vin ent  iu Radules u 

3399

Juan-Bos o Romero Marquez 

Paul Bra ken

3388

Josep Rubio-Masseg  u 

Shi Changwei

3381

Toshio Seimiya

Jose  Luis Daz-Barrero

3313, 3331, 3354, 3362, 3382, 3397

3301, 3334, 3357, 3366, 3375, 3386, 3387

Ovidiu Furdui

3331

Jose  Gibergans-Baguena 

3398

Gord Sinnamon

2557

3376

Panos E. Tsaousoglou

3359

3335

3382

3338, 3339, 3340, 3351, 3352, 3358, 3363, 3364, 3365, 3377

Bru e Shawyer

D.J. Smeenk

3368, 3396

Neven Juri 

3313

3307, 3308, 3318

D.E. Prithwijit

3302, 3303, 3304, 3314, 3316, 3317, 3326, 3327, 3328,

Manuel Benito Munoz ~

3374

Pham Huu Du

3311, 3312, 3320, 3321, 3336, 3337, 3360, 3361, 3383,

Mihaly  Ben ze

3309, 3310

Virgil Ni ula

3319, 3329, 3330, 3341, 3342

Mi hel Bataille

3322, 3323, 3324, 3332, 3350

George Tsintsifas

3369, 3370, 3371

Hung Pham Kim

3344, 3345

Vo Quo Ba Canh

3372, 3373

Va lav  Kone n  y 

3333

Stan Wagon

Ray Killgrove

3395

Tai hi Maekawa

Dragoljub Milosevi 

 Cristinel Morti i

3343 3355

Todor Yalamov

3393, 3394

Bin Zhao

3356

Li Zhou

3346

3367

Featured Solvers | Individuals Yakub N. Aliyev Arkady Alt

3400

3343, 3391

Va lav  Kone n  y 

3330(d), 3380

George Apostolopoulos

3354, 3362

Kee-Wai Lau

3318, 3551

Miguel Amengual Covas

3363, 3364

Matti Lehtinen

3313, 3318, 3324, 3326(a), 3327(a), 3339, 3341,

Ralph Lozano

3343

3350, 3354, 3370, 3372, 3378, 3389

Phil M Cartney

 Sefket Arslanagi 

Tai hi Maekawa

3370, 3385, 3393

3354

Dionne Bailey

3306, 3329, 3553

Roy Barbara

Ri ardo Barroso Campos

3318, 3363

3317, 3326(a), 3331, 3332

Manuel Benito Munoz ~ Elsie Campbell

Li Chao

3317, 3318, 3324, 3342, 3350, 3370

2557

 Os ar Ciaurri Chip Curtis

Andrea Munaro

3367, 3373, 3375, 3385

3318

Khanh Bao Nguyen

3301, 3303, 3305, 3307, 3370

3354

Cao Minh Quang

Cristinel Morti i

Virgil Ni ula

3342, 3345, 3370

3552, 3354, 3363, 3383

Mahdav R. Modak

3353, 3354, 3357, 3370, 3377, 3379, 3380, 3383, 3389, 3395, 3399 Mihaly  Ben ze

3551

Dung Nguyen Manh

3302, 3304, 3310, 3315, 3320, 3327(a), 3338, 3343, 3346,

Mi hel Bataille

3354

3324, 3552, 3370

Salem Maliki 

3350

3310

Paolo Perfetti

3370, 3387, 3388

Gottfried Perz

3354

Pantelimon George Popes u

3301, 3303, 3305, 3307, 3370

Daniel Reisz

3303, 3327(a), 3354, 3366, 3370, 3385

Apostolis K. Demis

3318

Jose  Luis Daz-Barrero Charles R. Diminnie Emilio Fernandez

3313

3335, 3354

3313

Henry Ri ardo Luz Ron al

3303, 3305, 3307, 3370

Xavier Ros

3304, 3322, 3323, 3325(a), 3328, 3343, 3354

Joel S hlosberg

3301, 3303, 3305, 3307, 3370

Toshio Seimiya

3336, 3370, 3371 3358

Ovidiu Furdui

3354, 3362

D.J. Smeenk

3340, 3391

Oliver Geupel

3310, 3311, 3314, 3316, 3319, 3330(ab ), 3333, 3334, 3337,

Digby Smith

3343

3342, 3344, 3347, 3354, 3356, 3358, 3359, 3361, 3368, 3369, 3370, 3382,

Albert Stadler

3387

3390, 3394, 3396, 3397, 3398

Babis Stergiou

3350

Jose  Hernandez  Santiago

3354

Son Hong Ta

3350

Ri hard I. Hess

3343, 3354, 3363, 3384

Tran Thanh Nam

John G. Heuver

3339, 3365

Daniel Tsai

Joe Howard

3307, 3308, 3354, 3370

Peter Hurthig

Steven Karp

Peter Y. Woo

3304, 3328, 3354, 3370, 3386, 3399

3304, 3312

Gerhard Kir hner

3349

3304

George Tsapakidis

3354

Walther Janous

3313

3339

3307, 3309, 3318, 3332

3321, 3336, 3340, 3343, 3348, 3354, 3355, 3363, 3364,

3374, 3381(a), 3392 Titu Zvonaru

3310, 3322, 3350, 3361, 3370, 3385

3386

Featured Solvers | Groups Missouri State University Problem Solving Group

3354, 3360, 3362

542 Other Solvers | Individuals Mohammed Aassila Arkady Alt

3304, 3306, 3308, 3323, 3325(a), 3334

3302, 3303, 3304, 3307, 3308, 3309, 3317, 3319, 3323, 3324,

3359

David Koster

3353(b)

Rodolfo Larrea

3329, 3330(ab ), 3341, 3342, 3349, 3371, 3379, 3381(a), 3382, 3384, 3389,

Kee-Wai Lau

3390, 3394, 3395

3374, 3377, 3386, 3397 3321, 3340, 3352, 3358, 3377

Miguel Amengual Covas George Apostolopoulos

3302, 3303, 3304, 3307, 3308, 3309, 3310, 3316,

3302, 3308, 3324, 3331, 3340, 3342, 3351, 3358, 3363, 3364,

Kathleen E. Lewis

William M Neary

3346, 3351, 3352, 3353(a), 3358, 3362, 3363, 3364, 3365, 3371, 3373, 3374,

Tai hi Maekawa

3376, 3377, 3379, 3380, 3381(a), 3382, 3388, 3390, 3391, 3395, 3399

Thanos Magkos

 Sefket Arslanagi 

Salem Maliki 

3302, 3303, 3304, 3305, 3307, 3308, 3313, 3316, 3318,

3343

3362

Phil M Cartney

3321, 3322, 3323, 3331, 3332, 3335, 3336, 3337, 3340, 3342, 3344, 3345,

3390

3338, 3340, 3352, 3358, 3361, 3363, 3376, 3377, 3395 3302, 3308, 3341, 3342, 3350, 3363, 3364

3302, 3303, 3304, 3308, 3311, 3313, 3318, 3321, 3323, 3331,

3323, 3324, 3331, 3332, 3338, 3340, 3341, 3342, 3345, 3350, 3351, 3352,

3332, 3337, 3338, 3339, 3340, 3341, 3342, 3344, 3345, 3348, 3351, 3358,

3358, 3363, 3364, 3365, 3373, 3376, 3377, 3378, 3380, 3382, 3394, 3395

3359, 3363, 3364, 3365, 3373, 3379, 3380, 3382

3303, 3323, 3324, 3362, 3390

Dionne T. Bailey

3302, 3304, 3308, 3309, 3312, 3321, 3322, 3323, 3324, 3331,

Roy Barbara

3307, 3311, 3312, 3331, 3385

David E. Manes

3303, 3304, 3344, 3350, 3376, 3379, 3380, 3386

Dung Nguyen Manh

3336, 3340, 3341, 3351, 3351, 3352, 3358, 3359, 3360, 3361, 3363, 3364,

D.P. Mehendale

3365, 3367, 3389, 3390, 3399

Dragoljub Milosevi 



3336, 3337, 3351, 3352, 3358, 3361, 3365, 3376

Ri ardo Barroso Campos

3303, 3305, 3307, 3309, 3311, 3312, 3313, 3316, 3318,

Mi hel Bataille

3350

3366, 3376, 3377, 3386, 3389, 3390, 3391 3356, 3367, 3370, 3379, 3390

3321, 3322, 3323, 3324, 3326(a), 3331, 3332, 3336, 3337, 3339, 3340, 3341,

Cristinel Morti i

3342, 3345, 3347, 3350, 3351, 3352, 3355, 3356(a), 3358, 3359, 3360, 3361,

Nan y Mueller

3362, 3363, 3364, 3365, 3371, 3373, 3375, 3376, 3378, 3382, 3384, 3385,

Andrea Munaro

3390, 3391, 3392, 3394, 3397

Nguyen Thanh Liem

3306, 3359, 3390

Brian D. Beasley

Fran is o Bellot Rosada Mihaly  Ben ze

3302, 3303, 3307, 3308, 3323

3302, 3303, 3304, 3314, 3316, 3327(a), 3332, 3348, 3349,

Virgil Ni ula

3350

3347, 3373

3386

Nguyen Van Vinh

3379, 3380, 3389, 3390, 3394, 3395

3309

3375

3350, 3390

Jennifer Pajda

3340

Mi hael Parmenter

3334, 3335, 3350, 3381(a), 3388

Paul Bra ken

Ri ard Peiro 

3342

S ott Brown

3301, 3303, 3304, 3307, 3309, 3362, 3366, 3373, 3378, 3379,

3380, 3386, 3388 3303, 3323, 3324, 3362, 3390

Elsie M. Campbell

3303, 3304, 3313, 3316, 3322, 3323, 3331, 3332, 3338,

Cao Minh Quang

3318

Paolo Perfetti

3351

Chris Broyles

3386

Moubinool Omarjee

3304, 3306, 3308, 3309, 3310, 3311, 3312, 3325(a),

Manuel Benito Munoz ~

3358, 3390

Khanh Bao Nguyen

3339

Pham Huu Du

3374

D.E. Prithwijit

3307, 3308, 3318, 3306, 3315

3341, 3345, 3352, 3358, 3376, 3379, 3380, 3394, 3395

Stanley Rabinowitz

 Os ar Ciaurri

Vin ent  iu Radules u 

Chip Curtis

3393, 3394, 3395

3338, 3340, 3341, 3350, 3351, 3353(a), 3356(a), 3358, 3365,

M.R. Modak

3304, 3306, 3308, 3309, 3310, 3311, 3312, 3375

3302, 3306, 3307, 3308, 3309, 3310, 3312, 3316, 3318, 3320,

3399

3304

Henry Ri ardo

3335, 3342

3321, 3322, 3323, 3324, 3326(a), 3328(a), 3329, 3331, 3332, 3336, 3341,

Juan-Bos o Romero Marquez 

3342, 3344, 3348, 3350, 3351, 3352, 3353, 3358, 3360, 3361, 3363, 3364,

Luz Ron al

3304, 3306, 3308, 3309, 3310, 3311, 3312, 3353(b), 3375

3365, 3371, 3373, 3374, 3375, 3376, 3377, 3378, 3379, 3380, 3382, 3383,

Xavier Ros

3307, 3329, 3331, 3342, 3362

3384, 3388, 3389, 3390, 3391, 3395, 3396, 3397

Josep Rubio-Masseg  u 

Apostolis K. Demis

3305, 3307, 3320, 3321, 3391, 3392, 3399 3307, 3331, 3362, 3382, 3397

Jose  Luis Daz-Barrero Charles R. Diminnie

3302, 3303, 3304, 3308, 3316, 3322, 3323, 3324, 3331,

3343, 3362, 3389, 3390 Joseph DiMuro

Emilio Fernandez Ovidiu Furdui

3385

3338, 3339, 3340, 3351, 3352, 3363, 3364, 3365, 3377

3307, 3340

Bob Serkey

3350, 3390

Bikram Kumar Sitoula D.J. Smeenk

3304, 3306, 3308, 3309, 3310, 3311, 3312, 3375

3301, 3304, 3334, 3357, 3366, 3375, 3386, 3387 3308, 3318, 3340, 3350, 3351

Ian June L. Gar es

Fran is o Javier Gar  a Capitan 

\Solver X"

3358

3318, 3321, 3337, 3352, 3358, 3361, 3365, 3376, 3395

3340

Son Hong Ta Alex Song

3308

3338, 3340

3351 3376, 3377, 3379, 3381, 3382, 3384, 3385, 3386, 3388, 3389,

Albert Stadler

3302, 3303, 3304, 3307, 3308, 3309, 3315, 3318, 3320, 3321,

Oliver Geupel

3365, 3376, 3384, 3385, 3390, 3391, 3392, 3395, 3398 Toshio Seimiya

3394, 3395

Oleh Faynshteyn

3308, 3311, 3312, 3318, 3321, 3348, 3352, 3358, 3359,

3363

Rebe

a Everding

3382

3358

Joel S hlosberg

Robert P. Sealy

3333

Gerald Edge omb

K.C. Sandeep

3390, 3395, 3397, 3398, 3399 3358, 3390

3322, 3323, 3324, 3325(a), 3328(a), 3329, 3331, 3332, 3335, 3336, 3338,

Seth Stahlheber

3340, 3341, 3343, 3348, 3350, 3351, 3353(a), 3355, 3360, 3362, 3363, 3364,

Julie Steele

3365, 3366, 3367, 3371, 3372(a), 3373, 3375, 3376, 3377, 3378, 3379, 3380,

Matthew Stein

3351

David R. Stone

3305, 3307, 3311, 3312

3381, 3383, 3384, 3385, 3387, 3388, 3389, 3391, 3392, 3395, 3400 Jose  Gibergans-Baguena  Douglass L. Grant

Adam Strzebonski

3343

Daniel Tsai

3362

John Hawkins

3302, 3304, 3331, 3332, 3341, 3342, 3345, 3350, 3352, 3358,

3362, 3373, 3395 Peter Hurthig

3302, 3303, 3308, 3310, 3321, 3322, 3323, 3331, 3352,

George Velisaris

3323, 3324

Vo Quo Ba Can

3372(a), 3373

Stan Wagon

3322, 3323, 3324, 3325(a), 3329, 3343, 3350, 3378, 3381, 3388 3377

Haohao Wang

3351, 3358

Walther Janous

George Tsapakidis

3358, 3363, 3364

3318, 3331, 3338, 3351, 3352, 3358, 3363, 3376, 3377,

3379, 3391, 3395 Joe Howard

3302, 3303, 3304, 3307, 3308, 3316, 3322, 3323,

3324, 3332, 3350

3352, 3353(a), 3358, 3361, 3362, 3366, 3375, 3378, 3381,

3385, 3388, 3398 John G. Heuver

3303, 3308, 3309,

Panos E. Tsaoussoglou

3305, 3307, 3311, 3312

Ri hard I. Hess

3329

3345, 3350

Tran Thanh Nam

3390

Cody Guinan Karl Havlak

3331

3363

Jerzy Wojdylo

3302, 3306, 3307, 3312, 3313, 3316, 3318, 3319, 3320,

Peter Y. Woo

3377

3302, 3303, 3304, 3305, 3308, 3309, 3310, 3312, 3313, 3316,

3321, 3322, 3323, 3324, 3325(a), 3327(a), 3329, 3331, 3332, 3333, 3334(a),

3318, 3319, 3320, 3323, 3329, 3331, 3332, 3333, 3335, 3338, 3339, 3341,

3335, 3336, 3337, 3341, 3342, 3343, 3344, 3346, 3350, 3351, 3352, 3353(a),

3344, 3346, 3347, 3350, 3351, 3352, 3356(a), 3358, 3361, 3362, 3365, 3369,

3358, 3359, 3362, 3363, 3365, 3366, 3367, 3371, 3373, 3374, 3375, 3376,

3370, 3371, 3373, 3376, 3377, 3382, 3383, 3385, 3388, 3391, 3399

3377, 3378, 3379, 3380, 3381, 3382, 3384, 3385, 3387, 3388, 3389, 3390,

Todor Yalamov

3391, 3393, 3394, 3395, 3397

Roger Zarnowski

Wei-Dong Jiang Neven Juri 

3350

3368, 3396

Giannis G. Kalogerakis Steven Karp

Bin Zhao

3351, 3358

Li Zhou

3289, 3292, 3294, 3296, 3297, 3298, 3300, 3308, 3309, 3311

Ray Killgrove

3359

Hung Pham Kim John Klassen

3389, 3390 3318, 3321

3346

3367

Titu Zvonaru

3302, 3303, 3304, 3308, 3309, 3312, 3317, 3318, 3321, 3323,

3331, 3332, 3336, 3340, 3341, 3351, 3352, 3353(a), 3356(a), 3358, 3360,

3344, 3345

3363, 3364, 3365, 3371, 3373, 3377, 3378, 3380, 3382, 3390, 3391, 3394,

3343

Va lav  Kone n  y 

3355

Konstantine Zelator

3395

3308, 3315, 3318, 3321, 3336, 3351, 3352, 3363, 3364,

3365, 3376, 3377, 3392

Other Solvers | Groups Missouri State University Problem Solving Group

3384

3351, 3352, 3359, 3360,

Skidmore College Problem Solving Group

3363, 3390

Southeast Missouri State Univeristy Math Club

3358, 3377