No title

1

SKOLIAD

No. 122

Lily Yen and Mogens Hansen

Please send your solutions to problems in this Skoliad by 1 July, 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 for this number of the Skoliad is the 27th New Brunswi k Mathemati s Competition, 2009, Grade 9, Part C. We thank Daryl Tingley, Department of Mathemati s and Statisti s, University of New Brunswi k, and Paul Deguire, Departement  de mathematiques  et de statistique, Fa ulte des s ien es, Universite de Mon ton, for providing us with this ontest and for permission to publish it. Complete justi ation is required in order for a solver to obtain redit for his/her solution. th

27

New Brunswi k Mathemati s Competition, 2009 Grade 9, Part C Approximately 30 minutes allowed

. If you write all integers from 1 to 100, how many even digits will be written? (When you write the number 42, two even digits are written.) (A) 50 (B) 71 (C) 80 (D) 89 (E) 91

1

. In a farm there are hens (no hump, two legs), amels (two humps, four legs) and dromedaries (one hump, four legs). If the number of legs is four times the number of humps, then the number of hens divided by the number of amels will be? (A) 12 (B) 1 (C) 32 (D) 2 (E) Not enough information 2

. A ubi box of side 1 m is pla ed on the oor. A se ond ubi box of side m is pla ed on top of the rst box so that the entre of the se ond box is dire tly above the entre of the rst box. A painter paints all of the surfa e area of the two boxes that an be rea hed without moving the boxes. What is the total area of surfa e that is painted?

3

2 3

(A) 4

49 9

m2

(B)

57 9

m2

(C)

61 9

. What is the ones digit of 22009 ? (A) 0 (B) 2 (C) 4

m2

(D)

72 9

(D) 6

m2

(E) None of these (E) 8

2 . The numbers 1, 2, 3, 4, 5, and 6 are to be arranged in a row. In how many ways an this be done if 2 is always to the left of 4, and 4 is always to the left of 6? (For example 2, 5, 3, 4, 6, 1 is an arrangement with 2 to the left of 4 and 4 to the left of 6.) (A) 20 (B) 36 (C) 60 (D) 120 (E) 240 5

. The square ABCD is ins ribed in a ir le with diameter BD of length 2. If AB is the diameter of the semi ir le on top of the square, what is the area of the shaded region? 6

(A) 4 −4 π (D) 1

2 (B) π − (C) 4 (E) Not enough information e

27

1 2

.............................. ......... ..... ..... .... . . . . .... ............................................. ..... . . A ........................................................................................ B ....... ......... ..... .. ... ... ... .... .... ..... . . .... .... . ... . . ... .... ..... ... .. ... .. .... . . . ..... ..... ... .... . . . . . .. .. ... .. .... 2 .. . . ... ... . . . ... ... . . ... ... . . ... ... ....... ... ... . . . .... ... .... ............................................................................... . ...... . . . D ..................................................... C

Con ours de Mathematiques  du Nouveau-Brunswi k, 2009 e

9

annee,  Partie C

Duree  : environ 30 minutes

. Si vous e rivez  tous les entiers de 1 a 100, ombien de hires pairs seront e rits  ? (Quand vous e rivez  le nombre 42, vous e rivez  deux hires pairs.) (B) 71 (C) 80 (D) 89 (E) 91 (A) 50

1

. Dans une ferme il y a des poules (pas de bosse, deux pattes), des hameaux (deux bosses, quatre pattes) et des dromadaires (une bosse, quatre pattes). Si le nombre de pattes est quatre fois le nombre de bosses, alors le nombre de poules, divise par le nombre de hameaux sera de ? 2

(A)

1 2

(B) 1

(C)

3 2

(D) 2

(E) Information insuÆsante

. Une bo^te ubique dont le ot ^ e mesure 1 m est pla ee  sur le sol. Une se onde bo^te ubique dont le ot ^ e mesure 23 m est pla ee  sur la premiere  de maniere  a e que son entre soit exa tement au dessus du entre de la premiere  bo^te. Un peintre peint alors les surfa es des deux bo^tes qu'il peut rejoindre sans bouger les bo^tes. Quelle est la surfa e totale qui est peinte ? 3

(A) 4

49 9

m2

(B)

57 9

m2

(C)

61 9

m2

. Quel est le hire des unites  de 22009 ? (A) 0 (B) 2 (C) 4

(D)

72 9

(D) 6

m2

(E) Au une de es reponses  (E) 8

3 5. Les nombres 1, 2, 3, 4, 5 et 6 sont pla es  en ligne. De ombien de fa ons

ela peut-il e^ tre fait si 2 est toujours a la gau he de 4 et 4 est toujours a la gau he de 6 ? (2, 5, 3, 4, 6, 1 est un tel pla ement ave 2 a la gau he de 4 et 4 a la gau he de 6). (B) 36 (C) 60 (D) 120 (E) 240 (A) 20

. Le arre ABCD est ins rit dans un er le dont le diametre  BD est de  du longueur 2. Si AB est le diametre demi- er le au-dessus du arre,  quelle est l'aire de la region  ombragee  ? 6

(A) 4 −4 π (D) 1

2 (B) π − (C) 4 (E) Information insuÆsante

1 2

......................... ........... ..... ..... .... . . . .. .. ..... ......................................... ..... . . . . . . . . . . . . . . . . . A ........................................................................... B .. ......... ....... ..... .. ... ... .. .... .... .... . . ... .... . ... . . .... ..... ..... ... .. ... .... .... . . . ..... ..... .. .. . .... .. .. ... .. . . . . .. .. 2 . ... ... . . . ... .. . . ... ... . . ... ... ....... ... ... . . . . .... .. .... ................................................................................. . ...... . . .. D ...................................................... C

Next follow solutions to the Swedish Junior High S hool Mathemati s Contest, Final Round, 2007/2008 [2009 : 129{131℄. 1. Values are assigned to a number of ir les, and these values are written in the ir les. When two or more ir les overlap, the sum of the values of the overlapping ir les is written in the ommon region. In the example on the left below, the three ir les have been assigned the values 1, 3, and 8. Where the ir le with value 1 overlaps the ir le with value 3 we write 4 (= 1 + 3). In the region in the middle, we add all three values and write 12. .............................................................................................. ........ ...... ..... ..... ..... .... ..... ....... .... .... . . ... . .... ... ... ... ... . . ... . . .. ... ... .. .. ... .. ... ... .. .. .... ... ... ... ................................... .... ... .......... ........ .. ... . . . . . ...... ... ... .... .... . . . . . . . . .. . ..... . .. ... . . . . . ... . . ... ... ... ... ... ... ... .... .... .... .. ..... .... .... ... ..... .... .. .... ....... ......... ........ ............ . . . ................. . . . . . . ........................... ....................................... ... ... ... .. .. ... .. .. .. ... .. . ... . .. .... .... ..... ..... ...... ....... .......... .....................................

1

3

4

9

12

8

11

.............................................................................................. ........ ...... ..... ..... ..... .... ..... ....... .... .... . . ... . .... ... ... ... ... . . ... . . .. ... ... .. .. ... .. ... ... .. .. .... ... ... . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ......... ... ........................... ... ................ ... . . . . . . .. ..... .. ..... .. ... . ... ....... . . . . . . . . . . .. ... .... .. .... .. .. ..... ...... . . ..... . . .... .... .. .. ... ... .... .. ... .. .... ? .... ... .... ... .. .... .. ........ ....... ... .... .... .. ......... ... ........... ..... . . ... .. . .......... . . .... . . . . . . . .. ................................... ........................................ ... . . ... ... .. .. .. .. .. .. .. . . .. . . .. . . ... . . . . . ... ... . .... ... .... ... .... .... .... .... ..... ....... .... ... . .. ........ ..... ............... .......................................... ...................... ........... ..........

In the gure on the right above are four ir les and, thus, thirteen regions. Find the number in the middle if the sum of all thirteen numbers is 294. Solution by Alison Tam, student, Burnaby South Se ondary S hool, Burnaby, BC. Assign the values a, b, c, and d to the four ir les, as shown in the diagram on the following page. Then the overlapping regions get the values shown. The sum of the thirteen regions is 7a + 7b + 7c + 7d. Sin e this

4

c

b+

+

d

a+d

c+

d

+

b+

a+b+ c+d

d

a

+

c+

c

a

a

b+c

. This is the 20th edition of the Swedish Junior High S hool Mathemati s Contest. The rst quali ation round was held in the fall of 1988, and this year's nal is held in2008. That is twenty-one alendar years, 1988{2008, but the table at right has room for only eighteen of them. Whi h three must be omitted if the digit sum in every row and every olumn must be divisible by 9? (Two solutions exist.) 2

a+b

b

b+

Also solved by CINDY CHEN, student, Burnaby North Se ondary S hool, Burnaby, BC; EMILY HUANG, student, Burnaby Central Se ondary S hool, Burnaby, BC; JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; LENA CHOI, student,  E ole Banting Middle S hool, Coquitlam, BC; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; and THOMAS HSU, student, Mos rop Se ondary S hool, Burnaby, BC.

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

a

sum is required to be 294, it follows that the value in the middle region is 294 a+b+c+d= = 42. 7

c+d

d

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

Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. The digits in the rst olumns of ea h of the two parts of the hart are 1's and 2's. Say one su h olumn has k opies of 1 and 9 − k opies of 2. Then the sum is k + 2(9 − k) = 18 − k. If this sum is a multiple of 9, then k is a multiple of 9, so either k = 0 or k = 9. Thus, one rst olumn ontains only 1's while the other rst olumn ontains only 2's. Sin e only nine years begin with 2, all of them must be used and they must all be in the same part of the hart. See the left-hand diagram below. .............................................. .............................................. ......1............9............................... ......2............0.............0............0...... .. 1 ..... 9 ..... ..... .. ... 2 ..... 0 ..... 0 ..... 1 .. .............................................. .............................................. ... 1 .... 9 .... .... . ... 2 .... 0 .... 0 .... 2 . .................................................. .................................................. ......1...........9............................ ......2...........0............0...........3..... ......1.............9.................................. ......2.............0..............0.............4....... ... .... .... .... .. ... .... .... .... .. ......1...........9............................. ......2...........0............0...........5...... ... 1 .... 9 .... .... . ... 2 .... 0 .... 0 .... 6 . ............................................... ............................................... ......1...........9............................ ......2...........0............0...........7..... ......1............9................................ ......2............0.............0............8.......

.............................................. .............................................. ......1............9.............9............6...... ......2............0.............0............0...... .. 1 ..... 9 ..... 9 ..... 5 .. ... 2 ..... 0 ..... 0 ..... 1 .. .............................................. .............................................. ... 1 .... 9 .... 9 .... 4 . ... 2 .... 0 .... 0 .... 2 . .................................................. .................................................. ......1...........9............9...........3..... ......2...........0............0...........3..... ......1.............9..............9.............2....... ......2.............0..............0.............4....... ... .... .... .... .. ... .... .... .... .. ......1...........9............9...........1...... ......2...........0............0...........5...... ... 1 .... 9 .... 9 .... x. ... 2 .... 0 .... 0 .... 6 . ............................................... ............................................... ......1...........9............9...........8..... ......2...........0............0...........7..... ......1............9.............9............7....... ......2............0.............0............8.......

Here x is either 0 or 9

The third olumn in the 1900's part of the hart ontains 8's and 9's.

5 Sin e there are at most two 8's and the olumn sum is divisible by 9, the

olumn must onsist of only 9's. Thus the years 1988 and 1989 must be ex luded from the hart. Using that the row sums are divisible by 9, it is now easy to ll in the hart as in the right-hand diagram. Hen e the ex luded years are either 1988, 1989, and 1990; or 1988, 1989 and 1999.  Also solved by LENA CHOI, student, E ole Banting Middle S hool, Coquitlam, BC.

. The line segments DE , CE , BF , and CF divide the re tangle ABCD into a number of smaller regions. Four of these, two triangles and two quadrilaterals, are shaded in the gure at right. The areas of the four shaded regions are 9, 35, 6, and x (see the gure). Determine the value of x. 3

B ................................................................................................................................................... C E

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

9

x

35

A

6

F

D

 Solution by Lena Choi, student, E ole Banting Middle S hool, Coquitlam, BC. Sin e the base of △BCF is |BC| and its height is |CD|, the area of B C △BCF is half the area of re tana gle ABCD. But then △ABF and 9 △CDF take up the other half of E x b the re tangle. Similarly, the area of △CDE equals half the area of the d re tangle as does the ombined area of 35 c △BCE and △ADE . 6 Assign letters to the areas of the A D F four unshaded regions as in the gure. Then the paragraph above amounts to saying that a+x+c, 9+b+35+d+6, b+x+d, and 9+a+35+c+6 are all equal. In parti ular, a + x + c = 9 + a + 35 + c + 6, so x = 9 + 35 + 6 = 50. ............................................................................................................................................................... ... .... ..... . ............. ......... .... ...... ............. . .... ... .............. .. ... .... .............. . . ... . . .. ... . . . . . . . . .... ...... . .. .. ... ...................... .... .. ... .. .... . .... ................................ . ... .... .............. .. . .... .. .... ................ ..... ... .. ......... .... .... .. ......... ...... .. . ......... .... .... .... ......... .... .. ... . ... . ............ . .... . ... ........... .. ......... .... .... . . ............. . ... . ... .... .......... .... .......... .... .... .... ........... .... .... .... .......... .... . ... ... . .... .. .......... ... . . . . . . ... ... . ......................................................................................................................................................................

Also solved by ALISON TAM, student, Burnaby South Se ondary S hool, Burnaby, BC; CINDY CHEN, student, Burnaby North Se ondary S hool, Burnaby, BC; JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; and NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia. 4. A goody bag ontains a two-digit number of goodies. Lisa adds the two digits and then removes as many goodies as the sum yields. Lisa repeats this pro edure until the number of goodies left is a single digit number larger than zero. Find this single digit number.

Solution by the editors. Say the number of goodies is 10a + b. Then Lisa removes a + b goodies and is left with 9a goodies. It follows that on e Lisa has removed goodies at

6 least on e, then the number of goodies left is divisible by nine. Thus, when the number of goodies left rea hes a single-digit number, that number must be nine. Several solvers found that Lisa is left with nine goodies with several hoi es of the initial number of goodies. However, that does not prove that Lisa always ends up with nine goodies.

. In how many ways an the list [1, 2, 3, 4, 5, 6] be permuted if the produ t of neighbouring numbers must always be even?

5

Solution by Natalia Desy, student, SMA Xaverius 1, Palembang, Indonesia. If the produ t of neighbouring numbers is even, then the parity of the numbers must follow one of the patterns eoeoeo , oeoeoe , oeoeeo , or oeeoeo , where e is an even number and o is an odd number. On e you have hosen one of the four patterns, you an arrange the three even numbers into the even slots in six ways, and you an arrange the odd numbers in six ways. Thus the total number of permutations is 4 · 6 · 6 = 144. Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.

. The digits of a ve-digit number are abcde. Prove that abcde is divisible by 7 if and only if the number abcd − 2 · e is divisible by 7. 6

Solution by the editors. Let A and B denote the two numbers abcde and abcd − 2e, respe tively. Let x denote the four-digit number abcd. Then A = 10x + e and B = x − 2e. Therefore, 2A + B = 20x + 2e + (x − 2e) = 21x . If A is divisible by 7, then A = 7m for some integer m, so B = 21x − 2A = 21x − 14m = 7(3x − 2m) , whi h is divisible by 7. If B is divisible by 7, then B = 7n for some integer n, so 2A = 21x − B = 21x − 7n = 7(3x − n) , whi h is divisible by 7. But if 2A is divisible by 7, then so is A. Thus, A is divisible by 7 if and only if B is divisible by 7. This issue's prize of one opy of CRUX with MAYHEM for the best solutions goes to Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. We would very mu h appre iate re eiving more solutions from our readers. Solutions to just some of the problems are also very wel ome.

7

MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by . 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). High S hool and University Students

Mayhem Problems Please send your solutions to the problems in this edition by 1 May 2010. Solutions re eived after this date will only be onsidered if there is time before publi ation of the solutions. The Mayhem Sta ask that ea h solution be submitted on a separate page and that the solver's name and onta t information be in luded with ea h solution. Ea h problem is given in English and Fren h, the oÆ ial languages of Canada. In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8, Fren h will pre ede English. The editor thanks Jean-Mar Terrier of the University of Montreal for translations of the problems.

. Proposed by the Mayhem Sta. Riley is a poor starving university student, but is mathemati ally astute. He noti es that ve suppers in residen e ost the same as seven lun hes. After one week of skipping supper most nights, he noti es that ve lun hes and one supper ost $48 in total. How mu h do 16 suppers ost? M420

. Proposed by Ne ulai Stan iu, George Emil Palade Se ondary S hool, Buzau, Romania. Let ⌊x⌋ be the greatest integer less than or equal to the real number x. Determine all real numbers x su h that M421

› ž

1

x

› ž

+

3

x

= 4.

M422. Proposed by Adnan Arapovi , student, University of Waterloo, Waterloo, ON. Prove that n X k(k + 1) k=1

2

=

n(n + 1)(n + 2) 6

.

8 M423. Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. The tens digit of a perfe t square S is three greater than the ones digit of S . Determine all possible remainders when S is divided by 100.

. Proposed by Margo Kondratieva, Memorial University of Newfoundland, St. John's, NL. In the diagram, line segments AB , A ........................................................ B ... ...... CDE , and F GH are parallel. Also, line .. ... ... ... ... ... ... segments ACF and BDG are perpendi ular ... ...... .. ... .... ... ... ... to AB . Suppose that the area of re tangle ... ... .. E .... C D . ................................................................................................ ABDC is x, the area of re tangle CDGF is . ... .... . . . y , and the area of △BDE is z . Determine ..................................................................................................... F H the area of DEHG in terms of x, y, and z . G M424

. Proposed by Titu Zvonaru, Comane  sti, Romania. In △ABC , ∠BAC = 90◦ and I is the in entre. The interior bise tor of angle C meets AB at D. The line segment through D perpendi ular to BI interse ts BC at E . The line segment through D parallel to BI meets AC at F . Prove that E , I , and F are ollinear. M425

.................................................................  . Propose par l'Equipe de Mayhem. Ri hard est un etudiant  pauvre et aame,  mais mathematiquement  doue.  Il a remarque qu'a la residen e, 

inq soupers outent ^ le m^eme prix que sept lun hs. Apres  avoir saute les soupers presque tous les soirs pen48 au dant une semaine, il onstate que inq lun hs et un souper outent ^ total. Combien outent ^ 16 soupers ? M420

 . Propose par Ne ulai Stan iu, E ole se ondaire George Emil Palade, Buzau, Roumanie. Soit ⌊x⌋ le plus grand entier plus petit ou egal  au nombre reel  x. Trouver tous les nombres reels  tels que M421

› ž

› ž

1 3 + x x

= 4.

. Propose par Adnan Arapovi , etudiant,  Universite de Waterloo, Waterloo, ON. Montrer que

M422

n X k(k + 1) k=1

2

=

n(n + 1)(n + 2) 6

.

9 M423. Propose  par John Grant M Loughlin, Universite du NouveauBrunswi k, Frederi ton, NB. La dieren e  entre le hire des dizaines et elui des unites  d'un arre parfait S est de trois. Trouver tous les restes de la division de S par 100.

. Propose par Margo Kondratieva, Universite Memorial de TerreNeuve, St. John's, NL. Dans la gure i- ontre, les segments de droite AB , CDE , et F GH sont pa- A .................................................................B ... .... ...... ... ralleles.  De plus, les segments ACF et ... ... ...... .. ... ... BDG sont perpendi ulaires a AB . Suppo... .... ... ... . ... .. E sons que les aires respe tives des re tangles C .................................................D .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................. ... . . ... ... . ABDC , CDGF , et △BDE sont x, y et z . . ... . .. Trouver l'aire de DEHG en fon tion de x, F ................................................................................................. H G y et z . M424

. Propose par Titu Zvonaru, Comane  sti, Roumanie. ◦ Dans le triangle ABC , ∠BAC = 90 et soit I le entre du er le ins rit. La bisse tri e interieure  de l'angle C oupe AB en D. La droite passant par D et perpendi ulaire a BI oupe BC en E . La droite passant par D et parallele  a BI oupe AC en F . Montrer que E , I et F sont olineaires.  M425

Mayhem Solutions Last year we re eived some late solutions that did not appear in the De ember issue. We therefore a knowledge a orre t solution to M383 by Mridul Singh, student, Kendriya Vidyalaya S hool, Shillong, India, and orre t solutions to problems M383, M384, and M386 by Hugo Luyo San hez,  Ponti ia Universidad Catoli a  del Peru, Lima, Peru. . Proposed by Kyle Sampson, St. John's, NL. A sequen e is generated by listing (from smallest to largest) for ea h positive integer n the multiples of n up to and in luding n2 . Thus, the sequen e begins 1, 2, 4, 3, 6, 9, 4, 8, 12, 16, 5, 10, 15, 20, 25, 6, 12, . . . . Determine the 2009th term in the sequen e. M388

Solution by Kristof  Huszar,  Valeria  Ko h Grammar S hool, Pe s,  Hungary. First, we noti e that there are k positive integral multiples of k less than or equal to k2 . If we group the terms of the sequen e as the multiples of 1, then the multiples of 2, then the multiples 3, and so on, we noti e that the groups have 1 term, then 2 terms, then 3 terms, and so on.

10 If n is a positive integer, then the sum of the rst n positive integers‹is  th

equal to 1 + 2 + 3 + · · · + n = n(n2+ 1) . Therefore, n2 is the n(n2+ 1) term of the sequen e. th  Hen e, 632 = 3969 is the 63 2· 64 = 2016th term. Sin e 3969 o

urs 2016 − 2009 = 7 terms after the 2009th term, we nd that the 2009th term is 632 − 7(63) = 3528.

Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina ; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB ; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia ; ANTONIO GODOY  TOHARIA, Madrid, Spain ; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA ; RICARD PEIRO, IES \Abastos", Valen ia, Spain ; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON ; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. One in orre t solution was submitted.

. Proposed by Lino Demasi, student, Simon Fraser University, Burnaby, BC. There are 2009 students and ea h has a ard with a dierent positive integer on it. If the sum of the numbers on these ards is 2020049, what are the possible values for the median of the numbers on the ards ? M389

Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Let a be the smallest of the numbers on the ards. The smallest possible sum of the 2009 numbers on the ards is then S

= a + (a + 1) + (a + 2) + · · · + (a + 2008) = 2009a +

1 (2008)(2009) = 2009a + 2017036 . 2

If a ≥ 2 then S ≥ 2021054 > 2020049. Therefore, a = 1. Next, onsider the sequen e of numbers 1, 2, 3, . . . , 2009. The sum of these numbers is 2019045 (whi h is 1004 less than the desired sum of 2020049). Also, their median is 1005. In order to obtain the desired sum of 2020049, some of the numbers in this sequen e need to be in reased. When a ertain term in the sequen e is in reased, then every greater term must be in reased as well in order for the terms of the sequen e to remain distin t. If a term that is less than 1006 is in reased, then every larger term will also have to in rease, resulting in an in rease of the initial sum 2019045 by at least 1005 (sin e at least 1005 terms are in reased), yielding a new sum of at least 2019045 + 1005 = 2020050, whi h is too large. Therefore, only terms that are 1006 or greater may be in reased. When only terms greater than or equal to 1006 are in reased, then the median 1005 remains un hanged. Thus, 1005 is the only possible value for the median.

11 Also solved by NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia ;  IES \Abastos", RICHARD I. HESS, Ran ho Palos Verdes, CA, USA ; RICARD PEIRO, Valen ia, Spain ; and EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON. There was one in orre t solution submitted.

. Proposed by Ne ulai Stan iu, George Emil Palade Se ondary S hool, Buzau, Romania. A Pythagorean triangle is a right-angled triangle with all three sides of integer length. Let a and b be the legs of a Pythagorean triangle and let h be the altitude to the hypotenuse. Determine all su h triangles for whi h M390

1 a

+

1 b

+

1 h

= 1.

Solution by D.J. Smeenk, Zaltbommel, the Netherlands. Let A be the area of the triangle and c the length of its hypotenuse. Then A equals both 12 ab and 12 ch, and so ab = ch. Also, 1 = =

1 1 1 bh + ah + ab ah + bh + ch + + = = a b h abh abh (a + b + c)h a+b+c = , abh ab

hen e ab = a + b + c. √ By the Pythagorean Theorem, c = a2 + b2 . Sin e a + b + c = ab, we then obtain the equivalent equations p ab = a + b + a2 + b2 , p ab − a − b = a2 + b2 , 2 2 (ab − a − b) = a + b2 , a2 b2 + a2 + b2 − 2a2 b − 2ab2 + 2ab = a2 + b2 , a2 b2 − 2a2 b − 2ab2 + 2ab = 0 , ab(ab − 2a − 2b + 2) = 0 . Sin e ab > 0, then ab −2a − 2b+ 2 = 0, or b(a − 2) = 2a − 2, whi h implies −2 2 that b = 2a =2+ . a−2 a−2 Sin e a and b are positive integers, then a − 2 is a positive divisor of 2 ; that is, a − 2 = 2 or a − 2 = 1. If a − 2 = 2, then a = 4, b = 3, and c = 5. If a − 2 = 1, then a = 3, b = 4, and c = 5. There is therefore just one Pythagorean triangle for whi h a1 + 1b + h1 = 1, namely the triangle with legs 3 and 4, and hypotenuse 5. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina ; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam ; ANTONIO GODOY TOHARIA, Madrid, Spain ; RICHARD I. HESS, Ran ho Palos  HUSZAR,  Verdes, CA, USA ; KRISTOF Valeria  Ko h Grammar S hool, Pe s,  Hungary ;  IES \Abastos", Valen ia, Spain ; EDWARD T.H. WANG, Wilfrid Laurier RICARD PEIRO,

12 University, Waterloo, ON ; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA.

. Proposed by Ne ulai Stan iu, George Emil Palade Se ondary S hool, Buzau, Romania. M391

Determine all pairs (a, b) of positive integers for whi h both b+2 a

are positive integers.

a+1 b

and

Solution by Edin Ajanovi , student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina, modi ed by the editor. 1 2 Let x = a + and y = b + , where x and y are positive integers. b a Rearranging, we obtain a = bx − 1 and b = ay − 2. Substituting for a in the se ond equation, we obtain b = (bx − 1)y − 2 y+2 and so y + 2 = bxy − b or b = xy . −1

2 y−1+3 3 . Sin e b and y are If x = 1, then b = yy + = = 1+ −1 y−1 y−1 positive integers, then y = 2 or y = 4 (giving b = 4 and b = 2, respe tively). y+2 If x = 2, then b = 2y . Sin e b is a positive integer, then we must −1 have y + 2 ≥ 2y − 1 and so y ≤ 3. Che king y = 1, y = 2, and y = 3 shows that y = 1 and y = 3 give positive integer values for b (namely b = 3 and b = 1, respe tively). y+2 If x = 3, then b = 3y . Sin e b is a positive integer, then we must −1

have y + 2 ≥ 3y − 1 and so y ≤ 32 . The only possible value of y is y = 1, whi h does not give an integer value for b. y+2 If x = 4, then b = 4y . Sin e b is a positive integer, then we must −1 have y + 2 ≥ 4y − 1 and so y ≤ 1. If y = 1, then b = 1. If x ≥ 5, then xy − 1 ≥ 5y − 1 > y + 2 for all positive integers y, so y+2 b=

annot be a positive integer. xy − 1 We nish by al ulating the values of a that go with the values of b to obtain the pairs (a, b) = (3, 4), (1, 2), (5, 3), (1, 1), (3, 1). Also solved by ANTONIO GODOY TOHARIA, Madrid, Spain ; RICHARD I. HESS, Ran ho  HUSZAR,  Valeria Palos Verdes, CA, USA ; KRISTOF  Ko h Grammar S hool, Pe s,  Hungary ;  IES \Abastos", Valen ia, Spain ; JOSE JAIME SAN JUAN CASTELLANOS, RICARD PEIRO, student, Universidad te hnologi a  de la Mixte a, Oaxa a, Mexi o ; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON ; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA.

. Proposed by the Mayhem Sta. Determine, with justi ation, the fra tion pq , where p and q are positive

M392

integers and q < 1000, that is losest to, but not equal to,

19 . 72

13 Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON, modi ed by the editor. In order to nd the desired fra tion, we need to minimize the value of p 19 |72p − 19q| − = q 72 72q

,

where p and q are positive integers and q < 1000. To do this, we attempt to make the numerator of this dieren e as small as possible, while at the same time keeping the denominator as large as possible, hen e by making q as large as possible. To minimize the numerator, we try to make 72p−19q equal to 1 or −1. Consider 72p − 19q modulo 19. Sin e 72 ≡ −4 (mod 19), then 72p − 19q ≡ −4p (mod 19), so we try to nd p su h that −4p ≡ ±1 (mod 19). Solving this ongruen e, we obtain p ≡ 14 (mod 19), or p ≡ 5 (mod 19), and so p = 14 + 19k for some integer k ≥ 0 or p = 5 + 19k for some integer k ≥ 0. In the rst ase, 72p − 19q = 1, so q = 72p19− 1 = 53 + 72k ; sin e q < 1000, then k ≤ 13 ; when k = 13, q = 989. In the se ond ase, 72p + 1 72p − 19q = −1, so q = = 19 + 72k ; sin e q < 1000, then k ≤ 13 ; 19 when k = 13, q = 955. 1 In either of these ases, the dieren e is equal to 72q , and so is minimized when q is maximized. Thus, in these ases, the minimum possible dieren e o

urs when q is as large as possible, or q = 989 (and so k = 13 and p = 261). This dieren e is 72 ·1989 . It is not possible to a hieve a smaller dieren e when |72p − 19q| ≥ 2 and q < 1000, sin e this dieren e would always be at least 72 · 21000 whi h is larger than the dieren e that we have already found. under the given Therefore, the fra tion losest to, but not equal to 19 72 p 261

onditions is q = 989 .  Also solved by RICHARD I. HESS, Ran ho Palos Verdes, CA, USA ; and KRISTOF  Valeria HUSZAR,  Ko h Grammar S hool, Pe s,  Hungary ; One in orre t solution was also submitted. A very similar problem appeared in the 2006 Canadian Open Mathemati s Challenge (problem 4(a), Part B).

. Proposed by the Mayhem Sta. Inside a large ir le of radius r two smaller ir les of radii a and b are drawn, as shown, so that the smaller ir les are tangent to the larger ir le at P and Q. The smaller ir les interse t at S and T . If P , S , and Q are ollinear (that is, they lie on the same straight line), prove that r = a + b. M393

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

T

P

S

Q

14 Solution by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain and Georey A. Kandall, Hamden, CT, USA (independently). Let O be the entre of the large ir le of radius r. Let O1 be the entre of the smaller ir le of radius a tangent to the large ir le at point P , and let O2 be the

entre of the smaller ir le of radius b tanO gent to the large ir le at point Q. T Sin e the ir les entred at O1 and O1 O2 are tangent to the large ir le, then O , O2 O1 , P are ollinear, as are O , O2 , Q. Triangle OP Q is isos eles with P S Q OP = OQ, triangle O1 P S is isos eles with O1 P = O1 S , and triangle O2 QS is isos eles with O2 Q = O2 S (sin e ea h of these triangles has two radii of one of the ir les as sides). Therefore, ∠OP Q = ∠OQP , ∠O1 P S = ∠O1 SP , and ∠O2 QS = ∠O2 SQ. Sin e P , S , and Q are ollinear, then ... ....................... ................................. .......... ......... ......... ....... ...... ....... . . . . ..... .. .... ..... ... .... . . ... .. . . ... ... ... . . ... . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... .. . ....... . . . . ... . . . . . . ... . ..... .... . .. . . . . ... . .... .. .. . . . ... ... . . ... ... .. .... . . ... .... ... ... . ... . .. .. .... ... ... . ... . . . .... .......................... .. ... .. .. . . . . . . . . . . . . . . . . ... .. ..... . .. ........ ... . . . . . . . .. .. . . . ... . ..... .. ... ... ...... .... ... .. .. .. .... ... .. .... .. .. .. ..... .... ...... ... ...... .. .... .... .. ... ...... .. .. ........... .... .. .... ..... . . . . . . . . . . ... ....... . .. .. .... ...... ....... ...... .. ... ..... .. .... .... ......... ................................................................ .. .... ................................................................... ...... ....... ... ......... .................................................................. . . . . . . ................... . ............................ ....................................... ....... ............. .................................................

∠P SO1 = ∠O1 P S = ∠OP Q = ∠P QO ,

whi h tells us that O1 S and OQ are parallel. Similarly, ∠QSO2 = ∠O2 QS = ∠OQP = ∠QP O ,

whi h tells us that O2 S and OP are parallel. Therefore, quadrilateral OO1 SO2 is a parallelogram. Thus, OO1 = SO2 . But SO2 = b and OO1 = OP − O1 P = r − a, and so r − a = b, or r = a + b.

Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina ; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long,  HUSZAR,  Vietnam ; ANTONIO GODOY TOHARIA, Madrid, Spain ; KRISTOF Valeria  Ko h  IES \Abastos", Valen ia, Spain ; Grammar S hool, Pe s,  Hungary ; RICARD PEIRO, D.J. SMEENK, Zaltbommel, the Netherlands ; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.

Problem of the Month Ian VanderBurgh

To start the new year of Problems of the Month, we'll look at a problem that relies on a on ept that we learn early on { addition { but requires us to think in some fairly deep ways to ome up with a omplete solution and total understanding of what is going on.

15 Sin e we've had a short break sin e the last issue, we should start with a warm-up problem. Your task is to ompute the following sum : 2 + 22 + 222 + 2222 + 22222 + 222222 + 2222222 . But before you start, there are two rules : no al ulators are allowed, and you have to ompute the sum aloud. If you're as out of pra ti e on this sort of thing as some of us are, this isn't all that easy. We an rewrite the sum rst in a form that makes it easier to ompute :

+

2

2 2 2 2 2

2 2 2 2

2 2 2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2

Here's an attempt to do this in words : The sum in the units' olumn is 14. Put down the 4 ; arry the 1. The sum in the tens' olumn is 12 plus the arry of 1 gives 13. Put down the 3 ; arry the 1. The sum in the hundreds' olumn is 10 plus the arry of 1 gives 11. Put down the 1 ; arry the 1. The sum in the thousands' olumn is 8 plus the arry of 1 gives 9. Put down the 9 ; no arry. The sum in the ten thousands' olumn is 6. Put down the 6. The sum in the hundred thousands' olumn is 4. Put down the 4. The sum in the millions' olumn is 2. Put down the 2. The nal sum is thus 2469134. That's a bit of a workout, isn't it ? We should larify the role of the digit and

arry. If the sum in a olumn is 14, we write this as 14 = 10(1) + 4 ; the units digit (the 4) is the digit that we write down, while the quotient when dividing by 10 (the 1) is the arry. (The units digit is also the remainder when we divide the sum by 10.) Let's have a look at our Problem of the Month, then. Problem

(2009 Fryer Contest)

The addition shown below, representing

2+22+222+2222+· · · , has 101 rows and the last term onsists of 101 2's :

2

2 2

2 2

2 2

2 2

2 2

··· C

B

A

.. .

+ 2

2 2 2 2

··· ···

2 2 2 2

2 2 2

.. .

16 (a) Determine the value of the ones digit A . (b) Determine the value of the tens digit B and the value of the hundreds digit C . ( ) Determine the middle digit of the sum. This problem looks pretty s ary at rst glan e. Despite this, at least (a) and (b) an be answered exa tly as in our warm-up problem. Let's do these parts and then think a bit about part ( ). We pro eed exa tly as we did above. The units' olumn

onsists of 101 opies of the digit 2. Therefore, the sum in the units' olumn is 101 × 2 = 202. We put down the 2 and arry 20. The tens' olumn onsists of 100 opies of the digit 2 plus the arry of 20. Therefore, the sum in the tens' olumn is 100 × 2 + 20 = 220. We put down the 0 and arry 22. The hundreds' olumn onsists of 99 opies of the digit 2 plus the arry of 22. Therefore, the sum in the hundreds' olumn is 99 × 2 + 22 = 220. We put down the 0 and arry 22. We an stop at this point, sin e we have determined the hundreds, tens, and units digits of the sum. Therefore, A = 2, B = 0, and C = 0. Solution to (a) and (b)

Great { that was mu h less s ary than it looked like it ould be. Now we need to try to ta kle ( ), whi h a tually is quite s ary. One approa h would be to pro eed by \brute for e" and work our way systemati ally olumn by olumn from the units' olumn towards the left. Of

ourse, we don't need to go all of the way to the leftmost olumn, sin e we

an stop when we get to the middle digit of the sum. Whi h digit will this be ? In order to answer this, we need to know how many digits the nal sum has. How many digits do you think that it has ? My best guess is 101 digits, sin e it seems pretty unlikely that the single 2 in the leftmost olumn is going to have enough of a arry from the olumn to its right to reate two-digit sum in this leftmost olumn. How do we know for sure that this is orre t ? If we knew this for sure, then the middle digit would be the 51st digit, sin e there would be 50 digits to its left and 50 digits to its right. Now, this 51st olumn onsists of 51 opies of the digit 2, so its sum is 102 plus whatever

arry omes from the olumn to the right. The olumn to the right onsists of 52 opies of 2, so its sum is 104 plus the arry from the olumn to its right, whose sum is at least 106 (that is, 106 from the 2's plus the arry). This is getting ompli ated ! Let's try this again with a bit of agreement on our terminology. We'll denote the leftmost olumn C1 and the rightmost olumn C101 ; we label the olumns in between in the logi al way. We also use sn to represent the sum in the nth olumn, in luding the arry. We saw above that the sum of the digits in C51 is 102, in C52 is 104, and in C53 is 106. Therefore, s53 ≥ 106. (We haven't in luded any arry here from C54.) Therefore, the arry from C53 to C52 is at least 10, so

17 s52 ≥ 104 + 10 = 114. Therefore, the arry from C52 to C51 is at least 11, so s51 ≥ 102 + 11 = 113. If s51 = 113, then the middle digit is 3, and we're done. But is it a tually the ase that s51 = 113 ? Could it be bigger ? If s51 was at least 114, then the arry from C52 to C51 would be at least 114 − 102 = 12, whi h would mean that s52 ≥ 120. If s52 ≥ 120, then the arry from C53 to C52 would be at least 120 − 104 = 16, so s53 ≥ 160. If s53 ≥ 160, then arry from C54 to C53 would be at least 160 − 106 = 54, so s54 ≥ 540. If s54 ≥ 540, then the arry from C55 to C54 would be at least 540 − 108 = 432, whi h is getting just plain silly, given that in parts (a) and (b) the arries that we got from the \largest olumns" were 22 only. So it seems pretty lear that s51 should be 113, so the middle digit should be 3. Now, I don't know about you, but I'm just about onvin ed. However, I'm not sure if \the middle digit should be 3" is all that rigorous and \just plain silly" ounts as a solid mathemati al proof. So we should prove some restri tion on the arries. Let's do this, and also write out a ohesive solution to part ( ). We'll use a bit of algebrai notation to simplify things.

We label the olumns as above and let sn be the sum in the nth

olumn, in luding the arry from the (n + 1)th olumn ; we denote this arry by cn+1 . Column n onsists of n opies of the digit 2, so sn = 2n + cn+1 . From our solution to (a) and (b), we know that c101 = 20 and that c100 = c99 = 22. Let's argue that cn ≤ 22 for all n with 1 ≤ n ≤ 101. We use an informal ba kwards indu tion. Suppose that cn+1 ≤ 22. (We know that this is true for n = 98, 99, 100.) Then sn = 2n + cn+1 is at most 2(101) + 22 = 224 and so cn ≤ 22. Thus, if cn+1 ≤ 22, then cn ≤ 22. Sin e c101 ≤ 22, then we an arry this hain along to show that cn ≤ 22 for all n with 1 ≤ n ≤ 101. We an use this to show that the sum has exa tly 101 digits. For the sum to have more than 101 digits, we would need to have s1 ≥ 10. But s1 = 2+ c2 , so this would mean that c2 ≥ 8 and so s2 ≥ 80. But s2 = 4+ c3 , so this would mean that c3 ≥ 76, whi h is not possible. Therefore, the sum has exa tly 101 digits. Finally, we an determine the 51st digit. We know that s51 = 102 + c52 and s52 = 104 + c53 and s53 = 106 + c54 . Sin e 0 ≤ c54 ≤ 22, then 106 ≤ s53 ≤ 128. Thus, 10 ≤ c53 ≤ 12. Sin e 10 ≤ c53 ≤ 12, then 114 ≤ s52 ≤ 116. Thus, c52 = 11, whi h means that s51 = 113, and so the 51st (that is, the middle) digit of the sum is 3. Solution to ( )

Let's make a ouple of observations to nish this o. First, a little bit of algebra and notation helped to save us a large number of words and onvoluted explanations. Se ond, a relatively simple topi like addition gave us a problem that requires some pretty high-level thinking. To me, one of the great beauties of mathemati s is that simpli ity and omplexity an be so

ompletely interwoven.

18

THE OLYMPIAD CORNER No. 283 R.E. Woodrow

Another year, and it is time to thank the many people who have ontributed solutions, problems, omments, and advi e during 2009. Arkady Alt Pavlos Maragoudakis Miguel Amengual Covas Robert Morewood George Apostolopoulos Bill Sands S efket Arslanagi D.J. Smeenk Ri ardo Barroso Campos Daniel Tsai Mi hel Bataille George Tsapakidis Jose Luis Daz-Barrero Edward T.H. Wang Oliver Geupel Konstantine Zelator Jean-David Houle Titu Zvonaru John Grant M Loughlin Thank you to both Jill Ainsworth, who produ ed the text for the Corner for the rst four issues, and to Joanne Canape, who ontinued the rest of the issues, for their skilled and tireless eorts to translate my s ribbles and notes into a ni e presentation. To start the New Year for the Corner, we give the problems proposed but not used at the 2007 IMO in Vietnam. My thanks go to Bill Sands, Canadian Team Leader, for olle ting them for our use. 2007 IMO in VIETNAM Problems Proposed But Not Used Contributing Countries. Austria, Australia, Belgium, Bulgaria, Canada, Croatia, Cze h Republi , Estonia, Finland, Gree e, India, Indonesia, Iran, Japan, Korea (North), Korea (South), Lithuania, Luxembourg, Mexi o, Moldova, Netherlands, New Zealand, Poland, Romania, Russia, Serbia, South Afri a, Sweden, Thailand, Taiwan, Turkey, Ukraine, United Kingdom, and the United States of Ameri a

. Ha Huy Khoai, Ilya Bogdanov, Tran Nam Dung,

Problem Sele tion Committee

Le Tuan Hoa, Geza  Kos. 

Algebra

. Consider those fun tions f

A1

: N → N whi h €

satisfy the ondition Š

f (m + n) ≥ f (m) + f f (n) − 1

for all m, n ∈ N. Find all possible values of f (2007). (Here N denotes the set of positive integers.)

19 A2. Let n be a positive integer, and let x and y be positive real numbers su h that xn + yn = 1. Prove that n X 1 + x2k k=1

1+

x4k

. Find all fun tions f

A3

€

!

n X 1 + y 2k k=1

1+

: R+ → R+ Š

f x + f (y)

!

y 4k


2, and let a(1), a(2), . . . , be a sequen e of nonnegative real numbers su h that (a) a(m + n) ≤ 2a(m) + 2a(n) for all m, n ≥ 1, and A4

(b)

€

Š

a 2k ≤

1 (k + 1)c

for all k ≥ 0.

Prove that the sequen e {a(n)}∞ n=1 is bounded. . Let a1 , a2 , . . . , a100 be nonnegative real numbers satisfying the relation 12 a21 + a22 + · · · + a2100 = 1. Prove that a21 a2 + a22 a3 + · · · + a2100 a1 < . A5

25

Combinatori s

C1

. Let n > 1 be an integer. Find all sequen es a1 , a2 , . . . , an

(a) (b)

ai ∈ {0, 1}

for all 1 ≤ i ≤ n2 + n, and

2 +n

ai+1 + ai+2 + · · · + ai+n < ai+n+1 + ai+n+2 + · · · + ai+2n

all 0 ≤ i ≤ n2 − n.

su h that holds for

C2. A unit square is disse ted into n > 1 re tangles whose sides are parallel to the sides of the square. Any line, parallel to a side of the square and interse ting its interior, also interse ts the interior of some re tangle. Prove that one of the re tangles has no point on the boundary of the square.

. Determine all positive integers n for whi h the numbers in the set S = {1, 2, . . . , n} an be oloured red and blue so that S × S × S ontains exa tly 2007 ordered triples (x, y, z) with these two properties :

C3

(a) x, y, z are of the same olour ; and (b) x + y + z is divisible by n. . Let A0 = (a1 , a2 , . . . , an ) be a sequen e of real numbers. For ea h integer k ≥ 0, form a new sequen e Ak+1 from Ak = (x1 , . . . , xn ) as follows :

C4

20 (a) Choose a partition {1, 2, . . . , n}

= I∪J

P

su h that

i∈I

xi −

P



xj

j∈J

is minimized (if I or J is empty, then the orresponding sum is 0 ; if several su h partitions exist, then hoose one arbitrarily). (b) Set Ak+1 = (y1 , y2 . . . , yn ), where yi = xi + 1 if i ∈ I , and yi = xi − 1 if i ∈ J . Prove that for some k, the sequen e Ak has a term x with |x| ≥ n2 . . In the Cartesian oordinate plane let Sn = {(x, y) : n ≤ x < n + 1} for ea h integer n, and paint ea h region Sn either red or blue. Prove that any re tangle whose side lengths are distin t positive integers may be pla ed in the plane so that its verti es lie in regions of the same olour.

C5

√

. Let α < 3 −2 5 be a positive real number. Prove that there exist positive integers n and p > α2n for whi h one an sele t 2p pairwise distin t subsets S1 , . . . , Sp , T1 , . . . , Tp of the set {1, 2, . . . , n} su h that Si ∩ Tj 6= ∅ for all 1 ≤ i, j ≤ p. C6

C7. A onvex n-gon P in the plane is given. For every three verti es of P , the triangle determined by them is good if all its sides are of unit length. Prove that P has at most 2n good triangles. 3 Geometry

. An isos eles triangle ABC with AB = AC is given. The midpoint of side BC is denoted by M . Let X be a variable point on the shorter ar M A of the ir um ir le of triangle ABM . Let T be the point in the angle domain BM A, for whi h ∠T M X = 90◦ and T X = BX . Prove that ∠M T B − ∠CT M does not depend on X . G1

. The diagonals of a trapezoid ABCD interse t at point P . Point Q lies between the parallel lines BC and AD su h that ∠AQD = ∠CQB , and line CD separates points P and Q. Prove that ∠BQP = ∠DAQ.

G2

G3. Let ABC be a xed triangle, and let A1 , B1 , C1 be the midpoints of sides BC , CA, AB , respe tively. Let P be a variable point on the ir um ir le of ABC . Let lines P A1 , P B1 , P C1 meet the ir um ir le again at A′ , B ′ , C ′ respe tively. Assume that the points A, B , C , A′ , B ′ , C ′ are distin t, and that the lines AA′ , BB ′ , CC ′ form a triangle. Prove that the area of this triangle does not depend on P .

. Let ABCD be a onvex quadrilateral, and let points A1 , B1 , C1 , and lie on sides AB , BC , CD, and DA, respe tively. Consider the areas of triangles AA1 D1 , BB1 A1 , CC1 B1 , and DD1 C1 ; let S be the sum of the two smallest ones, and let S1 be the area of quadrilateral A1 B1 C1 D1 . G4

D1

21 Find the smallest positive real number k su h that kS1 the ase.

≥S

is always

. Triangle ABC is a ute with ∠ABC > ∠ACB , in entre I , and ir umradius R. Point D is the foot of the altitude from vertex A, point K lies on line AD su h that AK = 2R, and D separates A and K . Finally, lines DI and KI meet sides AC and BC at E and F , respe tively. Prove that if IE = IF , then ∠ABC > 3∠ACB . G5

. Point P lies on side AB of a onvex quadrilateral ABCD. Let ω be the in ir le of triangle CP D, and let I be its in entre. Suppose that ω is tangent to the in ir les of triangles AP D and BP C at points K and L, respe tively, Let lines AC and BD meet at E , and let lines AD and BL meet at F . Prove that points E , I , and F are ollinear. G6

Number Theory

. Find all pairs of positive integers (k, n) su h that

N1

€

Š

€

7k − 3n | k4 + n2

Š

.

. Let b, n > 1 be integers. Suppose that for ea h k > 1 there exists an € Š n integer ak su h that k | b − ak . Prove that b = An for some integer A. N2

. Let X be a set of 10, 000 integers, none of them divisible by 47. Prove that there exists a 2007-element subset Y of X su h that a − b + c − d + e is not divisible by 47 for any a, b, c, d, e ∈ Y . N3

. For every integer k ≥ 2, prove that 23k divides the number

N4



2k+1 2k





−

2k



2k−1

but 23k+1 does not. . Find all surje tive fun tions f : N → N su h that for every m, n ∈ N and every prime p, the number f (m + n) is divisible by p if and only if f (m) + f (n) is divisible by p. (N is the set of all positive integers.) N5

. For a prime p and a positive integer n, denote by νp (n) the exponent of p in the prime fa torization of n!. Given a positive integer d and a nite set of primes {p1 , p2 , . . . , pk }, show that there are in nitely many positive integers n su h that d | νp (n) for all 1 ≤ i ≤ k. N6

i

As a nal pair of ontests for your puzzling pleasure, we give two rounds of the Bundeswettbewerb Mathematik. Thanks go to Bill Sands, Canadian Team Leader to the 2007 IMO in Vietnam, for olle ting them for our use.

22 BUNDESWETTBEWERB MATHEMATIK 2006 Se ond Round

1. A ir le is divided into 2n ongruent se tors, n of them oloured bla k and n of them oloured white. Starting with an arbitrarily hosen se tor, the white se tors are numbered lo kwise from 1 to n. Subsequently, the bla k se tors are numbered ounter lo kwise from 1 to n, again starting at an arbitrary se tor. Prove that there exist n onse utive se tors ontaining all of the numbers from 1 to n.

. Let Q+ (resp. R+ ) denote the set of positive rational (resp. real) numbers. Find all fun tions f : Q+ → R+ that satisfy

2

f (x) + f (y) + 2xyf (xy) =

f (xy) f (x + y)

for all

x, y ∈ Q+ .

. The point P lies inside the a ute-angled triangle ABC and C ′ , A′ , B ′ are the feet of the perpendi ulars from P to AB , BC , CA. Find all positions of P su h that ∠BAC = ∠B ′ A′ C ′ and ∠CBA = ∠C ′ B ′ A′ . 3

4. A positive integer n is de ient if there are at most nine dierent digits in the de imal representation of n. (Leading zeroes are not ounted.) Prove that for any nite set S of de ient numbers, the sum of the re ipro als of its elements is less than 180. BUNDESWETTBEWERB MATHEMATIK 2007 First Round

. Show that one an distribute the integers from 1 to 4014 on the verti es and the midpoints of the sides of a regular 2007-gon so that the sum of the three numbers along any side is onstant.

1

. Ea h positive integer is oloured either red or green so that (a) The sum of three (not ne essarily distin t) red numbers is red. (b) The sum of three (not ne essarily distin t) green numbers is green. ( ) There is at least one green number and one red number. Find all olourings satisfying these onditions.

2

3. In triangle ABC the points E and F lie in the interiors of sides AC and BC (respe tively) so that |AE| = |BF |. Furthermore, the ir le through A, C and F and the ir le through B , C and E interse t in a point D 6= C . Prove that the line CD is the bise tor of ∠ACB .

. Let a be a positive integer. How many nonnegative integers x satisfy

4

j k j k x x = a a+1

?

23 Our solutions in the New Year begin with solutions from our readers to problems of the Bulgarian National Olympiad 2006 given in the Corner at [2008 : 409{410℄. . (Emil Kolev) Let △ABC be su h that ∠BAC = 30◦ and ∠ABC = 45◦ . −→ Consider all pairs of points X and Y su h that X is on the ray AC , Y is on − − → the ray BC , and OX = BY , where O is the ir um enter of △ABC . Prove that the perpendi ular bise tors of the segments XY pass through a xed point. Solution by Titu Zvonaru, Comane  sti, Romania. Without loss of generality, supM pose that OA = OB = OC = 1. Sin e ∠BAC = 30◦ , we have that B ... ............ ∠BOC = 60◦ and BC = 1, that is, ... .......... ........ ... ........ ... △BOC is equilateral. O ........ ... ........ . ◦ ........ . ... Sin e ∠ABC = 45 , it follows . . ........ ... ........ ... ........ that △AOC has a right angle at O, ... ........ ... ........ ◦ hen e ∠OCA = ∠OAC = 45 and ........ . . ... √ ........ . ........ . AC = 2. ........ Y .... ........ ... ..... ... If X = A, then OX = 1 and ............................................................................................................................... BY = 1, and hen e Y = C ; as a A C X result, the xed point we seek lies on the perpendi ular bise tor of AC . Let M be the point on the same side of AC as O su h that △AM C is equilateral. Then ∠OCM = 60◦ − 45◦ = 15◦ , ∠M CB = 45◦ , and by the Law of Cosines in △M CB we obtain 5

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

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

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

BM 2

= M C 2 + BC 2 − 2M C · BC cos ∠M CB √ 1 = 2 + 1 − 2 2 · √ = 1, 2

hen e △M CB has a right angle B . Let AX = α. By the Law of Cosines we have √ OX 2 = OA2 + AX 2 − 2OA · AX cos ∠OAX = 1 + α2 − α 2 , √ M X 2 = M A2 + AX 2 − 2M A · AX cos ∠M AC = 2 + α2 − α 2 ,

(1)

and by the Pythagorean Theorem we have √ M Y 2 = M B 2 + BY 2 = M B 2 + OX 2 = α2 − α 2 + 2 .

(2)

By (1) and (2) it follows that the point M belongs to the perpendi ular bise tors of the segments XY .

24 We ontinue with solutions from our readers to problems of the Indian Mathemati al Olympiad 2006 (Team Sele tion Problems) given in the Corner at [2008 : 410{412℄. . Let n be a positive integer divisible by 4. Find the number of permutations of (1, 2, 3, . . . , n) whi h satisfy the ondition σ(j) + σ−1 (j) = n + 1 for all j ∈ {1, 2, 3, . . . , n}. 1

σ

Solution by Mi hel Bataille, Rouen, Fran e. Let Sn be the set of all permutations of [n] = {1, 2, . . . , n} and let Qn

= =

{σ ∈ Sn : ∀j ∈ [n] , σ(j) + σ −1 (j) = n + 1} {σ ∈ Sn : ∀k ∈ [n] , σ ◦ σ(k) = n + 1 − k} .

We show that if n = 4m, then |Qn | = (2m)! = 2m Nm where Nm denotes m! the produ t 1 × 3 × · · · × (2m − 1) of the rst m odd natural numbers. Let σ ∈ Qn , k ∈ [n] and let j = σ(k). Then, j 6= k (otherwise k = n + 1 − k, ontradi ting the fa t that n + 1 is odd) and σ(j) σ(n + 1 − k) σ(n + 1 − j)

= = =

σ ◦ σ(k) = n + 1 − k , σ ◦ σ(j) = n + 1 − j , σ ◦ σ(n + 1 − k) = n + 1 − (n + 1 − k) = k .

Thus, the y le ontaining k is (k, j, n + 1 − k, n + 1 − j). Sin e k is arbitrary, the standard expression of σ as a produ t of disjoint y les (the y le de omposition of σ) onsists of m 4- y les of the form (k, j, n + 1 − k, n + 1 − j ). Conversely, if σ ∈ Sn has su h a y le de omposition, then learly σ ∈ Qn . Consider now the set S = {pk : k ∈ [2m]} where pk = {k, n + 1 − k} and let σ ∈ Qn . Using its y le de omposition, we asso iate with σ in a natural way a partitionS{P1 , P2 , . . . , Pm } where ea h Pk is a 2-subset of S (Pk ∩ Pl = ∅ if k 6= l ; m i=1 Pi = S ). There are Nm su h partitions (see the lemma below) and ea h of them is obtained from exa tly 2m elements of Qn , sin e any 2-subset P = {pk , pj } of S is obtained from two distin t 4- y les, namely (j, k, n + 1 − j, n + 1 − k) and (k, j, n + 1 − k, n + 1 − j). It follows that |Qn | = 2m Nm . If S is a set with |S| = 2m, then the number of partitions of S formed by 2-subsets of S is Nm = (2m − 1)(2m − 3) · · · (3)(1) = (2m)! . 2m m! Lemma

Proof. Fix s ∈ S . We partition S into 2-subsets by rst hoosing one of the 2m − 1 2-subsets {s, t}, where t ∈ S and t 6= s, and then augmenting this subset with a partition into 2-subsets of S − {s, t} (and there are Nm−1 su h partitions). Thus, Nm = (2m − 1)Nm−1 when m > 1, and the result follows sin e N1 = 1.

25 3. (Short list, IMO 2005) There are n markers, ea h with one side white and the other side bla k, aligned in a row with their white sides up. In ea h step (if possible) we pi k a marker with the white side up that is not an outermost marker, remove it, and turn over the losest marker to the left and the losest marker to the right of it. Prove that one an rea h a terminal state of exa tly two markers if and only if (n − 1) is not divisible by 3.

Solution by Oliver Geupel, Bruhl,  NRW, Germany. We prove that if 3 ∤ (n − 1), then two markers an be left behind. This is lear if n = 2, 3. Now assume that by starting with n markers we an leave behind two markers, and that n + 3 markers are given. By su

essively

hoosing the se ond, third, then se ond marker from the left, we obtain n markers all white side up, and then we an nish with two markers. This

ompletes the proof by indu tion. It only remains to prove that if two markers an be left behind, then 3 ∤ (n − 1). In a xed state S , we let b(m) be the number of bla k to P markers the left of marker m. Further, we assign the number T (S) = (−1)b(m) m white

to the state S . Let B or W denote a marker with the bla k or white side up, respe tively ; then the admissible redu tion steps are (i) (ii)

BW B → W W , BW W → W B ,

(iii) (iv)

W W B → BW , W W W → BB .

Note that the parity of the number of B 's is invariant, hen e it is always even, and also we have T (BB) = 0, T (W W ) = 2, and for the initial state I with n white markers T (I) = n. It therefore suÆ es to prove that T (S) modulo 3 is invariant under the transitions (i)-(iv), be ause this implies that if F ∈ {BB , W W } is rea hable from I , then n = T (I) ≡ T (F ) 6≡ 1 (mod 3)

.

In the ase of transition (i), if the white marker m that is pi ked has value (−1)b(m) = v ∈ {−1, 1} in state S , then the two markers m′1 and m′2 whi h are turned over have values (−1)b(m ) = (−1)b(m ) = −v in the new state S ′ , whereas the values of all other markers remain un hanged, thus ′ 1

′ 2

T (S ′ ) = T (S) − v − 2v = T (S) − 3v ≡ T (S) (mod 3)

.

The transitions (ii) to (iv) are analyzed similarly. This ompletes the proof. . Let ABC be a triangle with inradius r, ir umradius R, a = BC , b = AC , and c = AB . Prove that

7

R 2r

≥

€

4a2 − (b − c)2

Š€

64a2 b2 c2 4b2 − (c − a)2

and with sides !2

Š€

4c2 − (a − b)2

Š

.

26 Commentary by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain. Solved by George Apostolopoulos, Messolonghi, Gree e. We give the omment of Amengual Covas. This problem appears as Problem 11195 of the Ameri an Mathemati al Monthly, Vol. 113 (January 2006), p. 79. The solution appears on p. 648 of the August{September 2007 issue of the Ameri an Mathemati al Monthly, Vol. 114, and in ludes two generalizations of the given inequality. Next we give the write-up of Apostolopoulos.

È

By Heron's formula, the area of △ABC is s(s − a)(s − b)(s − c), abc and also the same area is given by abc . Hen e, R = p . 4R 4 s(s − a)(s − b)(s − c) Furthermore, theparea of △ABC is rs, and a omparison with Heron's formula yields r = s(s − a)(ss − b)(s − c) , where s = a + 2b + c . Therefore, R

abc

=

2r

8(s − a)(s − b)(s − c)

.

(1)

By using the AM{GM Inequality, we obtain 4a2 − (b − c)2

= a2 + a2 + a2 + 4(s − b)(s − c) ≥ 4

È 4

= 4a

a2 a2 a2 4(s − b)(s − c)

È 4

4a2 (s − b)(s − c) .

Similarly, we have the inequality 4b2 −(c−a)2 ≥ 4b È and 4c2 − (a − b)2 ≥ 4c 4c2 (s − a)(s − b), so that 4

4a2 4a2 ≤

− (b −

c)2

4b2 4b2

a

È 4

·

4a2 (s

− (c −

a)2

·

È 4

4b2 (s − c)(s − a)

4c2 4c2

− (a − b)2 b

c · È · È 4 4 2 2 − b)(s − c) 4b (s − c)(s − a) 4c (s − a)(s − b)

Finally, we have €

4a2 − (b − c)2

Š€

64a2 b2 c2 4b2 − (c − a)2

!2 Š€

4c2 − (a − b)2

Š

a2 b2 c2

≤

È

≤

abc R = 8(s − a)(s − b)(s − c) 2r

64a2 b2 c2 (s − a)2 (s − b)2 (s − c)2

as desired, where the last equality is from (1).

.

.

27 8

. The positive divisors d1 , d2 , . . . , dl of a positive integer n are ordered 1 = d1 < d2 < · · · < dl = m .

Suppose it is known that d27 + d215 = d216 . Find all possible values of d17 . Solution by Titu Zvonaru, Comane  sti, Romania. It is known (see the omment at the end of this solution) that if a, b, c are positive integers su h that a2 = b2 + c2 , then abc is divisible by 60 and one of a, b, c is divisible by 4. Let D = {d1 , . . . , dl } be the set of divisors of n. Sin e d27 + d215 = d216 , it follows that n is divisible by 60, hen e d1 = 1, d2 = 2, d3 = 3, d4 = 4, d5 = 5, d6 = 6 and 10 ∈ D . We dedu e that d7 ∈ {7, 8, 9, 10}. 2 2 Case 1. Suppose that d7 = 10. Then we have d16 − d15 = 100, and fa toring yields (d16 − d15 )(d16 + d15 ) = 100. Sin e d16 + d15 and d16 − d15 have the same parity, we must have d16 − d15 = 2 and d16 + d15 = 50, and hen e d15 = 24, d16 = 26. This means that 8 ∈ D , ontradi ting d7 = 10. Case 2. Suppose that d7 = 9. Then (d16 − d15 )(d16 + d15 ) = 81. Sin e d15 + d16 ≥ 15 + 16 = 31, we must have d16 − d15 = 1 and d16 + d15 = 81, and hen e d15 = 40, d16 = 41. This means that 8 ∈ D, ontradi ting d7 = 9. Case 3. Suppose that d7 = 8. Sin e 7 6∈ D , we dedu e d15 > 15, d16 > 16, hen e d15 + d16 ≥ 33. Then (d16 − d15 )(d16 + d15 ) = 64 has no solution, be ause d15 + d16 ≥ 33 and d16 + d15 and d16 − d15 have the same parity. Case 4. Suppose d7 = 7. Then ne essarily d16 − d15 = 1 and d16 + d15 = 49, and hen e d15 = 24, d16 = 25. If n is divisible by 9, 11, or 13, then d15 < 24, a ontradi tion. As a result, the positive divisors of n are i di

1 1

2 3 2 3

4 5 4 5

6 7 6 7

8 9 8 10

10 12

11 14

12 15

13 20

14 21

15 24

16 25

and we dedu e that d17 = 28. Comment. The fa t that 60 | abc whenever a2 = b2 + c2 an be seen as follows. There are positive integers m, n with m > n su h that a = m2 +n2 , b = m2 − n2 , c = 2mn. Let Ri = {x ∈ Z : x ≡ ±i (mod 5)}. • If m ∈ R0 or n ∈ R0 , then c ≡ 0 (mod 5). • If m, n ∈ R1 or m, n ∈ R2 , then b ≡ 0 (mod 5). • If m ∈ R1 and n ∈ R2 (or vi e-versa), then a ≡ 0 (mod 5). Hen e, 5 | abc. The divisibility by 3 and 4 is proved similarly. Next we present solutions from our readers to problems given in the November 2008 number of the Corner, and the Third Round, Senior Division, of the 2004 South Afri an Mathemati al Olympiad given at [2008 : 412{413℄. . Let a = 1111 · · · 1111 and b = 1111 · · · 1111, where a has forty ones and has twelve ones. Determine the greatest ommon divisor of a and b.

1

b

28 Solved by George Apostolopoulos, Messolonghi, Gree e ; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. We give Wang's version. Š Š 1 € 40 1 € 12 10 −1 and b = 10 −1 . 9 € 9 Š Š 1 − 1 and 1012 − 1 , and we also have 9 Š€ Š

Let d = gcd(a, b). Observe that a = 1 € 40 10 €9

Sin e d divides both €

Š

104 − 1 = 1040 − 1 − 1028 + 1016 + 104 d|

1012 − 1

, then

Š 1€ 4 10 − 1 . 9 €

(1)

Š10

€

From (1) and (2), it follows that d =

Š 1€ 4 10 − 1 = 1111. 9

Š3

Conversely, sin e 1040 −1 = 104 −1 and 1012 −1 = 104 −1 are € Š both divisible by 104 −1, we see that both a and b are divisible by 19 104 −1 . Hen e, Š 1€ 4 10 − 1 | d . (2) 9

. Let n ≥ 2 be an integer. Find the number of integers x with 0 ≤ x < n and su h that x2 leaves a remainder of 1 when divided by n.

5

Solution by Mi hel Bataille, Rouen, Fran e. For ea h positive integer n let S(n) be the set of all integers x su h that 0 ≤ x < n and x2 ≡ 1 (mod n) and let s(n) denote its ardinality. We will prove that if λ(n) is the number of odd prime divisors of n and µ(n) is the greatest nonnegative integer su h that 2m divides n, then 8 < 2λ(n)

s(n) =

2λ(n)+1 : λ(n)+2 2

if if if

µ(n) = 0, 1 ; µ(n) = 2 ; µ(n) > 2 .

We rst show that s is a multipli ative fun tion. Clearly, s(1) = 1. Suppose that n = ab where a, b are oprime positive integers. If x ∈ S(n), then x2 ≡ 1 (mod ab), hen e x2 ≡ 1 (mod a) and x2 ≡ 1 (mod b) and therefore x ≡ y (mod a) and x ≡ z (mod b) for some unique pair of integers (y, z) ∈ S(a) × S(b). Conversely, if (y, z) ∈ S(a) × S(b), then the Chinese Remainder Theorem ensures that the system x ≡ y (mod a), x ≡ z (mod b) has a unique solution x with 0 ≤ x < ab = n. Then both a, b divide x2 − 1, hen e x2 − 1 ≡ 0 (mod ab) (sin e a, b are oprime) and so x ∈ S(n). Thus, S(n) and S(a) × S(b) orrespond bije tively and s(n) = s(a)s(b). It just remains to determine s(pr ) and s(2r ) where p is an odd prime and r is a positive integer. If x2 − 1 ≡ 0 (mod pr ), then p divides x − 1 or x + 1 but not both, or else it would divide (x + 1) − (x − 1) = 2. It follows that either x − 1 ≡ 0 (mod pr ) or x + 1 ≡ 0 (mod pr ). As a result, we have S(pr ) = {1, pr − 1} and s(pr ) = 2.

29 We readily verify that s(2) = 1, s(22 ) = 2. If r ≥ 3, then 2 is the highest power of 2 that an possibly divide both x − 1 and x + 1 whenever x2 − 1 ≡ 0 (mod 2r ). Hen e, in addition to the obvious solutions 1, 2r − 1 to the latter, we also have 2r−1 + 1 and 2r−1 − 1, orresponding to the ases 2r−1 |(x − 1), 2|(x + 1) and 2|(x − 1), 2r−1 |(x + 1), respe tively. Thus, s(2r ) = 4 whenever r ≥ 3. The result follows. Now we turn to the 2006 Vietnamese Mathemati al Olympiad given at [2008 : 413{414℄. . Find all real solutions of the system of equations

1

p

x2 − 2x + 6 · log3 (6 − y) = x ,

È p

y 2 − 2y + 6 · log3 (6 − z) = y ,

z 2 − 2z + 6 · log3 (6 − x) = z .

Solved by Jose Luis Daz-Barrero, Universitat Polite ni a  de Catalunya, Bar elona, Spain ; and Titu Zvonaru, Comane  sti, Romania. We give the solution of Daz-Barrero, modi ed by the editor. First, we observe that x, y, z ∈ (−∞, 6). To nd all real solutions we rst rewrite the system in the form y = 6−3 z = 6−3 x = 6−3

Let f (t) = 6 − 3

√

d dt

t t2 −2t+6

x x2 −2x+6

√

√

y

y 2 −26+6

z z 2 −2z+6

√

, , .

for t ∈ (−∞, 6). Sin e on the domain of f we have

t (6 − t) = € √ Š3/2 > 0 , t2 − 2t + 6 t2 − 2t + 6

then f is a stri tly de reasing fun tion, when e f (f (f (t))) is also a de reasing fun tion for t ∈ (−∞, 6). However, f (x) = y, f (y) = z , f (z) = x ; and thus f (f (f (x))) = x. We laim that if f (f (f (x))) = x, then f (x) = x. Indeed, let f n denote the n-fold omposition of f , and note that both f 3 (x) = x and f (x) < x yield f (x) = f (f 3 (x)) = f 3 (f (x)) > f 3 (x) = x, a ontradi tion. The ase f (x) > x similarly leads to a ontradi tion, and our laim is established. √ Now, 6 − x is a de reasing fun tion and 3 is in reasing for x ∈ (−∞, 6). Hen e, f (x) = x has at most one root. Sin e f (3) = 3, it follows that (x, y, z) = (3, 3, 3) is the unique real solution of the system. x

x2 −2x+6

30 2. Let ABCD be a given onvex quadrilateral. A point M moves on the line AB but does not oin ide with A or B . Let N be the se ond point of interse tion (distin t from M ) of the ir les (M AC) and (M BD), where (XY Z) denotes the ir le passing through the points X , Y , Z . Prove that

(a)

N

moves on a xed ir le,

(b) the line M N passes through a xed point. Solution to part (a) by Mi hel Bataille, Rouen, Fran e. Solution to part (b) by J. Chris Fisher, University of Regina, Regina, SK. (a) We denote by ∠(T U, V W ) the dire ted angle of the lines T U and V W . Let the diagonals AC and BD interse t at O. The point N moves on the ir le (CDO) as it immediately follows from the hara terization of on y li ity in terms of angles and the following al ulation : = = = =

∠(N C, N D) ∠(N C, N M ) + ∠(N M, N D) ∠(AC, AM ) + ∠(BM, BD) ∠(AC, BD) ∠(OC, OD) .

(b) Be ause of the use of dire ted angles, onvexity was not required in part (a). Likewise in part (b), A, B , C , and D an be any four points in the plane, no three ollinear. Let Γ be the ir le (CDO) from part (a). For any two positions of M on AB , say M1 and M2 , we know that the ir les (AM C) meet Γ at the orresponding points N , say N1 and N2 . It suÆ es to prove that the lines M1 N1 and M2 N2 interse t at a point of Γ. To this end, we de ne P to be the point where these two lines interse t, and we apply Miquel's theorem to triangle M1 M2 P and the points A on M1 M2 , N1 on M1 P , and N2 on M2 P . By onstru tion the ir les (AM1 N1 ) and (AM2 N2 ) meet at C . By Miquel's theorem the ir le (P N1 N2 ) also passes through C . Sin e Γ = (N1 N2 C), we on lude that P lies on Γ, as laimed. . Consider the fun tion

4

f (x) = −x +

È

(x + a)(x + b)

where a and b are distin t positive real numbers. Prove that for every real number s in the interval (0, 1), there exists a unique positive real number α su h that 

f (α) =

as + bs 2

‹1/s

.

31 Solution by Mi hel Bataille, Rouen, Fran e. Let v, w lie in (0, ∞) with w > v. Then f (w) − f (v)

È

È

(w + a)(w + b) −

=

(v + a)(v + b) − (w − v)

(w + a)(w + b) − (v + a)(v + b)

=

È

=

(w − v)

(w + a)(w + b) + 

 N −1 D È

where N = v + w + a + b and D = By the AM{GM Inequality, D
f (v) and tinuous on (0, ∞) and

f

lim f (x) =

x→0+

È

.

(w + a)(w + b) +

+

2v + a + b 2

is stri tly in reasing on √

− (w − v)

(v + a)(v + b)

ab ,

= N

È

(v + a)(v + b).

,

(0, ∞).

Also,

a+b 2

lim f (x) =

x→∞

f

is on-

,

the latter holding be ause a + b + (ab/x) (x + a)(x + b) − x2 È f (x) = È = (x + a)(x + b) + x 1 + 1 + (a + b)/x + (ab)/x2

Thus,

.

√

is a bije tion from (0, ∞) onto (m0 , m1 ), where m0 = ab and  s  a+b a + bs 1/s m1 = . The s-mean of a and b is ms = , and we know 2 2 from the Power Mean Inequality that m0 < ms < m1 whenever 0 < s < 1. Sin e f is bije tive, ms = f (α) for some unique α ∈ (0, ∞). 5

f

. Find all polynomials P (x) with real oeÆ ients satisfying €

Š

P (x2 ) + x 3P (x) + P (−x)

= P (x)2 + 2x2 ,

for all real numbers x. Solution by Titu Zvonaru, Comane  sti, Romania. First we will prove a lemma. 2 2 Lemma Let f be a polynomial that satis es f (x ) = f (x) for all x. Then m either f = 0 or f (x) = x . € Š Proof : We will prove by indu tion that f x2 = f (x)2 . For p = 1 the identity holds by hypothesis. Assume the identity holds for p − 1 ≥ 1, then p

€

p

f x2

Š

= f

€€

p−1

x2

Š2 Š

€

p−1

= f x2

Š2

=

p

€

p−1

f (x)2

Š2

p

= f (x)2

,

32 whi h ompletes the indu tion. Now we hoose a natural number p su h that g = 2p > deg f . Let x0 be any omplex root of f and let z1 , z2 , . . . , zg be the (distin t) omplex roots of z g = x0 . We then have €

Šg

f (zi )

= f (zig ) = f (x0 ) = 0 ,

so that f (zk ) vanishes for ea h zk . We dedu e that the polynomial f has g > deg f distin t roots, or x0 = 0. Therefore, f = 0 or f (x) = axm for some number a. In the latter ase f (x2 ) = f (x)2 be omes ax2m = a2 x2m , hen e f ≡ 0 or f (x) = xm . The given equation, for x and −x, is €

Š

€

Š

P (x2 ) + x 3P (x) + P (−x)

=

P (x)2 + 2x2 ,

P (x2 ) − x 3P (−x) + P (x)

=

P (−x)2 + 2x2 .

Subtra ting, we obtain €

Š

4x P (x) + P (−x)

=

€

Š€

P (x) − P (−x)

Š

P (x) + P (−x)

,

whi h leads to two ases. Case 1. P (x) + P (−x) = 0. Here the following are equivalent : P (x2 ) + 2xP (x) P (x2 ) − x2 P (x2 ) − x2

= P (x)2 + 2x2 , = P (x)2 − 2xP (x) + x2 , =

€

Š2

P (x) − x

.

Thus, f (x) = P (x) − x satis es f (x2 ) = f (x)2 , and by the Lemma we have either f ≡ 0 and P (x) = x, or f (x) = xm and P (x) = xm + x. In the latter ase, sin e P (x) + P (−x) = 0, we have xm + x + (−x)m − x = xm + (−x)m = 0 ,

that is, m is odd. We on lude that P (x) = x or P (x) = x2n+1 +x, with n a nonnegative integer. Case 2. P (x) − P (−x) = 4x. Here the following are equivalent : €

Š

P (x2 ) + x 3P (x) + P (x) − 4x

P (x2 ) − 2x2 P (x2 ) − 2x2

= P (x)2 + 2x2 , = P (x)2 − 4xP (x) + 4x2 , =

€

Š2

P (x) − 2x

.

Thus, f (x) = P (x) − 2x satis es f (x2 ) = f (x)2 , and by similar reasoning as in Case 1, we dedu e that either P (x) = 2x or P (x) = xm + 2x.

33 In the latter ase, sin e P (x) − P (−x) = 4x, we have xm + 2x − (−x)m + 2x = 4x ,

or xm − (−x)m = 0, that is, m is even. So, in this ase, we obtain the polynomials x2n + 2x with n a nonnegative integer.

P (x) = 2x

and

P (x) =

n = 0 we have x2n+1 + x = 2x, so a omplete list of polynomials P (x) is x, x2n+1 + x, and x2n + 2x, where n is a nonnegative integer.

For

6. A set of integers T is alled sum-free if for every two (not ne essarily distin t) elements u and v in T , their sum u + v does not belong to T . Prove that (a) a sum-free subset of S = {1, 2, . . . , 2006} has at most 1003 elements, (b) any set S onsisting of 2006 positive integers has a sum-free subset

onsisting of 669 elements. Solution by Oliver Geupel, Bruhl,  NRW, Germany. For part   (a), £ we laim that a sum free subset T of S = {1, 2, . . . , n} has at most n2 elements. To prove this fa t, let m = max S . If m is even  © then ea h of the sets {1, m − 1}, {2, m − 2}, . . . , m 2− 2 , m 2+ 2 ontains n at most one element of T ; hen e |T | ≤ m ≤ . Now suppose that m is odd. 2 2  © Then ea h of the sets {1, m − 1}, {2, m − 2}, . . . , m 2− 1 , m 2+ 1 ontains at most one element of S ; thus |S| ≤ m 2+ 1 . If n is even, then m 2+ 1 ≤ n2 ,  n£ 1 = while if n is odd, then m 2+ 1 ≤ n + . This ompletes the proof of 2 2 part (a). For part (b), we laim that any   set £ S onsisting of n positive integers has a sum free subset onsisting of n3 elements. To prove this, hoose a prime number p > max S su h that 3 | (p + 1), whi h is possible by Diri hlet's theorem. Let

§

I =

k∈N:

p+1 2p − 1 ≤ k ≤ 3 3

ª

.

Consider the bipartite graph with sets of nodes P = {1, 2, . . . , p − 1} and I , where a node k ∈ P is adja ent to a node l ∈ I if and only if there exists a number a ∈ S su h that ka ≡ l (mod p). Note that for ea h edge, the number a ∈ S is unique, and label the respe tive edge with the number a. 1 For ea h a ∈ S there are exa tly |I| = p + numbers k ∈ P su h that ka 3 has a residue modulo p whi h belongs to I , hen e our graph has n(p 3+ 1) edges. By the Pigeonhole Prin iple, there is a node k ∈ P with degree not

34   £

+ 1) less than n(p , hen e not less than n3 . Let T ⊆ S be the set of labels 3(p − 1) of its respe tive edges.   £ We laim that T has the required properties. Clearly, |T | ≥ n3 . Assume that a + b = c for some a, b, c ∈ T . Then ka, kb, kc are ea h ongruent modulo p to numbers in I . Moreover, ka + kb = kc, so that ka + kb is

ongruent modulo p to a number in I . On the other hand, it is easy to he k that for all l, l′ ∈ I , the residue modulo p of the number l + l′ does not belong to I . This ontradi ts the hyothesis a + b = c, and ompletes our proof.

Next we move to solutions from our readers to problems proposed but not used for the 47th International Mathemati al Olympiad 2006 in Slovenia given at [2008 : 459{464℄. . Given an arbitrary real number a0 , de ne a sequen e of real numbers by the re ursion

A1

a0 , a1 , a2 , . . .

ai+1 = ⌊ai ⌋ · {ai } ,

i ≥ 0,

where ⌊ai ⌋ is the greatest integer not ex eeding ai , and {ai } Prove that ai = ai+2 for suÆ iently large i.

= ai − ⌊ai ⌋.

Solution by Oliver Geupel, Bruhl,  NRW, Germany. We onsider the three ases a0 = 0, a0 > 0, and a0 < 0 separately. If a0 = 0, then the sequen e (ai ) is onstant. Next, suppose a0 > 0. Then all ai are nonnegative. Let ⌊ai ⌋ = n and {ai } = r . If n = 0, then ai+1 = ai+2 = ai+3 = · · · = 0. Otherwise we have n ≥ 1 and ai − ai+1 = (n + r) − nr = (n − 1)(1 − r) + 1 ≥ 1 ; hen e we obtain 0 ≤ ai+n < 1, whi h redu es to the previous ase. Lastly, suppose a0 < 0. Then a0 ≤ a1 ≤ a2 ≤ · · · ≤ 0. Hen e, the integer sequen e (⌊ai ⌋) is non-de reasing and bounded above by 0. If just one term of this sequen e is 0, then (ai ) terminates in zeros and we are done. The other possibility is that there exist positive integers i0 and n su h that for all i ≥ i0 we have ⌊ai ⌋ = −n. Let {ai } = r. We will prove by indu tion that for ea h nonnegative integer k, 0

‚

ai0 +k = −

n2 + (n − nr − r)(−n)k n+1

Œ

.

(1)

The equation (1) is immediate if k = 0. For the indu tive step, assuming (1),

35 we obtain ai0 +k+1

−n(ai0 +k) + n)

=

‚

2

−n + n

=

‚

−

=

n2 + (n − nr − r)(−n)k n+1

2

n + (n − nr − r)(−n)k+1 n+1

Œ

Œ

,

thus ompleting the proof of (1) by indu tion. If n > 1, then we obtain from (1) that lim |ai +k | = ∞. But this k→∞

ontradi ts our hypothesis that ⌊ai ⌋ = −n for i ≥ i0 . Consequently, n = 1. Then ai = −1+r, ai +1 = −r = −1+(1−r), and ai +2 = −(1−r) = ai , whi h ompletes the proof. 0

0

0

0

0

. Let a1 , a2 , . . . , an be positive real numbers. Prove that

A4

X i<j

X ai aj n ≤ ai aj . ai + aj 2(a1 + a2 + · · · + an ) i<j

Solution by Mi hel Bataille, Rouen, Fran e. Equality holds for n = 2, so we will suppose that n ≥ 3. We laim that if a, b, c are positive real numbers, then 1 1 1 1 + + ≤ b+c c+a a+b 2



1 1 1 + + a b c

2 is the harmoni mean of Proof : Observe that x + y y ). Using the AM{HM Inequality, it follows that

2 b+c

+

2 c+a

+

2 a+b

1 x

and

1 y

‹

.

(1)

(for positive x,

1 1 1 1 1 1 + + + b c c a a b = 1 + 1 + 1 ≤ + +

2

2

2

a

b

c

.

Ba k to the problem. The proposed inequality is equivalent to (a1 + a2 + · · · + an )

whose left-hand side is

P

ai aj +

i<j

Thus, (2) is equivalent to L ≤

X

ai aj

i<j

ai + aj

P i<j

ai aj ai + aj

n−2X 2

i<j

≤ ‚

nX 2

P k6=i,j

ai aj .

ai aj ,

(2)

i<j

Œ

ak

=

P i<j

ai aj + L.

(3)

36 Note that L =



P

ai aj ak

i,j,k≤n

1 1 1 + + ai + aj aj + ak ak + ai

‹

, the sum taken

over distin t positive integers i, j , k. By (1) above, ≤

L

X 1≤i<j