257Contributor Pro�les:V �alav Kone�n �y

V�alav Kone�n �y was born in B�rest, Czehoslo-vakia in 1934. Both of his parents were elemen-tary shool teahers. He used to wath his motherteahing basi arithmeti and he beame fasinatedwith mathematis at an early age.After the end of the o

upation of Czehoslo-vakia in 1945, his parents sent him and his brotherAnton��n to obtain their high shool eduation at theArhbishop Seminary and Gymnasium in Krom�e�r���z.There he exelled in Mathematis and Physis, butdid poorly in Latin and Greek. The Communist gov-ernment losed the shool in 1950, and he om-pleted his high shool eduation in Gottwaldov in1952, where he ompeted in his �rst Math Olympiad. After a year's work ata fatory he studied physis at Masaryk University in Brno. He graduated in1958, his thesis being on Lommel's problem in Optis (Prof. Alb �eri Boivinof Laval University veri�ed V �alav's orretion of one of his formulas).He was hired by the Physis Department of the Tehnial University ofBrno on the strength of his thesis and his involvement in Physis Olympiads.He began by preparing physis labs, but soon beame a Leturer in Physis.In 1959 he married and subsequently had two hildren.The Vie Chanellor of the University of Khartoum hired V �alav as aLeturer in mathematis in 1964. There he began preparing students for theCambridge exams until 1970. He also started publishing in the areas of Dif-ferential Equations, Combinatoris, Solid State Physis, Optis, Mehanisand problem solving. He reeived a Ph.D. in Mathematis in 1968 and anM.S. in Computer Siene in 1984, all while he ontinued to work.He was a Post-Dotoral Fellow in mathematis at the University ofSaskathewan for six months in 1970, thenmoved with his family to Hawkins,Texas, to teah mathematis and physis at Jarvis Christian College. His wifeand hildren returned to Czehoslovakia in 1973. V �alav followed, but threeyears later ed bak to Hawkins. He was �nally reunited with his family inTexas in 1980, the year he beame a professor at Ferris State University.He retired in 2001, but still promotes CRUX by presenting problemsto faulty and students at Ferris nearly every year. When time permits hesolves CRUX problems, espeially the geometry problems. He likes sym-phony musi, but he listens to the Beatles too. He enjoys reading books inforeign languages, espeially Spanish, and El onde de Monteristo is one ofhis favourites.V �alav's sinerest hope is for world peae.

258EDITORIALV�alav Linek

The all goes out for nominations for CRUX editors, in partiular thesearh is on for Problems Editors, Artiles Editors, Olympiad Editors, and anew Mayhem Editor. Though suh an extensive transition is a hallenge, itis also an exiting time as new people ome on board with fresh ideas.Of ourse, a new Editor-in-Chief is needed as soon as possible! Nomi-nations an be made either to the existing Editor-in-Chief, or to the Chair ofthe CMS Publiations Committee.The all for new editors omes with a strengthening of our preferenefor submissions in TEX or LATEX. We now strongly enourage our readers tosubmit their solutions in these languages, in order to save time and e�ortin the preparation of material for publiation. As mentioned in this year'sMarh editorial, the ideal submission onsists of a LATEX �le with a PDF �leof the output.For those not familiar with it, LATEX is a typesetting language ideal forpreparing mathematial douments, and the language that we use to typesetthe CRUX journal. For general information about LATEX, see the webpagehttp://en.wikipedia.org/wiki/Latex_%28markup_language%29Readers not uent in TEX or LATEX an begin learning it immediately,via the weblink http://www.arachnoid.com/latex/index.html to an onlineLATEX editor. (Note that this editor is set to \math mode", so material utand pasted from it should be enlosed between two $ signs or two $$ signs.)For those who are uent in LATEX, it is worth mentioning that suh an ed-itor is useful for latexing \on the go", for example, when one is visiting afar away, exoti ountry and only has internet a

ess in a lobby, or if oneis using an eletroni devie too small to support LATEX. LATEX pakages arealso available on the web for downloading, and one an install this softwareat no ost. Finally, there are some programs available for onverting someommon formats to LATEX.In the meantime, we will ontinue to proess other �le formats (eventhe o

asional solution written in lipstik on a napkin), but over time thepreferene for LATEX submissions will likely strengthen.We are happy to present our third Contributor Pro�le of the year in thisissue, and look forward to presenting pro�les in future issues.Finally, the all goes out for a variety of problem proposals, in partiu-lar send us your favourite problems in Geometry, Algebra, Number Theory,Game Theory, Logi, Calulus, and Probability. Our poliy of allowing eithernew or well-established proposers up to two (2) expedited problem proposalsremains in plae until the end of 2010.V �alav (Vazz) Linek

259SKOLIAD No. 126

Lily Yen and Mogens HansenPlease send your solutions to problems in this Skoliad by 1 Deember, 2010.A opy of CRUX with Mayhem will be sent to one pre-university reader whosends in solutions before the deadline. The deision of the editors is �nal.Our ontest this month is the Final Round of the Swedish Junior HighShool Mathematis Contest 2009/2010 (3 hours are allowed for writing).Our thanks go to Paul Vaderlind, Stokholm University, Stokholm, Swedenfor providing us with this ontest and for permission to publish it.We also thank Rolland Gaudet, University College of Saint Bonifae,Winnipeg, MB, for translating this ontest into Frenh.Swedish Junior High Shool Mathematis ContestFinal Round, 2009/20101. A 2009 × 2010 grid is �lled with the numbers 1 and −1. For eah row,alulate the produt of the entries in that row. Do likewise for the olumns.Show that the sum of all the row produts and all the olumn produts annotbe zero.2. The square ABCD has side length 6. The point P splits side AB suhthat |AP | : |PB| = 2 : 1. A point Q inside the square is hosen suh that

|AQ| = |PQ| = |CQ|. Find the area of △CPQ.3. The produt of three positive integers is 140. Determine the sum of thethree integers if the seond integer is seven times the �rst one.4. Five points are plaed at the intersetions of a retangular grid. Then themidpoint of eah pair of points is marked. Prove that at least one of thesemidpoints lands on an intersetion point of the grid.5. Points K and L on segment AM areplaed suh that |AK| = |LM |. Plaethe points B and C on one side of AMand point D on the other side of AM suhthat |BK| = |KM |, |CM | = |KL|, and|DL| = |LM |, and suh that BK, CM ,and DL are all perpendiular to AM .Prove that ABCD is a square. .................................................................................................................................................................................................................................................................................................

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2606. Let N be a positive integer. Ragnhild writes down all the divisors of Nother than 1 and N . She then notes that the largest divisor is 45 times thesmallest one. Whih positive integers satisfy this ondition?Conours math �ematique su �edoisNiveau �eole interm�ediaireRonde Finale, 2009/20101. Un tableau 2009 × 2010 est rempli de nombres 1 et −1. Pour haquerang �ee, on alule le produit de tous les nombres dans la rang �ee. On faitde même pour haune des olonnes. D �emontrer que la somme de tous lesproduits en rang �ees et de tous les produits en olonne ne peut pas être nul.2. Le arr �e ABCD a ôt �es de longueur 6. Le point P s �epare AB de fa�on �ae que |AP | : |PB| = 2 : 1. Un point Q �a l'int �erieur du arr �e est hoisi defa�on �a e que |AQ| = |PQ| = |CQ|. D �eterminer la surfae de △CPQ.3. Le produit de trois entiers positifs donne 140. D �eterminer la somme destrois entiers si le seond �egale sept fois le premier.4. Cinq points sont pla �es aux intersetions d'un quadrillage retangulaire.Ensuite, le point milieu est d �etermin �e pour haque paire de points parmi lesinq points. D �emontrer qu'au moins un de es points milieu se trouve �a unpoint d'intersetion du quadrillage.5. Les points K et L sont pla �es sur le seg-ment AM de fa�on �a e que |AK| = |LM |.Pla�ons maintenant des points B et C d'un ôt �ede AM et le point D de l'autre ôt �e de AMde fa�on �a e que |BK| = |KM |, |CM | =|KL|, et |DL| = |LM |, ave en plus queBK, CM , et DL soient tous perpendiulaires�a AM . D �emontrer que ABCD est un arr �e. ............................................................................................................................................................................................................................................................

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6. Soit N un entier positif. Ragnhild �erit alors la liste de tous les diviseursde N sauf 1 et N . Elle note que le plus grand de es diviseurs �egale 45 foisle plus petit des es diviseurs. Lesquels entiers satisfont es onditions ?Next follow solutions to theMaritimeMathematis Competition, 2009,given in Skoliad 120 at [2009 : 417{418℄.1. Two ars leave ity A at the same time. The �rst ar drives to ity B at

40 km/hr and then immediately returns to ity A at the same speed. Theseond ar drives to ity B at 60 km/hr and then returns to ity A at a on-stant speed, arriving at the same time as the �rst ar. What was the seondar's speed on its return trip?

261Solution by Alison Tam, student, Burnaby South Seondary Shool, Burnaby,BC. Let x denote the distane (in km) between the two ities. Then it takesthe �rst ar x

40hours to travel from ity A to ity B. The time to travel bakis also x

40hours, so the total time spent by the �rst ar is x

20hours.Similarly, the seond ar takes x

60hours to travel from ity A to ity B.If s denotes the return speed (in km/h) of the seond ar, then the return triptakes x

shours. Therefore the seond ar travels for a total of x

60+

x

shours.Sine the two ars arrive simultaneously, x

60+

x

s=

x

20. Therefore

1

60+

1

s=

1

20, so 1

s=

1

20− 1

60=

2

60=

1

30. Thus s = 30, so the returnspeed of the seond ar is 30 km/h.Also solved by IAN CHEN, student, Centennial Seondary Shool, Coquitlam, BC; PAULCHEN, student, Burnaby North Seondary Shool, Burnaby, BC; KRISTIAN HANSEN, student,Burnaby North Seondary Shool, Burnaby, BC; ROWENA HO, student, �Eole Banting MiddleShool, Coquitlam, BC; TIFFNEY HSIEH, student, Burnaby North Seondary Shool, Burnaby,BC; ETHAN LIN, student, Burnaby Mountain Seondary Shool, Burnaby, BC; and KENRICKTSE, student, Quilhena Elementary Shool, Vanouver, BC.2. The perimeter of a regular hexagon H is idential to that of an equilateraltriangle T . Find the ratio of the area of H to the area of T .Solution by Ian Chen, student, Centennial Seondary Shool, Coquitlam, BC.The area of an equilateral triangle with side length x is √3

4x2.Let a be the side length of the hexagon. Then the side length of theequilateral triangle is 2a, sineH and T have the same perimeter. Therefore,the area of T is √3

4(2a)2 =

√3a2. Cut H into six ongruent equilateraltriangles like this: .................................................................

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.................................... . Eah of the six triangles has area √34

a2, so H hasarea 3√32

a2. Therefore, the desired ratio is 3√32

a2 :√

3a2 =3

2: 1 = 3 : 2.Also solved by LENACHOI, student, �Eole Dr. Charles Best Seondary Shool, Coquitlam,BC; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; KRISTIAN HANSEN,student, Burnaby North Seondary Shool, Burnaby, BC; TIFFNEY HSIEH, student, BurnabyNorth Seondary Shool, Burnaby, BC; and KENRICK TSE, student, Quilhena ElementaryShool, Vanouver, BC.To prove our solver's formula for the area of an equilateral triangle,draw in the height, h, and use the Pythagorean Theorem: h2 +(x

2)2 = x2,so h2 = x2 − x2

4= 3

4x2, so h = √3

2x. The area is then 1

2hx =

√3

4x2.However, one an also solve the problem without alulating the twoareas: As our solver notes, you an ut H into triangles, ................................................................

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.................................... . Note thatthe perimeter of H is six times the side length of the small triangles. Youan also ut T into ongruent equilateral triangles: ............................................................................................................................................................................................................. ...................................................................................................................................................................................................................................................................................................................................................

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h

x

x/2Note that the perimeter of T is also six times the side length of the small triangles.Therefore the small triangles in H are ongruent with the small triangles in T . Sine H ontainssix triangles while T ontains four, the ratio of their areas is 6 : 4 = 3 : 2.

2623. Some integers may be expressed as the sum of onseutive odd positiveintegers. For example, 64 = 13 + 15 + 17 + 19. Is it possible to express2009 as the sum of onseutive odd positive integers? If so, �nd all suhexpressions for 2009.Solution by Kristian Hansen, student, Burnaby North Seondary Shool,Burnaby, BC.Sine 2009 is odd, but the sum of an even number of odd integers iseven, the number of integers in the sum must be odd. Thus the sum hasa entral number, c. The neighbours of c are c − 2 and c + 2. Note thattheir sum is 2c. The sum of their neighbours in turn is also 2c, and so on.Thus, the sum is an odd multiple of c. That is, c is a divisor of 2009. Sine2009 = 72 · 41, the divisors are 1, 7, 41, 49, 287, and 2009. If c = 2009,the sum only has one term|a very silly ase. If c = 287, the sum has seventerms: 281 + 283 + 285 + 287 + 289 + 291 + 293. If c = 49, the sum has41 terms: 9 + 11 + · · · + 49 + · · · + 87 + 89. If c = 41, the sum has 49terms: −7 − 5 − · · · + 41 + · · · + 87 + 89, but the terms were supposed tobe positive. If c = 7 or c = 1, even more terms are negative. Hene, onlythree sums work out: 9 + · · · + 89, 281 + · · · + 293, and just 2009 by itself.Several solvers found the sum 281+ · · ·+293 but not the other two. We an agree withour solver that a sum of a single term is silly. You an also onsider the sum of zero terms; thevalue is zero. This is an example of the non-disriminatory nature of mathematis: borderlineases are a

epted as long as a sensible de�nition is possible. The degenerate triangle with sidelengths 1, 2, and 3 and that 0! = 1 are other examples.4. The diagram shows threesquares and angles x, y, and z.Find the sum of the angles x,y, and z.

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x y zSolution by Natalia Desy, student, SMA Xaverius 1, Palembang, Indonesia.We'll use the trigonometri identitytan(α + β) =

tan α + tan β

1 − tan α tan β .Now, tan y = 12and tan x = 1

3, so

tan(x + y) =

1

2+

1

3

1 − 12

· 13

=

5

6

1 − 16

= 1 .Therefore x + y = 45◦. Clearly z = 45◦, so x + y + z = 90◦.The proof of our solver's trigonometri identity is too involved to present here. For-tunately, there is an easier solution: Several solvers noted that sine tan y = 1

2and

tan x = 13, it follows that y ≈ 26.6◦ and x ≈ 18.4◦ , whene x + y is around

26345◦ . This of ourse does not prove that x + y is exatly 45◦ .However, the Pythagorean Theorem yields that the length ofthe “y-diagonal" is √5 and that the length of the \x-diagonal"is √10.Keeping these two lengths, the Pythagorean Theorem,and a suspeted angle of 45◦ in your head at the same time mayinspire you to draw the diagram on the right. By the PythagoreanTheorem, the isoseles triangle in the diagram is right angled.Thus x + y = 45◦ , and x + y + z = 90◦ .

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xy

√10

√5

√5

5. Suppose that x1, x2, x3, x4, and x5 are real numbers satisfying the fol-lowing equations.x1 + 4x2 + 9x3 + 16x4 + 25x5 = 1 ,

4x1 + 9x2 + 16x3 + 25x4 + 36x5 = 8 ,9x1 + 16x2 + 25x3 + 36x4 + 49x5 = 23 .Find the value of x1 + x2 + x3 + x4 + x5.Solution by Ian Chen, student, Centennial Seondary Shool, Coquitlam, BC.First, note that the oeÆients of eah xi in the three equations areonseutive squares, for example 4x2, 9x2, and 16x2. Seond, note thatonseutive squares have the form (n − 1)2, n2, and (n + 1)2 . Third, notethat (n − 1)2 + (n + 1)2 = n2 − 2n + 1 + n2 + 2n + 1 = 2n2 + 2, so that

(n − 1)2 + (n + 1)2 − 2n2 = 2. Therefore, if you add the �rst and the lastequation, and then subtrat twie the middle equation, then the oeÆientof eah xi is 2, and2x1 + 2x2 + 2x3 + 2x4 + 2x5 = 1 + 23 − 2 · 8 = 8 .Therefore, x1 + x2 + x3 + x4 + x5 = 4.Also solved by PAUL CHEN, student, Burnaby North Seondary Shool, Burnaby, BC;and NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia.6. A math teaher writes the equation x2 − Ax + B = 0 on the blakboardwhere A and B are positive integers and B has two digits. Suppose that astudent erroneously opies the equation by transposing the two digits of Bas well as the plus and minus signs. However, the student �nds that herequation shares a root, r, with the original equation. Determine all possiblevalues of A, B, and r.Solution by the editors.Sine B has two digits, B = 10b + c, where b and c are digits. Theteaher's polynomial is then x2 − Ax + (10b + c). You know that r is a rootof this polynomial, so r2 − Ar + 10b + c = 0.If you reverse the digits ofB, you get the number 10c+b. The student'spolynomial is then x2 + Ax − (10c + b). Again, you know that r is a root ofthis polynomial, so r2 + Ar − b − 10c = 0.

264If you subtrat the equation r2 − Ar + 10b + c = 0 from the equation

r2 + Ar − b − 10c = 0, you get that 2Ar − 11b − 11c = 0. Sine A, b, and care positive integers, it follows that r is a positive fration or an integer.If you add r2 − Ar +10b + c = 0 and r2 + Ar − b − 10c = 0 together,you get that 2r2 + 9b − 9c = 0, so2r2 = 9(c − b) .Thus 2r2 is an integer. Sine r is a fration, r must atually be an integer. (If

c − b is even, then r is both a fration and the square root of an integer, andin this ase r must be an integer. If c − b is odd, then there is no fration rthat solves the equation, and this an be shown by using the same argumentthat shows √2 is an irrational number.) It follows that 9(c − b) is twie asquare. Sine 9 is already a square, c − b must be twie a square. Sine band c are digits, the possibilities are c − b = 0, c − b = 2, or c − b = 8.If c − b = 0, then r = 0, and then from the original quadrati equationwe have B = 0. However, B is given to be positive, so this annot work.If c − b = 8, then c = 9 and b = 1. Sine 2r2 = 9(c − b) = 72, youhave that r = 6. Sine 2Ar − 11b − 11c = 0, you have that 12A − 110 = 0,ontraditing the fat that A is an integer.If c − b = 2, then B = 10b + c is one of these values: 13, 24, 35, 46,57, 68, or 79. Moreover, 2r2 = 9 · 2, so r = 3. Sine 2Ar − 11b − 11c = 0,you have that A = 11(b + c)

6. Only the values 24 and 57 yield an integerfor A as required, namely 11 and 22, respetively. You an easily verify thatboth possibilities work out:teaher: x2 − 11x + 24 = (x − 3)(x − 8) ,student: x2 + 11x − 42 = (x − 3)(x + 14) ,or the other possible pair of polynomialsteaher: x2 − 22x + 57 = (x − 3)(x − 19) ,student: x2 + 22x − 75 = (x − 3)(x + 25) .

The prize of a opy of CRUX with MAYHEM for the best solutions goesto Ian Chen, student, Centennial Seondary Shool, Coquitlam, BC.We invite readers to submit solutions to one or more of our problems.

265MATHEMATICAL MAYHEMMathematial Mayhem began in 1988 as a Mathematial Journal for and byHigh Shool and University Students. It ontinues, with the same emphasis,as an integral part of Crux Mathematiorum with Mathematial Mayhem.The Mayhem Editor is Ian VanderBurgh (University of Waterloo). Theother sta� members are Monika Khbeis (Our Lady of Mt. Carmel SeondaryShool, Mississauga, ON) and Eri Robert (Leo Hayes High Shool, Freder-iton, NB).

Mayhem ProblemsPlease send your solutions to the problems in this edition by 1 Deember 2010.Solutions reeived after this date will only be onsidered if there is time before pub-liation of the solutions. The Mayhem Sta� ask that eah solution be submitted ona separate page and that the solver's name and ontat information be inluded witheah solution.Eah problem is given in English and Frenh, the oÆial languages of Canada.In issues 1, 3, 5, and 7, English will preede Frenh, and in issues 2, 4, 6, and 8,Frenh will preede English.The editor thanks Jean-Mar Terrier of the University of Montreal for transla-tions of the problems.M445. Proposed by the Mayhem Sta�.The lines with equations y = x + 1, y = mx − 1, and y = −4x + 2mpass through the same point. Determine all possible values for m.M446. Proposed by J. Walter Lynh, Athens, GA, USA.Let a, b, and c be positive digits. Suppose that b equals the produtof a, b, and c, and ac = a + b + c. Determine a, b, and c. (Here ab is thetwo-digit positive integer with tens digit a and units digit b.)M447. Proposed by Yakub N. Aliyev, Qafqaz University, Khyrdalan,Azerbaijan.Let ABCD be a parallelogram. The sides AB and AD are extended topoints E and F (respetively) so that E, C, and F all lie on a straight line.Prove that BE · DF = AB · AD.M448. Proposed by the Mayhem Sta�.A polyhedron with exatly m + n faes has m faes that are quadrilat-erals and n faes that are triangles. Exatly four faes meet at eah vertex.Prove that n = 8.

266M449. Proposed by Neulai Staniu, George Emil Palade SeondaryShool, Buz�au, Romania.Let E(x) = 4x

4x + 2.(a) Prove that E(x) + E(1 − x) = 1.(b) Find the value of E� 1

2010

�+ E

�2

2010

�+ · · · + E

�2008

2010

�+ E

�2009

2010

�.M450. Proposed by Edward T.H. Wang, Wilfrid Laurier University,Waterloo, ON.Prove that if n is an odd positive integer, then nn+2 + (n + 2)n isdivisible by 2(n + 1)..................................................................M445. Propos �e par l' �Equipe de Mayhem.Les droites d' �equations y = x + 1, y = mx − 1 et y = −4x + 2mpassent toutes par le même point. Trouver toutes les valeurs possibles de m.M446. Propos �e par J. Walter Lynh, Athens, GA, �E-U.On suppose que les hi�res positifs a, b et c sont tels que b �egale leproduit de a, b et c et que ac = a + b + c. D �eterminer a, b et c. (Ii abd �enote l'entier positif de deux hi�res, a �etant elui des dizaines et b eluides unit �es.)M447. Propos �e par Yakub N. Aliyev, Universit �e de Qafqaz, Khyrdalan,Azerba�djan.On prolonge respetivement les ôt �es AB et AD du parall �elogrammeABCD jusqu'aux points E et F de sorte que E, C et F soient tous sur lamême droite. Montrer que BE · DF = AB · AD.M448. Propos �e par l' �Equipe de Mayhem.Un poly �edre �a exatement m + n faes en poss �ede m qui sont desquadrilat �eres et n qui sont des triangles. Chaque sommet est le point derenontre d'exatement quatre faes. Montrer que n = 8.M449. Propos �e par Neulai Staniu, �Eole seondaire George Emil Palade,Buz�au, Roumanie.Soit E(x) = 4x

4x + 2.(a) Montrer que E(x) + E(1 − x) = 1.(b) �Evaluer E� 1

2010

�+ E

�2

2010

�+ · · · + E

�2008

2010

�+ E

�2009

2010

�.

267M450. Propos �e par Edward T.H. Wang, Universit �e Wilfrid Laurier, Wa-terloo, ON.Montrer que si n est un entier positif impair, alors nn+2 + (n + 2)nest divisible par 2(n + 1).

Mayhem SolutionsM407. Proposed by Neven Juri�, Zagreb,Croatia.Determine whether or not the square atright an be ompleted to form a 4 × 4 magisquare using the integers from 1 to 16. (Ina magi square, the sums of the numbers ineah row, in eah olumn, and in eah of thetwo main diagonals are all equal.) ......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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2 15 8

16 1 10

12

Solution by Larry Rollins, student, Auburn University Montgomery, Mont-gomery, Alabama, USA, modi�ed by the editor.Sine the magi square is to ontain eah of the integers from 1 to 16,then the sum of the entries would be 1+2+· · ·+15+16 = 12(16)(17) = 136.Sine the sum of the entries in eah row is the same, then this sum shouldbe 1

4(136) = 34. Also, the sum of the entries in eah olumn and on eahdiagonal should be 34.Suppose that we an omplete the square to form a magi square. Inthis ase, the missing entry in the fourth row should be 34 − 2 − 15 − 8 = 9,the missing entry in the third row should be 34 − 16 − 1 − 10 = 7, and themissing entry in the fourth olumn should be 34 − 12 − 10 − 8 = 4. Also,the missing entry in the \northeast" diagonal should be 34 − 9 − 16 − 4 = 5and the missing entry in the third olumn should be 34 − 5 − 1 − 15 = 13.At this point, we would have the square on theright. The four integers unused are 3, 6, 11, and 14.To form amagi square, the missing entries in the se-ond olumn would total 34 − 16 − 2 = 16. No pair ofthe unused numbers totals 16. Therefore, the squareannot be ompleted to form a magi square. ..........................................................................................................................................................................................................................................................................................................................................................................................

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29 15 8

167 1 10

125

13 4

Also solved by JACLYN CHANG, student, Western Canada High Shool, Calgary, AB;BRUNO SALGUEIRO FANEGO, Viveiro, Spain; CARL LIBIS, Cumberland University, Lebanon,TN, USA; PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; RICARD PEIR �O, IES \Abas-tos", Valenia, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; JIXUANWANG, student, Don Mills Collegiate Institute, Toronto, ON; and GUSNADI WIYOGA, student,SMPN 8, Yogyakarta, Indonesia. There were three inomplete solutions submitted.

268M408. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.Determine all three-digit positive integers abc that satisfy the equationabc = ab + bc + ca. (Here abc denotes the three-digit positive integer withhundreds digit a, tens digit b, and units digit c.)Solution by Gusnadi Wiyoga, student, SMPN 8, Yogyakarta, Indonesia.Sine abc = 100a + 10b + c and ab = 10a + b and bc = 10b + c andca = 10c + a, then from the given equation, we have 100a + 10b + c =(10a + b) + (10b + c) + (10c + a), or 89a = 10c + b.Sine a, b, and c are digits, then 10c + b ≤ 99, so 89a is at most 89.Thus, a = 1 and so 10c + b = 89. Sine b and c are digits, then b = 9 andc = 8, so abc = 198.Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS,Messolonghi, Greee; CAO MINH QUANG, Nguyen Binh Khiem High Shool, Vinh Long,Vietnam; JACLYN CHANG, student, Western Canada High Shool, Calgary, AB; GEOFFREYA. KANDALL, Hamden, CT, USA; CARL LIBIS, Cumberland University, Lebanon, TN, USA;HUGO LUYO S�ANCHEZ, Ponti�ia Universidad Cat �olia del Peru, Lima, Peru; MITEAMARIANA, No. 2 Seondary Shool, Cugir, Romania; PEDRO HENRIQUE O. PANTOJA,student, UFRN, Brazil; RICARD PEIR �O, IES \Abastos", Valenia, Spain; BRUNO SALGUEIROFANEGO, Viveiro, Spain; MRIDUL SINGH, student, Kendriya Vidyalaya Shool, Shillong, India;EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; JIXUAN WANG, student, DonMills Collegiate Institute, Toronto, ON; and KONSTANTINE ZELATOR, University of Pittsburgh,Pittsburgh, PA, USA.M409. Proposed by Brue Shawyer, Memorial University of Newfound-land, St. John's, NL and the Mayhem Sta�.The three altitudes of a triangle lie along the lines y = x, y = −2x+3,and x = 1. If one of the verties of the triangle is at (5, 5), determine theoordinates of the other two verties.Solution by Cao Minh Quang, Nguyen Binh Khiem High Shool, Vinh Long,Vietnam.We label the verties of the triangle as A(5, 5), B, and C. We label thepoints D, E, and F on sides BC, AC, and AB, respetively, so that AD,BE, and CF are the altitudes of the triangle.Sine A(5, 5) is on the line y = x but not on the line x = 1 nor onthe line y = −2x + 3, then it is the altitude from A that lies along the liney = x.We will assume, without loss of generality, that the altitude from Blies along the line y = −2x + 3 and that the altitude from C lies along theline x = 1.Sine the altitude from B has a slope of −2 and BE is perpendiular toAC, then the slope of AC is 1

2. Sine A(5, 5) also lies on segment AC, thenthe equation of the line through A and C is y −5 = 1

2(x−5) or y = 1

2x+ 5

2.Sine C lies on this line and on the line x = 1, then C has x-oordinate 1and y-oordinate y = 1

2+ 5

2= 3. Thus, C has oordinates (1, 3).

269Sine the altitude from C is vertial and CF is perpendiular to AB,then AB is horizontal and passes through A(5, 5). Thus, the equation of theline through A and B is y = 5. Therefore, sine B lies on the line y = 5and on the line y = −2x + 3, then it is the point of intersetion of theselines. Sine the y-oordinate of B is 5, then its x-oordinate is the solutionof −2x + 3 = 5 or x = −1. Thus, the oordinates of B are (−1, 5).Finally, the other two verties have oordinates (1, 3) and (−1, 5).Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greee; JACLYN CHANG,student, Western Canada High Shool, Calgary, AB; HUGO LUYO S�ANCHEZ, Ponti�iaUniversidad Cat �olia del Peru, Lima, Peru; KONSTANTINOS AL. NAKOS, Agrinio, Greee;PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; RICARD PEIR �O, IES \Abastos",Valenia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; JIXUAN WANG, student, DonMills Collegiate Institute, Toronto, ON; and KONSTANTINE ZELATOR, University of Pittsburgh,Pittsburgh, PA, USA.M410. Proposed by Matthew Babbitt, home-shooled student, FortEdward, NY, USA.A ube with edge length a, a regular tetrahedron with edge length b,and a regular otahedron with edge length c all have the same surfae area.Determine the value of √bc

a.Solution by Cao Minh Quang, Nguyen Binh Khiem High Shool, Vinh Long,Vietnam.We use the fat that the area of an equilateral triangle with side length

s is √34

s2. (This an be derived by splitting the equilateral triangle into twoongruent 30◦-60◦-90◦ triangles and using the known ratios of side lengthsto alulate the area of the larger triangle.)A ube has six square faes. Thus, the surfae area of a ube with edgelength a is 6a2.A regular tetrahedron has four faes, eah of whih is an equilateraltriangle. Thus, the surfae area of a regular tetrahedron with edge length bis 4 × √34

b2 =√

3b2.A regular otahedron has eight faes, eah an equilateral triangle. Thus,the surfae area of a regular otahedron of edge length c is 8×√34

c2 =2√

3c2.Sine the surfae area of eah of these solids is the same, we then have6a2 =

√3b2 = 2

√3c2. Thus, (6a2)2 = (√3b2)(2√3c2) or 36a4 = 6b2c2.Therefore, b2c2

a4= 6, so that b2c2

a4

1/4= 4

√6 and √bc

a= 4

√6.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greee; SCOTT BROWN,Auburn University, Montgomery, AL, USA; JACLYN CHANG, student, Western Canada HighShool, Calgary, AB; G.C. GREUBEL, Newport News, VA, USA; GEOFFREY A. KANDALL,Hamden, CT, USA; HUGO LUYO S�ANCHEZ, Ponti�ia Universidad Cat �olia del Peru, Lima,Peru; PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; RICARD PEIR �O, IES \Abastos",Valenia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and JIXUAN WANG, student,Don Mills Collegiate Institute, Toronto, ON. There was one inorret solution submitted.

270M411. Proposed by Neulai Staniu, George Emil Palade SeondaryShool, Buz�au, Romania.Triangle ABC has side lengths a, b, and c. If

2a + 3b + 4c = 4√

2a − 2 + 6p

3b − 3 + 8√

4c − 4 − 20 ,prove that triangle ABC is right-angled.Solution by Pedro Henrique O. Pantoja, student, UFRN, Brazil.From the given equation, we obtain the three equivalent equations2a + 3b + 4c − 4

√2a − 2 − 6

p3b − 3 − 8

√4c − 4 + 20 = 0 ,

(2a − 2 − 4√

2a − 2 + 4) + (3b − 3 − 6p

3b − 3 + 9)+ (4c − 4 − 8

√4c − 4 + 16) = 0 ,

(√

2a − 2 − 2)2 + (p

3b − 3 − 3)2 + (√

4c − 4 − 4)2 = 0 .Sine x2 is nonnegative for any real number x, then we an onludethat √2a − 2 − 2 =

p3b − 3 − 3 =

√4c − 4 − 4 = 0 .Therefore, 2a − 2 = 22 (whih yields a = 3) and 3b − 3 = 32 (whihyields b = 4) and 4c − 4 = 42 (whih yields c = 5).Sine a = 3, b = 4, and c = 5, then a2 + b2 = c2, whih tells us thatthe triangle is right-angled.Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS,Messolonghi, Greee; GEOFFREY A. KANDALL, Hamden, CT, USA; HUGO LUYO S�ANCHEZ,Ponti�ia Universidad Cat �olia del Peru, Lima, Peru; MITEA MARIANA, No. 2 SeondaryShool, Cugir, Romania; KONSTANTINOS AL. NAKOS, Agrinio, Greee; RICARD PEIR �O,IES \Abastos", Valenia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARDT.H. WANG, Wilfrid Laurier University, Waterloo, ON; and KONSTANTINE ZELATOR, Univer-sity of Pittsburgh, Pittsburgh, PA, USA.M412. Proposed by Edward T.H. Wang, Wilfrid Laurier University,Waterloo, ON.For a real number x, let ⌊x⌋ denote the greatest integer less than orequal to x, and let {x} = x−⌊x⌋ denote the frational part of x. Determineall real numbers x for whih ⌊x⌋ · {x} = x.Solution by Bruno Salgueiro Fanego, Viveiro, Spain.If x = 0, then ⌊x⌋ · {x} = ⌊0⌋ · {0} = 0 · 0 = 0, so ⌊x⌋ · {x} = x.If x > 0, then 0 ≤ ⌊x⌋ ≤ x and 0 ≤ {x} < 1, so ⌊x⌋ · {x} < x · 1 = xand in this ase ⌊x⌋ · {x} 6= x.If x < 0, then there is an integer k with k ≥ 1 and −k ≤ x < −k + 1;that is, ⌊x⌋ = −k. Thus, {x} = x−⌊x⌋ = x−(−k) = x+k. Therefore, theequation ⌊x⌋ · {x} = x is equivalent to −k(x + k) = x, or (k + 1)x = −k2,or x = − k2

k + 1.

271Therefore, the solution set to the equation ⌊x⌋ · {x} = x is x = 0or x = − k2

k + 1for some positive integer k. These an be ombined to give

x = − k2

k + 1for some nonnegative integer k.Also solved by CAO MINH QUANG, Nguyen Binh Khiem High Shool, Vinh Long,Vietnam; SAMUEL G �OMEZ MORENO, Universidad de Ja �en, Ja �en, Spain; CARL LIBIS,Cumberland University, Lebanon, TN, USA; PEDRO HENRIQUE O. PANTOJA, student, UFRN,Brazil; RICARD PEIR �O, IES \Abastos", Valenia, Spain; PAOLO PERFETTI, Dipartimento diMatematia, Universit �a degli studi di Tor Vergata Roma, Rome, Italy; JIXUAN WANG, student,DonMills Collegiate Institute, Toronto, ON; GUSNADIWIYOGA, student, SMPN 8, Yogyakarta,Indonesia; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA. Therewere two inorret solutions submitted.

Problem of the MonthIan VanderBurgh

Some of the best problems are ones that are simple to understand, donot require a whole lot of mathematial bakground, but that send you nielydown the \garden path".Problem (2010 Pasal Contest) The produt of N onseutive four-digit pos-itive integers is divisible by 20102. What is the least possible value of N?(A) 5 (B) 12 (C) 19 (D) 6 (E) 7Sine we want to �nd a produt of onseutive integers divisible by20102, it makes good sense to �nd the prime fators of 2010. (It's a goodthing that we didn't ask this in 2011.) This isn't that diÆult sine 2010 isdivisible by 10 (sine its units digit is 0) and 3 (sine the sum of its digits is 3whih is a multiple of 3). Therefore,

2010 = 10 × 201 = 10 × 3 × 67 = 2 × 3 × 5 × 67 .But we want to �nd a produt of onseutive integers divisible by 20102, sowe'd better write out the prime fatorization of 20102:20102 = 22 × 32 × 52 × 672 .Now, let's try to solve the problem.Solution 1. Sine we are looking for a set of N onseutive four-digit posi-tive integers whose produt is divisible by 20102, then we need to look forintegers that are divisible by the prime fators of 20102.We start with the largest prime fator of 20102, namely 67. We wantthe produt of the integers in our set to inlude two fators of 67, and so

272either two di�erent integers in the set are multiples of 67 or one integer inthe set has two fators of 67. In the �rst ase, our set of onseutive integerswould ontain two di�erent multiples of 67 whih must be at least 67 apartfrom eah other; this would mean that the set ontains at least 68 positiveintegers. Sine none of the available answers are anywhere lose to 68, thismust not be the ase.Sine the integers in the set are four-digit positive integers, then ourset should ontain 672 = 4489. We also need the produt of the integersin the set to ontain two multiples of 5. There is no multiple of 52 that islose to 4489, so we try expanding the set to inlude 4490 and 4485, the twolosest multiples of 5 to 4489. (These are also the two multiples of 5 that wean inlude to minimize the total number of integers in the set thus far.)Up to this point our set is

{4485, 4486, 4487, 4488, 4489, 4490} .The produt of these integers inludes two fators of 67 (sine 4489 = 672),two fators of 5 (one in 4485 and one in 4490), two fators of 2 (one in 4486and one in 4488), and also two fators of 3 (one in 4485 and one in 4488).We an hek this last fat using a alulator (don't be tempted!) or by notingthat the sum of the digits of 4485 is 21 and the sum of the digits of 4488 is 24;eah of these sums is divisible by 3 so eah of the integers is divisible by 3.Therefore, the produt of the four-digit integers in the set{4485, 4486, 4487, 4488, 4489, 4490}is divisible by 20102 as required. This set ontains 6 integers, so the answerto the problem is (D).This solution is not too diÆult and makes good sense. In other words,it is a great solution exept for one small problem. Solution 1 is wrong!Can you see what is wrong with it? Take a few minutes and read through itritially to see if you an spot the aw. Don't be too alarmed if you an't �ndthe aw { a number of quite good mathematiians have missed this already!The ruial mistake is in the sentene \Sine the integers in the set arefour-digit positive integers, then our set should ontain 672 = 4489." Canyou see the aw now? The sentene an be orreted be re-writing it as\Sine the integers in the set are four-digit positive integers, then our setshould ontain a four-digit integer that is divisible by 672 = 4489." Let'sstart Solution 2, piking up from Solution 1 in the third paragraph.Solution 2. Sine the integers in the set are four-digit positive integers, thenour set should ontain a four-digit integer that is divisible by 672 = 4489.The four-digit multiples of 4489 are 4489 and 2 × 4489 = 8978. (Note that

3 × 4489 is too large, sine it has �ve digits.)In Solution 1, we saw that if the set inludes 4489 and has the desiredproperty, then the minimal size of the set is 6. So let's onsider a set thatinludes 8978.

273Let's look for multiples of 5 to inlude in the set. As in Solution 1, if weinlude two di�erent multiples of 5, then the set inludes at least 6 integers.(Can you see why?) Is it possible to inlude a multiple of 52 in our set inthis ase that is lose to 8978? Yes! We an inlude 8975, whih is divisibleby 52.So let's expand our set by inluding the intermediate integers to get

{8975, 8976, 8977, 8978} .How are we doing so far with respet to the desired property? The produtof these integers inludes two fators of 67 (sine 8978 is divisible by 672),two fators of 5 (sine 8975 is divisible by 52), and two fators of 2 (at leastone in eah of 8976 and 8978). How about fators of 3? Sine the sum ofthe digits of 8976 is 30, then 8976 is divisible by 3. However, 8976 is notdivisible by 32 (sine the sum of its digits is not divisible by 9) and none of theother three integers is divisible by 3. (Chek the sum of the digits of eah.)Therefore, we need to expand the set to inlude a seond multiple of 3.Is either 8974 or 8979 divisible by 3? Yes, 8979 is divisible by 3. Therefore,the set{8975, 8976, 8977, 8978, 8979}has the property that the produt of its elements is divisible by 20102. Basedon our reasoning, this set is the smallest set of four-digit integers with thisproperty. Therefore, the answer to the problem is (A).I really like this problem, beause it really tests reasoning skills with-out requiring a whole lot of formal knowledge. There was quite a debateamong the ontest reation team as to whether to inlude the possible an-swer \(D) 6", given the trap that it sets. Leaving it in undoubtedly trappeda number of ontestants, but really rewarded those who saw through this.On the other hand, leaving this answer out might have served as a good\teahing point" for students who got 6 as the answer and then found thattheir answer wasn't there and as a result persevered to �nd the real answer.Food for thought! In the end, in the ontext of the entire paper, the ontestreation team deided to leave the answer (D) in.

274THE OLYMPIAD CORNERNo. 287

R.E. WoodrowWe start this issuewith problems from the BulgarianNational Olympiad,National Round, 2007. My thanks to Bill Sands, Canadian Team Leader tothe IMO in Vietnam, for olleting them for our use.2007 BULGARIAN NATIONAL OLYMPIADNational RoundMay 12{13, 20071. (Emil Kolev, Alexandar Ivanov) The quadrilateral ABCD is suh that∠BAD + ∠ADC > 180◦ and is irumsribed around a irle of entre I.A line through I meets AB and CD at points X and Y , respetively. Provethat if IX = IY , then AX · DY = BX · CY .2. (Alexandar Ivanov, Emil Kolev) Find the largest positive integer n suhthat one an hoose 2007 distint integers from the interval [2 · 10n−1, 10n)with the property that whenever 1 ≤ i ≤ j ≤ n, then there exists a hosennumber with deimal representation a1a2 . . . an and aj ≥ ai + 2.3. (Nikolai Nikolov, Oleg Mushkarov) Find the least positive integer n forwhih cos π

n

annot be expressed in the form p+√q + 3√r, where p, q, r arerational numbers.4. (Emil Kolev, Alexandar Ivanov) Let k > 1 be a �xed integer. A set ofpositive integers S is alled good if all positive integers an be painted in kolours suh that no element of S is a sum of two distint numbers of thesame olour. Find the largest positive integer t for whih the set

S = {a + 1, a + 2, a + 3, . . . , a + t}is good for all positive integers a.5. (Oleg Mushkarov, Nikolai Nikolov) Find the least number m for whihany �ve equilateral triangles with ombined area m an over an equilateraltriangle of area 1.6. (Alexandar Ivanov, Emil Kolev) Let f(x) be a moni polynomial of evendegree with integer oeÆients. Prove that if there are in�nitely many in-tegers x for whih f(x) is a perfet square, then there is a polynomial g(x)with integer oeÆients suh that f(x) = g2(x).

275Next we ontinue with problems from the IMO Team Seletion Testsfor the Bulgarian Team. Thanks again to Bill Sands, Canadian Team Leaderto the IMO in Vietnam, for olleting them for our use.48th IMOBulgarian Team First Seletion TestMay 16-17, 20071. The sequene {ai}∞i=1 is suh that a1 > 0 and an+1 = an1 + a2n for n ≥ 1.(a) Prove that an ≤ 1√

2nfor n ≥ 2.

(b) Prove that there exists an n suh that an > 710

√n.

2. Let A1A2A3A4A5 be a onvex pentagon suh that the triangles A1A2A3,A2A3A4, A3A4A5, A4A5A1, A5A1A2 have the same area. Prove thatthere exists a point M suh that the triangles A1MA2, A2MA3, A3MA4,A4MA5, A5MA1 have the same area.3. Prove that there are no distint positive integers x and y suh that

x2007 + y! = y2007 + x! .4. Given a point P on the side AB of a triangle ABC, onsider all pairs ofpoints (X, Y ) suh that X ∈ BC, Y ∈ AC and suh that ∠PXB = ∠PY A.Prove that the midpoints of the segments XY lie on a straight line.5. The real numbers ai, bi for 1 ≤ i ≤ n are suh thatnX

i=1

a2i =nX

i=1

b2i = 1 and nXi=1

aibi = 0 .Prove that

nXi=1

ai

!2+

nX

i=1

bi

!2≤ n .

6. For a �nite set S denote by P(S) the set of all subsets of S. The funtionf : P(S) → R is suh that

f(X ∩ Y ) = min(f(X), f(Y ))for any two subsets X, Y ∈ P(S). Find the largest number of distint valuesthat f an take.

27648th IMOBulgarian Team Seond Seletion TestMay 26-27, 2007

1. is externally tangent to Γ1 at Q and to Γ2 at R. The lines PQ and PRmeet Γ3 again at points A and B, Two irles Γ1 and Γ2 with entres O1 andO2, respetively are externally tangent at point P . A irle Γ3 respetively.If AO2 meets BO1 at a point S, prove that

SP ⊥ O1O2 .2. Find all positive integers m suh that2mαm − (α + β)m − (α − β)m

3α2 + β2is an integer for all integers α and β with αβ 6= 0.3. Find all integers n ≥ 3 suh that for any two positive integers m < n −1and r < n − 1 there exist m distint elements of the set {1, 2, . . . , n − 1}whose sum is ongruent to r modulo n.4. Solve the systemx2 + yu = (x + u)n ,x2 + yz = u4 ,where x, y, and z are prime numbers and u is a positive integer.5. Find all pairs of funtions f , g : R → R suh that(a) f(xg(y + 1)) + y = xf(y) + f(x + g(y)) for any x, y ∈ R, and(b) f(0) + g(0) = 0.6. Prove that n = 11 is the least positive integer suh that for any olouringof the edges of a omplete graph of n verties with three olours there existsa monohromati yle of length 4.

Next we turn to the problems fromHelleni ompetitions and the prob-lems of the Mediterranean Mathematial Competition 2007. Thanks againare due to Bill Sands for olleting them for our use.

27710th MEDITERRANEAN MATHEMATICALCOMPETITION 20071. Let x ≤ y ≤ z be real numbers satisfying the relation xy + yz + zx = 1.Prove that xz < 1

2. Is it possible to improve the value of the onstant 1

2?2. The quadrilateral ABCD is onvex and yli, and the diagonals AC and

BD interset at the point E. Given that AB = 39, AE = 45, AD = 60 andBC = 56, determine the length of CD.3. In the triangle ABC the angle α = ∠A and the side a = BC are given.It is known that a = √rR, where r is the inradius and R is the irumradiusof ABC. Determine all suh triangles, that is, ompute the sides b and c ofall suh triangles.4. Let x > 1 be a real number that is not an integer. Prove that�

x + {x}⌊x⌋

− ⌊x⌋x + {x}

�+

�x + ⌊x⌋

{x}− {x}

x + ⌊x⌋

�>

9

2,where ⌊x⌋ and {x} are the integer and the frational part of x, respetively.

Continuing with the Helleni theme we give the problems of the 24thBalkanMathematial Olympiadwritten in Rhodes, Greee, April 2007. Thanksagain go to Bill Sands for obtaining them for our use.24th BALKAN MATHEMATICAL OLYMPIADApril 26, 20071. Let ABCD be a onvex quadrilateral with AB = BC = CD, AC 6= BDand let E be the intersetion point of its diagonals. Prove that AE = DE ifand only if ∠BAD + ∠ADC = 120◦.2. Find all funtions f : R → R suh that for all x, y ∈ R,f(f(x) + y) = f(f(x) − y) + 4f(x)y .3. Find all positive integers n for whih there exists a permutation σ of theset {1, 2, . . . , n} suh that the number below is a rational numberÊ

σ(1) +

Éσ(2) +

q· · · +

Èσ(n) .[Ed.: a permutation of the set {1, 2, . . . , n} is a one-to-one funtion of thisset to itself.℄

2784. For a given positive integer n > 2, let C1, C2, C3 be the boundaries ofthree onvex n-gons in the plane suh that the sets C1∩C2, C2∩C3, C3∩C1are �nite. Find the maximum ardinality the set C1 ∩ C2 ∩ C3 may have.

To omplete the olletion of problems for this number we give theIndian Team Seletion Test problems, 2007. Thanks again go to Bill Sandsfor obtaining them for the Corner.INDIAN TEAM SELECTION TEST 20071. Let ABC be a triangle with AB = AC, and let Γ be its irumirle. Theinirle γ of ABC moves (slides) on BC in the diretion of B. Prove thatwhen γ touhes Γ internally, it also touhes the altitude through A.2. Consider the quadrati polynomial p(x) = x2 + ax + b, where a, b are inthe interval [−2, 2]. Determine the range of the real roots of p(x) as a and bvary over [−2, 2].3. Let triangle ABC have side lengths a, b, c; irumradius R, and internalangle bisetor lengths wa, wb, wc. Prove thatb2 + c2

wa+

c2 + a2

wb+

a2 + b2

wc> 4R .

4. Let a1, a2, . . . , an be an ordering of the numbers 1, 2, . . . , n. Findmin

nXj=1

|aj − aj+1| and max nXj=1

|aj − aj+1| ,where an+1 = a1 and the extrema are taken over all suh possible orderings.5. Show that in a non-equilateral triangle, the following are equivalent:(a) The angles of the triangle are in arithmeti progression.(b) The ommon tangent to the nine-point irle and the inirle is parallelto the Euler line.6. Let X be the set of all bijetive funtions from S = {1, 2, 3, . . . , n} toitself. Let f0(x) = x and f (k)(x) = f(f (k−1)(x)) for k ≥ 1, and for eahf ∈ X de�ne

Tf(j) =

�1, if f (12)(j) = j ,0, otherwise .Determine X

f∈X

nXj=1

Tf(j) .

2797. Let a, b, and c be nonnegative real numbers suh that a + b ≤ c + 1,b + c ≤ a + 1, and c + a ≤ b + 1. Prove that

a2 + b2 + c2 ≤ 2abc + 1 .8. Given a �nite string S of symbols a and b, we write △(S) for the numberof a's in S minus the number of b's. (For example, △(abbabba) = −1.) Weall a string S balaned if every substring (of onseutive symbols) T of Shas the property that −1 ≤ △(T ) ≤ 2. (Thus, abbabba is not balaned, asit ontains the substring bbabb and △(bbabb) = −3.) Find, with proof, thenumber of balaned strings of length n.9. De�ne the funtions f , g, h on Z × Z × Z as follows:f(x, y, z) = (3x + 2y + 2z, 2x + 2y + z, 2x + y + 2z) ,g(x, y, z) = (3x + 2y − 2z, 2x + 2y − z, 2x + y − 2z) ,h(x, y, z) = (3x − 2y + 2z, 2x − y + 2z, 2x − 2y + z) .Given a primitive Pythagorean triple (x, y, z), with x > y > z, prove that

(x, y, z) an be uniquely obtained by repeated appliation of f , g, h to thetriple (5, 4, 3). For example, (697, 528, 455) = (f ◦ h ◦ g ◦ h)(5, 4, 3).10. (Short-List, IMO-2007) Cirles Γ1 and Γ2, with entres O1 and O2 areexternally tangent at the point D and internally tangent to a irle Γ at pointsE and F , respetively. Line ℓ is the ommon tangent to Γ1 and Γ2 at D. LetAB be the diameter of Γ perpendiular to ℓ so that A, E, O1 are on the sameside of the line ℓ. Prove that AO1, BO2 and EF are onurrent.11. Find all pairs of integers (x, y) suh that y2 = x3 − p2x, where p is aprime suh that p ≡ 3 (mod 4).12. Find all funtions f : R → R satisfying the equation

f(x + y) + f(x)f(y) = (1 + y)f(x) + (1 + x)f(y) + f(xy) ,for all x, y ∈ R.Next is an apology for having mis�led some solutions fromMohammedAassila with a group of solutions to problems for a later number of the Cor-ner. He should also appear as a solver for two problems disussed in the Aprilnumber of the Corner. These are Problem 3, Thai Mathematial Olympiad[2010 : 155; 2009 : 22℄ and Problem 2, The Italian Mathematial Olympiad[2010 : 164; 2009 : 25{26℄.

280Now we turn to solutions from our readers to problems of the Hungar-ian Mathematial Olympiad 2005{2006, National Olympiad, Grades 11{12,Seond Round given at [2009 : 211℄.1. Find the positive values of x that satisfy

x(2 sin x−cos 2x) <1

x.Solution by Oliver Geupel, Br uhl, NRW, Germany.Sin cos 2x = 1 − 2 sin2 x and x > 0, the given inequality is equiva-lent to x2 sin x(1+sin x) < 1. Write a = 2 sin x(1 + sin x). For x > 0, theinequality xa < 1 holds if and only if either(1) 0 < x < 1 and a > 0, or(2) x > 1 and a < 0.If 0 < x < 1, then we have sin x > 0; hene the ondition (1) issatis�ed.If x > 1, then a < 0 holds if and only if −1 < sin x < 0. Thus, (2) isequivalent to (2k +1)π < x < �2k + 3

2

�π or �2k + 3

2

�π < x < (2k +2)πfor some nonnegative integer k.Consequently, the solution set is the union of these open intervals:

(0, 1) ∪∞[

k=0

�(2k + 1)π,

�2k +

3

2

�π�

∪∞[

k=0

��2k +

3

2

�π, (2k + 2)π

� .2. For f(x) = ax2 − bx + c we know that 0 < |a| < 1, f(a) = −b, andf(b) = −a. Prove that |c| < 3.Solved by Arkady Alt, San Jose, CA, USA; George Apostolopoulos, Messo-longhi, Greee; Mihel Bataille, Rouen, Frane; Geo�rey A. Kandall, Ham-den, CT, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON;Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and TituZvonaru, Com�ane�sti, Romania. We give Kandall's write-up.First, suppose a = b. Then f(x) = ax2 − ax + c and −a = f(a) =a3 − a2 + c, that is, c = a2 − a3 − a. Then |c| ≤ |a|2 + |a|3 + |a| < 3.Next, suppose a 6= b. Then

−b = f(a) = a3 − ab + c , (1)−a = f(b) = ab2 − b2 + c . (2)Subtrating (2) from (1), we obtain su

essivelya − b = a(a2 − b2) − b(a − b) ,

1 = a(a + b) − b ,b(1 − a) = (a + 1)(a − 1) ,

b = −(a + 1) .

281From (2) we now su

essively dedue

−a = a(a + 1)2 − (a + 1)2 + c ,c = 1 − a2 − a3 ,

|c| ≤ 1 + |a|2 + |a|3 < 3 ,as desired.Next we look at solutions to problems of the Final Round of the Hun-garian Mathematial Olympiad 2005{2006 National Olympiad, Grades 11{12given at [2009 : 211-212℄.1. De�ne the funtion t(n) on the nonnegative integers by t(0) = t(1) = 0,

t(2) = 1, and for n > 2 let t(n) be the smallest positive integer whih doesnot divide n. Let T (n) = t(t(t(n))). Find the value of S ifS = T (1) + T (2) + T (3) + · · · + T (2006) .Solved by Mihel Bataille, Rouen, Frane; Oliver Geupel, Br uhl, NRW, Ger-many; and Titu Zvonaru, Com�ane�sti, Romania. We give Bataille's solution,modi�ed by the editor.We observe that T (n) = 0 if n = 2 or if n is an odd integer, hene

S = T (4) + T (6) + T (8) + · · · + T (2006).If n ∈ P = {4, 6, 8, . . . , 2006} and n is not a multiple of 3, thent(n) = 3 and T (n) = 1.Let P1 = {6, 18, 30, . . . , 6 · 333} be the set of elements of P of theform 6(2m−1), and let P2 = {12, 24, 36 . . . , 6 ·334} be the set of elementsof P of the form 12m.For eah n ∈ P1, we have t(n) = 4, hene T (n) = 2, and there are167 numbers in P1.If n ∈ P2 and n is not a multiple of 5 or n is not a multiple of 7, thent(n) = 5 or t(n) = 7, and then T (n) = 1. Otherwise T (n) is one of thefollowing: T (420) = 2, T (840) = 1, T (1260) = 2, or T (1680) = 1.Thus, T (n) = 1 for eah n ∈ P2 exept for two numbers where T takesthe value 2.In summary, T (n) = 2 for 167+2 = 169 numbers and T (n) = 1 for the1002−169 = 833 remaining numbers. Therefore, S = 833+2 ·169 = 1171.3. A unit irle k with entre K and a line e are given in the plane. Theperpendiular from K to e intersets e in point O and KO = 2. Let H bethe set of all irles entred on e and externally tangent to K.Prove that there is a point P in the plane and an angle α > 0 suh that∠APB = α for any irle in H with diameter AB on e. Determine α andthe loation of P .

282Solution by Titu Zvonaru, Com�ane�sti, Romania.Let M and N be points on e suh that OM = ON and the irleswith diameter OM and ON are in H. Let ST be the diameter of the irlebelonging to H and entred at O.

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K

M NOO1

P

S T

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

r

r rrr

r

r

The △MPN is isoseles sine ∠MPO = ∠NPO = α and O is themidpoint of MN , hene the point P lies on KO.Let O1 be the midpoint of OM , and let r = MO1 = O1O. In △KO1Owe have O1O2 + OK2 = O1K2, or r2 + 4 = (r + 1)2, and hene r = 32.Let OP = t; then in △PMO we have tan α = 3

t, and in △SPO wehave tan α

2=

1

t. Using the formula tan α = 2 tan α/2

1 − tan2 α/2 we obtain3

t=

2t

1 − 1t2

⇐⇒ 3t

=2t

t2 − 1⇐⇒ t2 = 3 ,

so OP = √3 and α = 60◦.It remains to prove that the point P and the angle α have the propertythat ∠APB = α for any irle in H with diameter AB on e............................................................................................................................................................

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A B

K

O O′

P

a x e............................................. . . . . . . . . . . . . . . . . . . . . .......................

r r

r

r r

r

Suppose that A lies between O and B. Let OA = a, AO′ = x, and letthe midpoint of AB be O′.

283In △KOO′ we obtain

OO′2 + OK2 = O′K2 ⇐⇒ (a + x)2 + 4 = (x + 1)2 ,hene, x = a2 + 32(1 − a) . It follows that OB = a + 2 · a2 + 32(1 − a) = a + 31 − a .Thus

tan ∠APB = tan(∠OPB − ∠OPA)

=tan ∠OPB − tan ∠OPA

1 + tan ∠OPB · tan ∠OPA

=

a + 3√3(1 − a)

− a√3

1 +a(a + 3)

3(1 − a)

=a2 + 3

√3(1 − a)

· 3(1 − a)3 + a2

=√

3 ,hene ∠APB = 60◦.For other loations of AB, the alulations are similar.

Next we move to the Hungarian Mathematial Olympiad 2005{2006,Speialized Mathematial Classes, First Round given at