RMC Brosura 2011 Final

ROMANIAN MATHEMATICAL COMPETITIONS 2011 Editorial Board of the RMC series Marian Andronache National College “Sf. Sava” Bucharest ˘ luna ˘ Mihai Ba National College “Mihai Viteazul” Bucharest Mircea Becheanu University of Bucharest Bogdan Enescu National College “ B.P. Ha¸sdeu” Buz˘ au Radu Gologan – Coordinator of the series Institute of Mathematics and University “Politehnica” Bucharest ˘ lin Popescu Ca Institute of Mathematics Dan Schwarz Intuitext Company ˘ nescu Dinu S ¸ erba National College “Sf. Sava” Bucharest

c 2011 Societatea de S Copyright  ¸ tiint¸e Matematice din Romˆ ania

All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher.

Societatea de S ¸ tiint¸e Matematice din Romˆ ania Str. Academiei 14, 010014 Bucure¸sti, Romˆ ania

http://www.rms.unibuc.ro tel/fax: +40213124072 ISSN 2066–6446

ROMANIAN MATHEMATICAL COMPETITIONS 2011

˘ ˘ and RADU GOLOGAN Edited by MIHAIL BALUN A

˘ ˘ , A NDREI E CKSTEIN Contributors M ARIAN A NDRONACHE , M IHAIL B ALUN A ˘ ˘ R ADU G OLOGAN , C EZAR L UPU , C ALIN P OPESCU , D INU S¸ ERB ANESCU

TABLE OF CONTENTS

Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v 1.1. 2011 Romanian Mathematical Olympiad – District Round . . . . . . . . . . . . . . . . . . . . . .1 1.2. 2011 Romanian Mathematical Olympiad – Final Round . . . . . . . . . . . . . . . . . . . . . . .13 1.3. Shortlisted problems for the 2011 Romanian NMO . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1.4. Selection tests for the 2011 BMO and IMO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 1.5. Selection tests for the 2011 JBMO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 1.6. 2011 Balkan Mathematical Olympiad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 1.7. 2010 – 2011 Local Mathematical Competitions 1.7.1. The 2010 DANUBE Mathematical Competition . . . . . . . . . . . . . . . . . . . . . . . . 63 1.7.2. The 2010 Eighth IMAR1 Mathematical Competition . . . . . . . . . . . . . . . . . . 67 1.7.3. The 2010 Fourth S TARS OF M ATHEMATICS Competition . . . . . . . . . . . . . . . 73 1.7.4. The 2011 Fourth ROMANIAN M ASTER OF M ATHEMATICS . . . . . . . . . . . . . 81 1.7.5. The 2011 C LOCK -T OWER S CHOOL Seniors Competition . . . . . . . . . . . . . . 93 1.6.6. The 2011 C LOCK -T OWER S CHOOL Juniors Competition . . . . . . . . . . . . . . 97

1 I NSTITUTE OF

M ATHEMATICS OF THE ROMANIAN ACADEMY. One-day IMO-type contest.

iii

FOREWORD

The 18th volume of the ”Romanian Mathematical Contests” series contains more than 200 problems, submitted at different stages of the Romanian Mathematical Olympiad, other Romanian Contests, and some international ones. Most of them are original, but some problems from other sources were used as well during competition. A significant part of the problems are discussed in detail, and alternative solutions or generalizations are given. Some of the solutions belong to students and were given while they sat the contest; we thank them all. We thank the Ministry of Education, Research, Youth and Sports for constant involvement in supporting the Olympiads and the participation of our teams in international events. Special thanks are due to the ”Dinu Patriciu” Foundation, to WBS, and to the Council of the District of Dˆambovit¸a – all sponsors of the Romanian IMO team. The Editors

v

THE 62nd ROMANIAN MATHEMATICAL OLYMPIAD DISTRICT ROUND 7th GRADE

Problem 1. In a square of side length 60, 121 distinct points are given. Show that among them there exists three points which are vertices of a triangle with an area not exceeding 30. Solution. Divide the square into 60 rectangles 5×12. By the pigeonhole principle, there are three points among the given ones inside one of the rectangles. The area of this triangle does not exceed half of the area of the rectangle, that is 5 · 12/2 = 30, as needed. Problem 2. Consider an isosceles trapezoid ABCD with perpendicular diagonals. The parallel from the intersection point of the diagonals meets the non-parallel sides [BC] and [AD] at points P and R respectively. Point Q is the mirror image of P across the midpoint of [BC]. Show that a) QR = AD; b) QR ⊥ AD. Claudiu Popa

Solution. Let the diagonals AC and BD meet at point O and let M be the midpoint of the line segment [BC]. a) Since OM joins the midpoints of two sides of the triangle P QR, M O  RQ and OM = RQ 2 . On the other hand, [OM ] is a median of the rightangled triangle BOC, hence OM = 12 BC = 1 2 AD. Consequently, RQ = AD. b) Let the lines M O and AD meet at T . Then ∠M BO = ∠M OB = ∠DOT . Since ∠OCB = ∠T DO, it follows that ∠OT D = ∠BOC = 90◦ , so M T ⊥ AD. 1

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

Problem 3. A positive integer N has the digits 1, 2, 3, 4, 5, 6 and 7, so that each digit i, i ∈ {1, 2, 3, 4, 5, 6, 7} occurs 4i times in the decimal representation of N . Prove that N is not a perfect square. Solution. N has 1·4 = 4 digits equal to 1, 2·4 = 8 digits equal to 2, . . ., 7·4 = 28 digits equal to 7, so the sum of its digits equals S = 4(12 + 22 + · · · + 72 ) = 560. Since 560 = 3 · 186 + 2, the number N is not a square, and the remainder left by a perfect square upon division by 3 cannot be equal to 2. Problem 4. Find the sum of the elements of the set n m  M= + | m, n = 0, 1, 2, . . . , 100 . 2 5

Gazeta Matematic˘a

Solution. Consider the set A = {2a + 5b | a, b = 1, 2, . . . , 100} and notice that 1∈ / A and 3 ∈ / A. The largest even number from A is equal to 700 and it is obtained for a = b = 100. The number 698 is obtained for a = 99, b = 100. The largest odd number from A is equal to 695 and it is obtained for b = 99, a = 100, implying that 697 ∈ / A and 699 ∈ / A. We claim that all the integers between 4 and 695 belong to A. Let y ≤ 500 and let r be the remainder left by y upon division by 5. Write y = 5c + r with 0 ≤ c ≤ 100 and 0 ≤ r ≤ 4. If r is even, then y = 5c + 2k, where r = 2k. If r is odd, then y ≥ 5, so c ≥ 1 and y = 5(c − 1) + 2(k + 3), where r = 2k + 1. For y > 500, write y = 500 + z with z ≥ 200. If z is even, then y = 5 · 100 + 2k, where z = 2k. If z is odd, then y ≤ 695, so z ≤ 195 and y = 5 · 99 + 2(k + 3), where z = 2k + 1. A quick inspection of all the above cases shows that the claim holds. The sum of all the elements of M is equal to S=

1 ((1 + 2 + · · · + 700) − (1 + 3 + 697 + 699) = 350 · 697) = 35 · 697. 10

8th GRADE

Problem 1. Find the real numbers x and y such that (x2 − x + 1)(3y 2 − 2y + 3) − 2 = 0.

Petre B˘atrˆanet¸u

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

Solution. From x2 −x+1 = (x− 21 )2 + 34 ≥ 34 , 3y 2 −2y +3 = 3(y − 13 )2 + 83 ≥ 83 we derive that (x2 − x + 1)(3y 2 − 2y + 3) ≥ 34 · 83 = 2, for any real numbers x and y. Equality holds for (3y − 1)2 = 0 and (x − 12 )2 = 0, that is x = 12 and y = 13 . Problem 2. a) Show that m2 −m+1 is an element of the set {n2 +n+1 | n ∈ N}, for any positive integer m. b) Let p be a perfect square, p > 1. Prove that there exists positive integers r and q such that p2 + p + 1 = (r2 + r + 1)(q 2 + q + 1). Mircea Fianu

Solution. a) Since m2 − m + 1 = (m − 1)2 + (m − 1) + 1 and m − 1 ≥ 0, it follows that m − 1 ∈ N and consequently m2 − m + 1 ∈ {n2 + n + 1 | n ∈ N}. b) Write p = k 2 , where k is an integer. Since p > 1, we have k ≥ 2. Now p2 + p + 1 = k 4 + k 2 + 1 = (k 2 + 1)2 − k 2 = (k 2 − k + 1)(k 2 + k + 1). Numbers r = k and q = k − 1, both positive integer, satisfy the claim. Problem 3. Consider a regular prism ABCA B  C  . A plane α containing point A meet the rays (BB  and (CC  at points E and F such that area [ABE] + area [ACF ] = area [AEF ]. Find the angle determined by the planes (AEF ) and (BCC  ). Gazeta Matematic˘a

Solution. Let M be the midpoint of BC and let u be the angle determined by the planes (AEF ) and (BCC  ). Since triangle M EF is the projection of the triangle AEF onto (BCC  ), we have cos u =

[BCF E] [BCF E] [M EF ] = = = [AEF ] 2[AEF ] 2([ABE] + [ACF ]) [BCF E] 1 [BCF E] = = , = 4([M BE] + [M CF ]) 2[BCF E] 2

hence u = 60◦ . Problem 4. Find all positive integers m such that √ √ { m} = { m + 2011}. Alexandru Blaga

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

√ √ √ √ Solution. Rewrite the equation m − [ m] = m + 2011 − [ m + 2011] to √ √ √ √ √ get m + 2011 − m = [ m + 2011] − [ m] = p ∈ N. Squaring m + 2011 = √ √ p + m yields 2011 = p2 + 2p m ∈ N, hence m = k 2 , k ∈ N∗ . The relation 2011 = p(p + 2k) gives p = 1 and p + 2k = 2011, since 2011 is prime. Hence k = 1005 and m = 10052 . 9th GRADE

Problem 1. Points M , N , P , Q are given on the sides AB, BC, CD, DA of a −−→ −−→ −→ −−→ −−→ −−→ parallelogram ABCD such that M N + QP = AC. Prove that P N + QM = DB. −−→ AQ BN DP Solution. Denote m = AM AB , n = BC , p = DC , q = AD , to get M N = −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ AN − AM = AB + BN − AM = (1 − m)AB + nAD, QP = pAB + (1 − q)AD. −−→ −−→ −→ −−→ −−→ The given condition becomes (1 − m + p)AB + (1 − q + n)AD = AC = AB + AD. −−→ −−→ Notice that vectors AB and AD have distinct directions to deduce that m = p and −−→ −−→ −−→ −−→ −−→ −−→ n = q. On the other hand, P N = (1− p)AB − (1−n)AD and QM = mAB − q AD, −−→ −−→ −−→ −−→ −−→ −−→ −−→ implying P N + QM = (1 − p + m)AB − (1 − n + q)AD = AB − AD = DB, as needed. Problem 2. For each positive integer n consider the set An of all the numbers obtained by choosing signs in ±1 ± 2 ± · · · ± n; for instance, A2 = {−3, −1, 1, 3} and A3 = {−6, −4, −2, 0, 2, 4, 6}. Find the cardinal of the set An . Solution. The largest element of the set is 1 + 2 + 3 · · · + n = n(n+1) , while the 2 n(n+1) smallest is −1 − 2 − · · · − n = − 2 . The difference of any two elements of the set is even, implying that all elements have the same parity. and n(n+1) , sharing the same parity We claim that all numbers between − n(n+1) 2 2 . If with them belong to the set An – and only them. Indeed, let x ∈ An , x < n(n+1) 2 (one of) its representation begins with −1, changing into +1 gives x + 2 ∈ An . Suppose that all its representations start with +1. There exists a first negative – otherwise ; denote it by −k: x = 1 + 2 + · · · + n = n(n+1) 2 x = +1 + 2 + · · · + (k − 1) − k ± · · · ± n. Switching the signs of the summands k − 1 and k we derive that x + 2 ∈ An . + 1 elements. Consequently, the set An has n(n+1) 2

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

Problem 3. Let f : R → R be a function such that its 2-fold composition is equal to the floor function, i.e. f (f (x)) = x, for any real number x. Prove that there exist distinct real numbers a and b such that |f (a) − f (b)| ≥ |a − b|. Gazeta Matematic˘a

Solution. We claim that f (n) ∈ Z, for any integer n. Indeed, write f (f (f (x))) = f ([x]) = [f (x)] to derive that f (n) = [f (n)] for any integer n, implying f (n) ∈ Z. Suppose that for all a, b ∈ Z we have |f (a) − f (b)| < |a − b|. Then |f (n + 1) − f (n)| < 1, for any integer n. Since both f (n) and f (n + 1) are integers we derive that f (n) = f (n + 1), hence f (n) = f (0), for any integer n. It follows that n = f (f (n)) = f (f (0)) = 0, a contradiction. Second solution. Notice that f (0) = f (1), otherwise 0 = f (f (0)) = f (f (1)) = 1, false. If |f (0) − f (1)| ≥ |0 − 1|, we are done; if else, from |0 − 1| = |f (f (0)) − f (f (1))| we obtain |f (f (0)) − f (f (1))| > |f (0) − f (1)|. The distinct numbers f (0) and f (1) fulfill the claim. Problem 4. Let a be a real number with {a} + { a1 } = 1. Prove that   1 = 1, {an } + an

for every positive integer n, where {x} denotes the fractional part of a real number x.

Dorel Mihet¸

1

Solution. Since a + a1 = a + a + 1, it follows that a + a1 ∈ Z. For each positive integer n let sn = an + a1n . We have s1 sn = sn+1 + sn−1 for any integer n ≥ 2. Notice that s2 = a2 + a12 = (a + a1 )2 − 2 ∈ Z, hence, by induction, sn ∈ Z, for all positive integers n.     This implies {an } + { a1n } = an + a1n − an  − a1n = sn − an  − a1n ∈ Z,   hence {an } + a1n ∈ {0, 1}, for {x} ∈ [0, 1).   If {an } + a1n = 0, then both numbers an and a1n are integers, implying an = 1   and furthermore a = ±1. Then {a} + a1 = 0 = 1, a contradiction. Consequently   {an } + a1n = 1, for any positive integer n. Remark. A number under the given condition is a solution to a + a1 = m ∈ Z, √ m± m2 −4 , with |m| > 2. An alternative solution can be obtained implying a = 2 noticing that the Newton sums sn = an1 +an2 of the roots of the equation a2 −ma+1 = 0 satisfy the relation sn+2 − msn+1 + sn = 0, which, in fact, is an equivalent form of s1 sn = sn+1 + sn−1 .

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

10th GRADE

Problem 1. Prove that, if a, b, c are positive real numbers, then the equation a + bx = cx has at most a real solution. x

Gazeta Matematic˘a

 a x

 b x

Solution. Rewrite the equation as f (x) : = c + c = 1. If ( ac − 1)( cb − 1) ≤ 0, the equation has neither positive, nor negative solutions. If ( ac −1)( cb −1) > 0, the function f is monotone, therefore the equation f (x) = 1 has at most one solution. Problem 2. a) Let z1 , z2 , z3 , z4 be distinct complex numbers of zero sum, having equal absolute values. Prove that the points of affixes z1 , z2 , z3 , z4 are the vertices of a rectangle. b) Let x, y, z, t be real numbers such that sin x + sin y + sin z + sin t = 0 and cos x + cos y + cos z + cos t = 0. Prove that, for every integer n, sin(2n + 1)x + sin(2n + 1)y + sin(2n + 1)z + sin(2n + 1)t = 0. Aurel Bˆarsan

Solution. a) The equality z1 + z2 + z3 + z4 = 0 implies z¯1 + z¯2 + z¯3 + z¯4 = 0 and furthermore z11 + z12 + z13 + z14 = 0 (1), for |z1 | = |z2 | = |z3 | = |z4 | = 0. Suppose z1 + z2 = −z3 − z4 = 0. The relation (1) gives z1 z2 = z3 z4 , so {z1 , z2 } = {−z3 , −z4 }. On the other hand, if z1 + z2 = 0, then z3 + z4 = 0. In both cases the numbers z1 , z2 , z3 , z4 form two pair of equal sum, hence the conclusion. b) Let z1 = cos x + i sin x, z2 = cos y + i sin y, z3 = cos z + i sin z and z4 = cos t + i sin t to get z1 + z2 + z3 + z4 = 0 and |z1 | = |z2 | = |z3 | = |z4 | = 1. As before, the numbers z1 , z2 , z3 , z4 form two pairs of opposite numbers, so the same goes for numbers z12n+1 , z22n+1 , z32n+1 , z42n+1 . Therefore z12n+1 + z22n+1 + z32n+1 + z42n+1 = 0, implying the claim. Problem 3. Let a and b be two complex numbers. Prove that the following statements are equivalent: 1) The absolute values of the roots of the equation x2 −ax+b = 0 are respectively equal to the absolute values of the roots of the equation x2 − bx + a = 0. 2) a3 = b3 or b = a. Mihail B˘alun˘a

2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

7

Solution. Let |x1 | = |x3 |, |x2 | = |x4 | (1) and notice that |a| = |x3 x4 | = |x1 x2 | = |b| to derive that |x1 + x2 | = |x3 + x4 | (2). The relations (1) and (2) show that there exists a number k ∈ C such that x2 = kx1 , x4 = kx3 or x2 = kx1 , x4 = kx3 . In the first case we have a = kx23 = (1 + k)x1 and b = kx21 = (1 + k)x3 , so a3 = k(1 + k)2 x21 x23 = b3 . In the latter case we have a = kx23 = (1 + k)x1 and b = kx21 = (1 + k)x3 . It follows that x21 x1 = x3 x23 , so x1 = x3 or a = b = 0, and furthermore x2 = x4 , hence a = b. Conversely, if b = a, then x1 +x2 = x3 +x4 , x1 x2 = x3 x4 , implying {x1 , x2 } = {x3 , x4 }. If a3 = b3 , then a = εb, ε3 = 1. The roots satisfy the relations x1 + x2 = ε(x3 + x4 ), x1 x2 = ε2 x3 x4 . Both cases lead to {|x1 |, |x2 |} = { |x3 |, |x4 |}, as needed. Problem 4. a) Show that if a, b > 1 are distinct real numbers then loga (loga b) > logb (loga b). b) Let a1 > a2 > . . . > an > 1 be real numbers, n ≥ 2. Prove that loga1 (loga1 a2 )+loga2 (loga2 a3 )+. . .+logan−1 (logan−1 an )+logan (logan a1 ) > 0. Solution. a) For a < b, loga (loga b) = (loga b)(logb (loga b)) > logb (loga b) because logb (loga b) > 0 and loga b > 1. For a > b, the claim is reached from loga b < 1 and logb (loga b) < 0. b) Induct on n. For n = 2, loga1 (loga1 a2 )+loga2 (loga2 a1 ) > loga2 (loga1 a2 )+loga2 (loga2 a1 ) = loga2 1 = 0. Assume now that the claim holds for some n and consider the numbers a1 > a2 > . . . > an+1 > 1. Then loga1 (loga1 a2 ) + . . . + logan (logan an+1 ) + logan+1 (logan+1 a1 ) = = loga1 (loga1 a2 ) + . . . + logan−1 (logan−1 an ) + logan (logan a1 )+ + logan (logan an+1 ) + logan+1 (logan+1 a1 ) − logan (logan a1 ) > > logan+1 (logan an+1 ) + logan+1 (logan+1 a1 ) − logan (logan a1 ) = = logan+1 (logan a1 ) − logan (logan a1 ) > 0, because logan a1 > 1 and an+1 < an .

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

11th GRADE

Problem 1. Let α be an irrational number. For any n ∈ N∗ let an = {nα} and define the sequence (xn )n≥1 by xn = (a2 − a1 )(a3 − a2 ) · · · (an+1 − an ). Show that the sequence is convergent and find its limit. Marian Cucoanes¸

Solution. We claim that {x + y} − {y} is equal to {x} or {x} − 1, for any real numbers x and y. Write x + y = [y] + [x] + {x} + {y} to get {x + y} − {y} = {{x} + {y}} − {y}. Hence if {x} + {y} < 1 then {x + y} − {y} = {x}, while if 2 > {x} + {y} ≥ 1 then {x + y} − {y} = {x} − 1, as claimed. Apply the above result to infer that either |an+1 − an | = {α} or |an+1 − an | = 1 − {α}. Set b = max{{α}, 1 − {α}} and notice that 0 < b < 1 (for α is irrational) to derive that |xn | ≤ bn , implying lim |xn | = 0 and furthermore lim xn = 0. n→∞

n→∞

Problem 2. Consider the matrices A ∈ Mm,n (C), B ∈ Mn,m (C) with n ≤ m. Given that rank (AB) = n and (AB)2 = AB, find BA. Marian Andronache

Solution. Left–multiply with B and right–multiply by A the equality (AB)2 = AB to obtain (BA)3 = (BA)2 . Recall that the rank of a matrix product does not exceed the rank of its factors to derive from ABAB = AB that rankBA ≥ n, hence rankBA = n. The square matrix BA of order n is thus invertible, hence (BA)3 = (BA)2 , which implies BA = In . Problem 3. Let A, B ∈ M2 (C) be two non-zero matrices with AB + BA = O2 and det(A + B) = 0. Prove that tr(A) = tr(B) = 0. Gazeta Matematic˘a

Solution. From AB + BA = O2 we obtain (A + B)2 = A2 + B 2 and (A − B)2 = A + B 2 . Consequently, det(A − B) = 0. The characteristic equations for A + B and A − B yield 2

A2 + B 2 − tr(A + B)(A + B) = O2 ,

A2 + B 2 − tr(A − B)(A − B) = O2 ,

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

so, subtracting we get tr(A)B = tr(B)A. If tr(A) = 0, then tr(B) = 0 for else A = O2 , a contradiction. Hence A = λB, so AB + BA = O2 leads to λB 2 = O2 . Therefore λ = 0, which yields tr(A) = tr(B) = 0. Problem 4. Find all functions f : [0, 1] → R satisfying for all x, y ∈ [0, 1] the inequality |x − y|2 ≤ |f (x) − f (y)| ≤ |x − y|. Nicoale Bourb˘acut¸

Solution. The condition |f (x) − f (y)| ≤ |x − y| is sufficient for the continuity of f . From |x − y|2 ≤ |f (x) − f (y)| we derive that f is one-to-one, implying that f is a strictly monotonic function. We may assume that f is strictly increasing, because f can be replaced by −f . Set x = 0 and y = 1 to derive that 0 ≤ f (1) − f (0) ≤ 1, hence f (1) = f (0) + 1.

For x ≥ y we get f (x) − f (y) ≤ x − y or y − f (y) ≤ x − f (x), implying that the function g(x) = x − f (x) + f (0) is increasing.

Since g(0) = g(1) = 0, the function g is the identically null. Consequently, the functions are fa± given by fa± (x) = ±x + a, with a ∈ R.

Alternative solution. For x = 0 and y = 1 we get |f (1)−f (0)| = 1. For x ∈ [0, 1] we obtain 1 = |f (1) − f (0)| ≤ |f (1) − f (x)| + |f (x) − f (0)| ≤ |1 − x| + |1 − x| = 1 − x + x − 0 = 1.

Thus f (x) lies between f (0) and f (1); moreover, we have |f (1) − f (x)| = 1 − x and |f (x) − f (0)| = x. It follows that points (0, f (0)), (x, f (x)) and (1, f (1)) are collinear, hence f (x) = f (0) ± x. 12th GRADE

1 Problem 1. Show that π



cos sin

π 13

π 13



1 − x2 dx is a rational number.

Solution. Consider the function F : [0, π/2] → R, F (t) =

The function is differentiable and the derivative is equal to F  (t) = (− sin t)

Gazeta Matematic˘a



cos t sin t



1 − x2 dx.

  1 − (cos t)2 −(cos t) 1 − (sin t)2 = −(sin t)2 −(cos t)2 = −1.

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

Thus F (t) = −t+k, k ∈ C. Since F (π/4) = 0, it follows that k = π/4, implying F (t) = π/4 − t. Consequently, F (π/13)/π = (π/4 − π/13)/π = 9/52, which is a rational number. Problem 2. Let G be the set of matrices   a b , a, b ∈ Z7 , a = ˆ0. ˆ 0 ˆ 1 a) Show that G is a group with respect to the matrix multiplication. b) Prove that there exists no proper homomorphism from G to Z7 . 

a ˆ 0



b ˆ 1



x ˆ0

y ˆ1



Solution. a) Consider A = and B = two elements of G. Then   ax ay + b AB = ∈ G, for ax = ˆ 0. ˆ ˆ 0 1 To end the proof, recall that matrix multiplicationis an associative  law, the group a−1 −a−1 b unit is I2 ∈ G and, finally, the inverse matrix of A is ∈ G. ˆ0 ˆ1   a b b) Let A = be an element of G with a = ˆ1. Since ˆ 0 ˆ 1   ak b(ak−1 + ak−2 + · · · + 1) k , it follows that A6 = I2 . A = ˆ ˆ 0 1 Let f be a group homomorphism mapping G to Z7 . Notice that ˆ0 = f (I2 ) = ˆ (A) to derive that f (A) = 0. ˆ Since ker(f ) = {X ∈ G | f (X) = 0} ˆ f (A6 ) = 6f

is a subgroup of G with at least 36 elements, and G has 42 elements, we obtain that ker(f ) = G. Therefore f (X) = ˆ 0, for any X ∈ G. Problem 3. Consider an increasing continuous function f : [0, 1] → R and define    2n the sequence (an )n≥1 by an = 21n k=1 f 2kn , for all integers n ≥ 1. a) Prove that the sequence (an )n≥1 is increasing.

b) Given that there exists p ∈ N such that ap = ∗

constant function.



1

f (x)dx, prove that f is a 0

Nicoale Bourb˘acut¸

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

Solution. a) Notice that an+1 = and f

1 2n+1

 2k−1  2n+1

≤f

n+1 2

f

k=1



k 2n





k 2n+1



=

1 2n+1

 2n     2n   k 2k − 1 f f + 2n 2n+1 k=1

k=1

to derive that   2 1  k = an . ≤ n f 2 2n n

an+1

k=1

b) Consider the numbers k ∈ {1, 2, . . . , 2p }, c ∈

the closed interval [0, 1]

∆=



 k−1 k , , and a division of 2p 2p

 k−1 k 1 0, p , · · · , p , c, p , · · · , 1 . 2 2 2



Denote by S the upper Darboux sum with relative to ∆ to get        1 k k−1 k k ap − S = p f −c − f (c) c − p −f 2 2p 2 2p 2p      k−1 k − f (c) c− p ≥ 0. = f p 2 2  1  1 f (x)dx ≤ S ≤ ap = f (x)dx, it follows that ap = S and, furtherSince 0 0   k more, f (c) = f . Therefore, f is a constant function on each of the intervals p 2   k−1 k , k = 1, 2, . . . , 2p , whose union is (0, 1]. Hence f is a constant function , 2p 2p on [0, 1], because the function f is continuous. Problem 4. Let A be a ring and let a be an element of A. Prove that a) If A is commutative and a is nilpotent, then a + x is invertible for any invertible element of x ∈ A. b) If A is finite and a + x is invertible for any invertible element x ∈ A, then a is nilpotent. (An element a of a ring is called nilpotent if there exists some positive integer n such that an = 0.) Marian Andronache

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – D ISTRICT ROUND

Solution. a) Let x be an invertible element and let n be a positive integer such that an = 0. Since a + x = x(x−1 a + 1), it is enough to show that x−1 a + 1 is invertible. Set b = x−1 a. Then bn = x−n an = 0, (because A is commutative), hence b2n+1 = 0. Consequently, 1 = b2n+1 + 1 = (b + 1)(b2n − b2n−1 + · · · − b + 1), i. e., b + 1 is invertible. b) Induct on n, n ≥ 1, to prove that an − 1 is invertible. For x = −1 we get that a − 1 is invertible. Assume that b = an − 1 is invertible. From the hypothesis it follows that a − b−1 is also invertible, and so is ab − 1 = (a − b−1 )b. Moreover, an+1 − 1 = a + (a(an − 1) − 1) = a + (ab − 1) is invertible. Since A is finite, there exist two integers q > p ≥ 1 such that ap = aq , i.e. ap (aq−p − 1) = 0. Since aq−p − 1 is invertible, it follows that ap = 0.

THE 62nd ROMANIAN MATHEMATICAL OLYMPIAD FINAL ROUND

7th GRADE

Problem 1. Find all positive integers r with the property that there exists positive prime numbers p and q so that p2 + pq + q 2 = r2 . Aurel Bˆarsan 2

Solution. The given relation is equivalent to (p + q) = r2 + pq, which can be written (p + q + r) (p + q − r) = pq . The divisors of pq are 1, p, q and pq. Since p + q > max {p, q}, it follows that p + q − r = 1 and p + q + r = pq. Adding the last two equalities yields 2p+2q = pq+1 , that is (p − 2) (q − 2) = 3. This leads to (p, q) ∈ {(3, 5), (5, 3)}; in both cases r = 7. Problem 2. The numbers x, y, z, t, a and b are positive integers, so that xt − yz = 1 and xy > ab > zt . Prove that ab ≥ (x + z) (y + t).

Dan Nedeianu

Solution. From xy > ab follows that xb > ya, hence xb−ya ≥ 1. In the same way, at − bz ≥ 1. Multiplying the first inequality by t, the second one by y and adding the two relations yields bxt − byz ≥ t + y, that is b ≥ t + y. In the same way, a ≥ x + z. These two inequalities lead to the conclusion. Second solution. If d = g.c.d.(a, b), then a = da1 and b = db1 , with a1 , b1 ∈ N∗ , (a1 , b1 ) = 1 and xy > ab11 . It follows that b1 x − a1 y = u ≥ 1 and a1 t − b1 z = v ≥ 1, with u, v ∈ N. The first relation gives b1 xz − a1 yz = uz, and the second gives a1 xt − b1 zx = vx. Adding these relations yields a1 = uz + vx ≥ z + x. In the same way, b1 = ut + vy ≥ t + y. Since ab ≥ a1 b1 , the conclusion is proven. 13

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

Problem 3. The convex quadrilateral ABCD has ∠BCD = ∠ADC ≥ 90◦ . The bisectors of the angles ∠BAD and ∠ABC meet at a point M , placed on the line CD. Prove that M is the midpoint of the segment [CD]. Maria Mihet¸

Solution. Case I: AD and BC have a common point E. Then M is the incenter of the triangle ABE, hence (EM is the bisector of the angle ∠AEB. Since ∠ECD = ∠EDC, the triangle EDC is isosceles with base [DC]. Therefore [EM ] is a median in triangle EDC, so M is the midpoint of the segment CD.

Case II: AD  BC. Then ∠CAB = ∠ABD = 90◦ , hence ∠M AB + ∠M BA = 90 , that is triangle M AB has a right angle in M . Denote N the midpoint of the segment [AB]. Then triangle N AM is isosceles with base [AM ], so ∠N M A = ∠N AM = ∠M AD, hence M N  AD. It follows that M N is the central median of the trapezoid ABCD, whence the conclusion. ◦

Problem 4. Triangle ABC has ∠ABC = 60◦ . Points M and D are placed on the sides (AC) and (AB) respectively, so that ∠BCA = 2∠M BC and BD = M C. Find the measure of the angle ∠DM B. Gheorghe Bumb˘acea

Solution. Take G so that CG = CM , C ∈ (BG) and denote x = ∠M BC. Since ∠M CB is exterior to the isosceles triangle M CG, ∠M GC = x. This means that MB MG BG triangle M BG is isosceles and similar to triangle M CG. Thus, M G = M C = BD . Denote H the common point of the line AB and the perpendicular bisector of the segment [BG]. Then triangle HBG is isosceles and has an angle of 60◦ , so it is equilateral. This yields GH MG BG = = . MG MB BD

2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

15

Since ∠M GH = ∠M BD = 60◦ − x, triangles M GH and DBM are similar, therefore we get the answer ∠DM B = ∠M HG = 30◦ . 8th GRADE

Problem 1. Find all real numbers x, y, z, t ∈ [0, ∞) so that x + y + z ≤ t, x2 + y 2 + z 2 ≥ t and x3 + y 3 + z 3 ≤ t.

Cristian Laz˘ar

Solution. Adding x + y + z ≤ t, −2x2 − 2y 2 − 2z 2 ≤ −2t and x3 + y 3 + z 3 ≤ t, 2 2 2 one gets x (1 − x) + y (1 − y) + z (1 − z) ≤ 0. Since x, y, z ∈ [0, ∞), it follows x, y, z ∈ {0, 1}. If x = y = z = 0, then t = 0. If exactly two of the numbers x, y, z are nil, then 1 ≤ t and 1 ≥ t, therefore t = 1, (x, y, z) ∈ {(1, 0, 0) , (0, 1, 0) , (0, 0, 1)}. If exactly one of the numbers x, y, z is nil, then 2 ≤ t and 2 ≥ t, so t = 2, (x, y, z) ∈ {(1, 1, 0) , (1, 0, 1) , (0, 1, 1)}. If x = y = z = 1, then t = 3. Problem 2. Let a, b, c be distinct positive integers. a) Prove that a2 b2 + a2 c2 + b2 c2 ≥ 9. b) If, moreover, ab + ac + bc + 3 = abc > 0, show that (a − 1)(b − 1) + (a − 1)(c − 1) + (b − 1)(c − 1) ≥ 6.

Gabriel Popa

Solution. a) At least one of the numbers a, b, c has modulus at least 2, whence a2 b2 + a2 c2 + b2 c2 ≥ 1 · 1 + 1 · 4 + 1 · 4 = 9. b) The required inequality can be successively written ab + ac + bc − 2(a + b + c) + 3 ≥ 6 ab + ac + bc − 3 ≥ 2(a + b + c) (ab + ac + bc − 3)(ab + ac + bc + 3) ≥ 2abc(a + b + c) (ab + ac + bc)2 − 9 ≥ 2abc(a + b + c) a2 b2 + a2 c2 + b2 c2 ≥ 9,

that is exactly a).

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

Problem 3. Let V ABC be a regular pyramid with the base ABC having center O. Denote I and H the incenter and the orthocenter of the triangle V BC and suppose that AH = 3OI. Find the measure of the angle made by a lateral edge of the pyramid with the plane of the base. Mircea Fianu

Solution. Denote M the midpoint of the edge [BC]. Since triangle V BC is isosceles with V B = V C, the points V, H, I and M are collinear. From AC⊥OB and AC⊥OV follows AC⊥(BOV ), hence V B⊥AC, which, together with V B⊥CH leads to V B⊥(ACH), whence AH⊥V B. In the same way AH⊥V C, therefore AH⊥(V BC), so AH⊥V M . Denote I  the projection of O onto the plane (V BC); then I  belongs to V M and OI   AH. This leads to MO 1 OI  = = , AH MA 3 hence AH = 3OI  . This shows that OI equals the distance from O to (V BC), therefore I = I  . From OI⊥(V BC) and I = incenter of the triangle V BC follows that O has equal distances to the lines V B and BC. Denote J the projection of O onto the line V B. Then OJ = OM , hence the right triangles OJB and OM B are congruent. This leads to m(∠ (V B, (ABC))) = m(∠V BO) = m(∠M BO) = 30◦ . Problem 4. A positive integer will be called typical if the sum of its decimal digits is a multiple of 2011. a) Show that there are infinitely many typical numbers, each having at least 2011 multiples which are also typical numbers. b) Does there exist a positive integer such that each of its multiples is typical? ***

Solution. a) All the numbers made up by 2011n digits 1 (where n is a positive integer) are typical. Each such number, with 2011k digits, has among its multiples the numbers made up by 2011kn digits, which are typical. b) We will show that the answer is ’NO’.

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

We notice that each positive integer A has a multiple of the form 99 . . . 900 . . . 0. Indeed, if the decimal expansion of 1/A has a period of p digits and the initial nonperiodical part has q digits, then A1 = B C , where C = 99 . . . 900 . . . 0 has p digits 9 and q digits 0. Suppose now that n is a positive integer. Take a multiple of n of the form m = 99 . . . 900 . . . 0, with p nines and q zeros and consider the multiple of n M = (10p+1 − 1)m = 10p+1 m − m = 99 . . . 98900 . . . 0100 . . . 0, with p − 1 nines at the beginning. Then the sums of the digits of these numbers are s(M ) = s(m) + 9, so they cannot be both multiples of 2011. 9th GRADE

Problem 1. Let n be a positive integer and a1 , a2 , . . . , an be real numbers such that am + am+1 + · · · + an ≥ m + (m + 1) + · · · + n, for every m = 1, 2, . . . , n. n(n + 1)(2n + 1) . Prove that a21 + · · · + a2n ≥ 6

Romeo Raicu

Solution. Denote bk = ak −k, for k = 1, 2, . . . , n. Then bm +bm+1 +· · ·+bn ≥ 0 for m = 1, 2, . . . , n. It follows n 

a2i =

i=1

because

n  i=1

n  i=1

b2i +

n  i=1

2ibi +

n  i=1

i2 ≥

n  i=1

i2 =

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

ibi = (b1 + · · · + bn ) + (b2 + · · · + bn ) + · · · + bn ≥ 0.

Problem 2. Let n be a positive integer. Prove that every integer between 1 and n! can be written as the sum of at most n distinct positive divisors of n!. ***

Solution. We use induction on n; the base case n = 1 is obvious. Suppose now that the conclusion holds for some positive integer n and consider m ≤ (n + 1)!. Then m = (n + 1)q + r, q, r ∈ N, 0 ≤ r ≤ n. It is easily noticed that q ≤ n!, therefore q can be written as a sum of (at most) n divisors of n! – denote them d1 , . . . , dk , k ≤ n, whence m = (n + 1)d1 + (n + 1)d2 + · · · + (n + 1)dk + r

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

and (n + 1)di are divisors of (n + 1)!. If r = 0, the induction step is finished. Otherwise r > 0, r divides (n + 1)! and r < n + 1 ≤ (n + 1)di , therefore m can be written as the sum of k + 1 distinct divisors of (n + 1)!, with k + 1 ≤ n + 1 and we are done. √ Problem 3. Given an integer n > 0, prove that at least one of the numbers [2n 2], √ √ [2n+1 2], . . . , [22n 2] is even ([a] denotes the integral part of the real number a).

Radu Gologan

Solution. Suppose, by contradiction, that all the numbers are odd. Then there exists a positive integer a such that √ 2a − 1 < 2n 2 < 2a.

(1)

√ √ It follows that 4a − 2 < 2n+1 2 < 4a and, since [2n+1 2] is odd, √ 4a − 1 < 2n+1 2 < 4a.

√ Continuing in the same way we arrive at 2n+1 a − 1 < 22n 2 < 2n+1 a, whence √ 2n+1 a − 22n 2 < 1, which leads to √ 2n+1 (a2 − 22n−1 ) < a + 2n−1 2.

(2)

√ On the other hand, 2a > 2n 2 gives a2 > 22n−1 , implying a2 − 22n−1 ≥ 1. Thus (2) leads to √ √ √ √ a > 2n+1 − 2n−1 2 = 2n−1 (4 − 2) > 2n−1 ( 2 + 1) > 2n−1 2 + 0.5, which contradicts (1). Problem 4. Let ABC be a triangle and Ia be the excenter corresponding to A. Let P and Q be the tangency points of the excircle of center Ia with the lines AB and AC. Line P Q meets lines Ia B and Ia C at the points D and E respectively. Denote A1 the common point of the lines DC and BE; define in the same way points B1 and C1 . Prove that the lines AA1 , BB1 , CC1 are concurrent. Severius Moldoveanu

2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

19

Solution. We firstly prove that A1 is the orthocenter of the triangle Ia BC. In order to do this, we prove that the quadrilateral BEIa P is cyclic. Indeed, ∠P EIa = 180◦ − ∠P QA − ∠ECQ     ∠A ∠C ◦ ◦ ◦ − 90 − = 180 − 90 − 2 2 1 ∠A ∠C + = 90◦ − ∠B = 2 2 2 = ∠P BIa . So the quadrilateral BEIa P is cyclic and, since ∠BP Ia = 90◦ , BE⊥CIa . This leads to BA1  CI, where I is the incenter of triangle ABC. In the same way CA1  BI, therefore BICA1 is a parallelogram. Analogously AIBC1 is a parallelogram, hence AC1 A1 C is also a parallelogram, that is the segments [AA1 ] and [CC1 ] have the same midpoint. In the same way, segments [BB1 ] and [AA1 ] have the same midpoint, whence the conclusion. 10th GRADE

Problem 1. Let f : R → R be a function such that, for every x, y ∈ R: |f (x + y) + sin x + sin y| ≤ 2.

a) Prove that |f (x)| ≤ 1 + cos x, for every x ∈ R. b) Give an example of such a function, which vanishes nowhere in the interval (−π, π). ***

Solution. a) x = t − π2 , y = π2 yields f (t) − cos t + 1 ≤ 2, for every t ∈ R. Also, x = t + π2 , y = − π2 leads to f (t) + cos t − 1 ≥ −2, for every t ∈ R. The conclusion is obtained putting toghether the above. b) An example is f (x) = 2 − 2| sin x2 |. A heuristic approach: the hypothesis is equivalent to   t u t 2 ≥ max f (t) + 2 sin cos = f (t) + 2| sin | u∈R 2 2 2   u t t = f (t) − 2| sin |, −2 ≤ min f (t) + 2 sin cos u∈R 2 2 2

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

that is |f (x)| ≤ 2 − 2| sin x2 |, ∀x ∈ R. Problem 2. Find all positive integers n for which there exists three complex roots of order n of the unity, not necessarily different, adding up to 1. Mihai Piticari

Solution. If n is odd, then −1, 1, 1 are three complex roots of order n of the unity, adding up to 1. On the other hand, if x, y, z ∈ C, xn = y n = z n = 1 and x + y + z = 1, then |x| = |y| = |z|, hence x ¯ + y¯ + z¯ = 1/x + 1/y + 1/z = 1, which leads to xy + xz + yz = xyz. Replacing z = 1 − x − y gives (x + y)(1 − x)(1 − y) = 0, whence one of the numbers x, y, z is 1 and the other two are opposite. In the case n = odd, there are no opposite roots, so the final answer is: n = even. Problem 3. Let a, b, c be three positive real numbers. Prove that the function ax bx cx f : R → R, f (x) = x + + is increasing on [0, ∞) and b + cx ax + cx a x + bx decreasing on (−∞, 0]. Solution. We use straightforward computation: if x ≤ y are real numbers, then  ay (bx + cx ) − ax (by + cy ) f (y) − f (x) = (bx + cx )(by + cy ) cyc    1 1 y x x y = − (a b − a b ) (bx + cx )(by + cy ) (ax + cx )(ay + cy ) cyc    = ax bx ay−x − by−x · cyc

·

((ax+y − bx+y ) + cx (ay − by ) + cy (ax − bx )) , (bx + cx )(by + cy )(ax + cx )(ay + cy )

so f (y) − f (x) can be written as a sum of products of the form (ap − bp )(aq − bq ) multiplied with positive coefficients, where p = y − x ≥ 0 and all the q-s have the same sign as x, y. We finish now by noticing that    a q  ≥ 0 if q ≥ 0  a p (ap − bp )(aq − bq ) = bp+q −1 −1 ≤ 0 if q ≤ 0 b b hence f (y) − f (x) ≥ 0 if y ≥ x ≥ 0 and f (y) − f (x) ≤ 0 if 0 ≥ y ≥ x.

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

Problem 4. a) Show that, for every positive integer n, there exists uniquely deter√ √ mined positive integers xn , yn such that (1 + 33)n = xn + yn 33. b) Prove that if xn , yn are defined as above and p is a positive prime, then at least one of the numbers yp−1 , yp , yp+1 is divisible by p. Dorel Mihet¸

 n

 n

  2 n

Solution. a) The equality is obtained for xn = 0 + 33 2 + 33 4 + . . . and       yn = n1 + 33 n3 + 332 n5 + . . .. √ √ √ Also, if a + b 33 = c + d 33, a, b, c, d ∈ N and b = d, then 33 = a−c d−b is rational – false – hence b = d and a = c, which proves that the above representation is unique. b) If p = 2, 3 or 11, then yp is divisible by p. In the other cases, we start noticing that xn+1 = xn + 33yn , yn+1 = xn + yn . p−1 Also, xp ≡ 1 (mod p) and yp ≡ 33 2 (mod p), henceforth yp2 ≡ 1 (mod p). This gives p | x2p − yp2 = (xp − yp )(xp + yp ) = 32yp−1 yp+1 and, since p is prime and p = 2, p | yp−1 or p | yp+1 . 11th GRADE

Problem 1. Call a row of a matrix in Mn (C) permutable if, for any permutation of its entries, the value of the determinant does not change. Prove that any matrix that has two permutable rows is singular. Marian Andronache

Solution. Consider A ∈ Mn (C) such that row l is permutable. Denote by Γli the algebraic complement of ali , i = 1, ..., n. Suppose i, j, k, p ∈ {1, 2, ..., n} are such that ali = alj and Γlk = Γlp . Consider matrices B and C obtained from A by permuting elements in row l, such that B contains ali in place (l, k) and element alj in place (l, p) and, matrix C contains alj in place (l, k) and ali in place (l, p); the two matrices having in the other positions the initial elements. Developing determinants along row l, we get det(B)−det(C) = (ali −alj )(Γlk − Γlp ) = 0, in contradiction with the hypothesis. We deduce that row l of A is permutable if and only if all its elements are equal or, if all algebraic complements of elements in line l are equal. Consider now a matrix A that admits two permutable rows. Then:

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

• if both rows are constant det(A) = 0; • if both rows have constant algebraic complements then A∗ has two constant rows; that is det(A∗ ) = 0, which implies det(A) = 0; • if one row has all elements equal to a, and another row has all algebraic complements equal to b, then from AA∗ = det(A)In we deduce ab = 0, so a = 0 or b = 0. Thus, in every case, det(A) = 0. Problem 2. Let u : [a, b] → R be a continuous function which has at each point x ∈ (a, b] a finite left derivative, denoted by us (x). Prove that the function u is increasing if and only if us (x) ≥ 0, for all x ∈ (a, b]. Mihail B˘alun˘a

Solution. We shall use the following results which can be viewed as generalizations of the theorems of Rolle and Lagrange. Lemma 1. If a function u : [α, β] → R has left and right finite derivatives at any point in [α, β] andi u(α) = u(β), then there is c ∈ [α, β] such that us (c) ≤ 0. Proof. It is obvious that u is continous on [α, β], so it has a global minimum which belongs to the interior of the interval or coincides with on of the extremities; this point can be chosen as c. Lemma 2. If a function u : [α, β] → R has right and left finite derivatives at each point in the interval [α, β], then there is c ∈ [α, β] such that u(β) − u(α) ≥ (β − α)us (c). x. The function v satisfies Proof. Define v : [α, β] → R, v(x) = u(x) − u(β)−u(α) β−α the conditions in Lemma 1, so there is a point c ∈ [α, β] such that vs (c) ≤ 0. Returning to the solution of the problem, let a ≤ x < y ≤ b. By Lemma 1, there exists c ∈ [x, y] such that u(y) − u(x) ≥ (y − x)us (c), which implies that u is increasing. The reverse condition is easily implied by the definition of the left derivative.

Problem 3. Let g : R → R be a continuous, decreasing function, such that g(R) = (−∞, 0). Prove that there are no continuous functions f : R → R such that the equality f ◦ f ◦ ... ◦ f = g is true for some integer k ≥ 2.    k times

Dorel Mihet¸

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

Solution. Suppose that such a function f exists. Injectivity of g implies the injectivity of the continuous function f , which in turn is strictly monotone. As g is increasing we conclude that f is increasing and k is an odd number. Moreover, f is not surjective. Denote by f [k] = f ◦ f ◦ ... ◦ f (k times f ). Because f (R) is an interval with (−∞, 0) = g(R) = f (f k−1 (R)) ⊂ f (R), we deduce f is bounded from above. Let m ∈ R be such that f (x) < m, ∀ x ∈ R. Then f [k−1] (x) < m, for all x ∈ R, so g(x) = f (f [k−1] (x)) > f (m), for all x ∈ R, in contradiction with g(R) = (−∞, 0). Problem 4. Let A, B ∈ M2 (C) such that A2 + B 2 = 2AB. Prove AB = BA and tr A = tr B. Nicolae Bourb˘acut¸

Solution. a) We firstly prove that (AB − BA)2 = 0. Define the quadratic function g by f (x) = det(A2 + B 2 + x(AB − BA)) = det(A2 + B 2 ) + mx + x2 det(AB − BA). As f (−i) = det((A + iB)(A − iB)), f (i) = det((A − iB)(A + iB)), so f (−i) = f (i), we get m = 0. Moreover, from f (0) = det(A2 + B 2 ) = det(2AB) and f (−2) = det(2BA) = det(2AB) = f (0), we deduce that f is constant and det(AB − BA) = 0. The characteristic equation for AB − BA (tr(AB) = tr(BA)) gives then immediately (AB − BA)2 = 0. As AB − BA = (A − B)2 by hypothesis, we obtain 0 = (AB − BA)2 = (A − B)4 . As the matrices are of order two we must have (A − B)2 = O2 , so AB = BA. b) From (A − B)2 = O2 and 2 det(A − B) = det(A − B), the characteristic equation for A − B gives (tr(A) − tr(B))(A − B) = O2 , which finishes the solution. 12th GRADE

Problem 1. Let m be a positive integer and p a prime. Suppose A is a unitary ring having exactly m invertible elements and such that 1 + 1 + · · · + 1 = 0.    p times

Prove that A has non-zero nilpotent elements if and only if p divides m.

Mihai Piticari, Sorin R˘adulescu k

Solution. Let a ∈ A, a = 0, be nilpotent element and k ∈ N∗ , such that ap = 0. k k From (a + 1)p = ap + 1 = 1, we have that a + 1 is invertible and the order of

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

a + 1 in the group of units U (A) is a power of p. Because the order of a + 1 divides the order m of U (A), we deduce that p divides m. Conversely, by Cauchy’s theorem, there is an element a of order p in U (A). Due to (a − 1)p = ap − 1 = 0 and a = 1, we conclude that a˘ a − 1 is a non-zero nilpotent. Problem 2. Consider n ∈ N, n ≥ 2, and A a unitary ring with n elements , such that the equation xn+1 + x = 0 has in A \ {0} the unique solution x = 1. Prove that A is a field. Ioan B˘aetu

Solution. Because 1 satisfies xn+1 + x = 0, we have 1 + 1 = 0, so all non-zero elements of the additive group (A, +) are of order 2. By Cauchy’s theorem n = 2m , m ∈ N∗ .

Let a ∈ A, a = 0. The ring A being finite, there exists p < q, such that ap = aq . By succesive multiplications with aq−p , aq = a(k+1)q−kp , k ≥ 1. Take k ∈ N∗ , such that r = (k + 1)q − kp > 2q. Multiplication by ar−2q gives ar−q = a2(r−q) . Let b = ar−q . As b2 = b, we get bn+1 = b, so bn+1 + b = 0, that is b ∈ {0, 1}.

If b = ar−q = 0, take s ∈ N, s ≥ 2, such that as−1 = 0 s¸i as = 0. Consider m c = as−1 . Then c2 = 0 and, as a consequence (c + 1)n+1 = (c + 1)2 (c + 1) = m (c2 + 1)(c + 1) = c + 1. By hypothesis c + 1 ∈ {0, 1}, which is a contradiction. In conclusion ar−q = 1, i. e. a is invertible. Problem 3. Let f : [0, 1] → (0, ∞) be a continuous function. For n ∈ N, n ≥ 2, consider 0 = t0 < t1 < · · · < tn = 1, such that 

Compute

t1

f (t) dt =

t0

lim

n→∞



t2

t1

f (t) dt = · · · =



tn

f (t) dt.

tn−1

n . 1 1 1 + + ··· + f (t1 ) f (t2 ) f (tn ) Viorel Vˆajˆaitu

x 1 Solution. Define F : [0, 1] → [0, I], by F (x) = 0 f (t) dt, where I = 0 f (t) dt. As f is pozitive, F is strictly increasing, so one-to-one. Because F is continuous, F (0) = 0 and F (1) = I, F is surjective also and F −1 (kI/n) = tk , k = 0, 1, · · · , n, n ≥ 2.

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Denote by (xn )n≥2 the given sequence. We have n

n

n

k=1 n 

k=1

k=1

1 1 1 1 1 1 1 = = = −1 xn n f (tk ) n f (F (kI/n)) n F  (F −1 (kI/n)) = As (F

1 n

(F −1 ) (kI/n).

k=1

) is continous

−1 

1 = n→∞ xn lim



1

(F −1 ) (Ix) dx =

0

1 1 −1 1  F (Ix) = . I I 0

Problem 4. Let f : R → R be a non-decreasing function and F : R → R a function having right and left finite derivatives at any point in R and F (0) = 0. Suppose that lim f (x) ≤ Fs (x0 ) and lim f (x) ≥ Fd (x0 ), for any x0 ∈ R. Prove x↓x0 0 x↑x x that F (x) = 0 f (t) dt, x ∈ R. Mihail B˘alun˘a

Solution. Consider the function G : R → R, given by G(x) = F (x)−

Let x0 ∈ R, x = x0 . Then

F (x) =



x

f (t) dt. 0

F (x) − F (x0 ) · (x − x0 ) + F (x0 ). x − x0

(x0 ) As F has right and left finite derivatives in x0 , the function x → F (x)−F , x = x0 , x−x0 is bounded around x0 , so lim F (x) = F (x0 ). So G is continuous. x→x0

Let us show that G has finite right and left derivatives in x0 ∈ R, Gs (x0 ) ≥ 0 and  Gd (x0 ) ≤ 0. Let x > x0 . Because limx↓x0 f (x) ≤ f (t) ≤ f (x), x0 < t ≤ x, we get  x 1 f (t) dt ≤ f (x), lim f (x) ≤ x↓x0 x − x0 x 0

so

1 x − x0



G(x) − G(x0 ) = lim lim x↓x0 x↓x0 x − x0



lim

x↓x0

Consequently

x

f (t) dt = lim f (x). x↓x0

x0

1 F (x) − F (x0 ) − x − x0 x − x0

= Fd (x0 ) − lim f (x) ≤ 0. x↓x0





x

f (t) dt x0

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2011 ROMANIAN M ATHEMATICAL O LYMPIAD – F INAL ROUND

G(x) − G(x0 ) = Fs (x0 ) − lim f (x) ≥ 0. x↑x0 x↑x0 x − x0 In conclusion, 0 ≤ Gs (x) < ∞ and −∞ < Gd (x) ≤ 0, for any real x, therefore

Analogously, lim

lim ε↓0

1 G(x − ε) − G(x − ε/2) = − Gs (x) ≤ 0, ε 2

and lim ε↓0

1 G(x + ε) − G(x + ε/2) = Gd (x) ≤ 0, ε 2

x ∈ R, x ∈ R.

Let us show that G is constant. Consider a < b, such that G(a) = G(b). If G(a) < G(b), consider λ > (G(a) − G(b))/(b − a). The function H : [a, b] → R, H(x) = G(x) + λx, is continuous, H(a) < H(b) and lim ε↓0

1 λ λ H(x + ε) − H(x + ε/2) = Gd (x) + ≤ < 0, ε 2 2 2

x ∈ R.

So, for ay real x, there is δ(x) > 0, such that H(x + ε) < H(x + ε/2), 0 < ε < δ(x). Let c ∈ [a, b], such that H(c) = min {H(x) : a ≤ x ≤ b}. Because H(a) < H(b), we have a ≤ c < b. Consider a number ε < min(b − c, δ(c)). Then H(c) ≤ H(c + ε) < H(c + ε/2) < · · · < H(c + ε/2n ) < · · · ≤ H(c), (the last inequality is a consequence of the continuity of H), which is a contradiction. If G(a) > G(b), proceed analogously: consider λ < (G(a) − G(b))/(b − a). In this case the corresponding continuous function H(x) = G(x) + λx, satisfies H(a) > H(b) and lim ε↓0

1 λ λ H(x − ε) − H(x − ε/2) = − Gs (x) − ≤ − < 0, ε 2 2 2

x ∈ R.

That is, for any real x, there exists δ(x) > 0, such that H(x − ε) < H(x − ε/2), 0 < ε < δ(x). Let c ∈ [a, b], such that H(c) = min {H(x) : a ≤ x ≤ b}. Because H(a) > H(b), we get a < c ≤ b. Fix a real positive number ε < min(c − a, δ(c)). Then H(c) ≤ H(c − ε) < H(c − ε/2) < · · · < H(c − ε/2n ) < · · · ≤ H(c), which is a contradiction. We thus proved that G is constant. Taking into account G(0) = 0, we get G(x) = 0, for any realx .

SHORTLISTED PROBLEMS FOR THE 62nd NMO

JUNIORS

1. Prove that if n and p are integers, 1 < p < n, then the number (2p + 1)n3 + 6n(12 + 22 + 32 + ... + p2 ) can be written as the sum of 2p + 1 different perfect cubes. 2. Given an equilateral triangle ABC and the points M ∈ [BC], N ∈ [AC], P ∈ [AB] such that BM = M C, 3AN = N C and 2BP = AP , find the measure of ∠N M P . 3. In a triangle ABC it is known that ∠A = 120◦ , ∠C = 20◦ , (AD is the bisector of ∠BAC (where D ∈ (BC)) and the point E is taken so that E ∈ (AC) and [BD] ≡ [CE]. Find the measure of ∠EBC. 4. The convex quadrilateral ABCD has [AB] ≡ [BC], m(∠DCB) < 90◦ and m(∠DCB) + m(∠DAB) = 180◦ . The point E is taken on the segment (DC) such that DA + DC = 2DE. Prove that BE ⊥ DC. 5. The convex pentagon ABCDE has m(∠A) = m(∠C) = 90◦ , AB = 3, BC = 5, CD = 10, DE = 8 and the reflection of C about BD is on the line AE. The line M A is perpendicular on the plane of the pentagon and M A = 6. Find the measure of the angle of the lines M E and AB. 6. The parallelogram ABCD has center O, AD = DB and BA > BC. The point V is outside the plane (ABC) such that V D ⊥ (ABC). Point T is the foot of the perpendicular from D onto AB and E, F, G are the feet of the bisectors of the angles ∠V DA, ∠V DB, ∠V DC respectively (where E ∈ V A, F ∈ V B, G ∈ CV ). Prove that: 27

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i) EF  (ABC); ii) m(∠((V AB), (ABC))) = 45◦ if and only if V D = DT ; iii) if V D = DT , then tan(∠((GEF ), (ABC))) = 12 if and only if

BA BC

= 65 .

7. Find all the perfect squares whose product of their decimal digits is a prime. 8. Given a prime number p, p ≥ 3 and d a square free number (that is d’s prime decomposition contains no repeated factors), find the number of the elements of the set    √ √ n Ap = x = n d + − {n d} | n ∈ N , p where {a} denotes the fractional part of the real number a. 9. Prove that, if x, y > 0, then 0 <

x y 1 < 1. + − 2 2 x + y4 y + x4 x y +1

10. Let x, y be real numbers such that x + y, x3 + y 3 , x5 + y 6 and y 5 are rational. Prove that x and y are rational. SENIORS

1. An infinite set A of real numbers contains at least an irrational number. Prove that for each positive integer n it is possible to find n elements of A whose sum is irrational. 2. Given integers n ≥ 3 and k ≥ 1, find all positive integers a1 < a2 < . . . < an 1 1 1 such that k + k + . . . + k = 1. an a1 a2 3. Let p be a positive integer and Ap = {x ∈ R | px{x} = (p + 1)[x]}. Find the cardinals of the sets Ap and A1 ∪ A2 ∪ ... ∪ Ap . x  4. Solve the equation (ax + bx )2011 = a2011 + b2011 , where a, b are positive reals. 5. Find all the functions f : Z → N fulfilling the conditions: i) f (m + n) = f (m) + f (n) + 2mn, for every integers m, n; ii) f (f (x)) − f (1) is a perfect square for each integer x. 6. Find all positive integers n for which there exists a set S ⊂ C with n, such  that i) each element of S has modulus 1, ii) z = 0 and iii) z + w = 0, for each z, w ∈ S.

z∈S

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29

7. Given in the plane equilateral triangles ABC and BDE, with C ∈ (BD) and A, E are on different sides of the line BD. Denote M, N ,P the midpoints of the segments (AB), (CD), (BE) respectively. Find the measures of the angles of the triangle M N P . PUTNAM SENIORS

1. Find all polynomials P, Q with real coefficients, such that, for infinitely many positive integers n, P (1)P (2) . . . P (n) = Q(n!). 2. The sequence (an )n≥1 of real numbers is such that the sequence (xn )n≥1 defined by xn = max{an , an+1 , an+2 } is convergent and the sequence (yn )n≥1 defined by yn = an+1 − an has limit 0. Prove that the sequence (an )n is convergent. 3. Let A be a n × n matrix with complex entries. a) Prove that there exists a n×n matrix with complex entries B such that AB = 0n and rank A + rank B = n. b) If 1 < rank A < n, prove that there exists a n × n matrix with complex entries C such that AC = 0n , CA = 0n and rank A + rank C = n. 4. Let f : R → R be a continuous function such that, on each non degenerated interval I, the function reaches its maximum or its minimum in an interior point of I. Prove that f is a constant. 5. Let f : R → R be a function with the property: |f (x)−f (y)| ≤ | sin x−sin y|, for each x, y ∈ R. a) Prove that there exists an unique c ∈ R such that f (c) = c. b) Consider the sequence (xn )n∈N with x0 = 0 and xn+1 = f (xn ) for every n ∈ N. Prove that lim xn = c. n→∞

6. Given a continuous function f : R → R, denote, for each interval [a, b], mab = minx∈[a,b] f (x) and Mab = maxx∈[a,b]  f (x). Find all the continuous funcmab + Mab a+b = . tions f : R → R such that, for every a < b, f 2 2 7. The matrix A ∈ Mk (R) (k ≥ 2) has the property: for every positive integer n, det(n2 Ik − A2 ) ≥ nk (det(nIk − A) + det(nIk + A) − 1) . Prove that tr(A) = 0.

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S HORTLISTED PROBLEMS FOR THE 2011 ROMANIAN NMO

8. The function f : [0, 1] → R is differentiable and 

1

f (x) dx = 0



1

xf (x) dx = 0. 0

Prove that there exists c ∈ (0, 1) such that f  (c) = 0. 9. Let f : [0, 1] → R be a function which is continuous on [0, 1] and differentiable in 0. Consider the function s : [0, 1] → R,  x s(x) = sup{c ∈ [0; x] | f (t) dt = xf (c)}. 0

a) Prove that the function f ◦ s is differentiable in 0 and (f ◦ s) (0) = 21 f  (0). b) If f  (0) = 0, prove that s is differentiable in 0 and s (0) = 12 .   10. Find all the functions f : − π2 ; π2 → R such that:  π π a) f is differentiable on − 2 , 2 ;   b) there exists a antiderivative F of f such that F (x) + f  (x) ≤ 0, ∀x ∈ − π2 , π2 π  π and F − 2 = F 2 = 0.

11. Let (K, +, ·) be a finite field. Prove that: a) if K has 4k + 1 elements, then the polynomial f = X 4 + 4 has four roots in K; b) the polynomial g = X 8 − 16 has at least a root in K. 12. Prove that



1 0

x

ee dx ≥

e(e2 − 3) . 2

Contributors

Gheorghe Molea, Dan Nedeianu, Petre B˘atrˆanet¸u, Cosmin Manea & Dragos¸ Petric˘a, Dan Nedeianu, Alexandru Blaga, Aurel Bˆarsan, Cecilia Deaconescu & Dumitru Dobre, Dan Nedeianu, C˘at˘alin Cristea, Dinu Teodorescu, Dorel Mihet¸, Cecilia Deaconescu & Dumitru Dobre, Aurel Bˆarsan, Marian Ionescu, Nicolae Bourb˘acut¸, Dan Nedeianu, Ovidiu Furdui, Cristinel Mortici, Vasile Pop, Radu Gologan, Nicolae Bourb˘acut¸, ***, Nelu Chichirim, Cezar Lupu, Dan Marinescu, Dorel Mihet¸, Dorel Mihet¸, Dan Nedeianu.

THE 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

FIRST SELECTION TEST

Problem 1. Determine all real-valued functions f on the set of real numbers satisfying the condition 2f (x) = f (x + y) + f (x + 2y) for all real numbers x and all non-negative real numbers y. SEEMOUS 2011 Short List

Solution. Clearly, any constant function is a solution to the problem. Conversely, assume without loss of generality that f (0) = 0. Fix a positive real number y and let x = ny, where n is a non-negative integer number, to obtain 2f (ny) = f ((n + 1)y) + f ((n + 2)y) and thereby get f (ny) = f (y)(1 − (−2)n )/3. Consequently, f (y) = −f (2y) = f (4y) = −5f (y), which shows that f vanishes identically on the non-negative real ray. Finally, if x is any real number, write 2f (x) = f (x + |x|) + f (x + 2|x|) = 0 to conclude that f vanishes identically on the set of real numbers. Problem 2. Prove that the set S = {nπ : n = 0, 1, 2, 3, · · · } contains arithmetic progressions of any finite length, but no infinite arithmetic progressions. Vasile Pop, SEEMOUS 2011 Short List

Solution. If x is a real number, let {x} = x − x denote the fractional part of x. Given an integer number m ≥ 3, there exists a positive integer number n such 31

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that {nπ} < 1/m, for the set {{kπ} : k = 0, 1, 2, 3, · · · } is dense in the closed unit interval [0, 1]. Consequently, knπ = knπ + k{nπ} = knπ + k{nπ} = knπ,

k = 1, 2, · · · , m,

so nπ, 2nπ, · · · , mnπ are m numbers in S in arithmetic progression with ratio nπ. Suppose, if possible, that S contains an infinite arithmetic progression nk π, k = 0, 1, 2, 3, · · · , with (integral) ratio r, where the nk form a strictly increasing sequence of positive integer numbers. Write nk π = n0 π + kr + {nk π} to deduce that r is positive and   r−1 r+1 r + {nk+1 π} − {nk π} ∈ , . nk+1 − nk = π π π The length of this interval is less than 1, so nk+1 − nk = n for some positive integer k→∞ n and all indices k. Hence, nk = n0 + kn, so nk /k −→ n. On the other hand, k→∞ nk /k −→ r/π, so π = r/n wich contradicts irrationality of π. Problem 3. Let ABC be a triangle such that AB < AC. The perpendicular bisector of the side BC meets the side AC at the point D, and the (interior) bisectrix of the angle ADB meets the circumcircle ABC at the point E. Prove that the (interior) bisectrix of the angle AEB and the line through the incentres of the triangles ADE and BDE are perpendicular. Solution. The lines BC and DE are parallel, so the angles BED and DAE are equal. Then so are the angles AED and DBE. Let I and J be the incentres of the triangles ADE and BDE, respectively. It follows that the triangles DIE and DJB are similar, so DI/DE = DJ/DB. Since the angles IDJ and EDB are equal, the triangles DIJ and DEB are similar, so the angles DIJ and DEB are equal. Let the line BE meet the line IJ at the point F . Notice that the quadrangle DIEF is cyclic to deduce that the angles EF I and EDI are both equal to one half of the angle ACB. Consequently, the line IJ is parallel to the exterior bisectrix of the angle AEB. The conclusion follows. Problem 4. Given an integer number n ≥ 2, evaluate the sum  (sgn σ)n(σ) , σ∈Sn

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IMO

where Sn is the set of all n-element permutations, and (σ) is the number of disjoint cycles in the standard decomposition of σ. SEEMOUS 2011 Short List

Solution. The sum in question is fn (n) = n!, where fn (x) =



σ∈Sn

(sgn σ)x(σ) = x(x − 1) · · · (x − n + 1).

The latter is a straightforward consequence of the following recurrence formula: fn (x) = xfn−1 (x) − (n − 1)fn−1 (x) = (x − n + 1)fn−1 (x),

n ≥ 3,

where f2 (x) = x(x − 1). To establish the recurrence formula, consider the decomposition of a permutation σ in Sn into disjoint cycles, σ = (· · · ) · · · (· · · )(n)

or σ = (· · · ) · · · (· · · n · · · ) · · · (· · · ),

along with the permutation σ  in Sn−1 , obtained from σ by deleting n : n) σ  = (· · · ) · · · (· · · )(ˆ

or

σ = (· · · ) · · · (· · · n ˆ · · · ) · · · (· · · ).

Clearly, sgn σ  = sgn σ and (σ  ) = (σ) − 1 in the former case, and sgn σ  = −sgn σ and (σ  ) = (σ) in the latter. Conversely, any given permutation σ  in Sn−1 extends to a unique permutation σ in Sn which fixes n (the first case above), and to exactly n − 1 permutations σ in Sn which do not fix n — for the latter can be inserted in precisely n − 1 ways into σ  (the second case above). The conclusion follows. SECOND SELECTION TEST

Problem 5. A square with side length  is contained in a unit square whose centre is not interior to the former. Show that  ≤ 1/2. Marius Cavachi

Solution. Proof from The Book. Notice that there is a line through the centre of the unit square separating the square with side length  and a standard quarter of the unit square (i. e., a square with side length 1/2 and one vertex at the centre of the unit

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square). To conclude, apply the celebrated theorem of Erd¨os stating that the sum of the side lengths of two squares packed into a unit square does not exceed 1. Solution 1. The square with side length  is the intersection of two strips, S and S  , of breadth . Since the centre O of the unit square is not interior to the square with side length , it is not interior to at least one of the two strips, say S. Clearly, the distance from O to the farthest component  of the boundary of S is at least . Suppose, if possible, that  > 1/2. Then the line  meets the boundary of the unit square at precisely two points, X and Y , situated on consecutive sides. Let A denote the vertex of the unit square shared by those sides, and let M and N denote their midpoints. Without loss of generality, we may (and will) assume that the circular labelling around the boundary of the unit square is M , X, A, Y , N . Since dist (O,  ) ≥  > 1/2, the line t, parallel to  and tangent to the quarter M N of the incircle of the unit square, meets the segments M X and N Y : the former at X  , and the latter at Y  . Clearly, X  Y  > XY ≥ . Finally, to reach a contradiction, let T denote the point of contact of t and the quarter M N of the incircle of the unit square, and write successively: 1 = 1/2 + 1/2 = M A + AN = (M X  + X  A) + (AY  + Y  N ) = M X  + (X  A + AY  ) + Y  N = X  Y  + (X  A + AY  ) > 2 · X  Y  > 2 · XY ≥ 2. Solution 2. Begin by noticing that if the vertices M , N , P , Q of a rectangle lie on the boundary of a triangle ABC ( M ∈ AB, N ∈ AC and P, Q ∈ BC), then M N/BC + M Q/AA = AM/AB + BM/AB = 1, where A is the perpendicular foot dropped from the vertex A onto BC. Hence if a triangle ABC, whose internal angles at B and C are not obtuse, contains a square with side length  with a pair of opposite sides parallel to BC, then 1/BC + 1/AA ≤ 1/, where A is the perpendicular foot dropped from the vertex A onto BC : extend the farthest side of the square, which is parallel to BC, to meet the sides AB and AC at M and N , respectively, drop the perpendicular feet M  and N  from M and N , respectively, onto the side BC, notice that  ≤ min (M N, M M  ) and apply the above result. Back to the problem, let O be the centre of the unit square and let A, B, C, D be a circular labelling of its vertices around the boundary. Draw a line d through O, parallel to a pair of opposite sides of the square with side length . The latter lies in

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one of the closed half-planes determined by d. Clearly,  ≤ 1/2 if d is parallel to a pair of opposite sides of the unit square. Otherwise, we may assume that d meets the side AB at E and the extension of the side AD beyond D at F , and the triangle AEF contains the square with side length . We prove that  < 1/2. To this end, drop the perpendicular foot A from A onto the segment EF . By the preceding, 1/AA + 1/EF ≤ 1/, so it is sufficient to show that 1/AA + 1/EF > 2.

(∗)

Let O be the perpendicular foot dropped from O onto the side AB, and let EO = √ √ x/2 and ∠AEF = θ. Clearly, sin θ = 1/ 1 + x2 and cos θ = x/ 1 + x2 , so √ (1 + x) 1 + x2 1+x AE = and AA = AE sin θ = √ , EF = cos θ 2x 2 1 + x2 √ √ and (∗) is equivalent to x/(1 + x) 1 + x2 + 1 + x2 /(1 + x) > 1, x > 0, which is easily established. Problem 6. Let A0 A1 A2 be a non-equilateral triangle. The incircle of the triangle A0 A1 A2 touches the side Ai Ai+1 at the point Ti+2 (indices are reduced modulo 3). Let Xi be the perpendicular foot dropped from the point Ti onto the line Ti+1 Ti+2 . Show that the lines Ai Xi are concurrent at a point situated on the Euler line of the triangle T0 T1 T2 . Gazeta Matematic˘a

Solution. The lines Ai Ai+1 and Xi Xi+1 are parallel, for they are both antiparallel to the line Ti Ti+1 . Hence the triangles A0 A1 A2 and X0 X1 X2 are homologus: the three lines Ai Xi are concurrent at the homology centre which lies on the homology line. The latter passes through the incentres of the two triangles: one is the circumcentre of the triangle T0 T1 T2 and the other the orthocentre. The conclusion follows. Problem 7. Given a positive integer number n, determine the maximum number of edges a simple graph on n vertices may have in order that it contain no cycles of even length. American Mathematical Monthly

Solution. The required maximum is 3(n − 1)/2. It is achieved, for instance, by an (n − 1)/2 arm wind-mill if n is odd, and an (n − 2)/2 arm wind-mill with an extra edge joined at the hub if n is even.

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To show that a simple graph on n vertices with no cycles of even length has at most 3(n − 1)/2 edges, let G be one such with a maximal edge-set E. By maximality, G is connected. Let T be a spanning tree (that is, a maximal connected acyclic subgraph) of G, and let E  denote the edge-set of T ; it is well-known that |E  | = n − 1. The end-points of any edge e in E \ E  are joined by a unique simple path α in T . Since T is acyclic, if e and e are distinct edges in E \ E  , then the edges the corresponding paths in T , α and α , may share form a path; and since G has no cycles of even length, the paths α and α are actually edge-disjoint — otherwise, at least one of the cycles α + e, α + e and α + e + α + e (mod 2) would have an even length. It follows that the total length of the cycles α + e, e ∈ E \ E  , does not exceed |E|. There are |E \ E  | = |E| − |E  | = |E| − n + 1 such cycles, each of length at least 3, so 3(|E| − n + 1) ≤ |E|; that is, |E| ≤ 3(n − 1)/2. Problem 8. Show that: a) There are infinitely many positive integer numbers n such that there exists a square equal to the sum of the squares of n consecutive positive integer numbers. (For instance, 2 and 11 are such: 52 = 32 + 42 and 772 = 182 + 192 + · · · + 282 .) b) If n is a positive integer number which is not a perfect square and if x0 is an integer number such that x20 + (x0 + 1)2 + · · · + (x0 + n − 1)2 is a perfect square, then there are infinitely many positive integer numbers x such that x2 + (x + 1)2 + · · · + (x + n − 1)2 is a perfect square.

American Mathematical Monthly

Solution. a) We must show that there are infinitely many positive integer numbers n such that the equation y 2 = x2 + (x + 1)2 + · · · + (x + n − 1)2 has positive integral solutions. To this end, rewrite the equation in the form y 2 = n((x + (n − 1)/2)2 + (n2 − 1)/12).

(1)

We show that if n is a square greater than 25 not divisible by 2 or 3, then (1) has a positive but finite number of solutions in positive integers x and y. Since n is a perfect square, the expression in the outmost parantheses in (1) must be a perfect square z 2 , i. e., there is a positive integer z such that z 2 − (x + (n − 1)/2)2 = (n2 − 1)/12.

(2)

Thus, z ± (x + (n − 1)/2) must be complementary even divisors of (n2 − 1)/12 differing by more than n − 1; hence there are only finitely many solutions to (2) in

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positive integers. One such is obtained by taking z − (x + (n − 1)/2) = 2 and z + (x + (n − 1)/2) = (n2 − 1)/24, i. e., by letting x and z have the positive integer values x = (n − 25)(n + 1)/48

and z = 1 + (n2 − 1)/48.

b) Rewrite the equation y 2 = x2 + (x + 1)2 + · · · + (x + n − 1)2 in the form (2y)2 − n(2x + n − 1)2 = (n − 1)n(n + 1)/3

(3)

and let Tn (x) = x2 + (x + 1)2 + · · · + (x + n − 1)2 . We may assume n > 1. Since Tn (x) = Tn (−x − n + 1), we may further assume that x0 ≥ −(n − 1)/2. Suppose Tn (x0 ) = y02 , i. e., (2y0 )2 − n(2x0 + n − 1)2 = (n − 1)n(n + 1)/3, where we may assume y0 > 0 (and hence y0 > (n − 1)/2). By the theory of the Pell equation, there are infinitely many pairs of positive integers u, v such that u2 − nv 2 = 1.

(4)

Clearly, u is odd if n is even, so that (n − 1)(u − 1) is even. We now use the identity √ √ √ (2y0 + (2x0 + n − 1) n)(u + v n) = 2y + (2x + n − 1) n, (5) where and y = y0 u + x0 nv + (n − 1)nv/2. (6) √ √ Multiplying (5) by the identity obtained from (5) by replacing n by − n, we find that if x and y are given by (6) and u and v satisfy (4), then x = x0 u + y0 v + (n − 1)(u − 1)/2

(2y)2 − n(2x + n − 1)2 = ((2y0 )2 − n(2x0 + n − 1)2 )(u2 − nv 2 )

= (2y0 )2 − n(2x0 + n − 1)2 = (n − 1)n(n + 1)/3.

Thus, if x, y are given by (6), they satisfy (3), so that Tn (x) is a perfect square. Further, since x0 ≥ −(n − 1)/2 and y0 > (n − 1)/2, it follows that y ≥ y0 u > 0 and x ≥ y0 v − u(n − 1)/2 + (n − 1)(u − 1)/2 = y0 v − (n − 1)/2 ≥ y0 − (n − 1)/2 > 0. This ends the proof. Remark. Write Tn (x) = n(x + (n − 1)/2)2 + (n − 1)n(n + 1)/12 to derive a) from b). To this end, notice that if (n + 1)/12 is a perfect square, say n = 12m2 − 1, then (1) has the obvious solution x = −6m2 + m + 1 and y = m(12m2 − 1), so b) applies to show that (1) has infinitely many solutions in positive integers.

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FOURTH – ALL GEOMETRY – SELECTION TEST2

Problem 9. Let ABCD be a cyclic quadilateral so that BC and AD meet at a point P . Consider a point Q, different from P , on the line BP so that P Q = BP and construct the parallelograms CAQR and DBCS. Prove that the points C, Q, R, S are concyclic. BMO Shortlist 2011, United Kingdom

Solution. It is enough to prove that ∠RQC = ∠RSC.

(†)

Now ∠RQC = ∠ACQ = ∠ACB = ∠ADB. (1) −→ −→ −→ −−→ Take T such that QABT is a parallelogram. Then BT = AQ = CR and BD = −→ CS imply ∆BT D ≡ ∆CRS, so ∠RSC = ∠T DB. (2) On the other hand, since P is the midpoint of the segment BQ and ABT Q is a parallelogram, P is also the midpoint of the segment AT , so ∠T DB = ∠ADB. Combining this with (1) and (2), we get (†). Problem 10. Let ABCD be a convex quadrangle such that AB = AC = BD (vertices are labelled in circular order). The lines AC and BD meet at point O, the circles ABC and ADO meet again at point P , and the lines AP and BC meet at point Q. Show that the angles COQ and DOQ are equal. BMO Shortlist 2011, Bulgaria

Solution. We shall prove that the circles ADO and BCO meet again at the incentre I of the triangle ABO, so the line IO is the radical line of the circles ADO and BCO. Noticing further that the lines AP and BC are the radical lines of the pairs of 2 The

third test was the Balkan Mathematical Olympiad

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circles (ABC, ADO) and (ABC, BCO), respectively, it follows that the lines AP , BC and IO are concurrent (at point Q), whence the conclusion. To show that the point I lies on the circle ADO, notice that 1 1 1 ∠AIO = 90◦ + ∠ABO = 90◦ + ∠ABD = 90◦ + (180◦ − 2∠ADB) 2 2 2 = 180◦ − ∠ADB = 180◦ − ∠ADO. Similarly, the point I lies on the circle BCO, for 1 1 1 ∠BIO = 90◦ + ∠BAO = 90◦ + ∠BAC = 90◦ + (180◦ − 2∠ACB) 2 2 2 = 180◦ − ∠ACB = 180◦ − ∠BCO.

Remark. We may consider the corresponding configuration derived from four generic points in the plane, A, B, C, D, subject only to AB = AC = BD. The argument applies mutatis mutandis to show that the point Q always lies on one of the two bisectrices of the angle COD. Problem 11. Given a triangle ABC, let D be the midpoint of the side AC and let M be the point that divides the segment BD in the ratio 1/2; that is, M B/M D = 1/2. The rays AM and CM meet the sides BC and AB at points E and F , respectively. Assume the two rays perpendicular: AM ⊥ CM . Show that the quadrangle AF ED is cyclic if and only if the line of support of the median from A in triangle ABC meets the line EF at a point situated on the circle ABC. BMO Shortlist 2011, Saudi Arabia

Solution. Denote by a, b, c the sidelengths, and by ma , mb , mc the lengths of the medians of the triangle ABC. Since M D is median in the right-angled triangle AM C, it follows that 2mb /3 = M D = AD = CD = b/2, so mb = 3b/4, whence (3b/4)2 = m2b = (a2 + c2 )/2 − b2 /4; that is, 13b2 = 8(a2 + c2 ).

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Next, apply the Menelaus theorem to get EC/EB = 4 = F A/F B and deduce thereby that the lines AC and EF are parallel. The quadrangle AF ED is therefore a trapezium; it is cyclic if and only if AF = DE. Express the two in terms of a, b and c. Recall that F A/F B = 4 to obtain AF = 4c/5. Next, apply Stewart’s theorem in triangle BCD to get DE 2 = b2 /2 − 4a2 /25. By the preceding, the quadrangle AF ED is cyclic if and only if 25b2 − 8a2 = 32c2 . √ Recall that 13b2 = 8(a2 + c2 ) to express b and c in terms of a: b = 2a 2/3 and c = 2a/3.

Finally, let N be the midpoint of the side BC and let the lines AN and EF meet at P . Notice that EN = a/2 − a/5 = 3a/10, and the triangles AN C and P N E are similar, to obtain N P = 3ma /5, so N A · N P = 3m2a /5 = 3(2(b2 + c2 ) − a2 )/20 = a2 /4 = N B · N C. The conclusion follows. FIFTH SELECTION TEST

Problem 12. Show that there are infinitely many positive integer numbers n such that n2 + 1 has two positive divisors whose difference is n. Solution. Define the sequence (ak )k≥0 by a0 = 1, a1 = 2 and ak+2 ak = a2k+1 + 1, k = 0, 1, 2, · · · , and check inductively that the ak are all positive integer numbers, the nk = ak+1 − ak form a strictly increasing sequence of positive integer numbers, and ak and ak+1 both divide n2k + 1.

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Problem 13. Let n be an integer number greater than 2, let x1 , x2 , · · · , xn be n positive real numbers such that n  i=1

1 = 1, xi + 1

and let α be a real number greater than 1. Show that n  i=1

1 n ≥ α+1 xα + 1 (n − 1) i

and determine the cases of equality. BMO Shortlist 2011, Serbia

Solution. Let yi = 1/(xi + 1), i = 1, 2, · · · , n, so the yi are positive real numbers that add up to 1. Upon substitution, the left-hand member of the required inequality becomes n n   yiα yiα  α = ,  (1 − yi )α + yiα + yiα i=1 i=1 j=i yj

the latter on account of y1 + y2 + · · · + yn = 1. Apply Jensen’s inequality to the convex function t → tα , t > 0, to get   α  yj ≤ (n − 1)α−1 yjα , i = 1, 2, · · · , n, j=i

so

n  i=1

j=i

n

 yiα yiα   α ≥ α α. α−1 (n − 1) α j=i yj + yi y + y i=1 i j=i j

Now write zi = yiα , i = 1, 2, · · · , n, z = z1 + z2 + · · · + zn and a = (n − 1)α−1 to transform the right-hand member of the above inequality to n  i=1

zi . (1 − a)zi + az

Finally, notice that the function t → conclude by Jensen’s inequalty: n  i=1

t , t < az/(a − 1), is convex, to (1 − a)t + az

n 1 zi zi n n i=1 = ≥n· . n (1 − a)zi + az (n − 1)a + 1 (1 − a) n1 i=1 zi + az

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Clearly, equality holds if and only if the zi are all equal; tracing back, this is the case if and only if the xi are all equal to n − 1. Remark. Since the inequality in the statement can be rewritten as

n  i=1

n 

xi +1 x +1 xi + 1 1 i=1 i α · ≥ n , xi + 1 xα  xi i +1 +1 x +1 i=1 i

it is tempting to consider the real-valued function f (x) = (x + 1)/(xα + 1), x ≥ 0, and try to apply Jensen’s inequality. Unfortunately, f is concave on the closed unit interval [0, 1] and convex on the ray x ≥ 1, as shown by the second derivative: f  (x) =

αxα−2 

(xα + 1)

3

The sign of f  is given by

 (α − 1)xα+1 + (α + 1)xα − (α + 1)x − (α − 1) .

g(x) = (α − 1)xα+1 + (α + 1)xα − (α + 1)x − (α − 1),

x ≥ 0,

whose first two derivatives are

and

  g  (x) = (α + 1) (α − 1)xα + αxα−1 − 1 g  (x) = α(α2 − 1)xα−2 (x + 1) > 0,

x > 0.

Hence g  is strictly increasing. Since g  (0) = −(α + 1) < 0, and g  (1) = 2(α2 − 1) > 0, it follows that g  has a unique zero x0 ∈ (0, 1). Consequently, g is strictly decreasing on the closed interval [0, x0 ] and strictly increasing on the ray x ≥ x0 . Since g(0) = 1 − α < 0, and g(1) = 0, the conclusion follows. Problem 14. Given a set L of lines in general position in the plane (no two lines in L are parallel and no three lines are concurrent) and another line , show that the total number of edges of all faces in the corresponding arrangement interesected by  is at most 6|L|. Chazelle et al., Edelsbrunner et al.

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Solution. Assume without loss of generality that  is horizontal and does not pass through any vertex of the arrangement of lines in L. First, we shall bound the total number of edges of the upper parts of all faces intersected by , that is, those parts that lie above . The boundary of the upper part of such a face K consits of two convex chains of edges, the left and right chain, and a portion of . If the upper part of K is bounded, then the left and right chains meet at the topmost vertex of K. Otherwise, the last (topmost) edges of these chains are halflines. The edges belonging to the left (respectively, right) chain, with the exception of the topmost edge, are called the left (respectively, right) edges of K. We claim that every line in L contains at most one left edge. Suppose, if possible, that some line in L has two portions e and e that are left edges of K and K  , respectively, where e is above e. Then the line supporting the topmost edge of the left chain of K would cross K  , contradicting the fact that K  is a face of the arrangement of lines in L. Hence, the total number of left (respectively, right) edges of the upper parts of the faces intersected by  is at most |L|. Taking into account the topmost edges of the chains, the total number of edges of the upper parts of the faces intersected by  is at most 4|L|. The same argument applies verbatim for the lower parts of these faces. Consequently, the total number of edges of the faces intersected by  is at most 8|L| − 2|L| = 6|L|; the second term is due to the fact that the edges crossed by  are counted twice: once in the upper parts and once in the lower parts. SIXTH SELECTION TEST

Problem 15. Given a positive integer number k, define the function f on the set of all positive integer numbers to itself by  1, if n ≤ k + 1, f (n) = f (f (n − 1)) + f (n − f (n − 1)), if n > k + 1.

Show that preimage of every positive integer number under f is a finite non-empty set of consecutive positive integer numbers.

American Mathematical Monthly

Solution. It is sufficient to show that the difference ∆(n) = f (n) − f (n − 1) is 0 or 1 for all integer numbers n ≥ 2, and f is unbounded. Clearly, ∆(n) = 0, 2 ≤ n ≤ k + 1, provides the basis for an inductive proof. If n > k + 1, apply the recurrence. By the induction hypothesis, ∆(n − 1) is 0 or 1 and

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1 ≤ f (n − 1) ≤ n − 1. Notice that ∆(n) = ∆(n − f (n − 1)) if ∆(n − 1) = 0 and ∆(n) = ∆(f (n − 1)) if ∆(n − 1) = 1 to conclude that ∆(n) is indeed 0 or 1. To show that f is unbounded, suppose, if possible, that f achieves its maximum value N for the first time at m to deduce recursively that N = f (m + n) = f (N ) + f (m − N + n) for all positive integers n; that is, f (m − N + n) = N − f (N ), for all positive integers n. Setting n ≥ N yields a contradiction: N = N − f (N ). The conclusion follows. Problem 16. Given a prime number p congruent to 1 modulo 5 such that 2p + 1 is also prime, show that there exists a matrix of zeros and ones containing exactly 4p (respectively, 4p + 2) ones no submatrix of which contains exactly 2p (respectively, 2p + 1) ones. American Mathematical Monthly

Solution. Let p = 5q + 1, q ≥ 2, and write 4p = 5(4q + 1) − 1. Form a 5by-(4q + 1) matrix consisting of ones only except for a single entry. Such a matrix has exactly 4p ones. A submatrix comprising r rows, 1 ≤ r ≤ 5, and s columns, 1 ≤ s ≤ 4q + 1, contains rs or rs − 1 ones. In the former case, rs = 2p implies r ≥ p or s ≥ p, because p is prime; in the latter, rs − 1 = 2p implies rs = 2p + 1, so r = 2p + 1 or s = 2p + 1, because 2p + 1 is prime. Both cases contradict the size of the matrix, so no submatrix contains exactly 2p ones. Next, write 4p + 2 = 5(4q + 1) + 1. Now form a 5-by-(4q + 1) matrix consisting of ones only except for one column that contains only a single one. Such a matrix has exactly 4p + 2 ones. A submatrix comprising r rows, 1 ≤ r ≤ 5, and s all-one columns, 0 ≤ s ≤ 4q + 1, contains rs or rs + 1 ones. In the former case, rs = 2p + 1 implies r = 2p+1 or s = 2p+1 because 2p+1 is prime; in the latter, rs+1 = 2p+1, so either r ≥ p or s ≥ p since p is prime. Again, this contradicts the size of the matrix and the conclusion follws. Problem 17. The incircle of a triangle ABC touches the sides BC, CA, AB at points D, E, F , respectively. Let X be a point on the incircle, different from the points D, E, F . The lines XD and EF , XE and F D, XF and DE meet at points J, K, L, respectively. Let further M , N , P be points on the sides BC, CA, AB, respectively, such that the lines AM , BN , CP be concurrent. Prove that the lines JM , KN and LP are concurrent. Dinu S¸erb˘anescu

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Solution. Let the lines KL and N P , LJ and P M , JK and M N meet at points Q, R, S, respectively. By Desargues’ theorem on perspective triangles, the lines JM , KN and LP are concurrent if and only if the points Q, R and S are collinear.

Without loss of generality, we may assume that X lies on the arc F D of the incircle that does not contain E. Consequently, K lies on the side F D, while L and J lie on the respective extensions of the sides DE and EF . Consider the cyclic quadrangle DEF X : The diagonals meet at K, and the extensions of the opposite sides meet at L and J, respectively, so the line LJ is the polar of K with respect to the incircle – in what follows, all polar lines are considered with respect to the incircle. Since the line F D is the polar of B, and K lies on the line F D, it follows that B lies on the line LJ. Similarly, C lies on the line JK, and A lies on the line KL. Consequently, the lines AQ and CS meet at K. Projectively, the lines F D and LJ meet at some point T . Notice that T lies on the polar lines of B and K to deduce that the line BK is the polar of T , so the cross-ratio (T F KD) is harmonic. Let further the lines BK and P M meet at U . Read from B, the cross-ratio (RP U M ) equals (T F KD), so it is harmonic; that is, U is the harmonic conjugate of R relative to M and N . Consequently, the points Q, R and S

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are collinear if and only if the lines M Q, N U and P S are concurrent. To prove the lines M Q, N U and P S concurrent, simply check that QN U P SM · · = 1. QP U M SN To this end, write QN UP SM

= N A · sin(∠N AQ), QP = P B · sin(∠P BU ), U M = M C · sin(∠M CS), SN

= P A · sin(∠P AQ), = M B · sin(∠M BU ), = N C · sin(∠N CS)

to get upon rearrangement of factors   QN U P SM NA PB MC · · = · · QP U M SN NC PA MB   sin(∠N AQ) sin(∠P BU ) sin(∠M CS) · · . sin(∠P AQ) sin(∠M BU ) sin(∠N CS) Finally, notice that the lines in both triples (AM, BN, CP )

and

(AQ, BU, CS)

are concurrent (the former by hypothesis, and the latter concur at K) to infer that the products in the parantheses above both equal 1 and thereby conclude the proof. Remark. Clearly, the problem is of projective character: all we need is a tritangent conic and two pairs of suitably perspective triangles: Let ABC be a triangle and let γ be a conic tangent at points D, E, F to the lines BC, CA, AB, respectively. Let further J, K, L be points on the lines EF , F D, DE, respectively, and let M , N , P be points on the lines BC, CA, AB, respectively. If triangles JKL and DEF are perspective from some point on γ, and triangles M N P and ABC are perspective, then triangles JKL and M N P are perspective. The proof goes along the same lines.

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THE 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

FIRST SELECTION TEST

Problem 1. Call a positive integer balanced if the number of its distinct prime factors is equal to the number of its digits in the decimal representation; for example, the number 385 = 5 · 7 · 11 is balanced, while 275 = 52 · 11 is not. Prove that there exist only a finite number of balanced numbers. Solution. Let p1 = 2, p2 = 3, p3 = 5, . . . be the sequence of primes. Any balanced number a with n digits satisfies a ≥ p1 p2 · · · pn . Since p1 p2 · · · p11 = 2 · 3 · 5 · · · 29 · 31 => 1011 and pk > 10, for any k > 11, it follows that there are no balanced numbers having more than 10 digits. Problem 2. Consider a convex pentagon A0 A1 A2 A3 A4 so that the rays (Ai Ai+1 and (Ai+3 Ai+2 meet at Bi+4 , for each i = 0, 1, 2, 3, 4 – indices are considered modulo 5. Show that 4 4   Ai Bi+3 = Ai Bi+2 . i=0

i=0

Solution. Let Ci be the projection of Bi on the line Ai+2 Ai+3 , i = 0, 1, 2, 3, 4. Notice that the right–angled triangles (possibly degenerate) Ai Bi+3 Ci+3 and Ai Bi+3 Bi+3 Ci+3 Ai Bi+2 Ci+2 are similar, for angles Ai are vertical. Hence = , Ai Bi+2 Bi+2 Ci+2 implying 4 4   Ai Bi+3 Bi+3 Ci+3 = = 1, AB B C i=0 i i+2 i=0 i+2 i+2

as claimed.

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Remark. The claim holds for any n-gon with n ≥ 5 sides, within the given conditions. Problem 3. Let n be a positive integer and let x1 , x2 , . . . , xn and y1 , y2 , . . . , yn be real numbers. Prove that there exists a number i, i = 1, 2, . . . , n, such that n  j=1

|xi − xj | ≤

n  j=1

|xi − yj |.

Solution. Without the loss of generality, suppose x1 ≤ x2 ≤ . . . ≤ xn . For each k = 1, 2, . . . , n we have |x1 − xk | + |xn − xk | = |x1 − xn | ≤ |x1 − yk | + |xn − yk |, hence n 

k=1

|x1 − xk | +

n 

k=1

|xn − xk | ≤

The claim holds for i = 1 or i = n.

n 

k=1

|x1 − yk | +

n 

k=1

|xn − yk |.

Problem 4. Let k and n be integer numbers with 2 ≤ k ≤ n − 1. Consider a set A of n real numbers such that the sum of any k distinct elements of A is a rational number. Prove that all elements of the set A are rational numbers. Solution. The difference of any two elements from A is a rational number. To show this, let x = y ∈ A and choose other k − 1 elements of A – the choice can be made, for k − 1 ≤ n − 2. Denote s the sum of the k − 1 elements and apply the hypothesis to infer that x + s and y + s are both rational numbers. Subtracting w get x − y ∈ Q, as claimed. Now let α ∈ A. Rewrite the elements of A as α + qi , qi ∈ Q. The sum of k  distinct elements is equal to kα + qi ∈ Q, hence α ∈ Q, as needed. Problem 5. Consider n persons, each of them speaking at most 3 languages. From any 3 persons there are at least two which speak a common language. i) For n ≤ 8, exhibit an example in which no language is spoken by more than two persons. ii) For n ≥ 9, prove that there exists a language which is spoken by at least three persons.

Solution. i) Split the 8 persons in two groups of 4. Set any pair of persons in each group to speak a different language for a total of 6 + 6 = 12 languages, each spoken by 2 persons, each person speaking 3 languages.

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For n ≤ 7, just remove 8 − n persons. ii) Assume by contrary that each language is spoken by at most two persons. Then each person A can speak with at most three others, for otherwise, by pigeon–hole principle, there exists a language spoken by other two persons besides A, a contradiction. Let B, C, D the persons which whom A can speak. Likewise, E can speak with (at most) three others, namely F, G, H. There is left at least another person, say Z, and in the group A, E, Z no language is spoken in common, a contradiction. SECOND SELECTION TEST

Problem 6. Determine a) the smallest number b) the biggest number n ≥ 3 of non-negative integers x1 , x2 , ... , xn , having the sum 2011 and satisfying: x1 ≤ | x2 − x3 | , x2 ≤ | x3 − x4 | , ... , xn−2 ≤ | xn−1 − xn | , xn−1 ≤ | xn − x1 | and xn ≤ | x1 − x2 | . Solution. Let x1 , x2 , ... , xn non-negative integers satisfying the conditions from the statement. Let us imagine them as being arranged on a circle, in this order. We denote with M the biggest of the numbers x1 , ... , xn . Without loss of generality, we may suppose that x1 = M . Obviously M = 0. From x1 ≤ | x2 − x3 | and the choice of x1 it follows that {x2 , x3 } = {M, 0}. Therefore any M is followed by an M and a 0 on the circle, either in the succession M − M − 0 or the succession M − 0 − M . By induction after the last M known on the circle, we find that on the circle there are only the numbers 0 and M . Moreover • there are no the neighboring zeros (or else the number preceding them would also be a 0 and, inductively, all the numbers on the circle would be zeros, which contradicts the fact that their sum is 2011) • there are no three consecutive M -s. Since 2011 is prime and x1 +x2 +...+xn = kM , where k is the number of the M -s, we find that M = 1 and k = 2011. a) The smallest value of n is obtained when the number of zeros is the smallest possible. Out of three consecutive numbers on the circle, at least one is 0, therefore we have at least 1006 zeros, which means at least 3017 numbers. A possible configuration fulfilling the conditions of the statement is x3k = 0, k = 1, 2, ..., 1005, x3017 = 0, and the rest of the xj = 1.

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b) The biggest n is obtained when the number of zeros is the largest possible. Since there can be no neighboring zeros, we have at most 2011 zeros. This happens if we place alternatively 0 and 1 on the circle, configuration that satisfies the conditions from the statement. Thus, the biggest value of n is 4022. Problem 7. We consider an n × n (n ∈ N, n ≥ 2) square divided into n2 unit squares. Determine all the values of k ∈ N for which we can write a real number in each of the unit squares such that the sum of the n2 numbers is a positive number, while the sum of the numbers from the unit squares of any k × k square is a negative number. Solution. We will prove that the desired numbers k are those that are not factors of n. If k | n that we can tile the n × n square with k × k squares and the total sum should simultaneously be positive and negative. If k  | n then n = kq + r, where 0 < r < k. We fill the unit squares with a (to be chosen conveniently later on) in the positions (ik, jk) with i, j = 1, ..., q and with 1 in the other positions. Every k × k square contains exactly one unit square of the form (ik, jk), therefore the sum in every k × k square is a + k 2 − 1. The total is q 2 a +n2 − q 2 . We will choose a arbitrarily from the non-empty interval  sum n2 1 − q2 , 1 − k2 . Remark: For the case k  | n there are many other ways of choosing the numbers from the unit squares. Another choice is to fill all the unit squares of the columns jk, j = 1, ..., q, with a convenient a and the other ones with 1. Problem 8. a) Prove that if the sum of the non-zero digits a1 , a2 , ... , an is a multiple of 27, then it is possible to permute these digits in order to obtain an n-digit number that is a multiple of 27. b) Prove that if the non-zero digits a1 , a2 , ... , an have the property that every ndigit number obtained by permuting these digits is a multiple of 27, then the sum of these digits is a multiple of 27. Andrei Eckstein

. Solution. a) Obviously n ≥ 3. If n = 3 then a1 = a2 = a3 = 9 and 999 .. 27. Suppose n ≥ 4. Having the sum of its digits a multiple of 9, each of the numbers formed with the digits a1 , a2 , ... , an is a multiple of 9, therefore division with 27 will yield one of the remainders 0, 9 or 18.

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• If all the digits give the same remainder r when divided by 3: N a1 a 2 a n = ... has − if r = 0 then we choose N = a1 a2 ...an . We have that 3 3 3 3 N is a multiple of 9, hence the sum of its digits a multiple of 9, therefore the number 3 N is a multiple of 27. . − if r ∈ {1, 2}, then from a + a + ... + a .. 27 and a ≡ 0 (mod 3), for all i 1

2

n

i

it follows that nr ≡ 0 (mod 3), hence n is a multiple of 3. Then: a1 a2 ...an − (a1 +   a2 + ... + an ) = 9 a1 a1 ...a1 + a2 a2 ...a2 + ... + an−1 .       n−1 cifre



n−2 cifre

 M





 Then M ≡ (n−1)a1 +(n−2)a2 +...+an−1 ≡ r (n−1)+(n−2)+...+2+1 ≡ n(n − 1) ≡ 0 (mod 3). Therefore, in this case, a1 a2 ...an and a1 + a2 + ... + an r· 2 give the same remainder at the division by 27. • If not all of the digits a1 , a2 , ... , an have the same reminder when divided by 3, then we choose three of them such that their sum is not a multiple of 3. Let a, b, c be these three digits. Then the numbers x = ...abc, y = ...bca and z = ...cab give different reminders at the division by 27. Indeed, x − y = abc − bca = 108a − 81b − 9(a+b+c) = M27 −9(a+b+c) = M27 . Since the possible reminders at the division by 27 are only 0, 9 and 18, it follows that (exactly) one of the numbers x, y, z is a multiple of 27. b) The differece between two numbers obtained one from the other by swapping two neighboring digits a and b is 9(a − b) · 10k and needs to be a multiple of 27. We obtain that each of the digits a1 , ..., an has to give the same remainder when divided by 3. From here one continues as by a) in order to show that, in this case, the numbers a1 a2 ...an and a1 + a2 + ... + an give the same remainder when divided by 27.  of the acute triangle ABC is 60◦ , and Problem 9. The measure of the angle A HI = HB, where I and H are the incenter and the orthocenter of the triangle ABC.  Find the measure of the angle B.

Solution. We have m(∠BIC) ≡ m(∠BHC) = 120◦ , hence B, H, I, C are situated on a circle. If m(∠B) > 60◦ then m(∠HBI) = m(∠ABI) − m(∠ABH) = 21 m(∠B) − 30◦ . But m(∠HBI) = m(∠HIB) = m(∠HCB) = 90◦ −m(∠B), hence m(∠B) =

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S ELECTION TESTS FOR THE 2011 JBMO

80◦ . It is easy to see that the case m(∠B) ≤ 60◦ is not possible. THIRD SELECTION TEST

Problem 10. It is said that a positive integer n > 1 has the property (p) if in its prime factorization α

j 1 n = pα 1 · ... · pj

at least one of the prime factors p1 , ... , pj has the exponent equal to 2. a) Find the largest number k for which there exist k consecutive positive integers that do not have the property (p). b) Prove that there is an infinite number of positive integers n such that n, n + 1 and n + 2 have the property (p). Dorel Mihet¸

Solution. a) Among any 8 consecutive integers there exists one of the form 8j + 4. This number has the property (p) because the factor 2 from its prime factorization has the exponent 2. Therefore there can be at most 7 consecutive positive integers that do not have the property (p). Since none of the numbers 29, 30, 31, 32, 33, 34, 35 has the property (p), the largest k is k = 7. b) Since the numbers 98 = 2 · 72 , 99 = 32 · 11, and 100 = 22 · 52 have the property (p), so will the numbers 98 + (7 · 3 · 2)3 · k, 99 + (7 · 3 · 2)3 · k, and 100 + (7 · 3 · 2)3 · k. Alternative solution. By the Chinese remainder theorem, there exist an infinite number of solutions for the system of simultaneous congruences: n ≡ 4 (mod 8), n ≡ 8 (mod 27), n ≡ 23 (mod 125). Then n, n + 1, and n + 2 all have the property (p) because the factor 2, 3, and 5, respectively, has the exponent 2 in their prime factorizations. Problem 11. Find all the finite sets A of real positive numbers having at least two elements, with the property that a2 + b2 ∈ A for every a, b ∈ A with a = b. Solution. If a1 < a2 < ... < an are the elements of A, then a21 + a22 , a21 + a23 , a21 + a2n , a22 + a2n ,..., a2n−1 + a2n belong to A, and a21 + a22 < a21 + a23 < ... < a21 + a2n < a22 + a2n < ... < a2n−1 + a2n , which implies that 2n − 3 ≤ n; hence A has at most 3 elements.

S ELECTION TESTS FOR THE 2011 JBMO

53

If n = 3 and A = {a, b, c} with a < b < c, then a2 + b2 = a, b2 + c2 =

c, a2 + c2 = b, hence a2 − c2 = a − c, and a + c = 1. Substituting in a2 + c2 = b, we 1 1 obtain b = 2a2 − 2a + 1 = 2(−b2 ) + 1, and therefore b = . Then a2 − a + = 0 2 4 1 leads to a = , which contradicts a < b. 2 Hence n = 2, and A = {a, b} with a2 + b2 = a, i.e., for a ∈ (0, 1), we get √ √ b = a − a2 . In conclusion, the desired sets are {a, a − a2 }, with a ∈ (0, 1), 1 a = . 2 Problem 12. Let ABC be a triangle, Ia the center of the excircle at side BC, and M its reflection across BC. Prove that AM is parallel to the Euler line of the triangle BCIa . Solution. Let I be the incenter of ABC, Ha the orthocenter of Ia BC, and Oa the midpoint of the segment [IIa ]. Then BOa = IOa = Ia Oa = COa , therefore Oa is the circumcenter of triangle Ia BC. Moreover, IB  CHa (both are perpendicular on BIa ) and, similarly, IC  BHa , hence BICHa is a parallelogram. Let P denote the projection of Ia onto BC and T be the midpoint of [AIa ]. If r, ra are the inradius and the radius of the excircle at side BC, respectively, then we have: Ha Ia IIa 2Ia Oa Ia Oa r Ha P IA , hence = = = . This proves that = = AIa ra Ia P Ia P Ia A 2Ia T Ia T Oa Ha is parallel to T P . But T P  AM , therefore AM is parallel to Oa Ha , i.e. it is parallel to the Euler line of the triangle Ia BC. Problem 13. Let m be a positive integer. Determine the smallest positive integer n for which there exist real numbers x1 , x2 , . . . , xn ∈ (−1, 1) such that |x1 | + |x2 | + · · · + |xn | = m + |x1 + x2 + · · · + xn |. Solution. Let us consider x1 , x2 , ... , xn a solution of the equation above. We may suppose, without loss of generality, that x1 + x2 + ... + xn ≥ 0 (otherwise we change the signs of all the numbers) and that x1 ≤ ... ≤ xp ≤ 0 < xp+1 ≤ ... ≤ xn . Then the equation becomes −2(x1 +x2 +...+xp ) = m. From xi ∈ (−1, 1) we obtain m m that m < 2p, i.e. p > . Moreover, n − p > xp+1 + ... + xn ≥ −x1 − ... − xp = . 2 2 m m + 1, n − p ≥ + 1, and therefore n ≥ m + 2. A convenient If m is even then p ≥ 2 2 m m choice for n = m + 2 is x1 = ... = x m2 +1 = − , x m +2 = ... = xn = . m+2 2 m+2 m+1 m+1 If m is odd, from p ≥ ,n−p≥ , we obtain n ≥ m + 1. A convenient 2 2

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m m , x m+1 +1 = ... = xn = . 2 m+1 m+1 In conclusion, the smallest value of n for which the equation has solutions is n = m + 1 if m is odd and n = m + 2 if m is even.

choice for n = m+1 is x1 = ... = x m+1 = − 2

FOURTH SELECTION TEST

Problem 14. For every positive integer n let τ (n) denote the number of its positive n factors. Determine all n ∈ N that satisfy the equality τ (n) = . 3 n Solution. If n ∈ N satisfies the condition τ (n) = , then 3 | n. Put n = 3k, 3 n k = is a factor of n. Even if all the positive numbers k ∈ N. If k is even then 2 6 n n n n smaller than are factors of n and the numbers , , ..., are also factors of n, we 6 5 4 1 n n have = τ (n) ≤ + 5. Hence n ≤ 30. Checking the numbers 6, 12, 18, 24 and 30 3 6 we find that 18 and 24 satisfy the desired condition. In the case when k is odd we can proceed similarly, obtaining n = 9, or, alternatively, we can use the fact that a number having an odd number of positive factors is a m2 . We perfect square. If n = m2 , then n has at most m + 1 factors, hence m + 1 ≥ 3 obtain that m ≤ 3 and the conclusion. In conclusion, the problem admits three solutions: 9, 18 and 24. Alternative solution: Since 3 | n, it follows that n has a prime factorization of αj 1 the form n = 3a · pα 1 · ... · pj and the number of his positive factors is τ (n) = n αj 1 (a + 1)(α1 + 1)...(αj + 1). The condition τ (n) = leads to 3a−1 · pα 1 · ... · pj = 3 αi i (a + 1)(α1 + 1)...(αj + 1). Since pα ≥ αi + 1, in order for n to satisfy the i ≥ 2 equation from the statement, it is necessary that a + 1 ≥ 3a−1 , hence a = 1 or a = 2. αi i If in the prime factorization of n there is a prime pi > 3, then pα ≥ 2αi +2 i >4 n and the equality τ (n) = can not take place. If a = 1 then n = 3 · 2m and the 3 n equality τ (n) = reduces to 2(m + 1) = 2m . We find m = 3 (for m ≥ 4 we have 3 2m > 2m + 2), hence n = 24. Similarly, if a = 2 then n = 32 · 2m and from 3(m + 1) = 3 · 2m we obtain m ∈ {0, 1}, i.e. n ∈ {9, 18}. Problem 15. Let ABC be a triangle with circumcenter O. The points P and Q

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are interior points of the sides CA and AB respectively. Let K, L and M be the midpoints of the segments BP , CQ and P Q, respectively, and let Γ be the circle passing through K, L and M . Suppose that the line P Q is tangent to the circle Γ. Prove that OP = OQ. Sergei Berlov, Russia, IMO 2009

Solution. From AB  KM and AC  LM we obtain that ∠KM Q ≡ ∠AQP and ∠LM P ≡ ∠AP Q. If P Q is tangent to the circumcircle of the triangle KLM then ∠KLM ≡ ∠KM Q, hence ∠KLM ≡ ∠AQP . Then ∆AP Q ∼ ∆M KL, AP AQ AP AQ = . It follows that = , i.e. AQ · BQ = AP · P C. This hence ML MK PC BQ means that P and Q have equal powers with respect to the circumcircle of the triangle ABC. As they are both situated inside the circle, they are at equal distance from the center of the circle, O. Problem 16. a) Find the largest possible value of the number x1 x2 + x2 x3 + ... + xn−1 xn , if x1 , x2 , ... , xn (n ≥ 2) are non-negative integers and their sum is 2011. b) Find the numbers x1 , x2 , ... , xn for which the maximum value determined at a) is obtained. Dorel Mihet¸

Solution. a) Let x1 , x2 , ... , xn be non-negative integers satisfying the conditions from the statement. We put M = max xi . If xj = M then x1 x2 + x2 x3 + ... + 1≤i≤n

xn−1 xn ≤ x1 xj + x2 xj + ... + xj−1 xj + xj xj+1 + xj xj+2 + ... + xj xn = xj (2011 − xj ) = M (2011 − M ) ≤ 1005 · 1006. Indeed, the last inequality comes to (M − 1005)(M − 1006) ≥ 0 which is true for any integer M . The largest possible value is 1005 · 1006 because this value can be obtained by choosing, for example, x1 = 1005, x2 = 1006 and xk = 0 for k ≥ 3. b) For n = 2 we have x1 x2 = 1005 · 1006 ⇔ x1 (2011 − x1 ) = 1005 · 1006 ⇔ (x1 − 1005)(x1 − 1006) = 0 ⇔ (x1 , x2 ) ∈ {(1005, 1006), (1006, 1005)}. For n = 3, x1 x2 + x2 x3 = 1005 · 1006 ⇔ x2 (x1 + x3 ) = 1005 · 1006 comes, as above, to x2 = 1005, x1 + x3 = 1006 or x2 = 1006, x1 + x3 = 1005. We obtain (x1 , x2 , x3 ) ∈ {(k, 1005, 1006 − k) | k = 0, 1, ..., 1006} ∪ {(k, 1006, 1005 − k) | k = 0, 1, ..., 1005}.

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For n ≥ 4, we denote by j the smallest index for which xj > 0. Then, replacing xj by 0 and xj+2 by xj+2 + xj , increases the value of the sum by xj xj+3 . Using this remark it is easy to see that if x1 x2 + x2 x3 + ... + xn−1 xn = 1005 · 1006, then at most three of the terms can be non-zero. We obtain (x1 , ..., xn ) ∈ {(0, ..., 0, k, 1005, 1006−k, 0, ..., 0) | k = 0, 1, ..., 1006}∪{(0, ..., 0, k, 1006, 1005− k, 0, ..., 0) | k = 0, 1, ..., 1005}, where the group of the three non-zero components can be located anywhere. Problem 17. Show that there is an infinite number of positive integers t such that none of the equations x2 + y 6 = t, x2 + y 6 = t + 1, x2 − y 6 = t, x2 − y 6 = t + 1 has solutions (x, y) ∈ Z × Z.

Dorel Mihet¸

Solution. If x is a positive integer, then either x12 ≡ 0 (mod 13) or x12 ≡ 1 (mod 13), hence x6 is congruent with −1, 0 or 1 modulo 13. Therefore, if t is congruent with 6 (mod 13), then ±x6 + t is congruent with 5, 6 or 7 (mod 13), while ±x6 + t + 1 is congruent with 6, 7 or 8 (mod 13). On the other hand, perfect squares are congruent with 0, 1, 4, 9, 3, 12 or 10 (mod 13). Therefore a perfect square can not be equal to a number of the form ±x6 + t or ±x6 + t + 1 if t ≡ 6 (mod 13). In conclusion, all the numbers that are congruent with 6 modulo 13 have the required property.

THE 28th BALKAN MATHEMATICAL OLYMPIAD Ias¸i, Romania, May 4th – May 8th , 2011

Problem 1. Let ABCD be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at E. The midpoints of AB and CD are F and G respectively and  is the line through G parallel to AB. The feet of the perpendiculars from E onto  and CD are H and K, respectively. Prove that the lines EF and HK are perpendicular. United Kingdom

Solution. The points E, K, H, G are on the circle of diameter GE, so the angles EHK and EGK are equal.

57

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2011 BALKAN M ATHEMATICAL O LYMPIAD

Also, from ∠DCA = ∠DBA and CE/CD = BE/BA follows 2CE 2BE BE CE = = = , CG CD BA BF so the triangles CGE and BF E are similar. In particular, the angles EGC and BF E are equal, and therefore so are the angles EHK and BF E. But the lines EH and BF are perpendicular and so, since EF and HK are obtained by rotations of these lines by the same (directed) angle, the lines EF and HK also perpendicular. Problem 2. Given three real numbers x, y, z such that x + y + z = 0, show that x(x + 2) y(y + 2) z(z + 2) + + ≥ 0. 2x2 + 1 2y 2 + 1 2z 2 + 1 When does equality hold? Greece

Solution. The inequality is clear if xyz = 0, in which case equality holds if and only if x = y = z = 0. Henceforth assume xyz = 0 and rewrite the inequality as (2y + 1)2 (2z + 1)2 (2x + 1)2 + + ≥ 3. 2 2 2x + 1 2y + 1 2z 2 + 1 Notice that (exactly) one of the products xy, yz, zx is positive, say yz > 0, to get (2y + 1)2 (2z + 1)2 2(y + z + 1)2 + ≥ 2y 2 + 1 2z 2 + 1 y2 + z2 + 1 2(x − 1)2 = 2 x − 2yz + 1 2(x − 1)2 . ≥ x2 + 1

(by Jensen) (for x + y + z = 0) (for yz > 0)

Here equality holds if and only if x = 1 and y = z = −1/2. Finally, since 2(x − 1)2 2x2 (x − 1)2 (2x + 1)2 + − 3 = ≥ 0, 2x2 + 1 x2 + 1 (2x2 + 1)(x2 + 1)

x ∈ R,

the conclusion follows. Clearly, equality holds if and only if x = 1, so y = z = −1/2. Therefore, if xyz = 0, equality holds if and only if one of the numbers is 1, and the other two are −1/2.

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Solution 2. Rewrite the inequality as in Solution 1, (2y + 1)2 (2z + 1)2 (2x + 1)2 + + ≥ 3, 2x2 + 1 2y 2 + 1 2z 2 + 1 and use the condition x + y + z = 0 to replace 3 in the right-hand member by 4 2 3 (x

(2y + 1)2 (2z + 1)2 (2x + 1)2 + + , 4 4 2 2 2 2 2 2 + y2 + z2 ) + 1 3 (x + y + z ) + 1 3 (x + y + z ) + 1

and thereby deduce that it is sufficient to show that (2x + 1)2 ≥ 2x2 + 1

(2x + 1)2 4 2 2 2 3 (x + y + z ) + 1

(∗)

and the like. Unless x = −1/2, in which case equality holds trivially, use again the condition x + y + z = 0 to transform (∗) into the equivalent and obvious inequality (y−z)2 ≥ 0, which yields the equality case y = z. This proves the required inequality and shows that equality holds if and only if the variables all vanish simultaneously or two equal −1/2 and the third equals 1. Problem 3. Let S be a finite set of positive integer numbers which has the following property: if x is a member of S, then so are all positive divisors of x. Call a subset T of S good (respectively, bad ) if it is non-empty and, whenever x and y are members of T , and x < y, the ratio y/x is (respectively, is not) a power of a prime number; agree that a singleton (one-element) subset is both good and bad. Show that a maximal good subset of S has as many elements as a minimal partition of S into bad subsets. Bulgaria

Solution. Notice first that a bad subset of S contains at most one element from a good one, to deduce that a partition of S into bad subsets has at least as many members as a maximal good subset. Notice further that the elements of a good subset of S must be among the terms a geometric sequence whose ratio is a prime: if x < y < z are elements of a good subset of S, then y = xpα and z = yq β = xpα q β for some primes p and q and some positive integers α and β, so p = q for z/x to be a power of a prime. Next, let P = {2, 3, 5, 7, 11, · · · } denote the set of all primes, let m = max {expp x : x ∈ S and p ∈ P },

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where expp x is the exponent of the prime p in the canonical decomposition of x, and notice that a maximal good subset of S must be of the form {a, ap, · · · , apm } for some prime p and some positive integer a which is not divisible by p. Consequently, a maximal good subset of S has m + 1 elements, so a partition of S into bad subsets has at least m + 1 members. Finally, notice by maximality of m that the sets  expp x ≡ k (mod m + 1)}, k = 0, 1, · · · , m, Sk = {x : x ∈ S and p∈P

form a partition of S into m + 1 bad subsets. The conclusion follows. Problem 4. The opposite sides of a convex hexagon of unit area are pairwise parallel. The lines of support of three alternate sides meet in pairs to determine the vertices of a triangle. Similarly, the lines of support of the other three alternate sides meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is greater than or equal to 3/2. Bulgaria

Solution. Unless otherwise stated, throughout the proof indices take on values from 0 to 5 and are reduced modulo 6. Label the vertices of the hexagon in circular order, A0 , A1 , · · · , A5 , and let the lines of support of the alternate sides Ai Ai+1 and Ai+2 Ai+3 meet at Bi . To show that the area of at least one of the triangles B0 B2 B4 , B1 B3 B5 is greater than or equal to 3/2, it is sufficient to prove that the total area of the six triangles Ai+1 Bi Ai+2 is at least 1: 5  i=0

area Ai+1 Bi Ai+2 ≥ 1.

To begin with, reflect each Bi through the midpoint of the segment Ai+1 Ai+2 to get the points Bi . We shall prove that the six triangles Ai+1 Bi Ai+2 cover the hexagon. To this end, reflect A2i+1 through the midpoint of the segment A2i A2i+2 to get the points A2i+1 , i = 0, 1, 2. The hexagon splits into three parallelograms, A2i A2i+1 A2i+2 A2i+1 , i = 0, 1, 2, and a (possibly degenerate) triangle, A1 A3 A5 . Notice first that each parallelogram A2i A2i+1 A2i+2 A2i+1 is covered by the pair of   triangles (A2i B2i+5 A2i+1 , A2i+1 B2i A2i+2 ), i = 0, 1, 2. The proof is completed by showing that at least one of these pairs contains a triangle that covers the trian ≥ A2i A2i+5 and gle A1 A3 A5 . To this end, it is sufficient to prove that A2i B2i+5

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 ≥ A2j+2 A2j+3 for some indices i, j ∈ {0, 1, 2}. To establish the first A2j+2 B2j inequality, notice that  = A2i+1 B2i+5 , A2i B2i+5

to get

A2i A2i+5 = A2i+4 A2i+5 ,

A0 B5 A1 B5 = A4 A5 A5 B 3

2   A2i B2i+5

i=0

Similarly,

whence the conclusion.

and

A2i A2i+5

i = 0, 1, 2,

A3 B1 A2 A3 = , A0 A1 A0 B 5 = 1.

2   A2j+2 B2j = 1, A A j=0 2j+2 2j+3

Solution 2. With reference to Solution 1, suppose, if possible, that the six tri fail to cover the hexagon A0 A1 · · · A5 . Then some point X of angles Ai Ai+1 Bi+5 the hexagon falls outside each of those triangles. Up to an affine transformation, we may (and will) assume that the triangles B0 B2 B4 and B1 B3 B5 are both equilateral, so the hexagon is equiangular. We may further assume that the angle XA0 A1 does not exceed 60◦ . Since X is not covered by the triangle A0 A1 B5 , the angle XA1 A0 exceeds 60◦ , so XA0 > XA1 and the angle XA1 A2 is less than 60◦ . Next, the triangle A1 A2 B0 does not cover X, so the angle XA2 A1 exceeds 60◦ , XA1 > XA2 and the angle XA2 A3 is less than 60◦ . Continuing, we obtain XAi > XAi+1 , i = 0, 1, · · · , 5, and thereby reach the contradiction XA0 > XA1 > · · · > XA5 > XA0 . Solution 3. In the setting of Solution 2 – the triangles B0 B2 B4 and B1 B3 B5 are both equilateral etc. – let ai denote the length of the segment Ai Ai+1 , let 0 and 1

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respectively denote the side lengths of the equilateral triangles B0 B2 B4 and B1 B3 B5 and notice that ai−1 + ai + ai+1 = i mod 2 to get a0 + a2 + a4 = 21 − 0 and a1 + a3 + a5 = 20 − 1 . Incidentally, these relations show that the given configuration is geometrically possible and non-degenerate if and only if 1/2 < 0 /1 < 2. Now let A, A0 , A1 respectively denote the areas of the hexagon A0 A1 · · · A5 and the triangles B0 B2 B4 and B1 B3 B5 . Then √  3 2 A0 − A = (a1 + a23 + a25 ) area A2i+1 B2i A2i+2 = 4 i=0,1,2 √ √ 3 3 (a1 + a3 + a5 )2 = (20 − 1 )2 . ≥ 12 12 √ Similarly, A1 − A ≥ (21 − 0 )2 3/12, so √  3 (20 − 1 )2 + (21 − 0 )2 A0 + A1 − 2A ≥ 12 √ √   3 2 3 2 2 2 0 + 1 + 4(0 − 1 ) ≥ 0 + 21 = 12 12 1 = (A0 + A1 ) . 3 Consequently, A0 + A1 ≥ 3A and the conclusion follows. Solution 4. With reference again to the notation and conventions in Solution 1, let r = B2i B2i+2 /B2i+3 B2i+5 denote the ratio of similarity of the triangles B0 B2 B4 and B3 B5 B1 , let ri = Ai+1 Ai+2 /Bi−2 Bi+2 denote the ratio of similarity of the triangles Ai+1 Bi Ai+2 and Bi−2 Bi Bi+2 , and notice that rr2i = 1 − r2i−1 − r2i+1 and r2i+1 = r(1−r2i −r2i+2 ) to obtain r0 +r2 +r4 = 2−1/r and r1 +r3 +r5 = 2−r. As already noticed in Solution 3, these relations show that the given configuration is geometrically possible and non-degenerate if and only if 1/2 < r < 2. Now let A0 and A1 respectively denote the areas of the triangles B0 B2 B4 and B1 B3 B5 and recall that the hexagon A0 A1 · · · A5 has unit area, to write  2 1 1 1 1 2− = 1 − (r02 + r22 + r42 ) ≤ 1 − (r0 + r2 + r4 )2 = 1 − , A0 3 3 r 1 1 1 2 = 1 − (r12 + r32 + r52 ) ≤ 1 − (r1 + r3 + r5 )2 = 1 − (2 − r) , A1 3 3 and thereby conclude that A0 ≥ 3/2 if 1 ≤ r < 2, and A1 ≥ 3/2 if 1/2 < r ≤ 1.

THE DANUBE MATHEMATICAL COMPETITION C˘al˘aras¸i, October 2010

Problem 1. Determine all integer numbers n ≥ 3 such that the regular n-gon can be decomposed into isosceles triangles by noncrossing diagonals. Solution. The required numbers are of the form n = 2r (2s + 1), where r and s are nonnegative integer numbers which do not vanish simultaneously. Clearly, any such n works. To establish the converse, let K be a regular n-gon, n ≥ 4, which can be decomposed into isosceles triangles by noncrossing diagonals. Begin by noticing that each edge e of K must be an edge of a unique isosceles triangle Te in the decomposition. Two cases are possible: either e is opposite the apex of Te or e and one of the adjacent edges of K are the edges of Te issuing from the apex. (Since n ≥ 4, Te cannot be equilateral, so the apex is well defined.) If n is even, no vertex of K lies on the perpendicular bisector of an edge of K, so the first case is ruled out. Consequently, the decomposition must contain exactly one of the two bracelets of n/2 isosceles triangles clipped off by short diagonals joining consecutive vertices of K of likewise parity. These short diagonals are the edges of a regular n/2-gon which is also decomposed into isosceles triangles by noncrossing diagonals and the conclusion follows by induction. If n is odd, then K has a unique edge e opposite the apex of Te : Since n is odd and each short diagonal clips off two edges of K, at least one such e exists. The apex of Te lies on the perpendicular bisector of e, so it must be the vertex of K opposite e. Uniqueness of e should now be clear: were there another such e , the interiors of Te and Te would overlap. Consequently, e is unique and K splits into Te and two 63

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polygons L and L which are reflections of one another in the perpendicular bisector of e. To complete the proof, it is sufficient to show that the number of vertices of L is one plus a power of 2. Begin by noticing that L inherits by restriction a decomposition into isosceles triangles by noncrossing diagonals. Let x0 , . . . , xm be a circular labelling of the vertices of L around the boundary, where x0 is the vertex of K opposite e and xm is a vertex of e. Since dist(xi , xj ) < dist(x0 , xm ) if {i, j} = {0, m}, it follows that x0 xm is an edge of an isosceles triangle with apex at some xk , 0 < k < m. Notice that dist(x0 , xi ) < dist(xi , xm ) if 0 < i < m/2 and dist(x0 , xi ) > dist(xi , xm ) if m/2 < i < m, to deduce that m must be even, k = m/2, and L splits into an isosceles triangle, x0 xm/2 xm , and two polygons, x0 · · · xm/2 and xm/2 · · · xm , which are reflections of one another in the perpendicular bisector of the segment x0 xm . Now we are essentially back in the situation that arose above. Repeat the same argument verbatim to infer that m/2 must be even and so on all the way down to conclude that m must be a power of 2. Problem 2. Given a triangle ABC, let A , B  and C  be the perpendicular feet dropped from the centroid G of the triangle ABC on the sides BC, CA and AB, respectively. Reflect A , B  and C  through G to A , B  and C  , respectively. Prove that the lines AA , BB  and CC  are concurrent. Solution. Let A0 be the perpendicular foot dropped from A on the line BC, let A1 be the midpoint of the side BC, and let A2 be the point where the line AA meets the line BC; define the points Bi and Ci , i = 0, 1, 2, similarly. We shall prove that (AA0 , AA2 ), (BB0 , BB2 ) and (CC0 , CC2 ) are pairs of isotomic lines; that is, A0 and A2 are reflections of one another through A1 and the like. It will then follow that the lines AA = AA2 , BB  = BB2 and CC  = CC2 are concurrent at the isotomic conjugate of the orthocentre of the triangle ABC. Clearly, it is sufficient to show the lines AA0 and AA2 isotomic. To this end, notice first that A A1 /A0 A1 = 1/3. On the other hand, Menelaus’ theorem applied to triangle GA A1 and transversal AA A2 yields A1 A2 /A A2 = 3/4, so A A1 /A2 A1 = 1/3. Consequently, A0 and A2 are indeed reflections of one another through A1 . Problem 3. All sides and diagonals of a convex n-gon, n ≥ 3, are coloured one of two colours. Show that there exist (n + 1)/3 pairwise disjoint monochromatic

2010 DANUBE M ATHEMATICAL C OMPETITION

65

segments. (Two segments are disjoint if they do not share an endpoint or an interior point.) Solution. If all sides are monochromatic, then the assertion is clearly true. Otherwise, delete a vertex incident with two sides of different colours together with its neighbours, delete all sides and diagonals incident with these three vertices and apply induction. Problem 4. Given a prime number p congruent to 3 modulo 4, show that w2p + x2p + y 2p = z 2p for no integer numbers w, x, y, z whose product is not divisible by p. Solution. Suppose there are four such numbers. Without loss of generality, we may (and will) assume that they are jointly coprime: (w, x, y, z) = 1. Reduction modulo 4 shows that z and exactly one of the numbers w, x, y, say y, must be odd. Write   p−1 2(p−k−1)  2k   2 2p 2p 2p 2p 2 2k 2(p−1) to z − pz z − y w +x = z −y = z −y k=1 deduce that the second factor above is congruent to 3 modulo 4 and infer thereby that in its decomposition into prime factors some prime q ≡ 3 (mod 4) occurs with an odd exponent. Since −1 is a quadratic non-residue modulo q, it follows that w and x are both divisible by q, and in the decomposition of w2p + x2p into prime factors, q occurs with an even exponent. Hence z 2 − y 2 is divisible by q, and therefore so is pz 2(p−1) . Notice that p = q (for q divides w, but p does not by assumption) to deduce that z is divisible by q. Then so is y. Consequently, w, x, y, z all share the common factor q, in contradiction with their joint coprimality. Problem 5. Given an integer number n ≥ 3, determine the real numbers x1 , x2 , · · · , xn minimising (n − 1)(x21 + x22 + . . . + x2n ) + nx1 x2 · · · xn , subject to xk ≥ 0, k = 1, 2, . . . , n, and x1 + x2 + · · · + xn = n. Solution. The minimum is n2 and is achieved when all xk = 1 or when exactly one of the xk is 0 and other ones are all n/(n − 1). Notice that if some xk = 0, then the Cauchy-Schwarz inequality hints at an educated guess of the minimum; the guess is further supported by the fact that the same value is achieved when all xk = 1. Let f (x1 , x2 , · · · , xn ) = (n − 1)(x21 + x22 + . . . + x2n ) + nx1 x2 · · · xn . Assuming 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn , we first show that f (x1 , x2 , · · · , xn ) ≥ f (x1 , (x2 + xn )/2, x3 , · · · , xn−1 , (x2 + xn )/2).

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To begin with, notice that xn ≥ 1, to get f (x1 , x2 , . . . , xn ) − f (x1 , (x2 + xn )/2, x3 , . . . , xn−1 , (x2 + xn )/2) 1 = (x2 − xn )2 (2(n − 1) − nx1 x3 · · · xn−1 ) 4 1 ≥ (x2 − xn )2 (2(n − 1) − nx1 x3 · · · xn−1 xn ) . 4 We now show that x1 x3 · · · xn−1 xn ≤ 2(n − 1)/n, so the difference above is indeed non-negative. Notice first that n = x1 + x2 + · · · +xn ≥ 2x1 +x3 +· · · +xn ≥ (n− 1) (2x1 x3 · · · xn−1 xn )

1/(n−1)

,

so x1 x3 · · · xn−1 xn ≤ nn−1 /(2(n − 1)n−1 ). Since nn−1 /(2(n − 1)n−1 ) ≤ 2(n − 1)/n, n ≥ 2, the conclusion follows. (The latter follows from the well-known fact that the sequence un = (1 + 1/n)n+1 , n ≥ 1, is decreasing, so un ≤ u1 = 4, n ≥ 1; that un decreases is easily seen by Bernoulli’s inequality: n2(n+1) /(n2 − 1)n+1 = (1 + 1/(n2 − 1))n+1 ≥ 1 + (n + 1)/(n2 − 1) = n/(n − 1), n ≥ 2.) Consequently, the value of f does not increase when the largest and the second smallest of the xk is replaced by their arithmetic mean. It follows that f achieves its minimum when n − 1 of the xk are all equal to some x, and the smallest of the xk is n − (n − 1)x. (That f actually achieves a minimum subject to the stated conditions follows by a standard compactness argument which may be assumed without proof.) Since the xk are all non-negative, and n − (n − 1)x is the smallest among them, it follows that 1 ≤ x ≤ n/(n − 1). So we have to minimise the polynomial function   g(x) = (n − 1) (n − (n − 1)x)2 + (n − 1)x2 + n(n − (n − 1)x)xn−1

on the closed interval 1 ≤ x ≤ n/(n − 1). To this end, recall the guess made in the beginning, to evaluate g(x) − n2 = n(n − (n − 1)x)(x − 1)2

n−2  k=1



 xk−1 + xk−2 + · · · + 1 ≥ 0,

1 ≤ x ≤ n/(n − 1). Equality holds if and only if x = 1 or x = n/(n − 1), whence the conclusion.

THE Eighth IMAR MATHEMATICAL COMPETITION Bucures¸ti, S. Haret National College, November 2010

Problem 1. Show that a sequence (εn )n∈N of plus and minus ones is periodic with period a power of 2, if and only if εn = (−1)P (n) , n ∈ N, where P is an integer-valued polynomial with rational coefficients. Solution. A polynomial P of degree at most k with complex coefficients is integervalued if and only if P =

k  j=0

   k aj X = X(X − 1) · · · (X − j + 1), aj j j! j=0

where the aj are all integer numbers, so its coefficients are rational. We show that for such a P , the sequence ((−1)P (n) )n∈N is periodic with period 2r , where r = min {s : 2s > k}. To this end, it suffices to show that if j < 2s and m is   m+2s  integer, then m mod 2. These are the coefficients of X j in the expansions j j ≡ s s m m+2s , respectively. The congruence (1 + X)2 ≡ 1 + X 2 of (1 + X) and (1 + X)  s s mod 2 follows easily by induction on s. Hence (1 + X)m+2 = (1 + X)m 1 + X 2 mod 2. Since j is less than 2s , it is immediate that the coefficients of X j in (1 + X)m s and (1 + X)m+2 have the same parity. Conversely, let α0 , · · · , α2r −1 be arbitrary integers, and let β0 , · · · , β2r −1 be the solution to the lower triangular system of linear equations r 2 −1

i=0

  j = αj , βi i

j = 0, · · · , 2r − 1. 67

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Since the coefficient of βj in the j-th equation is 1, the βi are all integers. The polynomial r r   2 2 −1 −1 βi X X(X − 1) · · · (X − i + 1) βi = i! i i=0 i=0

realizes the sequence (−1)αj , j = 0, · · · , 2r − 1, and its extension with period 2r .

Solution 2. Given an integer-valued polynomial P with rational coefficients, we show that the sequence ((−1)P (n) )n∈N is periodic with period a power of 2. Clearly, it is sufficient to show that, for some non-negative integer s, the integer numbers P (n) and P (n + 2s ) both have the same parity, whatever n ∈ N. To this end, consider a positive integer m such that Q = mP is a polynomial with integral coefficients (e.g., let m be the least common multiple of the denominators of the coefficients of P when written in lowest terms), let 2r be the highest power of 2 dividing m, and let s be an integer greater than r. Write m = 2r (2m + 1) and fix a positive integer n. Since 2s divides the difference Q(n + 2s ) − Q(n) = m(P (n + 2s ) − P (n)) = 2r (2m + 1)(P (n + 2s ) − P (n)), and s > r, the conclusion follows. Conversely, given a sequence (εn )n∈N of plus and minus ones which is periodic with period 2r , let  0, if εk = 1, k = 0, · · · , 2r − 1, ak = 1, if εk = −1,

and consider the polynomial (Lagrange) P =

r 2 −1

k=0

=

r 2 −1

k=0

=

r 2 −1

k=0

ak

 X −j k−j j=k 

  1 1 (−1)k+1 ak  (X − j)  r k! (2 − k − 1)! (−1)

k+1

0≤j 0. Then BQ

(1 − P A tan 4x)/ tan 4x   (t2 + 2t − 1)(t2 + 1) 4t(1 − t2 ) 1 − 6t2 + t4 · = 1− 2 4 (t + 1)(3t2 − 1) 4t(1 − t2 ) 1 −26t + t  4t(1 − t)(t2 + 1) (t + 2t − 1)(t2 − 2t − 1) = 1− (3t2 − 1)(t2 − 2t − 1) 4t(1 − t2 ) t2 + 2t − 1 (7t4 − 10t3 − 2t + 1). = 4t(1 − t2 )(3t2 − 1) =

What we look for is Q ≡ B, i.e. BQ = 0 (as a matter of fact occurring for x = π/8, and seemingly no more realized). But 7t4 − 10t3 − 2t + 1 = (t2 + 2t − 1)(7t2 − 24t + 55) − 136t + 56 = (1 − x)(2 − √ √ 7x3 ) − (3x3 + 1), thus decreasing on [0, 2 − 1], while 56 < 136( 2 − 1), hence the √ polynomial considered above has just one root 0 < τ ≈ 0.3447 < 0.4142 ≈ 2 − 1, corresponding to just one angle 0 < ξ ≈ 0.3319 < 0.3927 ≈ π/8.5 4 An

alternative proof uses the fact that such a configuration, with P ≡ A, can occur if (and only if) x = π/8 (hence P cannot bypass A for angles 0 ≤ x ≤ π/8). Limit cases are P ≡ D for x = 0 and P ≡ A for x = π/8, so by continuity P ranges over the entire segment [DA]. The monotony of the variation of M , N and P must be argued by stronger methods than just continuity. 5 For angle x ∈ (ξ, π/8) the line P Q does not meet anymore the segment (AB), going beyond B, since Q does not vary monotonically. This phenomenon, I humbly reckon, can only be fathomed through analytic methods.

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Thus the two solutions x ∈ [0, π/8] are ξ and π/8. Problem 3. Find the largest constant K ≥ 0 such that for any 0 ≤ k ≤ K, and for any non-negative real numbers a, b, c, satisfying a2 + b2 + c2 + kabc = k + 3, to have a + b + c ≤ 3. Solution. Let us work from first principles. Whenever (at least) one variable is  zero, say c, it follows a2 + b2 = k + 3, hence a + b + c ≤ 2(k + 3) ≤ 3 for k ≤ 3/2, with equality holding for a = b = 3/2. We thus have a starting tentative limiting bound of K = 3/2. Let also notice that a = b = c = 1 checks for any value of k, while providing a maximal admissible value for a + b + c = 3. Assume σ = a + b + c > 3. Notice that then (for some k > 0) k(1 − abc) = 1 1 2 a + b2 + c2 − 3 ≥ (a + b + c) − 3 = σ 2 − 3 > 0, by Cauchy-Schwarz, and 3 3 so abc < 1. Now is the time for the key step. Since a + b + c > 3, assuming some ordering of our variables, say 0 ≤ a ≤ b ≤ c, it will follow c > 1, and so we can compute a 2 + b2 + c 2 − 3 k = 1 − abc (a + b)2 + c2 − 3 − 2ab = 1 − abc (σ − c)2 + c2 − 3 − 2ab = 1 − abc c(2c2 − 2σc + σ 2 − 3) − 2 + 2(1 − abc) = c(1 − abc) 2 2c3 − 2σc2 + (σ 2 − 3)c − 2 + . = c c(1 − abc) 2

Denote f (c) = 2c3 − 2σc2 + (σ 2 − 3)c − 2. As (x − y)(f (x)  − f (y)) 2  2 1 1 3 = (x − y)2 x + y − σ + (x − y)2 + (σ 2 − 9) ≥ 0, f is increasing; 2 3 2 3 and as f (1) = (σ + 1)(σ − 3) > 0, it follows f (c) > 0. Therefore the minimal value 1 for k is reached when ab = 0, for a = 0, whence k ≥ b2 + c2 − 3 ≥ (b + c)2 − 3 = 2 3 1 2 σ −3> . 2 2 The issue of the extremal points, leading to equality when k = 3/2, will better be addressed within the next alternative solution.

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77

Alternative Solution. Let us apply the trusted method of Lagrange multipliers. Define L(a, b, c) = a + b + c − λ(a2 + b2 + c2 + kabc − k − 3). The analysis of the values on the border of the domain of L has been done above, leading to K ≤ 3/2 (we could now simplify our work to just k = 3/2, but it is enlightening to see it in full generality). The system of partial derivatives is  ∂L   = 1 − λ(2a + kbc)   ∂a       ∂L = 1 − λ(2b + kca)  ∂b          ∂L = 1 − λ(2c + kab) ∂c

Equaling the partial derivatives to zero forbids λ = 0. Then, from pairwise equalities, we get λ(a − b)(2 − kc) = λ(b − c)(2 − ka) = λ(c − a)(2 − kb) = 0.

One possibility is a = b = c = x, thus (from the constraint) 3x2 + kx3 = k + 3, or (x−1)(kx2 +(k+3)x+(k+3)) = 0, with only non-negative real solution x = 1, since the coefficients of the quadratic factor are non-negative. Then a + b + c = 3x = 3, the admissible maximum. The other possibility is for two variables to be equal, say a = b, but not equal to the third, hence needing a = b = 2/k, thus (from the constraint) 8/k2 + c2 + 4c/k = k +√ 3, or k 2 c2 + 4kc − (k 3 + 3k 2 − 8) = 0, with non-negative real solution (k + 2) k − 1 − 2 c= for k 3 + 3k 2 − 8 ≥ 0, i.e. k ≥ κ ≈ 1.3553. Then a + b + c = k √ (k + 2) k − 1 + 2 < 3 for k < 2, since it is equivalent to (k − 2)3 < 0. As k needs k be at most K ≤ 3/2, these points are critical, but not global maxima (in fact they turn to be global minima).6 √  (k + 2) k − 1 + 2 the record, for the values k3 + 3k2 − 8 ≥ 0 we also have ≤ 2(k + 3), k  2 k3 + 3k2 − 8 since it turns to be equivalent to 2  ≥ 0. Let κ ≈ 1.3553 be the unique real root k 2(k + 3) + 4 of k3 + 3k2 − 8 = 0, with κ ∈ (1, 3/2). Then for k ∈ (κ, 3/2) we have the points on the border being local maxima, the other being global minima, with point (1, 1, 1) as unique global maximum. 6 For

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Putting it all together, the largest admissible value for K turns to be 3/2, with the maximum value a+b+c = 3 being reached only at points (3/2, 3/2, 0), (3/2, 0, 3/2), (0, 3/2, 3/2) and (1, 1, 1). For 0 ≤ k < 3/2 the unique maximum is reached at (1, 1, 1). Notice the importance of examining the values on the border; without that, the other critical interior points found but (1, 1, 1) (which works for any k) achieve a larger value than 3 for a + b + c only starting with k > 2, so would induce the erroneous bound K = 2. Alternative Solution. The fact the value 3 for a + b + c is reached for k = 3/2 both at a = b = c = 1, and at a = b = 3/2 and c = 0 et al., suggests this is a  Schur-type inequality. Indeed, assume 0 ≤ k ≤ 3/2 and a > 3. Then  2 >k + 3 = a + kabc  2 3kabc ≥ a +    a   k  2 k  2 9abc a + a +  = 1− a 3 3  k  2 2k  ≥ 1− a + ab 3  3    2k  2 k  2 = 1− a + a + 2 ab 3 3  2k  2 k  2 a + ( a) = 1− 3     3 2k k  2 1 1− + ( a) ≥ 3 3 3 k+3  2 ( a) = 9 >k+3  2     2 9abc a ( a) + since inequality a +  ≥ 2 ab is in turn equivalent to a      a2 b + ab2 , at last a(a − 9abc ≥ 2 ( ab) ( a), then a3 + 3abc ≥  2  1 a by Cauchy-Schwarz. b)(a−c) ≥ 0, a basic form of Schur; while a2 ≥ 3 Problem 4. Let a, b, c be given positive integers. Prove there exists some positive integer N such that a | N bc + b + c b | N ca + c + a

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c | N ab + a + b if and only if, denoting d = gcd(a, b, c) and a = dx, b = dy, c = dz, the positive integers x, y, z are pairwise co-prime, and also gcd(d, xyz) | x + y + z. Solution. The necessity of having x, y, z be pairwise co-prime is proved by say, assuming gcd(x, y) > 1. Then a | N bc+b+c becomes x | dN yz+y+z, and so we must have gcd(x, y) | z, absurd, since under this assumption it then follows gcd(x, y) | gcd(x, y, z) = 1. On the other hand, if the co-primality condition holds, consider the integers      xyz − x < 2xyz − x < ··· < xy xyz − x.

  These xy integer numbers will yield different remainders modulo xy, since if     ixyz − x ≡ jxyz − x (mod xy), then also xy | |i − j|xyz, whence   i = j, since we have 0 ≤ |i − j| < xy and gcd(xyz, xy) = 1. Therefore    there will exist some (unique) 1 ≤ t ≤ xy such that xy | txyz − x, i.e.     txyz − x = C xy for some positive integer C, therefore txyz = C xy + x, so x | Cyz + y + z et al. We found a suitable value C for the triplet x, y, z (for similar relations with the ones sought for a, b, c). Then all the other suitable values must be  of the form C  = C + Mxyz, since we need have xyz | (C  − C) xy, while  gcd(xyz, xy) = 1.

Now the time has come to analyze the last condition. In order to have a | N bc + b+c, and the similar others, equivalent to x | dN yz +y +z et al., we need have dN = C + M xyz for some non-negative integer M . Denote e = gcd(d, xyz); then e | d,   so e | C + M xyz. But then we also must have e | xyz | (C + M xyz) xy + x,  hence e | x. Conversely, if e |



x, then e | xyz | (C + M xyz)





x, so e | C



xy. d Since clearly gcd(e, xy) = 1, this means e | C. Therefore we need have N = e   xyz d xyz C  xyz −1 C +M , and since clearly gcd , = 1, take M ≡ − e e e e e e C xyz C + M xyz d divides +M . Take now N = (of (mod d/e), wherefore e e e d course, N  = N + Mabc also works). 

xy +

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R EMARKS . Notice there exist easy counterexamples for gcd(x, y, z) = 1, but x, y, z not pairwise co-prime; just take x = pq, y = qr, z = rp, with p, q, r prime. Notice also there exist counterexamples to contradict the last condition, for exam  ple (a, b, c) = (6, 9, 15), when d = 3, e = 3, x = 10, and so e  x. For all such examples thus there exists no solution.

THE Fourth ROMANIAN MASTER OF MATHEMATICS Bucures¸ti, T. Vianu National College, February 2011

FIRST DAY

Problem 1. Prove that there exist two functions f, g : R → R, such that f ◦ g is strictly decreasing and g ◦ f is strictly increasing.

(Poland) Andrzej Komisarski & Marcin Kuczma

Solution. Let      2k 2k+1  2 ,2 ; −22k+1 , −22k • A= k∈Z

• B=

 

k∈Z

−22k , −22k−1



22k−1 , 22k



.

Thus A = 2B, B = 2A, A = −A, B = −B, A∩B = ∅, and finally A∪B ∪{0} = R. Let us take   for x ∈ A  x f (x) =

  

−x for x ∈ B 0

and g(x) = 2f (x).

for x = 0

Then f (g(x)) = f (2f (x)) = −2x and g(f (x)) = 2f (f (x)) = 2x. Problem 2. Determine all positive integers n for which there exists a polynomial f (x) with real coefficients, with the following properties: 1) for each integer k, the number f (k) is an integer if and only if n does not divide k; 2) the degree of f is less than n. 81

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Solution. We will show that such polynomial exists if and only if n = 1 or n is a power of a prime. We will use two known facts stated in L EMMATA 1 and 2. L EMMA 1. If pa is a power of a prime and k is an integer, then (k − 1)(k − 2) . . . (k − pa + 1) (pa − 1)! is divisible by p if and only if k is not divisible by pa . Proof. First suppose that pa | k and consider k−1 k−2 k − pa + 1 (k − 1)(k − 2) · · · (k − pa + 1) = · · · · . (pa − 1)! pa − 1 pa − 2 1 In every fraction on the right-hand side, p has the same maximal exponent in the numerator as in the denominator. Therefore, the product (which is an integer) is not divisible by p. Now suppose that pa  k. We have pa k(k − 1) · · · (k − pa + 1) (k − 1)(k − 2) · · · (k − pa + 1) = · . (pa − 1)! k (pa )! The last fraction is an integer. In fraction

pa , denominator k is not divisible by pa . k

L EMMA 2. If g(x) is a polynomial with degree less than n then n  =0

  n g(x + n − ) = 0. (−1)  

Proof. Apply induction on n. For n = 1 then g(x) is a constant and     1 1 g(x) = g(x + 1) − g(x) = 0. g(x + 1) − 1 0 Now assume that n > 1 and the L EMMA holds for n − 1. Let h(x) = g(x + 1) − g(x); the degree of h is less than the degree of g, so the induction hypothesis applies

2011 ”M ASTER ” M ATHEMATICAL C OMPETITION

83

for g and n − 1: n−1  =0



 n−1 h(x + n − 1 − ) = 0 

  n−1  (−1) g(x + n − ) − g(x + n − 1 − ) = 0  =0     n−1  n−1 n−1 + (−1) g(x + n) + −1 0 =1     n−1 n−1 g(x) = 0 g(x + n − ) − (−1)n−1 n−1    n   n g(x + n − ) = 0. (−1) 

n−1 





(−1)



=0

L EMMA 3. If n has at least two distinct prime divisors then the greatest common      n  divisor of n1 , n2 , . . . , n−1 is 1.

   n  Proof. Suppose to the contrary that p is a common prime divisor of n1 , . . . , n−1 . n In particular, p | 1 = n. Let a be the exponent of p in the prime factorization of n.     Since n has at least two prime divisors, we have 1 < pa < n. Hence, pan−1 and pna  n  n  n   n are listed among 1 , . . . , n−1 and thus p | pa and p | pa −1 . But then p divides  n   n   n−1  pa − pa −1 = pa −1 , which contradicts L EMMA 1. Next we construct the polynomial f (x) when n = 1 or n is a power of a prime. For n = 1, f (x) = 12 is such a polynomial. If n = pa where p is a prime and a is a positive integer then let   1 x−1 1 (x − 1)(x − 2) · · · (x − pa + 1) = · . f (x) = p pa − 1 p (pa − 1)!

The degree of this polynomial is pa − 1 = n − 1. a +1) The number (k−1)(k−2)···(k−p is an integer for any integer k, and, by L EMMA (pa −1)! 1, it is divisible by p if and only if k is not divisible by pa = n. Finally we prove that if n has at least two prime divisors then no polynomial f (x) satisfies (1, 2). Suppose that some polynomial f (x) satisfies (1,2), and apply L EMMA

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2 for g = f and x = −k where 1 ≤ k ≤ n − 1. We get that      n k− n f (0) = f (−k + ). (−1) k  0≤≤n, =k

Since f (−k), . . . , f (−1) and f (1), . . . , f (n − k) are all integers, we conclude that  n k f (0) is an integer for every 1 ≤ k ≤ n − 1.  n      is 1. By dint of L EMMA 3, the greatest common divisor of n1 , n2 , . . . , n−1  n Hence, there will exist some integers u1 , u2 , . . . , un−1 for which u1 1 + · · · +  n  = 1. Then un−1 n−1 n−1   n−1   n n f (0) = f (0) f (0) = uk uk k k k=1

k=1

is a sum of integers. This contradicts the fact that f (0) is not an integer. So such polynomial f (x) does not exist.

Alternative Solution. (I. Bogdanov) We claim the answer is n = pα for some prime p and nonnegative α. L EMMA . For every integers a1 , . . . , an there exists an integer-valued polynomial P (x) of degree < n such that P (k) = ak for all 1 ≤ k ≤ n. Proof. Induction on n. For the base case n = 1 one may set P (x) = a1 . For the induction step, suppose that the polynomial P1 (x) satisfies the desired property for all k−1 x−1 ; since n−1 = 0 for 1 ≤ k ≤ n − 1. Then set P (x) = P1 (x) + (an − P1 (n)) n−1 n−1 1 ≤ k ≤ n − 1 and n−1 = 1, the polynomial P (x) is a sought one.

Now, if for some n there exists some polynomial f (x) satisfying the problem conditions, one may choose some integer-valued polynomial P (x) (of degree < n − 1) coinciding with f (x) at points 1, . . . , n − 1. The difference f1 (x) = f (x) − P (x) also satisfies the problem conditions, therefore we may restrict ourselves to the polynomials vanishing at points 1, . . . , n − 1 — that are, the polynomials of the form n−1 f (x) = c i=1 (x − i) for some (surely rational) constant c. Let c = p/q be its d α irreducible form, and q = j=1 pj j be the prime decomposition of the denominator. 1. Assume that a desired polynomial f (x) exists. Since f (0) is not an integer, we α have q  (−1)n−1 (n − 1)! and hence pj j  (−1)n−1 (n − 1)! for some j. Hence n−1  i=1

α

(pj j − i) ≡ (−1)n−1 (n − 1)! ≡ 0

α

(mod pj j ),

2011 ”M ASTER ” M ATHEMATICAL C OMPETITION

85

αi i therefore f (pα i ) is not integer, too. By the condition (i), this means that n | pi , and hence n should be a power of a prime.

2. Now let us construct a desired polynomial f (x) for any power of a prime n = pα . We claim that the polynomial     n x 1 x−1 = f (x) = p n−1 px n fits. Actually, consider some integer x. From the first representation, the denominator of the irreducible form of f (x) may be 1 or p only. If pα  x, then the prime decomposition of the fraction n/(px) contains p with a nonnegative exponent; hence f (x) is integer. On the other hand, if n = pα | x, then the numbers x−1, x−2, . . . , x−(n−1) contain the same exponents of primes as the numbers n − 1, n − 2, . . . , 1 respectively; hence the number   n−1 x−1 i=1 (x − i) = n−1 n−1 i=1 (n − i)

is not divisible by p. Thus f (x) is not an integer.

Problem 3. A triangle ABC is inscribed in a circle ω. A variable line  chosen parallel to BC meets segments AB, AC at points D, E respectively, and meets ω at points K, L (where D lies between K and E). Circle γ1 is tangent to the segments KD and BD and also tangent to ω, while circle γ2 is tangent to the segments LE and CE and also tangent to ω. Determine the locus, as  varies, of the meeting point of the common inner tangents to γ1 and γ2 . (Russia) Vasily Mokin & Fedor Ivlev

Solution. Let P be the meeting point of the common inner tangents to γ1 and γ2 . Also, let b be the angle bisector of ∠BAC. Since KL  BC, b is also the angle bisector of ∠KAL. Let H be the composition of the symmetry S with respect to b and the inversion √ I of centre A and ratio AK · AL (it is readily seen that S and I commute, so since S2 = I2 = id, then also H2 = id, the identical transformation). The elements of the configuration interchanged by H are summarized in Table I. Let O1 and O2 be the centres of circles γ1 and γ2 . Since the circles γ1 and γ2 are determined by their construction (in a unique way), they are interchanged by H, therefore the rays AO1 and AO2 are symmetrical with respect to b. Denote by 1 and 2 the radii of γ1 and γ2 . Since ∠O1 AB = ∠O2 AC, we have 1 / 2 = AO1 /AO2 . On

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the other hand, from the definition of P we have O1 P/O2 P = 1 /2 = AO1 /AO2 ; this means that AP is the angle bisector of ∠O1 AO2 and therefore of ∠BAC. The limiting, degenerated, cases are when the parallel line passes through A – when P coincides with A; respectively when the parallel line is BC – when P coincides with the foot A ∈ BC of the angle bisector of ∠BAC (or any other point on BC). By continuity, any point P on the open segment AA is obtained for some position of the parallel, therefore the locus is the open segment AA of the angle bisector b of ∠BAC. point K line KL ray AB point B point C segment BD arc BK arc CL

←→ ←→ ←→ ←→ ←→ ←→ ←→ ←→

point L circle ω ray AC point E point D segment EC segment EL segment DK Table 1: Elements interchanged by H.

SECOND DAY

Problem 4. Given a positive integer n = number

s 

s 

i pα i , we write Ω(n) for the total

i=1

αi of prime factors of n, counted with multiplicity. Let λ(n) = (−1)Ω(n)

i=1

(so, for example, λ(12) = λ(22 · 31 ) = (−1)2+1 = −1). Prove the following two claims: i) There are infinitely many positive integers n such that λ(n) = λ(n + 1) = +1; ii) There are infinitely many positive integers n such that λ(n) = λ(n + 1) = −1. (Romania) Dan Schwarz

Solution. Notice that we have Ω(mn) = Ω(m) + Ω(n) for all positive integers m, n (Ω is a completely additive arithmetic function), translating into λ(mn) = λ(m)·

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2011 ”M ASTER ” M ATHEMATICAL C OMPETITION

λ(n) (so λ is a completely multiplicative arithmetic function), hence λ(p) = −1 for any prime p, and λ(k 2 ) = λ(k)2 = +1 for all positive integers k.7 The start (first 100 terms) of the sequence S = (λ(n))n≥1 is +1, −1, −1, +1, −1, +1, −1, −1, +1, +1, −1, −1, −1, +1, +1, +1, −1, −1, −1, −1,

+1, +1, −1, +1, +1, +1, −1, −1, −1, −1, −1, −1, +1, +1, +1, +1, −1, +1, +1, +1,

−1, −1, −1, −1, −1, +1, −1, −1, +1, −1, +1, −1, −1, +1, +1, +1, +1, +1, −1, +1,

−1, +1, −1, +1, +1, −1, −1, −1, +1, −1, −1, −1, −1, +1, −1, −1, +1, −1, −1, −1,

+1, +1, −1, +1, +1, +1, +1, +1, −1, +1, +1, −1, +1, +1, +1, +1, −1, −1, −1, +1.

i) The Pell equation x2 −6y 2 = 1 has infinitely many solutions in positive integers; √ √ all solutions are given by (xn , yn ), where xn +yn 6 = (5+2 6)n . Since λ(6y 2 ) = 1 and also λ(6y 2 + 1) = λ(x2 ) = 1, the thesis is proven. Alternative Solution. Take any existing pair with λ(n) = λ(n + 1) = 1. Then λ((2n+1)2 −1) = λ(4n2 +4n) = λ(4)·λ(n)·λ(n+1) = 1, and also λ((2n+1)2 ) = λ(2n + 1)2 = 1, so we have built a larger (1, 1) pair. ii) The equation 3x2 − 2y 2 = 1 (again Pell theory) has also infinitely many solu√ √ √ √ tions in positive integers, given by (xn , yn ), where xn 3 + yn 2 = ( 3 + 2)2n+1 . Since λ(2y 2 ) = −1 and λ(2y 2 + 1) = λ(3x2 ) = −1, the thesis is proven. Alternative Solution. Assume (λ(n − 1), λ(n)) is the largest (−1, −1) pair, therefore λ(n + 1) = 1 and λ(n2 + n) = λ(n) · λ(n + 1) = −1, therefore again λ(n2 + n + 1) = 1. But then λ(n3 − 1) = λ(n − 1) · λ(n2 + n + 1) = −1, and also λ(n3 ) = λ(n)3 = −1, so we found yet a larger such pair than the one we started with, contradiction. Alternative Solution. Assume the pairs of consecutive terms (−1, −1) in S are finitely many. 7 Also see Sloane’s Online Encyclopædia of Integer Sequences (OEIS), sequence A001222 for Ω and  sequence A008836 for λ, which is called Liouville’s function. Its summatory function λ(d) is equal to d|n

1 for a perfect square n, and 0 otherwise. P´olya conjectured that L(n) :=

n 

k=1

λ(k) ≤ 0 for all n, but this

has been proven false by Minoru Tanaka, who in 1980 computed that for n = 906, 151, 257 its value was n  λ(k) positive. Tur´an showed that if T (n) := ≥ 0 for all large enough n, that will imply Riemann’s k k=1 Hypothesis; however, Haselgrove proved it is negative infinitely often.

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Then from some rank on we only have subsequences (1, −1, 1, 1, . . . , 1, −1, 1). By ”doubling” such a subsequence (like at point ii)), we will produce (−1, ?, 1, ?, −1, ?, −1, ?, . . . , ?, −1, ?, 1, ?, −1). According with our assumption, all ?-terms ought to be 1, hence the produced subsequence is (−1, 1, 1, 1, −1, 1, −1, 1, . . . , 1, −1, 1, 1, 1, −1), and so the ”separating packets” of 1’s contain either one or three terms. Now assume some far enough (1, 1, 1, 1) or (−1, 1, 1, −1) subsequence of S were to exist. Since it lies within some ”doubled” subsequence, it contradicts the structure described above, which thus is the only prevalent from some rank on. But then all the positions of the (−1)-terms will have the same parity. However though, we have λ(p) = λ(2p2 ) = −1 for all odd primes p, and these terms have different parity of their positions. A contradiction has been reached.8 Alternative Solution. (I. Bogdanov) Take ε ∈ {−1, 1}. There obviously exist infinitely many n such that λ(2n + 1) = ε (just take 2n + 1 to be the product of an appropriate number of odd primes). Now, if either λ(2n) = ε or λ(2n + 2) = ε, we are done; otherwise λ(n) = −λ(2n) = −λ(2n + 2) = λ(n + 1) = ε. Therefore, for such an n, one of the three pairs (n, n + 1), (2n, 2n + 1) or (2n + 1, 2n + 2) fits the bill. We have thus proved the existence in S of infinitely many occurrences of all possible subsequences of length 1, viz. (+1) and (−1), and of length 2, viz. (+1, −1), (−1, +1), (+1, +1) and (−1, −1).9 Problem 5. For every n ≥ 3, determine all the configurations of n distinct points X1 , X2 , . . . , Xn in the plane, with the property that for any pair of distinct points Xi , Xj there exists a permutation σ of the integers {1, . . . , n}, such that d(Xi , Xk ) = d(Xj , Xσ(k) ) for all 1 ≤ k ≤ n. (We write d(X, Y ) to denote the distance between points X and Y .) (United Kingdom) Luke Betts 8 Using

the same procedure for point i), we only need notice that λ((2k + 1)2 ) = λ((2k)2 ) = 1, and these terms again are of different parity of their position. 9 Is this true for subsequences of all lengths  = 3, 4, etc.? If no, up to which length  ≥ 2?

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2011 ”M ASTER ” M ATHEMATICAL C OMPETITION

Solution. Let us first prove that the points must be concyclic. Assign to each point Xk the vector xk in a system of orthogonal coordinates whose origin is the point of n 1 xk = 0. mass of the configuration, thus n k=1

2 Then d2 (Xi , Xk ) = ||xi − xk ||2 = xi − k  = ||xi || − 2 xi , xk  +  xk , xni − x n n    ||xk ||2 , hence d2 (Xi , Xk ) = n||xi ||2 − 2 xi , xk + ||xk ||2 = n||xi ||2 + n 

k=1

k=1

||xk ||2 = n||xj ||2 +

n 

k=1

||xσ(k) ||2 =

n 

k=1

k=1

k=1

d2 (Xj , Xσ(k) ), therefore ||xi || = ||xj ||

for all pairs (i, j). The points are thus concyclic (lying on a circle centred at O(0, 0)). Let now m be the least angular distance between any two points. Two points situated at angular distance m must be adjacent on the circle. Let us connect each pair of such two points with an edge. The graph G obtained must be regular, of degree n  deg(Xk ) = n deg(G) = 2|E|, we must have deg(G) = 1 or 2. If n is odd, since k=1

deg(G) = 2, hence the configuration is a regular n-gon. If n is even, we may have the configuration of a regular n-gon, but we also may have deg(G) = 1. In that case, let M be the next least angular distance between any two points; such points must also be adjacent on the circle. Let us connect each pair of such two points with an edge, in order to get a graph G . A similar reasoning yields deg(G ) = 1, thus the configuration is that of an equiangular n-gon (with alternating equal side-lengths). Problem 6. The cells of a square 2011 × 2011 array are labelled with the integers 1, 2, . . . , 20112 , in such a way that every label is used exactly once. We then identify the left-hand and right-hand edges, and then the top and bottom, in the normal way to form a torus (the surface of a doughnut). Determine the largest positive integer M such that, no matter which labelling we choose, there exist two neighbouring cells with the difference of their labels at least M .10 (Romania) Dan Schwarz

Preamble. For a planar N × N array, it is folklore that this value is M = N , with some easy models shown below. As such, the problem is mentioned in [B E´ LA ´ - The Art of Mathematics], 21. Neighbours in a Matrix. B OLLOB AS 10 Cells

with coordinates (x, y) and (x , y  ) are considered to be neighbours if x = x and y − y  ≡ ±1 (mod 2011), or if y = y  and x − x ≡ ±1 (mod 2011).

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2011 ”M ASTER ” M ATHEMATICAL C OMPETITION

1 N+1 .. . (N-1)N+1

2 N+2 .. . (N-1)N+2

... ... .. .

N 2N .. . N2

...

A planar parallel N × N model array. 1 3 6 .. .

2 5

4

.. . N(N+1)/2 - 1

.. .

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

N(N+1)/2

N(N-1)/2 + 1 N(N-1)/2 + 2 .. .

.. .

N2 - 1

N2 - 2 N2

A planar diagonal N × N model array. Solution. For the toroidal case, it is clear the statement of the problem is referring to the cells of a ZN × ZN lattice on the surface of the torus, labeled with the numbers 1, 2, . . . , N 2 , where one has to determine the least possible maximal absolute value M of the difference of labels assigned to orthogonally adjacent cells. The toroidal N = 2 case is trivially seen to be M = 2 (thus coinciding with the planar case). 1 3

2 4

The unique 2 × 2 toroidal array. For N ≥ 3 we will prove that value to be at least M ≥ 2N − 1. Consider such a configuration, and color all cells of the square in white. Go along the cells labeled 1, 2, etc. coloring them in black, stopping just on the cell bearing the least label k which, after assigned and colored in black, makes that all lines of a same orientation (rows, or columns, or both) contain at least two black cells (that is, before coloring in black the cell labeled k, at least one row and at least one column contained at most one black cell). Wlog assume this happens for rows. Then at most one row is all black, since

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2011 ”M ASTER ” M ATHEMATICAL C OMPETITION

if two were then the stopping condition would have been fulfilled before cell labeled k (if the cell labeled k were to be on one of these rows, then all rows would have contained at least two black cells before, while if not, then all columns would have contained at least two black cells before). Now color in red all those black cells adjacent to a white cell. Since each row, except the potential all black one, contained at least two black and one white cell, it will now contain at least two red cells. For the potential all black row, any of the neighbouring rows contains at least one white cell, and so the cell adjacent to it has been colored red. In total we have therefore colored red at least 2(N −1)+1 = 2N −1 cells. The least label of the red cells has therefore at most the value k + 1 − (2N − 1). When the white cell adjacent to it will eventually be labeled, its label will be at least k + 1, therefore their difference is at least (k + 1) − (k + 1 − (2N − 1)) = 2N − 1. The models are kind of hard to find, due to the fact that the direct proof offers little as to their structure (it is difficult to determine the equality case during the argument involving the inequality with the bound, and then, even this is not sure to be prone to being prolonged to a full labeling of the array). The weaker fact the value M is not larger than 2N is proved by the general model exhibited below (presented so that partial credits may be awarded). N+1 3N+1 .. . (2-1)N+1 .. . 2kN+1 .. . 2N+1 1

N+2 3N+2 .. . (2-1)N+2 .. . 2kN+2 .. . 2N+2 2

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

2N 4N .. . 2N .. . (2k+1)N .. . 3N N

A general model for M = 2N in a N × N array. By examining some small N > 2 cases, one comes up with the idea of spiral models for the true value M = 2N − 1. The models presented are for odd N (since 2011 is odd); similar models exist for even N (but are less symmetric).

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2011 ”M ASTER ” M ATHEMATICAL C OMPETITION

7 3 8

2 1 4

6 5 9

The spiral 3 × 3 array. 23 17 9 18 24

16 8 3 10 19

7 2 1 4 11

15 6 5 12 20

22 14 13 21 25

The spiral 5 × 5 array. 47 41 31 19 32 42 48

40 30 18 9 20 33 43

29 17 8 3 10 21 34

16 7 2 1 4 11 22

28 15 6 5 12 23 35

39 27 14 13 24 36 44

46 38 26 25 37 45 49

The spiral 7 × 7 array. (2n+1)2 -2 (2n+1)2 -8 . . .

(2n+1)2 -9 . . . 2n2

2n2 + 1 . . . (2n+1)2 -7 (2n+1)2 -1

2n2 +2 . . . (2n+1)2 -6

... ... ..

.

... ... ... ..

.

... ...

n(2n-1)+1 n(2n-1)+2 . . . 8 3 10 . . . n(2n+1)

. . . 2 1 4 . . . n(2n+1)+1

n(2n-1) . . . 6 5 12 . . . n(2+1)+2

... ... ..

.

... ... ... ..

.

... ...

(2n+1)2 -10

2n(n+1)+3 2n(n-1)+2 2n(n+1) . . . (2n+1)2 -5

The general spiral N × N array for N = 2n + 1 ≥ 5.

(2n+1)2 -3 (2n+1)2 -11 . . . 2n(n+1)+2 2n(n+1)+1 . . . (2n+1)2 -4 (2n+1)2

THE CLOCK-TOWER SCHOOL SENIORS COMPETITION Rm. Vˆalcea, March 2011

Problem 1. Let n be a positive integer and a1 ≤ a2 ≤ · · · ≤ an be positive real numbers. Show that  n   n  n   n     2 2 ak kak ≤ ak kak . k=1

k=1

k=1

k=1

Solution. The hypothesis leads to (i−j)(ai −aj )ai aj ≥ 0 for every i and j, hence

0≤ =

n 

i,j=1 n  i,j=1

=



=2

(i − j)(ai − aj )ai aj 

ia2i aj − iai a2j − ja2i aj + jai a2j

n 

ia2i

i=1



n 

k=1



ka2k

n  j=1



aj −

n 

k=1



n  i=1

ak − 2



iai



n 

n 

which leads to the required relation.

a2j

j=1

kak

k=1





−

n 



n  i=1

a2i



n 

jaj +

j=1



n 

ai

i=1



n 

ja2j

j=1

a2k ,

k=1

Problem 2. Let f : R → R be a function such that: f (xy + x + y) + f (xy − x − y) = 2(f (x) + f (y)), ∀ x, y ∈ R.

(1)

Prove that f fulfills the relation: f (x + y) = f (x) + f (y), ∀ x, y ∈ R. 93

(2)

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2011 ”C LOCK -T OWER ” S ENIOR M ATHEMATICAL C OMPETITION

Solution. x = y = 0 yields f (0) = 0 and y = 0 yields f (−x) = −f (x), x ∈ R. Then y = 1 ⇒ f (2x + 1) = 2f (x) + f (1) y = −1 ⇒ f (2x − 1) = 2f (x) − f (1)

(3) (4)

Denote x ∗ y = xy + x + y and notice that (ASOC) : x ∗ (y ∗ z) = (x ∗ y) ∗ z. For x = 1, y ∗ 1 = 2y + 1, hence f (y ∗ 1) = 2f (y) + f (1), ∀ y ∈ R

(5)

therefore f (x ∗ (y ∗ 1)) = f (x ∗ (2y + 1)) (1)

= f (x(2y + 1) − x − 2y − 1) + 2f (x) + 2f (2y + 1) = f (2(xy − y) − 1) + 2f (2y + 1) + 2f (x)

(3,4)

= 2f (xy − y) + f (x) + 2f (y) + f (1)

(6)

and (5)

(1)

f ((x ∗ y) ∗ 1) = 2f (x ∗ y) + f (1) = 2(2f (x) + 2f (y)) + 2f (xy − x − y) + f (1) = 2f (xy − x − y) + 2f (x) + 2f (y) + f (1).

(7)

Relations (6), (7) and (ASOC) yield: f (xy − x − y) + f (x) = f (xy − y), ∀ x, y ∈ R. The last relation means that f (u + v) = f (u) + f (v) for all u, v ∈ R which can be written in the form u = x, v = xy − x − y, that is there exists x, y ∈ R so that x = u and y = v+u u−1 ; this happens if u = 1 or u = 1 = −v. It remains to check that f (1 + v) = f (1) + f (v), ∀ v ∈ R. Indeed, (4) and x → x + 1 imply (3)

f (2x + 1) = 2f (x + 1) − f (1) = 2f (x) + f (1), whence f (x + 1) = f (x) + f (1), ∀ x ∈ R. Problem 3. A certain language uses an alphabet containing three letters. Some sequences of two ore more letters are forbidden, and every two forbidden sequences have different lengths. Prove that there exists admissible words of every length.

2011 ”C LOCK -T OWER ” S ENIOR M ATHEMATICAL C OMPETITION

95

Solution. Let an be the number of admissible words with n letters; then a0 = 1 (the empty word), a1 = 3 and a2 = 8. Then, if we add a letter at the end of a correct word with n letters, we obtain either a correct word with n + 1 letters, or a forbidden word of the form XY , with Y a forbidden sequence with k letters, 2 ≤ k ≤ n + 1 and X a correct word with n − k + 1 letters. So, the forbidden words with n+1 letters are at most a0 +a1 +a2 +. . .+an−1 , hence an+1 ≥ 3an − (a0 + a1 + a2 + . . . + an−1 ). The above relation allows to prove inductively that an+1 > 2an (∗ ), for every n ≥ 1. Indeed, the base case is obvious, and if (∗ ) is true for all the numbers from 0 to n − 1, n ≥ 2, then an ≥ 2k an−k , 0 ≤ k ≤ n − 1, whence    1 1 1 > 2an . + + . . . + an+1 ≥ an 3 − 2n 2n−1 2 This shows that there are at least 2n words of length n. Problem 4. A point P is on the side AB of a convex quadrilateral ABCD. Let ω be the incircle of the triangle CPD and I be its center. It is known that ω touches the incircles of the triangles APD and BPC at K, respectively L. Let E, respectively F be the meetincg points of the lines AC and BD, respectively AK and BL. Prove that the points E, I and F are collinear.

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2011 ”C LOCK -T OWER ” S ENIOR M ATHEMATICAL C OMPETITION

Solution. Denote J the center of the circle k placed in the half-plane (AB, C and tangent to AB, DA and BC. Denote a, respectively b the incircles of the triangles ADP and BCP. We will firstly prove that F ∈ IJ. The point A is the center of the homothety which transforms a into k; K is the center of the homothety with negative ratio which transforms a into ω. Denote F¯ the center of the homothety which transforms ω into k. It is known that A, K and F¯ are collinear. In the same way, F¯ ∈ BL, hence F = F¯ , so F ∈ IJ. We prove now that E ∈ IJ. Comparing the lengths of the tangents from A, P , C, D to the circles k and a gives AP + DC = AD + P C. This shows that the quadrilateral APCD has an incircle d. Let X be the center of the homothety which transforms a into ω. The above reasoning, applied to the circles a, d and ω shows that A, C and X are collinear. Consider now the circles a, ω and k. The point A is the center of the homothety which transforms a into k and X is the center of the homothety ¯ of the homothety which which transforms a into ω, hence XA contains the center E ¯ ∈ AC. In the same way E ¯ ∈ BD, therefore E ¯ = E, transforms ω into k, whence E so E ∈ IJ.

THE CLOCK-TOWER SCHOOL JUNIORS COMPETITION Rm. Vˆalcea, March 2011

Problem 1. Show that the number aaa . . . a 0a  

2011 times

is not a perfect square for any non-nil digit a.

Solution. The last digit of a perfect square cannot be 2, 3, 7 or 8. If a = 5, the next to the last digit must be 2 – false. If a = 6, the number is of the form 4k + 2, k ∈ N, which cannot be a perfect square. The cases a = 4 and a = 9 reduce to a number of the form 111 . . . 01 beeing a pefect square, which is impossible, because such a number is of the form 8k + 5. Problem 2. Find all positive integers a, b for which there exists sets A, B of positive integers so that A ∩ B = ∅, A ∪ B = N∗ and aA = bB (if x is a number and M is a set of numbers, xM = {xm | m ∈ M }).

Solution. We can assume that 1 ∈ A. Then a ∈ bB, hence there exists p ∈ B such that a = pb. Moreover, p ≥ 2, because 1 ∈ A. Every pair (pb, b) , with b ∈ N∗ , is a solution: we use the partition  2n  A = p q | n ∈ N, q ∈ N∗ , p  q  2n+1  B = p q | n ∈ N, q ∈ N∗ , p  q

Problem 3. A set D of n straight lines in a plane has the property that each line of the set intersects exactly 2011 straight lines of D. 97

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2011 ”C LOCK -T OWER ” J UNIOR M ATHEMATICAL C OMPETITION

Find n. Solution. Let d be a line from D and k be the number of the lines from D parallel to d. Then n = 2012 + k. Since every line a from D different from d meets 2011 lines from D, a is parallel to n − 2012 = k lines from D. This shows that each line from D is parallel to exactly k other lines, whence the n lines can be divided into groups of k + 1 mutually parallel lines. So k + 1 | n, hence k + 1 | k + 2012, therefore k ∈ {0, 2010} , that is n can be 2012 or 4022. A possible configuration for n = 2012 is made up by the supports of the sides of a regular 2012-gon. A possible configuration for n = 4022 is made up by 2011 parallel lines, meeting other 2011 parallel lines. Problem 4. The triangle ABC and the points M ∈ (BC), N ∈ (AC), P ∈ (AB) fulfill the conditions ∠BM P ≡ ∠CN M ≡ ∠AP N and BM = CN = AP . Prove that the triangle ABC is equilateral.   Solution. We start noticing that m(P M N ) = 180◦ − BM P −N M C = 180◦ −      CN M −N MC = N CM and, in the same way, M NP = N AP , so ∆ABC ∼ ∆N P M.

(1)

≥A  ≥ B.  (2) Suppose now, without loss of generality, that C Then AB ≥ BC ≥ AC, whence, from (1), N P ≥ P M ≥ M N. These, toghether ≥ B  ≥ with ∠BM P ≡ ∠CN M ≡ ∠AP N and BM = CN = AP , lead to A  (3) C. ≡B  ≡ C,  whence the conclusion. Relations (2) and (3) show that A

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Since 2009, the Dinu Patriciu Foundation has been one of the main sponsors of the Gazeta Matematică journal and of the Romanian National Mathematical Olympiad.

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ISSN 2066-6446