RMC_2014

ROMANIAN MATHEMATICAL COMPETITIONS 2014

˘ ˘ With the cooperation of MARIEAN ANDRONACHE, MIHAIL BALUN A, ˘ ˘ CATALIN GHERGHE, RADU GOLOGAN, ANDREI ECKSTEIN, ˘ ˘ MARIUS PERIANU, CALIN POPESCU, DINU S ¸ ERBANESCU

Technical Editor Alexandru Negrescu

TABLE OF CONTENTS

Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v 1.1.

2014 Romanian Mathematical Olympiad – District Round . . . . . . . . . . . . . 1

1.2.

2014 Romanian Mathematical Olympiad – Final Round . . . . . . . . . . . . . . 19

1.3.

Shortlisted problems for the 2014 Romanian NMO . . . . . . . . . . . . . . . . . . . 41

1.4.

Selection tests for the 2014 BMO and IMO . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1.5.

Selection tests for the 2014 JBMO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

1.6.

2013 – 2014 Local Mathematical Competitions 1.6.1.

The 2013 Danube Mathematical Competition . . . . . . . . . . . . . . . . . 85

1.6.2.

The 2013 Tenth IMAR1 Mathematical Competition . . . . . . . . . . . 91

1 Institute

of Mathematics of the Romanian Academy. One-day IMO-type contest.

iii

FOREWORD

The 21st 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 the competition. Most 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. Special thanks are due to C˘ alin Popescu, one of the editors, who produced the final form of the majority of the qualification tests for the IMO team. Alexandru Negrescu did a great job in putting all the materials together. A remarkable effort came from former multiple IMO-participants Tudor ¨ P˘adurariu and Omer Cerrahoˇ glu, who contributed to the training of this year Romanian IMO Team. We thank the Ministry of National Education for constant involvement in supporting the Olympiads and the participation of our teams in international events. Special thanks are due to companies who were involved in sponsoring the Romanian Olympiad and the participation of the Romanian teams to international competitions: BCR Bank, eMAG Foundation and T ¸ uca Zbˆarcea & Associates. The printing of this volume has been supported by The Group company. The Editors

THE 65th ROMANIAN MATHEMATICAL OLYMPIAD DISTRICT ROUND

5th GRADE

Problem 1. Find, with proof, all positive integers abc satisfying b · ac = c · ab + 10. ***

Solution. The given condition rewrites as b(10a + c) = c(10a + b) + 10, from which a(b − c) = 1, and hence a = b − c = 1. The numbers are 110, 121, 132, 143, 154, 165, 176, 187, 198. Problem 2. Let M be the set of palindromic numbers of the form 5n + 4, where n ≥ 0 is an integer. (A positive integer is a palindromic number if it remains the same when its digits are reversed. For instance, the numbers 7, 191, 23532, 3770773 are palindromic numbers.) a) If we write the elements of M in increasing order, which is the 50th number? b) Among all numbers in M, written with nonzero digits who sum up to 2014, which is the greatest one and which the smallest one? Mircea Fianu

Solution. a) The last (and hence, the first) digit of a number from M equals 4 or 9. A direct count shows that M contains 2 one digit numbers, 2 two digit numbers, 20 three digit and 20 four digit numbers, hence the 50th number is the 6th five digit number, that is, 40504. b) The greatest number in M has the maximum number of digits. Therefore, we put 4 as the first and last digit and complete the decimal representation 1

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2014 Romanian Mathematical Olympiad – District Round

with 2006 digits 1, obtaining thus 411 . . . 1 4. Similarly, the smallest number in   2006 digits 1

M has the least number of digits. The answer is 9899 . . . 989.   220 digits 9

Problem 3. Let A = {1, 3, 32 , 33 , . . . , 32014 }. We obtain a partition of A if A is written as a disjoint union of nonempty subsets. a) Prove that there is no partition of A such that the product of elements in each subset is a square. b) Prove that there exists a partition of A such that the sum of elements in each subset is a square. Traian Preda

Solution. a) Assume that such a partition exists. Then the product of all elements of A must be a square as well. But this equals 31+2+3+···+2014 = 32015·1007 , obviously not a square. b) Observe that 32n + 32n+1 = (3n · 2)2 , hence a possible partition is A = {1, 3} ∪ {32 , 33 } ∪ · · · ∪ {32012 , 32013 } ∪ {32014 }. Problem 4. A 10 digit positive integer is called a cute number if its digits are from the set {1, 2, 3} and every two consecutive digits differ by 1. a) Prove that exactly 5 digits of a cute number are equal to 2. b) Find the total number of cute numbers. c) Prove that the sum of all cute numbers is divisible by 1408. Csap¨ o Hajnalka

Solution. a) In the decimal representation of a cute number, the parity of the digits clearly alternates, hence exactly 5 digits are even, that is, equal to 2. b) There are 25 = 32 cute numbers of the form 2a2b2c2d2e and another 32 of the form a2b2c2d2e2, therefore the requested number is 64. c) If a1 a2 . . . a10 is a cute number, then (4 − a1 )(4 − a2 ) . . . (4 − a10 ) is also cute and distinct from the previous one. They add up to 4444444444. Grouping the cute numbers in 32 such pairs, we obtain that their sum equals 32 · 4444444444 = 27 · 11 · 101010101 = 1408 · 101010101, hence the conclusion.

3

2014 Romanian Mathematical Olympiad – District Round 6th GRADE

Problem 1. Prove that:  3  3  3 2 5 1 + + = 1; a) 2 3 6 b) 333 + 433 + 533 < 633 . Damian Marinescu

Solution. a) A straightforward computation proves the claim. b) It suffices to show that 433 533 333 + 33 + 33 < 1, 33 6 6 6 that is,

But

 33  33  33 1 2 5 + + < 1. 2 3 6  33  33  33  3  3  3 1 2 5 1 2 5 + + < + + = 1. 2 3 6 2 3 6

Problem 2. We call a nonempty set M good if its elements are positive integers, each having exactly 4 divisors. If the good set M has n elements, we denote by SM the sum of all 4n divisors of its members (the sum may contain repeating terms). a) Prove that A = {2 · 37, 19 · 37, 29 · 37} is good and SA = 2014. b) Prove that if the set B is good and 8 ∈ B, then SB �= 2014. Relu Ciupea

Solution. a) Any number equal to the product of two distinct primes p, q, has exactly 4 divisors: 1, p, q and pq, hence A is good. A short computation shows that SA = 2014. b) A number n having exactly 8 divisors is either the product of two distinct primes, or the cube of a prime. It is not difficult to see that the divisor’s sum of such a number is even, except for n = 8. Therefore, if 8 ∈ B, then SB must be odd, hence SB �= 2014.

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2014 Romanian Mathematical Olympiad – District Round

Problem 3. The points M, N, and P are chosen on the sides BC, CA and AB of the triangle ABC such that BM = BP and CM =CN . The perpendicular dropped from B onto M P and the perpendicular dropped from C onto   M N intersect at I. Prove that the angles IP A and IN C are congruent. Gabriel Popa

A

P

B

N I

M

C

Solution. Since CM = CN and CI ⊥ M N , the line CI is the perpendicular bisector of the line segment M N , hence IM = IN . Similarly, we have IM =   IP . Triangles IM C and IN C are equal, so IM C ≡ IN C, and, in a similar     way, we deduce that IM B ≡ IP B. It follows that IP A = IM C, and, finally,   that IP A ≡ IN C. Problem 4. Determine all positive integers a for which there exist exactly a 2014 positive integers b such that 2 ≤ ≤ 5. b D. M. B˘ atinet¸u-Giurgiu, Neculai Stanciu

a ≤ 5 as 2a ≤ 10b ≤ 5a. It follows that the Solution. Rewrite 2 ≤ b sequence 2a, 2a + 1, . . . , 5a contains 2014 multiples of 10, hence it contains at least 2013 and at most 2015 groups of 10 consecutive numbers. We deduce that 2013 · 10 ≤ 5a − 2a < 2015 · 10, which leads to a ∈ {6710, 6711, . . . , 6716}. By inspection, we conclude that the required values of a are: 6710, 6712, and 6713. 7th GRADE

Problem 1. a) Prove that for any real numbers a and b the following inequality holds:    2 a + 1 b2 + 1 + 50 ≥ 2 (2a + 1) (3b + 1) .

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2014 Romanian Mathematical Olympiad – District Round

b) Find all positive integers n and p such that: 



n2 + 1

 p2 + 1 + 45 = 2 (2n + 1) (3p + 1) .

Nicolae Papacu

Solution. a) The given condition rewrites as (ab−6)2 +(a−2)2 +(b−3)2 ≥ 0, obviously true. b) Similarly, we obtain (np−6)2 +(n−2)2 +(p−3)2 = 5, hence the numbers (np − 6)2 , (n − 2)2 , and (p − 3)2 are equal to 0, 1, and 4, in some order. By inspection, we find (n, p) ∈ {(2, 4), (2, 2)}. Problem 2. Let a, b, c be real numbers such that: |a − b| ≥ |c|,

|b − c| ≥ |a|,

|c − a| ≥ |b|.

Prove that one of the numbers a, b, c equals the sum of the other two. George Stoica

Solution. Squaring the first inequality gives (a − b)2 ≥ c2 , hence (a − b + c)(b + c − a) ≥ 0. Multiplying the latter with the other two similar inequalities implies (a + b − c)2 (b + c − a)2 (c + a − b)2 ≤ 0, hence one of a, b, c is the sum of the other two. ˆ = 135◦ . The perpenProblem 3. Let ABC be a triangle in which m(A) dicular to the line AB erected at A intersects the side [BC] at D, and the angle  bisector of ∠B intersects the side [AC] at E. Find the measure of BED. Traian Preda

A E I B

D

C

 We deSolution. Let I ∈ (BE) such that IA bisects the angle DAB.  A short computation shows duce that ID is the bisector of the angle ADB. ◦  that m(DIB) = 135 , hence triangles ABE and IBD are similar. It follows BE AB BI that AB IB = BD , so that EB = BD . Thus, triangles ABI and DBE are similar  = m(BAI),  we infer that m(BED)  = 45◦ . as well and, since m(BED)

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2014 Romanian Mathematical Olympiad – District Round

Problem 4. Let ABCD be a square and consider the points K ∈ (AB), L ∈ (BC), and M ∈ (CD) such that KLM is a right isosceles triangle, with the right angle at L. Prove that the lines AL and DK are perpendicular to each other. Bogdan Enescu

B

L

C

M K

A

D

Solution. It is not difficult to observe that △KLB ≡ △LM C, hence KB = LC. Because AB = BC, it follows that AK = BL. But then, △AKD ≡ △BLA, and since AK ⊥ BL and AD ⊥ BA, we deduce that AL ⊥ KD, as well. 8th GRADE

√ Problem 1. In the right parallelepiped ABCDA′ B ′ C ′ D′ , with AB = 12 3 cm and AA′ = 18 cm, we consider the points P ∈ [AA′ ] and N ∈ [A′ B ′ ] such that A′ N = 3B ′ N. Determine the length of the line segment [AP ] such that for any position of the point M ∈ [BC] , the triangle M N P is right angled at N. Damian Marinescu

Solution. We have BC ⊥ (ABB ′ ) , therefore BC ⊥ P N . Since P N ⊥ N M, it follows that P N ⊥ (N BC) , hence P N ⊥ N B, that is, triangle N BP is right 2 angled at N. Let AP = x; we obtain BP 2 = x2 + 432, P N 2 = (18 − x) + 243 and BN 2 = 351. But BP 2 = P N 2 + BN 2 , and hence x = 13, 5 cm. Problem 2. For each positive integer n we denote by p (n) the greatest square less than or equal to n.

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2014 Romanian Mathematical Olympiad – District Round

a) Find all pairs of positive integers (m, n) , with m ≤ n, for which p (2m + 1) · p (2n + 1) = 400. b) Determine the set

  p (n + 1)  ∈ /N . n ∈ N n ≤ 100 and p (n)



∗

Lucian Dragomir

Solution. a) Since 400 = 1 · 400 = 4 · 100 = 16 · 25, we analyze three cases.

If p (2m − 1) = 1, p (2n − 1) = 400, we obtain 1 ≤ 2m − 1 < 4 and 400 ≤ 2n − 1 < 441, hence m ∈ {1, 2} and n ∈ {200, 201, . . . , 219} , giving 40 pairs (m, n) . Similarly, in the second case we obtain 20 pairs, and in the last case, 24 pairs, leading to a total of 84 pairs. 2

b) Let n ∈ N, and let p (n) = k 2 ; it follows k 2 ≤ n ≤ (k + 1) − 1, 2 hence k 2 < n + 1 ≤ (k + 1) . We deduce that p (n + 1) ∈ {k, k + 1} . Then p (n + 1) 2 ∈ / N if and only if p (n) = k 2 �= 1 and p (n + 1) = (k + 1) , that is, p (n) 2 n = (k + 1) − 1, k ∈ N, k ≥ 2. Since n ≤ 100, it results k ≤ 9. The requested set is {8, 15, 24, 35, 48, 63, 80, 99} . Problem 3. Let ABCDEF be a regular hexagon with side length a. At √ point A, the perpendicular AS, with length 2a 3, is erected on the hexagon’s plane. The points M, N, P, Q, and R are the projections of point A onto the lines SB, SC, SD, SE, and SF, respectively. a) Prove that the points M, N, P, Q, R lie in the same plane. b) Find the measure of the angle between the planes (M N P ) and (ABC). Petru Braica

Solution. a) Using the Three Perpendiculars Theorem, from SA ⊥ (ABC) and AB ⊥ BD, it results SB ⊥ BD. Since BD ⊥ AB and BD ⊥ SB, it follows that BD ⊥ (SAB) , hence BD ⊥ AM.

Since AM ⊥ SB, it results AM ⊥ (SBD) , hence AM ⊥ SD. We also have SD ⊥ AP, therefore we obtain that SD ⊥ (AM P ). In a similar way one can show that SD ⊥ (ARP ), SD ⊥ (AN P ) and SD ⊥ (AQP ) , therefore the points M, N, P, Q, R lie in the same plane.

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2014 Romanian Mathematical Olympiad – District Round

S P

Q

D

E

R

N

F

M

d A

C B

b) Because M R � BF, it follows that the intersection between the planes (M N P ) and (ABC) is the line d parallel to BF and passing through A. Since d ⊥ SA and d ⊥ AD, it results that d ⊥ (SAD) , hence d ⊥ AP. Therefore the  angle between the planes (M N P ) and (ABC) equals P AD.  Using the Pythagorean Theorem, one obtains SD = 4a, hence m(P DA) = ◦ ◦  60 and m(P AD) = 30 . Problem 4. Let n ≥ 2 be a positive integer. Determine all possible values of the sum S = ⌊x2 − x1 ⌋ + ⌊x3 − x2 ⌋ + . . . + ⌊xn − xn−1 ⌋ , where x1 , x2 , . . . , xn are real numbers whose integer parts are 1, 2, . . . , n, respectively. Ion Pˆır¸se

Solution. Let a and b be arbitrary reals. We have ⌊b⌋ − ⌊a⌋ − 1 < b − a < ⌊b⌋ − ⌊a⌋ + 1, hence ⌊b⌋ − ⌊a⌋ − 1 ≤ ⌊b − a⌋ ≤ ⌊b⌋ − ⌊a⌋. Applying this to some consecutive terms of the sequence, we obtain 0 ≤ ⌊xk − xk−1 ⌋ ≤ 1, for all k = 2, 3, . . . , n. Thus, 0 ≤ S ≤ n − 1. We claim that the set of all possible values of S is {0, 1, 2, . . . , n − 1} . The value S = n − 1 can be obtained, for instance, for xk = k, k = 1, 2, . . . , n . For the value S = p, with 0 ≤ p ≤ n − 2, one can take, for instance,   k+ 1 , k+1 xk =  k,

if 1 ≤ k ≤ n − 1 − p if n − p ≤ k ≤ n

.

2014 Romanian Mathematical Olympiad – District Round

9

9th GRADE

Problem 1. Find the irrational numbers x with the property that x2 + x and x3 + 2x2 are integer numbers. George Stoica

Solution. Denote x2 + x = a and x3 + 2x2 = b. Then b − ax = x2 = a − x, hence x (a − 1) = b − a. Since x is an irrational√number and a, b are integers, we deduce that a = b = 1, and, finally, x = −1±2 5 . Problem 2. Let ABC be a triangle and let the points D ∈ (BC), E ∈ (AC), F ∈ (AB) be such that EC FA DB = = . DC EA FB The halflines (AD, (BE, and (CF intersect the circumcircle of ABC at points M, N and P. Prove that the triangles ABC and M N P share the same centroid if and only if the areas of the triangles BM C, CN A and AP B are equal. Marin Ionescu

A

N E

P

F B

C

D

M −−→ −−→ −−→ Solution. First, observe that AD + BE + CF = 0 (∗) . Denoting the areas of the triangles ABC, BM C, CN A and AP B by s, sa , sb , and sc , respectively, −−→ −→ sa s+sa s+sa − AM · AD, and we have DM AD = s , hence AD = s , which implies AM = s the analogous equalities. Triangles ABC and M N P share the same centroid −−→ −−→ −−→ if and only if AM + BN + CP = 0, hence if and only if s + sa −−→ s + sc −−→ s + sc −−→ · AD + · BE + · CF = 0, s s s or, equivalently,

−−→ −−→ −−→ sa · AD + sb · BE + sc · CF = 0.

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2014 Romanian Mathematical Olympiad – District Round

Using (∗) , the latter rewrites as −−→ −−→ (sa − sc ) · AD + (sb − sc ) · BE = 0, −−→ −−→ and since AD and BE are non collinear vectors, this holds if and only if sa = sb = sc . Problem 3. The medians AD, BE and CF of triangle ABC intersect at G. Let P be a point lying in the interior of the triangle, not belonging to any of its medians. The line through P parallel to AD intersects the side BC at A1 . Similarly one defines the points B1 and C1 . Prove that −−→ −−→ −−→ 3 −−→ A1 D + B1 E + C1 F = P G. 2 Bogdan Enescu

A CA BA P

CB

B

AB D

BC

A1

AC C

Solution. Let us draw through point P parallels to the triangle’s sides. If the parallels to AB and AC intersect the side BC at points AB and AC , then the triangles ABC and P AB AC are similar, and since P A1 is parallel to the median AD, it follows that A1 is the midpoint of the line segment AB AC . With the notations in the figure, we have −−→ −−→ −−→ −−−→ −−−→ −−−→ −−−→ 2A1 D = A1 B + A1 C = AB B + AC C = P CB + P BC . −−→ −−→ −−→ and the similar equalities. We deduce that the sum 2(A1 D + B1 E + C1 F ) equals −−−→ −−−→ −−−→ −−−→ −−−→ −−−→ → − v = P C B + P B C + P AC + P C A + P B A + P AB . −−−→ −−−→ −→ But P CA + P BA = P A and adding up all similar equalities yields −→ −−→ −−→ −−→ → − v = P A + P B + P C = 3P G, hence the conclusion.

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2014 Romanian Mathematical Olympiad – District Round

Problem 4. Find all functions f : N∗ → N∗ with the properties: a) f (m + n) − 1 divides f (m) + f (n), for all m, n ∈ N∗ ; b) n2 − f (n) is a square, for all n ∈ N∗ .

Lucian Dragomir

Solution. It is not difficult to check that f (1) = 1 and f (2) = 3. We prove inductively that f (n) = 2n − 1, for all n. Indeed, assume that for some k > 1, f (k) = 2k − 1. Then f (k + 1) − 1 divides 2k, hence f (k + 1) ≤ 2k + 1. If the 2 2 inequality is strict, then (k + 1) − f (k + 1) > (k + 1) − (2k + 1) = k 2 , and 2 2 since it is a square, we have (k + 1) − f (k + 1) ≥ (k + 1) , a contradiction. We deduce that f (k + 1) = 2k + 1, as desired. 10th GRADE

Problem 1. Solve in complex numbers the equation |z − |z + 1|| = |z + |z − 1||. Dan Negulescu

Solution. Writing the equation as |z − |z + 1||2 = |z + |z − 1||2 , and using |w| = w · w, for any complex number w, yields the equivalent form 2

(z + z) (|z − 1| + |z + 1| − 2) = 0. We deduce that either z + z = 2 Re z = 0, hence z = ia, for some real a, or |z − 1| + |z + 1| = 2. In this second case, we deduce that, in the complex plane, the sum of distances from the point z to the points −1 and 1 equals 2, which is possible if and only if z is a real number, lying on the line segment [−1, 1] . Problem 2. Solve in real numbers the equation       5x 25x x + log2 1 + = 4 + log1/2 1 + . 3x + 4x 7x + 24x Cristinel Mortici

Solution. Rewrite the equation as       1 1 = 4 + log1/2 1 + , x + log2 1 + x x x x (3/5) + (4/5) (7/25) + (24/25)

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2014 Romanian Mathematical Olympiad – District Round

and observe that the left hand side is an increasing function, while the right hand side is a decreasing one. We conclude that the equation has at most a solution, and it is not difficult to guess it: x = 2. Problem 3. Let p and n be positive integers, with p ≥ 2, and let a be a real number such that 1 ≤ a < a + n ≤ p. Prove that the set     ⌊log2 x⌋ + ⌊log3 x⌋ + · · · + logp x  x ∈ R, a ≤ x ≤ a + n has exactly n + 1 elements.

Ioan B˘ aetu

p Solution. Let f (x) = k=2 ⌊logk x⌋ and let M = {f (x) | x ∈ [a, a+n]}. It is easy to show that if k ≥ 2 is a positive integer, then ⌊logk ⌊x⌋⌋ = ⌊logk x⌋ . This implies that f (x) = f (⌊x⌋), for all x ∈ [1, ∞), and hence M = {f (x) | x ∈ S}, where S = {⌊a⌋ , ⌊a⌋ + 1, . . . , ⌊a⌋ + n} has n + 1 elements. On the other hand, for s ∈ S, s < ⌊a⌋ + n ≤ p, we have s + 1 ∈ {2, 3, . . . , p}, and f (s+1)−f (s) =

p 

    (⌊logk (s + 1)⌋−⌊logk s⌋) ≥ logs+1 (s + 1) − logs+1 s = 1,

k=2

therefore f (s + 1) > f (s), and this proves that M has exactly n + 1 elements. Problem 4. Find all functions f : Q → Q such that f (x + 3f (y)) = f (x) + f (y) + 2y, for all x, y ∈ Q.

Nicu¸sor Berbecel

Solution. We claim that the solutions are f1 (x) = x and f2 (x) = −2x/3, for all real x. Set x = y − 3f (y) to obtain f (y − 3f (y)) = −2y, y ∈ Q. Replacing y with y − 3f (y) in the initial equation gives f (x − 6y) = f (x) − 2y + 2(y − 3f (y)), hence f (x − 6y) = f (x) − 6f (y), for all x, y ∈ Q. Set now x = y = 0 to get f (0) = 0 and replace x = 6y to obtain f (6y) = 6f (y), for all y ∈ Q. We derived that f (x−6y) = f (x)−f (6y), which, for u = 6y and v = x−6y, yields f (u + v) = f (u) + f (v), for all u, v ∈ Q. The solution of this classical functional equation is f (x) = xf (1), x ∈ Q. On the other hand, by setting x = y = 1 in the initial equation we obtain 3f 2 (1) − f (1) − 2 = 0, hence f (1) = 1 or f (1) = −2/3, thus proving our claim.

13

2014 Romanian Mathematical Olympiad – District Round 11th GRADE

Problem 1. a) Give an example of matrices A and B from M2 (R), such that   2 3 2 2 . A +B = 3 2 2

2

b) Let A and B be matrices from M2 (R), such that A + B =



2 3

3 2



.

Prove that AB �= BA. Vladimir Cerbu, Mihai Piticari

   1 1 0 1 Solution. a) An example is A = 3/2 and B = . 1 1 −1 0 b) If the matrices A and B commute, then A2 + B 2 = (A + iB)(A − iB), hence 



|det(A + iB)|2 = det(A + iB) det(A − iB) = det(A2 + B 2 )   2 3 = det = −5, 3 2 a contradiction. Problem 2. a) Let f : R → R be a function such that g : R → R, g(x) = f (x) + f (2x), and h : R → R, h(x) = f (x) + f (4x), are continuous functions. Prove that f is also continuous. b) Give an example of a discontinuous function f : R → R, with the following property: there exists an interval I ⊂ R, such that, for any a in I, the function ga : R → R, ga (x) = f (x) + f (ax), is continuous. Dorel Mihet¸

Solution. a) Since g and h are continuous, and f (x) = (g(x) − g(2x) + h(x)) /2, x ∈ R, it follows that f is continuous, as well.

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2014 Romanian Mathematical Olympiad – District Round

b) The function f : R → R,

   −1, f (x) = 0,   1,

x < 0, x = 0, x > 0,

is discontinuous at 0, but ga : R → R, ga (x) = f (x) + f (ax) = 0, is continuous on R, for all a < 0. Problem 3. a) Let A be a matrix from M2 (C), A �= aI2 , for any a ∈ C. Prove that the matrix X from M2 (C) commutes with A, that is, AX = XA, if and only if there exist two complex numbers α and α′ , such that X = αA+α′ I2 . b) Let A, B and C be matrices from M2 (C), such that AB �= BA, AC = CA and BC = CB. Prove that C commutes with all matrices from M2 (C). Adrian Troie

Solution. if X = � αA+α′ I2 , then � a) Clearly, � � X and A commute. Conversely, a 1 a2 x1 x 2 let A = and X = . The equality AX = XA implies ′ ′ a1 a2 x′1 x′2 a2 x′1 = a′1 x2 ,

(1)

(a1 − a′2 )x2 + a2 (x′2 − x1 ) = 0,

(2)

(a1 − a′2 )x′1 + a′1 (x′2 − x1 ) = 0.

(3)

Since A �= aI2 , a ∈ C, either one of a2 , a′1 is non-zero, or a2 = a′1 = 0 and a1 �= a′2 . In the first case, if, for instance, a2 �= 0, we obtain � x2 a1 � X= A + x1 − x2 I2 , a2 a2 while in the second case

X=

x1 − x′2 a1 x′2 − a′2 x1 A + I2 . a1 − a′2 a1 − a′2

b) We prove that C = γI2 , for some complex number γ, hence C commutes with all matrices in M2 (C). Suppose the contrary. Since A and C commute, there exist α and α′ , such that A = αC + α′ I2 . Similarly, there exist β and β ′ , such that B = βC + β ′ I2 , therefore AB = BA, a contradiction.

2014 Romanian Mathematical Olympiad – District Round

15

Problem 4. Let f : N → N∗ be a strictly increasing function. Prove that:

a) there exists a decreasing sequence of positive real numbers, (yn )n∈N , converging to 0, such that yn ≤ 2yf (n) , for all n ∈ N; b) if (xn )n∈N is a decreasing sequence of real numbers, converging to 0, then there exists a decreasing sequence of real numbers (yn )n∈N , converging to 0, such that xn ≤ yn ≤ 2yf (n) , for all n ∈ N.

George Stoica

Solution. a) Since f (0) > 0 and f is strictly increasing, it follows that f (n) > n, for all n ∈ N. Consider the sequence of non-negative integers (nk )k∈N , defined by n0 = 0 and nk = f (nk−1 ), k ∈ N∗ . The properties of f imply that the sequence is strictly increasing. We define the decreasing sequence (yn )n∈N , by yn = 2−k , for all n with nk ≤ n < nk+1 , k ∈ N, obviously convergent to 0. It suffices to prove that yn ≤ 2yf (n) , n ∈ N, for nk ≤ n < nk+1 , k ∈ N. Since f strictly increasing, nk+1 = f (nk ) ≤ f (n) < f (nk+1 ) = nk+2 , hence yf (n) = 2−k−1 = yn /2. b) Obviously, xn ≥ 0, for all n ∈ N. Using the previously defined sequence (nk ) , we define the decreasing sequence of positive reals (zk )k∈N , as follows: z0 = x1 and zk = max (xnk , zk−1 /2), k ∈ N∗ . The monotony of this sequence follows inductively. Moreover, (zk )k∈N converges to 0: if zk = xnk , for infinitely many k’s, then zk → 0 because it is decreasing and xn → 0; if zk = xnk , only for finitely many k’s, then zk = zk−1 /2 from some k onwards, and again, zk → 0. Finally, we define the sequence (yn )n∈N by yn = zk , nk ≤ n < nk+1 , k ∈ N. Clearly, the sequence decreases to 0. In order to prove the inequalities xn ≤ yn ≤ 2yf (n) , n ∈ N, it suffices to check them for nk ≤ n < nk+1 , k ∈ N. Obviously, xn ≤ xnk ≤ zk = yn . On the other hand, nk+1 = f (nk ) ≤ f (n) < f (nk+1 ) = nk+2 , since f is strictly increasing, hence yf (n) = zk+1 ≥ zk /2 = yn /2.

12th GRADE

Problem 1. For each positive integer n the function fn : [0, n] → R is defined by fn (x) = arctg (⌊x⌋). Prove that fn is a Riemann integrable function

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2014 Romanian Mathematical Olympiad – District Round

and find

1 n→∞ n lim



n

fn (x)dx. 0

***

Solution. The function fn is locally constant, hence Riemann integrable. Next, we have 

n

fn (x)dx = 0

k−1   i+1 i=0

fn (i)dx =

i

n−1 

arctg i.

i=0

Applying Stolz-Ces` aro theorem, we obtain lim

n→∞

π arctg 1 + arctg 2 + . . . + arctg n = lim arctg n = . n→∞ n 2

Problem 2. Let f : [0, 1] → R be a derivable function, with continuous   n derivative, and let sn = k=1 f nk . 1 Prove that the sequence (sn+1 − sn )n∈N∗ converges to 0 f (x)dx.

***

Solution. Using the mean value theorem, we obtain n+1 

n k   k − f n+1 n k=1 k=1   n     k  k = f (1) − −f f n n+1

sn+1 − sn =

f



k=1

n

 1 = f (1) − kf ′ (xk ), n(n + 1) k=1

for some xk ,

k n+1

k n,

< xk < k = 1, 2, . . . , n. ′ kf ′ (xk ) xk f ′ (xk ) x f (x ) k k ≤ ≤ , k = 1, 2, . . . , n, hence If f ′ ≥ 0, then n+1 n(n + 1) n n n n  1 1 1  xk f ′ (xk ) ≤ kf ′ (xk ) ≤ xk f ′ (xk ). Because 0 ≤ n+1 n(n + 1) n k=1 k=1 k=1 x1 ≤ 1/n ≤ x2 ≤ . . . ≤ xn ≤ n/n is a tagged partition of the interval [0, 1], it follows that  1  1 n  1 1 lim kf ′ (xk ) = xf ′ (x)dx = xf (x)0 − f (x)dx, n→∞ n(n + 1) 0 0 k=1

17

2014 Romanian Mathematical Olympiad – District Round

hence the conclusion. If f ′ takes negative values, replace f with g : x �→ f (x) + M x, where M = sup |f ′ |. As above, for tn =

n k  g n

k=1

we have (tn+1 − tn )n → and



1

g(x)dx = 0



1

f (x)dx + 0

tn+1 − tn = sn+1 − sn +

M 2

M , 2

therefore the conclusion holds in this case as well. Problem 3. Let (A, +, ·) be an unit ring with the property: for all x ∈ A, x + x2 + x3 = x4 + x5 + x6 . a) Let x ∈ A and let n ≥ 2 be an integer such that xn = 0. Prove that x = 0. b) Prove that x4 = x, for all x ∈ A. George Stoica

Solution. a) From the given equality we derive that xn−1 = xn (x4 + x3 + x2 − x − 1) = 0 and, step by step, that xn−2 = xn−3 = . . . = x = 0. b) Rewrite the given equation as It follows that

x(x3 − 1)(x2 + x + 1) = 0.

(x4 − x)2 = x2 (x − 1)(x3 − 1)(x2 + x + 1) = 0, hence x4 − x = 0. Problem 4. Let (G, ·) be a group with no elements of order 4, and let f : G → G be a group morphism such that f (x) ∈ {x, x−1 }, for all x ∈ G. Prove that either f (x) = x for all x ∈ G, or f (x) = x−1 for all x ∈ G. Nicolae Bourb˘ acut¸

18

2014 Romanian Mathematical Olympiad – District Round

Solution. Assume, by way of contradiction, that there exist a, b ∈ G such that f (a) = a �= a−1 and f (b) = b−1 �= b. Then f (ab) = f (a)f (b) = ab−1 �= ab, therefore f (ab) = (ab)−1 = b−1 a−1 . It follows that ab−1 = b−1 a−1 , which implies b−1 = ab−1 a. Next, f (ab2 ) = f (a)f 2 (b) = ab−2 . If f (ab2 ) = ab2 , then ab−2 = ab2 , therefore either b2 = e, which contradicts b �= b−1 , or ord(b) = 4, which is impossible. Thus, f (ab2 ) = (ab2 )−1 = b−2 a−1 , hence ab−2 = b−2 a−1 . It follows that ab−2 a = b−2 = (b−1 )2 = ab−1 aab−1 a, so a2 = e, again a contradiction.

THE 65th ROMANIAN MATHEMATICAL OLYMPIAD FINAL ROUND

5th GRADE

Problem 1. Prove that the product of every three odd consecutive positive integers can be written as the sum of three consecutive integers. Marian Ciuperceanu

Solution. Let the three odd consecutive numbers be 2p + 1, 2p + 3 and 2p + 5, where p is a positive integer. Then one of these numbers is divisible by 3: if p = 3k, with integer k, then 2p + 3 = 2(3k) + 3 = 6k + 3 = M3; if p = 3k + 1, with integer k, then 2p + 1 = 2(3k + 1) + 1 = 6k + 3 = M3; if p = 3k + 2, with integer k, then 2p + 5 = 2(3k + 2) + 5 = 6k + 9 = M3. In all the cases the product P is a multiple of 3, therefore P = 3a = (a − 1) + a + (a + 1), with integer a. Problem 2. We will say that a positive integer n is subject to an interesting change if it is multiplied by 2 and the result is increased by 4, a special change if it is multiplied by 3 and the result is increased by 9 and an awesome change if it is multiplied by 4 and the result is increased by 16. a) Show that there exists a positive integer which after three changes, the first – interesting, the second – special and the third – awesome, becomes 2020. b) Find all positive integers with the property that after two changes of different types, selected among the three above, becomes 2014. Lucian Dragomir

Solution. a) Before the last change the number must be (2020 − 16) : 4 = 501; the previous number must be (501 − 9) : 3 = 164 and the required number is (164 − 4) : 2 = 80. 19

20

2014 Romanian Mathematical Olympiad – Final Round

b) An interesting change produces a multiple of 2, a special change gives a multiple of 3 and an awesome change yields a multiple of 4. Since 2014 is neither a multiple of 3, nor a multiple of 3, the last change must be an interesting one. So, the second number must be (2014 − 4) : 2 = 1005. Since this number is odd, the first change must be special, and the initial number is (1005 − 9) : 3 = 332. Problem 3. Show that there exists a multiple of 2013 which ends in 2014. Mihaela Berindeanu

Solution. Consider the numbers a1 , a2 , . . . , a2014 , where ak is the 4k- digit number whose digits are k groups of the form 2014, 1 ≤ k ≤ 2014. Since there are only 2013 possible remainders after the division by 2013, two different as must give the same remainder in this division, hence there exists i, j, with 1 ≤ j < i ≤ 2014, so that ai − aj is divisible by 2013. Since . . . 0 = 20142014 . . . 0 = ai−j · 104j ai − aj = 20142014   . . . 2014 00  . . . 2014 ·100 i−j times 2014

4j times

i−j times 2014

4j times

and 104j and 2103 are relatively prime, 2013 must divide ai−j .

Problem 4. One hundred boxes are labeled from 1 to 100. Each box has at most 10 stones.The difference of the numbers of stones for every two boxes labeled with consecutive numbers is 1. The boxes labeled 1, 4, 7, 10, . . . , 100 contain a total of 301 stones. Find the maximum possible number of stones contained by the 100 boxes. Gabriel Popa

Solution. Since the difference of the number of stones in every two consecutive boxes is 1, two consecutive boxes contain at most 19 stones. We know the number of stones in the 34 boxes #1, #4, #7, . . . , #100 and if we group the remaining 66 boxes in pairs of consecutive boxes, we obtain at most 301 + 33 · 19 = 928 stones. This value is indeed obtained if we have 10 groups of the form (9, 10, 9, 8, 9, 10), then 6 groups of the form (9, 10, 9, 10, 9, 10) and, at the end, 9, 10, 9, 8.

2014 Romanian Mathematical Olympiad – Final Round

21

6th GRADE

Problem 1. Denote A = {1000, 1001, 1002, . . . , 2014}. Find the maximum number of elements of a subset of A which contains only perfect squares pairwise relatively prime. Marcel Neferu

Solution. If n ∈ A and n = p2 , then 1000 ≤ p2 ≤ 2014, that is 32 ≤ p ≤ 44. The largest subset of A whose elements are perfect squares is   B = 322 , 332 , 342 , 352 , 362 , 372 , 382 , 392 , 402 , 412 , 422 , 432 , 442 . We must choose among them the maximum number of pairwise prime numbers. Consider the partition of B into the sets C1 = {322 , 342 , 362 , 382 , 402 , 422 , 442 }, C2 = {332 , 392 }, C3 = {352 }, C4 = {372 }, C5 = {412 }, C6 = {432 }. If we choose 7 or more elements of B, then two of them are in the same Ci , so they are not co-prime. So we cannot take more than 6 elements; an example is {322 , 332 , 352 , 372 , 412 , 432 }. Problem 2. An integer n > 1 will be called p-periodic if n1 is a repeating decimal fraction, whose shortest period has length p and begins immediately 1 = after the decimal point. For instance, 19 = 0.111 . . . is 1-periodic and 11 0.090909 . . . is 2-periodic. a) Find all p-periodic numbers n so that the first digit of the period of n1 is not nil. b) Find the largest 4-periodic prime. Marius Perianu 1 Solution. a) The first digit of the period is at least 1 if and only if n1 ≥ 10 , that is n ≤ 10. Since the prime factors of n must be different from 2 and 5, it follows that n ∈ {3, 7, 9}; indeed, in this case 13 = 0,(3), 17 = 0,(142857) and 1 9 = 0,(1). m , with 1 ≤ m ≤ 9998. b) Let p be the largest 4-periodic prime; then p1 = 9999 2 This yields pm = 9999 = 3 · 11 · 101. So, the candidate for the largest such 1 = 0,(0099). prime is p = 101, which is acceptable, because 101

Problem 3. If n is a positive integer, we will call a triple (x, y, z) of positive integers of type n if x + y + z = n. Denote s(n) the number of triples of type n. a) Prove that there exists no positive integer n such that s(n) = 14.

22

2014 Romanian Mathematical Olympiad – Final Round

b) Find the smallest positive n so that s(n) > 2014. Nicu¸sor Berbecel

Solution. Let us count the triples (x, y, z) of positive integers such that x + y + z = n, where n ≥ 3 is a given integer. One can assign to x any value from 1 to n − 2. If x = 1, then, taking into account that z ≥ 1, y can take n − 2 values: from 1 to n − 2. If x = 2, then y can take n − 3 values, . . . , if x = n − 2, then y can take only the value 1. This shows that the number of triples of type n is s(n) = 1 + 2 + 3 + . . . + (n − 3) + (n − 2) =

(n − 2)(n − 1) . 2

a) If there exists n such that s(n) = 14, then (n − 2)(n − 1) = 28, which is impossible: if n ≤ 6, then (n−2)(n−1) ≤ 20 and if n ≥ 7, then (n−2)(n−1) ≥ 30. b) We have to find the smallest n such that (n − 2)(n − 1) > 4028. Since (n − 1)2 > (n − 2)(n − 1), n − 1 must be at least the smallest perfect square larger than 4028. Since 4028 = 22 · 1007 and 210 = 322 = 1024 > 1007 > 31.52 , it follows that n − 1 ≥ 64, that is n ≥ 65. From (65 − 2)(65 − 1) = 4032 > 4028 follows that n = 65. Problem 4. In triangle ABC points M, N ∈ (AB), P, Q ∈ (BC) and   and S, R ∈ (AC) are taken such that AM = CR, AN = CS, M QB ≡ RQC   N P B ≡ SP C. Show that if M Q + QR = N P + P S, then triangle ABC is isosceles.

Mircea Fianu, Traian Preda

Solution. Denote R′ and S ′ the reflection of R, respectively S, across the ′ QC. Since M ≡R    it follows QB ≡ RQC, line BC. Then RQ = R′ Q and RQC ′ QC, and because B, Q, C are collinear, so are M , Q, R′ .   that M QB ≡ R This yields M Q + QR = M Q + QR′ = M R′ . In the same way N P + P S = N P + P S ′ = N S ′ . Now M Q + QR = N P + P S implies M R′ = N S ′ . ′ CQ and SCP ′ CP , so RCQ  ≡S  = ≡R  The symmetry also implies RCQ ′ ′ ′ ′    SCP implies R CQ = S CP , whence the points C, R , S are collinear. Also

23

2014 Romanian Mathematical Olympiad – Final Round

CR = CR′ and CS = CS ′ and, since CR = AM and CS = AN , M N = R′ S ′ . A

M

S R

N

C B

Q

P

R' S' � Therefore △N S M ≡ △R M S (case S-S-S), whence N� M S′ ≡ M S ′ R′ . ′ � ≡ BCS �′ . Now BCS �′ ≡ BCA � This shows that AB � CS , therefore ABC � � implies ABC ≡ ACB, q.e.d. ′

′

′

7th GRADE

Problem 1. Find all primes p and q, with p ≤ q, so that   p (2q + 1) + q (2p + 1) = 2 p2 + q 2 .

Lucian Dragomir

2

Solution. The equality can be written p + q = 2 (p − q) , which shows that p is odd. If p ≥ 5, then p and q leave remainder 1 or 2 when divided by 3. We will show that in this case the equality is impossible. Indeed, if p and 2 q leave the same remainder mod 3, then 3 | 2 (p − q) and 3 ∤ p + q; if p and q 2 leave different remainders, then 3 ∤ 2 (p − q) and 3 | p + q. Finally, if p = 3, then q = 5. Problem 2. Construct outside the square ABCD the rhombus BCM N , with ∠BCM an obtuse angle. The straight lines BM and AN meet at P . Prove that DM ⊥ CP and triangle DP M is right and isosceles. Andrei Bud

� � and Solution. Isosceles triangles CDM and CBM yield CM D ≡ CDM ◦ � � � B)+ CM B ≡ CBM . In triangle M BD, the sum of the angles is 180 = m(DM

24

2014 Romanian Mathematical Olympiad – Final Round

 ) + m(DBM  ) = m(DM   ) + 45◦ + m(CBM ) = m(BDM B) + 45◦ + m(CDM ◦ ◦   B), whence m(DM B) = 45 . 90 + 2m(DM   Since quadrilateral ADM N is a parallelogram, m(M P N ) = m(DM B) = 45◦ . The point P is on the perpendicular bisector of the segment [CN ], so    M PN ≡ M P C, whence m(N P C) = 90◦ , that is CP ⊥ AN , implying CP ⊥ DM . A D

B

C P N

M

In the isosceles triangle CDM the straight line CP is the perpendicular from C onto DM , so it is the perpendicular bisector of the segment [DM ]. This  shows that triangle DP M is isosceles with vertex P . So, from m(DM P ) = 45◦ ◦  follows m(DP M ) = 90 . 2

2

Problem 3. Find all positive integers n so that 17n + 9n = 23n + 3n . Marius Perianu

Solution. Clearly n = 0 and n = 1 are solutions.   2 2 2 2 If n ≥ 2, then n2 ≥ 2n, hence 9n − 3n = 3n (3n − 1) ≥ 32n 32n − 1 = 81n − 9n . Since 81n − 9n > 23n − 17n , the equation does not have solutions n ≥ 2. Problem 4. Construct outside the square ABCD the right isosceles triangle ABE, with hypotenuse [AB]. Denote N the midpoint of the segment [AD] and {M } = CE ∩ AB, {P } = CN ∩ AB, {F } = P E ∩ M N . Take on the straight line F P the point Q so that [CE is the bisector of the angle ∠QCB. Prove that M Q ⊥ CF . Sorin Furtun˘ a

Solution. Clearly P A = P B/2, so P A = AB = BC. This leads to  ≡ AEP  . Now △AP E ≡ △BCE (SAS), which implies CE = P E and BEC

2014 Romanian Mathematical Olympiad – Final Round

25

 + m(AEP  ) = 90◦ , so m(CEP ) =  + m(AEC)  = 90◦ yields m(AEC) m(BEC) ◦  E) = 45◦ . 90 . This shows that ECP is a right isosceles triangle, so m(CP P

Q E

A

N

D

M B

C

F Also △CEQ ≡ △P EM (AAS), whence EQ = EM , so EM Q is a right   ) = 45◦ = m(EP C), therefore M Q � isosceles triangle. This gives m(EQM CP . Menelaus’ Theorem for the triangle CP E and the transversal N F yields CN FP ME N P · F E · M C = 1. If we denote R the orthogonal projection of E onto AB, E ER 1 then △CBM ∼ △ERM , whence M M C = BC = 2 .

FP From N P = CN follows F E = 2, hence P E = EF , therefore CE is altitude and median in △CF P , that is this triangle is isosceles. This gives  m(F CP ) = 90◦ , that is CP ⊥ CF , whence M Q ⊥ CF .

8th GRADE

Problem 1. Let a, b, c ∈ (0, ∞). Prove the inequality

√ √ √ b − ca c − ab a − bc + + ≥ 0. a + 2 (b + c) b + 2 (c + a) c + 2 (a + b) Nicolae Papacu

Solution. The AM-GM inequality leads to √ a − b+c 2a − b − c a − bc 2 ≥ = . a + 2 (b + c) a + 2 (b + c) 2 (a + 2b + 2c)

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2014 Romanian Mathematical Olympiad – Final Round

So, the left-hand part of the given inequality is at least 2b − c − a 2c − a − b not 2a − b − c + + = S. 2 (a + 2b + 2c) 2 (2a + b + 2c) 2 (2a + 2b + c) Denote a + 2b + 2c = 5x, 2a + b + 2c = 5y and 2a + 2b + c = 5z. Then a = −3x + 2y + 2z, b = 2x − 3y + 2z, c = 2x + 2y − 3z and  −10x + 5y + 5z 1 y z 2a − b − c = = + −2 . 2 (a + 2b + 2c) 10x 2 x x This and the two similar relations leads to   1 x z  1 y z 1 x y S= + −2 + + −2 + + −2 2 x x 2 y y 2 z z        y 1 x y x z z + + + + + − 6 ≥ 0, = 2 y x z x y z whence the conclusion. Problem 2. The cube ABCDA′ B ′ C ′ D′ has edges of length a. Take the points E ∈ (AB) and F ∈ (BC) so that AE + CF = EF . a) Find the measure of the angle of the planes (D′ DE) and (D′ DF ). b) Compute the distance from D′ to the straight line EF . Sorin Peligrad, Adrian T ¸ urcanu

D'

C'

A'

B'

D P HA

E

F

C

B

Solution. a) Extend the segment BA with AH ≡ F C. Then ∆DAH ≡  ≡ F   ) = 90◦ . Then [DH] ≡ ∆DCF (SAS), whence HDA DC, so m(HDF   = [DF ], leads to ∆DHE ≡ ∆DF E (SSS) and from here m(F DE) = m(HDE) ◦ ′ ′  , (D′ DF )) = 45 . Since F D ⊥ DD and ED ⊥ DD , it follows that m((D′ DE) ◦  m(F DE) = 45 .

2014 Romanian Mathematical Olympiad – Final Round

27

b) Denote P the orthogonal projection of the point D onto EF . The Three Perpendiculars Theorem yields D′ P ⊥ EF , hence d (D′ , EF ) = D′ P . The congruence ∆DHE ≡ ∆DF E gives DP = AD = a. Finally, from Pythagoras’ √ Theorem, D′ P = a 2. Problem 3. Find the smallest integer n for which the set A = {n, n + 1, n + 2, . . . , 2n} contains five elements a < b < c < d < e so that b c a = = . c d e Mircea Fianu

Solution. Let p, q ∈ N∗ , (p, q) = 1 so that ac = db = ec = pq . Obviously, p < q. Since a, b and c are divisible by p and c, d, e are divisible by q, there exists m ∈ N∗ so that c = mpq. Let us find the minimal value of the difference e − a, for given p, q. This is obtained when a, b, c are consecutive multiples of p and c, d, e are consecutive mpq multiples of q, that is a = mpq − 2p and e = mpq + 2q. From ec = mpq+2q = pq follows m (q − p) = 2, hence m ∈ {1, 2}. Condition n ≤ a < e ≤ 2n ≤ 2a implies 2a ≥ e, that is 2mpq − 4p ≥ mpq + 2q, or mpq ≥ 4p + 2q. (∗) 2 If m = 1, then q − p = 2, hence q = p + 2. Relation (∗) yields (p − 2) ≥ 8, whence p ≥ 5. For p = 5 and q = 7 we get a = 25, b = 30, c = 35, d = 42, e = 49 and, since n ≤ a < e ≤ 2n, n = 25. 2 If m = 2, then q − p = 1, so q = p + 1. Relation (∗) yields (p − 1) ≥ 2, whence p ≥ 3. For p = 3 and q = 4 we get a = 18, b = 21, c = 24, d = 28, e = 32 and, since n ≤ a < e ≤ 2n, n ∈ { 16, 17, 18 }. Therefore, n min = 16. Problem 4. Prove that three discs of radius 1 cannot cover entirely a square surface of side 2, but they can cover more than 99.75% of it. Traian Preda

Solution. Denote ABCD the square and S1 , S2 , S3 the discs. Suppose that S1 , S2 and S3 cover the whole square. Since there are three discs and the square has four vertices, one of the discs must cover two vertices of the square, say S1 covers A, B. Then [AB] is a diameter for S1 , so S1 cannot cover any point from (BC] ∪ [CD] ∪ [DA). So, C must be covered by

28

2014 Romanian Mathematical Olympiad – Final Round

one of the other discs, say S2 . Then S2 cannot cover any point from (AD), therefore (AD) ⊂ S3 . In this case [AD] is a diameter of S3 , so S3 cannot cover any point from (BC), therefore S2 must cover (BC). This shows that [BC] must be a diameter of S2 . But, in this case, no point from (CD) is covered – contradiction. b) Take M ∈ (AC) so that AM = 2. Denote P and R the orthogonal projections of M onto AB and onto AD. Let T ∈ BC and U ∈ DC be such that P T = RU = 2. R D A

P B

M

XU TC

We take as S1 , S2 , S3 the discs of diameters [AM ], [P T ], [RS]. Denote X the point on [AC] for which XT ⊥ BC (and XU ⊥ CD). Then S1 covers the square surface AP M R , S2 covers the pentagonal surface BP M XT and S3 covers the pentagonal surface DRM XU , so the points not covered by the three discs are inside the square CU XT . It is enough to prove that area[CU XT ] < 0.25% · area[ABCD], which is equivalent to CT < BC/20 = 0.1, or BT > 1.9. Simple considerations √ √ 2 √ √  yield AP = 2, BP = 2 − 2, BT 2 = 4 − 2 − 2 = 4 2 − 2. We are left √ √ to prove BT > 1.9, that is 4 2 − 2 > 1.92 , or 2 > 1.4025, which is true. 9th GRADE

Problem 1. Let n be a natural number. Find the integers x, y, z such that x2 + y 2 + z 2 = 2n (x + y + z) . Petre Simion, Mircea T ¸ eca

Solution. If n = 0, using the inequalities x2 ≥ x and its analogues, we deduce that x, y, z ∈ {0, 1}.

2014 Romanian Mathematical Olympiad – Final Round

29

If n ≥ 1, then 2 divides x2 + y 2 + z 2 , and hence, either the three numbers are even, or one is even and the others are odd. In the former case, if we take   x = 2x1 + 1, y = 2y1 + 1, z = 2z1 , we get 4 x21 + x1 + y12 + y1 + z12 + 2 = 4 (x1 + y1 + z1 + 1) , which is a contradiction. Let us consider the case when the numbers x, y, z are even. For x = 2x1 , y = 2y1 , z = 2z1 , we get x21 + y12 + z12 = 2n−1 (x1 + y1 + z1 ) , and thus, if n = 1, then x, y, z ∈ {0, 2}. For n > 1, using the same argument, we deduce that if x = 2n xn , y = 2n yn , z = 2n zn , then xn , yn , zn ∈ Z and x2n + yn2 + zn2 = xn + yn + zn , whence xn , yn , zn ∈ {0, 1} , and thus x, y, z ∈ {0, 2n }. Problem 2. Let a be a natural odd number which is not a perfect square. If m and n are strictly positive integers, prove that a) {m (a + b) [m (a +

√

a)} �= {n (a −

√

a)} ,

√ √ a)] �= [n (a − a)]. Vasile Pop

√ Solution. a) As ma, na are natural numbers, the equality implies {m a} = √ {−n a}. Two numbers have the same fractional part if and only if their difference is √ an integer, whence (m + n) a ∈ Z, which is absurd. b) Again, let us as suppose that there is a natural number N , for which there exist two not equal numbers m, n which are different from zero, such √ √ √ that N = [m(a + a] = [n(a − a]. Then N ≤ m(a + a) < N + 1 and √ N ≤ n(a − a) < N + 1; moreover, the inequalities are strict, because the terms in the middle are irrational numbers. We rewrite the inequalities as N +1 N N +1 N √ a

On the other side f has a minimum at a, because f (a) = a = inf I = inf{f (x) | x ∈ R},

(2)

36

2014 Romanian Mathematical Olympiad – Final Round

so, by the Theorem of Fermat, f ′ (a) = 0, in contradiction with (2). We conclude a = −∞. Analogously b = +∞. Problem 3. Consider a positive integer n and A, B two matrices in Mn (C) such that A2 + B 2 = 2AB. Prove that: a) The matrix AB − BA is not invertible. b) If the rank of A − B is 1, then matrices A and B commute. Leonard Giugiuc, C˘ at˘ alin Gherghe

Solution. a) The given relation can be written in each of the two forms: (A − B)2 = AB − BA,

(1)

A(A − B) = (A − B)B.

(2)

Suppose AB − BA is non singular. By (1), A − B is also non-singular, that is B = (A − B)−1 A(A − B), by (2). Then A − B = A − (A − B)−1 A(A − B), from where we get In = A(A − B)−1 − (A − B)−1 A. Considering traces we obtain   n = tr In = tr A(A − B)−1 − (A − B)−1 A = 0,

a contradiction. So AB − BA is singular; moreover by (1) the matrix A − B is also singular. b) It is known that for a matrix X of rank 1 in Mn (C), then X 2 = (tr X) X (each row of X is proportional with a nonzero fixed row X). Using (1) we get AB − BA = (A − B)2 = (tr (A − B))(A − B), so 0 = tr (AB − BA) = (tr (A − B))2 , i.e., tr (A − B) = 0. We conclude, AB − BA = On , i.e., AB = BA. Problem 4. Let A be an invertible matrix in M4 (R), such that tr A = tr A∗ �= 0, where A∗ is the adjugate of A. Prove that the matrix A2 + I4 is singular if and only if there exists a nonzero matrix B in M4 (R), so that AB = −BA. Mariean Andronache

2014 Romanian Mathematical Olympiad – Final Round

37

Solution. We show first that if A is a matrix from Mn (R) such that A2 + In is singular, then there exists a matrix B in Mn (R), such that AB = −BA. As A2 + In is singular, i is a proper value of A, and −i is a proper value of the transpose Aτ . There exist two proper vectors x and y in Mn,1 (C), so that Ax = ix and Aτ y = −iy. Because x and y are nonzero, B = xyτ is a nonzero matrix in Mn (C). The following relations show that A and B anticommute: τ

AB = Axyτ = ixyτ = −x(−iy)τ = −x (Aτ y) = −xyτ A = −BA. Taking conjugates and using that A is in Mn (R), we get that the conjugate ¯ B of B anticommutes with A. This proves that any linear combination with ¯ anticommutes with A. Consecomplex coefficients of the matrices B and B ¯ is a nonzero matrix in Mn (R) that anticommutes with quently, B or i(B − B) A. This proves the first part of the problem. We shall prove the converse. Let B be a nonzero matrix in M4 (R) that anticommutes with A. Then Ak B = (−1)k BAk ,

k ∈ N.

(∗)

Consider the characteristic polynomial f of A, f = λ4 −(tr A)λ3 +aλ2 −(tr A∗ )λ+det A = λ4 −(tr A)λ3 +aλ2 −(tr A)λ+det A, where a is a real number (the latter form of f is a consequence of tr A = tr A∗ ). By Hamilton-Cayley, f (A) = O4 . Taking into account (∗), we successively obtain:   O4 = f (A)B = B A4 + (tr A)A3 + aA2 + (tr A)A + (det A)I4   = B f (A) + 2(tr A)(A2 + I4 )A = 2(tr A)B(A2 + I4 )A. Because tr A �= 0 and A is invertible, we obtain that B(A2 + I4 ) = O4 . As B is nonzero we conclude that A2 + I4 is singular. 12th GRADE

Problem 1. Let (A, +, ·) be a ring with unity. For a ∈ A define functions sa : A → A and da : A → A by sa (x) = ax, da (x) = xa, for all x ∈ A.

38

2014 Romanian Mathematical Olympiad – Final Round

a) Suppose A is finite. Prove that for any a ∈ A, sa is injective if and only if da is injective. b) Give an example of an ring that contains an element a such that exactly one of the functions sa and da is injective. Dorel Mihet¸

Solution. a) Suppose sa is one to one. As A is finite sa is bijective, thus there exists b ∈ A such that ab = 1. As a consequence, if da (x) = da (y), we get in succession (x − y)a = 0, (x − y)ab = 0, that is x − y = 0, which proves the injectivity of da . The proof of the converse goes on the same lines. b) To construct an example, consider S = {(xn )n∈N | xn ∈ R} and the ring of additive functions f : S → S, endowed with the operations of addition and composition. For a one can consider the function defined by a ((xn )n ) = (xn+1 )n . As a is surjective, da is injective. It is clear that sa is not one to one. Problem 2. Let I, J be intervals and consider ϕ : J → R a continuous function which is nonzero on J. Let f, g : I → J be two differentiable functions such that f ′ = ϕ ◦ f and g ′ = ϕ ◦ g. Prove that if there exists x0 ∈ I such that f (x0 ) = g(x0 ), than f and g coincide. ***

Solution. As ϕ is non-zero and is continuous, the function 1/ϕ is correctly defined and continuous. Consider an anti-derivative F : J → R. The given relation for f can be then written (F ◦ f )′ (x) = 1, for all x ∈ I. Thus there exists a ∈ R such that F (f (x)) = x + a, for all x ∈ I. In the same manner, there exists b ∈ R such that F (g(x)) = x + b, for all x ∈ I. We deduce x0 + a = F (f (x0 )) = F (g(x0 )) = x0 + b, that is a = b. So F (f (x)) = F (g(x)), for all x ∈ J. (*) On the other side, from F ′ (x) �= 0, for all x ∈ J, F ′ has a constant sign on J, so F is injective. So, (*) implies f = g. Problem 3. Let f : [1, +∞) → (0, +∞) be a continuous function having the following properties: (i) The function g : [1, +∞) → (0, +∞) given by g(x) = +∞,

f (x) has limit at x

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2014 Romanian Mathematical Olympiad – Final Round

1 (ii) The function h : [1, +∞) → (0, +∞) given by h(x) = x finite limit at +∞.

x

f (t) dt has

1

a) Show that lim g(x) = 0. x→+∞

1 b) Show that lim 2 x→+∞ x

x

f 2 (t) dt = 0.

1

Mihai Piticari

Solution. a) Let ℓ =

lim g(x). If ℓ ∈ (0, +∞), let a > 0 be such that

x→+∞

g(x) > ℓ/2 for x ≥ a. Then h(x)

= =

 a     x 1 a ℓ x 1 f (t) dt + f (t) dt ≥ f (t) dt + t dt = x 1 x 1 2x a a  1 a ℓ(x2 − a2 ) −→ +∞, f (t) dt + x→+∞ x 1 4x

in contradiction with (ii). In the same way we can prove that ℓ = +∞ is in contradiction with (ii), so ℓ = 0. x b) Remark that 1 f (t)dt > 0, for x > 1, so 1 lim x→+∞ x2



x

2

f (t)dt

=

1

= where λ = lim h(x), u(x) = x→+∞

x 1

x   x 2 f (t)dt f (t)dt 1 1 = · lim x x→+∞ x x f (t)dt  x1 2 f (t)dt u(x) 1 λ lim , = λ lim x x→+∞ x x→+∞ v(x) f (t)dt 1

f 2 (t)dt, v(x) = x

x 1

f (t)dt.

u(x) = 0 we shall use l’Hospital Rule: v(x) • u and v are differentiable, To show that lim

x→+∞

•

lim v(x) = +∞ (this follows from v(x) ≥ m(x − 1) for x ≥ 2, where

x→+∞

m = inf x∈[1,2] f (x)),  x • v ′ (x) = f (t) dt + xf (x) �= 0, for all x ≥ 1, 1

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2014 Romanian Mathematical Olympiad – Final Round

• relations

u′ (x) = v ′ (x)

f 2 (x) f (x)  x ∈ (0, g(x)) and = g(x) · f (x) + h(x) f (t) dt xf (x) + 1

u′ (x) lim g(x) = 0 imply lim ′ = 0. x→+∞ x→+∞ v (x)

Problem 4. Suppose (G, ·) is a finite group with unity e, a is an element in G\{e} and p is a prime number such that xp+1 = a−1 xa, for all x ∈ G. a) Show that there is k ∈ N∗ such that ord(G) = pk . b) Prove that H = {x ∈ G | xp = e} is a subgroup of G and (ord(H))2 > ord(G). Ioan B˘ aetu

Solution. a) If x, y ∈ G, than (xy)p+1 = a−1 xya = a−1 xaa−1 ya = = xp+1 y p+1 . We can write x(yx)p y = xp+1 y p+1 , then (yx)p = xp y p . For x = a we get ap = e so by the preceding equality (ya)p = y p . Multiplying at left by ya we obtain yay p = (ya)p+1 = y p+1 a, that is ay p = p y a, for all y ∈ G. From the hypothesis we have y p(p+1) = a−1 y p a = y p , so 2 y p = e, for all y ∈ G. Because p is a prime, every element of the group has order 1, p or p2 and by the Cauchy theorem we deduce ord(G) = pk , for a k ∈ N∗ . b) For x, y ∈ H, we have (xy)p = y p xp = e, that is xy ∈ H, proving that H is a stable part of G, and, as it is finite, H is a subgroup. 2 Consider f : G → G, given by f (x) = xp . Because e = xp = (xp )p , the image of f is contained in H. Moreover, x, y ∈ G and f (x) = f (y) imply xp (y −1 )p = e, so (y −1 x)p = e, that is y −1 x ∈ H. This gives x ∈ Hy. We conclude that for every element in Imf , the number of its pre-images in G is ord(G) . exactly ord(H), so |Imf | = ord(H) Because a �= e we get a ∈ / Imf : for if not a = bp for some b ∈ G. This would imply bp+1 = b−p bbp , that is e = bp = a a contradiction. As a ∈ H, we ord(G) , which gives the conclusion. conclude ord(H) > |Imf | = ord(H)

SHORTLISTED PROBLEMS FOR THE 65th NMO

SMALL JUNIORS

1. a) Prove that, for every positive integers a, b, k such that a < b, a+k a ≤ . b b+k b) Prove that 4 7 148 1 + + + ... + ≥ 25. 100 101 102 149 Gabriel Vrˆınceanu, Ion Cicu

2. A 4 × 4 magic square has in its first line the non-nil digits a, b, c, d, written in this order (from left to right), and each of the other lines contains the same numbers, written in different orders. It is known that the sum of the eight four-digit numbers obtained by reading the columns downwards and by reading the lines from left to right equals 59994. Find the largest and the smallest possible value for abcd. Gabriel Vrˆınceanu

3. Find how many pairs (a, b), with a and b non-nil digits, have the property a+b are finite. that ab is irreducible and the decimal fractions ab and b(b+1) Gabriel Vrˆınceanu

4. Triangle ABC has a right angle in A and AD ⊥ BC, with D ∈ (BC). Points E ∈ (AB) and F ∈ (AC) are taken so that (DE is the bisector of the  and (DF is the bisector of the angle ADC.  Prove that triangle angle ADB AEF is isosceles. Petru Braica

41

42

Shortlisted problems for the 2014 Romanian NMO

5. The isosceles triangle ABC has AB = AC and points M and N are taken on (BC) such that M is between B and N . Prove that the following properties are equivalent: i) m(∠M AN ) = 12 m(∠BAC); ii) the segments [BM ], [M N ] and [M C] are the sides of a triangle in which the opposite angle of [M N ] has measure 180◦ − m(∠A). Mihai Dicu

6. Denote O the center of the square ABCD. The bisector of the angle ∠OAB meets OB in N and BC in P . Prove that P C = 2ON . George Stoica

7. Find all positive integers n with at least two digits, n’s digits are pairwise distinct and n equals the product of the sum of its digits with one of its digits. Cosmin Manea, Drago¸s Petric˘ a

8. Define an = 18 77 . . . 7 889 for every non-negative integer n.   n digits

a) Prove that an is divisible by 13, for every n ∈ N. b) Denote by cn the quotient of the division of an by 13. Find all n ∈ N so that s(an ) = 2s(cn ), where s(m) represents the sum of the digits of the number m. Marius Burtea

9. If n is a composite number, denote Dn its largest proper divisor. A number n will be called squarish if the number Dn + Dn+1 is a perfect square. a) Show that 35, 76 and 755 are squarish. b) Show that there are infinitely many squarish numbers. Marius Burtea JUNIORS

1. Solve in positive integers the equation 2n = 3m + 23. Andreea Dima, Traian Preda

Shortlisted problems for the 2014 Romanian NMO

43

2. Let n be a positive integer and M be the arithmetic mean of its positive √ divisors. Prove that M ≥ n. When does the equality hold? Nicolae Bourb˘ acut¸

3. Let ABC be a triangle and D, E, F be the second intersection of its medians with its circumcircle. Prove that if the triangles BDC, CEA, AF B have the same area, then triangle ABC is equilateral. Marian Ionescu

4. Let △ABC be isosceles, with AB = AC, and P, Q be points on the ) = m(P   If [AD] is an altitude, side (AC) so that m(ABP BQ) = m(QBC). D ∈ BC, BP ∩ AD = {M } , BQ ∩ AD = {N } and △ABN is isosceles, prove that: a) M is the orthocenter of triangle ABC; AB (AB − AD). b) M N = AD Andrei Bud

5. Let triangle ABC be so that m (∠B) = m (∠C) = 50◦ . Take D on the √ ray (AB so that BC = AD 3. Let (M C) be a segment perpendicular on the plane (ABC) so that CD = 2M C.   a) Prove that m((M AC) , (M CD)) = 4m((M BC) , (M CD)). b) Find the measure of the angle between the planes (M AB) and (ABC). Silvia & Ionel Brabeceanu

6. Find all real solutions of the system    |x + y + z| = 1 |x| + |y| + |z| = 1    |x − y| + |y − z| + |z − x| = 2.

Dumitru M. B˘ atinet¸u-Giurgiu, Neculai Stanciu

7. Denote M the midpoint of the edge [AD] of the cube ABCDA′ B ′ C ′ D′ . Denote α the plane perpendicular in B on the straight line BM . Find the tangent of the angle between the planes α and (ACB ′ ). Gabriel Popa

44

Shortlisted problems for the 2014 Romanian NMO

8. Find the minimum value of the expression    1 1 1 2 2 E = x + 2 + y + 2 + z2 + 2 , y z x

taken for all x, y, z ∈ R∗ .

Ion Safta

SENIORS

1. Let M be a nonempty set of positive reals so that, for every a, b, c in M , the number ab + bc + ca is rational. Prove that ab is rational for every a, b in M . Bogdan Enescu

2. Find all strictly increasing sequences (an )n of positive integers with the following two properties: a) a31 + a32 + · · · + a3n = (a1 + a2 + · · · + an )2 , for every integer n ≥ 2014; b) for every integer k ≥ 2015, the number a1 + a2 + · · · + a2014 cannot be written as a sum of k consecutive positive integers. Cristinel Mortici

3. A sequence (an )n of positive integers is such that the sequence ann is bounded and an − am is divisible by n − m for every positive integers m, n. Prove that (an )n is an arithmetic sequence.

4. Let x ∈ [0, 1] and a, b, c ≥ 0. Prove that a(a + 2b)x + b(b + 2c)x + c(c + 2a)x ≤ (a + b + c)x+1 .

George Stoica

5. Let a ∈ (0, 1). Find all functions f : R → (0, ∞) so that f (x) ≤ ax and f (x + y + z) ≤ f (x)f (y)f (z), for every x, y, z ∈ R.

Traian T˘ amˆ aian

6. Let n ≥ 2 be an integer  and z be acomplex number so that z n = 1. √ n n  1 + (−1)n 1+ 5 Prove that: . − (1 − z k − z 2k ) = 2 2 k=1

Mircea Merca

45

Shortlisted problems for the 2014 Romanian NMO PUTNAM SENIORS

1. Let f : [0, 1] → [0, 1] be a continuous function a and x0 , y0 be two points in [0, 1]. Define sequences (xn )n∈N and (yn )n∈N of points in [0, 1] by   f (x0 ) + · · · + f (xn ) y 0 + · · · + yn and yn+1 = f ; xn+1 = n+1 n+1 clearly, the definition is correct. Prove that the sequences (xn )n∈N and (yn )n∈N are convergent. Dan S ¸ tefan Marinescu, Mihai Monea

2. Let a be a positive real number, (an )n≥1 be a sequence of real numbers and (xn )n≥1 be the sequence defined by  a an xn + , xn+1 = 1 − n n where x1 is an arbitrary real n umber. Prove that lim xn = 0

n→∞

if and only if

lim

n→∞

a1 + a2 + . . . + a n = 0. n George Stoica

3. Find the infimum and the supremum of the set {sin 1 + sin 3 + . . . + sin(2n + 1) | n ∈ N}. Leonard Giugiuc

4. Find all differentiable functions f : R → R, whose derivative is bounded in a neighborhood of the origin and fulfill the condition xf (x) − yf (y) = (x2 − y 2 ) max(f ′ (x), f ′ (y)), for every real numbers x and y. Marcel Chirit¸˘ a

5. Let A and B be two matrices from M3 (C), such that (AB)2 = A2 B 2 and (BA)2 = B 2 A2 . Prove that (AB − BA)3 = O3 . Marian Cucoane¸s

46

Shortlisted problems for the 2014 Romanian NMO

6. Let A and B be two matrices from M3 (C), such that A2 = AB + BA. Prove that the matrix AB − BA is singular. Marian Cucoane¸s

7. Let n ∈ N∗ and (G, ·) be a group with the property that there exists an endomorphism f : G → G, so that f (xn y n+1 ) = xn+1 y n for every x, y ∈ G Prove that: a) x2n+1 = e, for every x ∈ G; b) the group (G, ·) is abelian. Traian T˘ amˆ aian

8. Let I ⊂ R be an interval and f : I → R be a continuous function. Prove that the following properties are equivalent: i) f (x) − f (y) < |x − y|, for every x, y ∈ I;  y 1 ii) f (t)dt − f (x)(y − x) ≤ (x − y)2 , for every x, y ∈ I. 2 x

Nicolae Bourb˘ acut¸

9. Find all polynomials P , Q with rational coefficients, with the property that the polynomial P + aQ is irreducible for every rational a. George Stoica

10. Let f : [0, 1] → [0, 1] be a continuous function and x0 ∈ [0, 1]. Define the sequence (xn )n∈N by xn+1 =



1 n+1 (x0 +x1 +...+xn )

f (x) dx. 0

Prove that the sequence (xn )n∈N is convergent. Mihai Monea, Dan S ¸ tefan Marinescu

THE 65th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

FIRST SELECTION TEST

Problem 1. Let ABC be a triangle, let A′ , B ′ , C ′ be the orthogonal projections of the vertices A, B, C on the lines BC, CA and AB, respectively, and let X be a point on the line AA′ . Let γB be the circle through B and X, centered on the line BC, and let γC be the circle through C and X, centered on the line BC. The circle γB meets the lines AB and BB ′ again at M and M ′ , respectively, and the circle γC meets the lines AC and CC ′ again at N and N ′ , respectively. Show that the points M , M ′ , N and N ′ are collinear. Petru Braica

A

B' X N C' M' M N' H A' B 1 B

C

Solution. Let H be the orthocenter of the triangle ABC. The line AH is the radical axis of the circles γB and γC , hence HM ′ · HB = HN ′ · HC and AM · AB = AN · AC, so the lines M ′ N ′ and M N are both antiparallel to BC. The circle γB meets the line BC again at B1 . Then the lines M M ′ and BC are antiparallel, since BM M ′ B1 is a cyclic quadrangle. The conclusion follows. 47

48

Selection tests for the 2014 BMO and IMO

Problem 2. Given an integer n ≥ 2, show that there exist n + 1 pairwise distinct numbers x1 , x2 , . . ., xn , xn+1 in Q \ Z such that {x31 } + {x32 } + · · · + {x3n } = {x3n+1 }, where {x} is the fractional part of the real number x. Dorel Mihet¸

Solution. Notice that, if w1 < w2 < · · · < wn+1 < wn+2 are positive integers such that 3 3 = wn+2 , (∗) w13 + w23 + · · · + wn+1 then the numbers xk = wk /wn+2 , k = 1, 2, . . . , n, and xn+1 = −wn+1 /wn+2 meet the required conditions. We now show by induction on n ≥ 2 that there exist integers 3 = w1 < w2 < · · · < wn+1 < wn+2 satisfying (∗). The equalities 33 + 43 + 53 = 63 and 33 + 153 + 213 + 363 = 393 settle the cases n = 2 and n = 3, respectively. Finally, for the induction step n �→ n + 2, notice that if 3 < w2 < · · · < wn+1 < wn+2 are n + 2 integers satisfying (∗), then 3 < 4 < 5 < 2w2 < · · · < 2wn+1 < 2wn+2 are n + 4 integers satisfying the corresponding condition. Remarks. By Andreescu, T., Andrica, D., and Cucurezeanu, I., An Introduction to Diophantine Equations, Birkhauser, 2010, Example 5, p. 44, if n is an integer greater than or equal to 412, there exist positive integers a1 , . . ., an such that 1/a31 + · · · + 1/a3n = 1. Now, given an integer n ≥ 412 and a positive real number ε, choose a positive integer a > 1/ε and set xk = 1/(aak ), k = 1, . . . , n, and xn+1 = 1/a, to obtain n + 1 positive rational numbers less than ε such that x31 + · · · + x3n = x3n+1 . Problem 3. Given a triangle A0 A1 A2 , determine the locus of the centers of the equilateral triangles X0 X1 X2 satisfying the condition that each of the lines Xk Xk+1 passes through Ak (all indices are reduced modulo 3). ***

Solution. Erect the three outer Napoleon triangles associated with the triangle A0 A1 A2 , and let γk , k = 0, 1, 2, be the corresponding circumcircles. From any point X0 on γ0 draw lines X0 A2 X1 and X0 A1 X2 , where X1 lies on γ1 and X2 lies on γ2 . The points X1 , A0 , X2 are collinear, and the triangle X0 X1 X2 is an equilateral triangle satisfying the conditions in the statement.

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Selection tests for the 2014 BMO and IMO

X1 A0

X M1 M0

X2

A1

A2 X0

Let Mk be the midpoint of the minor arc Ak+1 Ak+2 of the circle γk , and notice that the triangle M0 M1 M2 is equilateral, since the Mk are the centers of the inner Napoleon triangles associated with the triangle A0 A1 A2 . The center X of the triangle X0 X1 X2 is the intersection of X0 M0 and X1 M1 , which must intersect at 60◦ . Since the locus of X includes the three points Mk , it turns out that the locus of X is the circle M0 M1 M2 . Similarly, another circle is obtained by starting with the inner Napoleon triangles. The required locus is a pair of circles. Problem 4. Let k be a positive integer and let m be a positive odd integer. Show that there exists a positive integer n such that mn + nm has at least k distinct prime factors. ***

Solution. Design a set of k primes p1 < p2 < · · · < pk as follows. Begin by choosing p1 > 2m. Having selected pj , use Dirichlet’s theorem to choose a prime pj+1 ≡ −1 (mod p1 (p1 − 1)p2 (p2 − 1) · · · pj (pj − 1)). If i < j, then pi < pj , so pj does not divide pi − 1; further, pj − 1 ≡ −2 (mod pi (pi − 1)), so pi does not divide pj − 1. Next, use the Chinese Remainder Theorem, to choose a positive integer n such that n ≡ −1 (mod p1 p2 · · · pk ) and n ≡ 0 (mod (p1 − 1) · · · (pk − 1)). Finally, since p1 p2 · · · pk and m are coprime, and (p1 − 1)(p2 − 1) · · · (pk − 1) divides n, and m is odd, Euler’s Theorem applies to show that mn + nm ≡ 1 + (−1)m ≡ 0 (mod p1 p2 · · · pk ). The conclusion follows.

50

Selection tests for the 2014 BMO and IMO

Problem 5. Let n be an integer greater than 1 and let S be a finite set containing more than n + 1 elements. Consider the collection of all sets A of subsets of S satisfying the following two conditions: (a) Each member of A contains at least n elements of S; and (b) Each element of S is contained in at least n members of A. Determine maxA minB |B|, as B runs through all subsets of A whose members cover S, and A runs through the above collection. ***

Solution. The required number is m = |S| − n. We begin by showing that any set A of subsets of S satisfying the two conditions in the statement has a subcover of cardinality at most m. This is clear if S is a member of A. Assume henceforth that A does not contain S. If some member A of A has more than n elements, for each element of S \ A choose a containing member of A. The latter along with A form a subcover of A of cardinality |S \ A| + 1 ≤ |S| − (n + 1) + 1 = m. Assume henceforth that each member of A has exactly n elements. Fix a member A of A. If some member B of A contains more than one element of S \ A, for each element of S \ (A ∪ B) choose a containing member of A. The latter along with A and B form a subcover of A of cardinality |S \ (A ∪ B)| + 2 = |S \ A| − |(S \ A) ∩ B| + 2 ≤ |S \ A| = m. Finally, if no member of A contains more than one element of S \ A, write S \ A = {x1 , x2 , . . . , xm }, choose a member A1 of A containing x1 and notice that A \ A1 is a singleton set, say A \ A1 = {x}. Since x2 is contained in at least n members of A, each of which contains (exactly) n − 1 elements of A, we may choose a member A2 of A containing both x and x2 (recall that n ≥ 2). If n ≥ 3, continue choosing members Ai of A containing xi , i = 3, . . . , n, to form an m-element subcover of A consisting of A1 , A2 , . . ., Am . To complete the proof, we produce a set of subsets of S satisfying the two conditions in the statement, no subcover of which has less than m members. To this end, write S = {1, 2, . . . , m + n}, m ≥ 2, and let S1 , S2 , . . ., Sn be the (n − 1)-element subsets of the upper part {m + 1, m + 2, . . . , m + n} of S. The

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sets Si,j = Sj ∪ {i}, i = 1, 2, . . . , m, j = 1, 2, . . . , n, satisfy both conditions in the statement and at least m of them are needed to cover S. (The condition m ≥ 2 is required for an element in the upper part to lie in at least n of these sets.) SECOND SELECTION TEST

Problem 6. Let ABC be a triangle and let X, Y , Z be interior points on the sides BC, CA, AB, respectively. Show that the magnified image of the triangle XY Z under a homothety of factor 4 from its centroid covers at least one of the vertices A, B, C. ***

Solution. Since the problem is of an affine nature, we may (and will) assume that the triangle XY Z is equilateral. The triangle ABC has at least one vertex angle, say at A, greater than or equal to 60◦ , so A is covered by the closed circumdisc OY Z, where O is the center of the triangle XY Z. Since the latter is covered by the 4-fold blow-up of the triangle XY Z from O, the conclusion follows. Alternative Solution. Suppose, if possible, that none of the vertices A, B, C is covered by the 4-fold blow-up of the triangle XY Z from its centroid. Then the distance of the point A to the line Y Z is greater than the distance of the point X to this line, so the area of the triangle AY Z is greater than the area of the triangle XY Z. Similarly, the triangles BZX and CXY both have an area greater than that of the triangle XY Z, in contradiction with the well known fact that of the four triangles AY Z, BZX, CXY , XY Z, the latter has not the smallest area. Problem 7. Let a be a real number in the open interval (0, 1), let n be a positive integer and let fn : R → R, fn (x) = x + x2 /n. Show that an + a2 a(1 − a)n2 + 2a2 n + a3 . < (fn ◦ · · · ◦ fn )(a) < 2 2 2    (1 − a) n + a(2 − a)n + a (1 − a)n + a n

***

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Solution. Let ak = (fn ◦ · · · ◦ fn )(a), k ∈ N, and notice that    k

1/ak+1 = 1/ak − 1/(ak + n),

to deduce that 1/an = 1/a −

n−1 k=0

k ∈ N,

1/(ak + n), so

1/a − n/(a + n) < 1/an < 1/a − n/(an + n),

(∗)

since the ak form an increasing sequence of positive real numbers. The first inequality above yields the required upper bound, an <

an + a2 . (1 − a)n + a

Plugged into the rightmost expression in (∗), this upper bound yields the required lower bound, an >

a(1 − a)n2 + 2a2 n + a3 . (1 − a)2 n2 + a(2 − a)n + a2

Problem 8. Determine all positive integers n such that all positive integers less than n and coprime to n be powers of primes. ***

Solution. Let p1 = 2 < p2 = 3 < p3 = 5 < · · · be the sequence of primes and let q and r, q < r, be the first two primes which do not divide n. A necessary and sufficient condition that n be of the required type is that n < qr. Each of the primes less than r and different from q divides n, and so does their product. Therefore the product of all primes less than r does not exceed nq < q 2 r. If r = pm , then q ≤ pm−1 , so p1 p2 · · · pm−2 < pm−1 pm . Notice that 6 is the first index k such that p1 p2 · · · pk−2 > pk−1 pk . Now, if p1 p2 · · · pk−2 > pk−1 pk for some index k ≥ 6, then (by Bertrand-Tchebysheff) p1 p2 · · · pk−1 > p2k−1 pk > 2pk−1 · 2pk > pk pk+1 , so p1 p2 · · · pk−2 > pk−1 pk for all indices k ≥ 6. Consequently, m ≤ 5, r = pm ≤ p5 = 11, q ≤ p4 = 7, and n < qr ≤ p4 p5 = 7 · 11 = 77. Examination of the integers less than 77 quickly yields the required numbers: 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 18, 20, 24, 30, 42, 60.

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Problem 9. Let f be the function of the set of positive integers into itself, defined by f (1) = 1, f (2n) = f (n) and f (2n + 1) = f (n) + f (n + 1). Show that, for any positive integer n, the number of positive odd integers m such that f (m) = n is equal to the number of positive integers less than and coprime to n. BMO 2014 Short List

Solution. With reference to the recurrence for f , notice that if n is a positive even, respectively odd, integer, then f (n) < f (n + 1), respectively f (n) ≥ f (n + 1), so f (n) < f (n + 1) if and only if n is even. With reference again to the recurrence for f , an easy induction shows f (n) and f (n + 1) coprime for each positive integer n. Discarding the trivial case n = 1, given a positive integer n ≥ 2, it follows that if m is a positive odd integer such that f (m) = n, then f (m − 1) is a positive integer less than and coprime to n. Next, we prove that for every pair of coprime positive integers (k, n) there exists a unique positive integer m such that k = f (m) and n = f (m + 1). If, in addition, k < n, then m is even by the preceding, so m + 1 is a positive odd integer such that f (m + 1) = n and the conclusion follows. To prove the above claim, proceed by induction on k + n. The base case, k + n = 2, i.e. k = n = 1, is clear. If k + n > 2, apply the induction hypothesis to the pair (k, n−k) or (k −n, n), according as to k < n or k > n. In the former case, k = f (m) = f (2m) and n = k + f (m + 1) = f (m) + f (m + 1) = f (2m + 1) for some positive integer m; in the latter, n = f (m + 1) = f (2m + 2) and k = f (m) + n = f (m) + f (m + 1) = f (2m + 1) for some positive integer m. This establishes the existence of the desired positive integer. To prove uniqueness, write k = f (m) and n = f (m + 1) for some positive integer m, and consider again the two possible cases. If k < n, then m is even, say m = 2m′ , where m′ is a positive integer, so k = f (2m′ ) = f (m′ ) and n − k = f (2m′ + 1) − f (m′ ) = f (m′ + 1). The induction hypothesis applies to the pair (k, n − k) to imply uniqueness of m′ , hence uniqueness of m. If k > n, then m is odd, say m = 2m′ +1, where m′ is a non-negative integer, so k − n = f (2m′ + 1) − f (2m′ + 2) = f (m′ ) + f (m′ + 1) − f (m′ + 1) = f (m′ ) and n = f (2m′ + 2) = f (m′ + 1). The induction hypothesis applies now to the

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pair (k − n, n) to imply uniqueness of m′ , hence again uniqueness of m. This completes the induction step and ends the proof. THIRD SELECTION TEST

Problem 10. Let ABC be an isosceles triangle, AB = AC, and let M and N be points on the sides BC and CA, respectively, such that the angles BAM and CN M are equal. The lines AB and M N meet at P . Show that the internal angle bisectors of the angles BAM and BP M meet at a point on the line BC. Bogdan Enescu

C

D

M N

P

A

I

B

Solution. Denote I the intersection of the bisector of ∠BAM with BC and denote D the reflection of A about BC. Then ∠BM D ≡ ∠BM A ≡ ∠CM N , so P , M , D are collinear. On the other hand, DI is the bisector of ∠BDM – the reflection of ∠BAM – and BI is the bisector of ∠ABD, therefore I is the incenter of triangle P BD, whence the conclusion. Problem 11. For every positive integer n, let σ(n) denote the sum of all positive divisors of n (1 and n, inclusive). Show that a positive integer n, which has at most two distinct prime factors, satisfies the condition σ(n) = 2n − 2 if   and only if n = 2k 2k+1 + 1 , where k is a non-negative integer and 2k+1 + 1 is prime. ***

Solution. Sufficiency is a routine verification. To prove necessity, assume first that n = 2k pl , where k and l are non-negative integers, and p is an odd prime. Notice that l must be positive, so 1+

p − p1l σ(pl ) p 1 ≤ < , = p pl p−1 p−1

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whence (since σ is multiplicative) � � � k+1 � � � p σ(n) 1 2 ≤ l = 2k+1 − l < 2k+1 − 1 . 2 −1 1+ p p p p−1 � � By the first inequality, 2k+1 − 1 (1 + 1/p) < 2k+1 , so p > 2k+1 − 1, i.e., p ≥ 2k+1 + 1 since p is odd. On the other hand, p < 2k+1 + 2(p − 1)/pl , by the second inequality, so 2(p − 1) > pl , and consequently l = 1 and p = 2k+1 + 1. To rule out the case n = pk q l , where p and q are distinct odd primes, and k and l are positive integers, write 2−

σ(n) σ(pk ) σ(q l ) q 2 p = = · . · < n n pk ql p−1 q−1

Alternatively, but equivalently, 1 1 2 1 + + + > 1, p − 1 q − 1 (p − 1)(q − 1) n so min(p, q) = 3, say p = 3. Then 3/(q − 1) + 4/n > 1, and it follows that q = 5 and k = l = 1, i.e., n = 15 which does not satisfy the condition σ(n) = 2n − 2. This completes the proof. Remarks. A positive integer n satisfying the condition σ(n) = 2n − 2 is called a perfect-plus-two (pp2) number. The result in the problem shows that 3 = 20 (20+1 + 1), 10 = 2 · 5 = 21 (21+1 + 1), 136 = 23 · 17 = 23 (23+1 + 1), 128 · 257 = 27 (27+1 + 1) are all pp2 integers. It can be shown that a pp2 odd integer greater than 3, if any, must have at least four distinct prime factors, and if, in addition, it is not divisible by 3, then it must have at least seven distinct prime factors. Pp1 integers, i.e., positive integers n such that σ(n) = 2n − 1, have equally well been considered, but the powers of 2 are the only ones known. Problem 12. Determine the smallest real constant c such that  2 n k n � � � 1  xj  ≤ c x2k , k j=1 k=1

k=1

for all positive integers n and all positive real numbers x1 , . . ., xn .

Dumitru Barac

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Solution. The best constant is c = 4. We first show that, if n is a positive integer and x1 , . . ., xn are positive real numbers, then n �

k=1



2 � n �2 k n � � � 2 1  xj  + xk 0. The left-hand member is a quadratic form 1)¯ xn x ¯n+1 whose discriminant is −2n and the inequality follows. in x ¯n and x To show c ≥ 4, we prove that 

2 k n � � 1 1 1  √  >4 − 24. k j=1 j k k=1 k=1 n �

√ √ Divergence of the harmonic series settles the case. Write 1/ j > 2( j + 1 − √ j), to obtain 

2 � � k �2 � 2 8 4 �√ 4 4 1 1   √ 1− √ > 2 k+1−1 > = − √ . k j=1 j k k k k k k

√ √ √ Finally, notice that 1/(2k k) < 1/ k − 1 − 1/ k, k ≥ 2, to get n �

2 1 √ ≤ 3 − √ < 3, n k k k=1 and deduce thereby the desired inequality. Problem 13. Let n be a positive integer, and let An , respectively Bn , be the set of non-negative integers k < n such that the number of distinct prime factors of gcd(k, n) is even, respectively odd. Show that |An | = |Bn | if n is even, and |An | > |Bn | if n is odd. ***

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Solution. Since gcd(k, n) depends only upon the residue class of k modulo  n, |An | − |Bn | = k (−1)s(k,n) , where s(k, n) is the number of distinct prime factors of gcd(k, n), and k ranges over any complete residue system modulo n.  We shall prove that the above sum equals n p|n (1 − 2/p), p prime, whence the conclusion; the latter is precisely the number of positive integers k < n such that k and k + 1 are both coprime to n.  In the above notation, let e(k, n) = (−1)s(k,n) and let f (n) = k e(k, n). We show that f is a numerical multiplicative function — that is, if n1 and n2 are coprime positive integers, then f (n1 n1 ) = f (n1 )f (n2 ). If n1 , n2 are coprime positive integers, then e(k, n1 n2 ) = e(k, n1 )e(k, n2 ). Further, if ki ranges once over a complete residue system modulo ni , i = 1, 2, then k = k1 n2 + k2 n1 ranges once over a complete residue system modulo n1 n2 , and e(k, ni ) = e(ki , ni ), i = 1, 2. Hence e(k, n1 n2 ) = e(k, n1 )e(k, n2 ) = e(k1 , n1 )e(k2 , n2 ), and   f (n1 n2 ) = e(k, n1 n2 ) = e(k1 , n1 )e(k2 , n2 ) = f (n1 )f (n2 ). k

k1

k2

Finally, if p is a prime, and m is a positive integer, then f (pm ) equals the number of k’s coprime to p, which is pm − pm−1 , minus the number of k’s divisible by p, which is pm−1 , so f (pm ) = pm (1 − 2/p). This ends the proof. FOURTH SELECTION TEST

Problem 14. Let ABC be an acute-angled triangle and let O be its circumcenter. The tangents of the circumcircle ABC at vertices B and C meet at P , the circle of radius P B centered at P meets the internal angle bisector of the angle BAC at point Q lying in the interior of the triangle ABC, and the lines OQ and BC meet at D. Finally, let E and F be the orthogonal projections of Q on the lines AC and AB, respectively. Prove that the lines AD, BE and CF are concurrent. Cosmin Pohoat¸˘ a

Solution. The line AB and the circle of radius P B centered at P meet again at some point R. Standard angle-chasing shows that the angle BRC is the complement of the angle BAC. Hence the lines AC and CR are perpendicular, so the lines EQ and CR are parallel, and the angles CQE and QCR are equal.

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Since the points B, Q, C, R are concyclic, the angles QBF and QCR are equal, so the angles QBF and CQE are equal and the right-angled triangles BQF and QCE are similar: BQ/QC = BF/QE = QF/CE. Recall that AQ is the internal angle bisector of the angle BAC, so EQ = F Q, whence BQ2 /CQ2 = (BF/EQ)(F Q/CE) = BF/CE.

C E Q

O

D

A F

B

P

R

On the other hand, the line OQD is the Q-symmedian of the triangle BCQ, so BQ2 /CQ2 = BD/CD. Consequently, 1=

BD CE AF BD CE · = · · , CD BF CD AE BF

on account of AQ being the internal angle bisector of the angle BAC, so AE = AF . The conclusion follows by Ceva’s Theorem. Problem 15. Given an odd prime p, determine all polynomials f and g p−1 with integral coefficients satisfying the condition f (g(X)) = k=0 X k .

Cezar Lupu, Vlad Matei

Solution. More generally, let k be an integer greater than 1 and let P be a polynomial of degree at most k−2 with integral coefficients. The polynomials f and g with integral coefficients satisfying the condition f (g(X)) = X k +X k−1 + P (X) are: f (X) = (±X ∓a)k +(±X ∓a)k−1 +P (±X ∓a) and g(X) = ±X +a, where a is an integer, and f (X) = ±X +b and g(X) = ±X k ±X k−1 ±P (X)∓b, where b is an integer — in both cases, the signs correspond to one another; for instance, f (X) = (−X + a)k + (−X + a)k−1 + P (−X + a) and g(X) = −X + a are admissible, but f (X) = (−X + a)k + (−X + a)k−1 + P (−X + a) and g(X) = X + a are not. Clearly, deg f ≥ 1, deg g ≥ 1 and the leading coefficients of f and g are ±1.

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If deg g = 1, we immediately obtain the first pair of polynomials above. n m If deg g > 1, write f (X) = i=0 ai X i and g(X) = i=0 bi X i , and notice mn + mam bm−1 bn−1 X mn−1 + · · · . Identification of that f (g(X)) = am bm nX n m−1 bn−1 = 1. The latter forces m = 1 coefficients yields am bm n = 1 and mam bn which leads to the second pair of polynomials above. p−1 Alternative Solution. The polynomial Φp (X) = k=0 X k is the p-th cyclotomic polynomial. With the sign convention in the previous solution, the required polynomials are either f (X) = ±X ∓ a and g(X) = ±Φp (X) + a, where a is an integer, or f (X) = Φp (±X ∓ b) and g(X) = ±X + b, where b is an integer. The proof relies upon the well known fact that both Φp and its (formal) p−2 derivative, Φ′p (X) = k=0 (k +1)X k , are irreducible in Z[X]. This follows from Eisenstein’s irreducibility criterion upon substitution X �→ X + 1: Φp (X + 1) =  p  k p−2 p−1  p  k ′ k=0 k+1 X and Φp (X + 1) = k=0 (k + 1) k+2 X .

Now take the derivative both sides of the relation in the statement to obtain f (g(x))g ′ (X) = Φ′p (X). Since Φ′p (X) is irreducible in Z[X] and its coefficients are obviously jointly coprime, either f ′ (g(x)) = ±1 and g ′ (X) = ±Φ′p (X) or vice versa — as before, the signs correspond to one another. With the same convention for signs, in the former case, g(X) = ±Φp (X)+a and f (X) = ±X ∓ a, where a is an integer; in the latter, g(X) = ±X + b and f (X) = Φp (±X ∓ b), where b is an integer. ′

Alternative Solution. We show that if deg g > 1, then deg g = p − 1. The remaining details are easily filled in and hence omitted. To begin, let ζ = cos(2π/p) + i sin(2π/p), let α be a (complex) root of f , and let β be a (complex) root of g − α. Since Φp (β) = f (g(β)) = f (α) = 0, it follows that β is one of the ζ j , j = 1, . . . , p − 1. In particular, g − α has no multiple roots, for Φp has no such. Now let n = deg g > 1, and let ζ k1 , . . ., ζ kn be the roots of g − α, where 1 ≤ n k1 < · · · < kn ≤ p−1. Since g has integral coefficients, a = j=1 ζ kj is integral, n kj − a is a monic non-constant by the first Vieta relation, so h = j=1 X polynomial with integral coefficients. Since Φp is the minimal polynomial of ζ and h(ζ) = 0, it follows that Φp divides h, so deg h ≥ deg Φp = p − 1. On the other hand, deg h = kn ≤ p − 1, so deg h = p − 1, and since h is monic, it

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follows that h = Φp . Hence a = −1 and kj = j, j = 1, . . . , p − 1. Consequently, the ζ j , j = 1, . . . , p − 1, are the roots of g − α and deg g = p − 1. Problem 16. Let n be a positive integer, let Sn be the set of all permutations of the set {1, 2, . . . , n}, and, for each σ in Sn , let I(σ) = {i : σ(i) ≤ i}. Evaluate the sum � � 1 (i + σ(i)). |I(σ)| σ∈Sn

i∈I(σ)

***

Solution. Consider the involution of Sn which sends a permutation σ to the permutation σ ∗ defined by σ ∗ (i) = j if and only if σ(n − j + 1) = n − i + 1. Notice that, for each σ in Sn , the assignment i �→ n−σ(i)+1 defines a bijection from I(σ) to I(σ ∗ ) whose inverse sends j to n − σ ∗ (j) + 1. Consequently, � � 1 (i + σ(i)) = |I(σ)| σ∈Sn i∈I(σ)   � � 1 1 �  1 (i + σ(i)) + (i + σ ∗ (i)) = 2 |I(σ)| |I(σ ∗ )| ∗ σ∈Sn

i∈I(σ)

i∈I(σ )

� 1 � 1 = (i + σ(i) + (n − σ(i) + 1) + (n − i + 1)) 2 |I(σ)| σ∈Sn

i∈I(σ)

= (n + 1)! . FIFTH SELECTION TEST

Problem 17. Let A0 A1 A2 be a triangle and let O be its circumcenter. The lines OAk and Ak+1 Ak+2 meet at Bk , and the tangent of the circumcircle A0 A1 A2 at Ak meets the line Bk+1 Bk+2 at Ck , k = 0, 1, 2, indices being reduced modulo 3. Show that the points C0 , C1 , C2 are collinear. ***

Solution. Let the tangents at Ak and Ak+1 meet at A′k+2 . We shall prove that the lines A′k Bk are concurrent, whence the conclusion by Desargues’ theorem. We show that the lines A′k Bk are concurrent at the centroid of the triangle A′0 A′1 A′2 . More precisely, we prove that the line A′k Bk passes through the midpoint of the segment A′k+1 A′k+2 .

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A0 A'2 C1

C2

C0 B2

O

A1 B0

B1 A2

To this end, let the parallel through O to the tangent A′k+1 A′k+2 meet the tangents A′k A′k+1 and A′k A′k+2 at Xk and Yk , respectively, and notice that: (1) the angles OXk Bk and OAk+2 Bk are equal, since the points Ak+2 , Bk , O, Xk all lie on the circle on diameter OXk ; (2) the angles OAk+2 Bk and OAk+1 Ak+2 are equal, since OAk+1 = OAk+2 ; and (3) the angles OAk+1 Ak+2 and OYk Bk are equal, since the points Ak+1 , Bk , O, Yk all lie on the circle on diameter OYk . Consequently, the angles OXk Bk and OYk Bk are equal. Since the lines Xk Yk and A′k+1 A′k+2 are parallel, and the latter is perpendicular to the line OAk = OBk , it follows that Bk is the midpoint of the segment Xk Yk , whence the conclusion. 2 Alternative Solution. We will show that k=0 (Ck Bk+1 /Ck Bk+2 ) = 1, so the conclusion follows by Menelaus’ theorem. To this end, we will prove that Ak Ak+1 Ak Bk+1 Ck Bk+1 = · . Ck Bk+2 Ak Ak+2 Ak Bk+2

(∗)

Multiplying the three yields the desired relation, since, by Ceva’s theorem, 2 k=0 (Ak Bk+1 /Ak Bk+2 ) = 1.

We now turn to prove (∗). To avoid directed angles, assume, without any loss, that the triangle A0 A1 A2 is acute-angled, fix an index k and let Ak Ak+1 ≤

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Ak Ak+2 (the case Ak Ak+2 ≤ Ak Ak+1 is dealt with similarly). Apply the law of sines in the triangles Ak Bk+1 Ck and Ak Bk+2 Ck to get Ak Bk+1 sin(∠Ck Ak Bk+1 ) Ck Bk+1 . = · Ck Bk+2 Ak Bk+2 sin(∠Ck Ak Bk+2 ) Since the angle Ck Ak Bk+1 is the supplement of the internal angle at Ak+1 of the triangle A0 A1 A2 , and the angle Ck Ak Bk+2 is equal to the internal angle at Ak+2 of the triangle A0 A1 A2 , the above ratio of sines equals Ak Ak+1 /Ak Ak+2 and (∗) follows. Remarks. The problem was inspired by the Sharyghin configuration below: Let ABC be a triangle, let P be a point on one of the bisectors of the angle BAC, let X, Y , Z be the orthogonal projections of P on the lines BC, CA, AB, respectively, and let the lines P X and Y Z meet at Q. Then the line AQ passes through the midpoint of the segment BC. The proof goes along the lines in the first solution. Problem 18. Let m be a positive integer and let A, respectively B, be an alphabet with m, respectively 2m letters. Let n be an even integer greater than or equal to 2m. Let an be the number of words of length n made of letters from A such that every letter in A occurs a positive even number of times. Let bn be the number of words of length n made of letters from B such that every letter in B occurs an odd number of times. Determine the ratio bn /an . ***

Solution. The required ratio is 2n−m . To prove this, let An and B n denote the sets of words described in the statement, so an = |An | and bn = |B n |, and think of B as an extension of A = {a1 , . . . , am } by a disjoint copy A¯ = ¯m }. {¯ a1 , . . . , a

Deletion of all bars in a word in B n produces a word in An . Now let α be a word in An and let ai occur ki times in α; the ki are positive even integers which add up to n. Since there are exactly 2ki −1 distinct ways to bar ai an odd number of times in α, the preimage of α under deletion of all bars has exactly 2k1 −1 · · · · · 2km −1 = 2n−m elements. The conclusion follows.

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Alternative Solution. The number an is the coefficient of xn /n! in the formal expansion m  � �2m � � �m x2k /(2k)! = (ex + e−x )/2 − 1 = 2−m ex/2 − e−x/2 . A(x) =  k≥1

Similarly, bn is the coefficient of xn /n! in the formal expansion 

B(x) = 

�

k≥1

2m

x2k−1 /(2k − 1)!

� �2m = 2−2m ex − e−x .

Finally, notice that A(2x) = 2m B(x) to conclude that bn /an = 2n−m . Problem 19. Given a positive integer n and an increasing real-valued function f on the closed unit interval [0, 1], determine the maximum value the sum n � f (|xk − (2k − 1)/(2n)|) k=1

may achieve subject to 0 ≤ x1 ≤ · · · ≤ xn ≤ 1.

***

Solution. Let ak = (2k − 1)/(2n), k = 1, . . . , n. The required maximum �n is k=1 f (ak ) and is achieved, for instance, at x1 = · · · = xn = 0 or at x1 = · · · = xn = 1. �n To show that k=1 f (ak ) is an upper bound for the sum under consideration subject to the given constraint, fix an n-tuple (x1 , . . . , xn ) such that 0 ≤ x1 ≤ · · · ≤ xn ≤ 1, write [n] = {1, . . . , n}, define an increasing function ϕ : [n] → [n] by ϕ(k) = max {j : aj − 1/(2n) ≤ xk }, and notice that |xk − ak | ≤ |xk − aϕ(k) | + |aϕ(k) − ak | ≤ 1/(2n) + |ϕ(k) − k|/n = a|ϕ(k)−k|+1 , for k = 1, . . . , n. We shall prove that there exists a permutation σ of [n] such that |ϕ(k) − k| + 1 ≤ σ(k) for all k, so a|ϕ(k)−k|+1 ≤ aσ(k) and the conclusion follows: n �

k=1

f (|xk − ak |) ≤

n �

k=1

f (a|ϕ(k)−k|+1 ) ≤

n �

k=1

f (aσ(k) ) =

n �

k=1

f (ak ).

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We now show by induction on n that, for any increasing function ψ : [n] → [n], there exists a permutation σ of [n] such that |ψ(k) − k| + 1 ≤ σ(k) for all k. The base case, n = 1, is clear. For the induction step, let n > 1 and distinguish two cases. If ψ(n) < n, then the restriction of ψ to [n − 1] is an increasing function of [n−1] into itself, so |ψ(k)−k|+1 ≤ σ(k), k = 1, . . . , n−1, for some permutation σ of [n − 1]. Since |ψ(n) − n| + 1 = n − ψ(n) + 1 ≤ n, the permutation σ extends to a permutation of [n] satisfying the required condition by letting σ(n) = n. If ψ(n) = n, consider the increasing function ψ ′ : [n − 1] → [n − 1] defined by ψ ′ (k) = ψ(k) if ψ(k) < n and ψ ′ (k) = n − 1 if ψ(k) = n. By the induction hypothesis, there exists a permutation π of [n − 1] such that |ψ ′ (k) − k| + 1 ≤ π(k), k = 1, . . . , n − 1, so |ψ(k) − k| + 1 ≤ |ψ ′ (k) − k| + 2 ≤ π(k) + 1, k = 1, . . . , n − 1. Finally, since |ψ(n) − n| + 1 = 1, setting σ(k) = π(k) + 1, k = 1, . . . , n − 1, and σ(n) = 1 defines the required permutation.

THE 65th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

FIRST SELECTION TEST

Problem 1. Let x, y, z > 0 be real numbers such that xyz+xy+yz+zx = 4. Prove that x + y + z ≥ 3. Lucian Petrescu



3

2

(x + y + z) x+y+z , and xy + yz + zx ≤ 3 3 3 2 s s 2 putting s = x+y+z, we get that + ≥ 4. This leads to (s − 3) (s + 6) ≥ 0, 27 3 so s ≥ 3. The equality holds for x = y = z = 1. Solution. Since xyz ≤

Problem 2. Determine all pairs (a, b) of integers which satisfy the equality 6 a+2 a+1 + =1+ . b+1 b+2 a+b+1 Lucian Dragomir

Solution. Obviously, b �=−2  and b �= −1.Adding 2 in both members of the  6 a+1 a+2 +1 + +1 =3+ , hence equality, we get b+1 b+2 a+b+1 (a + b + 3)



1 1 + b+1 b+2



=

3 (a + b + 3) . a+b+1

Case 1. If a + b + 3 = 0, then every pair (a, b) = (−3 − u, u), where u ∈ Z \ {−2, −1}, is a solution. 65

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Case 2. If a + b + 3 �= 0, then

1 3 1 + = , which leads to b+1 b+2 a+b+1

a=

b2 + 4b + 3 . 2b + 3

Since a ∈ Z, it results that 2b + 3 | b2 + 4b + 3. It follows that 2b + 3 | 3, so b ∈ {−3, −2, −1, 0}. We get one additional solution, namely (a, b) = (1, 0) . Problem 3. Consider six points in the interior of a square of side length 3. Prove that among the six points, there are two whose distance is less than 2. ***

Solution. Let us split the square into five rectangles, as shown in the figure √ above. The sizes of the three rectangles at the bottom are 1 × 3, so the length of their diagonal is 2. The sizes of the two rectangles at the top are √ 2 √    1.5 × 3 − 3 . Since 1.52 + 3 − 3 < 4, the length of their diagonal is less than 2. By the pigeonhole principle, there are two points among the six given which are situated within or on the sides of one of the five rectangles, so the distance between them is at most equal to the length of the diagonal, which is at most 2. However, the distance can be 2 only if the two points are opposite vertices of one of the bottom rectangles, which implies that one of the six points is not inside, but on one side of the square – contradiction. Remarks. 1. Splitting the squares into three 1 × 1.7 rectangles and two 1.5 × 1.3 rectangles, we may prove that the result is still true if the six given points are inside or on the sides of the square. 2. Let us notice that five points are not enough to draw the same conclusion √ (taking the vertices and center of the square, the shortest distance is 3/ 2 > 2). The strongest result known it literature is that the shortest distance (for 6 √ points) is 13/2 ≈ 9/5 < 2.

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Problem 4. Let ABCD be a quadrilateral with ∠A + ∠C = 60◦ and AB · CD = BC · AD. Prove that AB · CD = AC · BD. Leonard Giugiuc

E

D'

D

A' C

C'

A

B

Solution. Let us construct the equilateral triangle BCE in the half-plane determined by line BC and point D. BC CE AB = = and ∠DAB ≡ ∠DCE, so ∆DAB ∼ ∆DCE. Then AD CD CD AD DB Therefore = and ∠ADB ≡ ∠CDE, hence ∠ADC ≡ ∠BDE. DC DE AD AC AC It follows that ∆ADC ∼ ∆BDE, so = = , which leads to BD BE BC AC · BD = BC · AD = AB · CD. Remark. Let A′ , C ′ and D′ be the images of points A, C and D by an inversion of center B. The hypothesis reduces to A′ D′ = C ′ D′ and ∠A′ D′ C ′ = 60◦ . The equality to prove is equivalent with A′ C ′ = C ′ D′ , which is obvious. Problem 5. Let D and E be the midpoints of sides [AB] and [AC] of the triangle ABC. The circle of diameter [AB] intersects the line DE on the opposite side of AB than C, in X. The circle of diameter [AC] intersects DE on the opposite side of AC than B in Y . Prove that the orthocenter of triangle XY T lies on BC. Marius Bocanu

Solution. Let XM and Y N be altitudes in XY T and H be their intersection. Since AY ⊥ Y T , we have AY � XM . Similarly, we have AX � Y N , so AXHY is a parallelogram. Its center O lies on the midline DE, so H, which is the reflection of A about O, is situated on BC.

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A

D

X

O

E

Y

H B

C N

M T

SECOND SELECTION TEST

Problem 6. Find all positive integers a and b such that a2 + b b2 − a

and

b2 + a a2 − b

are both integers. Asian-Pacific M.O., 2002

b2 + a ∈ N∗ , hence b2 + a ≥ a2 − b, a2 − b which leads to (a + b) (a − b − 1) ≤ 0. Consequently, a = b or a = b + 1. a+1 a2 + b = ∈ N∗ if and only if a − 1 | a + 1, or a − 1 | 2, If a = b, then 2 b −a a−1 which comes to a ∈ {2, 3} . b2 + 3b + 1 a2 + b = 2 ∈ N∗ , if and only if b2 − b − 1 | If a = b + 1, then 2 b −a b −b−1 b2 + 3b + 1, which means that b2 − b − 1 | 4b + 2, hence b2 − b − 1 ≤ 4b + 2. It follows immediately that b ≤ 5. Checking these values, we find that only b = 1 and b = 2 satisfy the given condition. In conclusion, the solutions of the problem are: (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2). Solution. The assumption a ≥ b, yields

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Problem 7. Determine all real numbers x, y, z ∈ (0, 1) that satisfy simultaneously the conditions:  √ 2 2 2   (x + y )√ 1 − z ≥ z (y 2 + z 2 ) 1 − x2 ≥ x  �   2 (z + x2 ) 1 − y 2 ≥ y .

Lucian Petrescu

Solution. The first inequality is equivalent to � z2 ≤ z 1 − z2. x2 + y 2

Since z

�

1 − z2 =

�

z 2 (1 − z 2 ) ≤

1 z2 + 1 − z2 = , 2 2

z2 ≤ 2, and therefore x2 + y 2 ≥ 2z 2 . Writing the other x2 + y 2 two similar inequalities and adding them together yields 2(x2 + y 2 + z 2 ) ≥ equality must hold in all the inequalities above, 2(x2 + y 2 + z 2 ). Consequently, √ 2 . Clearly this triple satisfies all the requirements. hence x = y = z = 2 it follows that

Alternative Solution. As x, y, z ∈ (0, 1), we can write x = sin α, y = sin β, z = sin γ, with α, β, γ ∈ (0, π/2). The first inequality becomes sin2 α + sin2 β ≥ tan γ ≥ 2 sin2 γ, because it reduces to sin γ(2 sin γ cos γ − 1) = sin γ(sin 2γ − 1) ≤ 0. Adding this inequality with� the two similar �ones obtained from the second and third condition, we get 2 sin2 α ≥ 2 sin2 α. Consequently, equality must hold in all the preceding inequalities, therefore α = β = γ = π/4, which translates √ into x = y = z = 2/2. Problem 8. Let ABC be an acute triangle and D ∈ (BC) , E ∈ (AD) be mobile points. The circumcircle of triangle CDE meets the median from C of the triangle ABC at F. Prove that the circumcenter of triangle AEF lies on a fixed line. ***

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Solution. The quadrilateral EF DC is cyclic, hence ∠F ED ≡ ∠F CD. Let P be the reflection of point C in the midpoint of the segment line [AB] ; clearly AP � BC and ∠F CD ≡ ∠F P A. It follows that ∠F P A ≡ ∠F ED, which means that the quadrilateral F EAP is cyclic. This shows that the circumcenter AEF lies on the perpendicular bisector of the segment line [AP ] (which is a fixed line).

A

P E F B

D

C

Problem 9. Let n ≥ 6 be an integer. We have at our disposal n colors. We color each of the unit squares of an n × n board with one of the n colors. a) Prove that, for any such coloring, there exists a path of a chess knight from the bottom-left to the upper-right corner, that does not use all the colors. b) Prove that, if we reduce the number of colors to ⌊2n/3⌋ + 2, then the statement from a) is true for infinitely many values of n and it is false also for infinitely many values of n. Marius Bocanu

Solution. We associate coordinates to each unit square, the square in the bottom-left corner having coordinates (1, 1), while the one in the upper-right corner has coordinates (n, n). Notice the one can get from (k, ℓ) to (k + 3, ℓ + 3) in two moves: (k, ℓ) → (k + 2, ℓ + 1) → (k + 3, ℓ + 3). • n ≡ 1 (mod 3). Using moves similar to the ones described above, the 2n + 1 < ⌊2n/3⌋ + 2 < n unit path (1, 1) → · · · → (n, n) passes through 3 squares, hence it does not use all the colors, whether there are n colors, or only ⌊2n/3⌋ + 2 colors. • n ≡ 0 (mod 3). The path that starts with (1, 1) → (2, 3) → (3, 5) → (5, 4) → (6, 6) → · · · → (n, n)

Selection tests for the 2014 JBMO

71

and continues with moves similar to the ones described above, will pass through exactly 2n + 3 < ⌊2n/3⌋ + 2 ≤ n 3 squares, hence it will not pass through squares of all possible colors, and this for both the situations, a) and b); • n ≡ 2 (mod 3). The path that starts with (1, 1) → (2, 3) → (4, 2) → (3, 4) → (5, 5) → · · · → (n, n) and continues with moves similar to the ones described above, will pass through exactly 2n + 5 = ⌊2n/3⌋ + 2 < n 3 squares, hence it does not pass through squares of all possible colors, under the conditions from a). We have proven a) and we have seen that for all n ≡ 0 (mod 3) and all n ≡ 1 (mod 3) (n ≥ 6), the statement from a) remains true even if there are only ⌊2n/3⌋ + 2 colors. We prove that the statement from a) is false if we have ⌊2n/3⌋ + 2 colors and n ≡ 2 (mod 3). It remains to exhibit a coloring with ⌊2n/3⌋ + 2 colors such that any path (1, 1) → (n, n) contains squares of all the colors. We give several such examples: Example 1. We use the color 1 of N = ⌊2n/3⌋ + 2 colors (note that N is odd) for the bottom-left corner; then we use colors 2 ≤ k ≤ (N −1)/2 for all the squares to which the knight can get in k − 1 moves, but not less. We use color number N for the upper-right corner; then we use color (N + 3)/2 ≤ k ≤ N − 1 for all the squares to which the knight can get (starting from the upper-right corner) in N − k moves, but not less. It is easy to see that there is no conflict in this coloring because the squares that use the first (N − 1)/2 colors and the ones that use the last (N − 1)/2 colors are separated by the diagonal {(m, n − m + 1) | 1 ≤ m ≤ n}. Finally, we color with (N + 1)/2 all the remaining squares. The absolute value of the difference between the color numbers of two consecutive squares in the knight’s path is at most 1, hence, for getting from the

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square with color number 1 to the square with color number N it has to pass through squares of all the other colors. 8 7 6 5 4 3 2 1

5 5

5 6

5 3 3

3 3 2

1 1

2

6 5

5 3

7

5

5 5

3 2 3

3

3

4

3 5

6

7

8

Example 1 for n = 8, the smallest with n ≡ 2 (mod 3). (The squares left white are to be colored with color number 4.) Example 2. (given in the contest by Tudor Plopeanu) Consider the sequence (am )m≥1 given by: 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, . . .. Color row number k, from left to right, with ak , ak+1 , . . . , an+k−1 . It is easy to check the upper-right corner has color a2n−1 = N = ⌊2n/3⌋ + 2. From a square with color aj , the knight can jump to a square having one of the colors aj−3 , aj−1 , aj+1 , aj+3 . By the way the sequence is constructed, aj−3 , aj−1 , aj+1 , aj+3 ∈ {aj − 1, aj + 1}, so again, in its path from color 1 to color N , the knight can not possibly skip a color. Again, we show the model for n = 8: 8 7 6 5 4 3 2 1

4 3 4 3 2 3 2 1

5 4 3 4 3 2 3 2

4 5 4 3 4 3 2 3

5 4 5 4 3 4 3 2

6 5 4 5 4 3 4 3

5 6 5 4 5 4 3 4

6 5 6 5 4 5 4 3

7 6 5 6 5 4 5 4

1

2

3

4

5

6

7

8

Selection tests for the 2014 JBMO

73

Example 3. Let n = 3j + 2, j ∈ N∗ . Color the bottom-left corner with 1, and use color k for all the squares to which the knight can get in k − 1 moves, but not less. We prove that the upper-right corner has the color N = ⌊2n/3⌋+2 = 2j +3. Once we have shown this, it is clear that, again, the knight can not skip a color, so any path has to contain squares of all the N colors. For each square, the color number indicates the length of the shortest path (number of squares contained), from the square in the bottom-left corner to the given square. We prove that the shortest path to the upper-right corner contains exactly N squares. We have already seen an example of a path containing N squares. To prove that this is the shortest one possible, notice that each move changes the sum of coordinates by 1 or by 3, so in order to get from the bottom-left corner (having the sum of coordinates 2) to the upper-right corner (sum of coordinates 2n = 6j + 4), one needs at least 2j + 1 moves. But if we color the board in a chessboard pattern, the knight changes color each time it moves; the bottom-left and the upper-right corners are of the same color, so any path between the two of them will consist of an even number of moves, therefore the minimum number of moves is at least 2j + 2; the upper-right corner will have the color 2j + 3 = N . THIRD SELECTION TEST

Problem 10. Let a, b, c, d be positive real numbers so that abc + bcd + cda + dab = 4. Prove that a2 + b2 + c2 + d2 ≥ 4. ***

Solution. We successively have: 4 = abc + bcd + cda + dab = ab (c + d) + cd (a + b) ≤ a2 + b 2  2 c2 + d2  · 2 (c + d2 ) + · 2 (a2 + b2 ) = ≤ 2 2     a 2 + b2 c 2 + d2 2 2 2 2 + ≤ = (a + b ) (c + d ) · 2 2  2    a + b2 + c 2 + d 2  2 ≤ · (a + b2 ) + (c2 + d2 ). 2

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 1 3 It follows that 4 ≤ · (a2 + b2 + c2 + d2 ) , i.e. a2 + b2 + c2 + d2 ≥ 4. 2 Equality holds when a = b = c = d = 1. Alternative Solution. Adding a2 + b2 ≥ 2ab with the other five similar inequalities and using MacLaurin’s inequality we obtain 

2

a ≥4



ab 6



≥4



abc 4

2/3

≥ 4,

with equality if and only if a = b = c = d = 1. Remark. The statement of the problem is equally easy to prove for a, b, c, d not necessarily positive real numbers. Problem 11. Determine the prime numbers p and q that satisfy the equality p3 + 107 = 2q (17q + 24) . Lucian Petrescu

Solution. For q = 2 we obtain p = 5, which is a prime. For q ≥ 3, reducing modulo 4, we get p3 ≡ 3 (mod 4) , which leads to p ≡ 3 (mod 4) . The equation reduces to p3 + 125 = 34q 2 + 48q + 18, or, equivalently,     2 (p + 5) p2 − 5p + 25 = 2 q 2 + (4q + 3) .

As p2 − 5p + 25 ≡ 3 (mod 4) , it follows that p2 − 5p + 25 must have at least 2 one prime factor d ≡ 3 (mod 4) and, because of d | q 2 +(4q + 3) , it follows that d | q and d | 4q + 3. In conclusion, d | 3, which means that d = q = 3 and p = 7, which is a prime. The solutions to the equation are (p, q) ∈ {(5, 2) , (7, 3)} . Problem 12. Consider two integers n ≥ m ≥ 4 and A = {a1 , a2 , . . . , am } a subset of the set {1, 2, ..., n} such that: for all a, b ∈ A, a �= b, if a + b ≤ n, then a + b ∈ A. Prove that:

n+1 a1 + a2 + . . . + am ≥ . m 2 ***

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Solution. Assume 1 ≤ a1 < a2 < . . . < am ≤ n. For even m we can group the elements of A in pairs of the form (ai , am+1−i ), m . We prove that the sum of the numbers in each pair is at with 1 ≤ i ≤ 2 least n + 1. Assuming the contrary to be true, it would exist an i for which ai + am+1−i ≤ n. But, as i < m + 1 − i, the following i distinct numbers a1 + am+1−i < a2 + am+1−i < . . . < ai + am+1−i have to belong to the set {am+2−i , am+3−i , . . . , am }, which has only i − 1 elements, a contradiction. Adding ai + am+1−i ≥ n + 1,

for 1 ≤ i ≤

m , 2

the conclusion follows immediately. For m = 2k − 1, k > 2, it can be shown, as above, that ai + am+1−i ≥ n + 1, 1 ≤ i ≤ k − 1. n+1 . Suppose 2ak < n + 1. Consider We now prove that ak ≥ 2 ak−1 < a1 + ak−1 < a1 + ak < a2 + ak < . . . < ak−1 + ak < n + 1. It follows that a1 + ak−1 , a1 + ak < a2 + ak , . . . , ak−1 + ak must all belong to A, hence they must be equal to ak , ak+1 , . . . , am , respectively. We obtain that n+1 ≤ a1 +a2k−1 = a1 +(ak−1 +ak ) = (a1 +ak−1 )+ak = 2ak , which contradicts the assumption we have made. Problem 13. In the acute triangle ABC, with AB �= BC, let T denote the midpoint of the side [AC], A1 and C1 denote the feet of the altitudes drawn from A and C, respectively. Let Z be the point of intersection of the tangents in A and C to the circumcircle of triangle ABC, X be the point of intersection of lines ZA and A1 C1 and Y be the point of intersection of lines ZC and A1 C1 . a) Prove that T is the incircle of triangle XY Z. b) The circumcircles of triangles ABC and A1 BC1 meet again at D. Prove that the orthocenter H of triangle ABC is on the line T D. c) Prove that the point D lies on the circumcircle of triangle XY Z. Marius Bocanu

Solution. a) From AZ = ZC, it follows immediately that [ZT ] is the angle bisector of ∠AZC. Notice that ∠XAB ≡ ∠ACB ≡ ∠BC1 A1 ≡ ∠AC1 X and

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AT = C1 T, which means that X and T are on the perpendicular bisector of the segment line [AC1 ]. It follows that (XT is the angle bisector of angle Y XZ. Z

A

X

C1 H

T

D B

A1

C

Y b) We may assume that AB < BC; in this case, D is on the minor arc AB. Let O denote the circumcenter of triangle ABC and let L be the midpoint of the segment line [BH]. The radical axis of two circles being perpendicular to the line connecting the centers of the circles, it follows that BD ⊥ LO. As BD ⊥ DH, we get that LO � DH. OLHT is a parallelogram, therefore OL � HT , and now it is clear that points D, H, T are collinear. c) The quadrilateral AT DX is cyclic because ∠ADT = ∠BDA − ∠BDH = 180◦ − ∠ACB − 90◦ = 90◦ − ∠XAB = ∠AXT . It is easy to prove (in a similar manner to a)) that T Y is the perpendicular bisector of the segment line [A1 C] . From ∠CDT = ∠HDB − ∠CDB = 90◦ − ∠CAB = 90◦ − ∠BCY = ∠CY T , it follows that the quadrilateral CT DY is cyclic. This leads to ∠XDY + ∠XZY = ∠XDT + ∠T DY + ∠XZY = = 180◦ − ∠XAT + 180◦ − ∠T CY + ∠XZY = = ∠ZAT + ∠ZCT + ∠XZY = 180◦ ,

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77

which means that the quadrilateral DXZY is also cyclic. FOURTH SELECTION TEST

Problem 14. We call a composite positive integer n nice if it is possible to arrange its factors that are larger than 1 on a circle such that two neighboring numbers are not coprime. How many of the elements of the set {1, 2, 3, . . . , 100} are nice? ***

Solution. We prove that a composite number is nice if and only if it is not a product of two distinct prime numbers. If n = pq, where p, q are distinct primes, it is clear that we can not arrange p, q and pq without p and q being neighbors, therefore pq is not nice. αk 1 α2 If n is not a product of two distinct primes, then n = pα 1 p2 . . . pk , where k ≥ 1 and αi ≥ 1. A convenient way of arranging the factors of n, larger than 1, on a circle is the following: we write the numbers in a succession of the form: n, S1 , S2 , . . . , Sk−1 , Sk where: • S1 is a sequence of numbers that contains all the factors of n, other than n, that are multiples of p1 , the last one in the sequence being p1 p2 ; • S2 is a sequence of numbers that contains all the factors of n that are multiples of p2 , but not of p1 , the last number in the sequence being p2 p3 ; • S3 is a sequence of numbers that contains all the factors of n that are multiples of p3 , but are multiples of neither p1 nor p2 , the last number in the sequence being p3 p4 ; ..................................... • Sk is a sequence of numbers that contains all the factors of n that are multiples of pk , but not multiples of any of the numbers p1 , p2 , . . . , pk−1 . The set {1, 2, . . . , 100} contains 74 composite numbers, 30 of which being of the form pq, with p, q distinct primes. This leaves 44 nice numbers. Remark. A stronger fact can be proven: the factors of a nice number can actually be arranged on a circle such that, for every two neighboring numbers, the smaller one is a factor of the larger one.

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Problem 15. Let x1 , x2 . . . , x5 be real numbers. Find the least positive integer n with the following property: if there exist n distinct sums of the form xp + xq + xr (with 1 ≤ p < q < r ≤ 5) which are equal to 0, then x1 = x2 = . . . = x5 = 0. Bulgaria, 2003

Solution. We prove that the smallest convenient value of n is 7. If the numbers are 1, 1, 1, 1, −2, (or 1, 1, 1, −2, −2) there are 6 sums that are equal to 0 without the numbers themselves being 0. Therefore, knowing 6 sums (or less) to be 0 is not enough to conclude that the numbers are all 0. Now we prove that 7 is enough. Suppose we are given that 7 sums are equal to 0. These sums contain, in total, 21 terms, while there are only 5 numbers. By the Pigeonhole Principle it follows that there is a number that appears (at least) 5 times. Suppose x1 appears (at least) 5 times. x1 is involved in 6 sums, so there is only one sum, say x1 +x4 +x5 , that is not necessarily 0. All the other ones are 0: x1 + x2 + x3 = x1 + x2 + x4 = x1 + x2 + x5 = x1 + x3 + x4 = x1 + x3 + x5 = 0. The first three equations simplify to x3 = x4 = x5 , and then comparing the first and the last equation gives x2 = x3 = x4 = x5 . Finally, x1 only participates in at most 6 sums; the seventh one gives x2 = x3 = x4 = x5 = 0, and it follows immediately that x1 = x2 = x3 = x4 = x5 = 0. Problem 16. Let n ≥ 5 be an integer. Prove that n is prime if and only if for any representation of n as a sum of four positive integers n = a + b + c + d, it is true that ab �= cd.

***

Solution. The statement of the problem is equivalent to saying that an integer n ≥ 5 is composite if and only if there exists a representation of n as the sum of four positive integers such that ab = cd. (⇐) If ab = cd and n = a + b + c + d, then an = a2 + ab + ac + ad = a2 + cd + ac + ad = (a + c) (a + d) , (∗) hence n | (a + c) (a + d) . Supposing that n was prime, it would follow that n | a + c or n | a + d. but neither of these two conditions can hold because n > a + c and n > a + d.

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(⇒) If n = pq, with p, q ≥ 2, is a composite number, we can chose a = 1, b = (p − 1) (q − 1) , c = p − 1, d = q − 1. This choice can easily be guessed by writing (∗) with a = 1. Problem 17. In a circle, consider two chords [AB], [CD] that intersect at E. The lines AC and BD meet at F. Let G be the projection of E onto AC. We denote by M, N, K the midpoints of the segment lines [EF ] , [EA] , and [AD], respectively. Prove that the points M, N, K, G are concyclic. Marius Bocanu

D B E

K

M N C F

P

G

A

Solution. Let P be the midpoint of [AF ] ; points N, G, P, M are on the Euler circle of triangle AEF, which means they are concyclic (∗). As N K � DE and N M � AF, we have: ∠KN M = ∠KN E + ∠EN M = ∠DEB + ∠BAC = ∠DEB + ∠BDE = ∠ABF , and, as KP � DF and M P � AE, we have ∠KP M = 180◦ − ∠KP A − ∠M P F = 180◦ − ∠DF A − ∠EAF = ∠ABF . It follows that the quadrilateral KN P M is cyclic, hence K, N, P, M are concyclic. Combining this with (∗) gives the conclusion.

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FIFTH SELECTION TEST

Problem 18. Let n be a positive integer and x1 , x2 , . . . , xn > 0 be real numbers so that x1 + x2 + . . . + x n =

1 1 1 + 2 + ... + 2 . 2 x1 x2 xn

Show that for each positive integer k ≤ n, there are k numbers among x1 , x2 , . . . , xn whose sum is at least k. ***

Solution. Arguing by contradiction, suppose that every sum of k numbers from x1 , x2 , . . . , xn is strictly less than k. Then the numbers aj = xj + xj+1 + . . . xj+k−1 , j = 1, 2, . . . , n are also less than k (where the indices from the sums aj are considered to be taken modulo n). Adding up yields a1 + a2 + . . . + an < nk and, since a1 + a2 + . . . + an = k (x1 + x2 + . . . + xn ) , we get

But n2 ≤



n 

xk

k=1



n 

x1 + x2 + . . . + xn < n.  n n n    1 1 1 n. It follows that xk xk xk

k=1

k=1

n  1 1 xk = ≥ x2k n

k=1

k=1



k=1

n  1 xk

k=1

2

>

1 2 · n = n, n

contradiction. Alternative Solution. Suppose that x1 ≤ x2 ≤ . . . ≤ xn ; it is enough to prove that xn−k+1 + . . . + xn−1 + xn ≥ k, for each k = 1, 2, . . . , n.

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If xn−k+1 +. . .+xn−1 +xn < k, for a certain k ∈ {1, 2, .., n}, then xn−k+1 < 1, hence x1 < x2 < . . . < xn−k < 1. This leads to x1 + x2 + . . . + xn−k < n − k, so x1 + x2 + . . . + xn = (x1 + x2 + . . . + xn−k ) + (xn−k+1 + . . . + xn−1 + xn ) < n. What remains to be done in order to get the contradiction is to show that x1 + x2 + . . . + xn ≥ n. This can be achieved as follows:   n n n    1 1 2· xk = xk + 3 x2 k=1 k=1 k=1 k     n n  1 1 1 3 xk + xk + 2 ≥ = xk · xk · 2 = n. 3 xk xk k=1

k=1

Problem 19. Solve, in the positive integers, the equation 5m + n2 = 3p . ***

Solution. Obviously n is even, so let n = 2a, a ∈ N∗ . Reducing the equation p modulo 4 yields 1 = (−1) , so p is even, p = 2b, b ∈ N∗ .  b   Then we have 3 − 2a 3b + 2a = 5m , hence 

3b − 2a = 5x , 3b + 2a = 5y

where x, y ∈ N, x < y, x + y = m.

Adding the above equations, we get 2 · 3b = 5x + 5y = 5x (1 + 5y−x ) , so x = 0 and y = m. Therefore, 5m + 1 = 2 · 3b . Reducing modulo 6, it results that m is an odd number. m Suppose that b ≥ 2; working modulo 9, we get that (−4) + 1 = 0, which implies 4m ≡ 1 (mod 9). Since 43 ≡ 1 (mod 9), we have 3 | m. But m is odd, so it exists t ∈ N so that m = 6t + 3. It follows that: 2 · 3b = 5m + 1 = 2t+1 2t+1 + 1 ≡ (−1) + 1 ≡ 0 (mod 7), 56t+3 + 1 = 1252t+1 + 1 = (7·18 − 1) absurd. Hence b = 1, so m = 1, n = 2, p = 2.

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Alternative Solution. By doing the same as above, we obtain 5m + 1 = 2 · 3b , where m is an odd number. Since 3 | 5 + 1, using Lifting the Exponent Lemma, we conclude that v3 (5m + 1) = v3 (5 + 1) + v3 (m) , so v3 (m) = b − 1. Hence, m ≥ 3b−1 . If b ≥ 2, we have 3b−1 ≥ 3. Using the inequality 5t + 1 > 6t, for each t ∈ N, b−1 + 1 > 6 · 3b−1 = 2 · 3b , absurd. t ≥ 3, it follows that 2 · 3b = 5m + 1 ≥ 53 Therefore, the only possibility is b = 1, which leads to m = 1, n = 2, p = 2. Problem 20. Let O be the circumcenter of the acute triangle ABC. An arbitrary diameter intersects side [AB] in D and side [AC] in E. If F is the midpoint of [BE] and G is the midpoint of [CD] , show that ∠F OG = ∠BAC. ***

A'

A

M

D

E' O

F

E

D' D''

N

G

C

B Solution. We shall make use of the following:

Lemma. Let ABC be a triangle and M N be a chord of its circumcircle which BM CM DM EM : = : . intersects side [AB] in D and side [AC] in E. Then DN EN BN CN Proof of the lemma. Using the law of sines in triangles BDM and BDN, we have BM DM = sin(∠ABM ) sin(∠BDM )

and

DN BN = . sin(∠ABN ) sin(∠BDN )

Selection tests for the 2014 JBMO

83

Since ∠BDM + ∠BDN = 180◦ , we have sin(∠BDM ) = sin(∠BDN ), so BM sin(∠ABM ) DM = · . DN BN sin(∠ABN ) CM sin(∠ACM ) EM = · . But ∠ABM = ∠ACM Analogously, we get EN CN sin(∠ACN ) BM CM DM EM : = : . � and ∠ABN = ∠ACN, so DN EN BN CN Returning to our problem, let D′ and E ′ the reflections of D and E about O and A′ be the second intersection point of BE ′ with the circumcircle of ABC. Also, let D′′ be the intersection of lines A′ C and M N . From the lemma, we have BM CM D′′ M E ′ M DM EM : = : = ′′ : ′ . DN EN BN CN D N EN Since DM = D′ N, DN = D′ M, EM = E ′ N and EN = E ′ M, we conclude D′′ M D′ M that ′ = ′′ , hence D′ = D′′ . DN D N Since [OF ] is a midsegment of the triangle BEE ′ and [OG] is a midsegment of the triangle CDD′ , we get that OF � BA′ and OG � CA′ , so ∠F OG = ∠BA′ C = ∠BAC. Alternative Solution. Let B ′ the point diametrically opposed to B and C ′ the point diametrically opposed to C; the triangle ABC being acute, B ′ is on the minor arc AC, and C ′ is on the minor arc AB. Let X be an arbitrary point on the minor arc BC. Applying Pascal’s theorem to the hexagram ABB ′ XC ′ C shows that points O = BB ′ ∩CC ′ , D = AB ∩C ′ X, E = AC ∩B ′ X are collinear – on Pascal’s line, which is the support line of a diameter. Conversely, if a diameter intersects the sides [AB] and [AC] at D and E, respectively, then the lines B ′ E and C ′ D will meet at a point X situated on the minor arc BC (consider X only as the intersection point of the line B ′ E with the circle, and apply Pascal’s theorem; D′ = AB ∩ C ′ X will be collinear with O, E, hence, it will be the intersection of the diameter with the line AB, which means that D′ is in fact D). Now, OF is midsegment in △BB ′ E, hence ∠BOF = ∠BB ′ X; OG is midsegment in △CC ′ D, hence ∠COG = ∠CC ′ X. But clearly ∠BB ′ X +∠CC ′ X = ∠BAC and ∠BOC = 2∠BAC, therefore ∠F OG = ∠BOC − (∠BOF + ∠COG) = 2∠BAC − ∠BAC = ∠BAC,

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and we are done. Problem 21. On each side of an equilateral triangle of side n ≥ 1 consider n − 1 points that divide the sides into n equal segments. Through these points draw parallel lines to the sides of the triangles, obtaining a net of equilateral triangles of side length 1. On each of the vertices of the small triangles put a coin head up. A move consists in flipping over three mutually adjacent coins. Find all values of n for which it is possible to turn all coins tail up after a finite number of moves. Colombia 1997

Solution. We shall prove that the admissible values of n (those for which our goal can be achieved) are all the positive integers that are not divisible by 3. Obviously, such turning is possible for n = 1. For n = 2, flip each of the four 1-sided equilateral triangles once and all the coins will be tail-up. We shall use now induction of step 3. Assume that n is an admissible value. Flipping the coins of each unit sided triangle of an equilateral triangle of side length n + 3, the coins from the vertices of the big triangle will turn one time, those along the sides three times and the interior coins will turn six times each. Consequently, all the exterior coins are turned tail up and all the interior coins are heads up. But the interior coins form the net corresponding to an n - sided triangle, so the induction works. If 3 | n, then color the coins in red, yellow and blue so that any three adjacent coins have different colors. Also, any three coins in a row will have different colors. In this case the corners will all have the same color, say red. (n + 1) (n + 2) ≡ 1 (mod 3) coins, then there will be Since there are, in total, 2 exactly one more red coin than yellow or blue ones. Thus, at the beginning, the parity of the number of red heads is different than the parity of the number of yellow heads. Since each move changes the parity of the number of heads of each color, we cannot end up with the parity of red heads equal to that of yellow or blue heads, which would be the case if all coins showed tails. Thus the coins cannot all be inverted, so n is not an admissible value.

THE DANUBE MATHEMATICAL COMPETITION C˘ al˘ ara¸si, October 2013

JUNIORS

Problem 1. Find all integers n ≥ 2 for which there exist x1 , x2 , . . . , xn ∈ R such that ∗

x1 + x2 + . . . + x n =

1 1 1 + + ... + = 0. x1 x2 xn ***

Solution. Let M the set of the numbers n that satisfy the conditions above. We shall show that M = N \ {0, 1, 3} . We can observe that 2p ∈ M, for any p ≥ 1, the relations being satisfied e.g. for x1 = x2 = . . . = xp = 1 and xp+1 = xp+2 = . . . = x2p = −1. 1 1 1 + +...+ = 0, taking Moreover, if n ∈ M and x1 + x2 + . . . + xn = x1 x2 xn xn+1 = 1 and xn+2 = −1, we conclude that n + 2 ∈ M, because x1 + x2 + . . . + xn + xn+1 + xn+2 =

1 1 1 1 1 + + ... + + + = 0. x1 x2 xn xn+1 xn+2

Therefore, all that is left to be done is to find the smallest odd element of M. Suppose that 3 ∈ M ; then there exist x1 , x2 , x3 ∈ R∗ so that x 1 + x2 + x3 =

1 1 1 + + = 0. x1 x2 x3

This leads to x21 + x1 x2 + x22 = 0, which is possible if and only if x1 = x2 = 0, contradiction. It follows that 3 ∈ / M. 85

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2011 Danube Mathematical Competition

Let us search for an example to prove that 5 ∈ M. Take x √1 = x2 = x3 = −1; 3± 5 . then x4 + x5 = 3 and x4 x5 = 1, which leads to x4,5 = 2 Problem 2. Consider 64 distinct natural numbers less than or equal to 2012. Prove that among them there are four numbers, denoted by a, b, c, d, such that a + b − c − d is a multiple of 2013. ***

Solution. Using the 64 given numbers, there can be formed 2016 pairs (a, b) , with a < b; these pairs generate 2016 sums a + b which give 2016 remainders when divided by 2013. Consequently, there are two different pairs (a, b) and (c, d) having the same remainder when divided by 2013, hence 2013 | (a + b) − (c + d) . Assuming that (a, b) and (c, d) have a common component, e.g. a = c, then 2013 | b − d, and, since |b − d| ≤ 2012, we get b = d, contradiction. The conclusion now follows. Problem 3. Find all natural numbers m, n so that 85m − n4 = 4. ***

Solution. The equation can be written as 85m = (n2 − 2n + 2)(n2 + 2n + 2). Obviously, n is odd and cannot be equal to 1. Since (n2 −2n+2, n2 +2n+2) = 1, it follows that n2 − 2n + 2 = 5m and n2 + 2n + 2 = 17m . Consequently, (n − 1)2 = 5m − 1

and

(n + 1)2 = 17m − 1.

For m > 1 there are many (more than one) perfect squares between 5m − 1 and 17m − 1 (e.g. 9m and 16m ), therefore m = 1 and n = 3. Problem 4. Consider ABCD a rectangle of center O with AB �= BC. The perpendicular dropped from O to BD intersects lines AB and BC in E and F. Let M and N be the midpoints of segments [CD] and [AD] . Prove that F M ⊥ EN. ***

Solution. Let P be the midpoint of [BC] and Q be the intersection point of EO and CD. As [P M ] is the midsegment of the triangle BCD, it results

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2011 Danube Mathematical Competition

that P M � BD and OQ ⊥ BD, so QF ⊥ P M. But P C ⊥ M Q, so F is the orthocenter of the triangle M P Q, which leads to P Q ⊥ M F. Since the quadrilateral EN QP is a parallelogram, we have P Q � EN. Consequently, F M ⊥ EN.

M

D

C

Q

F N

P O

E

A

B

SENIORS

Problem 1. Given six points on a circle, A, a, B, b, C, c, show that the Pascal lines of the hexagrams AaBbCc, AbBcCa, AcBaCb are concurrent. ***

c

A

D'

F'

E' E

a Z

Y

X B D F b C

Solution. The lines Aa and bC meet at D, and the lines Bb and cA meet at D′ to determine the Pascal line of the hexagram AaBbCc; similarly, the

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2011 Danube Mathematical Competition

lines Bc and aA meet at E, and the lines Ca and bB meet at E ′ to determine the Pascal line of the hexagram AbBcCa; finally, the lines Cb and cB meet at F , and the lines Ac and aC meet at F ′ to determine the Pascal line of the hexagram AcBaCb. By Desargues’ theorem, the lines DD′ , EE ′ , F F ′ are concurrent if and only if the pairs of lines DE and D′ E ′ , EF and E ′ F ′ , F D and F ′ D′ meet at three collinear points. Since the latter lie on the Pascal line of the hexagram AcBbCa, the conclusion follows. Problem 2. Let a, b, c, n be four integers, where n ≥ 2, and let p be a prime dividing both a2 +ab+b2 and an +bn +cn , but not a+b+c ; for instance, a ≡ b ≡ −1 (mod 3), c ≡ 1 (mod 3), n a positive even integer, and p = 3 or a = 4, b = 7, c = −13, n = 5, and p = 31 satisfy these conditions. Show that n and p − 1 are not coprime. ***

Solution. Throughout the proof congruences are taken modulo p. Begin by ruling out the case p = 2. If p = 2, then a2 + ab + b2 and an + bn + cn are both even, and a + b + c is odd. The first condition forces both a and b even, so c is also even by the second, contradicting the third. Consequently, p must be odd, and the conclusion follows unless n is odd. Henceforth assume n odd. It is easily seen from the conditions in the statement that a �≡ 0, so a−1 exists modulo p, and the hypotheses yield B 2 + B + 1 ≡ 0, B n + C n + 1 ≡ 0 and B + C + 1 �≡ 0, where B = a−1 b and C = a−1 c. The first congruence yields B 3 ≡ 1, where B �≡ 1. (Otherwise, 3 ≡ 0, so p = 3, C n ≡ 1 and C �≡ 1 which is impossible since n is odd.) Hence 3 is a factor of p − 1; in particular, p ≥ 7 and p − 1 is divisible by 6, so the conclusion follows unless n = 6m ± 1. Let n = 6m ± 1 and recall that B 3 ≡ 1 to deduce that B 2n + B n + 1 ≡ n  B ±2 + B ±1 + 1 ≡ 0, so C n ≡ B 2n ≡ (−B − 1)n ; that is, −C(B + 1)−1 ≡ 1, since B + 1 �≡ 0. The condition B + C + 1 �≡ 0 shows that −C(B + 1)−1 �≡ 1, so the multiplicative order of −C(B + 1)−1 in Z∗p is a divisor d of n, greater than 1. Since d is also a divisor of p − 1, the conclusion follows. Remarks. The argument shows that if n is a prime greater than 3, then p − 1 is divisible by 6n. This is precisely the case in the second example in the statement.

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2011 Danube Mathematical Competition

Problem 3. Show that, for every integer r ≥ 2, there exists an r-chromatic simple graph (no loops, nor multiple edges) which has no cycle of less than 6 edges. ***

Solution. The case r = 2 is clear: Any cycle of even length works. In the other cases, define a sequence of graphs Gr , r ≥ 3, as follows. The graph G3 is a cycle of just 7 edges. (Any larger odd number would do.) When Gr is defined, with nr vertices say, construct Gr+1 as follows. Consider   rnr − r + 1 nr disjoint copies of Gr . Adjoin rnr − r + 1 extra vertices. Set up a one-toone correspondence between the copies of Gr and the nr -element sets of extra vertices. Join each copy of Gr to the members of the corresponding nr -element set of extra vertices by nr disjoint new edges (no two have a common end). The resulting graph is Gr+1 . The construction ensures that no graph Gr has a cycle of less than 6 edges. Clearly, G3 is 3-chromatic. If r ≥ 3 and Gr+1 has a coloring C in r or fewer colors, then some nr of the extra vertices in Gr+1 must share the same color in C, so the corresponding copy of Gr must be colored in r − 1 or fewer colors. It follows, by descending induction, that G3 must be colored in 2 or fewer colors which is, of course, impossible. Consequently, no Gr can be colored in less than r colors. This does not prove that Gr is r-chromatic, but if it is not, deletion of some monochromatic classes of vertices together with their incident edges yields one such. Problem 4. Show that there exists a proper non-empty subset S of the set of real numbers such that, for every real number x, the set {nx + S : n ∈ N} is finite, where nx + S = {nx + s : s ∈ S}. ***

Solution. Let H be a Hamel basis; that is, H is a set of real numbers such

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2011 Danube Mathematical Competition

that every real number x can uniquely be written in the form  q(x, h) · h, x=

(∗)

h∈H

where the q(x, h) are all rational and vanish for all but a finite number (depending on x) of h’s. The existence of Hamel bases can be proved via Zorn’s lemma or Zermelo’s well ordering theorem or any other statement equivalent to the axiom of choice. We are now going to prove that the set S of those real numbers x whose q(x, h) in (∗) are all integral satisfies the required condition. To this end, fix a real number x. Since the conclusion is clear if x = 0, let x be different from 0 and let m(x) be the least common multiple of the denominators of the non-vanishing q(x, h) in (∗). Finally, notice that m(x) · x is a member of S, to conclude that any set of the form nx + S, where n is a non-negative integer, must be one of the sets rx + S, r = 0, 1, . . . , m(x) − 1.

THE Tenth IMAR MATHEMATICAL COMPETITION Bucure¸sti, S. Haret National College, November 2013

Problem 1. Given a prime p ≥ 5, show that there exist at least two distinct primes q and r in the range 2, 3, . . ., p − 2 such that q p−1 �≡ 1 (mod p2 ) and rp−1 �≡ 1 (mod p2 ). ***

Solution. In what follows, all congruences are to be understood modulo p2 . An integer n coprime to p will be called proper if np−1 ≡ 1, and improper otherwise. Our solution is based on the following two simple facts: (1) An improper integer greater than 1 has at least one improper prime divisor; and (2) If k is an integer coprime to p and n is a proper integer, then kp − n is improper. The first claim follows from the fact that the product of two proper integers is again proper. For the second, notice that p does not divide knp−2 to deduce that (kp − n)p−1 ≡ np−1 − (p − 1)kpnp−2 ≡ 1 + kpnp−2 �≡ 1, so kp − n is indeed improper (since n is coprime to p, so is kp − n). Since ±1 are both proper, letting k ∈ {1, 2} and n = ±1 in (2) shows that p ± 1 and 2p ± 1 are all improper, so each has an improper prime divisor by (1). 91

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2011 IMAR Mathematical Competition

Since p ≥ 5, the prime factors of p ± 1 are all less than p − 1; and since 2 is the highest common factor of p − 1 and p + 1, the conclusion follows, provided that 2 is proper. Otherwise, look for an improper odd prime in the required range. To this end, notice that one of the numbers 2p ± 1 is divisible by 3, so its prime factors are all less than p − 1, for p ≥ 5; clearly, they are all odd, and the conclusion follows. Alternative Solution. If p = 5, the primes q = 2 and r = 3 satisfy the required conditions, so let p ≥ 7. In the setting of the previous solution, distinguish the following two cases: If p−2 is improper, it has an improper prime divisor q by (1). On the other hand, since 1 is proper, setting k = n = 1 in (2) shows that p − 1 is improper, so it has an improper prime divisor r by (1). Clearly, q and r both lie in the required range, and they are distinct since p − 2 and p − 1 are coprime. If p − 2 is proper, set k = 1 and n = p − 2 in (2) to deduce that 2 is improper. On the other hand, since (p − 2)2 is proper, so is −4p + 4. Setting k = −3 and n = −4p + 4 in (2) shows p − 4 improper, so it has an improper prime divisor s by (1). Finally, since p − 4 is odd, so is s, and it should now be clear that the primes 2 and s satisfy the required conditions. Problem 2. For every non-negative integer n, let sn be the sum of the digits in the decimal expansion of 2n . Is the sequence (sn )n∈N eventually increasing? ***

Solution. The answer is in the negative. To prove this, begin by noticing that the sequence is periodic modulo 9, of period 6, the first block of values it takes on being 1, 2, 4, 8, 7, 5. Suppose, if possible, that the sequence is eventually increasing, say from some rank n0 on. Fix a non-negative integer m such that 6m ≥ n0 to write s6m+1 ≥ s6m + 1,

s6m+2 ≥ s6m+1 + 2,

s6m+3 ≥ s6m+2 + 4,

s6m+4 ≥ s6m+3 + 8,

s6m+5 ≥ s6m+4 + 7,

s6m+6 ≥ s6m+5 + 5,

and deduce thereby that s6m+6 ≥ s6m + 27, so s6m+6n ≥ s6m + 27n,

n ∈ N.

(∗)

2011 IMAR Mathematical Competition

93

On the other hand, the number of non-vanishing digits in the decimal expansion of 26m+6n does not exceed ⌈(6m + 6n) log10 2⌉ < 2m + 2n, so s6m+6n ≤ 18m + 18n, contradicting (∗) for n large enough. The conclusion follows. Problem 3. The closure (interior and boundary) of a convex quadrangle is covered by four closed discs centered at each vertex of the quadrangle each. Show that three of these discs cover the closure of the triangle determined by their centers. ***

Solution. Suppose, if possible, that the conclusion does not hold. Then no three discs meet, and each disc contains points of the closure of the triangle determined by the centers of the other three discs, not covered by the latter. Amongst the four discs, choose one, say ∆0 , containing the point O where the diagonals of the quadrangle cross one another. Let A0 be the center of ∆0 , label the other three centers in circular order, A1 , A2 , A3 , so that the opposite angles A0 OA1 and A2 OA3 be not obtuse, and let ∆i denote the disc centered at Ai . Before proceeding, we take time out to state a simple, but quite useful lemma whose proof is postponed for the sake of clarity. Lemma. Let ABCD be a convex quadrangle, let ∆ be a disc centered at A, and let E be the point where the ray AC emanating from A crosses the boundary of ∆. If the orthogonal projection of B on the line AC falls on the closed ray EA emanating from E, then dist (B, [ACD] \ ∆) ≥ BE, where [ACD] is the closure of the triangle ACD. We now apply the lemma to show that O is also covered by ∆1 . To this end, let B0 be the point where the ray A0 O emanating from A0 crosses the the boundary of ∆0 . Since ∆0 contains O, the latter lies on the closed segment A0 B0 , and since the angle A0 OA1 is not obtuse, it follows that A1 O ≤ A1 B0 . On the other hand, ∆1 contains points of the closure [A0 A2 A3 ] of the triangle A0 A2 A3 not covered by ∆0 , so the radius of ∆1 is greater than or equal to dist (A1 , [A0 A2 A3 ] \ ∆0 ), which in turn is greater than or equal to A1 B0 by the lemma. Consequently, O is indeed covered by ∆1 .

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2011 IMAR Mathematical Competition

Recall that no three discs meet to deduce that neither ∆2 , nor ∆3 contains O. It follows, for i = 2, 3, that the open segment Ai O crosses the boundary of ∆i at some point Bi . The open segments A2 B3 and A3 B2 cross each other, so r2 + r3 = A2 B2 + A3 B3 < A2 B3 + A3 B2 , where ri is the radius of the disc ∆i , i = 2, 3. We are presently going to show that r2 ≥ A2 B3 and r3 ≥ A3 B2 and reach thereby the contradiction we were heading for. Only the first inequality will be dealt with; the argument applies mutatis mutandis to the other. Since the angle A2 OA3 is not obtuse, the orthogonal projection A′2 of A2 on the line A1 A3 falls on the closed ray OA3 emanating from O. If A′2 fell on the closed segment B3 O, then the image of the line A2 A′2 under a slight rotation about the midpoint of the segment A2 A′2 would separate the disc ∆3 and the closure [A0 A1 A2 ] of the triangle A0 A1 A2 , in contradiction with the second remark in the opening paragraph. Hence A′2 lies on the open ray B3 A3 emanating from B3 , so dist (A2 , [A0 A1 A3 ] \ ∆3 ) ≥ A2 B3 by the lemma. Finally, recall that ∆2 covers points in [A0 A1 A3 ] \ ∆3 , to conclude that r2 ≥ A2 B3 . Proof of the lemma. Since the quadrangle ABCD is convex, the whole configuration of points lies on one side of the line AB, say H. Let F be the point where the ray AD emanating from A crosses the boundary of ∆, let α denote the arc EF of the boundary of ∆ situated in H, and let r be the ray emanating from E along the line AC, not containing A. Notice that, if X is a point in [ACD] \ ∆, then the closed segment BX meets either α or r (this fails to hold if the quadrangle ABCD is not convex at D), so it is sufficient to consider only points X in α ∪ r. Now, as a point X traces α from E to F , the length of the segment BX varies increasingly by the cosine law in the triangle ABX (this fails to hold if the quadrangle ABCD is not convex at A or at B), so BX ≥ BE. Finally, since the orthogonal projection of B on the line AC is not interior to r, the length of the segment BX varies again increasingly, as X runs along r away from E, so BX ≥ BE again. This ends the proof of the lemma and completes the solution. Remarks. Since the distance to the empty set may take on any value, the conclusion of the lemma still holds if ∆ covers [ACD]. Under the conditions in the lemma, it may very well happen that

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2011 IMAR Mathematical Competition

dist (B, [ACD] \ ∆) > BE, in which case C is certainly interior to ∆. Such configurations are easily produced. Finally, it is not hard to see that the conclusion of the lemma may fail to hold if the quadrangle ABCD is not convex at one of the vertices A, B, D or the projection of B on the line AC does not fall on the closed ray EA emanating from E. Problem 4. Given a triangle ABC, a circle centered at some point O meets the segments BC, CA, AB in the pairs of points X and X ′ , Y and Y ′ , Z and Z ′ , respectively, labeled in circular order: X, X ′ , Y , Y ′ , Z, Z ′ . Let M be the Miquel point of the triangle XY Z (i.e., the point of concurrence of the circles AY Z, BZX, CXY ), and let M ′ be that of the triangle X ′ Y ′ Z ′ . Prove that the segments OM and OM ′ have equal lengths. ***

A Y' Z Z'

M' O

Y

M B

X

C X'

Solution. We begin by reviewing some basic facts on conics. For an ellipse Σ with center N , foci M and M ′ , semiaxes a and b, it is known that the orthogonal projections P and P ′ of M and M ′ on any line t tangent to Σ lie on the major auxiliary circle of Σ, so that N P = a = N P ′ . Application to triangle M N P (respectively, M ′ N P ′ ) of a rotation θ (respectively, −θ) about M (respectively, M ′ ) and a homothety of ratio sec θ with center M (respectively, M ′ ) yields triangle M OX (respectively, M ′ OX ′ ), where M O = M ′ O, N O = 21 · M M ′ · tan θ, OX = a sec θ = OX ′ , and X, X ′ both lie on t. If t varies and θ is constant, the locus of X and X ′ is then a circle Γ centered at O.

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2011 IMAR Mathematical Competition

By Cartesian geometry it is readily checked that Γ and Σ are bitangent, and the line ℓ supporting their common chord is also the radical axis of the circles Γ and OM M ′ , with this real geometrical significance even if the bitangency is not real. Since ℓ and the circle OM M ′ are mutually inverse in Γ, the inverse points of M and M ′ in Γ both lie on ℓ. Finally, the distance d between the parallel lines ℓ and M M ′ is given by d · M M ′ = 2b2 tan θ. Similar considerations hold for a hyperbola Σ. Consider now an isopair M , M ′ (two isogonally conjugate in the triangle ABC, the foci of a conic Σ touching its sides), and take points X, Y , Z (respectively, X ′ , Y ′ , Z ′ ) on lines BC, CA, AB, respectively, so that the lines M X, M Y , M Z (respectively, M ′ X ′ , M ′ Y ′ , M ′ Z ′ ) make the same directed angle θ (respectively, −θ) with the perpendiculars to BC, CA, AB, respectively; then the isopedal triangles XY Z, X ′ Y ′ Z ′ of angles θ, −θ for the isopair M , M ′ have their Miquel points at M , M ′ and are inscribed in a common isopedal circle Γ bitangent to Σ, centered at a point O on the perpendicular bisector of the segment M M ′ . Conversely, for any pair of triangles inscribed in a triangle ABC and in a circle Γ (as in the statement of the problem), the Miquel points M , M ′ are an isopair and Γ is an isopedal circle of M , M ′ . (If M , M ′ are the Brocard points, Σ is the Brocard ellipse and Γ is a Tucker circle.)

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