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ROMANIAN MATHEMATICALCOMPETITIONS 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 . A remarkable effort came from former multiple IMO-participants Tudor ¨ P˘adurariu and Omer Cerrahoˇ glu. eMAG Foundation and T ¸ uca Zbˆarcea & Associates. Most of them are original. Most of the problems are discussed in detail. who contributed to the training of this year Romanian IMO Team. submitted at different stages of the Romanian Mathematical Olympiad. we thank them all.FOREWORD The 21st volume of the ”Romanian Mathematical Contests” series contains more than 200 problems. and alternative solutions or generalizations are given. Some of the solutions belong to students and were given while they sat the contest. who produced the final form of the majority of the qualification tests for the IMO team. one of the editors. 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. Special thanks are due to C˘ alin Popescu. The Editors . We thank the Ministry of National Education for constant involvement in supporting the Olympiads and the participation of our teams in international events. but some problems from other sources were used as well during the competition. The printing of this volume has been supported by The Group company. Alexandru Negrescu did a great job in putting all the materials together. other Romanian contests. and some international ones. . Find. 121. we put 4 as the first and last digit and complete the decimal representation 1 . Therefore. which is the 50th number? b) Among all numbers in M. 40504. 132. Let M be the set of palindromic numbers of the form 5n + 4. hence the 50th number is the 6th five digit number. 3770773 are palindromic numbers. a) The last (and hence. (A positive integer is a palindromic number if it remains the same when its digits are reversed. from which a(b − c) = 1. 187. 20 three digit and 20 four digit numbers. where n ≥ 0 is an integer. 154. The given condition rewrites as b(10a + c) = c(10a + b) + 10. The numbers are 110. 2 two digit numbers. and hence a = b − c = 1. the first) digit of a number from M equals 4 or 9. 165.) a) If we write the elements of M in increasing order. that is. all positive integers abc satisfying b · ac = c · ab + 10. 23532. Problem 2. b) The greatest number in M has the maximum number of digits. which is the greatest one and which the smallest one? Mircea Fianu Solution. with proof. *** Solution. 143.THE 65th ROMANIAN MATHEMATICAL OLYMPIAD DISTRICT ROUND 5th GRADE Problem 1. the numbers 7. 198. 191. 176. A direct count shows that M contains 2 one digit numbers. For instance. written with nonzero digits who sum up to 2014. b) Prove that there exists a partition of A such that the sum of elements in each subset is a square. 3} ∪ {32 . 3. a10 is a cute number. 3} and every two consecutive digits differ by 1. . Problem 4. a) Prove that exactly 5 digits of a cute number are equal to 2. 33 . . . . . 2. hence exactly 5 digits are even. hence the conclusion. 32013 } ∪ {32014 }. equal to 2. obtaining thus 411 . The answer is 9899 . a) In the decimal representation of a cute number. 220 digits 9 Problem 3. . But this equals 31+2+3+···+2014 = 32015·1007 . . Then the product of all elements of A must be a square as well. . . . 1 4. Let A = {1. Csap¨ o Hajnalka Solution. b) There are 25 = 32 cute numbers of the form 2a2b2c2d2e and another 32 of the form a2b2c2d2e2. hence a possible partition is A = {1. Grouping the cute numbers in 32 such pairs. b) Observe that 32n + 32n+1 = (3n · 2)2 . that is. A 10 digit positive integer is called a cute number if its digits are from the set {1. . We obtain a partition of A if A is written as a disjoint union of nonempty subsets. . we obtain that their sum equals 32 · 4444444444 = 27 · 11 · 101010101 = 1408 · 101010101. therefore the requested number is 64. a) Assume that such a partition exists. the parity of the digits clearly alternates. 33 } ∪ · · · ∪ {32012 . Traian Preda Solution. Similarly. the smallest number in 2006 digits 1 M has the least number of digits. c) Prove that the sum of all cute numbers is divisible by 1408. a) Prove that there is no partition of A such that the product of elements in each subset is a square.2 2014 Romanian Mathematical Olympiad – District Round with 2006 digits 1. 32 . 9 89. b) Find the total number of cute numbers. obviously not a square. 32014 }. (4 − a10 ) is also cute and distinct from the previous one. They add up to 4444444444. . c) If a1 a2 . then (4 − a1 )(4 − a2 ) . b) A number n having exactly 8 divisors is either the product of two distinct primes. If the good set M has n elements. hence SB �= 2014. except for n = 8. then SB �= 2014. Therefore. or the cube of a prime. if 8 ∈ B. b) It suffices to show that 433 533 333 + 33 + 33 < 1. A short computation shows that SA = 2014. q. b) Prove that if the set B is good and 8 ∈ B. p. Relu Ciupea Solution. has exactly 4 divisors: 1. a) A straightforward computation proves the claim. a) 2 3 6 b) 333 + 433 + 533 < 633 . . 19 · 37. a) Any number equal to the product of two distinct primes p. q and pq. We call a nonempty set M good if its elements are positive integers. 29 · 37} is good and SA = 2014. 2 3 6 2 3 6 Problem 2. 33 6 6 6 that is. Prove that: 3 3 3 2 5 1 + + = 1. But 33 33 33 1 2 5 + + < 1. each having exactly 4 divisors. then SB must be odd.3 2014 Romanian Mathematical Olympiad – District Round 6th GRADE Problem 1. Damian Marinescu Solution. hence A is good. 2 3 6 33 33 33 3 3 3 1 2 5 1 2 5 + + < + + = 1. 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. It is not difficult to see that the divisor’s sum of such a number is even. 6716}. in a similar way. 6712. 2a + 1. and. which leads to a ∈ {6710. and 6713. We deduce that 2013 · 10 ≤ 5a − 2a < 2015 · 10. the line CI is the perpendicular bisector of the line segment M N . It follows that IP A = IM C. hence it contains at least 2013 and at most 2015 groups of 10 consecutive numbers. . Triangles IM C and IN C are equal. Since CM = CN and CI ⊥ M N . Prove that the angles IP A and IN C are congruent. finally. . The points M. . CA and AB of the triangle ABC such that BM = BP and CM =CN . . 6711. . and P are chosen on the sides BC. Rewrite 2 ≤ b sequence 2a. . Neculai Stanciu a ≤ 5 as 2a ≤ 10b ≤ 5a. Gabriel Popa A P B N I M C Solution. Problem 4. N. . The perpendicular dropped from B onto M P and the perpendicular dropped from C onto M N intersect at I. It follows that the Solution. . we deduce that IM B ≡ IP B. Determine all positive integers a for which there exist exactly a 2014 positive integers b such that 2 ≤ ≤ 5. we conclude that the required values of a are: 6710. 7th GRADE Problem 1. . hence IM = IN .4 2014 Romanian Mathematical Olympiad – District Round Problem 3. 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) . so IM C ≡ IN C. By inspection. B˘ atinet¸u-Giurgiu. 5a contains 2014 multiples of 10. and. Similarly. we have IM = IP . that IP A ≡ IN C. M. b D. 1. 4). Prove that one of the numbers a. ˆ = 135◦ . Multiplying the latter with the other two similar inequalities implies (a + b − c)2 (b + c − a)2 (c + a − b)2 ≤ 0. Let I ∈ (BE) such that IA bisects the angle DAB. ◦ that m(DIB) = 135 . p) ∈ {(2. (n − 2)2 . The perpenProblem 3. a) The given condition rewrites as (ab−6)2 +(a−2)2 +(b−3)2 ≥ 0. we find (n. |b − c| ≥ |a|. Squaring the first inequality gives (a − b)2 ≥ c2 . Traian Preda A E I B D C We deSolution. c is the sum of the other two. (2. and (p − 3)2 are equal to 0. b. Find the measure of BED. hence (a − b + c)(b + c − a) ≥ 0. we infer that m(BED) = 45◦ . By inspection.5 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) . It follows BE AB BI that AB IB = BD . obviously true. we obtain (np−6)2 +(n−2)2 +(p−3)2 = 5. A short computation shows duce that ID is the bisector of the angle ADB. c be real numbers such that: |a − b| ≥ |c|. George Stoica Solution. in some order. |c − a| ≥ |b|. 2)}. as well and. c equals the sum of the other two. so that EB = BD . Problem 2. and 4. b. Thus. Nicolae Papacu Solution. Let ABC be a triangle in which m(A) dicular to the line AB erected at A intersects the side [BC] at D. hence one of a. b) Similarly. hence the numbers (np − 6)2 . triangles ABI and DBE are similar = m(BAI). b. and the angle bisector of ∠B intersects the side [AC] at E. since m(BED) . Let a. hence triangles ABE and IBD are similar. the triangle M N P is right angled at N. we deduce that AL ⊥ KD. For each positive integer n we denote by p (n) the greatest square less than or equal to n.6 2014 Romanian Mathematical Olympiad – District Round Problem 4. Problem 2. △AKD ≡ △BLA. therefore BC ⊥ P N . Since P N ⊥ N M. and hence x = 13. L ∈ (BC). Prove that the lines AL and DK are perpendicular to each other. We have BC ⊥ (ABB ′ ) . Bogdan Enescu B L C M K A D Solution. Determine the length of the line segment [AP ] such that for any position of the point M ∈ [BC] . it follows that AK = BL. hence P N ⊥ N B. . Let AP = x. we consider the points P ∈ [AA′ ] and N ∈ [A′ B ′ ] such that A′ N = 3B ′ N. triangle N BP is right 2 angled at N. Damian Marinescu Solution. as well. Let ABCD be a square and consider the points K ∈ (AB). Because AB = BC. P N 2 = (18 − x) + 243 and BN 2 = 351. with the right angle at L. and since AK ⊥ BL and AD ⊥ BA. it follows that P N ⊥ (N BC) . It is not difficult to observe that △KLB ≡ △LM C. In the right parallelepiped ABCDA′ B ′ C ′ D′ . and M ∈ (CD) such that KLM is a right isosceles triangle. that is. 5 cm. hence KB = LC. But BP 2 = P N 2 + BN 2 . But then. 8th GRADE √ Problem 1. with AB = 12 3 cm and AA′ = 18 cm. we obtain BP 2 = x2 + 432. the perpendicular AS. SE. In a similar way one can show that SD ⊥ (ARP ). 24 pairs. 35. and R are the projections of point A onto the lines SB. k + 1} . . from SA ⊥ (ABC) and AB ⊥ BD. Since AM ⊥ SB. leading to a total of 84 pairs. and SF. 24. . . that is. it follows that BD ⊥ (SAB) . Petru Braica Solution. Q. N. for which p (2m + 1) · p (2n + 1) = 400. P. SD ⊥ (AN P ) and SD ⊥ (AQP ) . is erected on the hexagon’s plane. Q. it follows k 2 ≤ n ≤ (k + 1) − 1. k ∈ N. We deduce that p (n + 1) ∈ {k. R lie in the same plane. 63. therefore the points M. 2 b) Let n ∈ N. The requested set is {8. it results k ≤ 9. 201. p (2n − 1) = 400. respectively. 15. Let ABCDEF be a regular hexagon with side length a. giving 40 pairs (m. R lie in the same plane. Since BD ⊥ AB and BD ⊥ SB. in the second case we obtain 20 pairs. 2} and n ∈ {200. Problem 3. Q. N. a) Since 400 = 1 · 400 = 4 · 100 = 16 · 25. N. SD. If p (2m − 1) = 1. and in the last case. . k ≥ 2. it results AM ⊥ (SBD) . a) Prove that the points M. and let p (n) = k 2 . n) . hence m ∈ {1. p (n) 2 n = (k + 1) − 1. we obtain 1 ≤ 2m − 1 < 4 and 400 ≤ 2n − 1 < 441. SC. hence BD ⊥ AM. a) Using the Three Perpendiculars Theorem. The points M.7 2014 Romanian Mathematical Olympiad – District Round a) Find all pairs of positive integers (m. b) Find the measure of the angle between the planes (M N P ) and (ABC). Similarly. 80. We also have SD ⊥ AP. hence AM ⊥ SD. it results SB ⊥ BD. we analyze three cases. 219} . P. 2 hence k 2 < n + 1 ≤ (k + 1) . 48. At √ point A. with length 2a 3. Since n ≤ 100. . 99} . P. Then p (n + 1) 2 ∈ / N if and only if p (n) = k 2 �= 1 and p (n + 1) = (k + 1) . n ∈ N n ≤ 100 and p (n) ∗ Lucian Dragomir Solution. therefore we obtain that SD ⊥ (AM P ). n) . b) Determine the set p (n + 1) ∈ /N . with m ≤ n. k = 1. . we obtain 0 ≤ ⌊xk − xk−1 ⌋ ≤ 1. n. hence ⌊b⌋ − ⌊a⌋ − 1 ≤ ⌊b − a⌋ ≤ ⌊b⌋ − ⌊a⌋. Therefore the angle between the planes (M N P ) and (ABC) equals P AD. For the value S = p. 2. one obtains SD = 4a. Ion Pˆır¸se Solution. . hence d ⊥ AP. 2. . Thus. 2. Let a and b be arbitrary reals. n. if 1 ≤ k ≤ n − 1 − p if n − p ≤ k ≤ n . . Determine all possible values of the sum S = ⌊x2 − x1 ⌋ + ⌊x3 − x2 ⌋ + . . We have ⌊b⌋ − ⌊a⌋ − 1 < b − a < ⌊b⌋ − ⌊a⌋ + 1. . . Since d ⊥ SA and d ⊥ AD. for all k = 2. . respectively. xn are real numbers whose integer parts are 1. for xk = k. The value S = n − 1 can be obtained. 1. . Problem 4. . . . with 0 ≤ p ≤ n − 2. n − 1} . k+ 1 . . Applying this to some consecutive terms of the sequence. . . it follows that the intersection between the planes (M N P ) and (ABC) is the line d parallel to BF and passing through A. . + ⌊xn − xn−1 ⌋ . for instance. . where x1 . 0 ≤ S ≤ n − 1. . hence m(P DA) = ◦ ◦ 60 and m(P AD) = 30 . 3. . . . it results that d ⊥ (SAD) . n . Using the Pythagorean Theorem. Let n ≥ 2 be a positive integer. .8 2014 Romanian Mathematical Olympiad – District Round S P Q D E R N F M d A C B b) Because M R � BF. one can take. . x2 . We claim that the set of all possible values of S is {0. for instance. k+1 xk = k. Denote x2 + x = a and x3 + 2x2 = b. b are integers. (BE. and. and (CF intersect the circumcircle of ABC at points M. George Stoica Solution. N and P. −−→ −−→ −−→ sa · AD + sb · BE + sc · CF = 0. hence if and only if s + sa −−→ s + sc −−→ s + sc −−→ · AD + · BE + · CF = 0. finally. sa . hence x (a − 1) = b − a. F ∈ (AB) be such that EC FA DB = = . Problem 2. 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 by s. s s s or. CN A and AP B are equal. E ∈ (AC). Find the irrational numbers x with the property that x2 + x and x3 + 2x2 are integer numbers. and we have DM AD = s .2014 Romanian Mathematical Olympiad – District Round 9 9th GRADE Problem 1. 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. respectively. and sc . equivalently. hence AD = s . Since x is an irrational√number and a. . Then b − ax = x2 = a − x. Let ABC be a triangle and let the points D ∈ (BC). Denoting the areas of the triangles ABC. observe that AD + BE + CF = 0 (∗) . −−→ −→ sa s+sa s+sa − AM · AD. sb . BM C. we deduce that a = b = 1. Marin Ionescu A N E P F B C D M −−→ −−→ −−→ Solution. First. DC EA FB The halflines (AD. x = −1±2 5 . 10 2014 Romanian Mathematical Olympiad – District Round Using (∗) . 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. With the notations in the figure. the latter rewrites as −−→ −−→ (sa − sc ) · AD + (sb − sc ) · BE = 0. −−→ −−→ −−→ and the similar equalities. Problem 3. Similarly one defines the points B1 and C1 . 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 . this holds if and only if sa = sb = sc . If the parallels to AB and AC intersect the side BC at points AB and AC . it follows that A1 is the midpoint of the line segment AB AC . BE and CF of triangle ABC intersect at G. The line through P parallel to AD intersects the side BC at A1 . −−−→ −−−→ −→ But P CA + P BA = P A and adding up all similar equalities yields −→ −−→ −−→ −−→ → − v = P A + P B + P C = 3P G. −−→ −−→ and since AD and BE are non collinear vectors. not belonging to any of its medians. we have −−→ −−→ −−→ −−−→ −−−→ −−−→ −−−→ 2A1 D = A1 B + A1 C = AB B + AC C = P CB + P BC . and since P A1 is parallel to the median AD. . Let P be a point lying in the interior of the triangle. hence the conclusion. then the triangles ABC and P AB AC are similar. Prove that −−→ −−→ −−→ 3 −−→ A1 D + B1 E + C1 F = P G. The medians AD. and using |w| = w · w.11 2014 Romanian Mathematical Olympiad – District Round Problem 4. we have (k + 1) − f (k + 1) ≥ (k + 1) . or |z − 1| + |z + 1| = 2. Then f (k + 1) − 1 divides 2k. Writing the equation as |z − |z + 1||2 = |z + |z − 1||2 . 1] . It is not difficult to check that f (1) = 1 and f (2) = 3. Problem 2. and 2 2 since it is a square. Dan Negulescu Solution. n ∈ N∗ . Solve in complex numbers the equation |z − |z + 1|| = |z + |z − 1||. f (k) = 2k − 1. a contradiction. We deduce that either z + z = 2 Re z = 0. In this second case. as desired. Find all functions f : N∗ → N∗ with the properties: a) f (m + n) − 1 divides f (m) + f (n). yields the equivalent form 2 (z + z) (|z − 1| + |z + 1| − 2) = 0. 3x + 4x 7x + 24x Cristinel Mortici Solution. for some real a. b) n2 − f (n) is a square. then (k + 1) − f (k + 1) > (k + 1) − (2k + 1) = k 2 . Lucian Dragomir Solution. We deduce that f (k + 1) = 2k + 1. for all n. the sum of distances from the point z to the points −1 and 1 equals 2. in the complex plane. for any complex number w. hence f (k + 1) ≤ 2k + 1. Indeed. 10th GRADE Problem 1. hence z = ia. for all m. for all n ∈ N∗ . assume that for some k > 1. which is possible if and only if z is a real number. we deduce that. We prove inductively that f (n) = 2n − 1. Rewrite the equation as 1 1 = 4 + log1/2 1 + . If the 2 2 inequality is strict. lying on the line segment [−1. Solve in real numbers the equation 5x 25x x + log2 1 + = 4 + log1/2 1 + . x + log2 1 + x x x x (3/5) + (4/5) (7/25) + (24/25) . while the right hand side is a decreasing one. ⌊a⌋ + n} has n + 1 elements. for all y ∈ Q. for all u. It is easy to show that if k ≥ 2 is a positive integer. k=2 therefore f (s + 1) > f (s). x ∈ Q. . y ∈ Q. Problem 4. Let p and n be positive integers. and hence M = {f (x) | x ∈ S}. y ∈ Q. by setting x = y = 1 in the initial equation we obtain 3f 2 (1) − f (1) − 2 = 0. then ⌊logk ⌊x⌋⌋ = ⌊logk x⌋ . a+n]}. . The solution of this classical functional equation is f (x) = xf (1). for all x. v ∈ Q.12 2014 Romanian Mathematical Olympiad – District Round and observe that the left hand side is an increasing function. for all real x. On the other hand. ⌊a⌋ + 1. Set now x = y = 0 to get f (0) = 0 and replace x = 6y to obtain f (6y) = 6f (y). hence f (x − 6y) = f (x) − 6f (y). We conclude that the equation has at most a solution. and let a be a real number such that 1 ≤ a < a + n ≤ p. for all x. y ∈ Q. p}. On the other hand. Problem 3. This implies that f (x) = f (⌊x⌋). Replacing y with y − 3f (y) in the initial equation gives f (x − 6y) = f (x) − 2y + 2(y − 3f (y)). . which. We derived that f (x−6y) = f (x)−f (6y). Find all functions f : Q → Q such that f (x + 3f (y)) = f (x) + f (y) + 2y. s < ⌊a⌋ + n ≤ p. 3. Nicu¸sor Berbecel Solution. ∞). . Set x = y − 3f (y) to obtain f (y − 3f (y)) = −2y. a ≤ x ≤ a + n has exactly n + 1 elements. and this proves that M has exactly n + 1 elements. . yields f (u + v) = f (u) + f (v). for s ∈ S. . . thus proving our claim. for all x ∈ [1. Prove that the set ⌊log2 x⌋ + ⌊log3 x⌋ + · · · + logp x x ∈ R. and f (s+1)−f (s) = p (⌊logk (s + 1)⌋−⌊logk s⌋) ≥ logs+1 (s + 1) − logs+1 s = 1. we have s + 1 ∈ {2. and it is not difficult to guess it: x = 2. with p ≥ 2. . for u = 6y and v = x−6y. Let f (x) = k=2 ⌊logk x⌋ and let M = {f (x) | x ∈ [a. . We claim that the solutions are f1 (x) = x and f2 (x) = −2x/3. Ioan B˘ aetu p Solution. hence f (1) = 1 or f (1) = −2/3. where S = {⌊a⌋ . the function ga : R → R. h(x) = f (x) + f (4x). g(x) = f (x) + f (2x). Prove that AB �= BA. Prove that f is also continuous. is continuous. 1 1 −1 0 b) If the matrices A and B commute. Mihai Piticari 1 1 0 1 Solution. with the following property: there exists an interval I ⊂ R. A +B = 3 2 2 2 b) Let A and B be matrices from M2 (R). a) Since g and h are continuous. and f (x) = (g(x) − g(2x) + h(x)) /2.13 2014 Romanian Mathematical Olympiad – District Round 11th GRADE Problem 1. Dorel Mihet¸ Solution. as well. b) Give an example of a discontinuous function f : R → R. for any a in I. such that 2 3 2 2 . Problem 2. Vladimir Cerbu. are continuous functions. such that. such that A + B = 2 3 3 2 . it follows that f is continuous. hence |det(A + iB)|2 = det(A + iB) det(A − iB) = det(A2 + B 2 ) 2 3 = det = −5. x ∈ R. and h : R → R. 3 2 a contradiction. then A2 + B 2 = (A + iB)(A − iB). . a) Let f : R → R be a function such that g : R → R. a) Give an example of matrices A and B from M2 (R). a) An example is A = 3/2 and B = . ga (x) = f (x) + f (ax). Adrian Troie Solution. hence C commutes with all matrices in M2 (C). Problem 3. a ∈ C. −1. Prove that C commutes with all matrices from M2 (C). a contradiction. a2 a2 while in the second case X= x1 − x′2 a1 x′2 − a′2 x1 A + I2 . there exist α and α′ . x < 0. Since A and C commute. such that B = βC + β ′ I2 . (2) (a1 − a′2 )x′1 + a′1 (x′2 − x1 ) = 0. that is. The equality AX = XA implies ′ ′ a1 a2 x′1 x′2 a2 x′1 = a′1 x2 . a1 − a′2 a1 − a′2 b) We prove that C = γI2 . if X = � αA+α′ I2 . . either one of a2 . if and only if there exist two complex numbers α and α′ . but ga : R → R. AC = CA and BC = CB. for instance. for some complex number γ. if. f (x) = 0. (3) Since A �= aI2 . B and C be matrices from M2 (C). a) Let A be a matrix from M2 (C). is discontinuous at 0. a′1 is non-zero. therefore AB = BA. Similarly. we obtain � x2 a1 � X= A + x1 − x2 I2 . such that AB �= BA. In the first case. or a2 = a′1 = 0 and a1 �= a′2 . then � a) Clearly. for all a < 0. (1) (a1 − a′2 )x2 + a2 (x′2 − x1 ) = 0. x > 0. there exist β and β ′ . Conversely. A �= aI2 . such that X = αA+α′ I2 .14 2014 Romanian Mathematical Olympiad – District Round b) The function f : R → R. AX = XA. Prove that the matrix X from M2 (C) commutes with A. x = 0. for any a ∈ C. Suppose the contrary. such that A = αC + α′ I2 . is continuous on R. a 1 a2 x1 x 2 let A = and X = . 1. � � X and A commute. b) Let A. ga (x) = f (x) + f (ax) = 0. a2 �= 0. as follows: z0 = x1 and zk = max (xnk . by yn = 2−k . n ∈ N. n] → R is defined by fn (x) = arctg (⌊x⌋). k ∈ N. k ∈ N. George Stoica Solution. k ∈ N. k ∈ N∗ . for all n ∈ N. We define the decreasing sequence (yn )n∈N . for all n with nk ≤ n < nk+1 . For each positive integer n the function fn : [0. for infinitely many k’s. if zk = xnk . then zk = zk−1 /2 from some k onwards. b) Obviously. Consider the sequence of non-negative integers (nk )k∈N . hence yf (n) = zk+1 ≥ zk /2 = yn /2. zk−1 /2). Since f strictly increasing. the sequence decreases to 0. The monotony of this sequence follows inductively. converging to 0. defined by n0 = 0 and nk = f (nk−1 ). The properties of f imply that the sequence is strictly increasing. Clearly. nk+1 = f (nk ) ≤ f (n) < f (nk+1 ) = nk+2 . zk → 0. and again. k ∈ N. nk ≤ n < nk+1 . (yn )n∈N . xn ≤ xnk ≤ zk = yn . it follows that f (n) > n. such that yn ≤ 2yf (n) . we define the decreasing sequence of positive reals (zk )k∈N . 12th GRADE Problem 1.2014 Romanian Mathematical Olympiad – District Round 15 Problem 4. for nk ≤ n < nk+1 . such that xn ≤ yn ≤ 2yf (n) . n ∈ N. xn ≥ 0. In order to prove the inequalities xn ≤ yn ≤ 2yf (n) . for all n ∈ N. Let f : N → N∗ be a strictly increasing function. only for finitely many k’s. k ∈ N∗ . then there exists a decreasing sequence of real numbers (yn )n∈N . Prove that fn is a Riemann integrable function . Obviously. then zk → 0 because it is decreasing and xn → 0. Using the previously defined sequence (nk ) . Finally. Prove that: a) there exists a decreasing sequence of positive real numbers. It suffices to prove that yn ≤ 2yf (n) . for all n ∈ N. obviously convergent to 0. Moreover. hence yf (n) = 2−k−1 = yn /2. since f is strictly increasing. (zk )k∈N converges to 0: if zk = xnk . converging to 0. b) if (xn )n∈N is a decreasing sequence of real numbers. a) Since f (0) > 0 and f is strictly increasing. nk+1 = f (nk ) ≤ f (n) < f (nk+1 ) = nk+2 . converging to 0. we define the sequence (yn )n∈N by yn = zk . for all n ∈ N. it suffices to check them for nk ≤ n < nk+1 . On the other hand. The function fn is locally constant. 1]. hence Riemann integrable. Let f : [0. n. we have n fn (x)dx = 0 k−1 i+1 i=0 fn (i)dx = i n−1 arctg i. . *** Solution. ′ kf ′ (xk ) xk f ′ (xk ) x f (x ) k k ≤ ≤ . . . n(n + 1) k=1 for some xk . 1 Prove that the sequence (sn+1 − sn )n∈N∗ converges to 0 f (x)dx. and let sn = k=1 f nk . 2. 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 ). ≤ xn ≤ n/n is a tagged partition of the interval [0. 2. i=0 Applying Stolz-Ces` aro theorem. with continuous n derivative. . 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 . n→∞ n 2 Problem 2. + arctg n = lim arctg n = . n. Because 0 ≤ n+1 n(n + 1) n k=1 k=1 k=1 x1 ≤ 1/n ≤ x2 ≤ . we obtain lim n→∞ π arctg 1 + arctg 2 + . . then n+1 n(n + 1) n n n n 1 1 1 xk f ′ (xk ) ≤ kf ′ (xk ) ≤ xk f ′ (xk ). 0 *** Solution. Using the mean value theorem. .16 2014 Romanian Mathematical Olympiad – District Round and find 1 n→∞ n lim n fn (x)dx. hence If f ′ ≥ 0. Next. . k = 1. . . < xk < k = 1. k n+1 k n. . 1] → R be a derivable function. . ·) be an unit ring with the property: for all x ∈ A. b) Rewrite the given equation as It follows that x(x3 − 1)(x2 + x + 1) = 0. . a) From the given equality we derive that xn−1 = xn (x4 + x3 + x2 − x − 1) = 0 and.17 2014 Romanian Mathematical Olympiad – District Round hence the conclusion. or f (x) = x−1 for all x ∈ G. where M = sup |f ′ |. b) Prove that x4 = x. x + x2 + x3 = x4 + x5 + x6 . Let (A. . Problem 3. replace f with g : x �→ f (x) + M x. Prove that either f (x) = x for all x ∈ G. and let f : G → G be a group morphism such that f (x) ∈ {x. a) Let x ∈ A and let n ≥ 2 be an integer such that xn = 0. Problem 4. Prove that x = 0. ·) be a group with no elements of order 4. Nicolae Bourb˘ acut¸ . George Stoica Solution. +. If f ′ takes negative values. Let (G. 2 therefore the conclusion holds in this case as well. for all x ∈ G. for all x ∈ A. (x4 − x)2 = x2 (x − 1)(x3 − 1)(x2 + x + 1) = 0. = x = 0. hence x4 − x = 0. x−1 }. step by step. 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 . that xn−2 = xn−3 = . As above. which implies b−1 = ab−1 a. Then f (ab) = f (a)f (b) = ab−1 �= ab.18 2014 Romanian Mathematical Olympiad – District Round Solution. which contradicts b �= b−1 . which is impossible. It follows that ab−2 a = b−2 = (b−1 )2 = ab−1 aab−1 a. f (ab2 ) = f (a)f 2 (b) = ab−2 . Assume. that there exist a. therefore either b2 = e. hence ab−2 = b−2 a−1 . or ord(b) = 4. again a contradiction. Next. therefore f (ab) = (ab)−1 = b−1 a−1 . . If f (ab2 ) = ab2 . f (ab2 ) = (ab2 )−1 = b−2 a−1 . by way of contradiction. Thus. It follows that ab−1 = b−1 a−1 . then ab−2 = ab2 . b ∈ G such that f (a) = a �= a−1 and f (b) = b−1 �= b. so a2 = e. a) Before the last change the number must be (2020 − 16) : 4 = 501. Lucian Dragomir Solution. if p = 3k + 1. Problem 2. with integer k. b) Find all positive integers with the property that after two changes of different types. selected among the three above. 2p + 3 and 2p + 5. becomes 2014. Let the three odd consecutive numbers be 2p + 1. then 2p + 3 = 2(3k) + 3 = 6k + 3 = M3. with integer k. where p is a positive integer. In all the cases the product P is a multiple of 3. therefore P = 3a = (a − 1) + a + (a + 1). then 2p + 1 = 2(3k + 1) + 1 = 6k + 3 = M3. the second – special and the third – awesome. becomes 2020. if p = 3k + 2. a) Show that there exists a positive integer which after three changes. 19 . with integer a. the previous number must be (501 − 9) : 3 = 164 and the required number is (164 − 4) : 2 = 80. Prove that the product of every three odd consecutive positive integers can be written as the sum of three consecutive integers. Then one of these numbers is divisible by 3: if p = 3k. with integer k. Marian Ciuperceanu Solution. then 2p + 5 = 2(3k + 2) + 5 = 6k + 9 = M3.THE 65th ROMANIAN MATHEMATICAL OLYMPIAD FINAL ROUND 5th GRADE Problem 1. 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. the first – interesting. Each box has at most 10 stones. 9. 8. #7. 10. . 9. Find the maximum possible number of stones contained by the 100 boxes. Show that there exists a multiple of 2013 which ends in 2014. . One hundred boxes are labeled from 1 to 100. 9. two different as must give the same remainder in this division. . nor a multiple of 3. so that ai − aj is divisible by 2013. . 9. . 9. #4. a2014 . . . . . . the first change must be special. 10. Since 2014 is neither a multiple of 3. . 2013 must divide ai−j . 10. . 2014 ·100 i−j times 2014 4j times i−j times 2014 4j times and 104j and 2103 are relatively prime. . Gabriel Popa Solution. Since the difference of the number of stones in every two consecutive boxes is 1. Mihaela Berindeanu Solution. 10. . 10) and. two consecutive boxes contain at most 19 stones. j. with 1 ≤ j < i ≤ 2014. Consider the numbers a1 . we obtain at most 301 + 33 · 19 = 928 stones. the last change must be an interesting one. Problem 3. Since this number is odd. a special change gives a multiple of 3 and an awesome change yields a multiple of 4. . Problem 4. . a2 . then 6 groups of the form (9. . 1 ≤ k ≤ 2014.20 2014 Romanian Mathematical Olympiad – Final Round b) An interesting change produces a multiple of 2. So. 9. and the initial number is (1005 − 9) : 3 = 332. 4. 7. at the end. Since there are only 2013 possible remainders after the division by 2013. Since . 0 = ai−j · 104j ai − aj = 20142014 . 0 = 20142014 . . We know the number of stones in the 34 boxes #1. where ak is the 4k. #100 and if we group the remaining 66 boxes in pairs of consecutive boxes. This value is indeed obtained if we have 10 groups of the form (9. . The boxes labeled 1.digit number whose digits are k groups of the form 2014. 2014 00 . . 8. the second number must be (2014 − 4) : 2 = 1005. 10). 100 contain a total of 301 stones. . 10. hence there exists i.The difference of the numbers of stones for every two boxes labeled with consecutive numbers is 1. z) of positive integers of type n if x + y + z = n. we will call a triple (x. is 2-periodic.2014 Romanian Mathematical Olympiad – Final Round 21 6th GRADE Problem 1. is 1-periodic and 11 0. . 422 . a) Find all p-periodic numbers n so that the first digit of the period of n1 is not nil. m .111 . Denote A = {1000. 402 . If n ∈ A and n = p2 . that is n ≤ 10. 432 }. b) Find the largest 4-periodic prime. 412 . because 101 Problem 3. If we choose 7 or more elements of B. 402 . a) Prove that there exists no positive integer n such that s(n) = 14. 352 . We must choose among them the maximum number of pairwise prime numbers. 362 . indeed. 412 . . 1002. Find the maximum number of elements of a subset of A which contains only perfect squares pairwise relatively prime. Consider the partition of B into the sets C1 = {322 .(142857) and 1 9 = 0. 432 . whose shortest period has length p and begins immediately 1 = after the decimal point. 2014}. So. 392 . 392 }. 332 . an example is {322 . then 1000 ≤ p2 ≤ 2014.(1). 372 . For instance. . 342 . that is 32 ≤ p ≤ 44. . prime is p = 101. b) Let p be the largest 4-periodic prime. The largest subset of A whose elements are perfect squares is B = 322 . 372 . then p1 = 9999 2 This yields pm = 9999 = 3 · 11 · 101.(3). 1001. . then two of them are in the same Ci . C4 = {372 }. So we cannot take more than 6 elements. 362 . a) The first digit of the period is at least 1 if and only if n1 ≥ 10 . 382 . Marius Perianu 1 Solution. C5 = {412 }.(0099). which is acceptable. 19 = 0. . C6 = {432 }. y.090909 . . 17 = 0. 352 . 342 . Problem 2. Denote s(n) the number of triples of type n. in this case 13 = 0. so they are not co-prime. 9}. An integer n > 1 will be called p-periodic if n1 is a repeating decimal fraction. 7. it follows that n ∈ {3. 422 . Marcel Neferu Solution. the candidate for the largest such 1 = 0. . 442 . Since the prime factors of n must be different from 2 and 5. 382 . C3 = {352 }. with 1 ≤ m ≤ 9998. C2 = {332 . . 332 . 442 }. If n is a positive integer. Let us count the triples (x. ′ CQ and SCP ′ CP . y. that M QB ≡ R This yields M Q + QR = M Q + QR′ = M R′ . P. Now M Q + QR = N P + P S implies M R′ = N S ′ . In triangle ABC points M. Also . and because B. Show that if M Q + QR = N P + P S. so are M . that is n ≥ 65. Since M ≡R it follows QB ≡ RQC. Since 4028 = 22 · 1007 and 210 = 322 = 1024 > 1007 > 31. AN = CS. . then y can take only the value 1. taking into account that z ≥ 1. then triangle ABC is isosceles. R ∈ (AC) are taken such that AM = CR. . then (n−2)(n−1) ≥ 30. Problem 4.52 . where n ≥ 3 is a given integer. Since (n − 1)2 > (n − 2)(n − 1). . + (n − 3) + (n − 2) = (n − 2)(n − 1) . C are collinear. M QB ≡ RQC N P B ≡ SP C. This shows that the number of triples of type n is s(n) = 1 + 2 + 3 + . . whence the points C. Q ∈ (BC) and and S. One can assign to x any value from 1 to n − 2. N ∈ (AB).22 2014 Romanian Mathematical Olympiad – Final Round b) Find the smallest positive n so that s(n) > 2014. y can take n − 2 values: from 1 to n − 2. it follows that n − 1 ≥ 64. Mircea Fianu. 2 a) If there exists n such that s(n) = 14. Denote R′ and S ′ the reflection of R. . S are collinear. n − 1 must be at least the smallest perfect square larger than 4028. If x = 1. Nicu¸sor Berbecel Solution. z) of positive integers such that x + y + z = n. then. line BC. If x = 2. R′ . In the same way N P + P S = N P + P S ′ = N S ′ . . then y can take n − 3 values. across the ′ QC. R . Traian Preda Solution. Q. if x = n − 2. b) We have to find the smallest n such that (n − 2)(n − 1) > 4028. then (n−2)(n−1) ≤ 20 and if n ≥ 7. so RCQ ≡S = ≡R The symmetry also implies RCQ ′ ′ ′ ′ SCP implies R CQ = S CP . Then RQ = R′ Q and RQC ′ QC. respectively S. which is impossible: if n ≤ 6. Q. From (65 − 2)(65 − 1) = 4032 > 4028 follows that n = 65. then (n − 2)(n − 1) = 28. which shows that p is odd. Construct outside the square ABCD the rhombus BCM N . In triangle M BD. therefore ABC � � implies ABC ≡ ACB. Andrei Bud � � and Solution. if p and 2 q leave the same remainder mod 3.d. q. The straight lines BM and AN meet at P . then 3 | 2 (p − q) and 3 ∤ p + q.e. ′ ′ ′ 7th GRADE Problem 1. with p ≤ q. Lucian Dragomir 2 Solution. then 3 ∤ 2 (p − q) and 3 | p + q. Isosceles triangles CDM and CBM yield CM D ≡ CDM ◦ � � � B)+ CM B ≡ CBM . then q = 5. The equality can be written p + q = 2 (p − q) . if p and q 2 leave different remainders. ′ � ≡ BCS �′ . since CR = AM and CS = AN . so that p (2q + 1) + q (2p + 1) = 2 p2 + q 2 . then p and q leave remainder 1 or 2 when divided by 3. Problem 2. whence N� M S′ ≡ M S ′ R′ . Indeed. We will show that in this case the equality is impossible. A M S R N C B Q P R' S' � Therefore △N S M ≡ △R M S (case S-S-S). the sum of the angles is 180 = m(DM . Find all primes p and q. Prove that DM ⊥ CP and triangle DP M is right and isosceles. Finally. If p ≥ 5. M N = R′ S ′ . if p = 3.23 2014 Romanian Mathematical Olympiad – Final Round CR = CR′ and CS = CS ′ and. with ∠BCM an obtuse angle. Now BCS �′ ≡ BCA � This shows that AB � CS . 2 2 2 2 If n ≥ 2. the equation does not have solutions n ≥ 2.24 2014 Romanian Mathematical Olympiad – Final Round ) + m(DBM ) = m(DM ) + 45◦ + m(CBM ) = m(BDM B) + 45◦ + m(CDM ◦ ◦ B). hence 9n − 3n = 3n (3n − 1) ≥ 32n 32n − 1 = 81n − 9n . that is CP ⊥ AN . So. The point P is on the perpendicular bisector of the segment [CN ]. whence m(N P C) = 90◦ . whence m(DM B) = 45 . Denote N the midpoint of the segment [AD] and {M } = CE ∩ AB. then n2 ≥ 2n. Sorin Furtun˘ a Solution. This shows that triangle DP M is isosceles with vertex P . Problem 4. Since 81n − 9n > 23n − 17n . {P } = CN ∩ AB. Marius Perianu Solution. so it is the perpendicular bisector of the segment [DM ]. implying CP ⊥ DM . m(M P N ) = m(DM B) = 45◦ . so M PN ≡ M P C. Now △AP E ≡ △BCE (SAS). so P A = AB = BC. with hypotenuse [AB]. 2 2 Problem 3. This leads to ≡ AEP . Prove that M Q ⊥ CF . {F } = P E ∩ M N . Clearly n = 0 and n = 1 are solutions. which implies CE = P E and BEC . Construct outside the square ABCD the right isosceles triangle ABE. A D B C P N M In the isosceles triangle CDM the straight line CP is the perpendicular from C onto DM . from m(DM P ) = 45◦ ◦ follows m(DP M ) = 90 . Find all positive integers n so that 17n + 9n = 23n + 3n . Take on the straight line F P the point Q so that [CE is the bisector of the angle ∠QCB. 90 + 2m(DM Since quadrilateral ADM N is a parallelogram. Clearly P A = P B/2. ∞). c ∈ (0. E ER 1 then △CBM ∼ △ERM . This shows that ECP is a right isosceles triangle. therefore M Q � isosceles triangle. so m(CEP ) = + m(AEC) = 90◦ yields m(AEC) m(BEC) ◦ E) = 45◦ . whence M Q ⊥ CF . If we denote R the orthogonal projection of E onto AB. 8th GRADE Problem 1. FP From N P = CN follows F E = 2.2014 Romanian Mathematical Olympiad – Final Round 25 + m(AEP ) = 90◦ . so m(CP P Q E A N D M B C F Also △CEQ ≡ △P EM (AAS). hence P E = EF . a + 2 (b + c) a + 2 (b + c) 2 (a + 2b + 2c) . that is CP ⊥ CF . whence EQ = EM . Prove the inequality √ √ √ b − ca c − ab a − bc + + ≥ 0. Menelaus’ Theorem for the triangle CP E and the transversal N F yields CN FP ME N P · F E · M C = 1. This gives m(F CP ) = 90◦ . b. whence M M C = BC = 2 . 90 . so EM Q is a right ) = 45◦ = m(EP C). Let a. that is this triangle is isosceles. a + 2 (b + c) b + 2 (c + a) c + 2 (a + b) Nicolae Papacu Solution. This gives m(EQM CP . therefore CE is altitude and median in △CF P . The AM-GM inequality leads to √ a − b+c 2a − b − c a − bc 2 ≥ = . (D′ DF )) = 45 . 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) 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 ≥ 0.26 2014 Romanian Mathematical Olympiad – Final Round So. whence HDA DC. a) Extend the segment BA with AH ≡ F C. Adrian T ¸ urcanu D' C' A' B' D P HA E F C B Solution. it follows that m((D′ DE) ◦ m(F DE) = 45 . c = 2x + 2y − 3z and −10x + 5y + 5z 1 y z 2a − b − c = = + −2 . Problem 2. Take the points E ∈ (AB) and F ∈ (BC) so that AE + CF = EF . so m(HDF = [DF ]. Then a = −3x + 2y + 2z. Then [DH] ≡ ∆DCF (SAS). = 2 y x z x y z whence the conclusion. Then ∆DAH ≡ ≡ F ) = 90◦ . 2a + b + 2c = 5y and 2a + 2b + c = 5z. Since F D ⊥ DD and ED ⊥ DD . leads to ∆DHE ≡ ∆DF E (SSS) and from here m(F DE) = m(HDE) ◦ ′ ′ . 2 (a + 2b + 2c) 2 (2a + b + 2c) 2 (2a + 2b + c) Denote a + 2b + 2c = 5x. b) Compute the distance from D′ to the straight line EF . b = 2x − 3y + 2z. . The cube ABCDA′ B ′ C ′ D′ has edges of length a. Sorin Peligrad. a) Find the measure of the angle of the planes (D′ DE) and (D′ DF ). (∗) 2 If m = 1. 2n} contains five elements a < b < c < d < e so that b c a = = . c = 35. S2 and S3 cover the whole square. or mpq ≥ 4p + 2q. Since a. that is 2mpq − 4p ≥ mpq + 2q. 17. e are divisible by q. Therefore.75% of it. hence d (D′ . whence p ≥ 3. This is obtained when a. . hence m ∈ {1. d. n min = 16. c are consecutive multiples of p and c. p < q. for given p. From ec = mpq+2q = pq follows m (q − p) = 2. b and c are divisible by p and c. n + 2. B. Relation (∗) yields (p − 1) ≥ 2. Find the smallest integer n for which the set A = {n. For p = 5 and q = 7 we get a = 25. one of the discs must cover two vertices of the square. c = 24. there exists m ∈ N∗ so that c = mpq. D′ P = a 2. (p. S3 the discs. 18 }. n ∈ { 16. b = 30. then q − p = 2. b. Relation (∗) yields (p − 2) ≥ 8. The Three Perpendiculars Theorem yields D′ P ⊥ EF . Traian Preda Solution. say S1 covers A. since n ≤ a < e ≤ 2n. e = 49 and. Prove that three discs of radius 1 cannot cover entirely a square surface of side 2. from Pythagoras’ √ Theorem. so q = p + 1. d. q. So. c d e Mircea Fianu Solution. hence q = p + 2. then q − p = 1. b = 21. q) = 1 so that ac = db = ec = pq . Since there are three discs and the square has four vertices. Suppose that S1 . Problem 4. e are consecutive mpq multiples of q. C must be covered by . Problem 3. since n ≤ a < e ≤ 2n. S2 . Then [AB] is a diameter for S1 . d = 28. Condition n ≤ a < e ≤ 2n ≤ 2a implies 2a ≥ e. Let p. n = 25. EF ) = D′ P . 2}. but they can cover more than 99. Let us find the minimal value of the difference e − a.2014 Romanian Mathematical Olympiad – Final Round 27 b) Denote P the orthogonal projection of the point D onto EF . 2 If m = 2. Finally. e = 32 and. n + 1. For p = 3 and q = 4 we get a = 18. so S1 cannot cover any point from (BC] ∪ [CD] ∪ [DA). . The congruence ∆DHE ≡ ∆DF E gives DP = AD = a. d = 42. that is a = mpq − 2p and e = mpq + 2q. q ∈ N∗ . whence p ≥ 5. . . Denote ABCD the square and S1 . Obviously. so the points not covered by the three discs are inside the square CU XT . But. Let n be a natural number. S2 covers the pentagonal surface BP M XT and S3 covers the pentagonal surface DRM XU . R D A P B M XU TC We take as S1 . Denote P and R the orthogonal projections of M onto AB and onto AD. z such that x2 + y 2 + z 2 = 2n (x + y + z) . say S2 . in this case. therefore S2 must cover (BC).25% · area[ABCD]. Then S1 covers the square surface AP M R . therefore (AD) ⊂ S3 .4025. 9th GRADE Problem 1. [RS].1.9. so S3 cannot cover any point from (BC). Then S2 cannot cover any point from (AD). It is enough to prove that area[CU XT ] < 0. S2 . Denote X the point on [AC] for which XT ⊥ BC (and XU ⊥ CD). y. that is 4 2 − 2 > 1. [P T ]. This shows that [BC] must be a diameter of S2 . Let T ∈ BC and U ∈ DC be such that P T = RU = 2. Simple considerations √ √ 2 √ √ yield AP = 2. Mircea T ¸ eca Solution. we deduce that x. which is true. If n = 0. b) Take M ∈ (AC) so that AM = 2. We are left √ √ to prove BT > 1. no point from (CD) is covered – contradiction. or BT > 1. Petre Simion.28 2014 Romanian Mathematical Olympiad – Final Round one of the other discs. which is equivalent to CT < BC/20 = 0. z ∈ {0. BP = 2 − 2. BT 2 = 4 − 2 − 2 = 4 2 − 2. or 2 > 1. 1}. Find the integers x. S3 the discs of diameters [AM ]. . using the inequalities x2 ≥ x and its analogues. In this case [AD] is a diameter of S3 .9. y.92 . we get x21 + y12 + z12 = 2n−1 (x1 + y1 + z1 ) . z are even. a−1 a−1 a−1 (m + n) < N + 1. y. √ √ a)] �= [n (a − a)]. If m and n are strictly positive integers. 2}. either the three numbers are even. 2n }. Two numbers have the same fractional part if and only if their difference is √ an integer. z = 2z1 . zn ∈ {0. Vasile Pop √ Solution. na are natural numbers. we deduce that if x = 2n xn . prove that a) {m (a + b) [m (a + √ a)} �= {n (a − √ a)} . 1} . √ <n< √ . z ∈ {0. we get N 2 2 < m + n < (N + 1) . Problem 2. we get 4 x21 + x1 + y12 + y1 + z12 + 2 = 4 (x1 + y1 + z1 + 1) . z = 2z1 . N < 2 the term in the middle is natural (a is odd). the inequalities are strict. y. which is a contradiction. . then x. zn ∈ Z and x2n + yn2 + zn2 = xn + yn + zn . y = 2y1 + 1. n which are different from zero. y = 2n yn . yn . a+ a a+ a a− a a− a and thus.2014 Romanian Mathematical Olympiad – Final Round 29 If n ≥ 1. We rewrite the inequalities as N +1 N N +1 N √ <m< √ . by addition. whence xn . if we take x = 2x1 + 1. or one is even and the others are odd. a) As ma. z = 2n zn . such √ √ √ that N = [m(a + a] = [n(a − a]. y = 2y1 . because From here. for which there exist two not equal numbers m. then 2 divides x2 + y 2 + z 2 . let us as suppose that there is a natural number N . moreover. and hence. because the terms in the middle are irrational numbers. which is a contradiction. In the former case. and thus x. z ∈ {0. whence (m + n) a ∈ Z. yn . Let a be a natural odd number which is not a perfect square. then xn . y. which is absurd. Let us consider the case when the numbers x. if n = 1. using the same argument. the equality implies {m a} = √ {−n a}. For x = 2x1 . b) Again. For n > 1. Then N ≤ m(a + a) < N + 1 and √ N ≤ n(a − a) < N + 1. and thus. 30 2014 Romanian Mathematical Olympiad – Final Round Remark. whence we −→ −→ get GS + 2GR = 0. Prove that AB = AD. Prove that R ∈ (P Q) and S ∈ (AC) such that MC NC RQ SC the centroid of the triangle AM N lies on the segment [RS]. Andrei Bud Solution. DN PR AS BM = = = = k. Bogdan Enescu Solution. inscribed in the circle of diameter AC. 0 = GA + GM + GN = GA + 1+k 1+k −→ −−→ −−→ −−→ from where. We suppose that there exist the points E ∈ (CD) and F ∈ (BC) such that the lines AE and DF are perpendicular on the lines AF and BE respectively. −→ −−→ −→ −→ −−→ AF · BE = 0 ⇔ AF · (AE − AB) = 0. From the perpendicularity conditions. . GS = 1+k On the other hand. Then we have −−→ −−→ −→ −−→ −−→ −−→ −→ GP + k GQ GB + GD + k(GA + GC) = GR = 1+k 2 (1 + k) and −→ −−→ −→ GA + k GC . Let G be the centroid of the triangle AM N . We could get the problem by using Beatty Theorem. Problem 4. Consider the points M ∈ (BC). which states 1 1 + = 1 then the sets that if α and β are positive irrational numbers and α β [nα] and [nβ] determine a partition of N. Problem 3. −→ −→ −→ −−→ AE · AF = AF · AB. we have −→ −−→ −→ −→ −−→ AE · DF = 0 ⇔ AE · (AF − AD) = 0. Let P and Q be the midpoints of the diagonals BD and AC of the quadrilateral ABCD. we deduce that (1 + k) GA + GB + 2k GC + GD = 0. Let ABCD a cyclic quadrilateral. and thus G. N ∈ (CD). whence −→ −→ −→ −−→ AE · AF = AE · AD. R and S are collinear points. −−→ −−→ −−→ −−→ −→ −−→ −−→ −→ GB + k GC GD + k GC + . −−→ −−→ −−→ −−→ But ∠ADE = ∠ABF = 90◦ . Equivalently. . A(d2 ) and only to them. we have −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ AD + DE · AD = AB + BF · AB ⇔ AD2 + AD · DE = AB 2 + AB · BF . For any natural number k. 2. . n} belongs to the sets A(d). n2 Compute the sum a(k). . c) f (p1 p2 · · · pn ) = f (p1 ) + f (p2 ) + · · · + f (pn ) for any prime numbers. . a(k) = k=1 n k=1 k=1 d = n(n + 1)/2. For 1 ≤ k ≤ n2 . We will get the result by doubly-counting the elements of the set S(n) = {(k. d) | d divides k. n2 n2 It follows that the contribution of each d in the sum a(k) = |A(k)| is 1 + 1 + · · · + 1 = d.31 2014 Romanian Mathematical Olympiad – Final Round −→ −−→ −→ −−→ and thus AE · AD = AF · AB. that is AB = AD. we denote by a(k) the number of natural divisors d of k such that k ≤ d2 ≤ n2 . . A(2d). d terms n2 Thus. . b) f (p) = 1 + f (p − 1) for any prime number p. k ≤ d2 ≤ n2 . 10th GRADE Problem 1. Let n be a strictly positive integer. k=1 *** Solution. Consider the function f : N∗ → N∗ which satisfies the properties: a) f (1) = 1. and hence AD · DE = AB · BF = 0. 1 ≤ k ≤ n2 }. . not necessary distinct. d=1 Problem 2. let A(k) be the set of the natural divisors d of k such that k ≤ d2 ≤ n2 . A natural number d ∈ {1. . . . We remark that f (k + 1) − f (k) = ak ∈ {0. n ≥ 2. George Stoica Solution. using the induction hypothesis. y ∈ A. i. then n − 1 is an even number and n−1 n−1 n−1 = 1+f (2)+f = 3+f . f (n) = 1+f (n−1) = 1+f 2 2 2 2 Then f (n) = 3 + f n−1 2 ≤ 3 + 3 log2 n−1 = 3 log2 (n − 1) ≤ 3 log2 n 2 and f (n) = 3 + f n−1 2 ≥ 3 + 3 log3 n−1 3(n − 1) = 3 log3 ≥ 3 log3 n.e. n− 1. . k = 1. i. whence 2 ≤ 23 ≤ 32 . n ≥ 3. for any x. 1}. Then f (3) = 1 + f (2) = 3. Vladimir Cerbu. .32 2014 Romanian Mathematical Olympiad – Final Round Prove that 2f (n) ≤ n3 ≤ 3f (n) . Mihai Piticari not Solution. we have f (2) = 1 + f (1) = 2. . n}. k = 1. y ∈ A. 2. Moreover. . and let n = p1 p2 · · · pk be its prime factorization (not necessarily distinct factors). If n is not a prime number. is . . any function f which satisfies the conditions in the problem. as ak ∈ {0. 1}. If n is a prime number. We will prove the inequalities 3 log3 n ≤ f (n) ≤ 3 log2 n 2 by mathematical induction on n. Therefore. . for any natural number n.e. . and thus 23 < 33 = 33 . Find the number of the increasing functions f : A → A which satisfy the property |f (x) − f (y)| ≤ |x − y|. Let n be a natural number. f (n) = k k f (p1 ) + f (p2 ) + · · · + f (pk ) ≥ 3 log2 pi = log3 pi = 3 log3 n and f (n) ≤ 3 i=1 i=1 3 log2 n. k ≥ 2. 2 2 Problem 3. Let n be a strictly positive integer and let A = {1. As 2 is a prime number. 2. we have |f (x) − f (y)| ≤ |x − y| for any x. with the required property. . then an |an z n + an−1 z n−1 + · · · + a1 z + a0 | = |an z n + a0 | ≤ ≤ |an z n | + |a0 | = |an | + |a0 | = |an + a0 |. b) = Cn−1 Thus. an be complex numbers with an = � 0. ∞). . for every complex number z of modulus 1. . . a2 . ∞). for every complex number z of modulus 1. b) = 0≤a≤b≤n b−a Cn−1 = 0≤a≤b≤n n−1 k=0 k (n − k)Cn−1 . (Q) a1 = a2 = · · · = an−1 = 0 and a0 ∈ [0. is equal with the number of choices of b − a terms equal to 1 in the sum a1 + · · · + an−1 . that is b−a . 1} × · · · × {0. . Let n be an integer with n ≥ 2. . n(a. an−1 ) ∈ A × {0. Prove that the following statements are equivalent: (P) |an z n + an−1 z n−1 + · · · + a1 z + a0 | ≤ |an + a0 |. . .33 2014 Romanian Mathematical Olympiad – Final Round completely determined by the n-tuple (f (0). If we fix f (0) = a and f (n) = b. If a1 = a2 = · · · = an−1 = 0 and a0 ∈ [0. 1} which satisfies the condition f (0) + a1 + · · · + an−1 ≤ n. (Q)⇒(P). and let a0 . . the number of functions is n(a. a1 . Summing up we have a number of functions equal to n−1 k=0 k nCn−1 − n−1 k kCn−1 = n2n−1 − = n−1 k=1 n2 n−1 k=1 k−1 (n − 1)Cn−2 = − (n − 1)2n−2 = (n + 1)2n−2 Problem 4. a1 . an Dan S ¸ tefan Marinescu Solution. and thus the number of the functions f . we have b − a = a1 + · · · + an−1 . a2 . From the above. M .34 2014 Romanian Mathematical Olympiad – Final Round (P)⇒(Q). We shall prove that all such functions are of the form f (x) = −x + c. and thus the points M . that satisfy simultaneously the conditions: a) The function f 0 + f 1 is increasing. Because g1 is increasing. which implies that g(ε) = 0. summing As up. we get |an z n +a0 | ≤ |an +a0 | for every complex number z of modulus a0 and t = z n . we have g(ε) = 0. . We get that −c ≤ 0. respectively 1. For a non-negative integer n the n-th iterate of a function f : R → R is f n = f ◦ · · · ◦ f . . z ∈ C and w = a0 + an . k = 1. gm is decreasing. Let P . |w + g(ε)| ≤ |w| or |g(ε)|2 + wg(ε) + wg(ε) k ε = 0. If we denote c = an number z of modulus 1. . whence. Determine the n times continuous functions f : R → R. . As the point A belongs to both circles. O and A are collinear and M A ≥ OA = 1. and f 0 is the identity function. As g1 (x) − g1 (y) = x − y = gm (x) − gm (y). . n ∈ N. . n − 1. we have |t + c| ≤ |1 + c| for every complex 1. For any ε ∈ Un = {z ∈ C | z n = 1} we have |an + g(ε) + a0 | ≤ |an + a0 |. . we get . |g(ε)|2 +wg(ε)+ wg(ε) ε∈Un ε∈Un for every ε ∈ Un . 2. where c is a real constant. we get ak = 1 g(ε) = 0. n−1. n εk ε∈Un Thus. b) There is a positive integer m such that the function f 0 + · · · + f m is decreasing. Dorel Mihet¸ Solution. . 11th GRADE Problem 1. and thus c ∈ [0. we get that they are tangent in the point A. It is clear that such functions verify the given conditions. A be the points in the complex plane with complex coordinates t. k = 1. Let x and y be real numbers such that f (x) = f (y) and define gn = f 0 + · · · + f n . −c. ε∈Un ε∈Un ¯ = |g(ε)|2 ≤ 0. We shall first prove that f is one to one. for every point P on the unit circle. Then P M ≤ M A. therefore the unit circle is interior to the circle with center M and radius M A. that is ¯ ≤ 0. Let g(z) = an−1 z n−1 + · · · + a1 z. ∞). 2. so all its iterates of the form f 2k are increasing. (1) We shall show that a = −∞ and b = +∞. Problem 2. As gm is decreasing we deduce that m is odd and gm is constant. By the given condition we deduce that the restriction of f to the interval (a. I = R and f is the identity function. Mihai Piticari. where a. that satisfy the equality f ◦ f = f . b) is the identity: f (x) = x. Because f is continuous. as g1 is increasing and all even iterates of f are increasing. Suppose a is a finite number. Finally. from the equality n/2−1 � g1 ◦ f 2k + f n . As g1 is increasing. We shall show that only the identity function and the constant functions satisfy the conditions of the problem. Marcel T ¸ ena Solution. so lim f ′ (a) = fd′ (a) = x→a x>a x−a f (x) − f (a) = x→a lim = 1. It is clear that these functions verify indeed the conditions. k=0 gn = (n−1)/2 � g1 ◦ f 2k . so x = y. If I is non-degenerate. if n is even. (2) . because f (a) = a = inf I = inf{f (x) | x ∈ R}. let a = inf I < sup I = b.2014 Romanian Mathematical Olympiad – Final Round 35 (x − y)2 = (g1 (x) − g1 (y))(gm (x) − gm (y)) ≤ 0. if n odd. b ∈ R. By the continuity of f in a and by (1) we get f (a) = a. If I is degenerate at a point then f is constant. we conclude that g1 is constant. The fact that f is one to one and continuous implies that it is strictly monotonic. its range {f (x) | x ∈ R} is an interval I ⊆ R. i. Determine all differentiable functions f : R → R. x−a x−a x>a On the other side f has a minimum at a. k=0 gn is strictly increasing for even n and increasing for odd n. a < x < b.e. This gives the result. A − B is also non-singular. Problem 3. (1) A(A − B) = (A − B)B. Mariean Andronache . so that AB = −BA. We conclude.36 2014 Romanian Mathematical Olympiad – Final Round so. from where we get In = A(A − B)−1 − (A − B)−1 A. Analogously b = +∞. i. moreover by (1) the matrix A − B is also singular.. by (2). B two matrices in Mn (C) such that A2 + B 2 = 2AB. by the Theorem of Fermat. f ′ (a) = 0. Using (1) we get AB − BA = (A − B)2 = (tr (A − B))(A − B). where A∗ is the adjugate of A. then X 2 = (tr X) X (each row of X is proportional with a nonzero fixed row X). Prove that the matrix A2 + I4 is singular if and only if there exists a nonzero matrix B in M4 (R). a contradiction. AB − BA = On . Prove that: a) The matrix AB − BA is not invertible. Then A − B = A − (A − B)−1 A(A − B). so 0 = tr (AB − BA) = (tr (A − B))2 . (2) Suppose AB − BA is non singular. Problem 4. then matrices A and B commute. b) It is known that for a matrix X of rank 1 in Mn (C). b) If the rank of A − B is 1. tr (A − B) = 0. Leonard Giugiuc. i. Considering traces we obtain n = tr In = tr A(A − B)−1 − (A − B)−1 A = 0. in contradiction with (2).e. that is B = (A − B)−1 A(A − B). So AB − BA is singular. a) The given relation can be written in each of the two forms: (A − B)2 = AB − BA. AB = BA. Consider a positive integer n and A. By (1). C˘ at˘ alin Gherghe Solution.. We conclude a = −∞. such that tr A = tr A∗ �= 0. Let A be an invertible matrix in M4 (R).e. we get that the conjugate ¯ B of B anticommutes with A.1 (C). This proves that any linear combination with ¯ anticommutes with A. . Then Ak B = (−1)k BAk . f (A) = O4 . Consecomplex coefficients of the matrices B and B ¯ is a nonzero matrix in Mn (R) that anticommutes with quently. for all x ∈ A. There exist two proper vectors x and y in Mn. so that Ax = ix and Aτ y = −iy. This proves the first part of the problem. We show first that if A is a matrix from Mn (R) such that A2 + In is singular. Taking conjugates and using that A is in Mn (R). By Hamilton-Cayley. As B is nonzero we conclude that A2 + I4 is singular. such that AB = −BA. B = xyτ is a nonzero matrix in Mn (C). 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. f = λ4 −(tr A)λ3 +aλ2 −(tr A∗ )λ+det A = λ4 −(tr A)λ3 +aλ2 −(tr A)λ+det A. we obtain that B(A2 + I4 ) = O4 . Let B be a nonzero matrix in M4 (R) that anticommutes with A. (∗) Consider the characteristic polynomial f of A.2014 Romanian Mathematical Olympiad – Final Round 37 Solution. i is a proper value of A. We shall prove the converse. +. The following relations show that A and B anticommute: τ AB = Axyτ = ixyτ = −x(−iy)τ = −x (Aτ y) = −xyτ A = −BA. ·) be a ring with unity. k ∈ N. and −i is a proper value of the transpose Aτ . 12th GRADE Problem 1. Taking into account (∗). As A2 + In is singular. where a is a real number (the latter form of f is a consequence of tr A = tr A∗ ). Because tr A �= 0 and A is invertible. For a ∈ A define functions sa : A → A and da : A → A by sa (x) = ax. B or i(B − B) A. da (x) = xa. then there exists a matrix B in Mn (R). Because x and y are nonzero. Let (A. As A is finite sa is bijective. As ϕ is non-zero and is continuous. that is x − y = 0. there exists b ∈ R such that F (g(x)) = x + b. for all x ∈ I. (x − y)ab = 0. As a is surjective. The given relation for f can be then written (F ◦ f )′ (x) = 1. b) To construct an example. +∞) given by g(x) = +∞. (*) implies f = g. thus there exists b ∈ A such that ab = 1. (*) On the other side. F ′ has a constant sign on J. for all x ∈ J. Let f. J be intervals and consider ϕ : J → R a continuous function which is nonzero on J. endowed with the operations of addition and composition. So F (f (x)) = F (g(x)). Consider an anti-derivative F : J → R. *** Solution.38 2014 Romanian Mathematical Olympiad – Final Round a) Suppose A is finite. +∞) be a continuous function having the following properties: (i) The function g : [1. Problem 3. consider S = {(xn )n∈N | xn ∈ R} and the ring of additive functions f : S → S. The proof of the converse goes on the same lines. g : I → J be two differentiable functions such that f ′ = ϕ ◦ f and g ′ = ϕ ◦ g. For a one can consider the function defined by a ((xn )n ) = (xn+1 )n . Problem 2. for all x ∈ I. from F ′ (x) �= 0. As a consequence. so F is injective. for all x ∈ I. We deduce x0 + a = F (f (x0 )) = F (g(x0 )) = x0 + b. In the same manner. Let f : [1. f (x) has limit at x . if da (x) = da (y). than f and g coincide. +∞) → (0. for all x ∈ J. which proves the injectivity of da . Let I. a) Suppose sa is one to one. It is clear that sa is not one to one. sa is injective if and only if da is injective. Prove that if there exists x0 ∈ I such that f (x0 ) = g(x0 ). Prove that for any a ∈ A. So. b) Give an example of an ring that contains an element a such that exactly one of the functions sa and da is injective. Thus there exists a ∈ R such that F (f (x)) = x + a. that is a = b. Dorel Mihet¸ Solution. +∞) → (0. we get in succession (x − y)a = 0. the function 1/ϕ is correctly defined and continuous. da is injective. x b) Remark that 1 f (t)dt > 0. let a > 0 be such that x→+∞ g(x) > ℓ/2 for x ≥ a. If ℓ ∈ (0. 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 ) −→ +∞. 1 . +∞). = λ lim x x→+∞ x x→+∞ v(x) f (t)dt 1 f 2 (t)dt. for all x ≥ 1. for x > 1.2] f (x)). so ℓ = 0. x→+∞ 1 b) Show that lim 2 x→+∞ x x f 2 (t) dt = 0.39 2014 Romanian Mathematical Olympiad – Final Round 1 (ii) The function h : [1. x f (t) dt has 1 a) Show that lim g(x) = 0. x • v ′ (x) = f (t) dt + xf (x) �= 0. so 1 lim x→+∞ x2 x 2 f (t)dt = 1 = where λ = lim h(x). f (t) dt + x→+∞ x 1 4x in contradiction with (ii). +∞) → (0. +∞) given by h(x) = x finite limit at +∞. v(x) = x x 1 f (t)dt. where x→+∞ m = inf x∈[1. u(x) = 0 we shall use l’Hospital Rule: v(x) • u and v are differentiable. a) Let ℓ = lim g(x). 1 Mihai Piticari Solution. 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 . In the same way we can prove that ℓ = +∞ is in contradiction with (ii). To show that lim x→+∞ • lim v(x) = +∞ (this follows from v(x) ≥ m(x − 1) for x ≥ 2. g(x)) and = g(x) · f (x) + h(x) f (t) dt xf (x) + 1 u′ (x) lim g(x) = 0 imply lim ′ = 0. so 2 y p = e. then (yx)p = xp y p . x→+∞ x→+∞ v (x) Problem 4. 2 Consider f : G → G. a) If x. This would imply bp+1 = b−p bbp . Because p is a prime. that is y −1 x ∈ H. for all x ∈ G. that is ay p = p y a. as it is finite. a is an element in G\{e} and p is a prime number such that xp+1 = a−1 xa. which gives the conclusion. that is xy ∈ H. We conclude that for every element in Imf . we ord(G) . we have (xy)p = y p xp = e. exactly ord(H). and. a) Show that there is k ∈ N∗ such that ord(G) = pk . for all y ∈ G. for a k ∈ N∗ . Because e = xp = (xp )p . so |Imf | = ord(H) Because a �= e we get a ∈ / Imf : for if not a = bp for some b ∈ G. We can write x(yx)p y = xp+1 y p+1 . Suppose (G. given by f (x) = xp . p or p2 and by the Cauchy theorem we deduce ord(G) = pk . Moreover.40 2014 Romanian Mathematical Olympiad – Final Round • relations u′ (x) = v ′ (x) f 2 (x) f (x) x ∈ (0. Ioan B˘ aetu Solution. y ∈ G. so (y −1 x)p = e. x. This gives x ∈ Hy. From the hypothesis we have y p(p+1) = a−1 y p a = y p . proving that H is a stable part of G. than (xy)p+1 = a−1 xya = a−1 xaa−1 ya = = xp+1 y p+1 . conclude ord(H) > |Imf | = ord(H) . the image of f is contained in H. b) For x. H is a subgroup. ·) is a finite group with unity e. b) Prove that H = {x ∈ G | xp = e} is a subgroup of G and (ord(H))2 > ord(G). every element of the group has order 1. y ∈ H. y ∈ G and f (x) = f (y) imply xp (y −1 )p = e. For x = a we get ap = e so by the preceding equality (ya)p = y p . the number of its pre-images in G is ord(G) . that is e = bp = a a contradiction. As a ∈ H. Multiplying at left by ya we obtain yay p = (ya)p+1 = y p+1 a. for all y ∈ G. Prove that triangle angle ADB AEF is isosceles. Gabriel Vrˆınceanu 3. 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. b). + ≥ 25. with a and b non-nil digits. with D ∈ (BC). Triangle ABC has a right angle in A and AD ⊥ BC.. for every positive integers a. Petru Braica 41 . that ab is irreducible and the decimal fractions ab and b(b+1) Gabriel Vrˆınceanu 4. 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. d. b b+k b) Prove that 4 7 148 1 + + + . k such that a < b.. a+k a ≤ . Find how many pairs (a. 100 101 102 149 Gabriel Vrˆınceanu.SHORTLISTED PROBLEMS FOR THE 65th NMO SMALL JUNIORS 1. Find the largest and the smallest possible value for abcd. b. and each of the other lines contains the same numbers. A 4 × 4 magic square has in its first line the non-nil digits a. written in this order (from left to right). b. have the property a+b are finite. a) Prove that. written in different orders. c. Ion Cicu 2. Marius Burtea 9. A number n will be called squarish if the number Dn + Dn+1 is a perfect square. for every n ∈ N. Find all n ∈ N so that s(an ) = 2s(cn ). 7 889 for every non-negative integer n. Define an = 18 77 . Andreea Dima.42 Shortlisted problems for the 2014 Romanian NMO 5. The bisector of the angle ∠OAB meets OB in N and BC in P . ii) the segments [BM ]. George Stoica 7. [M N ] and [M C] are the sides of a triangle in which the opposite angle of [M N ] has measure 180◦ − m(∠A). Marius Burtea JUNIORS 1. Prove that the following properties are equivalent: i) m(∠M AN ) = 12 m(∠BAC). 76 and 755 are squarish. n digits a) Prove that an is divisible by 13. Prove that P C = 2ON . Cosmin Manea. denote Dn its largest proper divisor. Traian Preda . . a) Show that 35. The isosceles triangle ABC has AB = AC and points M and N are taken on (BC) such that M is between B and N . b) Show that there are infinitely many squarish numbers. Find all positive integers n with at least two digits. Drago¸s Petric˘ a 8. Solve in positive integers the equation 2n = 3m + 23. where s(m) represents the sum of the digits of the number m. . n’s digits are pairwise distinct and n equals the product of the sum of its digits with one of its digits. b) Denote by cn the quotient of the division of an by 13. Denote O the center of the square ABCD. If n is a composite number. Mihai Dicu 6. and P. Let △ABC be isosceles. Denote α the plane perpendicular in B on the straight line BM . Gabriel Popa . Take D on the √ ray (AB so that BC = AD 3. (M CD)). a) Prove that m((M AC) . b) Find the measure of the angle between the planes (M AB) and (ABC). F be the second intersection of its medians with its circumcircle. When does the equality hold? Nicolae Bourb˘ acut¸ 3. prove that: a) M is the orthocenter of triangle ABC. side (AC) so that m(ABP BQ) = m(QBC). B˘ atinet¸u-Giurgiu. Silvia & Ionel Brabeceanu 6. Dumitru M. Marian Ionescu 4. Q be points on the ) = m(P If [AD] is an altitude. AF B have the same area. Denote M the midpoint of the edge [AD] of the cube ABCDA′ B ′ C ′ D′ . Find all real solutions of the system |x + y + z| = 1 |x| + |y| + |z| = 1 |x − y| + |y − z| + |z − x| = 2. BQ ∩ AD = {N } and △ABN is isosceles. b) M N = AD Andrei Bud 5. Find the tangent of the angle between the planes α and (ACB ′ ). Prove that if the triangles BDC. (M CD)) = 4m((M BC) . Let triangle ABC be so that m (∠B) = m (∠C) = 50◦ . Neculai Stanciu 7. CEA. with AB = AC. Prove that M ≥ n. then triangle ABC is equilateral. Let ABC be a triangle and D. Let (M C) be a segment perpendicular on the plane (ABC) so that CD = 2M C. BP ∩ AD = {M } . E. AB (AB − AD). Let n be a positive integer and M be the arithmetic mean of its positive √ divisors.Shortlisted problems for the 2014 Romanian NMO 43 2. D ∈ BC. Prove that a(a + 2b)x + b(b + 2c)x + c(c + 2a)x ≤ (a + b + c)x+1 . z ∈ R∗ . 1] and a. Let a ∈ (0. b. 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. Find all strictly increasing sequences (an )n of positive integers with the following two properties: a) a31 + a32 + · · · + a3n = (a1 + a2 + · · · + an )2 . ∞) so that f (x) ≤ ax and f (x + y + z) ≤ f (x)f (y)f (z). Let n ≥ 2 be an integer and z be acomplex number so that z n = 1. Bogdan Enescu 2. for every x. c ≥ 0. z ∈ R. √ n n 1 + (−1)n 1+ 5 Prove that: . c in M .44 Shortlisted problems for the 2014 Romanian NMO 8. y. b in M . Prove that ab is rational for every a. n. Find the minimum value of the expression 1 1 1 2 2 E = x + 2 + y + 2 + z2 + 2 . for every a. the number ab + bc + ca is rational. Traian T˘ amˆ aian 6. Ion Safta SENIORS 1. Let x ∈ [0. Let M be a nonempty set of positive reals so that. 4. − (1 − z k − z 2k ) = 2 2 k=1 Mircea Merca . George Stoica 5. b. Find all functions f : R → (0. Cristinel Mortici 3. 1). Prove that (an )n is an arithmetic sequence. y. b) for every integer k ≥ 2015. y z x taken for all x. for every integer n ≥ 2014. the number a1 + a2 + · · · + a2014 cannot be written as a sum of k consecutive positive integers. xn+1 = 1 − n n where x1 is an arbitrary real n umber. . 1] by f (x0 ) + · · · + f (xn ) y 0 + · · · + yn and yn+1 = f . (an )n≥1 be a sequence of real numbers and (xn )n≥1 be the sequence defined by a an xn + . Find all differentiable functions f : R → R. . + sin(2n + 1) | n ∈ N}. y0 be two points in [0. whose derivative is bounded in a neighborhood of the origin and fulfill the condition xf (x) − yf (y) = (x2 − y 2 ) max(f ′ (x). + a n = 0. for every real numbers x and y. xn+1 = n+1 n+1 clearly. Prove that the sequences (xn )n∈N and (yn )n∈N are convergent. Mihai Monea 2. . Dan S ¸ tefan Marinescu. 1] be a continuous function a and x0 . Leonard Giugiuc 4. the definition is correct. Let f : [0. Let A and B be two matrices from M3 (C). Define sequences (xn )n∈N and (yn )n∈N of points in [0. such that (AB)2 = A2 B 2 and (BA)2 = B 2 A2 . Prove that lim xn = 0 n→∞ if and only if lim n→∞ a1 + a2 + . . Marian Cucoane¸s . 1]. Prove that (AB − BA)3 = O3 . f ′ (y)).45 Shortlisted problems for the 2014 Romanian NMO PUTNAM SENIORS 1. Find the infimum and the supremum of the set {sin 1 + sin 3 + . Marcel Chirit¸˘ a 5. 1] → [0. n George Stoica 3. Let a be a positive real number. Define the sequence (xn )n∈N by xn+1 = 1 n+1 (x0 +x1 +. Traian T˘ amˆ aian 8.. so that f (xn y n+1 ) = xn+1 y n for every x. Let f : [0. y ∈ G Prove that: a) x2n+1 = e.46 Shortlisted problems for the 2014 Romanian NMO 6. with the property that the polynomial P + aQ is irreducible for every rational a.+xn ) f (x) dx. Mihai Monea. for every x ∈ G. for every x. Let I ⊂ R be an interval and f : I → R be a continuous function. Q with rational coefficients. y ∈ I.. Let n ∈ N∗ and (G. 1] be a continuous function and x0 ∈ [0. Let A and B be two matrices from M3 (C). for every x. ·) be a group with the property that there exists an endomorphism f : G → G. Prove that the matrix AB − BA is singular. y ∈ I. y 1 ii) f (t)dt − f (x)(y − x) ≤ (x − y)2 . 1]. ·) is abelian. 1] → [0. such that A2 = AB + BA. Find all polynomials P . Dan S ¸ tefan Marinescu . b) the group (G. 2 x Nicolae Bourb˘ acut¸ 9. George Stoica 10. 0 Prove that the sequence (xn )n∈N is convergent. Prove that the following properties are equivalent: i) f (x) − f (y) < |x − y|. Marian Cucoane¸s 7. B ′ . 47 . centered on the line BC. C on the lines BC.THE 65th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS FIRST SELECTION TEST Problem 1. and the circle γC meets the lines AC and CC ′ again at N and N ′ . C ′ be the orthogonal projections of the vertices A. The circle γB meets the line BC again at B1 . M ′ . hence HM ′ · HB = HN ′ · HC and AM · AB = AN · AC. and let X be a point on the line AA′ . The conclusion follows. Then the lines M M ′ and BC are antiparallel. The circle γB meets the lines AB and BB ′ again at M and M ′ . so the lines M ′ N ′ and M N are both antiparallel to BC. Let γB be the circle through B and X. Let ABC be a triangle. since BM M ′ B1 is a cyclic quadrangle. and let γC be the circle through C and X. The line AH is the radical axis of the circles γB and γC . let A′ . Show that the points M . centered on the line BC. CA and AB. respectively. Let H be the orthocenter of the triangle ABC. Petru Braica A B' X N C' M' M N' H A' B 1 B C Solution. N and N ′ are collinear. respectively. respectively. B. . . Andrica. 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). Remarks. and Cucurezeanu. Dorel Mihet¸ Solution. Finally. k = 1. . Erect the three outer Napoleon triangles associated with the triangle A0 A1 A2 . 2. . there exist positive integers a1 . show that there exist n + 1 pairwise distinct numbers x1 . The equalities 33 + 43 + 53 = 63 and 33 + 153 + 213 + 363 = 393 settle the cases n = 2 and n = 3. 1. From any point X0 on γ0 draw lines X0 A2 X1 and X0 A1 X2 . . Birkhauser. k = 1. Problem 3. X2 are collinear. . .. xn+1 in Q \ Z such that {x31 } + {x32 } + · · · + {x3n } = {x3n+1 }. . An Introduction to Diophantine Equations. The points X1 . Given an integer n ≥ 2. . . *** Solution. n. 2. where X1 lies on γ1 and X2 lies on γ2 . and let γk . and xn+1 = −wn+1 /wn+2 meet the required conditions. (∗) w13 + w23 + · · · + wn+1 then the numbers xk = wk /wn+2 . . 2010. D. to obtain n + 1 positive rational numbers less than ε such that x31 + · · · + x3n = x3n+1 . n. 44. where {x} is the fractional part of the real number x... . respectively. notice that if 3 < w2 < · · · < wn+1 < wn+2 are n + 2 integers satisfying (∗). for the induction step n �→ n + 2. . I. xn . Notice that. an such that 1/a31 + · · · + 1/a3n = 1. . if w1 < w2 < · · · < wn+1 < wn+2 are positive integers such that 3 3 = wn+2 . By Andreescu. Example 5. A0 . p. then 3 < 4 < 5 < 2w2 < · · · < 2wn+1 < 2wn+2 are n + 4 integers satisfying the corresponding condition. choose a positive integer a > 1/ε and set xk = 1/(aak ). Now. k = 0. T. given an integer n ≥ 412 and a positive real number ε. and xn+1 = 1/a. x2 . ..48 Selection tests for the 2014 BMO and IMO Problem 2. be the corresponding circumcircles. . Given a triangle A0 A1 A2 . and the triangle X0 X1 X2 is an equilateral triangle satisfying the conditions in the statement. We now show by induction on n ≥ 2 that there exist integers 3 = w1 < w2 < · · · < wn+1 < wn+2 satisfying (∗). if n is an integer greater than or equal to 412. Euler’s Theorem applies to show that mn + nm ≡ 1 + (−1)m ≡ 0 (mod p1 p2 · · · pk ). use the Chinese Remainder Theorem. another circle is obtained by starting with the inner Napoleon triangles. The required locus is a pair of circles. then pi < pj . since the Mk are the centers of the inner Napoleon triangles associated with the triangle A0 A1 A2 . Finally. *** Solution. Having selected pj . and m is odd. so pj does not divide pi − 1. Design a set of k primes p1 < p2 < · · · < pk as follows. it turns out that the locus of X is the circle M0 M1 M2 . pj − 1 ≡ −2 (mod pi (pi − 1)). since p1 p2 · · · pk and m are coprime. Next.49 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 . Let k be a positive integer and let m be a positive odd integer. The conclusion follows. further. to choose a positive integer n such that n ≡ −1 (mod p1 p2 · · · pk ) and n ≡ 0 (mod (p1 − 1) · · · (pk − 1)). The center X of the triangle X0 X1 X2 is the intersection of X0 M0 and X1 M1 . Similarly. . so pi does not divide pj − 1. If i < j. use Dirichlet’s theorem to choose a prime pj+1 ≡ −1 (mod p1 (p1 − 1)p2 (p2 − 1) · · · pj (pj − 1)). and notice that the triangle M0 M1 M2 is equilateral. Begin by choosing p1 > 2m. Since the locus of X includes the three points Mk . Problem 4. and (p1 − 1)(p2 − 1) · · · (pk − 1) divides n. which must intersect at 60◦ . Show that there exists a positive integer n such that mn + nm has at least k distinct prime factors. n. . no subcover of which has less than m members. xm }. S2 . i = 3. and (b) Each element of S is contained in at least n members of A. . . and A runs through the above collection. *** Solution. . 2. if no member of A contains more than one element of S \ A. say A \ A1 = {x}. Assume henceforth that each member of A has exactly n elements. 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. 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. Determine maxA minB |B|. . . Since x2 is contained in at least n members of A. . we may choose a member A2 of A containing both x and x2 (recall that n ≥ 2). . . Assume henceforth that A does not contain S. This is clear if S is a member of A. . each of which contains (exactly) n − 1 elements of A. If n ≥ 3. x2 . Finally. m + n}. m ≥ 2. The . To complete the proof. write S \ A = {x1 . .50 Selection tests for the 2014 BMO and IMO Problem 5. . The required number is m = |S| − n. choose a member A1 of A containing x1 and notice that A \ A1 is a singleton set. . to form an m-element subcover of A consisting of A1 . m + 2. If some member B of A contains more than one element of S \ A. . . Fix a member A of A. . . If some member A of A has more than n elements.. we produce a set of subsets of S satisfying the two conditions in the statement. and let S1 . Let n be an integer greater than 1 and let S be a finite set containing more than n + 1 elements. Sn be the (n − 1)-element subsets of the upper part {m + 1. continue choosing members Ai of A containing xi . 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. . . . for each element of S \ (A ∪ B) choose a containing member of A. write S = {1.. . . for each element of S \ A choose a containing member of A. as B runs through all subsets of A whose members cover S. The latter along with A form a subcover of A of cardinality |S \ A| + 1 ≤ |S| − (n + 1) + 1 = m. m + n} of S. A2 . To this end. Am . The triangle ABC has at least one vertex angle. Since the latter is covered by the 4-fold blow-up of the triangle XY Z from O. 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 *** . fn (x) = x + x2 /n. *** Solution. Suppose. XY Z. so the area of the triangle AY Z is greater than the area of the triangle XY Z. where O is the center of the triangle XY Z. n. Since the problem is of an affine nature. CXY . . j = 1. Let a be a real number in the open interval (0. say at A. . greater than or equal to 60◦ . that none of the vertices A. . . let n be a positive integer and let fn : R → R. the latter has not the smallest area. Similarly. . respectively. . Problem 7. the triangles BZX and CXY both have an area greater than that of the triangle XY Z. Z be interior points on the sides BC. 2.Selection tests for the 2014 BMO and IMO 51 sets Si. Y . we may (and will) assume that the triangle XY Z is equilateral. C. 1). Alternative Solution.j = Sj ∪ {i}. so A is covered by the closed circumdisc OY Z. Let ABC be a triangle and let X. in contradiction with the well known fact that of the four triangles AY Z. B. m. CA. . i = 1. 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. 2. AB. BZX. . the conclusion follows. (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. B. 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. satisfy both conditions in the statement and at least m of them are needed to cover S. if possible. so p1 p2 · · · pk−2 > pk−1 pk for all indices k ≥ 6. 1/(ak + n). . (∗) since the ak form an increasing sequence of positive real numbers. 12. to deduce that 1/an = 1/a − n−1 k=0 k ∈ N. r = pm ≤ p5 = 11. *** Solution. Therefore the product of all primes less than r does not exceed nq < q 2 r. A necessary and sufficient condition that n be of the required type is that n < qr. 3.52 Selection tests for the 2014 BMO and IMO Solution. 6. Notice that 6 is the first index k such that p1 p2 · · · pk−2 > pk−1 pk . m ≤ 5. 4. an > a(1 − a)n2 + 2a2 n + a3 . Consequently. if p1 p2 · · · pk−2 > pk−1 pk for some index k ≥ 6. so 1/a − n/(a + n) < 1/an < 1/a − n/(an + n). so p1 p2 · · · pm−2 < pm−1 pm . then q ≤ pm−1 . k ∈ N. Let ak = (fn ◦ · · · ◦ fn )(a). q ≤ p4 = 7. 14. then (by Bertrand-Tchebysheff) p1 p2 · · · pk−1 > p2k−1 pk > 2pk−1 · 2pk > pk pk+1 . The first inequality above yields the required upper bound. and notice that k 1/ak+1 = 1/ak − 1/(ak + n). 18. 8. this upper bound yields the required lower bound. be the first two primes which do not divide n. 20. Determine all positive integers n such that all positive integers less than n and coprime to n be powers of primes. Now. 24. (1 − a)2 n2 + a(2 − a)n + a2 Problem 8. and n < qr ≤ p4 p5 = 7 · 11 = 77. and so does their product. If r = pm . 30. Each of the primes less than r and different from q divides n. Let p1 = 2 < p2 = 3 < p3 = 5 < · · · be the sequence of primes and let q and r. 10. (1 − a)n + a Plugged into the rightmost expression in (∗). 5. an < an + a2 . 60. q < r. 9. Examination of the integers less than 77 quickly yields the required numbers: 2. 42. In the former case.Selection tests for the 2014 BMO and IMO 53 Problem 9. k = n = 1. i. k + n = 2. say m = 2m′ . where m′ is a non-negative integer. Next. write k = f (m) and n = f (m + 1) for some positive integer m. we prove that for every pair of coprime positive integers (k. then m is even by the preceding. 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. If. proceed by induction on k + n. for any positive integer n. so m + 1 is a positive odd integer such that f (m + 1) = n and the conclusion follows. With reference again to the recurrence for f .e. respectively f (n) ≥ f (n + 1). in addition. n) there exists a unique positive integer m such that k = f (m) and n = f (m + 1). is clear. integer. n−k) or (k −n. Let f be the function of the set of positive integers into itself. k < n. notice that if n is a positive even. so k = f (2m′ ) = f (m′ ) and n − k = f (2m′ + 1) − f (m′ ) = f (m′ + 1). If k > n. it follows that if m is a positive odd integer such that f (m) = 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. With reference to the recurrence for f . respectively odd. The induction hypothesis applies to the pair (k. k = f (m) = f (2m) and n = k + f (m + 1) = f (m) + f (m + 1) = f (2m + 1) for some positive integer m. The base case. then m is odd. given a positive integer n ≥ 2. apply the induction hypothesis to the pair (k. say m = 2m′ +1. then f (n) < f (n + 1). and consider again the two possible cases. so f (n) < f (n + 1) if and only if n is even. according as to k < n or k > n. an easy induction shows f (n) and f (n + 1) coprime for each positive integer n. hence uniqueness of m. If k < n. then f (m − 1) is a positive integer less than and coprime to n. BMO 2014 Short List Solution. defined by f (1) = 1. The induction hypothesis applies now to the . To prove the above claim. f (2n) = f (n) and f (2n + 1) = f (n) + f (n + 1). To prove uniqueness. n). This establishes the existence of the desired positive 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). Show that. n − k) to imply uniqueness of m′ . If k + n > 2. then m is even. where m′ is a positive integer. Discarding the trivial case n = 1. DI is the bisector of ∠BDM – the reflection of ∠BAM – and BI is the bisector of ∠ABD. Problem 11. n) to imply uniqueness of m′ . such that the angles BAM and CN M are equal. satisfies the condition σ(n) = 2n − 2 if and only if n = 2k 2k+1 + 1 . Notice that l must be positive. let σ(n) denote the sum of all positive divisors of n (1 and n. THIRD SELECTION TEST Problem 10. Denote I the intersection of the bisector of ∠BAM with BC and denote D the reflection of A about BC. assume first that n = 2k pl . respectively. Show that a positive integer n. whence the conclusion. inclusive). where k is a non-negative integer and 2k+1 + 1 is prime. D are collinear. hence again uniqueness of m. Let ABC be an isosceles triangle. *** Solution. therefore I is the incenter of triangle P BD. For every positive integer n.54 Selection tests for the 2014 BMO and IMO pair (k − n. = p pl p−1 p−1 . M . On the other hand. AB = AC. so 1+ p − p1l σ(pl ) p 1 ≤ < . Then ∠BM D ≡ ∠BM A ≡ ∠CM N . and let M and N be points on the sides BC and CA. so P . Show that the internal angle bisectors of the angles BAM and BP M meet at a point on the line BC. which has at most two distinct prime factors. and p is an odd prime. where k and l are non-negative integers. Bogdan Enescu C D M N P A I B Solution. This completes the induction step and ends the proof. Sufficiency is a routine verification. To prove necessity. The lines AB and M N meet at P . . xn . p − 1 q − 1 (p − 1)(q − 1) n so min(p. positive integers n such that σ(n) = 2n − 1. A positive integer n satisfying the condition σ(n) = 2n − 2 is called a perfect-plus-two (pp2) number. so 2(p − 1) > pl . in addition. say p = 3. i. so p > 2k+1 − 1. It can be shown that a pp2 odd integer greater than 3. and it follows that q = 5 and k = l = 1. This completes the proof. i. . k j=1 k=1 k=1 for all positive integers n and all positive real numbers x1 . To rule out the case n = pk q l . it is not divisible by 3. q) = 3. write 2− σ(n) σ(pk ) σ(q l ) q 2 p = = · . Pp1 integers. have equally well been considered. i. 2 −1 1+ p p p p−1 � � By the first inequality. but the powers of 2 are the only ones known. by the second inequality. then it must have at least seven distinct prime factors. where p and q are distinct odd primes.. but equivalently. 1 1 2 1 + + + > 1.e. and consequently l = 1 and p = 2k+1 + 1.. Remarks.e. On the other hand. and k and l are positive integers. . must have at least four distinct prime factors. if any. Determine the smallest real constant c such that 2 n k n � � � 1 xj ≤ c x2k . p < 2k+1 + 2(p − 1)/pl . · < n n pk ql p−1 q−1 Alternatively.. p ≥ 2k+1 + 1 since p is odd. 2k+1 − 1 (1 + 1/p) < 2k+1 .. Then 3/(q − 1) + 4/n > 1. The result in the problem shows that 3 = 20 (20+1 + 1). Problem 12. n = 15 which does not satisfy the condition σ(n) = 2n − 2.55 Selection tests for the 2014 BMO and IMO whence (since σ is multiplicative) � � � k+1 � � � p σ(n) 1 2 ≤ l = 2k+1 − l < 2k+1 − 1 . 128 · 257 = 27 (27+1 + 1) are all pp2 integers. 136 = 23 · 17 = 23 (23+1 + 1). Dumitru Barac .e. and if. 10 = 2 · 5 = 21 (21+1 + 1). to get n � 2 1 √ ≤ 3 − √ < 3.56 Selection tests for the 2014 BMO and IMO Solution. Write 1/ j > 2( j + 1 − √ j). k ≥ 2. xn are positive real numbers. . k j=1 j k k k k k k √ √ √ Finally. in x ¯n and x To show c ≥ 4. The base case. this is equivalent to 2n(2n + 1)¯ x2n − 8n(n + Since xn+1 = (n + 1)¯ 2 2 ¯n+1 + (4n + 6n + 1)¯ xn+1 > 0. n) is even. x2n < 4x2n+1 . n k k k=1 and deduce thereby the desired inequality. k ≥ 1. k j=1 j k k=1 k=1 n � √ √ Divergence of the harmonic series settles the case. be the set of non-negative integers k < n such that the number of distinct prime factors of gcd(k. and |An | > |Bn | if n is odd. . and notice that it is sufficient to show that (2n + 3)¯ x2n+1 − 2n¯ xn+1 − n¯ xn . . is clear. respectively Bn . We first show that. if n is a positive integer and x1 . Show that |An | = |Bn | if n is even. let x ¯k = (x1 + · · · + xk )/k. k j=1 n k=1 k=1 so c ≤ 4. Problem 13. we prove that 2 k n � � 1 1 1 √ >4 − 24. to obtain 2 � � k �2 � 2 8 4 �√ 4 4 1 1 √ 1− √ > 2 k+1−1 > = − √ . The left-hand member is a quadratic form 1)¯ xn x ¯n+1 whose discriminant is −2n and the inequality follows. and let An . respectively odd. proceed by induction on n.. notice that 1/(2k k) < 1/ k − 1 − 1/ k. Let n be a positive integer. *** . then n � k=1 2 � n �2 k n � � � 2 1 xj + xk <4 x2k . n = 1. For the induction step. The best constant is c = 4. To prove the above inequality. then k = k1 n2 + k2 n1 ranges once over a complete residue system modulo n1 n2 . if n1 and n2 are coprime positive integers. n2 ) = e(k1 . n). Finally. We shall prove that the above sum equals n p|n (1 − 2/p). then f (pm ) equals the number of k’s coprime to p. |An | − |Bn | = k (−1)s(k. Hence e(k. respectively.n) and let f (n) = k e(k. the latter is precisely the number of positive integers k < n such that k and k + 1 are both coprime to n. n) = (−1)s(k. and the angles CQE and QCR are equal. n1 n2 ) = e(k. i = 1. n1 n2 ) = e(k. FOURTH SELECTION TEST Problem 14. n) depends only upon the residue class of k modulo n. n2 ) = f (n1 )f (n2 ). The tangents of the circumcircle ABC at vertices B and C meet at P . Further. n1 )e(k2 . If n1 . In the above notation. BE and CF are concurrent.n) . 2. and the lines OQ and BC meet at D. Since gcd(k. n2 ). . Prove that the lines AD. if ki ranges once over a complete residue system modulo ni . Cosmin Pohoat¸˘ a Solution. let e(k. We show that f is a numerical multiplicative function — that is. Let ABC be an acute-angled triangle and let O be its circumcenter. n1 )e(k. and f (n1 n2 ) = e(k. ni ) = e(ki . n1 )e(k. p prime. 2. Hence the lines AC and CR are perpendicular. which is pm−1 . 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. n2 are coprime positive integers. where s(k. i = 1. minus the number of k’s divisible by p. This ends the proof. if p is a prime. whence the conclusion. ni ). so the lines EQ and CR are parallel. n1 n2 ) = e(k1 . and m is a positive integer. and e(k. n2 ). The line AB and the circle of radius P B centered at P meet again at some point R. let E and F be the orthogonal projections of Q on the lines AC and AB. Standard angle-chasing shows that the angle BRC is the complement of the angle BAC. n) is the number of distinct prime factors of gcd(k. and k ranges over any complete residue system modulo n. then f (n1 n1 ) = f (n1 )f (n2 ).Selection tests for the 2014 BMO and IMO 57 Solution. k k1 k2 Finally. so f (pm ) = pm (1 − 2/p). n1 )e(k2 . then e(k. n). which is pm − pm−1 . Given an odd prime p. CD BF CD AE BF on account of AQ being the internal angle bisector of the angle BAC. let k be an integer greater than 1 and let P be a polynomial of degree at most k−2 with integral coefficients. . where a is an integer. but f (X) = (−X + a)k + (−X + a)k−1 + P (−X + a) and g(X) = X + a are not. Clearly. Q. 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. More generally. deg f ≥ 1.58 Selection tests for the 2014 BMO and IMO Since the points B. so AE = AF . Vlad Matei Solution. Cezar Lupu. so the angles QBF and CQE are equal and the right-angled triangles BQF and QCE are similar: BQ/QC = BF/QE = QF/CE. Consequently. The conclusion follows by Ceva’s Theorem. 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. C. R are concyclic. the signs correspond to one another. deg g ≥ 1 and the leading coefficients of f and g are ±1. 1= BD CE AF BD CE · = · · . Recall that AQ is the internal angle bisector of the angle BAC. the angles QBF and QCR are equal. f (X) = (−X + a)k + (−X + a)k−1 + P (−X + a) and g(X) = −X + a are admissible. determine all polynomials f and g p−1 with integral coefficients satisfying the condition f (g(X)) = k=0 X k . so EQ = F Q. whence BQ2 /CQ2 = (BF/EQ)(F Q/CE) = BF/CE. Problem 15. where b is an integer — in both cases. and f (X) = ±X +b and g(X) = ±X k ±X k−1 ±P (X)∓b. for instance. so BQ2 /CQ2 = BD/CD. it follows that β is one of the ζ j . The latter forces m = 1 coefficients yields am bm n = 1 and mam bn which leads to the second pair of polynomials above. In particular. deg h = kn ≤ p − 1. . We show that if deg g > 1. g(X) = ±X + b and f (X) = Φp (±X ∓ b). To begin. The proof relies upon the well known fact that both Φp and its (formal) p−2 derivative. ′ Alternative Solution. let α be a (complex) root of f .. . where a is an integer. Φ′p (X) = k=0 (k +1)X k . Now take the derivative both sides of the relation in the statement to obtain f (g(x))g ′ (X) = Φ′p (X). either f ′ (g(x)) = ±1 and g ′ (X) = ±Φ′p (X) or vice versa — as before. and notice mn + mam bm−1 bn−1 X mn−1 + · · · . so deg h ≥ deg Φp = p − 1. we immediately obtain the first pair of polynomials above. 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 . then deg g = p − 1. p − 1. Since Φ′p (X) is irreducible in Z[X] and its coefficients are obviously jointly coprime. where a is an integer. . the signs correspond to one another. Since Φp is the minimal polynomial of ζ and h(ζ) = 0. . Identification of that f (g(X)) = am bm nX n m−1 bn−1 = 1. The polynomial Φp (X) = k=0 X k is the p-th cyclotomic polynomial. The remaining details are easily filled in and hence omitted. n m If deg g > 1. g(X) = ±Φp (X)+a and f (X) = ±X ∓ a. in the latter. . so deg h = p − 1. g − α has no multiple roots. and let β be a (complex) root of g − α. On the other hand. Since Φp (β) = f (g(β)) = f (α) = 0. it follows that Φp divides h. and since h is monic. are irreducible in Z[X]. p−1 Alternative Solution. ζ kn be the roots of g − α. it . j = 1. where b is an integer. for Φp has no such. Since g has integral coefficients. so h = j=1 X polynomial with integral coefficients. and let ζ k1 . . With the same convention for signs. where b is an integer. in the former case. the required polynomials are either f (X) = ±X ∓ a and g(X) = ±Φp (X) + a. a = j=1 ζ kj is integral. write f (X) = i=0 ai X i and g(X) = i=0 bi X i . Now let n = deg g > 1. n kj − a is a monic non-constant by the first Vieta relation. With the sign convention in the previous solution. .Selection tests for the 2014 BMO and IMO 59 If deg g = 1. let ζ = cos(2π/p) + i sin(2π/p). where 1 ≤ n k1 < · · · < kn ≤ p−1. or f (X) = Φp (±X ∓ b) and g(X) = ±X + b. Evaluate the sum � � 1 (i + σ(i)). 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. . j = 1.60 Selection tests for the 2014 BMO and IMO follows that h = Φp . we prove that the line A′k Bk passes through the midpoint of the segment A′k+1 A′k+2 . j = 1. C1 . . Let the tangents at Ak and Ak+1 meet at A′k+2 . . 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. 2. . . Hence a = −1 and kj = j. 2. Show that the points C0 . FIFTH SELECTION TEST Problem 17. . . We shall prove that the lines A′k Bk are concurrent. Consequently. indices being reduced modulo 3. 1. Problem 16. . the ζ j . n}. . . More precisely. *** Solution. and. Let A0 A1 A2 be a triangle and let O be its circumcenter. let Sn be the set of all permutations of the set {1. � � 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)! . k = 0. The lines OAk and Ak+1 Ak+2 meet at Bk . . Let n be a positive integer. and the tangent of the circumcircle A0 A1 A2 at Ak meets the line Bk+1 Bk+2 at Ck . . let I(σ) = {i : σ(i) ≤ i}. for each σ in Sn . . Consequently. C2 are collinear. 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 . are the roots of g − α and deg g = p − 1. Notice that. |I(σ)| σ∈Sn i∈I(σ) *** Solution. for each σ in Sn . p − 1. p − 1. fix an index k and let Ak Ak+1 ≤ . (2) the angles OAk+2 Bk and OAk+1 Ak+2 are equal. O. Xk all lie on the circle on diameter OXk . Ck Bk+2 Ak Ak+2 Ak Bk+2 (∗) Multiplying the three yields the desired relation. the angles OXk Bk and OYk Bk are equal. To avoid directed angles. 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 . Consequently. since. Since the lines Xk Yk and A′k+1 A′k+2 are parallel. without any loss. since OAk+1 = OAk+2 . We now turn to prove (∗). so the conclusion follows by Menelaus’ theorem. since the points Ak+2 . we will prove that Ak Ak+1 Ak Bk+1 Ck Bk+1 = · . and the latter is perpendicular to the line OAk = OBk . and (3) the angles OAk+1 Ak+2 and OYk Bk are equal. and notice that: (1) the angles OXk Bk and OAk+2 Bk are equal. that the triangle A0 A1 A2 is acute-angled. by Ceva’s theorem. assume. Bk . it follows that Bk is the midpoint of the segment Xk Yk . whence the conclusion. O. Yk all lie on the circle on diameter OYk . Bk .61 Selection tests for the 2014 BMO and IMO A0 A'2 C1 C2 C0 B2 O A1 B0 B1 A2 To this end. To this end. We will show that k=0 (Ck Bk+1 /Ck Bk+2 ) = 1. respectively. 2 Alternative Solution. since the points Ak+1 . 2 k=0 (Ak Bk+1 /Ak Bk+2 ) = 1. . and let the lines P X and Y Z meet at Q. . be an alphabet with m.62 Selection tests for the 2014 BMO and IMO Ak Ak+2 (the case Ak Ak+2 ≤ Ak Ak+1 is dealt with similarly). let P be a point on one of the bisectors of the angle BAC. . Z be the orthogonal projections of P on the lines BC. respectively. . The required ratio is 2n−m . Then the line AQ passes through the midpoint of the segment BC. and the angle Ck Ak Bk+2 is equal to the internal angle at Ak+2 of the triangle A0 A1 A2 . Let m be a positive integer and let A. The proof goes along the lines in the first solution. Y . . and think of B as an extension of A = {a1 . . AB. {¯ a1 . . respectively B. The problem was inspired by the Sharyghin configuration below: Let ABC be a triangle. the above ratio of sines equals Ak Ak+1 /Ak Ak+2 and (∗) follows. *** Solution. am } by a disjoint copy A¯ = ¯m }. . the preimage of α under deletion of all bars has exactly 2k1 −1 · · · · · 2km −1 = 2n−m elements. Let n be an even integer greater than or equal to 2m. respectively 2m letters. The conclusion follows. 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. Since there are exactly 2ki −1 distinct ways to bar ai an odd number of times in α. Now let α be a word in An and let ai occur ki times in α. so an = |An | and bn = |B n |. 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. Problem 18. 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 . the ki are positive even integers which add up to n. CA. a Deletion of all bars in a word in B n produces a word in An . . let X. Remarks. = · 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 . To prove this. Determine the ratio bn /an . let An and B n denote the sets of words described in the statement. 63 Selection tests for the 2014 BMO and IMO 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 ). 64 Selection tests for the 2014 BMO and IMO 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 namely (a. *** Solution.52 + 3 − 3 < 4. Splitting the squares into three 1 × 1.5 × 1. which is at most 2. −1. The strongest result known it literature is that the shortest distance (for 6 √ points) is 13/2 ≈ 9/5 < 2. so the length of their diagonal is 2. Remarks. By the pigeonhole principle. It follows that 2b + 3 | 3. Since 1. 0}.3 rectangles. which implies that one of the six points is not inside. Prove that among the six points. The sizes of the two rectangles at the top are √ 2 √ 1. the shortest distance is 3/ 2 > 2). then 1 3 1 + = . 2.66 Selection tests for the 2014 JBMO Case 2. but on one side of the square – contradiction. −2. so the distance between them is at most equal to the length of the diagonal.7 rectangles and two 1. 0) . we may prove that the result is still true if the six given points are inside or on the sides of the square. as shown in the figure √ above. there are two whose distance is less than 2. The sizes of the three rectangles at the bottom are 1 × 3. there are two points among the six given which are situated within or on the sides of one of the five rectangles. the distance can be 2 only if the two points are opposite vertices of one of the bottom rectangles. . However.5 × 3 − 3 . If a + b + 3 �= 0. the length of their diagonal is less than 2. b) = (1. it results that 2b + 3 | b2 + 4b + 3. Consider six points in the interior of a square of side length 3. Let us notice that five points are not enough to draw the same conclusion √ (taking the vertices and center of the square. 2b + 3 Since a ∈ Z. Let us split the square into five rectangles. 1. so b ∈ {−3. which leads to b+1 b+2 a+b+1 a= b2 + 4b + 3 . Problem 3. We get one additional solution. 67 Selection tests for the 2014 JBMO 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. 68 Selection tests for the 2014 JBMO 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 69 Selection tests for the 2014 JBMO 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. *** b) Prove that. ℓ + 1) → (k + 3. This shows that the circumcenter AEF lies on the perpendicular bisector of the segment line [AP ] (which is a fixed line). that does not use all the colors. Marius Bocanu Solution. for any such coloring. We associate coordinates to each unit square. Let P be the reflection of point C in the midpoint of the segment line [AB] . a) Prove that. ℓ + 3). the 2n + 1 < ⌊2n/3⌋ + 2 < n unit path (1. hence it does not use all the colors. while the one in the upper-right corner has coordinates (n. ℓ + 3) in two moves: (k. ℓ) to (k + 3. Using moves similar to the ones described above. 1) → · · · → (n. The path that starts with (1. • n ≡ 1 (mod 3). or only ⌊2n/3⌋ + 2 colors. n). Let n ≥ 6 be an integer. 4) → (6. if we reduce the number of colors to ⌊2n/3⌋ + 2.70 Selection tests for the 2014 JBMO Solution. clearly AP � BC and ∠F CD ≡ ∠F P A. which means that the quadrilateral F EAP is cyclic. • n ≡ 0 (mod 3). ℓ) → (k + 2. 1). then the statement from a) is true for infinitely many values of n and it is false also for infinitely many values of n. 6) → · · · → (n. It follows that ∠F P A ≡ ∠F ED. The quadrilateral EF DC is cyclic. We color each of the unit squares of an n × n board with one of the n colors. Notice the one can get from (k. 1) → (2. We have at our disposal n colors. 5) → (5. n) passes through 3 squares. 3) → (3. whether there are n colors. hence ∠F ED ≡ ∠F CD. the square in the bottom-left corner having coordinates (1. A P E F B D C Problem 9. n) . there exists a path of a chess knight from the bottom-left to the upper-right corner. will pass through exactly 2n + 3 < ⌊2n/3⌋ + 2 ≤ n 3 squares. • n ≡ 2 (mod 3). We have proven a) and we have seen that for all n ≡ 0 (mod 3) and all n ≡ 1 (mod 3) (n ≥ 6). a) and b). hence. We prove that the statement from a) is false if we have ⌊2n/3⌋ + 2 colors and n ≡ 2 (mod 3). n) contains squares of all the colors. The absolute value of the difference between the color numbers of two consecutive squares in the knight’s path is at most 1. 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. We use the color 1 of N = ⌊2n/3⌋ + 2 colors (note that N is odd) for the bottom-left corner. We use color number N for the upper-right corner. under the conditions from a). but not less. n − m + 1) | 1 ≤ m ≤ n}. 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. 3) → (4. It remains to exhibit a coloring with ⌊2n/3⌋ + 2 colors such that any path (1. 5) → · · · → (n. 1) → (2. hence it does not pass through squares of all possible colors. n) and continues with moves similar to the ones described above. and this for both the situations. 2) → (3. 1) → (n. will pass through exactly 2n + 5 = ⌊2n/3⌋ + 2 < n 3 squares. We give several such examples: Example 1. the statement from a) remains true even if there are only ⌊2n/3⌋ + 2 colors.Selection tests for the 2014 JBMO 71 and continues with moves similar to the ones described above. for getting from the . Finally. 4) → (5. we color with (N + 1)/2 all the remaining squares. hence it will not pass through squares of all possible colors. The path that starts with (1. but not less. then we use colors 2 ≤ k ≤ (N −1)/2 for all the squares to which the knight can get in k − 1 moves. aj−3 . aj+1 .72 Selection tests for the 2014 JBMO square with color number 1 to the square with color number N it has to pass through squares of all the other colors. 4. aj+3 ∈ {aj − 1. 5. in its path from color 1 to color N . 3. so again. It is easy to check the upper-right corner has color a2n−1 = N = ⌊2n/3⌋ + 2. 4. 3. from left to right. 3. . an+k−1 . aj + 1}. 6. aj−1 . . Again. . aj−1 . 4. .) Example 2. 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 . 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 knight can not possibly skip a color. the smallest with n ≡ 2 (mod 3). By the way the sequence is constructed. aj+1 . . (given in the contest by Tudor Plopeanu) Consider the sequence (am )m≥1 given by: 1. . with ak . 5. aj+3 . the knight can jump to a square having one of the colors aj−3 . . (The squares left white are to be colored with color number 4. 2. From a square with color aj . ak+1 .. Color row number k. 2. THIRD SELECTION TEST Problem 10. We prove that the shortest path to the upper-right corner contains exactly N squares. c. one needs at least 2j + 1 moves.Selection tests for the 2014 JBMO 73 Example 3. b. For each square. Once we have shown this. 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. *** Solution. again. the knight can not skip a color. To prove that this is the shortest one possible. and use color k for all the squares to which the knight can get in k − 1 moves. notice that each move changes the sum of coordinates by 1 or by 3. it is clear that. Let a. so any path has to contain squares of all the N colors. But if we color the board in a chessboard pattern. 2 . d be positive real numbers so that abc + bcd + cda + dab = 4. the bottom-left and the upper-right corners are of the same color. Prove that a2 + b2 + c2 + d2 ≥ 4. 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 ). j ∈ N∗ . the upper-right corner will have the color 2j + 3 = N . but not less. Color the bottom-left corner with 1. We prove that the upper-right corner has the color N = ⌊2n/3⌋+2 = 2j +3. Let n = 3j + 2. 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). from the square in the bottom-left corner to the given square. We have already seen an example of a path containing N squares. the color number indicates the length of the shortest path (number of squares contained). the knight changes color each time it moves. Alternative Solution. . q) ∈ {(5. which is a prime.. it follows that d | q and d | 4q + 3. 2 (p + 5) p2 − 5p + 25 = 2 q 2 + (4q + 3) . 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. Remark.74 Selection tests for the 2014 JBMO 1 3 It follows that 4 ≤ · (a2 + b2 + c2 + d2 ) . The statement of the problem is equally easy to prove for a. The equation reduces to p3 + 125 = 34q 2 + 48q + 18. 3)} . Lucian Petrescu Solution. n} such that: for all a. . a2 . . As p2 − 5p + 25 ≡ 3 (mod 4) . For q = 2 we obtain p = 5.e. b ∈ A. Problem 11. + am ≥ . am } a subset of the set {1. it follows that p2 − 5p + 25 must have at least 2 one prime factor d ≡ 3 (mod 4) and. d not necessarily positive real numbers. which means that d = q = 3 and p = 7. m 2 *** .. then a + b ∈ A. or. In conclusion. (7. Consider two integers n ≥ m ≥ 4 and A = {a1 . Problem 12. we get p3 ≡ 3 (mod 4) . 2 Equality holds when a = b = c = d = 1. i. a �= b. because of d | q 2 +(4q + 3) . with equality if and only if a = b = c = d = 1. . c. which is a prime. reducing modulo 4. equivalently. .. For q ≥ 3. The solutions to the equation are (p. Determine the prime numbers p and q that satisfy the equality p3 + 107 = 2q (17q + 24) . d | 3. . if a + b ≤ n. 2) . which leads to p ≡ 3 (mod 4) . Prove that: n+1 a1 + a2 + . 2. a2 + b2 + c2 + d2 ≥ 4. b. . a1 + ak < a2 + ak . as above. am }. Prove that the orthocenter H of triangle ABC is on the line T D. Assuming the contrary to be true. let T denote the midpoint of the side [AC]. But. < ai + am+1−i have to belong to the set {am+2−i . . Marius Bocanu Solution. respectively. the following i distinct numbers a1 + am+1−i < a2 + am+1−i < . 2 the conclusion follows immediately. n+1 . We prove that the sum of the numbers in each pair is at with 1 ≤ i ≤ 2 least n + 1. . it would exist an i for which ai + am+1−i ≤ n. Suppose 2ak < n + 1. . . a contradiction. < ak−1 + ak < n + 1. with AB �= BC. ak−1 + ak must all belong to A. . . For m = 2k − 1.75 Selection tests for the 2014 JBMO Solution. A1 and C1 denote the feet of the altitudes drawn from A and C. am . which has only i − 1 elements. ak+1 . am+1−i ). 1 ≤ i ≤ k − 1. . a) From AZ = ZC. it can be shown. In the acute triangle ABC. it follows immediately that [ZT ] is the angle bisector of ∠AZC. for 1 ≤ i ≤ m . . c) Prove that the point D lies on the circumcircle of triangle XY Z. We obtain that n+1 ≤ a1 +a2k−1 = a1 +(ak−1 +ak ) = (a1 +ak−1 )+ak = 2ak . . < am ≤ n. . . Assume 1 ≤ a1 < a2 < . For even m we can group the elements of A in pairs of the form (ai . which contradicts the assumption we have made. Consider We now prove that ak ≥ 2 ak−1 < a1 + ak−1 < a1 + ak < a2 + ak < . m . . . b) The circumcircles of triangles ABC and A1 BC1 meet again at D. hence they must be equal to ak . . respectively. . . Adding ai + am+1−i ≥ n + 1. . a) Prove that T is the incircle of triangle XY Z. 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 . that ai + am+1−i ≥ n + 1. Problem 13. am+3−i . Let Z be the point of intersection of the tangents in A and C to the circumcircle of triangle ABC. . k > 2. It follows that a1 + ak−1 . as i < m + 1 − i. Notice that ∠XAB ≡ ∠ACB ≡ ∠BC1 A1 ≡ ∠AC1 X and . . It is easy to prove (in a similar manner to a)) that T Y is the perpendicular bisector of the segment line [A1 C] . and now it is clear that points D. we get that LO � DH. it follows that the quadrilateral CT DY is cyclic. D is on the minor arc AB.76 Selection tests for the 2014 JBMO AT = C1 T. This leads to ∠XDY + ∠XZY = ∠XDT + ∠T DY + ∠XZY = = 180◦ − ∠XAT + 180◦ − ∠T CY + ∠XZY = = ∠ZAT + ∠ZCT + ∠XZY = 180◦ . therefore OL � HT . H. The radical axis of two circles being perpendicular to the line connecting the centers of the circles. T are collinear. c) The quadrilateral AT DX is cyclic because ∠ADT = ∠BDA − ∠BDH = 180◦ − ∠ACB − 90◦ = 90◦ − ∠XAB = ∠AXT . 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. OLHT is a parallelogram. in this case. As BD ⊥ DH. it follows that BD ⊥ LO. From ∠CDT = ∠HDB − ∠CDB = 90◦ − ∠CAB = 90◦ − ∠BCY = ∠CY T . Let O denote the circumcenter of triangle ABC and let L be the midpoint of the segment line [BH]. which means that X and T are on the perpendicular bisector of the segment line [AC1 ]. . 30 of which being of the form pq. How many of the elements of the set {1. but not of p1 ... . that are multiples of p1 . Sk where: • S1 is a sequence of numbers that contains all the factors of n. • S3 is a sequence of numbers that contains all the factors of n that are multiples of p3 . 100} are nice? *** Solution.Selection tests for the 2014 JBMO 77 which means that the quadrilateral DXZY is also cyclic... the smaller one is a factor of the larger one... the last number in the sequence being p3 p4 . where k ≥ 1 and αi ≥ 1. .. . . We prove that a composite number is nice if and only if it is not a product of two distinct prime numbers. 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. but not multiples of any of the numbers p1 . . . but are multiples of neither p1 nor p2 . . pk .. This leaves 44 nice numbers.. S1 . other than n... FOURTH SELECTION TEST Problem 14. .... .. The set {1. p2 .. A stronger fact can be proven: the factors of a nice number can actually be arranged on a circle such that. ... q distinct primes. Remark.. then n = pα 1 p2 ... • S2 is a sequence of numbers that contains all the factors of n that are multiples of p2 ... on a circle is the following: we write the numbers in a succession of the form: n.. . αk 1 α2 If n is not a product of two distinct primes. larger than 1.. .. q are distinct primes. the last number in the sequence being p2 p3 . therefore pq is not nice. pk−1 . with p. S2 . . 100} contains 74 composite numbers. the last one in the sequence being p1 p2 ... q and pq without p and q being neighbors. . . . 2. Sk−1 . for every two neighboring numbers. • Sk is a sequence of numbers that contains all the factors of n that are multiples of pk . it is clear that we can not arrange p.. A convenient way of arranging the factors of n. ... . If n = pq.. 2. . .... 3. where p. All the other ones are 0: x1 + x2 + x3 = x1 + x2 + x4 = x1 + x2 + x5 = x1 + x3 + x4 = x1 + x3 + x5 = 0. Let n ≥ 5 be an integer. . 21 terms. . and then comparing the first and the last equation gives x2 = x3 = x4 = x5 . −2. say x1 +x4 +x5 . Suppose x1 appears (at least) 5 times. x1 is involved in 6 sums. then x1 = x2 = . 1. . 1. . . 1. Now we prove that 7 is enough. then an = a2 + ab + ac + ad = a2 + cd + ac + ad = (a + c) (a + d) . and it follows immediately that x1 = x2 = x3 = x4 = x5 = 0. Bulgaria. so there is only one sum. *** Solution. Supposing that n was prime. (∗) hence n | (a + c) (a + d) . (or 1. x5 be real numbers. −2. but neither of these two conditions can hold because n > a + c and n > a + d. 1. The first three equations simplify to x3 = x4 = x5 . in total. x2 . We prove that the smallest convenient value of n is 7. 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. Problem 16. Let x1 . it would follow that n | a + c or n | a + d. . = x5 = 0. the seventh one gives x2 = x3 = x4 = x5 = 0. 1. (⇐) If ab = cd and n = a + b + c + d. that is not necessarily 0. x1 only participates in at most 6 sums. knowing 6 sums (or less) to be 0 is not enough to conclude that the numbers are all 0. it is true that ab �= cd. By the Pigeonhole Principle it follows that there is a number that appears (at least) 5 times. 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. 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. Suppose we are given that 7 sums are equal to 0. Finally. These sums contain. while there are only 5 numbers. 2003 Solution. If the numbers are 1. Therefore.78 Selection tests for the 2014 JBMO Problem 15. −2) there are 6 sums that are equal to 0 without the numbers themselves being 0. Let G be the projection of E onto AC. P. K the midpoints of the segment lines [EF ] . M are on the Euler circle of triangle AEF. [CD] that intersect at E. K. P. We denote by M. is a composite number. points N. G are concyclic. Problem 17. c = p − 1. G. and [AD]. Prove that the points M.79 Selection tests for the 2014 JBMO (⇒) If n = pq. which means they are concyclic (∗). N. consider two chords [AB]. we have ∠KP M = 180◦ − ∠KP A − ∠M P F = 180◦ − ∠DF A − ∠EAF = ∠ABF . This choice can easily be guessed by writing (∗) with a = 1. The lines AC and BD meet at F. we can chose a = 1. . and. N. [EA] . b = (p − 1) (q − 1) . As N K � DE and N M � AF. q ≥ 2. with p. respectively. Let P be the midpoint of [AF ] . In a circle. as KP � DF and M P � AE. N. Marius Bocanu D B E K M N C F P G A Solution. hence K. It follows that the quadrilateral KN P M is cyclic. Combining this with (∗) gives the conclusion. d = q − 1. we have: ∠KN M = ∠KN E + ∠EN M = ∠DEB + ∠BAC = ∠DEB + ∠BDE = ∠ABF . M are concyclic. . . Adding up yields a1 + a2 + . n are also less than k (where the indices from the sums aj are considered to be taken modulo n). xn > 0 be real numbers so that x1 + x2 + . 2. . . . Alternative Solution. for each k = 1. xn whose sum is at least k. it is enough to prove that xn−k+1 + . + xn < n. . Let n be a positive integer and x1 . xn is strictly less than k. . xj+k−1 .80 Selection tests for the 2014 JBMO FIFTH SELECTION TEST Problem 18. Then the numbers aj = xj + xj+1 + . . since a1 + a2 + . *** Solution. . so > n. + xn ) . . . 2 x1 x2 xn Show that for each positive integer k ≤ n. . . . . Suppose that x1 ≤ x2 ≤ . . . n n n 1 1 1 <n . + xn−1 + xn ≥ k. . . . there are k numbers among x1 . . x2 . + x n = 1 1 1 + 2 + . . . .. 2. . . Arguing by contradiction. + an < nk and. j = 1. . n. + an = k (x1 + x2 + . . x2 . . . . suppose that every sum of k numbers from x1 . . . x2 . n contradiction. . . we get But n2 ≤ n xk k=1 n x1 + x2 + . + 2 . . 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. . . ≤ xn . . we get that (−4) + 1 = 0. working modulo 9. . Therefore. Hence b = 1.. p = 2b.. so x1 + x2 + . . . which implies 4m ≡ 1 (mod 9). 5m + 1 = 2 · 3b . b Then we have 3 − 2a 3b + 2a = 5m . . < xn−k < 1. Adding the above equations. + xn−1 + xn ) < n. . y ∈ N.+xn−1 +xn < k. hence x1 < x2 < . then xn−k+1 < 1. for a certain k ∈ {1. . so p is even. 3b + 2a = 5y where x. it results that m is an odd number. . x < y. Reducing the equation p modulo 4 yields 1 = (−1) . + xn−k ) + (xn−k+1 + . so m = 1. 3 xk xk k=1 k=1 Problem 19. p = 2. Solve. in the positive integers. This leads to x1 + x2 + . .81 Selection tests for the 2014 JBMO If xn−k+1 +. we have 3 | m. Since 43 ≡ 1 (mod 9). hence 3b − 2a = 5x . so let n = 2a. n}. + xn = (x1 + x2 + . 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. . . What remains to be done in order to get the contradiction is to show that x1 + x2 + . + xn ≥ n. . *** Solution. . . But m is odd. 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. m Suppose that b ≥ 2. Obviously n is even. x + y = m. so x = 0 and y = m. a ∈ N∗ . + xn−k < n − k. 2. so it exists t ∈ N so that m = 6t + 3. we get 2 · 3b = 5x + 5y = 5x (1 + 5y−x ) . b ∈ N∗ . Reducing modulo 6. the equation 5m + n2 = 3p . . . n = 2. it follows that 2 · 3b = 5m + 1 ≥ 53 Therefore. We shall make use of the following: Lemma. m ≥ 3b−1 . *** A' A M D E' O F E D' D'' N G C B Solution. Using the law of sines in triangles BDM and BDN. An arbitrary diameter intersects side [AB] in D and side [AC] in E. the only possibility is b = 1. so v3 (m) = b − 1. for each t ∈ N. we conclude that v3 (5m + 1) = v3 (5 + 1) + v3 (m) .82 Selection tests for the 2014 JBMO Alternative Solution. n = 2. Let O be the circumcenter of the acute triangle ABC. b−1 + 1 > 6 · 3b−1 = 2 · 3b . Using the inequality 5t + 1 > 6t. If F is the midpoint of [BE] and G is the midpoint of [CD] . t ≥ 3. where m is an odd number. which leads to m = 1. we have 3b−1 ≥ 3. show that ∠F OG = ∠BAC. absurd. p = 2. Then DN EN BN CN Proof of the lemma. Let ABC be a triangle and M N be a chord of its circumcircle which BM CM DM EM : = : . If b ≥ 2. sin(∠ABN ) sin(∠BDN ) . we obtain 5m + 1 = 2 · 3b . Problem 20. using Lifting the Exponent Lemma. Since 3 | 5 + 1. we have BM DM = sin(∠ABM ) sin(∠BDM ) and DN BN = . Hence. By doing the same as above. intersects side [AB] in D and side [AC] in E. Conversely. hence D′ = D′′ . hence ∠COG = ∠CC ′ X. DN BN sin(∠ABN ) CM sin(∠ACM ) EM = · . EM = E ′ N and EN = E ′ M. Let X be an arbitrary point on the minor arc BC. let D′′ be the intersection of lines A′ C and M N . . But ∠ABM = ∠ACM Analogously. we get that OF � BA′ and OG � CA′ . Let B ′ the point diametrically opposed to B and C ′ the point diametrically opposed to C. and apply Pascal’s theorem. we get EN CN sin(∠ACN ) BM CM DM EM : = : . so DN EN BN CN Returning to our problem. OF is midsegment in △BB ′ E. D = AB ∩C ′ X. therefore ∠F OG = ∠BOC − (∠BOF + ∠COG) = 2∠BAC − ∠BAC = ∠BAC. so ∠F OG = ∠BA′ C = ∠BAC. Now. which means that D′ is in fact D). it will be the intersection of the diameter with the line AB. respectively. we conclude D′′ M D′ M that ′ = ′′ . which is the support line of a diameter. 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. 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. DN = D′ M. Also. B ′ is on the minor arc AC. E. Applying Pascal’s theorem to the hexagram ABB ′ XC ′ C shows that points O = BB ′ ∩CC ′ . and C ′ is on the minor arc AB. But clearly ∠BB ′ X +∠CC ′ X = ∠BAC and ∠BOC = 2∠BAC. Alternative Solution. so BM sin(∠ABM ) DM = · . we have sin(∠BDM ) = sin(∠BDN ). From the lemma. DN EN BN CN D N EN Since DM = D′ N. E = AC ∩B ′ X are collinear – on Pascal’s line. D′ = AB ∩ C ′ X will be collinear with O. DN D N Since [OF ] is a midsegment of the triangle BEE ′ and [OG] is a midsegment of the triangle CDD′ . if a diameter intersects the sides [AB] and [AC] at D and E. we have BM CM D′′ M E ′ M DM EM : = : = ′′ : ′ . the triangle ABC being acute. hence ∠BOF = ∠BB ′ X. OG is midsegment in △CC ′ D. � and ∠ABN = ∠ACN.Selection tests for the 2014 JBMO 83 Since ∠BDM + ∠BDN = 180◦ . hence. those along the sides three times and the interior coins will turn six times each. which would be the case if all coins showed tails. Since each move changes the parity of the number of heads of each color. yellow and blue so that any three adjacent coins have different colors. obtaining a net of equilateral triangles of side length 1. But the interior coins form the net corresponding to an n .sided triangle. in total. Through these points draw parallel lines to the sides of the triangles. Flipping the coins of each unit sided triangle of an equilateral triangle of side length n + 3. 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. all the exterior coins are turned tail up and all the interior coins are heads up. then color the coins in red. We shall use now induction of step 3.84 Selection tests for the 2014 JBMO and we are done. such turning is possible for n = 1. 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. For n = 2. say red. If 3 | n. so n is not an admissible value. In this case the corners will all have the same color. Obviously. Colombia 1997 Solution. On each of the vertices of the small triangles put a coin head up. Assume that n is an admissible value. the coins from the vertices of the big triangle will turn one time. so the induction works. flip each of the four 1-sided equilateral triangles once and all the coins will be tail-up. 2 exactly one more red coin than yellow or blue ones. Thus the coins cannot all be inverted. Also. any three coins in a row will have different colors. the parity of the number of red heads is different than the parity of the number of yellow heads. Thus. then there will be Since there are. at the beginning. we cannot end up with the parity of red heads equal to that of yellow or blue heads. Problem 21. Consequently. . On each side of an equilateral triangle of side n ≥ 1 consider n − 1 points that divide the sides into n equal segments. (n + 1) (n + 2) ≡ 1 (mod 3) coins. = xp = 1 and xp+1 = xp+2 = . Let M the set of the numbers n that satisfy the conditions above.. we conclude that n + 2 ∈ M. for any p ≥ 1. It follows that 3 ∈ / M. = x2p = −1. 1.+ = 0. We shall show that M = N \ {0. . . the relations being satisfied e. 1 1 1 + +. taking Moreover. + xn = x1 x2 xn xn+1 = 1 and xn+2 = −1. x3 ∈ R∗ so that x 1 + x2 + x3 = 1 1 1 + + = 0. . if n ∈ M and x1 + x2 + . x1 x2 xn xn+1 xn+2 Therefore. x1 x2 xn *** Solution. . Suppose that 3 ∈ M . . . x2 .THE DANUBE MATHEMATICAL COMPETITION C˘ al˘ ara¸si. October 2013 JUNIORS Problem 1. x1 x2 x3 This leads to x21 + x1 x2 + x22 = 0. 85 . x2 . all that is left to be done is to find the smallest odd element of M.. + = 0. for x1 = x2 = . We can observe that 2p ∈ M. . + x n = 1 1 1 + + . . + + + = 0.. . . . . + xn + xn+1 + xn+2 = 1 1 1 1 1 + + . . xn ∈ R such that ∗ x1 + x2 + . Find all integers n ≥ 2 for which there exist x1 . contradiction. which is possible if and only if x1 = x2 = 0.g. 3} .. then there exist x1 .. because x1 + x2 + .. . n is odd and cannot be equal to 1. c. which leads to x4. Let M and N be the midpoints of segments [CD] and [AD] . these pairs generate 2016 sums a + b which give 2016 remainders when divided by 2013. contradiction. Problem 4. there are two different pairs (a. Obviously. denoted by a. hence 2013 | (a + b) − (c + d) . then 2013 | b − d. Let P be the midpoint of [BC] and Q be the intersection point of EO and CD. Since (n2 −2n+2. with a < b. therefore m = 1 and n = 3. Find all natural numbers m. since |b − d| ≤ 2012. *** Solution. b) and (c. b. The perpendicular dropped from O to BD intersects lines AB and BC in E and F. b) and (c.5 = 2 Problem 2. 9m and 16m ). d. such that a + b − c − d is a multiple of 2013. Consider 64 distinct natural numbers less than or equal to 2012. Prove that among them there are four numbers. Assuming that (a. *** Solution. (n − 1)2 = 5m − 1 and (n + 1)2 = 17m − 1. and. it results . Problem 3. *** Solution. The equation can be written as 85m = (n2 − 2n + 2)(n2 + 2n + 2). Consequently. 3± 5 . n2 +2n+2) = 1. e. n so that 85m − n4 = 4. we get b = d. Consider ABCD a rectangle of center O with AB �= BC. d) have a common component. The conclusion now follows.g.86 2011 Danube Mathematical Competition Let us search for an example to prove that 5 ∈ M. Prove that F M ⊥ EN. it follows that n2 − 2n + 2 = 5m and n2 + 2n + 2 = 17m .g. then x4 + x5 = 3 and x4 x5 = 1. As [P M ] is the midsegment of the triangle BCD. Consequently. b) . For m > 1 there are many (more than one) perfect squares between 5m − 1 and 17m − 1 (e. d) having the same remainder when divided by 2013. there can be formed 2016 pairs (a. Using the 64 given numbers. Take x √1 = x2 = x3 = −1. a = c. Since the quadrilateral EN QP is a parallelogram. A. AcBaCb are concurrent. F M ⊥ EN. Consequently. The lines Aa and bC meet at D. c. But P C ⊥ M Q. B. which leads to P Q ⊥ M F. *** c A D' F' E' E a Z Y X B D F b C Solution. so QF ⊥ P M.87 2011 Danube Mathematical Competition that P M � BD and OQ ⊥ BD. the . b. Given six points on a circle. AbBcCa. M D C Q F N P O E A B SENIORS Problem 1. so F is the orthocenter of the triangle M P Q. similarly. a. C. we have P Q � EN. show that the Pascal lines of the hexagrams AaBbCc. and the lines Bb and cA meet at D′ to determine the Pascal line of the hexagram AaBbCc. greater than 1. so the conclusion follows unless n = 6m ± 1. then a2 + ab + b2 and an + bn + cn are both even. F F ′ are concurrent if and only if the pairs of lines DE and D′ E ′ . If p = 2. B n + C n + 1 ≡ 0 and B + C + 1 �≡ 0. By Desargues’ theorem. The condition B + C + 1 �≡ 0 shows that −C(B + 1)−1 �≡ 1. b. Since d is also a divisor of p − 1. Henceforth assume n odd. and the lines Ca and bB meet at E ′ to determine the Pascal line of the hexagram AbBcCa. Show that n and p − 1 are not coprime. Consequently. n be four integers. the conclusion follows. F D and F ′ D′ meet at three collinear points. The argument shows that if n is a prime greater than 3. the lines Cb and cB meet at F . Problem 2. so a−1 exists modulo p. Let a. Since the latter lie on the Pascal line of the hexagram AcBbCa. and the hypotheses yield B 2 + B + 1 ≡ 0. so C n ≡ B 2n ≡ (−B − 1)n . since B + 1 �≡ 0.) Hence 3 is a factor of p − 1. . where B = a−1 b and C = a−1 c. the lines DD′ . EF and E ′ F ′ . so c is also even by the second. b = 7. *** Solution. The first condition forces both a and b even. (Otherwise. but not a+b+c . that is. for instance. n a positive even integer. EE ′ . and p = 3 or a = 4. and a + b + c is odd. 3 ≡ 0. so p = 3. It is easily seen from the conditions in the statement that a �≡ 0. and the lines Ac and aC meet at F ′ to determine the Pascal line of the hexagram AcBaCb. finally. contradicting the third. where B �≡ 1. the conclusion follows. where n ≥ 2. Throughout the proof congruences are taken modulo p. p ≥ 7 and p − 1 is divisible by 6. This is precisely the case in the second example in the statement. and p = 31 satisfy these conditions. n = 5. −C(B + 1)−1 ≡ 1. c ≡ 1 (mod 3). then p − 1 is divisible by 6n. The first congruence yields B 3 ≡ 1. c = −13. c. Remarks.88 2011 Danube Mathematical Competition lines Bc and aA meet at E. C n ≡ 1 and C �≡ 1 which is impossible since n is odd. and let p be a prime dividing both a2 +ab+b2 and an +bn +cn . Begin by ruling out the case p = 2. 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. p must be odd. so the multiplicative order of −C(B + 1)−1 in Z∗p is a divisor d of n. a ≡ b ≡ −1 (mod 3). and the conclusion follows unless n is odd. in particular. Consider rnr − r + 1 nr disjoint copies of Gr . but if it is not. Consequently. that G3 must be colored in 2 or fewer colors which is. The case r = 2 is clear: Any cycle of even length works. Problem 4. with nr vertices say. Clearly. Set up a one-toone correspondence between the copies of Gr and the nr -element sets of extra vertices. Adjoin rnr − r + 1 extra vertices. for every integer r ≥ 2. of course. (Any larger odd number would do.) When Gr is defined. by descending induction. In the other cases. no Gr can be colored in less than r colors. for every real number x. deletion of some monochromatic classes of vertices together with their incident edges yields one such. there exists an r-chromatic simple graph (no loops. that is. Show that. Show that there exists a proper non-empty subset S of the set of real numbers such that. G3 is 3-chromatic. the set {nx + S : n ∈ N} is finite. as follows. *** Solution. It follows. then some nr of the extra vertices in Gr+1 must share the same color in C. where nx + S = {nx + s : s ∈ S}. Let H be a Hamel basis. The resulting graph is Gr+1 . nor multiple edges) which has no cycle of less than 6 edges. so the corresponding copy of Gr must be colored in r − 1 or fewer colors. impossible. This does not prove that Gr is r-chromatic. H is a set of real numbers such . define a sequence of graphs Gr . *** Solution. 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). r ≥ 3.89 2011 Danube Mathematical Competition Problem 3. The construction ensures that no graph Gr has a cycle of less than 6 edges. The graph G3 is a cycle of just 7 edges. If r ≥ 3 and Gr+1 has a coloring C in r or fewer colors. construct Gr+1 as follows. . . To this end. h) · h. We are now going to prove that the set S of those real numbers x whose q(x. h) in (∗).90 2011 Danube Mathematical Competition that every real number x can uniquely be written in the form q(x. x= (∗) h∈H where the q(x. where n is a non-negative integer. h) in (∗) are all integral satisfies the required condition. . to conclude that any set of the form nx + S. 1. notice that m(x) · x is a member of S. 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. m(x) − 1. . 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. r = 0. Finally. must be one of the sets rx + S. h) are all rational and vanish for all but a finite number (depending on x) of h’s. . Since ±1 are both proper. Given a prime p ≥ 5. 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. show that there exist at least two distinct primes q and r in the range 2. . November 2013 Problem 1. S. 2} and n = ±1 in (2) shows that p ± 1 and 2p ± 1 are all improper. Haret National College. 3. all congruences are to be understood modulo p2 . and (2) If k is an integer coprime to p and n is a proper integer. so each has an improper prime divisor by (1).THE Tenth IMAR MATHEMATICAL COMPETITION Bucure¸sti. In what follows. Our solution is based on the following two simple facts: (1) An improper integer greater than 1 has at least one improper prime divisor. then kp − n is improper. and improper otherwise. The first claim follows from the fact that the product of two proper integers is again proper. p − 2 such that q p−1 �≡ 1 (mod p2 ) and rp−1 �≡ 1 (mod p2 ). 91 . *** Solution. An integer n coprime to p will be called proper if np−1 ≡ 1. .. so is kp − n). For the second. so kp − n is indeed improper (since n is coprime to p. letting k ∈ {1. and they are distinct since p − 2 and p − 1 are coprime. Clearly. set k = 1 and n = p − 2 in (2) to deduce that 2 is improper. On the other hand. begin by noticing that the sequence is periodic modulo 9. the prime factors of p ± 1 are all less than p − 1. setting k = n = 1 in (2) shows that p − 1 is improper. n ∈ N. Alternative Solution. s6m+5 ≥ s6m+4 + 7. The answer is in the negative. If p = 5. and deduce thereby that s6m+6 ≥ s6m + 27. and the conclusion follows. it has an improper prime divisor q by (1). look for an improper odd prime in the required range. Problem 2. q and r both lie in the required range. since p − 4 is odd. say from some rank n0 on. of period 6. so is −4p + 4. so it has an improper prime divisor r by (1). s6m+4 ≥ s6m+3 + 8. s6m+2 ≥ s6m+1 + 2. 2. clearly. the primes q = 2 and r = 3 satisfy the required conditions. For every non-negative integer n. Fix a non-negative integer m such that 6m ≥ n0 to write s6m+1 ≥ s6m + 1. 7. the first block of values it takes on being 1. and since 2 is the highest common factor of p − 1 and p + 1. In the setting of the previous solution.92 2011 IMAR Mathematical Competition Since p ≥ 5. so its prime factors are all less than p − 1. they are all odd. 5. Suppose. To this end. if possible. the conclusion follows. Setting k = −3 and n = −4p + 4 in (2) shows p − 4 improper. so it has an improper prime divisor s by (1). for p ≥ 5. provided that 2 is proper. Finally. If p − 2 is proper. so let p ≥ 7. On the other hand. and it should now be clear that the primes 2 and s satisfy the required conditions. Is the sequence (sn )n∈N eventually increasing? *** Solution. s6m+6 ≥ s6m+5 + 5. Otherwise. let sn be the sum of the digits in the decimal expansion of 2n . (∗) . so is s. distinguish the following two cases: If p−2 is improper. s6m+3 ≥ s6m+2 + 4. notice that one of the numbers 2p ± 1 is divisible by 3. so s6m+6n ≥ s6m + 27n. since (p − 2)2 is proper. To prove this. that the sequence is eventually increasing. since 1 is proper. 4. 8. and each disc contains points of the closure of the triangle determined by the centers of the other three discs. If the orthogonal projection of B on the line AC falls on the closed ray EA emanating from E. Problem 3. it follows that A1 O ≤ A1 B0 . if possible. On the other hand. let ∆ be a disc centered at A. Amongst the four discs. We now apply the lemma to show that O is also covered by ∆1 . not covered by the latter. the number of non-vanishing digits in the decimal expansion of 26m+6n does not exceed ⌈(6m + 6n) log10 2⌉ < 2m + 2n.2011 IMAR Mathematical Competition 93 On the other hand. containing the point O where the diagonals of the quadrangle cross one another. we take time out to state a simple. the latter lies on the closed segment A0 B0 . Show that three of these discs cover the closure of the triangle determined by their centers. A1 . To this end. contradicting (∗) for n large enough. which in turn is greater than or equal to A1 B0 by the lemma. Before proceeding. The conclusion follows. *** Solution. label the other three centers in circular order. then dist (B. choose one. so that the opposite angles A0 OA1 and A2 OA3 be not obtuse. so the radius of ∆1 is greater than or equal to dist (A1 . say ∆0 . O is indeed covered by ∆1 . and let ∆i denote the disc centered at Ai . where [ACD] is the closure of the triangle ACD. that the conclusion does not hold. so s6m+6n ≤ 18m + 18n. Let A0 be the center of ∆0 . [ACD] \ ∆) ≥ BE. ∆1 contains points of the closure [A0 A2 A3 ] of the triangle A0 A2 A3 not covered by ∆0 . Then no three discs meet. . [A0 A2 A3 ] \ ∆0 ). but quite useful lemma whose proof is postponed for the sake of clarity. A3 . A2 . let B0 be the point where the ray A0 O emanating from A0 crosses the the boundary of ∆0 . Suppose. and since the angle A0 OA1 is not obtuse. Lemma. and let E be the point where the ray AC emanating from A crosses the boundary of ∆. Consequently. Since ∆0 contains O. The closure (interior and boundary) of a convex quadrangle is covered by four closed discs centered at each vertex of the quadrangle each. Let ABCD be a convex quadrangle. Notice that. to conclude that r2 ≥ A2 B3 . Hence A′2 lies on the open ray B3 A3 emanating from B3 . It follows. the length of the segment BX varies again increasingly. then the closed segment BX meets either α or r (this fails to hold if the quadrangle ABCD is not convex at D). Finally. so BX ≥ BE. The open segments A2 B3 and A3 B2 cross each other. If A′2 fell on the closed segment B3 O. the whole configuration of points lies on one side of the line AB. since the orthogonal projection of B on the line AC is not interior to r. so it is sufficient to consider only points X in α ∪ r.94 2011 IMAR Mathematical Competition Recall that no three discs meet to deduce that neither ∆2 . so r2 + r3 = A2 B2 + A3 B3 < A2 B3 + A3 B2 . Since the distance to the empty set may take on any value. for i = 2. it may very well happen that . where ri is the radius of the disc ∆i . 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). the conclusion of the lemma still holds if ∆ covers [ACD]. This ends the proof of the lemma and completes the solution. not containing A. Under the conditions in the lemma. Finally. if X is a point in [ACD] \ ∆. that the open segment Ai O crosses the boundary of ∆i at some point Bi . so BX ≥ BE again. as a point X traces α from E to F . 3. Proof of the lemma. nor ∆3 contains O. Remarks. [A0 A1 A3 ] \ ∆3 ) ≥ A2 B3 by the lemma. and let r be the ray emanating from E along the line AC. We are presently going to show that r2 ≥ A2 B3 and r3 ≥ A3 B2 and reach thereby the contradiction we were heading for. 3. Now. in contradiction with the second remark in the opening paragraph. the argument applies mutatis mutandis to the other. i = 2. let α denote the arc EF of the boundary of ∆ situated in H. say H. recall that ∆2 covers points in [A0 A1 A3 ] \ ∆3 . Since the quadrangle ABCD is convex. Since the angle A2 OA3 is not obtuse. 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 . so dist (A2 . the orthogonal projection A′2 of A2 on the line A1 A3 falls on the closed ray OA3 emanating from O. Let F be the point where the ray AD emanating from A crosses the boundary of ∆. as X runs along r away from E. Only the first inequality will be dealt with. D or the projection of B on the line AC does not fall on the closed ray EA emanating from E. M ′ OX ′ ). Z. Z ′ . and let M ′ be that of the triangle X ′ Y ′ Z ′ . Given a triangle ABC. labeled in circular order: X. We begin by reviewing some basic facts on conics. *** A Y' Z Z' M' O Y M B X C X' Solution. the locus of X and X ′ is then a circle Γ centered at O. Such configurations are easily produced.e. 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. CA. foci M and M ′ . B. AB in the pairs of points X and X ′ . the point of concurrence of the circles AY Z. Y ′ . where M O = M ′ O. If t varies and θ is constant. Finally. Problem 4. Y and Y ′ . Let M be the Miquel point of the triangle XY Z (i. M ′ ) and a homothety of ratio sec θ with center M (respectively. a circle centered at some point O meets the segments BC. −θ) about M (respectively.95 2011 IMAR Mathematical Competition dist (B. Application to triangle M N P (respectively. Z and Z ′ .. BZX. M ′ N P ′ ) of a rotation θ (respectively. . 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. Y . For an ellipse Σ with center N . respectively. in which case C is certainly interior to ∆. semiaxes a and b. X ′ both lie on t. Prove that the segments OM and OM ′ have equal lengths. [ACD] \ ∆) > BE. N O = 21 · M M ′ · tan θ. M ′ ) yields triangle M OX (respectively. X ′ . so that N P = a = N P ′ . and X. OX = a sec θ = OX ′ . CXY ). Similar considerations hold for a hyperbola Σ. AB. respectively. Y . M ′ are an isopair and Γ is an isopedal circle of M .96 2011 IMAR Mathematical Competition By Cartesian geometry it is readily checked that Γ and Σ are bitangent. M ′ Y ′ . Since ℓ and the circle OM M ′ are mutually inverse in Γ. for any pair of triangles inscribed in a triangle ABC and in a circle Γ (as in the statement of the problem). the inverse points of M and M ′ in Γ both lie on ℓ. (If M . M ′ and are inscribed in a common isopedal circle Γ bitangent to Σ. −θ for the isopair M . M ′ . Z (respectively. and the line ℓ supporting their common chord is also the radical axis of the circles Γ and OM M ′ . AB. Conversely. the foci of a conic Σ touching its sides). respectively. M ′ are the Brocard points. Consider now an isopair M . Σ is the Brocard ellipse and Γ is a Tucker circle. X ′ . then the isopedal triangles XY Z. M ′ have their Miquel points at M . −θ) with the perpendiculars to BC. CA. M ′ (two isogonally conjugate in the triangle ABC. Y ′ . centered at a point O on the perpendicular bisector of the segment M M ′ . the distance d between the parallel lines ℓ and M M ′ is given by d · M M ′ = 2b2 tan θ. Finally. M ′ X ′ . Z ′ ) on lines BC. with this real geometrical significance even if the bitangency is not real. M ′ Z ′ ) make the same directed angle θ (respectively. M Z (respectively. M Y . so that the lines M X. and take points X.) . the Miquel points M . CA. 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