OLYMPIAD SOLUTIONS. OC16. Given a 1 1 and a k+1 a k + 1 for all k = 1, 2,..., n, show that

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1 THE OLYMPIAD CORNER / 135 OLYMPIAD SOLUTIONS OC16 Given a 1 1 and a k+1 a k + 1 for all k = 1, 2,, n, show that a a a 3 n (a 1 + a a n ) 2 (Originally question # 3 from 2010 Singapore Mathematical Olympiad, Senior Section, Round 2) Solved by Arkady Alt, San Jose, CA, USA ; Michel Bataille, Rouen, France ; Felix Boos, University of Kaiserslautern, Kaiserslautern, Germany ; Marian Dincă, Bucharest, Romania ; Oliver Geupel, Brühl, NRW, Germany ; Paolo Perfetti, Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, Rome, Italy ; Edward TH Wang, Wilfrid Laurier University, Waterloo, ON and Titu Zvonaru, Cománeşti, Romania We give the solution of Zvonaru Let s k = a 1 + a a k We first prove by induction that a n a n 1 2s n 1 The case n = 2 follows immediately from a 2 a , while the inductive step is the following: a n+1 a n (a n 1 + 2)a n = a n 1 a n + 2a n 2s n 1 + 2a n = 2s n Also, let s observe that equality can be obtained only if a n+1 = a n for all n, that is if a n = n Now, we prove the problem by induction For n = 1 the problem is equivalent to We now prove the inductive step: a 3 1 a 2 1 a 1 1 a a a 3 n+1 s 2 n + a 3 n+1 s 2 n + a 2 n+1(a n + 1) = s 2 n + a 2 n+1a n + a 2 n+1 s 2 n + 2s n a n+1 + a 2 n+1 = s 2 n+1 Equality only holds if a n = n for all n OC17 Prove that the vertices of a convex pentagon ABCDE are concyclic if and only if the following holds d(e, AB) d(e, CD) = d(e, AC) d(e, BD) = d(e, AD) d(e, BC) (Originally question # 6 from the 60 th national mathematical Olympiad Selection Tests for the Balkan and IMO, 2nd selection test) Copyright c Canadian Mathematical Society, 2013

2 136/ THE OLYMPIAD CORNER Solved by Oliver Geupel, Brühl, NRW, Germany Lemma 1: Let E, P, Q be distinct points on a circle Γ Let R, S, T be the feet of the perpendiculars from E onto the line P Q and onto the tangents to Γ in P and Q, respectively Then, ER 2 = ES ET E Γ S O T P R Q Proof: Since ESP + ERP = 180 the quadrilateral ESPR is cyclic Thus Since P S is tangent to Γ, we get Thus ESR = EP Q ; SRE = SP E SP E = P E 2 = EQP ESR = EP Q ; SRE = EQP The same argument now works in ET QR: ET Q + ERQ = 180, hence the quadrilateral ETQR is cyclic Thus ET R = EQP ; ERT = EQT = EP Q Thus, SRE RT E which implies our claim Corollary 2: If ABCDE is a convex cyclic pentagon, then the following holds: d(e, AB) d(e, CD) = d(e, AC) d(e, BD) = d(e, AD) d(e, BC) (1) Proof: Let A 0, B 0, C 0, D 0 be the feet of the perpendiculars from E onto the tangents to the the circumcircle of the pentagon in A, B, C, D, respectively By Lemma 1, we have d(e, AB) 2 = EA 0 EB 0, d(e, CD) = EC 0 ED 0 and similar relations for the remaining distances Hence, each expression in the equation (1) equals to EA 0 EB 0 EC 0 ED 0 Lemma 3: Let ABCDE be a convex pentagon such that the equation (1) holds Then, the quadrilateral ABCD is cyclic Crux Mathematicorum, Vol 38(4), April 2012

3 THE OLYMPIAD CORNER / 137 Proof: Let m = d(e, AB) d(e, CD), α = AEB, β = BEC, γ = CED Then, by the relation (1), we have m (AC BD AB CD AD BC) AE BE CE DE = d(e, AC) AC AE CE d(e, BD) BD BE DE d(e, BC) BC BE CE d(e, AB) AB AE BE d(e, AD) AD AE DE = sin(α + β) sin(β + γ) sin α sin γ sin(α + β + γ) sin β d(e, CD) CD CE DE = (sin α cos β + cos α sin β)(sin β cos γ + cos β sin γ) sin α sin γ = 0, (sin α cos β cos γ + cos α sin β cos γ + cos α cos β sin γ sin α sin β sin γ) sin β that is, AC BD AB CD AD BC = 0 By the converse of Ptolemy s Theorem, the points A, B, C, D are concylic Corollary 4: Let ABCDE be a convex pentagon such that the equation (1) holds Then, the pentagon is cyclic Proof: By Lemma 3, A, B, C, D lie on a common circle Γ It is enough to show that E also lies on Γ Let F be the foot of the perpendicular from E onto the line AB Let Γ be the unit circle in the complex plane and let a, b, c, d, e, f be the coordinates of A, B, C, D, E, F We have d(e, AB) 2 = (f e)(f e) = 1 (a + b e abe) 1 4 a + 1 b e e ab = 1 (a + b e abe)2 4ab and similar relations for the remaining distances Direct computation yields 16abcd d(e, AB) 2 d(e, CD) 2 = e 2 σe τe + (ab + cd)ee + (a + b)(c + d) 2, where σ = a+b+c+d and τ = abc+abd+acd+bcd, and two analogous identities Let z denote the principal value of the square root of the complex number 16abcd d(e, AB) 2 d(e, CD) 2 By (1), there are numbers ϑ 1, ϑ 2, ϑ 3 { 1, 1} such that z = ϑ 1 e 2 σe τe + (ab + cd)ee + (a + b)(c + d) = ϑ 2 e 2 σe τe + (ac + bd)ee + (a + c)(b + d) = ϑ 3 e 2 σe τe + (ad + bc)ee + (a + d)(b + c) By the Pigeonhole Principle, two of the numbers ϑ 1, ϑ 2, ϑ 3 must be equal If, for example, ϑ 1 = ϑ 2, then we obtain 0 = ϑ 1 [(ab + cd)ee + (a + b)(c + d) (ac + bd)ee (a + c)(b + d)] = ϑ 1 (ee 1)(a d)(b c), Copyright c Canadian Mathematical Society, 2013

4 138/ THE OLYMPIAD CORNER whence ee = 1, that is E Γ The other cases ϑ 2 = ϑ 3 and ϑ 3 = ϑ 1 are analogous From Corollary 2 and Corollary 4, the equivalence stated in the problem follows OC18 Suppose a 1, a 2,, a n are n non-zero complex numbers, not necessarily distinct, and k, l are distinct positive integers such that a k 1, a k 2,, a k n and a l 1, a l 2,, a l n are two identical collections of numbers Prove that each a j, 1 j n, is a root of unity (Originally question # 2 from the problems used in selection of the Indian team for IMO-2009) Solved by Oliver Geupel, Brühl, NRW, Germany Let 1 j n be arbitrary By hypothesis, we can construct inductively a sequence i 1 = j, i 2,, i k, so that for all m 1 we have a k i m = a l i m+1 Then, by induction, one can easily prove that a km i 1 = a lm i m Since 1 i m n for all m, there exists some q < r such that i q = i r Then This implies that a kr i 1 = a lr i r = a lr i q = a lq i qšl r q = a kq i 1Šl r q a kr j = a kq l r q j a kq l r q k r j = 1 Let s observe that k q l r q k r 0, since otherwise we would have k r q = l r q, and hence k = l Thus, we proved that there exists some integer p 0 so that This shows that a j is a root of unity a p j = 1 OC19 There were 64 contestants at a chess tournament Every pair played a game that ended either with one of them winning or in a draw If a game ended in a draw, then each of the remaining 62 contestants won against at least one of these two contestants There were at least two games ending in a draw at the tournament Show that we can line up all the contestants so that each of them has won against the one standing right behind him (Originally question # 3 from 53 rd national mathematical Olympiad in Slovenia, 3 rd selection Exam for the IMO 2009) Solved by Oliver Geupel, Brühl, NRW, Germany We prove by induction that the claim holds for any number n 4 of contestants Crux Mathematicorum, Vol 38(4), April 2012

5 THE OLYMPIAD CORNER / 139 Let us write A > B, A < B, or A B if A won against B, A lost against B, or A tied with B, respectively Let P denote the property that, if a game ended in a draw, then each of the remaining n 2 players won against at least one of these two contestants Observe that no player tied with more than one other player For, if A tied with B and C, where B C, then by hypothesis P, we have B > C and C > B, a contradiction We are now ready to prove our initial claim by mathematical induction In the base case there are four players A, B, C, and D Since no contestant had more than one tie, there is no loss of generality in assuming A B and C D By symmetry we may further assume A > C Condition P now yields C > B and B > D Then, A > C > B > D is a chain with the desired property, which completes the base case For the induction step, assume that the claim holds for each tournament of n = k 4 contestants Consider a tournament of k + 1 players A 1,, A k+1 If each contestant had a game that ended in a draw, then the total number of draws is k + 1 3, 2 and there are at least two draws in the subtournament of the players A 1,, A k Otherwise there is a player, say A k+1, without any draws In each case there are two draws in the subtournament of the contestants A 1,, A k By the induction hypothesis, these k contestants can be arranged in a descending chain, say, A 1 > > A k By the property P, the contestant A k+1 won against another player Let i be the least index such that A k+1 > A i If i = 1, then A k+1 > A 1 > A 2 > > A k is a descending chain of length k + 1 and we are done It remains to consider i > 1 If A i 1 A k+1 then A i lost against both A i 1 and A k+1, which contradicts P Thus, A i 1 > A k+1 We obtain the chain This completes the induction OC20 A 1 > > A i 1 > A k+1 > A i > > A k Given an integer n 2, determine the maximum value the sum x 1 + x x n may achieve, as the x i run through the positive integers subject to x 1 x 2 x n and x 1 + x x n = x 1 x 2 x n (Originally question #10 from 60 th national mathematical Olympiad selection tests for the Balkan and IMO, 4 th selection test) Solved by Michel Bataille, Rouen, France ; Oliver Geupel, Brühl, NRW, Germany ; Edward TH Wang, Wilfrid Laurier University, Waterloo, ON and Titu Zvonaru, Cománeşti, Romania We give the solution of Zvonaru Copyright c Canadian Mathematical Society, 2013

6 140/ THE OLYMPIAD CORNER We denote s n = x 1 + x x n We claim that the maximum value s n can achieve is 2n If x 1 = 0 then x x n = 0, thus s n = 0 Now, suppose x 1 1 If x n = 1, then x 1 = = x n = 1 and hence n = x x n = x 1 x n = 1, a contradiction to n 2 Hence, for the rest of the proof we can assume 1 x 1 x 2 x n and x n 2 For n = 2, the equality x 1 + x 2 = x 1 x 2 is equivalent to (x 1 1)(x 2 1) = 1, hence x 1 = x 2 = 2 and s 2 = 4 Suppose now that n 3 It is easy to see that the equality can be rewritten as x x n = x 1 x n (x 1 1)(x 2 1) + (x 1 x 2 1)(x 3 1) + + (x 1 x 2 x n 1 1)(x n 1) = n 1 Thus (x 1 x 2 x n 1 1)(x n 1) n 1 (1) Since (x 1 x 2 x n 1 1)x n = x x n 1 0, x n 1 n 1 x n n Now, substituting x 1 x n 1 = sn into (1), we get x n s n x n 1 1 (x n 1) n 1 s n n 1 + x n 1 x n x n Let s observe that s n x2 n + (n 2)x n x n 1 x 2 n + (n 2)x n x n 1 2n x 2 n (n + 2)x n + 2n 0 (x n 2)(x n n) 0, which is true since 2 x n n This shows that s n 2n Moreover, when x n = n, x n 1 = 2, x n 2 = = x 1 = 1, we have s n = 2n, which shows that 2n is the maximum value the sum can achieve Crux Mathematicorum, Vol 38(4), April 2012