The 2017 Danube Competition in Mathematics, October 28 th. Problema 1. Să se găsească toate polinoamele P, cu coeficienţi întregi, care

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1 The 017 Dnube Competition in Mthemtics, October 8 th Problem 1. ă se găsescă tote polinomele P, cu coeficienţi întregi, cre verifică relţi + b c P () + P (b) P (c), pentru orice numere întregi, b, c. Problem. Pe o tblă sunt scrise n numere rele, unde n este un număr nturl. Printr-o trnsformre, legem două dintre numere, le ştergem şi poi le înlocuim (pe fiecre) cu produsul lor. ă se găsescă numerele nturle n pentru cre, oricre r fi secvenţ de n numere scrise iniţil, este posibil c, după un număr finit de trnsformări, să obţinem pe tblă n numere egle. Problem. Fie H ortocentrul şi O centrul cercului circumscris triunghiului ABC. Fie F piciorul înălţimii din C şi M mijlocul segmentului CH. Fie N piciorul perpendiculrei din C pe prlel din H l OM, fie D A pe drept AB stfel încât CA = CD, şi fie P intersecţi dreptelor BN şi CD. Dcă Q este intersecţi dreptelor HP şi AC rătţi că QF este perpendiculră pe OF. Problem. În fiecre dintre pătrăţelele unitte le unei reţele infinite este scris un număr, stfel încât modulul sumei numerelor din orice pătrt le cărui lturi sunt determinte de liniile reţelei este mi mic su egl decât 1. ă se demonstreze că modulul sumei numerelor din orice dreptunghi cu lturile determinte de liniile reţelei este mi mic su egl decât.

2 The 017 Dnube Competition in Mthemtics, October 8 th Problem 1. Find ll polynomils P, with integer cofficients, such tht, for ll integers, b, c, + b c P () + P (b) P (c). olution. Tke = k, b = k, c = 5k, with integer k. Then 0 divides P (k) + P (k) P (5k), so P (k) + P (k) P (5k) = 0 for infinitely mny k, hence P (X) + P (X) P (5X) = 0. If P is not nil, sy tht P hs degree n nd leding coefficient n. Then n n + n n = 5 n n, so, with stndrd rgument, n =. Further 0 P (0) implies P = αx +βx nd P () for ll integer implies β = 0, therefore P = αx, with integer α. Clerly, every such polynomil fulfills the condition. Problem. There re written n rel numbers on blckbord, where n is nturl number. In trnsformtion we choose two numbers, erse them nd replce ech of them by their product. Find n such tht, for ny initil n-tuple, it is possible to obtin n equl numbers on the blckbord fter finite number of trnsformtions. olution. We will show tht every even n stisfies the condition. We shll proceed by induction. The clim is trivil for n =. For n = we pply the cse n = in four steps. Now we suppose tht the clim is true for ny even number k nd prove it for n = k +. By the induction hypothesis, we first construct n n-tuple of the form (,,...,, b, b). Now, for every i =,,..., k, we trnsform the pirs (i, k + 1), obtining thus the n-tuple (,, b, b, b,..., k b, k b, b). After selecting the lst two numbers, we get (,, b, b,..., k b, k b, k b ). Finlly, we use the trnsformtion step by step for the pirs t the positions (i, n + 1 i) for i = 1,,..., n, obtining the sequence ( k 1 b, k 1 b,..., k 1 b ). For odd n we consider the n-tuple (,,..., ). Let m be the number of occurrences of the mximum elements (fter ech step) in the n-tuple. Initilly we hve m = n 1. This number remins every time even, contrdicting the fct tht n is odd.

3 Problem. Denote H the orthocentre, O the circumcentre, F the foot of the ltitude from C nd M the midpoint of the segment CH in tringle ABC. Let N be the orthogonl projection of C onto the prllel from H to OM, let D A be on AB such tht CA = CD nd let P be the common point of BN nd CD. If HP intersects AC in Q, prove tht QF OF. E C' T B F D A P Q H O M N C olution. In order to get the conclusion, we prove tht F is the midpoint of the chord of C(ABC) through Q nd F. To this end, denote C the dimetrlly opposed point of C in C(ABC). Then OM is midline in tringle CHC, so HC OM, hence N HC. We show tht the qudrilterl CP HN is cyclic (1). Indeed, CP N = BCP + CBP = ADC ABC + CBN = BAC ABC CBN CHN = HCC + HC C = F CB BCC + CC N = 90 F BC 90 + BC C + CC N = BAC ABC + CBN therefore CP N = CHN, which proves (1). This yields P HC = P NC = BAC. Denote E = CH C(ABC) nd T = QF BE. Then HF Q EF T (AA), becuse HF = F E, HF Q = EF T nd F HQ = P HC = P NC = BNC = BAC = F ET. This shows tht QF = F T. () We will use now the following result, which is converse of The Butterfly Theorem (BT). Converse of BT. If the chords XY nd ZT of circle C meet t N nd line through N meets the chords t P nd Q such tht NP = NQ nd meets C t R nd, then NR = N. X Z R P N Q T Y

4 Proof of the lemm. If N is the center of the circle, everything is cler. Otherwise, consider the chord of C whose midpoint is N nd its intersections P, Q with XY nd ZT. Then BT shows tht NP = NQ. o Q belongs to the reflection of XT cross the perpendiculr bisector of this chord. ince this reflection is on line different from Y Z, the point Q is uniquely determined, so it must coincide with Q. Now the Lemm nd () shows tht F is the midpoint of the chord of C through Q, F nd T, whence the conclusion. Problem. In ech unit squre of n infinite plne grid we write rel number such tht the bsolute vlue of the sum of the numbers in ny squre whose sides re determined by the lines of the grid is less thn or equl to 1. Prove tht the bsolute vlue of the sum of the numbers in ny rectngle whose sides re determined by the lines of the grid is less thn or equl to. olution. uppose tht there is rectngle (the bolded rectngle in the figures) with sides nd b ( < b) such tht the bsolute vlue of the sum of its numbers is equl to + ɛ, (ɛ > 0). In the exterior of this rectngle we construct four squres with sides b (see the figures) to obtin new rectngle (the dotted rectngle in the figures) with sides b nd b. It is cler tht the cse b = is not possible. 1 b Cse 1 Cse Denote by i, i = 1, 9 the corresponding sums of the numbers in the nine rectngles. uppose tht the sum of the numbers in the initil rectngle (i.e ) is + ɛ, otherwise we chnge the signs of ll numbers. We consider two cses: Cse 1. b <. Then 5 = ( + 5 ) + ( ) ( ) ɛ = ɛ, + 8 = ( ) + ( ) ( ) ( + ɛ) = ɛ, nd thus ɛ. Cse. b >. Then 5 = ( ) 6 + ɛ 1 1 = + ɛ,

5 + 8 = ( 1 + ) + ( + ) + ( ) + ( ) ( ) ( )+ +( ) + ɛ, nd thus ɛ. In both cses, strting from rectngle ( 1 =, b 1 = b), we obtin new rectngle (, b ) with sides = b, b = b (in the first cse) nd = b, b = b (in the second cse), such tht the bsolute vlue of the sum of its numbers is greter thn or equl to + ɛ. We cn continue the procedure to get sequence ( n, ) of rectngles such tht the bsolute vlue of the sum of their numbers is greter thn or equl to + n 1 ɛ. We shll prove now tht ll rectngles ( n, ) in this sequence (for ny n n 0, where n 0 is lrge enough) re of the second type, tht is > n. First of ll it esy to check tht if n < 1 then n+1 +1 n bn > 1, then 1 n+1 +1 = 1 1 n 1 n n n = bn n n < 1 = >, n. On the other hnd, if nd thus fter finite number of steps we get 1 n > 1 n, tht is < 1. For simplicity we suppose now tht 1 b 1 < 1. Then = b = (b 1 1 ) (b 1 1 ) = 1 b = b = (b 1 1 ) (b 1 1 ) = b 1. In generl we obtin k+1 = k 1 nd b k+1 = k b 1. The bsolute vlue of the sum of the numbers in the rectngle ( k 1, k b 1 ) is greter thn + 9 k ɛ. This vlue, for lrge enough k, could be s lrge s we wish. The contrdiction is obtined now becuse ny rectngle (m 1, mb 1 ), with 1, b 1, m nturl numbers, cn be decomposed in 1 b 1 squres (with side m) nd thus, the bsolute vlue of the sum of the numbers in this rectngle is less thn or equl to 1 b 1.