MathDB

2010 IMO Shortlist

Part of IMO Shortlist

Subcontests

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x_1 + x_2 + ... + x_n >= 0

A sequence x1,x2,…x_1, x_2, \ldots is defined by x1=1x_1 = 1 and x2k=−xk,x2k−1=(−1)k+1xkx_{2k}=-x_k, x_{2k-1} = (-1)^{k+1}x_k for all k≥1.k \geq 1. Prove that ∀n≥1\forall n \geq 1 x1+x2+…+xn≥0.x_1 + x_2 + \ldots + x_n \geq 0.
Proposed by Gerhard Wöginger, Austria

IMO Shortlist 2010 - Problem N4

Let a,ba, b be integers, and let P(x)=ax3+bx.P(x) = ax^3+bx. For any positive integer nn we say that the pair (a,b)(a,b) is nn-good if n∣P(m)−P(k)n | P(m)-P(k) implies n∣m−kn | m - k for all integers m,k.m, k. We say that (a,b)(a,b) is very goodvery \ good if (a,b)(a,b) is nn-good for infinitely many positive integers n.n. * Find a pair (a,b)(a,b) which is 51-good, but not very good. * Show that all 2010-good pairs are very good.
Proposed by Okan Tekman, Turkey
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36 <= 4*sum(a^3) - sum(a^4) <= 48

Let the real numbers a,b,c,da,b,c,d satisfy the relations a+b+c+d=6a+b+c+d=6 and a2+b2+c2+d2=12.a^2+b^2+c^2+d^2=12. Prove that 36≤4(a3+b3+c3+d3)−(a4+b4+c4+d4)≤48.36 \leq 4 \left(a^3+b^3+c^3+d^3\right) - \left(a^4+b^4+c^4+d^4 \right) \leq 48.
Proposed by Nazar Serdyuk, Ukraine

There exist N flags forming a diverse set

On some planet, there are 2N2^N countries (N≥4).(N \geq 4). Each country has a flag NN units wide and one unit high composed of NN fields of size 1×1,1 \times 1, each field being either yellow or blue. No two countries have the same flag. We say that a set of NN flags is diverse if these flags can be arranged into an N×NN \times N square so that all NN fields on its main diagonal will have the same color. Determine the smallest positive integer MM such that among any MM distinct flags, there exist NN flags forming a diverse set.
Proposed by Tonći Kokan, Croatia

IMO Shortlist 2010 - Problem N2

Find all pairs (m,n)(m,n) of nonnegative integers for which m2+2⋅3n=m(2n+1−1).m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).
Proposed by Angelo Di Pasquale, Australia
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Least common multiple of steps of these progressions

Let P1,…,PsP_1, \ldots , P_s be arithmetic progressions of integers, the following conditions being satisfied:
(i) each integer belongs to at least one of them; (ii) each progression contains a number which does not belong to other progressions.
Denote by nn the least common multiple of the ratios of these progressions; let n=p1α1⋯pkαkn=p_1^{\alpha_1} \cdots p_k^{\alpha_k} its prime factorization.
Prove that s≥1+∑i=1kαi(pi−1).s \geq 1 + \sum^k_{i=1} \alpha_i (p_i - 1).
Proposed by Dierk Schleicher, Germany

IMO Shortlist 2010 - Problem G7

Three circular arcs γ1,γ2,\gamma_1, \gamma_2, and γ3\gamma_3 connect the points AA and C.C. These arcs lie in the same half-plane defined by line ACAC in such a way that arc γ2\gamma_2 lies between the arcs γ1\gamma_1 and γ3.\gamma_3. Point BB lies on the segment AC.AC. Let h1,h2h_1, h_2, and h3h_3 be three rays starting at B,B, lying in the same half-plane, h2h_2 being between h1h_1 and h3.h_3. For i,j=1,2,3,i, j = 1, 2, 3, denote by VijV_{ij} the point of intersection of hih_i and γj\gamma_j (see the Figure below). Denote by VijVkj^VklVil^\widehat{V_{ij}V_{kj}}\widehat{V_{kl}V_{il}} the curved quadrilateral, whose sides are the segments VijVil,V_{ij}V_{il}, VkjVklV_{kj}V_{kl} and arcs VijVkjV_{ij}V_{kj} and VilVkl.V_{il}V_{kl}. We say that this quadrilateral is circumscribedcircumscribed if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals V11V21^V22V12^,V12V22^V23V13^,V21V31^V32V22^\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}} are circumscribed, then the curved quadrilateral V22V32^V33V23^\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}} is circumscribed, too.
Proposed by Géza Kós, Hungary
[asy] pathpen=black; size(400); pair A=(0,0), B=(4,0), C=(10,0); draw(L(A,C,0.3)); MP("A",A); MP("B",B); MP("C",C); pair X=(5,-7); path G1=D(arc(X,C,A)); pair Y=(5,7), Z=(9,6); draw(Z--B--Y); struct T {pair C;real r;}; T f(pair X, pair B, pair Y, pair Z) { pair S=unit(Y-B)+unit(Z-B); real s=abs(sin(angle((Y-B)/(Z-B))/2)); real t=10, r=abs(X-A); pair Q; for(int k=0;k<30;++k) { Q=B+t*S; t-=(abs(X-Q)-r)/abs(S)-s*t; }
T T=new T; T.C=Q; T.r=s*t*abs(S); return T; } void g(pair Q, real r) { real t=0; for(int k=0;k<30;++k) { X=(5,t); t+=(abs(X-Q)+r-abs(X-A)); } } pair Z1=(1.07,6); draw(B--Z1); T T=f(X,B,Y,Z1); draw(CR(T.C,T.r)); T T=f(X,B,Y,Z); draw(CR(T.C,T.r)); g(T.C,T.r); path G2=D(arc(X,C,A)); T T=f(X,B,Y,Z1); draw(CR(T.C,T.r)); T=f(X,B,Y,Z); draw(CR(T.C,T.r)); g(T.C,T.r); path G3=D(arc(X,C,A)); pen p=black+fontsize(8); MC("\gamma_1",G1,0.85,p); MC("\gamma_2",G2,0.85,NNW,p); MC("\gamma_3",G3,0.85,WNW,p); MC("h_1",B--Z1,0.95,E,p); MC("h_2",B--Y,0.95,E,p); MC("h_3",B--Z,0.95,E,p); path[] G={G1,G2,G3}; path[] H={B--Z1,B--Y,B--Z}; pair[][] al={{S+SSW,S+SSW,3*S},{SE,NE,NW},{2*SSE,2*SSE,2*E}}; for(int i=0;i<3;++i) for(int j=0;j<3;++j) MP("V_{"+string(i+1)+string(j+1)+"}",IP(H,G[j]),al[j],fontsize(8));[/asy]
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f(f(x)^2*y)=x^3f*(xy)

Denote by Q+\mathbb{Q}^+ the set of all positive rational numbers. Determine all functions f:Q+↦Q+f : \mathbb{Q}^+ \mapsto \mathbb{Q}^+ which satisfy the following equation for all x,y∈Q+:x, y \in \mathbb{Q}^+: f(f(x)2y)=x3f(xy).f\left( f(x)^2y \right) = x^3 f(xy).
Proposed by Thomas Huber, Switzerland

Number of wins and losses of the ith player

n≥4n \geq 4 players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players badbad if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that there is no bad company in this tournament. Let wiw_i and lil_i be respectively the number of wins and losses of the ii-th player. Prove that ∑i=1n(wi−li)3≥0.\sum^n_{i=1} \left(w_i - l_i\right)^3 \geq 0.
Proposed by Sung Yun Kim, South Korea

IMO Shortlist 2010 - Problem G5

Let ABCDEABCDE be a convex pentagon such that BC∥AE,BC \parallel AE, AB=BC+AE,AB = BC + AE, and ∠ABC=∠CDE.\angle ABC = \angle CDE. Let MM be the midpoint of CE,CE, and let OO be the circumcenter of triangle BCD.BCD. Given that ∠DMO=90∘,\angle DMO = 90^{\circ}, prove that 2∠BDA=∠CDE.2 \angle BDA = \angle CDE.
Proposed by Nazar Serdyuk, Ukraine
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In a concert, 20 singers will perform

In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of other singers such that he wishes to perform later than all the singers from that set. Can it happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied?
Proposed by Gerhard Wöginger, Austria

IMO Shortlist 2010 - Problem G1

Let ABCABC be an acute triangle with D,E,FD, E, F the feet of the altitudes lying on BC,CA,ABBC, CA, AB respectively. One of the intersection points of the line EFEF and the circumcircle is P.P. The lines BPBP and DFDF meet at point Q.Q. Prove that AP=AQ.AP = AQ.
Proposed by Christopher Bradley, United Kingdom

IMO Shortlist 2010 - Problem N1

Find the least positive integer nn for which there exists a set {s1,s2,…,sn}\{s_1, s_2, \ldots , s_n\} consisting of nn distinct positive integers such that (1−1s1)(1−1s2)⋯(1−1sn)=512010. \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \frac{51}{2010}.
Proposed by Daniel Brown, Canada