Međunarodna matematička olimpijada 2010

[ 2010 | IMO ]
Find all function f:\mathbb{R}\rightarrow\mathbb{R} such that for all x,y\in\mathbb{R} the following equality holds f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor


where \left\lfloor a\right\rfloor is greatest integer not greater than a.

Proposed by Pierre Bornsztein, France
Let a_1, a_2, a_3, \ldots be a sequence of positive real numbers, and s be a positive integer, such that
a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.
Prove there exist positive integers \ell \leq s and N, such that
a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.

Proposed by Morteza Saghafiyan, Iran
Each of the six boxes B_1, B_2, B_3, B_4, B_5, B_6 initially contains one coin. The following operations are allowed

Type 1) Choose a non-empty box B_j, 1\leq j \leq 5, remove one coin from B_j and add two coins to B_{j+1};

Type 2) Choose a non-empty box B_k, 1\leq k \leq 4, remove one coin from B_k and swap the contents (maybe empty) of the boxes B_{k+1} and B_{k+2}.

Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes B_1, B_2, B_3, B_4, B_5 become empty, while box B_6 contains exactly 2010^{2010^{2010}} coins.

Proposed by Hans Zantema, Netherlands
Let P be a point interior to triangle ABC (with CA \neq CB). The lines AP, BP and CP meet again its circumcircle \Gamma at K, L, respectively M. The tangent line at C to \Gamma meets the line AB at S. Show that from SC = SP follows MK = ML.

Proposed by Marcin E. Kuczma, Poland
Given a triangle ABC, with I as its incenter and \Gamma as its circumcircle, AI intersects \Gamma again at D. Let E be a point on the arc BDC, and F a point on the segment BC, such that \angle BAF=\angle CAE < \dfrac12\angle BAC. If G is the midpoint of IF, prove that the meeting point of the lines EI and DG lies on \Gamma.

Proposed by Tai Wai Ming and Wang Chongli, Hong Kong
Find all functions g:\mathbb{N}\rightarrow\mathbb{N} such that
\left(g(m)+n\right)\left(g(n)+m\right)


is perfect square for all m,n\in\mathbb{N}.

Proposed by Gabriel Carroll, USA