Difference between revisions of "2010 IMO Problems"

Revision as of 09:00, 10 May 2012

Problems of the 51st IMO 2010 in Astana, Kazakhstan.

Day I

Problem 1.

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ 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 $a.$

Author: Pierre Bornsztein, France

Solution

Problem 2.

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 arc $BDC$ , and $F$ a point on the segment $BC$ , such that $\angle BAF=\angle CAE< \frac12\angle BAC$ . If $G$ is the midpoint of $IF$ , prove that the intersection of lines $EI$ and $DG$ lies on $\Gamma$ .

Authors: Tai Wai Ming and Wang Chongli, Hong Kong

Solution

Problem 3.

Find all functions $g:\mathbb{N}\rightarrow\mathbb{N}$ such that $\left(g(m)+n\right)\left(g(n)+m\right)$ is a perfect square for all $m,n\in\mathbb{N}.$

Author: Gabriel Carroll, USA

Solution

Day II

Problem 4.

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 $MK = ML$ .

Author: Unknown currently

Solution

Problem 5.

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.

Author: Unknown currently

Solution

Problem 6.

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 $a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.$