Školjka

%V0 Let $P$ be a convex polygon. Prove that there exists a convex hexagon that is contained in $P$ and whose area is at least $\frac34$ of the area of the polygon $P$. Alternative version. Let $P$ be a convex polygon with $n\geq 6$ vertices. Prove that there exists a convex hexagon with a) vertices on the sides of the polygon (or) b) vertices among the vertices of the polygon such that the area of the hexagon is at least $\frac{3}{4}$ of the area of the polygon. I couldn't solve this one, partially because I'm not quite sure of the statements' meaning (a) or (b) :) Obviously if a) is true then so is b).

%V0 In the Cartesian coordinate plane define the strips $S_n = \{(x,y)|n\le x < n + 1\}$, $n\in\mathbb{Z}$ and color each strip black or white. Prove that any rectangle which is not a square can be placed in the plane so that its vertices have the same color. IMO Shortlist 2007 Problem C5 as it appears in the official booklet:In the Cartesian coordinate plane define the strips $S_n = \{(x,y)|n\le x < n + 1\}$ for every integer $n.$ Assume each strip $S_n$ is colored either red or blue, and let $a$ and $b$ be two distinct positive integers. Prove that there exists a rectangle with side length $a$ and $b$ such that its vertices have the same color. Edited by Orlando Döhring Author: Radu Gologan and Dan Schwarz, Romania

%V0 Let $k$ and $n$ be integers with $0\le k\le n - 2$. Consider a set $L$ of $n$ lines in the plane such that no two of them are parallel and no three have a common point. Denote by $I$ the set of intersections of lines in $L$. Let $O$ be a point in the plane not lying on any line of $L$. A point $X\in I$ is colored red if the open line segment $OX$ intersects at most $k$ lines in $L$. Prove that $I$ contains at least $\dfrac{1}{2}(k + 1)(k + 2)$ red points. Proposed by Gerhard Woeginger, Netherlands

%V0 There is given a convex quadrilateral $ABCD$. Prove that there exists a point $P$ inside the quadrilateral such that $$\angle PAB + \angle PDC = \angle PBC + \angle PAD = \angle PCD + \angle PBA = \angle PDA + \angle PCB = 90^{\circ}$$ if and only if the diagonals $AC$ and $BD$ are perpendicular. Proposed by Dukan Dukic, Serbia

%V0 On a $999\times 999$ board a limp rook can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A non-intersecting route of the limp rook consists of a sequence of pairwise different squares that the limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting route is called cyclic, if the limp rook can, after reaching the last square of the route, move directly to the first square of the route and start over. How many squares does the longest possible cyclic, non-intersecting route of a limp rook visit? Proposed by Nikolay Beluhov, Bulgaria

%V0 Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P\subset R$ we have $$\frac{|R|}{|P|}\leq \sqrt 2$$ where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. Proposed by Witold Szczechla, Poland