Problem

Source: Iranian National Olympiad (3rd Round) 2008

Tags: geometry, rectangle, geometry proposed



a) Suppose that $ RBR'B'$ is a convex quadrilateral such that vertices $ R$ and $ R'$ have red color and vertices $ B$ and $ B'$ have blue color. We put $ k$ arbitrary points of colors blue and red in the quadrilateral such that no four of these $ k+4$ point (except probably $ RBR'B'$) lie one a circle. Prove that exactly one of the following cases occur? 1. There is a path from $ R$ to $ R'$ such that distance of every point on this path from one of red points is less than its distance from all blue points. 2. There is a path from $ B$ to $ B'$ such that distance of every point on this path from one of blue points is less than its distance from all red points. We call these two paths the blue path and the red path respectively. Let $ n$ be a natural number. Two people play the following game. At each step one player puts a point in quadrilateral satisfying the above conditions. First player only puts red point and second player only puts blue points. Game finishes when every player has put $ n$ points on the plane. First player's goal is to make a red path from $ R$ to $ R'$ and the second player's goal is to make a blue path from $ B$ to $ B'$. b) Prove that if $ RBR'B'$ is rectangle then for each $ n$ the second player wins. c) Try to specify the winner for other quadrilaterals.