Sides of triangle divided into 2002 congruent segments
Source: Moldova IMO-BMO TST 2003, day 1, problem 1.
August 14, 2008
modular arithmeticrationumber theoryrelatively prime
Problem Statement
Each side of an arbitrarly triangle is divided into congruent segments. After that, each vertex is joined with all "division" points on the opposite side.
Prove that the number of the regions formed, in which the triangle is divided, is divisible by .
Proposer: Dorian Croitoru