Problem 1159

Analysis: First, discuss the regular pentagon. When the five vertices of a regular pentagon are colored with two colors, there must be at least one monochromatic isosceles triangle. A regular 2005-gon contains 401 regular pentagons with no common vertices, so we can obtain 401 monochromatic isosceles triangles. Using the pigeonhole principle, we can prove that at least 101 of them are congruent and monochromatic.

Solution: First, discuss the regular pentagon. When the five vertices of a regular pentagon are colored with two colors, by the pigeonhole principle, there must be three points of the same color, meaning there must be a monochromatic triangle. In a regular pentagon, the distance between any two vertices is of only two types (sides and diagonals). The three sides of the triangle (items) can only be of two lengths (pigeonholes). Using the pigeonhole principle again, we know that at least two sides must be of the same length, so this monochromatic triangle must be a monochromatic isosceles triangle.

A regular 2005-gon contains 401 regular pentagons with no common vertices, so there must be at least 401 monochromatic isosceles triangles. The triangles can only be of two colors, so there must be at least 201 isosceles triangles of the same color. The triangles in a regular pentagon can only be of two types: two sides and one diagonal, or two diagonals and one side. Using the pigeonhole principle again, there must be at least 101 monochromatic isosceles triangles that are congruent and of the same color.