errrr, THEOREM >.< ... oops .... 1. The problem statement, all variables and given/known data Prove that when edges of a complete heptagon are colored with two different colors, there will be at least three pure triangles. 2. Relevant equations 3. The attempt at a solution i can do two pure triangles, but not three pick a vertex v. It has 6 edges incident to it, at least 3 of which are the same color. 1. Suppose 3 of these edges, connecting to vertices r, s and t, are blue. If any of the edges (r, s), (r, t), (s, t) are also blue then we have an entirely blue triangle. If not, then those three edges are all red and we have an entirely red triangle. Same goes for the other 3 edges incident to v, say red, that are connected to other 3 vertices, say a, b, c. Therefore we have 2 pure triangles. 2. Suppose at least 4 edges connecting v to r, s, t, u are blue. Then the least amount of pure triangles connecting r,s,t,u is 0 when the internal diagonals are blue and the outside square is red. But since we have two blue diagonals, each connected to v by two blue edges, we have 2 pure triangles. now any ideas as far as how to get the third?