Given a regular polygon of $2n$ edges, we color each edge black or white (with $n$ of each color). Prove that there exists a diagonal that divides the polygon into two sections such that all the black edges on each section correspond to a white edge on the other side, directly across from the diagonal, and vice versa.
How should I do this? Pigeonhole?
If I'm understanding the problem condition properly, I don't think this is true.
Given a regular $2n$-gon, there are $\binom{2n}n$ ways to color its sides black and white. However, if we fix a diagonal that should satisfy the above property, we only have $2^n$ colorings of the $2n$-gon that it can "cover" (i.e. the coloring satisfies the condition with this chosen diagonal) -- once you choose the way to color all sides on one side of the diagonal (each side can be black or white), all sides of the other half of the $2n$-gon are determined. There are $n$ ways to choose the diagonal, so you have at most $n2^n$ colorings of your $2n$-gon for which there exists a diagonal that satisfies the above condition. This implies that $n2^n\geq \binom{2n}n$, which is false for $n\geq 4$.