Pick one person, say Alex. Among the other 5, either at least 3 are friends with Alex or at least 3 are strangers to Alex. By focusing on that group of 3, you apply the pigeonhole principle again to force a monochromatic triangle in the friendship graph.
This is equivalent to showing every tournament has a Hamiltonian path. Use induction: Remove a vertex, find a path in the remaining tournament, then insert the vertex somewhere. Olympiad Combinatorics Problems Solutions
But here’s the secret:
Take a classic problem like “Prove that in any set of 10 integers, there exist two whose difference is divisible by 9.” Apply the pigeonhole principle. You’ve just taken the first step into a larger world. Pick one person, say Alex
Let’s break down the most common types of Olympiad combinatorics problems and the strategies to solve them. The principle is deceptively simple: If you put (n) items into (m) boxes and (n > m), at least one box contains two items. This is equivalent to showing every tournament has
Color the board black and white in the usual pattern. A knight always moves from a black square to a white square and vice versa. For a closed tour, the knight must make an equal number of black and white moves, but there are 64 squares. Since 64 is even, a closed knight’s tour is possible in theory—but parity alone doesn’t guarantee it; it’s a starting point for deeper invariants.