Olympiad Combinatorics Problems Solutions Apr 2026

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.

Whenever you see sums of numbers counting relationships, try counting the total number of pairs or triples in two ways. 4. Extremal Principle: Look at the Extreme Pick an object that maximizes or minimizes some quantity. Then show that if the desired condition isn’t met, you can find a contradiction by modifying that extreme object.

At a party, some people shake hands. Prove that the number of people who shake an odd number of hands is even.

Happy counting! 🧩 Do you have a favorite Olympiad combinatorics problem or a clever solution that blew your mind? Share it in the comments below! Olympiad Combinatorics Problems Solutions

But here’s the secret:

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.

When a problem says "prove there exist two such that…", think pigeonhole. 2. Invariants & Monovariants: Finding the Unchanging Invariants are properties that never change under allowed operations. Monovariants are quantities that always increase or decrease (but never go back). Pick one person, say Alex

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.

A finite set of points in the plane, not all collinear. Prove there exists a line passing through exactly two of the points.

A knight starts on a standard chessboard. Is it possible to visit every square exactly once and return to the start (a closed tour)? Whenever you see sums of numbers counting relationships,

In a tournament (every pair of players plays one game, no ties), prove there is a ranking such that each player beats the next player in the ranking.

Count the total number of handshakes (sum of all handshake counts divided by 2). The sum of degrees is even. The sum of even degrees is even, so the sum of odd degrees must also be even. Hence, an even number of people have odd degree.