Olympiad Combinatorics Problems Solutions < BEST · SECRETS >
Consider all lines through at least two points. Pick the line with the smallest positive distance to a point not on it. Show that line must contain exactly two points, otherwise you’d get a smaller distance.
When stuck, ask: “What’s the smallest/biggest/largest/minimal possible …?” 5. Graph Theory Modeling: Turn the Problem into Vertices & Edges Many combinatorial problems—about friendships, tournaments, networks, or matchings—are secretly graph problems.
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.
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. Olympiad Combinatorics Problems Solutions
At a party, some people shake hands. Prove that the number of people who shake an odd number of hands is even.
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).
A knight starts on a standard chessboard. Is it possible to visit every square exactly once and return to the start (a closed tour)? Consider all lines through at least two points
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.
A finite set of points in the plane, not all collinear. Prove there exists a line passing through exactly two of the points.
Happy counting! 🧩 Do you have a favorite Olympiad combinatorics problem or a clever solution that blew your mind? Share it in the comments below! A knight always moves from a black square
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.
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.
When a problem involves moves or transformations, look for what doesn’t change modulo 2, modulo 3, or some clever coloring. 3. Double Counting: Two Ways to Tell the Same Story One of the most elegant weapons in the Olympiad arsenal. Count the same set of objects in two different ways to derive an identity.