Consider 9 points in the plane with no alignment. What is the minimum value of $n$ such that if we color $n$ edges connecting two of the points in red or blue, we are sure to have a monochromatic triangle regardless of the coloring?

(IMO 1992)
The answer is 33. Indeed, there are $\binom{9}{2}=36$ drawable edges. If 33 are drawn, 3 different points are selected which are vertices of these edges. The other 6 therefore form a complete graph (each connected to each other). It is a classic small exercise to show the existence of a monochromatic triangle. Now, let's find a configuration with 32 edges without a monochromatic triangle. It is easy to draw a complete graph of 5 vertices without a monochromatic triangle. For example, imagine a regular pentagon where the sides are blue and the diagonals are red. Now, we select a vertex $S$ of the graph. We add a vertex $S^{\prime}$. We connect $S^{\prime}$ to all other vertices $A$ such that $S A$ and $S^{\prime} A$ are the same color. Thus, $S^{\prime} A B$ is not monochromatic since $S A B$ is not.

We start with 10 edges, then add $4+5+6+7$, which gives 32 edges without a monochromatic triangle.