Problem - 2105
Prove: any convex pentagon must have three vertices $A$, $B$, and $C$ satisfying $\angle{ABC} \le 36^\circ$.
The sum of a pentagon's five interior angles equals $540^\circ$. Hence, the smallest of them cannot be larger than $540\div 5=108^\circ$ by the pigeonhole principle. (The pigeonhole principle is discussed in the book Art of Thinking .)
Now this interior angle can be divided into three angles by two sides and two diagonals from its vertex. The smallest among them cannot be larger than $108\div 3=36^\circ$.