Three distinct vertices are randomly selected among the five vertices of a regular pentagon. Let $p$ be the probability that the triangle formed by the chosen vertices is acute. Compute $10 p$.
Solution: The only way for the three vertices to form an acute triangle is if they consist of two adjacent vertices and the vertex opposite their side. Since there are 5 ways to choose this and $\left(\begin{array}{l}5 \\ 3\end{array}\right)=10$ ways to choose the three vertices, we have $p=\frac{5}{10}=\frac{1}{2}$. $\fbox{5}$.