Folklore

Given a regular nonagon.

In how many ways can three of its vertices be chosen so that they form the vertices of an isosceles triangle?
For every two vertices of the nonagon, there exists exactly one vertex that is equidistant from them, so each of the resulting isosceles but not equilateral triangles is uniquely determined by its base. The number of ways to choose two points out of nine is $C_{9}^{2}=36$. However, with this method of counting, each of the three equilateral triangles is counted three times. Therefore, the desired number is $36-6=30$.

## Answer

30 ways