China Western Invitational Mathematical Competition

Let the vertices of the regular $(2n-1)$-gon be $A_0,A_1,\dots ,A_{2n-2}$, labeled in order around the circle.

Suppose we select any $n$ vertices from these $2n-1$ vertices. For each vertex $A_i$, consider the set of possible isosceles triangles with $A_i$ as a vertex and the other two vertices also among the $n$ chosen points.

For each $A_i$, the isosceles triangles with $A_i$ as a vertex are determined by choosing another vertex $A_j$ such that the third vertex is $A_k$, where $A_k$ is symmetric to $A_j$ with respect to $A_i$ (i.e., $A_k$ is the reflection of $A_j$ over $A_i$). Since the polygon is regular, for each pair $(A_i,A_j)$, there is a unique $A_k$ such that $A_i{A}_j=A_i{A}_k$ and $A_j\ne A_k$ unless $A_j$ is diametrically opposite $A_i$ (which cannot happen since $2n-1$ is odd).

Now, consider the set of all possible unordered pairs of chosen vertices. There are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ such pairs. For each such pair $(A_j,A_k)$, there is a unique center $A_i$ such that $A_j$ and $A_k$ are symmetric with respect to $A_i$ (i.e., $A_i$ is the midpoint of the arc $A_j{A}_k$). Since $2n-1$ is odd, for each pair $(A_j,A_k)$, there is a unique $A_i$ (possibly not among the chosen $n$ vertices).

Suppose, for contradiction, that among the $n$ chosen vertices, no three form an isosceles triangle. Then, for each chosen vertex $A_i$, the other $n-1$ chosen vertices must be distributed so that no two are symmetric with respect to $A_i$. For each $A_i$, the $2n-2$ other vertices are paired into $n-1$ pairs of symmetric points with respect to $A_i$. Since we have chosen $n-1$ vertices (excluding $A_i$), by the pigeonhole principle, at least one pair must have both points chosen. But then, together with $A_i$, these two points form an isosceles triangle.

Therefore, among any $n$ vertices of a regular $(2n-1)$-gon, there exist three that are the vertices of an isosceles triangle.