How many triangles are there whose vertices are among the vertices of a given regular $(2 n+1)$-gon, and the triangle contains the center of the $(2 n+1)$-gon in its interior?
I. solution. Let the vertices of the $(2 n+1)$-gon be denoted by $A_{0}, A_{1}, A_{2}, \ldots, A_{2 n}$, and its center by $O$. Consider the diagonal $A_{0} A_{k}$ $(k \leq n)$. The triangle $A_{0} A_{k} A_{l}$ contains $O$ if and only if $A_{l}$ is in the opposite quadrant to the one containing the segment $A_{0} A_{k}$, determined by the lines $O A_{0}$ and $O A_{k}$. The line $O A_{0}$ intersects the side $A_{n} A_{n+1}$ of the polygon, and the line $O A_{k}$ intersects the side $A_{n+k} A_{n+1+k}$, so $A_{0} A_{k} A_{l}$ is a good triangle if and only if $n+1 \leq l \leq n+k$ (Figure 1). Therefore, for a fixed $A_{0} A_{k}$, the number of good triangles is $(n+k)-n=k$. Thus, the number of good triangles containing $A_{0}$ is $1+2+\ldots+n=\frac{n(n+1)}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_185f5892d90f1c40dd27g-1.jpg?height=685&width=675&top_left_y=476&top_left_x=709)

Clearly, the same number of good triangles contains any fixed vertex of the $(2 n+1)$-gon. Therefore, $(2 n+1) \cdot \frac{n(n+1)}{2}$ equals three times the number of good triangles, since we counted each good triangle at each of its three vertices.

Thus, the number of good triangles is $\frac{n(n+1)(2 n+1)}{6}$.

II. solution. The vertices of the $(2 n+1)$-gon determine $\binom{2 n+1}{3}$ triangles in total. The number of good triangles is obtained by subtracting the number of (bad) triangles that do not contain the center of the polygon. We count these triangles as follows.

![](https://cdn.mathpix.com/cropped/2024_05_02_185f5892d90f1c40dd27g-1.jpg?height=596&width=590&top_left_y=1610&top_left_x=754)

Figure 2

Label the vertices of the triangle as $A, B, C$ such that $A C$ is the longest side, and the traversal of the triangle is negative (then the line $A C$ separates $B$ from the center of the polygon, Figure 2). Each bad triangle corresponds uniquely to a labeling. If we fix $A$, then by choosing any two of the next $n$ vertices in the negative traversal direction, we get a bad triangle, and we get every bad triangle with $A$ as a fixed vertex this way. The number of these triangles is therefore $\binom{n}{2} \cdot A$, and we can fix $A$ in $(2 n+1)$ ways, so the number of bad triangles is $(2 n+1) \cdot \binom{n}{2}$.

The number of good triangles is:

$$
\binom{2 n+1}{3}-(2 n+1) \cdot \binom{n}{2}=\frac{2 n+1}{6}[2 n(2 n-1)-3 n(n-1)]=\frac{2 n+1}{6}\left(n^{2}+n\right)=\frac{n(n+1)(2 n+1)}{6}
$$

Based on the work of András György (Budapest, Árpád Gym., II. grade)