The points $A_{1}, A_{2}, \ldots, A_{2 n+1}$ are the vertices of a regular polygon with $2 n+1$ sides. Find the number of triples $A_{i}, A_{j}, A_{k}$ such that the triangle $A_{i} A_{j} A_{k}$ is obtuse.

## Solution.



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Since $2n+1$ is odd, it is not possible to construct right-angled triangles. Let's observe that any obtuse triangle will leave the center $O$ (its circumcenter) outside of it. If we rotate it in a direct or inverse direction around $O$, we can achieve that one of its acute vertices is at $A_{1}$. The other two are either in the set $\left\{A_{2}, \ldots, A_{n+1}\right\}$ or in $\left\{A_{n+2}, \ldots, A_{2 n+1}\right\}$. The number we are looking for is $2\left(\begin{array}{c}n \\ 2\end{array}\right)$. Since we can do this with each of the $2n+1$ vertices, there will be $2(2 n+1)\left(\begin{array}{l}n \\ 2\end{array}\right)$ triangles. But each triangle has been counted

![](https://cdn.mathpix.com/cropped/2024_04_26_0ab64de9b6f941361bb2g-3.jpg?height=871&width=846&top_left_y=630&top_left_x=982)
twice, once for each acute vertex. Therefore, the solution we are looking for is $(2 n+1)\left(\begin{array}{l}n \\ 2\end{array}\right)$.

## Alternative Solution.

Fix the obtuse vertex at one vertex, for example, at $A_{1}$. The three sides of the triangle will cover $x, y$, and $z$ sides of the $(2n+1)$-sided polygon. It will be $x+y+z=2n+1$. The side opposite the obtuse angle, say $z$, must satisfy $z \geq n+1$. Let's calculate the number of positive integer solutions to the equation $x+y+z=2n+1$ with the fixed condition for $z$.

If $z=n+1$, we get $x+y=n$ which has $n-1$ solutions. If $z=n+2$, we get $x+y=n-1$ which has $n-2$ solutions. ... If $z=2n-1$, we get $x+y=1$ which has 1 solution.

In total, there are

$$
(n-1)+(n-2)+\cdots+2+1=\frac{n(n-1)}{2}=\left(\begin{array}{l}
n \\
2
\end{array}\right)
$$

solutions with the obtuse angle at $A_{1}$. If we consider the other possible positions for this angle, we get a total of

$$
(2 n+1)\left(\begin{array}{l}
n \\
2
\end{array}\right)
$$