2. Given the vertices of a regular 120-gon $A_{1}, A_{2}, A_{3}, \ldots, A_{120}$. In how many ways can three vertices be chosen from them to form an obtuse triangle? (10 points)
Solution. Let the vertices be numbered clockwise.

Denote the selected vertices clockwise as $K, L, M$, where angle $K L M$ is obtuse. If $K=A_{k}, L=A_{l}, M=A_{m}$, then $\alpha=\angle K L M=\frac{180^{\circ}}{120}(120-(m-k))>90^{\circ}, 0<m-k<60$.

The difference $m-k$ is taken modulo 120

(for example, $15-\left.90\right|_{\text {mod } 120}=45$ ).

Let's count the number of ways to choose vertices $K, L, M$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d45ee4ec72eb6c7fb975g-3.jpg?height=482&width=514&top_left_y=884&top_left_x=1362)

First, choose vertex $K$ in one of 120 ways. Then choose any two of the vertices $A_{k+1}, A_{k+2}, \ldots, A_{k+59}$ (vertex numbers are taken modulo 120). The closer of these vertices to $K$ will be $L$, and the farther one will be $M$. Thus, we have $120 \cdot$ C $_{59}^{2}=120 \cdot 59 \cdot 29=205320$.

## Answer: 205320.