2. Given the vertices of a regular 100-gon $A_{1}, A_{2}, A_{3}, \ldots, A_{100}$. 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 $\alpha=\angle K L M=\frac{180^{\circ}}{100}(100-(m-k))>90^{\circ}, 0<m-k<50$.

The difference $m-k$ is taken modulo 100

(for example, $15-\left.70\right|_{\bmod 100}=45$ ).

![](https://cdn.mathpix.com/cropped/2024_05_06_d45ee4ec72eb6c7fb975g-1.jpg?height=477&width=531&top_left_y=869&top_left_x=1322)

Let's count the number of ways to choose vertices $K, L, M$. First, choose vertex $K$ in one of 100 ways. Then choose any two of the vertices $A_{k+1}, A_{k+2}, \ldots, A_{k+49}$ (vertex numbers are taken modulo 100). The closer of these vertices to $K$ will be $L$, and the farther one will be $M$. Thus, we have $100 \cdot \mathrm{C}_{49}^{2}=100 \cdot 49 \cdot 24=117600$.

Answer: 117600.