Saudi Arabian Mathematical Competitions

We will solve the problem for a regular $n$-gon $A_{1} A_{2} \ldots A_{n}$, $n \geq 3$.

Solution 1.
Let $P(n)$ be the desired number of obtuse triangles and let $P_{1}(n)$ be the number of obtuse angles $A_{1} A_{i} A_{j}$, where $1 < i < j \leq n$. Clearly $P(n) = n \cdot P_{1}(n)$. Any of the considered angles $A_{1} A_{i} A_{j}$ is obtuse if and only if $j \leq \frac{n+1}{2}$, hence $P_{1}(n)$ equals the number of all two-elements of $\{2,3, \ldots, \left[\frac{n+1}{2}\right]\}$. Thus

$$\begin{gathered} P(n) = n \cdot P_{1}(n) = n\left(\left[\frac{n+1}{2}\right] - 1\right)\left(\left[\frac{n+1}{2}\right] - 2\right) \\ = \frac{n}{2} \cdot \left[\frac{n-1}{2}\right] \cdot \left[\frac{n-3}{2}\right] . \end{gathered}$$

Solution 2.
Let $\alpha = \widehat{A_{i} A_{j}}$ be the largest arc of the circumcircle that contains the vertices of an obtuse triangle, say $\triangle A_{i} A_{k} A_{j}$. The size of $\alpha$ is less than $180^{\circ}$ and the number of all obtuse triangles $A_{i} A_{k} A_{j}$ with the common longest side $A_{i} A_{j}$ is identical with the numbers $v(\alpha)$ of those vertices $A_{k}$ of the $n$-gon that are inner points of the arc $\alpha$. Clearly, $v(\alpha) \in \{1,2, \ldots, \left[\frac{n-3}{2}\right]\}$. Since there are exactly $n$ arcs $\alpha$ with the same value $v(\alpha)$, the desired number is

$$n\left(1+2+\ldots+\left[\frac{n-3}{2}\right]\right) = \frac{n}{2} \cdot \left[\frac{n-3}{2}\right] \cdot \left[\frac{n-1}{2}\right] .$$

In our problem $n = 2010$, hence the desired result is

$$\frac{2010}{2} \cdot \left[\frac{2007}{2}\right] \cdot \left[\frac{2009}{2}\right] = 1003 \cdot 1004 \cdot 1005$$