SAMC

We will solve the problem for a regular $n$-gon $A_1{A}_2\dots A_n$, $n\ge 3$.

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 $1<i<j\le n$. 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 $j\le \frac{n+1}{2}$, hence $P_1(n)$ equals the number of all two-elements of $\{2,3,\dots ,\left[\frac{n+1}{2}\right]\}$. Thus

$$\begin{array}{c}P(n)=n\cdot P_1(n)=n\left(\left[\frac{n+1}{2}\right]-1\right)\left(\left[\frac{n+1}{2}\right]-2\right)\\ =\frac{n}{2}\cdot \left[\frac{n-1}{2}\right]\cdot \left[\frac{n-3}{2}\right].\end{array}$$

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 $△A_i{A}_k{A}_j$. 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, $v(\alpha )\in \{1,2,\dots ,\left[\frac{n-3}{2}\right]\}$. Since there are exactly $n$ arcs $\alpha$ with the same value $v(\alpha )$, the desired number is

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

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

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