I'm too bad at Combinatorics so please ! Label the vertices of the polygon $A_0,A_1 \cdots A_{2007}$ We will explicitly find the number of obtuse angle triangles Let the triangle chosen be $A_iA_jA_k$ with $0 \le i < j < k \le 2007$ Then it is easy to see that the triangle is obtuse iff either of the three angles are obtuse Obviously the three cases so formed are disjoint CASE 1 $ \angle A_iA_kA_j$ is obtuse $\implies j-i>1004$ No. of such triplets $(i,j,k)$ are $\sum_{m=1}^{1002} \frac{m(m+1)}{2}$ CASE 2 $ \angle A_jA_iA_k$ is obtuse $\implies k-j>1004$ Again, No. of such triplets $(i,j,k)$ are $\sum_{m=1}^{1002} \frac{m(m+1)}{2}$ CASE 3 $ \angle A_iA_jA_k$ is obtuse $\implies k-i<1004$ No. of such triplets is $1004 \times \binom{1003}{3}$ After that I think it should be easy Hope I have made no mistakes