There's Too Many Triangles Here!

Consider the general case of an $\text{n-gon}$ (where $n$ is even). Consider any one of the $n$ points. From this point, if we join two other arbitrary points, we get a triangle. Our goal is to find the number of such triangles such that atleast one of the angles are obtuse.

From this one point, if one of the two other vertices are directly opposite to it (Only possible in even sided regular polygons, hence mentioned at the beginning) , then the angle subtended will be a right angle. thales' theorem

If the two vertices are on either side of this diagonal, then an acute angle is formed. Obtuse triangles will only be formed when both points are on the same side of the semi-polygon.

Hence, the total number of ways to choose the remaining points is $\binom{\frac n2 -1}{2}$ . Therefore considering for each of the $n$ vertices, in total there would be $n \times \binom{\frac n2 -1}{2}$ or $\frac n2 \times (\frac n2 -1) \times (\frac n2 -2)$

Substituting $n=18$ yields $504$

Vertices is only the numbered dot. If we choose 1 and 3 as the end, there will be 1 obtuse triangle (123). If we choose 1 and 4 as the end, there will be 2 obtuse triangle (124;134). Beside if we choose 1 and 10 as the end, it only create right angle. Hence if we choose 1 as first end and 2-9 as the second end, we can create 1+2+3+4+5+6+7 = 28 obtuse triangle. Because there is 18 vertices, then total obtuse triangle is $28 \times 18 = 504$