Question 835852
There are 20C3 = 1140 ways to choose three vertices that form a triangle.


Of these, there are 20 triangles with two side lengths on the perimeter of the regular polygon. Also, for each side, there are 16 points that can make a triangle with exactly one side length on the 20-gon. This makes *[tex 20 + 16 \cdot 20 = 340] triangles we have to subtract, since one of the side lengths is on the 20-gon.


1140 - 340 = 800.