How many non-congruent triangles can be formed by connecting three vertices of a regular 101-gon?
Any three vertices of a regular n-gon form a triangle, so we need to count the number of ways to choose three vertices without regard to order. There are C(101,3) = 17711 ways to choose the three vertices, but we must divide by 3 to account for overcounting, since a triangle is unchanged by rotation by 120∘. So the answer is C(101,3)/3=5903.
