Tough Geometry!

Consider a 20-sided convex polygon K, with vertices $A_1, A_2, . . . , A_{20}$ in that order. Find the number of ways in which three sides of K can be chosen so that every pair among them has at least two sides of K between them.

Clarification :

1. For example $(A_1 A_2, A_4 A_5, A_{11} A_{12})$ is an admissible triple while $(A_1 A_2, A_4 A_5, A_{19} A_{20})$ is not.

2. The language of the question is correct. The question is a modified form of one that appeared in CRMO 2011.