How many non-congruent triangles can be formed by connecting three vertices of a regular 101-gon?
Whats a 101-gon? and how do i approach this?
A 101-gon is just like an octagon but with 101 sides
Since no three vertices of an octagon are colinear, then any possible triplet forms a triangle. So the number of non-congruent triangles is equal to all possible ways to choose three vertices of an octagon, where rotations and reflections still count as the same object.
One simple way to count them is to realize that if you have three vertices, there are three gaps between them. Each of these three gaps contains 0 to 5 of the remaining vertices and all three together have 5 vertices.
And since there are only three of them, their order is irrelevant (it would be different if there were four of them).
So, the possibilities are:
There are $5$ non-congruent triangles that can be formed from vertices of a regular octagon.