Counting the triangles formed by the sides and diagonals of a regular hexagon

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In this regular convex Hexagon, how many triangles are possible if we consider the intersection points of the diagonals?

I've tried to count the triangles.

First, I counted all the vertices of the hexagon and its diagonals' intersection points(Here 19)
and tried to choose 3 points from the(19C3).
Then, each diagonal has 5 points on them and there are 5 diagonals. So, there should be 5*(5C3) ways that I am overcounting as they don't make any triangle.

My answer is : 19C3 – 5*(5C3). But it is not correct. Why? And what is the correct answer?

Best Answer

Just count the edges of the triangle:

  • you can't form a triangle inside this hexagon with three sides of the hexagon.
  • With two sides you can do it in 6 ways.
  • With one side the only nontriangle is if two diagonals perpendicular to the side. That gives $6\times(3\times 3-1)=48$ ways
  • With all diagonals: there are $\frac62(6-3)=9$ diagonals of which there are 3 pairs of parallels, so of the $\binom{9}{3}=84$ ways of selecting three diagonals, we have to exclude $3\times (9-2)=21$ parallel pair of diagonals plus another, and also the $6$ way of selecting all three diagonals from a vertex and $1$ way of three diagonals through the centre. This leaves $84-21-6-1=56$ ways.

So the total is $6+48+56=110$.

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