Making triangles with no common side from a polygon.

Question

How many triangles can be formed by joining vertices of polygon such that no two triangles share a common side?

I tried small cases

$n=3,4: 1 ; n=5,6: 2$

I first tried to calculate number of possible sides we can have in a polygon, which is $\frac{n(n-1)}{2}$, and I tried to divide it to $3$, however, I realized that some sides can't intersect at all, hence this shouldn't be true.

Another thing I tried was fix a triangle, and make new triangles from it, however new triangles themselves must have no common side from each other, so it makes things diffucult I guess

and I have no other ideas to find a closed formula for it, can someone give me a hindsight please?

Answer 1

Let the Triangle have $n$ Vertexes.

Select 1 Vertex X , which is going to be the common vertex of all triangles. Choose two other vertexes (consecutively) $A,B$ to get Triangle $XAB$ , where there is no common Side.
Choose two other vertexes (consecutively) $C,D$ to get Triangle $XCD$ , where there is no common Side.
Continuing like-wise , you may use all Vertexes or there will be 1 Vertex Y left over.

Hence , Number of triangles is $(n-1)/2$ [when all Vertexes were used up] or $(n-2)/2$ [when 1 Vertex was left over] or Equivalently , Integer Part of $(n-1)/2$ [in all Cases] where there are no common Vertexes.

In your Case , $N=3,4$ , we get $1$.
& $N=5,6$ , we get $2$.

Continuing , $N=7,8$ , we get $3$.
& $N=9,10$ , we get $4$.
& $N=11,12$ , we get $5$.
& $N=13,14$ , we get $6$.

Answer 2

Let the polygon have $n$ vertices.

We can pick $\triangle_{total} = {n \choose 3}$ combinations of 3 vertices from the polygon and draw triangles.

Two triangles may have either 0 common sides or at most one common side. If two triangles have two sides common then they are identical.

If two triangles have one common side then they share two vertices. The number of such triangles with a common side is $\triangle_{1} = {n \choose 2}$.

We have

$$\triangle_{total} = \triangle_{0} + \triangle_{1} \tag 1$$

We want $\triangle_{0}$ which can be calculated from Eqn. (1) above.

$$\triangle_{0} = {n \choose 3} - {n \choose 2}$$