given that the number of triangles is $T(n)$.since $T(6)=1,T(n+2)=T(n)+1\rightarrow T(n)=\left \lfloor \frac{n}{2} \right \rfloor-2$

mmaht wrote: given that the number of triangles is $T(n)$.since $T(6)=1,T(n+2)=T(n)+1\rightarrow T(n)=\left \lfloor \frac{n}{2} \right \rfloor-2$ I think you are wrong, solution should be $ \frac{n(n-4)(n-5) }{6} $

gobathegreat wrote: mmaht wrote: given that the number of triangles is $T(n)$.since $T(6)=1,T(n+2)=T(n)+1\rightarrow T(n)=\left \lfloor \frac{n}{2} \right \rfloor-2$ I think you are wrong, solution should be $ \frac{n(n-4)(n-5) }{6} $ can the triangles intersect each other?(because i thought that they can't intersect each other)

Yes, triangles can intersect

so the answer is $\binom{n}{3}-n(n-4)-n=\frac{n(n-4)(n-5)}{6}$ because we have $\binom{n}{3}$ triangles in total but n(n-4) triangle have one common side with polygon and n triangle have two common sides with polygon.

There is a minor mistake in your Latex mmaht

sorry i fixed that

The number of triangles is $\binom{n}{3}$.Now let's find the number of 'bad' triangles.Start from one of neigbour points and just continue.They build $n-2$ 'bad' triangles.Next pair build $n-3$ pairs because it already built a pair with the previous one.That hold for every pair except the last one because it has already built $2$ 'bad' triangles with the previous pair and with the next pair that has already been used so the number of bad triangles this pair forms is $n-4$ so we now have that the number of bad triangles is $(n-2)+(n-2)(n-3)+(n-4)=n(n-3)$.So the number of 'good' triangles is $\binom{n}{3} - n(n-3) = \frac{n(n-1)(n-2)}{6}-n(n-3) = \frac{n(n-1)(n-2)-6n(n-3)}{6} = \frac{n(n^2-9n+20)}{6} = \frac{n(n-4)(n-5)}{6}$.

Also i think that this problem is too easy too.It is even easier than the first problem.