# A combinatorics problem involving geometry

Find the number of triangles that can be formed in a regular polygon of $2n+1$ sides such that the center of polygon lies inside the triangle. (The triangles are to be formed using the vertices of the polygon).

• Are we to assume the vertices of the triangles are to be picked among the vertices of the polygon? Aug 14 '14 at 13:02
• yes , the triangles are to be formed using the vertices of the polygon
– raj
Aug 14 '14 at 13:06
• I think the answer should look something like this $\frac{(2n+1){ n+1\choose2}}{3}$,I'll post the solution after rechecking my logic. Aug 14 '14 at 13:21
• that's right -only that it should be over 6 (you've counted each triangle twice and then three times) Aug 14 '14 at 13:26
• but checking for the case n=1 it seems my answer is correct Aug 14 '14 at 13:27

First, pick any starting point $A$. (Suppose it is at the bottom of the circle.) This can be done in $(2n+1)$ ways. Number the points starting at $A=0$. Let the second point $B$ be at position $i\leq n$. (We know that at least one of $B$ and $C$ has to be at the left side of the line $AO$.) Now, for $C$ we know that $n+1\leq C< 2n+1$, because it has to be at the right of $AO$. It also has to be above $BO$, which implies $C\leq i+n<2n+1$. Now, $n+1\leq C\leq i+n$, thus the number of possibilities for $C$ is $i+n-(n+1)+1=i$. Summing over all possible $i$ gives the total number of possibilities for $B$ and $C$: $$\sum_{i=1}^n i=\binom {n+1}2$$ Now, let $A$ be a random point on the circle. This creates $2n+1$ as many solutions, but since we count each triangle three times now (with $A$ once at each corner), we have to divide by three, resulting in: $$\frac{2n+1}3\binom{n+1}2$$