# What is the probability that the center of a odd sided regular polygon lies inside a triangle formed by the vertices of the polygon?

Three distinct vertices are choose at random from the vertices of a given regular polygon of $2n+1$ sides. Each of those three vertices will determine a triangle. What is the probability that the center of the polygon lies inside the triangle determined by three vertices of the polygon?

Note: All vertices of a regular polygon lie on a common circle (the circumscribed circle), the center of this circle is the center of the polygon.

P.S: why is this question asked specially for a odd sided regular polygon, will the answer differ for a even sided regular polygon?

I have just found a difficult proof of a Much more general question (Probability of a fixed point in a convex region being inside a triangle formed by any three point form the convex region). I am not interested in such generality, can one give a much simpler proof this special case (the convex region being the odd-sided polygon and the fixed point the center of the polygon).

• With an even sided regular polygon, the centre lies on several diagonals - would that count as inside or outside? Jul 2, 2016 at 22:48
• For an odd regular polygon, the centre would always be exterior or interior to the triangle. For an even regular polygon, it could be on the boundary. Jul 2, 2016 at 22:48

Call the $(2n+1)$-gon $P$, and label an arbitrary vertex $A$. We assume without loss of generality that $A$ is one of the chosen vertices. Let the other two chosen vertices in clockwise order be $B$ and $C$, respectively, and suppose there are $k$ edges of $P$ contained within minor arc $BAC$. Because $\triangle ABC$ contains the center of $P$, we have $\angle BAC < 90^\circ$. This implies $k\cdot \frac{180}{2n+1} < 90$; hence $1\le k \le n$.
For a fixed $k$ between $1$ and $n$ inclusive, it is easy to verify that there are $k$ choices of $B$ and $C$ such that exactly $k$ edges of $P$ are contained within minor arc $BAC$ . This gives us a total of $\sum_{k=1}^n k = n(n+1)/2$ valid choices for $B$ and $C$. Because we have $\binom{2n}{2}$ ways to choose $B$ and $C$ from all the remaining vertices of $P$, the probability that a randomly chosen triangle contains the center of $P$ is $$\frac{\frac{n(n+1)}{2}}{\binom{2n}{2}} = \frac{n+1}{2(2n-1)}.$$
Note: This probability implies there are a total of $\frac{n(n+1)(2n+1)}{6} = \sum_{k=1}^n k^2$ triangles that contain the center of $P$. Is there also a bijective proof that counts this directly?
• I did not understand how you wrote $k \times \frac{180}{2n+1} \lt 90$ can you please explain Aug 30, 2017 at 2:52