## Problem of the Day #24: Product of areas of triangles in regular n-gon
*April 12, 2011*

*Posted by Mitchell in : potd , trackback*

Given a regular $n$-gon $\mathcal{P}$ with side length $1$, find the product of the areas of all the triangles whose vertices are vertices of $\mathcal{P}$.

Thus, the resulting product is \[\frac{\prod_{r=1}^{n-1} (A_n A_{r})^{s_r}}{(4R)^{\binom{n}{3}}},\] where $s_r$ is the number of $j-i$, $k-j$, or $i-k+n$ ($1 \leq i < j < k \leq n$) which are equal to $r$.

The number $s_r$ can also be computed: it is equal to $n(n-r-1)$ by some simple constructive counting. Thus, the needed product is \[\frac{\prod_{r=1}^{n-1} (A_n A_{r})^{n(n-r-1)}}{(4R)^{\binom{n}{3}}}.\] Let $P = \prod_{r=1}^{n-1} (A_n A_{r})^{n(n-r-1)}$. Then, by a index change $r \mapsto n – r$ in the product and a use of the fact that $A_n A_r = A_{n} A_{n-r}$, we find also that $P = \prod_{r=1}^{n-1} (A_n A_{r})^{n(r-1)}$. Multiplying these two expressions together, we find \[P^2 = \prod_{r=1}^{n-1} (A_n A_{r})^{n(n-2)} =\left( \prod_{r=1}^{n-1} A_n A_{r}\right)^{n(n-2)}.\] Because for a regular polygon inscribed in a unit circle, the product of the distances from one vertex to all the other vertices is the number of sides of the polygon (cf. this result), we have $\prod_{r=1}^{n-1} A_n A_{r} = nR^{n-1}$, so $P^2 = (n R^{n-1})^{n(n-2)}$, so $P = \sqrt{n}^{n(n-2)} R^{3 \binom{n}{3}}$.

The final answer is then \[\frac{\prod_{r=1}^{n-1} (A_n A_{r})^{n(n-r-1)}}{(4R)^{\binom{n}{3}}} = \frac{P}{(4R)^{\binom{n}{3}}}.\] Plugging in the known formulas for $R$ and $P$, this can be simplified to \[\boxed{\frac{\sqrt{n}^{n(n-2)}}{(4\sin\frac{\pi}{n})^{2 \binom{n}{3}}}.}\]

## Comments»

no comments yet - be the first?