Mean Diagonals

Observe that the mean of the lengths of all the diagonals of the $n$ -gon is equal to the mean of the lengths of only the diagonals that are incident to a particular vertex of the $n$ -gon.

The diagonals of the $n$ -gon can be imagined as chords of its circumcircle, whose radius is $\pi$ .

The length of a chord with a subtended angle $\theta$ and radius $r$ is given by $2r\sin{\frac{\theta}{2}}$ . (see the figure below)

http://mathcentral.uregina.ca/QQ/database/QQ.09.05/bes1.html

The lengths of the $n-3$ diagonals of the $n$ -gon incident to one vertex are $2\pi\sin{\frac{2\pi k}{2n}} = 2\pi\sin{\frac{\pi k}{n}}$ for integers $k \in [2,n-2]$ . Further, the mean of these lengths is $\displaystyle\frac{1}{n-3}\sum_{k=2}^{n-2}2\pi\sin{\frac{\pi k}{n}}$ .

The limit of this expression is $\lim_{n \to \infty} 2\pi \sum_{k=2}^{n-2} \frac{1}{n-3} \sin{\frac{\pi k}{n}}$

$=2\pi\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{n}\sin{\frac{\pi k}{n}}$

Notice that I can extend the index to $(0,n)$ because $\lim_{n\to\infty}\frac{1}{n}\sin{\frac{\pi m}{n}} = 0$ for all $m \in \{0,1,n-2,n-1\}$ . I also replaced $n-3$ with $n$ because $(n-3) \sim n$ .

By definition as the limit of a Riemann sum , the limit is equal to the definite integral $\displaystyle 2\pi\int_{0}^{1}\sin{\pi x}\ dx$

$= 2\pi\cdot\frac{1}{\pi}\cdot -\left(\cos{\pi} - \cos{0)}\right) = \boxed{4}$

Motivation: The problem is equivalent to finding the average chord length in a circle. We focus on a single point on the circle, specifically (as we will see) the origin.

Take the circle with radius 1 centered at (0,1), C. Parametrising C in terms of $\theta$ , we need to find the intersection of $y = x\tan \theta$ and $(y-1)^2 + x^2 = 1$ . This occurs at $x = \sin 2\theta$ . Hence the length of the chord from the origin to a point on C is $\sqrt{x^2 + y^2} = \sqrt{2y} = \sqrt{2\tan\theta \sin 2\theta} = |2\sin \theta|$ .

Thus the average length of a chord is

$\frac{1}{\pi} \int_0^\pi |2\sin \theta| \ d\theta = 4/\pi.$

Scaling up by $\pi$ we see the average chord length is 4.

(Credit to @Jake Lai.)

Let $w=e^{2\pi i/n}$ be a n-th primitive root of unity. Then, the vertices of the n-gon in the complex plane are given by $P_k=Rw^k$ , where $R$ is the circumradius and $0 \leq k \leq n-1$ with $k \in \mathbb{Z}$ . First let's find the sum of the lengths of all the diagonals from $P_0$ to the other vertices: $\sum_{k=2}^{n-2} \overline{P_0 P_k}=R\sum_{k=2}^{n-2} \lvert w^k-1 \rvert=2R\sum_{k=2}^{n-2} \left\lvert \sin\left(\dfrac{\pi k}{n}\right) \right\rvert=2R\sum_{k=2}^{n-2} \sin\left(\dfrac{\pi k}{n}\right)$ Now we have to multiply that by $n$ to count all the diagonals from all vertices, but we have counted twice every diagonal, so the sum (say $S$ ) of the length of every diagonal is then: $\displaystyle S=nR\sum_{k=2}^{n-2} \sin\left(\dfrac{\pi k}{n}\right)$ . Using the formula for the sum of sines of angles in A.P. we get: $S=\dfrac{nR\cos\left(\dfrac{3\pi}{2n}\right)}{\sin\left(\dfrac{\pi}{2n}\right)}$ .

We know that there are $n(n-3)/2$ diagonals, so $M_n=\dfrac{S}{n(n-3)/2}=\dfrac{2R\cos\left(\dfrac{3\pi}{2n}\right)}{(n-3)\sin\left(\dfrac{\pi}{2n}\right)}$ .

Finally, $L=\displaystyle \lim_{n \to \infty} M_n=\left(\displaystyle \lim_{n \to \infty} \dfrac{2R}{n-3}\right) \cdot \dfrac{\displaystyle \lim_{n \to \infty} \cos\left(\dfrac{3\pi}{2n}\right)}{\displaystyle \lim_{n \to \infty} \sin\left(\dfrac{\pi}{2n}\right)}=\left(\displaystyle \lim_{n \to \infty} \dfrac{2R}{n-3}\right) \cdot \dfrac{1}{\frac{\pi}{2n}}=\displaystyle \lim_{n \to \infty} \dfrac{4nR}{\pi(n-3)}=\dfrac{4R}{\pi}$ .

But $R=\pi$ , so $L=\boxed{4}$ .