The sum of the squares of the diagonals in a polygon

Question

The first question that got me here: A regular dodecagon $P_1 P_2 P_3 \dotsb P_{12}$ is inscribed in a circle with radius $1.$ Compute$ {P_1 P_2}^2 + {P_1 P_3}^2 + {P_1 P_4}^2 + \dots + {P_{10} P_{11}}^2 + {P_{10} P_{12}}^2 + {P_{11}P_{12}}^2.$ (The sum includes all terms of the form ${P_i P_j}^2$ where $1 \le i < j \le 12.$ We write $P_iP_j$ to mean the length of segment $\overline{P_iP_j}$.)

I solved the question and got 144. Since this is I wondered if the total would be true for any n-gon?. More specifically, for any n-gon would this total be equal to $n^2$.

This is Dodecagon:

Answer 1

For a regular $n$-gon of radius $1$, the sum of squares of diagonals from one vertex is

$S_\text{one}=\sum\limits_{k=1}^{n-1}\left(2\sin \left(\frac{k\pi}{n}\right)\right)^2$

If $n$ is even, $n=2m$:

$\begin{align} S_\text{one}&=4\sum\limits_{k=1}^{2m-1}\sin^2\left(\frac{k\pi}{2m}\right)\\ &=4\left(\sum\limits_{k=1}^{m-1}\sin^2\left(\frac{k\pi}{2m}\right)+1+\sum\limits_{k=m+1}^{2m-1}\sin^2\left(\frac{k\pi}{2m}\right)\right)\\ &=4\left(\sum\limits_{k=1}^{m-1}\sin^2\left(\frac{k\pi}{2m}\right)+1+\sum\limits_{k=m+1}^{2m-1}\cos^2\left(\frac{\pi}{2}-\frac{k\pi}{2m}\right)\right)\\ &=4\left(\sum\limits_{k=1}^{m-1}\sin^2\left(\frac{k\pi}{2m}\right)+1+\sum\limits_{k=m+1}^{2m-1}\cos^2\left(\frac{(m-k)\pi}{2m}\right)\right)\\ &=4\left(\sum\limits_{k=1}^{m-1}\sin^2\left(\frac{k\pi}{2m}\right)+1+\sum\limits_{k=1}^{m-1}\cos^2\left(\frac{k\pi}{2m}\right)\right)\\ &=4\left(\sum\limits_{k=1}^{m-1}1+1\right)\\ &=4m\\ &=2n \end{align}$

A similar argument shows that if $n$ is odd then $S_\text{one}=2n$.

To find the the sum of squares of all diagonals, we first multiply $S_\text{one}$ by the number of vertices, $n$. But then we have counted each diagonal twice, so we divide by $2$.

So the sum of squares of all diagonals is $n^2$.

Answer 2

The complex vertices of an $n$-gon inscribed in a circle of unit radius are

$ P_k = e^{ \dfrac{2 \pi i k}{n}} , k = 0, 1, \dots, n-1$

Therefore, the indicated sum is

$ S = \displaystyle \sum_{j = 0}^{n-1} \sum_{k = j+1}^{n-1} \left(e^{\dfrac{2 \pi j i}{n}} - e^{\dfrac{2 \pi k i}{n}} \right) \left(e^{- \dfrac{2 \pi j i}{n}} - e^{- \dfrac{2 \pi ki }{n} }\right) $

and this equals

$ S = \displaystyle \sum_{j = 0}^{n-1} \sum_{k = j+1}^{n-1} \left( 2 - 2 \cos \left( \dfrac{ 2 \pi (j - k) }{n} \right) \right) $

Let $ m = k - j $ , then

$ S = \displaystyle \sum_{j = 0}^{n-1} \sum_{m = 1}^{n - 1 - j} \left( 2 - 2 \cos \left( \dfrac{ 2 \pi m }{n} \right) \right) $

It can be shown that this summation is equal to

$ S = n^2 $

To prove this,

Let $z = e^{\dfrac{2 \pi i}{n}} $

Then

$ A = \displaystyle \sum_{j = 0}^{n-1} \sum_{m = 1}^{n - j -1} \cos \left( \dfrac{ 2 \pi m }{n} \right) = \text{Re}\left( \sum_{j = 0}^{n-1} \sum_{m = 1}^{n - j-1} z^m \right) = \text{Re}\left(\sum_{j = 0}^{n-1} \dfrac{ z }{z - 1} ( z^{n-j-1} - 1 )\right) $

Changing the index of summation from $j$ to $k = n - j-1$, this summation becomes

$A= \displaystyle \text{Re}\left( \dfrac{z}{z-1} \sum_{k = 0}^{n-1} ( z^{k} - 1 )\right) = \text{ Re } \left(\dfrac{z}{z-1} \left( z \dfrac{ z^n - 1}{z-1} - n \right) \right) $

Note that $z^n = 1 $, therefore,

$ A = (-n) \text{ Re } \left(\dfrac{z}{z-1} \right) $

Now,

$ \dfrac{z}{z-1} = \dfrac{ e^{\dfrac{2 \pi i}{n}}}{ e^{\dfrac{2 \pi i}{n}} - 1 } = \dfrac{ e^{\dfrac{2 \pi i}{n}}}{ e^{\dfrac{\pi i}{n}} \left(e^{\dfrac{\pi i}{n}} - e^{\dfrac{-\pi i}{n}} \right) } = \dfrac{ e^{\dfrac{ \pi i}{n}}}{ 2 i \sin\left( \dfrac{ \pi }{n} \right)} $

Expanding the numerator using Euler's formula,

$ \dfrac{z}{z-1} = \dfrac{ \cos \left(\dfrac{\pi}{n}\right) + i \sin \left(\dfrac{\pi}{n}\right) }{ 2 i \sin \left( \dfrac{ \pi }{n} \right)} = \dfrac{1}{2} - \dfrac{1}{2} i \cot \left( \dfrac{ \pi }{n} \right) $

Therefore,

$ \text{Re} \left( \dfrac{z}{z-1} \right) = \dfrac{1}{2} $

Therefore,

$ A = - \dfrac{n}{2} $

Recall that

$ S =\displaystyle \left( \sum_{j = 0 }^{n-1} \sum_{m=1}^{n-j-1} 2 \right) - 2 A =\left(2 \sum_{j = 0 }^{n-1} \sum_{m=1}^{n-j-1} 1 \right) + n \\ \displaystyle = \left( 2 \sum_{j = 0}^{n-1} (n-j-1) \right) + n = \left( 2 \sum_{k = 0 }^{n-1} k \right) + n = n(n-1) + n = n^2 $

So yes, the indicated summation of the square of the diagonals is indeed equal to $n^2$.

Answer 3

For any regular $n$-gon we have $P_k = \exp\left(\frac{2i\pi k}{n}\right) = x^k$ where $k \in \{0, 1,2,\dots,n-1\}$.

Suppose the sum you're looking for is $S$, then $$2S = \sum_{r=0}^{n-1} \sum_{j=0}^{n-1} |x^r-x^j|^2 = \sum_{r=0}^{n-1} \sum_{j=0}^{n-1} 4\sin^2\left (\frac{\pi (r-j)}{n}\right)= 2 \sum_{r=0}^{n-1} \sum_{j=0}^{n-1}\left(1 - \cos\left(2\frac{\pi(r-j)}{n}\right)\right) = 2n^2 - \sum_{r=0}^{n-1} \sum_{j=0}^{n-1} \left(\cos\left(2\frac{\pi r}{n}\right)\cos\left(2\frac{\pi j}{n}\right)+\sin\left(2\frac{\pi r}{n}\right)\sin\left(2\frac{\pi j}{n}\right)\right) = 2n^2 - \sum_{r=0}^{n-1} \left(\cos\left(2\frac{\pi r}{n}\right) \sum_{j=0}^{n-1} \cos\left(2\frac{\pi r}{n}\right)\right) - \sum_{r=0}^{n-1} \left(\sin\left(2\frac{\pi r}{n}\right)\sum_{j=0}^{n-1}\sin\left(2\frac{\pi j}{n}\right)\right) = 2n^2$$

Hence $S=n^2$.

Answer 4

There is no need of any trigonometry. Simple application of vectors plus symmetry argument suffices to solve this problem.

Let $p_1, \ldots, p_n \in \mathbb{R}^2$ be any $n$ points on the unit circle $|p|=1$.
The sum of squares of their diagonal lengths equals to

$$\begin{align} \sum_{1\le i < j\le n}|p_i-p_j|^2 &= \frac12 \sum_{i=1}^n\sum_{j=1}^n |p_i-p_j|^2 \\ &= \frac12\left[\sum_{i=1}^n |p_i|^2 \sum_{j=1}^n 1 + \sum_{i=1}^n 1\sum_{j=1}^n |p_j|^2 -2\left(\sum_{i=1}^np_i \right) \cdot \left( \sum_{j=1}^n p_j\right)\right]\\ &= n\sum_{i=1}^n |p_i|^2 - \left|\sum_{i=1}^n p_i\right|^2\\ &= n^2\left(1 - |p_G|^2\right)\end{align} $$ where $p_G = \frac1n\sum\limits_{i=1}^n p_i$ is the centroid of the $n$ points. So in general, sum of squares of diagonal lengths doesn't equal to $n^2$.

However, for the special case that $p_k$ are vertices of a regular $n$-gon, the set of points is invariant under a rotation with respect to origin for angle $\frac{2\pi}{n}$. This means $p_G$ is also invariant under such a rotation. This forces $p_G$ to be the origin. As a result,

$$|p_G| = 0 \quad\implies\quad \sum_{1\le i < j\le n}|p_i-p_j|^2 = n^2$$

For other configuration of points which has some sort of symmetry with respect to origin, same argument applies and the sum of squares again equals to $n^2$.