Question

Let A_1, A_2, ..., A_12 be twelve equally spaced points on a circle with radius 1. Find (A_1 A_2)^2 + (A_1 A_3)^2 + ... + (A_11 A_12)^2. (The sum includes the square of the distance between any pair of points, so the sum includes 12choose2 = 66 terms.)

Answer 1

Let $n$ be a fixed positive integer. Then $n$ points can be equally spaced along the boundary of the unit disk in the complex plane, and their positions described in the complex plane are given by $e^{i2\pi k/n}$, where $k=0,1,\ldots,n-1$.

So, for any $0\le k\le n-1$, we can write

$A_k=e^{i2\pi k/n}=\cos\dfrac{2\pi k}n+i\sin\dfrac{2\pi k}n\equiv\left(\cos\dfrac{2\pi k}n,\sin\dfrac{2\pi k}n\right)$

The distance between two points $A_j$ and $A_k$ is given by the norm of the difference of the points. We denote this by

$A_jA_k:=|A_j-A_k|$

We're interested in all the possible distinct pairings, which means $A_jA_k$ and $A_kA_j$ are to be considered the same, so to accomplish this we assume $0\le j<k\le n-1$.

Then the sum we're interested in finding is

$\displaystyle\sum_{0\le j<k\le n-1}(A_jA_k)^2=(A_0A_1)^2+\cdots+(A_0A_{n-1})^2+(A_1A_2)^2+\cdots+(A_1A_{n-1})^2+(A_2A_3)^2+\cdots+(A_{n-2}A_{n-1})^2$

where this sum is identical to the one in your original question, except the indices of each point are shifted down by one.

We have

$(A_jA_k)^2=|A_j-A_k|^2$
$=\left|\left(\cos\dfrac{2\pi j}n,\sin\dfrac{2\pi j}n\right)-\left(\cos\dfrac{2\pi k}n,\sin\dfrac{2\pi k}n\right)\right|^2$
$=\left(\cos\dfrac{2\pi j}n-\cos\dfrac{2\pi k}n\right)^2+\left(\sin\dfrac{2\pi j}n-\sin\dfrac{2\pi k}n\right)^2$
$=\cos^2\dfrac{2\pi j}n-2\cos\dfrac{2\pi j}n\cos\dfrac{2\pi k}n+\cos^2\dfrac{2\pi k}n+\sin^2\dfrac{2\pi j}n-2\sin\dfrac{2\pi j}n\sin\dfrac{2\pi k}n+\sin^2\dfrac{2\pi k}n$
$=2-2\cos\dfrac{2\pi}n(j-k)$
$=4\sin^2\dfrac\pi n(j-k)$

So we can express the sum as

$\displaystyle\sum_{0\le j<k\le n-1}4\sin^2\dfrac\pi n(j-k)=\sum_{0\le j<k\le n-1}4\sin^2\dfrac\pi n(k-j)=\sum_{k=1}^{n-1}\sum_{j=0}^{k-1}4\sin^2\dfrac\pi n(k-j)$

For example, if we take $n=4$ points, we can see the points make up the vertices of a square with inradius 1, which means the distance between adjacent points ($k=j+1$) is $\sqrt2$, while the distance between opposite points ($k=j+2$, which gets counted only once) is $2$. So we would expect the sum to be

$(A_0A_1)^2+(A_0A_2)^2+(A_0A_3)^2+(A_1A_2)^2+(A_1A_3)^2+(A_2A_3)^2=4(\sqrt2)^2+2(2)^2=16$

In terms of our sum, we would add

$\displaystyle\sum_{0\le j<k\le3}4\sin^2\dfrac\pi4(k-j)$
$=4\left(4\sin^2\dfrac\pi4\right)+2\left(4\sin^2\dfrac{2\pi}4\right)$
$=4(\text{distance between adjacent vertices})+2(\text{distance between opposite vertices})$
$=16$

as required.

This was just an example to convince you that the formula works.

For $n=12$, you would end up with

$4\sin^2\dfrac\pi{12}(1-0)+4\sin^2\dfrac\pi{12}(2-0)+\cdots+4\sin^2\dfrac\pi{12}(11-0)$
$+4\sin^2\dfrac\pi{12}(2-1)+4\sin^2\dfrac\pi{12}(3-1)+\cdots+4\sin^2\dfrac\pi{12}(11-1)$
$+4\sin^2\dfrac\pi{12}(3-2)+4\sin^2\dfrac\pi{12}(4-2)+\cdots+4\sin^2\dfrac\pi{12}(11-2)$
$+\cdots$
$+4\sin^2\dfrac\pi{12}(11-10)$
$=144$