Let $A_1$, $A_2$, $\dots$, $A_{12}$ be $12$ equally spaced points on a circle with radius $1$. Find
$$(A_1 A_2)^2 + (A_1 A_3)^2 + \dots + (A_{11} A_{12})^2.$$
The sum includes the square of the distance between any pair of points, so the sum includes $\binom{12}{2} = 66$ terms.
Is there a one-to-one correspondence that can be used here? I don't see anything that clicks, and I should be able to solve this without trigonometry.
Best Answer
By a rotation, we can assume that $A_i = \omega^i$ where $\omega$ is a primitive 12th root of 1. For ease of calculation, let us add the terms $(A_iA_i)^2$. Then, if $S$ is the sum we want, we have \begin{align*} 2S &= \sum_{i=1,j=1}^{12,12}(A_iA_j)^2 \end{align*} Now, $$(A_iA_j)^2 = |\omega^i - \omega^j|^2 = |\omega^{i-j}-1|^2 $$ Also, $$|1-\omega^i|^2 = (1-\omega^i)(1-\omega^{12-i}) = 1 - \omega^i - \omega^{12-i} + 1 $$ and hence \begin{align*} 2S &= 12\sum_{i=1}^{12}(2 - \omega^{i}- \omega^{12-i}) \\ &=288 - 24\sum_{i=1}^{12}\omega^i \end{align*} Now, $\omega^i$ are the roots of the equation $z^{12}-1 = 0$ and hence the second term above is zero. Thus $S=144$.