<s>[INST] Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3. Find
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2.$
 [/INST] Let $\omega = e^{2 \pi i/11}$, a primitive $11^{\text{th}}$ root of unity. We can assume that the circle is centered at the origin. We can also assume that $A_k$ is associated with the complex number $2 \omega^k$.
Let $p$ be complex number associated with the point $P$. Then
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \sum_{k = 0}^{10} |p - 2 \omega^k|^2.$
From the identity $z \cdot \overline{z} = |z|^2$,
\begin{align*}
\sum_{k = 0}^{10} |p - 2 \omega^k|^2 &= \sum_{k = 0}^{10} (p - 2 \omega^k)(\overline{p} - 2 \overline{\omega}^k) \\
&= \sum_{k = 0}^{10} (p \overline{p} - 2 \overline{\omega}^k p - 2 \omega^k \overline{p} + 4 \omega^k \overline{\omega}^k) \\
&= 11 p \overline{p} - 2p \sum_{k = 0}^{10} \overline{\omega}^k - 2 \overline{p} \sum_{k = 0}^{10} \omega^k + 4 \sum_{k = 0}^{10} \omega^k \overline{\omega}^k.
\end{align*}
The distance from $P$ to the origin is 3, so $11p \overline{p} = 11 \cdot |p|^2 = 11 \cdot 9 = 99$.
Since $\omega$ is a primitive $11^{\text{th}}$ root of unity, $\omega^{11} - 1 = 0$, which factors as
$(\omega - 1)(\omega^{10} + \omega^9 + \dots + \omega + 1) = 0.$
Since $\omega \neq 1$, we have $\omega^{10} + \omega^9 + \dots + \omega + 1 = 0$. Therefore,
$2 \overline{p} \sum_{k = 0}^{10} \omega^k = 0.$
Also, $|\omega| = 1$, so $\overline{\omega} = 1/\omega$, which means
$\sum_{k = 0}^{10} \overline{\omega}^k = 1 + \frac{1}{\omega} + \dots + \frac{1}{\omega^9} + \frac{1}{\omega^{10}} = \frac{\omega^{10} + \omega^9 + \dots + \omega + 1}{\omega^{10}} = 0.$
Finally, $\omega^k \overline{\omega}^k = \omega^k/\omega^k = 1$, so
$4 \sum_{k = 0}^{10} \omega^k \overline{\omega}^k = 4 \cdot 11 = 44.$
Therefore,
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = 99 + 44 = 143.$
</s>