For $A_1 A_2\cdots A_{11}$ a regular $11$-gon with circumradius $2$, and point $P$ at distance $3$ from the center, find $\sum PA_k^2$ 
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
$$PA_1^2 + PA_2^2 + \dots + PA_{11}^2$$

This kind of reminded me of roots of unity, because of a regular polygon inscribed in a circle, but I'm not sure on how to work this problem.

I like the idea of using the roots of unity.
$z_1, z_2, \cdots, z_{11}$ are the vertices of our 11=gon. 
$z_n = 2e^{(\frac {2n\pi}{11} + \alpha)i}$
$\alpha$ would represent a factor for the rotation of the 11-gon.  We will see if it turns out to be relevant.
$P = 3$ 
$|(3-z_1)|^2 + |(3-z_2)|^2 +\cdots +  |(3-z_{11})|^2\\
(3-z_1)(3-\bar z_1) + (3-z_2)(3-\bar z_2) +\cdots +  (3-z_{11})(3-z_{11})$ 
$(3-z_1)(3-\bar z_1) = 9 - 3 z_1 - 3\bar{z_1} + z_1\bar z_1$
$z_n\bar z_n = |z_n|^2 = 4$
$(3-z_1)(3-\bar z_1) + (3-z_2)(3-\bar z_2) +\cdots +  (3-z_{11})(3-z_{11}) = 13\cdot 11 - 3\sum_\limits{n=1}^{11}z_n - 3\sum_\limits{n=1}^{11}\bar z_n $ 
Since we have a regular polygon centered at the origin...
$3\sum_\limits{n=1}^{11}z_n = 0$ and $3\sum_\limits{n=1}^{11}\bar z_n = 0$
$\alpha$ turns out to not be relevant.
$13\cdot 11 = 143$
