Find $LK_1^2 + LK_2^2 + \dots + LK_{11}^2$. $K_1 K_2 \dotsb K_{11}$ is a regular $11$-gon inscribed in a circle, which has a radius of $2$. Let $L$ be a point, where the distance from $L$ to the circle's center is $3$. Find
$LK_1^2 + LK_2^2 + \dots + LK_{11}^2$.
Any suggestions as to how to solve this problem? I'm unsure what method to use. 
 A: Use complex coordinates. Let $L = 0$ (the origin), and let $K_n = 3 + 2e^{i\tfrac{2\pi n}{11}}$ for $n = 1,\ldots,11$. 
Clearly $K_1,\ldots,K_{11}$ are $11$ equally spaced points on the circle of radius $2$ centered at $3$.
Then for each $n$, you have $LK_n^2 = \left|3 + 2e^{i\tfrac{2\pi n}{11}}\right|^2 = \left(3 + 2e^{i\tfrac{2\pi n}{11}}\right)\left(3 + 2e^{-i\tfrac{2\pi n}{11}}\right) = \cdots$ (I'll let you finish expanding this). 
From here, computing the sum $LK_1^2+LK_2^2+\cdots+LK_{11}^2$ is easy (assuming you know how to sum a geometric series).

 I believe you should get $LK_1^2+LK_2^2+\cdots+LK_{11}^2 = 143$ as the answer.

EDIT: You don't really need to know how to sum a geometric series if you know that the sum of the $k$-th roots of unity is zero for any positive integer $k$. 
A: 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,
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = 99 + 44 = \boxed{143}.$
A: Use the law of cosine for each triangle $\triangle LOK_i, i = 1,\cdots, 11$, we have: $LK_i^2 = LO^2+OK_i^2 - 2LO\cdot OK_i\cos \theta_i, \theta_i = \angle LOK_i\Rightarrow LK_i^2 = 3^2+2^2-12\cos \theta_i=13-12\cos \theta_i\Rightarrow S = \displaystyle \sum_{i=1}^{11}LK_i^2=13\cdot 11-12\displaystyle \sum_{i=1}^{11}\cos \theta_i=143-12\displaystyle \sum_{i=1}^{11}\dfrac{\vec{OK_i}\cdot \vec{OL}}{|\vec{OK_i}||\vec{OL}|}=143-2\displaystyle \sum_{i=1}^{11}\vec{OL}\cdot\left(\displaystyle \sum_{i=1}^{11}\vec{OK_i}\right)=143-2\displaystyle \sum_{i=1}^{11}\vec{OL}\cdot \vec{0}=143-2\cdot 0 = 143$
A: Let $O$ be the center of the circle.  One would like to say $LO^2 + OK_j^2 = LK_j^2\quad$ --- a Pythagorean theorem of sorts, except that $LO$ is not at a right angle with $OK_j$.  So
\begin{align}
LO^2 + OK_1^2 & \ne LK_1^2 \\
LO^2 + OK_2^2 & \ne LK_2^2 \\
LO^2 + OK_3^2 & \ne LK_3^2 \\
& \,\,\,\vdots \\
LO^2 + OK_{11}^2 & \ne LK_{11}^2
\end{align}
So I ask myself whether perhaps the sum of the left sides equals the sum of the right sides.  Notice that every left side is $3^2+2^2 = 13$, so their sum is $11\times13 = 143$.  We will need the fact that the average of $K_1,\ldots,K_{11}$ is $O$.  Let $x_L$, $x_O$, $x_{K_j}$ be the respective $x$-coordinates.  One has
\begin{align}
(x_L-x_{K_j})^2 & = \Big((x_L - x_O) + (x_O - x_{K_j})\Big)^2 \\[10pt]
& = (x_L - x_O)^2 + 2(x_L - x_O)(x_O - x_{K_j}) + (x_O - x_{K_j})^2
\end{align}
The middle term is not zero, but if we sum this over all $11$ values of $j$ then the sum of the middle terms is zero because the sum of the deviations $x_{K_j} - x_O$ from the average is zero. (Here we are using the fact that the other factor in the middle term, $2(x_L-x_O)$, does not change as $j$ runs through the list $1,\ldots,11$.)
The same thing works with the $y$-coordinates, and we have $LK_j^2 = (x_L-x_{K_j})^2+(y_L - y_{K_j})^2$.
Another approach:
In an $11$-dimensional space, show that the two vectors $(OK_1,\ldots,OK_{11})$ and $(LO,\ldots,LO)$ are at right angles to each other and then apply the Pythagorean theorem.
