<s>[INST] Proving that $\sum \overline{PV_i}^2=\frac{nl^2}{4}\left(1+2\cot^2 \frac{\pi}{n}\right)$ Given that $P$ is any point on the incircle of a regular $n$-sided polygon with edge length $l$ and vertices $V_1,V_2...V_n$, how do we prove that $$\displaystyle\sum_{i=1}^{n} \overline{PV_i}^2=\dfrac{nl^2}{4} \left(1+2\cot^2 \dfrac{\pi}{n}\right)$$
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Here is a figure of what I understand 

I don't even know where to begin with this one.. We want to prove that $\displaystyle\sum_{i=1}^{n} \overline{PV_i}^2$ is independent of the chosen point $P$..
Hints in the right direction and answers appreciated.
Please avoid using vector algebra or complex numbers to solve this problem. 
 [/INST] Let $W_i$ be the midpoint of the segment $\overline{V_iV_{i+1}}$ (where $V_{n+1}=V_1$). Apply the Apollonius' theorem on the triangle $\triangle PV_iV_{i+1}$ with the midpoint $W_i$. We have
$$\overline{PV_i}^2+\overline{PV_{i+1}}^2=2(\overline{PW_i}^2+\overline{V_iW_i}^2)=2\overline{PW_i}^2+\frac{l^2}{2}$$
Summing this up with all $1\le i\le n$ gives
$$2\sum_{i=1}^n\overline{PV_i}^2=2\sum_{i=1}^n\overline{PW_i}^2+\frac{n}{2}l^2\tag{1}$$
Now we will show that if $T_1T_2\cdots T_n$ is a regular polygon with a circumcircle of radius $r$, we have $\displaystyle\sum_{i=1}^n\overline{PT_i}^2=2nr^2$ for every point $P$ on the circle.
Without the loss of generality, assume $P$ lie on the arc $T_1T_n$. From the law of cosines, $$\overline{PT_i}^2+\overline{PT_{i+1}}^2=2\overline{PT_i}\cdot\overline{PT_{i+1}}\cos\frac\pi n+(2r\sin\frac\pi n)^2=4\cot\frac\pi n|\triangle PT_iT_{i+1}|+(2r\sin\frac\pi n)^2$$ for all $1\le i\le n-1$. For $i=n$,\begin{align}\overline{PT_1}^2+\overline{PT_{n}}^2&=2\overline{PT_1}\cdot\overline{PT_{n}}\cos\frac{(n-1)\pi }n+(2r\sin\frac{(n-1)\pi} n)^2\\&=-4\cot\frac\pi n|\triangle PT_1T_n|+(2r\sin\frac\pi n)^2\end{align}
Sum this up by $i$ and get \begin{align}2\sum_{i=1}^n\overline{PT_i}^2&=4\cot\frac\pi n|T_1T_2T_3\cdots T_n|+4nr^2\sin^2\frac\pi n\\&=4\cot\frac\pi n\sum_{i=1}^n|\triangle OT_iT_{i+1}|+4nr^2\sin^2\frac\pi n\\&=4\cot\frac\pi n\sum_{i=1}^nr\sin\frac\pi nr\cos\frac\pi n+4nr^2\sin^2\frac\pi n\\&=4nr^2\end{align} Therefore $\displaystyle\sum_{i=1}^n\overline{PT_i}^2=2nr^2$
Now back to the original problem. The $W_1W_2\cdots W_n$ is a regular n-polygon lying on a circle with radius $r=\frac l2\cot\frac\pi n$. So $$\sum_{i=1}^n\overline{PW_i}^2=2nr^2=\frac{nl^2}{2}\cot^2\frac\pi n$$ Hence from $(1)$, we get $$\sum_{i=1}^n\overline{PV_i}^2=\frac{nl^2}4(1+2\cot^2\frac\pi n)$$ and we are done.
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