Proving that $\sum \overline{PV_i}^2=\frac{nl^2}{4}\left(1+2\cot^2 \frac{\pi}{n}\right)$

Question

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 $$\displaystyle\sum_{i=1}^{n} \overline{PV_i}^2=\dfrac{nl^2}{4} \left(1+2\cot^2 \dfrac{\pi}{n}\right)$$ ?

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.

Answer 1

Here is a sketch of the proof. Rescaling, you can assume $l=1$ without loss. The vertices of the regular polygon are $V_1,V_2,\ldots,V_n$ where $V_k$ has coordinates $(\cos(k\frac{2\pi}{n}),\sin(k\frac{2\pi}{n}))$. Denote by $I_k$ the middle point of the segment $[V_kV_{k+1}]$. Using the prosthaphaeresis formulas, we obtain

$$I_k\ \Bigg(\cos\bigg(k\frac{2\pi}{n}\bigg)\cos\bigg(\frac{(2k+1)\pi}{n}\bigg), \cos\bigg(k\frac{2\pi}{n}\bigg)\sin\bigg(\frac{(2k+1)\pi}{n}\bigg)\Bigg)\tag{1}$$

Since all the $I_k$ are on the incircle, we deduce $r=OI_k=|\cos\bigg(\frac{2\pi}{n}\bigg)|$.

If $P(x,y)$ is a point on the incircle, let us put $\theta=\frac{2\pi}{n}$ and $T=\sum_{k=1}^n PV_k^2$. Then :

$$ \begin{array}{lcl} T &=& \sum_{k=1}^n (x-\cos(k\theta))^2+(y-\sin(k\theta))^2 \\ &=& \sum_{k=1}^n x^2+y^2-2\cos(k\theta)x-2\sin(k\theta)y+1 \\ &=& n(r^2+1)-2x\Bigg(\sum_{k=1}^n\cos(k\theta)\Bigg) -2y\Bigg(\sum_{k=1}^n\sin(k\theta)\Bigg) \\ \end{array} $$

Now, each of the sums $C=\sum_{k=1}^n\cos(k\theta)$ and $S=\sum_{k=1}^n\sin(k\theta)$ are zero. For $S$, this is easy : the fact that the sine function is odd suffices. For $C$, this is very easy using complex numbers, and a little harder if you won’t use complex numbers (hint : use two different symmetries, according to whether $n$ is even or odd). So $T=n(r^2+1)$ is indeed independent of $P$. The rest is simple trigonometric manipulation.

Answer 2

Let $P \in \mathbb{C}$ and let your regular polygon by the points $R\omega^k = Re^{2\pi i k /n}$ in the complex plane.

$$ \sum_{i=0}^{n-1} |P - R\omega^k|^2 = \sum_{i=0}^{n-1} \bigg( |P|^2 - R(P\omega^{-k} +\overline{P}\omega^k) + R^2 \bigg) = n \big(|P|^2 + R^2\big) $$

$P$ is a point on the incircle so we need the radius, $|P| = r = \tfrac{l}{2}\csc \frac{\pi}{n}{n}$.

The radius of the circle in terms of the edge length is: $R = \tfrac{l}{2} \tan \frac{\pi}{n}$.

$\boxed{\sum \overline{PV}^2 = n \big( \tfrac{l^2}{4} \cot^2 \tfrac{\pi}{n} + \tfrac{l^2}{4}\csc^2 \tfrac{\pi}{n} \big) }$

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Alterantively we can use parallel axis theorem. Your sum is the moment of inertia ($I=\sum mr^2$) for unit masses ($m=1$) placed on the vertices $V$.

If your rotation axis were placed at the center of the polygon, the moment of inertia would be $m$ times the radius:

$$I_{CM} = \sum \overline{OV}^2 = \frac{nl^ 2}{4} \csc^2 \tfrac{\pi}{n} $$

Moving the center we add the term $md^2$ where $d$ is the distance from the origin to the point $\overline{OP}$ which is the radius of the incircle $|P| = 2l \cot n$

$$ I = I_{CM} + md^2 = \frac{nl^ 2}{4} \csc^2 \tfrac{\pi}{n} + \frac{nl^2 }{4} \cot^2 \tfrac{\pi}{n} = \boxed{\frac{nl^ 2}{4} \big( 1 + 2 \cot^2 \tfrac{\pi}{n} \big)} $$

For discussion of moment of inertia see Feynman Lectures I Ch 19

Answer 3

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.