# Very confusing polygon question. Can anyone help?

I was practising questions on principles on mathematics. I stumbled onto this question and I don't know where to start. Can anyone please help??

If $P_1P_2....P_n$ is a regular polygon in the $(x,y)$-plane, each side of length a (so the $P_i$ are the corners of an $n$-sided figure with sides of equal length $a$). Find the sum $$S = (P_1P_2)^2 + (P_1P_3)^2 + \dots + (P_1P_n)^2;$$ here $P_1P_j$ stands for the length of the line form the point $P_1$ to the point $P_j$ (your expression for $S$ will be a function of $a$, $n$ and also a well-known trigonometric function).

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I don't know the answer, but the point $P_1$ you begin with is relevant : think of a four-sided diamond ; it is a polygon with equal sides, but depending on which point you call $P_1$, you'll get different sums of squares. –  Patrick Da Silva Jan 7 '13 at 15:47
@PatrickDaSilva The point $P_1$ is not relevant by the symmetry of the regular polygon. –  Hagen von Eitzen Jan 7 '13 at 15:51
@Hagen : Man did I not read that it was regular! The problem is so trivial now compared to the one I was looking at. –  Patrick Da Silva Jan 7 '13 at 16:07

Let $\zeta\in\mathbb C$ be a primitive $n$th root of unity. Then in the specaial case that $a=|\zeta-1|$, we have $$S=\sum_{k=1}^n|\zeta^k-1|^2=\sum_{k=1}^n(2-\zeta^k-\bar\zeta^k)=2n.$$ Hence in general $$S = \frac{2a^2n}{|\zeta-1|^2}.$$ Note that $|\zeta-1|=2\sin\frac\pi n$, so that ultimately $$S = \frac{a^2n}{2\sin^2\frac\pi n}.$$

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Of course. Good job, +1! –  Patrick Da Silva Jan 7 '13 at 16:09

The more general situation (and nicer way) to state the problem is: In the circle $\Gamma$ with center $O$ and radius $r$, we have inscribed a regular $n-$gon $P_1 P_2 \ldots P_n$. If $P$ is any point such that $OP=d$, then $\sum | P_i P |^2 = n(r^2+d^2)$.

To show this, you can either use complex numbers, or vectors. I'd present the dot product of vectors.

Let $O$ be the origin. Then $\vec{P_1} + \ldots + \vec{P_n} = 0$. Hence,

\begin{align} \sum | P_i P |^2 & = \sum (\vec{P} - \vec{P_i}) \cdot (\vec{P} - \vec{P_i}) \\ & = \sum \left(\vec{P} \cdot \vec{P} + \vec{P} \cdot \vec{P} - 2 \vec{P} \cdot \vec{P_i}\right)\\ & = nd^2 + nr^2 - 2 \vec{P} \cdot \left( \sum \vec{P_i} \right) \\ & = n(d^2+r^2) \\ \end{align}

Finally, to answer your original question, express $r$ in terms of $a$ and $n$, and set $P=P_1$.

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Consider,

$\theta$= $360/k$

Where $k$ ->Number of sides of polygon.

$a = 2Rsin(\theta$/$2)$ = $2Rsin(360/(2*k))$

$R = a/(2*sin(180/k))$

$S=(P1P2)^2+(P1P3)^2+⋯+(P1Pn)^2$= $\sum(2Rsin(180n/k))^2$

=$4R^2 \sum sin^2(180*n/k)$

= $2R^2(n - cos(n(n-1)180/k))$ = $2 a^2 (n - cos(n(n-1)180/k))/(2*sin(180/k))^2$

$S= a^2 (n - cos(n(n-1)180/k))/(2*sin^2(180/k))$

We have $n = k-1$

$S= a^2 (k-1 - cos((k-1)*(k-2)*180/k)/(2*sin^2(180/k))$

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Please use $\LaTeX$ to format your answer. –  Parth Kohli Jan 7 '13 at 16:58