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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 $P_1\,P_2\,\ldots\,P_n$ is a regular polygon in the $(x,y)$-plane, each side of length $a>0$ (so the $P_i$ are the corners of an $n$-sided figure with sides of equal length $a>0$ ). Find the sum $$ S=\sum_{j=2}^{n}\;(\overline{P_1P_j})^2=(\overline{P_1P_2})^2 +(\overline{P_1P_3} )^2 +\ldots+(\overline{P_1P_n})^2 ; $$ here $\overline{P1 Pj}$ 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 a well-known trigonometric function).

Exemple for $n=4$: enter image description here

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Sounds like you might want to identify the points with complex numbers. Do you know that the nth roots of 1 form a regular polygon? – hardmath Jan 8 '13 at 11:52
@Elias I believe that if you have a picture which would improve the post, it is better to post a link in a comment (and the OP can use it, if he wants). To me it seemed like too radical change of the original post. But it seems that other users have different opinion, since your edit has been approved. – Martin Sleziak Jan 8 '13 at 12:22
@MartinSleziak The image is not hosted on any web site. I even did one in Inkscape vector drawing software that supports LaTeX code So I do not have a link to the image. – MathOverview Jan 8 '13 at 12:49
@Elias Obviously, it there exists an imgur link now. Posting an image in a post and then using the link without saving the edit is not uncommon:… – Martin Sleziak Jan 8 '13 at 14:30

Hardmath's hint is excellent. You may also want to use (prove) that :

1) We can express

$$P_{k+1}=ae^{\frac{2k\pi i}{n}}\;\;,\;\;k=0,1,2,...,n-1$$

2) If $\,z_1=r_1e^{it_1}\,\,,\,z_2=r_2e^{it_2}\,$ , then


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Here you have a geometric approach...

As showed in the picture, $O$ represents the (circum)center of the polygon, $\alpha=\frac{2\pi}{n}$ denotes the angle $\widehat{P_1OP_2}$ and $\beta=(k-1)\alpha$ is the angle $\widehat{P_1OP_k}$ where $2\leq k\leq n$ is fixed.

enter image description here

The Law of cosines on the triangle $P_1OP_2$ implies that $\overline{OP_1}^2=\frac{a^2}{2(1-\cos\alpha)}$, and the same reasoning on the triangle $P_1OP_k$ shows that $$ \overline{P_1P_k}^2=2\overline{OP_1}^2(1-\cos\beta) = \frac{a^2}{\left(1-\cos{\frac{2\pi}n}\right)}\left(1-\cos{\frac{(k-1)2\pi}n}\right) $$ Therefore the answer is $$ \sum_{k=2}^n{\overline{P_1P_k}^2} = \frac{a^2}{\left(1-\cos{\frac{2\pi}n}\right)} \sum_{k=1}^{n-1}{\left(1-\cos{\frac{2k\pi}n}\right)} $$ Using the fact that $\cos x$ is the real part of $e^{ix}$ and that $\sum_{k=0}^{n-1}e^{i\frac{2k\pi}{n}}=0$ you have that the above sum is equal to $$ \frac{na^2}{1-\cos{\frac{2\pi}{n}}} $$

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A little explanation about the last passages... $e^{i\frac{2k\pi}{n}}$ for $k=0\ldots n-1$ are the complex roots of the polynomial $x^n-1$. It is well known that the sum of all the roots of a polynomial is equal to the linear coefficient of the polynomial (i.e. of '$x$'), which in $x^n-1$ is $0$. Therefore $0=\sum_{k=0}^{n-1}e^{i\frac{2k\pi}n}=1+\sum_{k=1}^{n-1}e^{i\frac{2k\pi}n}$. – AndreasT Jan 8 '13 at 12:52

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