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Let $P_1P_2⋯P_7$ be a regular heptagon inscribed in a unit circle. Let $O$ be the center of the circle, and let $P$ be a point such that $OP=1/3$. Then find the sum $$\sum\limits_{i=1}^7PP_i^2$$ How do I find the distance in the triangle $OPP_1$( I don't know the angle, nor the second side)

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You can approach this by using complex numbers. Let $P_k$ be represented by the complex number $$z_k=e^{2k\pi i/7}\ ,$$ and let $P$ be given by $$w=\frac13e^{i\theta}\ .$$ Then $$\eqalign{\sum_{k=1}^7 PP_k^2 &=\sum_{k=1}^7 |z_k-w|^2\cr &=\sum_{k=1}^7 (z_k\overline{z_k}-z_k\overline{w}-w\overline{z_k} +w\overline w)\cr &=7-\overline w\Bigl(\sum_{k=1}^7z_k\Bigr) -w\Bigl(\sum_{k=1}^7\overline{z_k}\Bigr)+\frac79\cr &=\frac{70}{9}\ ,\cr}$$ since both the sums are zero.

Alternative, not using complex methods. Place $P_7$ on the $x$ axis, that is, at $(1,0)$, and then $P_1,P_2,\ldots$ in anticlockwise order around the circle. Let $\theta$ be the angle between $OP$ and the $x$ axis. Drawing a diagram, you can see that the angle $POP_k$ is $(2k\pi/7)-\theta$. By the cosine rule, $$PP_k^2=1+\frac19-\frac29\cos\Bigl(\frac{2k\pi}7-\theta\Bigr)\ .$$ Now adding up all the constant terms is easy. To see how to add up all the cosine terms (without using complex numbers), see my answer here - this is a sum of sine terms, but you can use similar methods.

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David
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  • Thanks, but I am not much familiar with complex numbers. Can you find it using euclidean geometry. –  Feb 24 '15 at 05:03
  • A better idea, Garvil, is to familiarize yourself more with complex numbers. Believe me, you'll be glad you did. – Gerry Myerson Feb 24 '15 at 06:02
  • @Garvil I have added a brief note to my answer on doing this without complex methods. – David Feb 25 '15 at 00:20