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This question is an exact duplicate of:

I have no idea how to solve this problem, but I'm pretty sure that it could be made easier by using Roots of Unity! Help would be appreciated!

Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3. Find $PA_1^2 + PA_2^2 + \dots + PA_{11}^2$.

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marked as duplicate by Pedro Tamaroff Feb 15 at 14:15

This question was marked as an exact duplicate of an existing question.

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    $\begingroup$ Interpret the symbols as corresponding complex numbers. $$\sum_{k=1}^{11} |P-A_k|^2 = \sum_{k=1}^{11}\left(|P|^2+ \bar{P}A_k + P \bar{A_k} + |A_k|^2\right) = 11(3^2+ 0 + 0 + 2^2)$$ $\endgroup$ – achille hui Jul 17 '17 at 5:15
  • $\begingroup$ @achillehui: why is $\sum (\bar P A_k + P \bar A_k) = 0$? $\endgroup$ – Robert Lewis Jul 17 '17 at 5:19
  • $\begingroup$ @RobertLewis $\frac{1}{11} \sum_{k=1}^{11} A_k$ is the complex number corresponds to center of the circle. (We can choose a coordinate system to make it the origin before we identify points on Euclidean plane with complex plane) $\endgroup$ – achille hui Jul 17 '17 at 6:18
  • $\begingroup$ @Achillehui: thanks. I'd monentarily forgotten that $x^{11} - 1$ has no term of degree $10$. 😏 $\endgroup$ – Robert Lewis Jul 17 '17 at 6:35
  • $\begingroup$ Also, if you are familiar with some physics, then this is just an application of the parallel axis theorem (moment of inertia). :D $\endgroup$ – Sawarnik Jul 17 '17 at 7:21
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You can take the $A_j$ to have coordinates $2\zeta$ where $\zeta$ runs through the eleventh roots of unity. Then $P$ will have coordinates $3u$ for some complex $u$ with $|u|=1$. So $$|PA_j|^2=|3u-2\zeta|^2 =(3u-2\zeta)(3\overline u-2\overline\zeta)=13-6(\zeta\overline u+\overline \zeta u).$$ It should be easy to add these up over all zeta.

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  • $\begingroup$ I think this can be pushed a lot further. Cheers! $\endgroup$ – Robert Lewis Jul 17 '17 at 5:18

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