# Geometrical Proof

Here is an interesting geometry problem-

Let $A_1$, $A_2 \ \cdots \ A_n$ be the vertices of a regular $n$-sided polygon $P$. Let $X$ be a random point on the incircle of $P$.

Prove that $\displaystyle\sum_{i=1}^{n} \overline{A_iX}^2=nR^2\left[1+\cos^2\left(\dfrac{\pi}{n}\right)\right]$.

Note : Here, $R$ denotes the circumradius of $P$.

Note by Pratik Shastri
6 years, 5 months ago

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## Comments

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@KARAN SHEKHAWAT Here is the solution :

Let $O$ represent the center of the polygon.

Now, \begin{aligned}\vec{XA_i}&=\vec{XO}+\vec{OA_i}\\ |\vec{XA_i}|^2&=|\vec{XO}|^2+|\vec{OA_i}|^2+2(\vec{XO} \cdot \vec{OA_i})\\ \sum_{i=1}^{n} |\vec{XA_i}|^2&=n(r^2+R^2)+2\sum_{i=1}^{n}(\vec{XO} \cdot \vec{OA_i})\\ &=n(r^2+R^2)+2\vec{XO} \cdot \left(\sum_{i=1}^{n} \vec{OA_i}\right)\end{aligned} Since $\displaystyle\sum_{i=1}^{n} \vec{OA_i}=0$, $\sum_{i=1}^{n} |\vec{XA_i}|^2=\sum_{i=1}^{n} \overline{A_iX}^2=n(R^2+r^2)$

All that's left to do is to express $r$ in terms of $R$. That's a matter of simple trigonometry.

Note 1 : $r$ is the inradius of $P$.

Note 2 : $\displaystyle\sum_{i=1}^{n}(\vec{XO} \cdot \vec{OA_i})=\vec{XO} \cdot \left(\sum_{i=1}^{n} \vec{OA_i}\right)$ due to the distributivity of the dot product over vector addition.

- 6 years, 5 months ago

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Awesome ! Thanks a lot ! It's is much better than complex number tecnique atleast for me :) You are genius !

- 6 years, 4 months ago

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Can we use particular case of polygon in which vertex are n'th roots of of unity To prove it ?

- 6 years, 5 months ago

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You could do that. We are to prove the general result which can also be, to my knowledge, proven that way.

- 6 years, 5 months ago

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I'm get stuck in that ! If you Don't mind Can you please Upload its solution So that we can learn from it ?

Thanks !

- 6 years, 5 months ago

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Well, I didn't solve it using complex numbers, I used vectors.

- 6 years, 5 months ago

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Anyways I will upload it within a day or two.

- 6 years, 5 months ago

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