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# 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
2 years, 6 months ago

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@KARAN SHEKHAWAT Here is the solution :

Let $$O$$ represent the center of the polygon.

Now, \begin{align}\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{align} 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. · 2 years, 5 months ago

Awesome ! Thanks a lot ! It's is much better than complex number tecnique atleast for me :) You are genius ! · 2 years, 5 months ago

Can we use particular case of polygon in which vertex are n'th roots of of unity To prove it ? · 2 years, 6 months ago

You could do that. We are to prove the general result which can also be, to my knowledge, proven that way. · 2 years, 6 months ago

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 ! · 2 years, 5 months ago

Anyways I will upload it within a day or two. · 2 years, 5 months ago