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 \(P\).

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TopNewest@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. – Pratik Shastri · 2 years, 3 months agoLog in to reply

– Karan Shekhawat · 2 years, 3 months ago

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

Can we use particular case of polygon in which vertex are n'th roots of of unity To prove it ? – Karan Shekhawat · 2 years, 4 months ago

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– Pratik Shastri · 2 years, 4 months ago

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

Thanks ! – Karan Shekhawat · 2 years, 3 months ago

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– Pratik Shastri · 2 years, 3 months ago

Anyways I will upload it within a day or two.Log in to reply

– Pratik Shastri · 2 years, 3 months ago

Well, I didn't solve it using complex numbers, I used vectors.Log in to reply