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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Pratik Shastri
3 years, 11 months ago
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Top Newest@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.
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Awesome ! Thanks a lot ! It's is much better than complex number tecnique atleast for me :) You are genius !
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Can we use particular case of polygon in which vertex are n'th roots of of unity To prove it ?
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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.
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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 !
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Anyways I will upload it within a day or two.
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Well, I didn't solve it using complex numbers, I used vectors.
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