Geometrical Proof

Here is an interesting geometry problem-

Let A1A_1, A2  AnA_2 \ \cdots \ A_n be the vertices of a regular nn-sided polygon PP. Let XX be a random point on the incircle of PP.

Prove that i=1nAiX2=nR2[1+cos2(πn)]\displaystyle\sum_{i=1}^{n} \overline{A_iX}^2=nR^2\left[1+\cos^2\left(\dfrac{\pi}{n}\right)\right].


Note : Here, RR denotes the circumradius of PP.

Note by Pratik Shastri
4 years, 11 months ago

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@KARAN SHEKHAWAT Here is the solution :

Let OO represent the center of the polygon.

Now, XAi=XO+OAiXAi2=XO2+OAi2+2(XOOAi)i=1nXAi2=n(r2+R2)+2i=1n(XOOAi)=n(r2+R2)+2XO(i=1nOAi)\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 i=1nOAi=0\displaystyle\sum_{i=1}^{n} \vec{OA_i}=0, i=1nXAi2=i=1nAiX2=n(R2+r2)\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 rr in terms of RR. That's a matter of simple trigonometry.

Note 1 : rr is the inradius of PP.

Note 2 : i=1n(XOOAi)=XO(i=1nOAi)\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 - 4 years, 11 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 !

Karan Shekhawat - 4 years, 11 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 ?

Karan Shekhawat - 4 years, 11 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.

Pratik Shastri - 4 years, 11 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 !

Karan Shekhawat - 4 years, 11 months ago

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

Pratik Shastri - 4 years, 11 months ago

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@Karan Shekhawat Anyways I will upload it within a day or two.

Pratik Shastri - 4 years, 11 months ago

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