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Here is an interesting geometry problem-
Let A1A_1A1, A2 ⋯ AnA_2 \ \cdots \ A_nA2 ⋯ An be the vertices of a regular nnn-sided polygon PPP. Let XXX be a random point on the incircle of PPP. Prove that ∑i=1nAiX‾2=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]i=1∑nAiX2=nR2[1+cos2(nπ)].
Let A1A_1A1, A2 ⋯ AnA_2 \ \cdots \ A_nA2 ⋯ An be the vertices of a regular nnn-sided polygon PPP. Let XXX be a random point on the incircle of PPP.
Prove that ∑i=1nAiX‾2=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]i=1∑nAiX2=nR2[1+cos2(nπ)].
Note : Here, RRR denotes the circumradius of PPP.
Note by Pratik Shastri 5 years ago
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@KARAN SHEKHAWAT Here is the solution :
Let OOO represent the center of the polygon.
Now, XAi⃗=XO⃗+OAi⃗∣XAi⃗∣2=∣XO⃗∣2+∣OAi⃗∣2+2(XO⃗⋅OAi⃗)∑i=1n∣XAi⃗∣2=n(r2+R2)+2∑i=1n(XO⃗⋅OAi⃗)=n(r2+R2)+2XO⃗⋅(∑i=1nOAi⃗)\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}XAi∣XAi∣2i=1∑n∣XAi∣2=XO+OAi=∣XO∣2+∣OAi∣2+2(XO⋅OAi)=n(r2+R2)+2i=1∑n(XO⋅OAi)=n(r2+R2)+2XO⋅(i=1∑nOAi) Since ∑i=1nOAi⃗=0\displaystyle\sum_{i=1}^{n} \vec{OA_i}=0i=1∑nOAi=0, ∑i=1n∣XAi⃗∣2=∑i=1nAiX‾2=n(R2+r2)\sum_{i=1}^{n} |\vec{XA_i}|^2=\sum_{i=1}^{n} \overline{A_iX}^2=n(R^2+r^2)i=1∑n∣XAi∣2=i=1∑nAiX2=n(R2+r2)
All that's left to do is to express rrr in terms of RRR. That's a matter of simple trigonometry.
Note 1 : rrr is the inradius of PPP.
Note 2 : ∑i=1n(XO⃗⋅OAi⃗)=XO⃗⋅(∑i=1nOAi⃗)\displaystyle\sum_{i=1}^{n}(\vec{XO} \cdot \vec{OA_i})=\vec{XO} \cdot \left(\sum_{i=1}^{n} \vec{OA_i}\right)i=1∑n(XO⋅OAi)=XO⋅(i=1∑nOAi) 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 !
Can we use particular case of polygon in which vertex are n'th roots of of unity To prove it ?
You could do that. We are to prove the general result which can also be, to my knowledge, proven that way.
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 – Well, I didn't solve it using complex numbers, I used vectors.
@Karan Shekhawat – Anyways I will upload it within a day or two.
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Top Newest@KARAN SHEKHAWAT Here is the solution :
Let O represent the center of the polygon.
Now, XAi∣XAi∣2i=1∑n∣XAi∣2=XO+OAi=∣XO∣2+∣OAi∣2+2(XO⋅OAi)=n(r2+R2)+2i=1∑n(XO⋅OAi)=n(r2+R2)+2XO⋅(i=1∑nOAi) Since i=1∑nOAi=0, i=1∑n∣XAi∣2=i=1∑nAiX2=n(R2+r2)
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 : i=1∑n(XO⋅OAi)=XO⋅(i=1∑nOAi) 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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