Here is an interesting geometry problem-
Let A1, A2 ⋯ An be the vertices of a regular n-sided polygon P. Let X be a random point on the incircle of P.
Prove that i=1∑nAiX2=nR2[1+cos2(nπ)].
Note : Here, R denotes the circumradius of P.
#Geometry
#Polygons
#Circles
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@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 !
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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