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Vertices to incircle (Posted on 2019-03-30) Difficulty: 3 of 5
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

A1X2 + A2X2 + ... + AnX2 = nR2(1 + cos2(pi/n))

Here, R denotes the circumradius of P.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Comment 1 of 1
The distance (AiX2) from point X to vertice Ai is igiven by the rule of cosine

AiX2 = R2 + r2 - 2Rr cos(ai) where aiX is the angle XOAi, O is the centre of both circles, R is the circumradius and r is the inradius

A1X2 + A2X2 + ... + AnX2 = nR2+nr- 2Rr [cos(a1)+cos(a2)+ ...+cos(an)]

The term with the sum of cosines is always 0 (with a square the cosines are opposed by pairs, with a triangle for a different reason). So:

A1X2 + A2X2 + ... + AnX2 = nR2+nr

But r=Rcos(pi/n). So: 

A1X2 + A2X2 + ... + AnX2 = nR2(1 + cos2(pi/n))




Edited on April 7, 2019, 4:59 pm
  Posted by armando on 2019-04-07 16:57:57

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