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theEricBest ResponseYou've already chosen the best response.1
Google says: " A hexagon is inscribed in a circle radius 1. Alternate sides have length 1. Show that the midpoints of the other three sides form an equilateral triangle" http://mks.mff.cuni.cz/kalva/putnam/putn67.html "Each side length 1 forms an equilateral triangle with the centre. Let the other three sides subtend angles 2A, 2B, 2C at the centre. The midpoints of the sides subtending 2A and 2B are distances cos A and cos B from the centre and the line joining them subtends an angle A+B+60 at the centre. So, using the cosine formula, the square of the distance between them is cos2A + cos2B  2 cos A cos B cos(A+B+60) (*). But A + B + C = 90, so cos(A+B+60) = cos(150C) = √3/2 cos C + 1/2 sin C. Hence we may write (*) as (cos2A + cos2B + cos2C) + √3 cos A cos B cos C  cos A cos B sin C  cos2C. But we can write cos C as sin(A + B) = sin A cos B + cos A sin B and hence cos2C as sin A cos B cos C + cos A sin B cos C. Thus (*) has the symmetrical form (cos2A + cos2B + cos2C) + √3 cos A cos B cos C  (cos A cos B sin C + cos A sin B cos C + sin A cos B cos C), which establishes that the triangle is equilateral."  http://mks.mff.cuni.cz/kalva/putnam/psoln/psol677.html
 one year ago

theEricBest ResponseYou've already chosen the best response.1
I didn't read it myself, but I hope this helps you understand the proof!
 one year ago

theEricBest ResponseYou've already chosen the best response.1
The only thing that seems different is that this link has the radius be 1, while the radius in your problem has the radius be r.
 one year ago

theEricBest ResponseYou've already chosen the best response.1
I don't follow the proof :P Good luck!
 one year ago

jayz657Best ResponseYou've already chosen the best response.0
ok thanks alot i read over this and try to work it out myself
 one year ago
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