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Hexagon in a square (Posted on 2013-11-10) Difficulty: 3 of 5
What is the size of the largest regular hexagon that can be constructed inside a square with side-length 's'?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution solution Comment 1 of 1

Playing with Geometers' Sketchpad, one can see by symmetry that the diagonal of the hexagon with endpoints not on the square is tilted at 45 to both the horizontal and the vertical (of an ordinarily oriented square). Since the radius toward a square-touching endpoint of a hex-diagonal is at a 60 angle to this line, it's 15 away from being perpendicular to its side of the square, and the side of the square s = 2*cosine(15)*h where h is the side length (also semidiameter) of the hexagon.

cos(30) = sqrt(3)/2

cos(15) = sqrt((1 + sqrt(3)/2) / 2)

h = s / (2*cos(15)) = s / sqrt((2*(1 + sqrt(3)/2)))

              ~= s * 0.517638090205041

So this, h, is the length of one side of the sought hexagon.


  Posted by Charlie on 2013-11-10 15:25:21
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