What is the size of the largest regular hexagon that can be constructed inside a square with side-length 's'?

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