What is the size of the largest regular hexagon that can be constructed inside a square with sidelength '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 squaretouching endpoint of a hexdiagonal 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 20131110 15:25:21 