What is the smallest square which can enclose a regular hexagon ?
I will use the notation r2 = square root(2) and r3 = square root(3).
Let S = side of the square, s = side of the hexagon.
Then the area of the hexagon = 6 * s/2 * s r3 / 2 = 3/2 s^2 r3.
If you put one diameter of the hexagon parallel
to one pair of sides of the square,
then the side of the hexagon will be half the side of the square.
S = 2s, and the area of the hexagon will be
3/2 (S/2)^2 r3 = 3 S^2 r3 / 8 =
S^2 * 3 r3 / 8 = 0.6495 times the area of the square.
If you put one diameter of the hexagon on a diagonal of the square,
as shown here, then,
We have:
BL = s/ r2
BC = 2*s/2 r3 = s r3
BC = MB r2 = s r3, so MB = s r3 / r2
S = MB + BL
S = s/r2 + s r3 / r2
S / s = (1+r3) / r2 ; S = s * 1.932
or s = S r2 / (1 + r3) = S * 0.5176
That is, the area of the hexagon is about 70% of the area of the square.
(While this is not a proof that this ratio, S/s, is the smallest possible,
I think it is.)
If you put one diameter of the hexagon on a diagonal of the square, 