The largest square that can be fitted inside a regular hexagon:

The square and the hexagon must share the same centre.

Proof: If the square within the hexagon is not centred [red], we can rotate it by a half turn about the hexagon centre [blue], and it will still fit in the hexagon.

We can translate the square from one position to the other staying within the hexagon, passing over the centre. Thus the largest square must be able to be drawn centred with the hexagon.

The square we will derive will have reflective symmetry about horizontal and vertical axes, and will have b < c. Any rotation about the centre will therefore make it smaller (consider the length of its diagonal). This is therefore the largest possible square inside the hexagon.

To derive the size of this square:

The sides of the hexagon are equal:

Adding some lines:

Taking the left hand edge of the square:

..and the top edge of the square:

Substituting for c:

Substituting for d and re-arrranging:

The standard formulae for sin(30), sin(60) (can be found by using Pythagoras on a right angled triangle):

Inserting these formulae and simplifying:

Dividing the earlier ('left hand edge of square') equation by a, substituting for b/a and simplifying:

page date: 26Oct05.      I enjoy correspondence stimulated by this site. You can contact me here.