Going, going, ‘gon

Surprise, surprise, a Catriona Shearer puzzle.

16.
Going, Going, ‘gon

Six identical squares and a smaller rectangle are fitted into this regular hexagon. What fraction of the hexagon do they cover?

Spoiler!

Is there any way to do this besides the hard way? I looked for a dissection solution but came up with nothing.

Look at just one twelfth of the hexagon, a right triangle, ABO:

BC and CD are two sides of a square. OA and DE are two line segments I added. The central rectangle isn’t shown here because it differs in the different triangles around the hexagon.

If AC = DE is defined to be length 1 and AB = CE = x then BC = CD = \sqrt{1+x^2} and BD = \sqrt{2+2x^2}. But DO = 2 and BD+DO = 2x, so

\sqrt{2+2x^2} = 2x - 2

2+2x^2 = 4(x-1)^2

x^2-4x+1=0

x = 2 + \sqrt{3} is the correct solution

Then the height of the triangle ABO is 3+ 2\sqrt{3} and its area is (3+2\sqrt{3})(2+\sqrt{3})/2 = (6+3\sqrt{3}+4\sqrt{3}+6)/2 = 6 + 7\sqrt(3)/2, and the hexagon’s full area is 72+42\sqrt{3}.

The sides of the squares are BC = 2\sqrt{2+\sqrt{3}}, so the six squares have area 48+24\sqrt{3}.

The central rectangle has area 8\times DEO = 8(\sqrt{3})/2 = 4\sqrt{3}.

Then the squares plus rectangle are 48 + 28\sqrt{3}, which is 2/3 of the hexagon’s area.

 

 

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