# 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.