Triangles on a Hexagon

Let $A$ be at $(0,0)$ and let $B$ be at $(1,0)$ on the cartesian plane. The cordinates of point $F$ can be found by drawing perpendicular lines at points $F$ and $A$ to form a 30-60-90 triangle as shown. Since the hypotenuse is of length 1, the shorter leg has length $\frac{1}{2}$ and the longer leg has length $\frac{1}{2}\cdot\sqrt{3}$ or $\frac{\sqrt{3}}{2}$ . This gives us the coordinates of $F$ which are $(-\frac{1}{2},\frac{\sqrt{3}}{2})$ . The coordinates of $F$ , $G$ , $H$ , $I$ , $J$ , and $B$ can all be found using various methods using properties of 30-60-90 and 45-45-90 triangles and circles.

The final coordinates (you'll have to take my word for it, or do it yourself) end up being the following.

$B$ : $\left(1,0\right)$

$F$ : $\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$

$G$ : $\left(-\frac{1}{2},\sqrt{3}\right)$

$H$ : $\left(\frac{1}{2},\sqrt{3}+\frac{1}{2}\right)$

$I$ : $\left(\frac{7}{4},\frac{3\sqrt{3}}{4}\right)$

$J$ : $\left(\frac{\left(5+\sqrt{3}\right)}{4},\frac{\left(-1+\sqrt{3}\right)}{4}\right)$

Using point-slope form gives us the equations for the lines $\overline { HB }$ , $\overline { GB }$ , $\overline { IF }$ , and $\overline { JF }$ .

$\overline { HB }$ : $y=-\frac{\left(\sqrt{3}+\frac{1}{2}\right)}{\frac{1}{2}}\left(x-1\right)$

$\overline { GB }$ : $y=\frac{\sqrt{3}}{-\frac{3}{2}}\left(x-1\right)$

$\overline { IF }$ : $y-\frac{\sqrt{3}}{2}=\frac{\frac{\sqrt{3}}{4}}{\frac{9}{4}}\left(x+\frac{1}{2}\right)$

$\overline { JF }$ : $y-\frac{\sqrt{3}}{2}=\frac{\frac{\left(-1+\sqrt{3}\right)}{4}-\frac{\sqrt{3}}{2}}{\frac{\left(5+\sqrt{3}\right)}{4}+\frac{1}{2}}\left(x+\frac{1}{2}\right)$

We can solve these equations to find the points $K$ , $L$ , $M$ , $N$ .

$K$ : $\left(\frac{1}{7},\frac{4\sqrt{3}}{7}\right)$

$L$ : $\left(\frac{9\sqrt{3}+110}{167},\frac{3}{167}\left(1+35\sqrt{3}\right)\right)$

$M$ : $\left(\frac{2\left(15+\sqrt{3}\right)}{37},\frac{1}{37}\left(12\sqrt{3}-5\right)\right)$

$N$ : $\left(\frac{3\sqrt{3}+26}{59},\frac{2}{59}\left(11\sqrt{3}-3\right)\right)$

The shoelace theorem is a useful theorem, which gives us the following. $KLMN = \frac{|\frac{1}{7}\cdot \frac{3}{167}\left(1+35\sqrt{3}\right)+\frac{9\sqrt{3}+110}{167}\cdot \frac{1}{37}\left(12\sqrt{3}-5\right)+\frac{2\left(15+\sqrt{3}\right)}{37}\cdot \frac{2}{59}\left(11\sqrt{3}-3\right)+\frac{3\sqrt{3}+26}{59}\cdot \frac{4\sqrt{3}}{7}-(\frac{4\sqrt{3}}{7}\cdot \frac{9\sqrt{3}+110}{167}+\frac{3}{167}\left(1+35\sqrt{3}\right)\cdot \frac{2\left(15+\sqrt{3}\right)}{37}+\frac{1}{37}\left(12\sqrt{3}-5\right)\cdot \frac{3\sqrt{3}+26}{59}+\frac{2}{59}\left(11\sqrt{3}-3\right)\cdot \frac{1}{7})|}{2}$

This simplifies to $KLMN=\frac{390789+591161\sqrt{3}}{2551927}$ .

The area of hexagon $ABCDEF$ is equivalent to 6 times the area of an equilateral triangle with side length 1. $ABCDEF=6\cdot\frac{\sqrt{3}}{4}$

The ratio of the areas of $ABCDEF$ to $KLMN$ is $\frac{6\cdot\frac{\sqrt{3}}{4}}{\frac{390789+591161\sqrt{3}}{2551927}}$ or about 9.373088985.

Rounding to the nearest thousandth gives us $\boxed{9.373}$ .