WMC 2018 Pursuit - Pack 1, A3

Since $\Delta ABC$ is a right triangle, we can place a coordinate system such that $BC$ lies on the x-axis and $AD$ lies on the y-axis. Then, points $A$ , $B$ and $C$ have coordinates $(0,4)$ , $(0,0)$ and $(8,0)$ . $AC$ lies on the line with equation $y = - \frac 1 2 x + 4$ .

To find $EF$ , we can also solve for $FG$ since $EFGH$ is a square.

The x-coordinate $x_0$ of F is given by the sum of the height of equilateral triangle $DEH$ and one side of square $EFGH$ , which is $x$ .

If we plug that into the equation for line $AC$ , we get

$- \frac 1 2 \left(\frac{\sqrt{3}}2 x + x\right) + 4 = x$ .

This is equal to $x$ because $x$ is also the y-coordinate of $F$ .

So now we only have to solve this equation.

$- \frac 1 2 \left(\frac{\sqrt{3}}2 x + x\right) + 4 = x \Leftrightarrow - \frac{x \sqrt{3}}4 - \frac 1{2}x + 4 = x \Leftrightarrow 4 = \left(\frac 3 2 + \frac{\sqrt{3}}4\right) x \Leftrightarrow x = \frac{4}{ \frac 3 2 + \frac{\sqrt{3}}{4} } = \frac{16}{6 + \sqrt{3}}$ .

But the solution must be in the form $\frac{a(b-\sqrt{c})}d$ to find the answer.

To get to this form, we rationalize the denominator by expanding the fraction by the conjugate of $6+\sqrt{3}$ , which is ${ \color{#D61F06} 6-\sqrt{3}}$ .

$x = \frac{16}{6 + \sqrt{3}} = \frac{16 ({ \color{#D61F06} 6-\sqrt{3}})} {(6+\sqrt{3})({ \color{#D61F06} 6-\sqrt{3}})} = \frac{16(6-\sqrt{3})} {6^2 - (\sqrt{3})^2} = \frac{16(6-\sqrt{3})} {33}$ .

So, we get $a=16, b=6, c=3, d=33$ which gives the answer of $16+6+3+33 = \boxed{58}$ .