WMC 2018 Senior Problem 7

Let the side length of square $ABCD$ be $a$ and $BE=DF=b$ . Since the area of $\triangle ABE$ is $\dfrac 13$ of the square, $\dfrac {ab}2 = \dfrac {a^2}3$ , $\implies b = \dfrac 23a$ .

Draw $EI || GH$ . We note that $AE || FC$ and hence $EI=GH=5$ also that $\triangle EIC$ is similar to $\triangle ABE$ and $\triangle CDF$ . Therefore,

$\begin{aligned} EI:IC:CE & = AB:BE:\color{#3D99F6}AE & \small \color{#3D99F6} \text{Since }\angle ABE = 90^\circ \\ & = 1 : \frac 23 : \color{#3D99F6}\frac {\sqrt {13}}3 & \small \color{#3D99F6} \text{As }\sqrt{1+\left(\frac 23\right)^2} = \frac {\sqrt{13}}3 \\ \implies CE & = \frac {\sqrt{13}}3 EI = \frac {5\sqrt{13}}3 & \small \color{#3D99F6} \text{Note that }CE = a-b = \frac a3 \\ \frac a3 & = \frac {5\sqrt{13}}3 \\ \implies AB & = a = \boxed{5\sqrt{13}} \end{aligned}$

Note that, since $\text{Area}(\triangle ABE) = \text{Area}(\triangle CDF)$ , $BE = FD$ and, therefore $\triangle ABE \cong\triangle CDF$ . It then follows that $\angle BEA \cong \angle DFC$ , which in turn implies that $\overline{AE}\parallel\overline{FC}$ . Therefore $AECF$ is a parallelogram and its area will be $GH\times AE$ .

Now, let $AB=s$ and let $BE=x$ . Since $\text{Area}(\triangle ABE) = \frac{1}{3}\text{Area}(ABCD)$ , we know that $\begin{aligned} &&\frac{1}{2}sx &= \frac{1}{3} s^2 \\ \Rightarrow && x &= \frac{2}{3}s & \color{#3D99F6} \text{Note: } s\neq 0 \end{aligned}$

Now, by the Pythagorean Theorem, $AE = \sqrt{s^2 + x^2} = \sqrt{\frac{13}{9}s^2} = \frac{\sqrt{13}s}{3}$ . Using the fact that $\text{Area}(\triangle ABE) = \frac{1}{3}\text{Area}(ABCD)$ : $\begin{aligned} &&5\times\frac{\sqrt{13}s}{3} &= \frac{1}{3}s^2 \\ \Rightarrow && \boxed{5\sqrt{13}} &= s \end{aligned}$