2015 AIME I #7

Problem:

In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$.

Solution:

Let the side lengths of $FGHJ$ and $ABCD$ be $x,y$. Then, after a bunch of similarity, we see that $JC=x\sqrt{\frac{5}{4}}$ and $BJ=\left(\frac{x}{2}\right)\left(\sqrt{\frac{4}{5}}\right)$. Our first equation is $BJ+JB=BC$, so$$x\sqrt{\frac{5}{4}}+\left(\frac{x}{2}\right)\left(\sqrt{\frac{4}{5}}\right)=y.$$After climbing up the ladder of $AB$, we see that $AN=y-x\sqrt{\frac{4}{5}}-\sqrt{99}\sqrt{\frac{5}{4}}$. We know that $NM=\sqrt{99}$, and $NM=2\sqrt{\frac{5}{4}}AN$, so$$2(y-x\sqrt{\frac{4}{5}}-\sqrt{99}\sqrt{\frac{5}{4}})=\sqrt{99}.$$Solving these two equations, we get $x=7\sqrt{11}$, and $x^2=\boxed{{\displaystyle 539}}$.

Similar triangles FTW!