Question

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 $FGHJ.$

Answer

Solution:

Let $AE = s,$ so the big square has side $2s$ and $CE = s\sqrt{5}.$ The right triangles $CDE,$ $JFC,$ $HBJ,$ $NKH,$ and $MAN$ are all similar, with legs in ratio $1 : 2.$ Let $x$ be the side of $FGHJ.$ In $\triangle HBJ$ the hypotenuse is $HJ = x,$ so $BJ = \frac{x}{\sqrt{5}}$ and $HB = \frac{2x}{\sqrt{5}};$ in $\triangle JFC$ the longer leg is $JF = x,$ so the hypotenuse is $JC = \frac{x\sqrt{5}}{2}.$ Then $2s = BC = BJ + JC = x\left(\frac{1}{\sqrt{5}} + \frac{\sqrt{5}}{2}\right) = \frac{7x}{2\sqrt{5}},$ so $x = \frac{4\sqrt{5}\,s}{7}.$

Next, $AH = 2s - HB = 2s - \frac{8s}{7} = \frac{6s}{7}.$ The identical decomposition along $\overline{AB}$ for the square $KLMN$ of side $y$ gives $\frac{6s}{7} = AH = AN + NH = y\left(\frac{1}{\sqrt{5}} + \frac{\sqrt{5}}{2}\right).$ Dividing the two equations, $\frac{x}{y} = \frac{2s}{6s/7} = \frac{7}{3}.$

The areas are therefore in ratio $\left(\frac{7}{3}\right)^2 = \frac{49}{9},$ so the area of $FGHJ$ is $99 \cdot \frac{49}{9} = 539.$

2015 AIME I Problem 7

Solution:

Problem 7 in Other Years