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Geometry Difficulty 4.7 AIME Find the answer Canada

In the diagram, $\$\triangle$
ABC $is right-angled at$ C$.
Points $D$ , $E$ , $F$
are on $AB$ , points $G$ , $H$ , $J$
are on $AC$ , point $K$ is on $EH$ , point $L$ is on $FJ$ , and point $M$ is on $BC$ so that $DKHG$ , $ELJH$ and $FMCJ$ are squares.

The area of $DKHG$ is $16$ and the area of $ELJH$ is $36$ . The area of square $FMCJ$ is

Pick one

Solution

Since squares $DKHG$ , $ELJH$ and $FMCJ$ have their bases along the same
line, then $DK$ , $EL$ and $FM$ are parallel.

Since $DK$ and $EL$ are parallel, then $\angle EDK = \angle FEL$ .

Since $\triangle EKD$ is right-angled
at $K$ and $\triangle FLE$ is right-angled at $L$ , then $\triangle EKD$ and $\triangle FLE$ are similar.

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Since the area of $DKHG$ is $16$ , then its side length is $\sqrt{16} = 4$ .

Since the area of $ELJH$ is $36$ , then its side length is $\sqrt{36} = 6$ .

Since $EH=6$ and $KH = 4$ , then $EK = 2$ .

Therefore, $\triangle EKD$ has $EK = 2$ and $DK
= 4 $; in other words,$ EK: DK =
1:2$.

Since $\triangle FLE$ is similar to
$\triangle EKD$ , then $FL:LE = 1:2$ .

Since $EL = 6$ , then $FL = 3$ . Since $LJ = 6$ and $FL
= 3 $, then$ FJ = FL + LJ =
9$.

Therefore, the area of square $FMCJ$
is $9^2$ or $81$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.