Rectangle $ABCD$ has $AB=9$ and $BC=27$. Point $M$ lies on $\overline{BC}$ and point $N$ lies on $\overline{CD}$ so that $\angle AMN=90^\circ$. The areas of $\triangle ABM$ and $\triangle ADN$ are equal. What is the area of $\triangle AMN$?

1. Let $BM=x$, so $MC=27-x$. Since $\angle AMN=90^\circ$, $\triangle ABM\sim\triangle MCN$, so $\dfrac{AB}{MC}=\dfrac{BM}{CN}$ and $CN=\dfrac{x(27-x)}{9}$.
2. $[ABM]=\frac12\cdot9x$ and $[ADN]=\frac12\cdot27\cdot DN$. They are equal, so $DN=\frac x3$.
3. $CN+DN=CD=9$ gives $\dfrac{x(27-x)}9+\dfrac x3=9$, i.e. $x^2-30x+81=0$, so $x=3$ ($x=27$ would put $M$ at $C$). Then $CN=8$, $DN=1$, $MC=24$.
4. $[AMN]=[ABCD]-[ABM]-[ADN]-[MCN]=243-\frac{27}2-\frac{27}2-\frac12\cdot24\cdot8=\boxed{120}$.