HMMT-Nov 2021 guts, Problem 21

Circle $\omega$ is inscribed in rhombus $H M_{1} M_{2} T$ so that $\omega$ is tangent to $\overline{H M_{1}}$ at $A, \overline{M_{1} M_{2}}$ at $I, \overline{M_{2} T}$ at $M$, and $\overline{T H}$ at $E$. Given that the area of $H M_{1} M_{2} T$ is 1440 and the area of $E M T$ is 405 , find the area of $A I M E$.

Solution: Solution: First, from equal tangents, we know that $T E=T M$. As the sides of a rhombus are also equal, this gives from SAS similarity that $E M T \sim T H M_{2}$. Further, the ratio of their areas is $\frac{405}{1440 / 2}=\frac{9}{16}$. This means that $T E=T M=\frac{3}{4} H T$. Then, we get that $M M_{2}=M I$, so $M_{2} M I \sim M_{2} T M_{1}$, and since $M M_{2}=\frac{1}{4} M_{2} T$, we get that $\left[M_{2} M I\right]=\frac{1}{16}\left[M_{2} T M_{1}\right]=\frac{720}{16}=45$. From here, \[ [A I M E]=\left[H M_{1} M_{2} T\right]-[E H A]-\left[A M_{1} I\right]-\left[I M_{2} M\right]-[M T E]=1440-2(405+45)=540 \] $\fbox{540}$.

Answer: 540