2016 AIME I #9

Problem:

Triangle $ABC$ has $AB=40,AC=31,$ and $\sin A=\frac{1}{5}$. This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$. Find the maximum possible area of $AQRS$.

Solution:

Diagram - https://www.geogebra.org/calculator/ws78ucvv.

Let $\mathrm{\angle }BAD=\theta$, and $\mathrm{\angle }BAC=\alpha$. 

Note that $AD=40\cos \theta$ and $AF=31\sin (\alpha +\theta )$. The area of the rectangle is then$$(40\cdot 31)(\cos \theta \cdot \sin (\alpha +\theta ))$$$$=(40\cdot 31)\left(\frac{1}{2}\right)(\sin (\alpha +2\theta )+\sin (\alpha ))$$$$=(20\cdot 31)(\sin (\alpha +2\theta )+\frac{1}{5})$$The maximum value of $\sin (\alpha +2\theta )$ is $1$, so the maximum value of the area is $(20\cdot 31)(1+1/5)=\boxed{{\displaystyle 744}}$.