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

Note that if angle $BAC$ is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where $A$ is obtuse. Therefore, angle A is acute. Let angle $CAS=n$ and angle $BAQ=m$ . Then, $\overline{AS}=31\cos(n)$ and $\overline{AQ}=40\cos(m)$ . Then the area of rectangle $AQRS$ is $1240\cos(m)\cos(n)$ . By product-to-sum, $\cos(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))$ . Since $\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}$ . The maximum possible value of $\cos(m-n)$ is 1, which occurs when $m=n$ . Thus the maximum possible value of $\cos(m)\cos(n)$ is $\frac{1}{2}(\frac{1}{5}+1)=\frac{3}{5}$ so the maximum possible area of $AQRS$ is $1240\times{\frac{3}{5}}=\fbox{744}$ . -AkashD

2016 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2016 AIME I Problems/Problem 9

Problem

Solution

See Also