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

Solution 1

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

Solution 2

As above, we note that angle $A$ must be acute. Therefore, let $A$ be the origin, and suppose that $Q$ is on the positive $x$ axis and $S$ is on the positive $y$ axis. We approach this using complex numbers. Let $w=\text{cis} A$ , and let $z$ be a complex number with $|z|=1$ , $\text{Arg}(z)\ge 0^\circ$ and $\text{Arg}(zw)\le90^\circ$ . Then we represent $B$ by $40z$ and $C$ by $31zw$ . The coordinates of $Q$ and $S$ depend on the real part of $40z$ and the imaginary part of $31zw$ . Thus $[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).$ We can expand this, using the fact that $z\overline{z}=|z|^2$ , finding $[AQRS]=620\left(\frac{z^2w-\overline{z^2w}+w-\overline{w}}{2i}\right)=620(\Im(z^2w)+\Im(w)).$ Now as $w=\text{cis}A$ , we know that $\Im(w)=\frac15$ . Also, $|z^2w|=1$ , so the maximum possible imaginary part of $z^2w$ is $1$ . This is clearly achievable under our conditions on $z$ . Therefore, the maximum possible area of $AQRS$ is $620(1+\tfrac15)=\boxed{744}$ .

Solution 3 (With Calculus)

Let $\theta$ be the angle $\angle BAQ$ . The height of the rectangle then can be expressed as $h = 31 \sin (A+\theta)$ , and the length of the rectangle can be expressed as $l = 40\cos \theta$ . The area of the rectangle can then be written as a function of $\theta$ , $[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos \theta = 1240 \sin (A+\theta) \cos \theta$ . For now, we will ignore the $1240$ and focus on the function $f(\theta) = \sin (A+\theta) \cos \theta = (\sin A \cos \theta + \cos A \sin \theta)(\cos \theta) = \sin A \cos^2 \theta + \cos A \sin \theta \cos \theta = \sin A \cos^2 \theta + \frac{1}{2} \cos A \sin 2\theta$ .

Taking the derivative, $f'(\theta) = \sin A \cdot -2\cos \theta \sin \theta + \cos A \cos 2\theta = \cos A \cos 2\theta - \sin A \sin 2\theta = \cos(2\theta + A)$ . Setting this equal to $0$ , we get $\cos(2 \theta + A) = 0 \Rightarrow 2\theta +A = 90, 270 ^\circ$ . Since we know that $A+ \theta < 90$ , the $270^\circ$ solution is extraneous. Thus, we get that $\theta = \frac{90 - A}{2} = 45 - \frac{A}{2}$ .

Plugging this value into the original area equation, $a(45 - \frac{A}{2}) = 1240 \sin (45 - \frac{A}{2} + A) \cos (45 - \frac{A}{2}) = 1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2}) = 1240\cdot \frac{1}{2}\cdot(\sin((45 + \frac{A}{2}) + (45 -\frac{A}{2}))+\sin((45 +\frac{A}{2})-(45 - \frac{A}{2}))) = 620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}$ .

2016 AIME I (Problems • Answer Key • Resources)
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2016 AIME I Problems/Problem 9

Contents

Problem

Solution

Solution 1

Solution 2

Solution 3 (With Calculus)

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