Revision as of 10:55, 5 March 2016

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

@@ Line 14: / Line 14: @@
 Now as <math>w=\text{cis}A</math>, we know that <math>\Im(w)=\frac15</math>. Also, <math>|z^2w|=1</math>, so the maximum possible imaginary part of <math>z^2w</math> is <math>1</math>. This is clearly achievable under our conditions on <math>z</math>. Therefore, the maximum possible area of <math>AQRS</math> is <math>620(1+\tfrac15)=\boxed{744}</math>.
-===Solution 3===
+===Solution 3 (With Calculus)===
+Let <math>/theta</math> be the angle
 =See Also=
 {{AIME box|year=2016|n=I|num-b=8|num-a=10}}
 {{MAA Notice}}

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