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2016 AIME I Problems/Problem 9

Revision as of 00:59, 29 December 2021 by Kingravi (talk | contribs) (→‎Solution 2)

Problem 9

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 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, $\cos(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))$. $\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}$. 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 $1240\times{\frac{3}{5}}=\fbox{744}$.

Solution 2

We start by drawing a diagram;

[asy] size(400); import olympiad; import geometry; pair A = (0, 20) ,B=(30,10) ,C=(15,0), Q=(30,20) ,R=(30,0), S=(0,0); draw(A--B--C--cycle); draw(A--Q); draw(Q--R); draw(R--S); draw(S--A); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$Q$", Q, E); label("$R$", R, E); label("$S$", S, W); label("$w$", (-1,10)); label("$l$", (15,21)); label("$x$", (7.5,-1)); label("$y$", (31,15)); label("$31$",(7.5,10), E); label("$40$",(15,15), N); markangle(Label("$\alpha$", Relative(0.5)), n=1, C, A, B); markangle(Label("$\beta$", Relative(0.5)), n=1, B, A, Q); markangle(Label("$\gamma$", Relative(0.5)), n=1, S, A, C); [/asy]

Solution in progress

~KingRavi

Solution 3

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 4 (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, $a(45 - \frac{A}{2}) = 1240 \sin (45 - \frac{A}{2} + A) \cos (45 - \frac{A}{2}) = 1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2})$. Using a product-to-sum formula, we get that: \[1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2}) =\] \[1240\cdot \frac{1}{2}\cdot(\sin((45 + \frac{A}{2}) + (45 -\frac{A}{2}))+\sin((45 +\frac{A}{2})-(45 - \frac{A}{2})))=\] \[620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}\].

Solution 5

Let $\alpha$ be the angle $\angle CAS$ and $\beta$ be the angle $\angle BAQ$. Then \[\alpha + \beta + \angle A = 90^\circ \Rightarrow \alpha + \beta = 90^\circ - \angle A\] \[\cos(\alpha + \beta) = \cos(90^\circ - \angle A)\] \[\cos(\alpha + \beta) = \sin(\angle A) = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sqrt{(1-\cos^2\alpha)(1-\cos^2\beta)} = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sqrt{1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta} = \frac{1}{5}\] However, by AM-GM: \[\cos^2\alpha+\cos^2\beta \ge 2\cos\alpha\beta\] Therefore, \[1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta \le 1-2\cos\alpha\beta+\cos^2\alpha\cos^2\beta = (1-\cos\alpha\cos\beta)^2\] \[\sqrt{1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta} \le 1-\cos\alpha\cos\beta\] So, \[\frac{1}{5} \ge \cos\alpha\cos\beta - (1-\cos\alpha\cos\beta) = 2\cos\alpha\cos\beta-1\] \[\frac{3}{5} \ge \cos\alpha\cos\beta\]. However, the area of the rectangle is just $AS \cdot AQ = 31\cos\alpha \cdot 40\cos\beta \le 31 \cdot 40 \cdot \frac{3}{5} = \boxed{744}$.


Note on Problem Validity

It has been noted that this answer won't actually work. Let angle $QAB = m$ and angle $CAS = n$ as in Solution 1. Since we know (through that solution) that $m = n$, we can call them each $\theta$. The height of the rectangle is $AS = 31\cos\theta$, and the distance $BQ = 40\sin\theta$. We know that, if the triangle is to be inscribed in a rectangle, $AS \geq BQ$.

\[AS \geq BQ\]

\[31\cos\theta \geq 40\sin\theta\]

\[\frac{31}{40} \geq \tan\theta\]

However, $\tan\theta = \tan(\frac{90-A}{2}) = \frac{\sin(90-A)}{\cos(90-A)+1} = \frac{\cos A}{\sin A + 1} = \frac{\frac{2\sqrt6}{5}}{\frac{6}{5}} = \frac{\sqrt6}{3} > \frac{31}{40}$, so the triangle does not actually fit in the rectangle: specifically, B is above R and thus in the line containing segment QR but not on the actual segment or in the rectangle.

[asy] size(200); pair A,B,C,Q,R,S; real r = (pi/2 - asin(1/5))/2; A = (0,0); B = 40*dir(r*180/pi); C = 31*dir(90-r*180/pi); draw(A--B--C--cycle); Q = (40*cos(r),0); R = (40*cos(r),31*cos(r)); S = (0, 31*cos(r)); draw(A--Q--R--S--cycle); label("$A$",A,SW); label("$B$",B,NE); label("$C$",C,N); label("$Q$",Q,SE); label("$R$",R,E); label("$S$",S,NW); [/asy]

The actual answer is a radical near $728$ (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer $744$ despite the invalid problem statement.

See Also

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

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