2011 AIME I Problems/Problem 9

Problem

Suppose $x$ is in the interval $[0, \pi/2]$ and $\log_{24\sin x} (24\cos x)=\frac{3}{2}$ . Find $24\cot^2 x$ .

Solution

We can rewrite the given expression as $\sqrt{24^3\sin^3 x}=24\cos x$ Square both sides and divide by $24^2$ to get $24\sin ^3 x=\cos ^2 x$ Rewrite $\cos ^2 x$ as $1-\sin ^2 x$ $24\sin ^3 x=1-\sin ^2 x$ $24\sin ^3 x+\sin ^2 x - 1=0$ Testing values using the rational root theorem gives $\sin x=\frac{1}{3}$ as a root, $\sin^{-1} \frac{1}{3}$ does fall in the first quadrant so it satisfies the interval. There are now two ways to finish this problem.

First way: Since $\sin x=\frac{1}{3}$ , we have $\sin ^2 x=\frac{1}{9}$ Using the Pythagorean Identity gives us $\cos ^2 x=\frac{8}{9}$ . Then we use the definition of $\cot ^2 x$ to compute our final answer. $24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}$ .

Second way: Multiplying our old equation $24\sin ^3 x=\cos ^2 x$ by $\dfrac{24}{\sin^2x}$ gives $576\sin x = 24\cot^2x$ So, $24\cot^2x=576\sin x=576\cdot\frac{1}{3}=\boxed{192}$ .

2011 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2011 AIME I Problems/Problem 9

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