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Difference between revisions of "2011 AIME I Problems/Problem 9"

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

Revision as of 14:44, 9 August 2018

Problem

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, $24\cot^2x=576\sin x=576\cdot\frac{1}{3}=\boxed{192}$.

See also

2011 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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