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

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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>.
  
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== Solution ==
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==Solution 1==
 
We can rewrite the given expression as
 
We can rewrite the given expression as
 
<cmath>\sqrt{24^3\sin^3 x}=24\cos x</cmath>
 
<cmath>\sqrt{24^3\sin^3 x}=24\cos x</cmath>
Line 20: Line 20:
 
<cmath>576\sin x = 24\cot^2x</cmath>
 
<cmath>576\sin x = 24\cot^2x</cmath>
 
So, <math>24\cot^2x=576\sin x=576\cdot\frac{1}{3}=\boxed{192}</math>.
 
So, <math>24\cot^2x=576\sin x=576\cdot\frac{1}{3}=\boxed{192}</math>.
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 +
==Solution 2==
 +
Like Solution 1, we can rewrite the given expression as
 +
<cmath>24\sin^3x=\cos^2x</cmath>
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Divide both sides by <math>\sin^3x</math>.
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<cmath>24 = \cot^2x\csc x</cmath>
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Square both sides.
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<cmath>576 = \cot^4x\csc^2x</cmath>
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Substitute the identity <math>\csc^2x = \cot^2x + 1</math>.
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<cmath>576 = \cot^4x(\cot^2x + 1)</cmath>
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Let <math>a = \cot^2x</math>. Then
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<cmath>576 = a^3 + a^2</cmath>.
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Since <math>\sqrt[3]{576} \approx 8</math>, we can easily see that <math>a = 8</math> is a solution. The answer is <math>24\cot^2x = 24a = 24 \cdot 8 = \boxed{192}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2011|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2011|n=I|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:00, 18 February 2019

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 1

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

Solution 2

Like Solution 1, we can rewrite the given expression as \[24\sin^3x=\cos^2x\] Divide both sides by $\sin^3x$. \[24 = \cot^2x\csc x\] Square both sides. \[576 = \cot^4x\csc^2x\] Substitute the identity $\csc^2x = \cot^2x + 1$. \[576 = \cot^4x(\cot^2x + 1)\] Let $a = \cot^2x$. Then \[576 = a^3 + a^2\]. Since $\sqrt[3]{576} \approx 8$, we can easily see that $a = 8$ is a solution. The answer is $24\cot^2x = 24a = 24 \cdot 8 = \boxed{192}$.

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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