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>. | ||
− | == Solution == | + | ==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 10: | Line 10: | ||
<cmath>24\sin ^3 x=1-\sin ^2 x</cmath> | <cmath>24\sin ^3 x=1-\sin ^2 x</cmath> | ||
<cmath>24\sin ^3 x+\sin ^2 x - 1=0</cmath> | <cmath>24\sin ^3 x+\sin ^2 x - 1=0</cmath> | ||
− | Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root, <math>\sin^{-1} \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. | + | Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root, <math>\sin^{-1} \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. |
+ | There are now two ways to finish this problem. | ||
+ | |||
+ | '''First way:''' Since <math>\sin x=\frac{1}{3}</math>, we have | ||
<cmath>\sin ^2 x=\frac{1}{9}</cmath> | <cmath>\sin ^2 x=\frac{1}{9}</cmath> | ||
Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>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}</math>. | Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>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}</math>. | ||
+ | |||
+ | '''Second way:''' Multiplying our old equation <math>24\sin ^3 x=\cos ^2 x</math> by <math>\dfrac{24}{\sin^2x}</math> gives | ||
+ | <cmath>576\sin x = 24\cot^2x</cmath> | ||
+ | So, <math>24\cot^2x=576\sin x=576\cdot\frac{1}{3}=\boxed{192}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Like Solution 1, we can rewrite the given expression as | ||
+ | <cmath>24\sin^3x=\cos^2x</cmath> | ||
+ | Divide both sides by <math>\sin^3x</math>. | ||
+ | <cmath>24 = \cot^2x\csc x</cmath> | ||
+ | Square both sides. | ||
+ | <cmath>576 = \cot^4x\csc^2x</cmath> | ||
+ | Substitute the identity <math>\csc^2x = \cot^2x + 1</math>. | ||
+ | <cmath>576 = \cot^4x(\cot^2x + 1)</cmath> | ||
+ | Let <math>a = \cot^2x</math>. Then | ||
+ | <cmath>576 = a^3 + a^2</cmath>. | ||
+ | Since <math>\sqrt[3]{576} \approx 8</math>, we can easily see that <math>a = 8</math> is a solution. Thus, the answer is <math>24\cot^2x = 24a = 24 \cdot 8 = \boxed{192}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/SXwcmdgoQpk | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2011|n=I|num-b=8|num-a=10}} | ||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 20:47, 10 December 2023
Problem
Suppose is in the interval
and
. Find
.
Solution 1
We can rewrite the given expression as
Square both sides and divide by
to get
Rewrite
as
Testing values using the rational root theorem gives
as a root,
does fall in the first quadrant so it satisfies the interval.
There are now two ways to finish this problem.
First way: Since , we have
Using the Pythagorean Identity gives us
. Then we use the definition of
to compute our final answer.
.
Second way: Multiplying our old equation by
gives
So,
.
Solution 2
Like Solution 1, we can rewrite the given expression as
Divide both sides by
.
Square both sides.
Substitute the identity
.
Let
. Then
.
Since
, we can easily see that
is a solution. Thus, the answer is
.
Video Solution
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.