Difference between revisions of "1983 AIME Problems/Problem 9"

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Therefore, the minimum value is <math>12</math> (when <math>x\sin{x}=\frac23</math>).
 
Therefore, the minimum value is <math>12</math> (when <math>x\sin{x}=\frac23</math>).
  
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* [[1983 AIME Problems|Back to Exam]]
 
{{AIME box|year=1983|num-b=8|num-a=10}}
 
  
 
== See also ==
 
== See also ==
* [[AIME Problems and Solutions]]
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{{AIME box|year=1983|num-b=8|num-a=10}}
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]

Revision as of 20:41, 20 December 2007

Problem

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$.

Solution

Let $y=x\sin{x}$.

We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$.

Since $x>0$ and $\sin{x}>0$ because $0< x<\pi$, we have $y>0$.

So we can apply AM-GM:

$9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12$

The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23$

Therefore, the minimum value is $12$ (when $x\sin{x}=\frac23$).


See also

1983 AIME (ProblemsAnswer KeyResources)
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