Difference between revisions of "1983 AIME Problems/Problem 9"
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== Problem == | == Problem == | ||
+ | Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>. | ||
== Solution == | == Solution == | ||
+ | Let <math>y=x\sin{x}</math>. | ||
+ | |||
+ | We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</math>. | ||
+ | |||
+ | Since <math>x>0</math> and <math>\sin{x}>0</math> because <math>0< x<\pi</math>, we have <math>y>0</math>. | ||
+ | |||
+ | So we can apply [[AM-GM]]: | ||
+ | |||
+ | <math>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</math> | ||
+ | |||
+ | The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math> | ||
+ | |||
+ | Therefore, the minimum value is <math>12</math> (when <math>x\sin{x}=\frac23</math>). | ||
+ | |||
+ | ---- | ||
+ | |||
+ | * [[1983 AIME Problems/Problem 8|Previous Problem]] | ||
+ | * [[1983 AIME Problems/Problem 10|Next Problem]] | ||
+ | * [[1983 AIME Problems|Back to Exam]] | ||
== See also == | == See also == | ||
− | * [[ | + | * [[AIME Problems and Solutions]] |
+ | * [[American Invitational Mathematics Examination]] | ||
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | [[Category:Intermediate Trigonometry Problems]] |
Revision as of 00:11, 24 July 2006
Problem
Find the minimum value of for
.
Solution
Let .
We can rewrite the expression as .
Since and
because
, we have
.
So we can apply AM-GM:
The equality holds when
Therefore, the minimum value is (when
).