Difference between revisions of "2003 AIME I Problems/Problem 11"
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== Solution == | == Solution == | ||
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Note that the three expressions are symmetric with respect to interchanging <math>\sin</math> and <math>\cos</math>, and so the probability is symmetric around <math>45^\circ</math>. Thus, take <math>0 < x < 45</math> so that <math>\sin x < \cos x</math>. Then <math>\cos^2 x</math> is the largest of the three given expressions and those three lengths not forming a [[triangle]] is equivalent to a violation of the [[triangle inequality]] | Note that the three expressions are symmetric with respect to interchanging <math>\sin</math> and <math>\cos</math>, and so the probability is symmetric around <math>45^\circ</math>. Thus, take <math>0 < x < 45</math> so that <math>\sin x < \cos x</math>. Then <math>\cos^2 x</math> is the largest of the three given expressions and those three lengths not forming a [[triangle]] is equivalent to a violation of the [[triangle inequality]] | ||
| − | < | + | <cmath>\cos^2 x > \sin^2 x + \sin x \cos x</cmath> |
This is equivalent to | This is equivalent to | ||
| − | < | + | <cmath>\cos^2 x - \sin^2 x > \sin x \cos x</cmath> |
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| − | <math> | + | and, using some of our [[trigonometric identities]], we can re-write this as <math>\cos 2x > \frac 12 \sin 2x</math>. Since we've chosen <math>x \in (0, 45)</math>, <math>\cos 2x > 0</math> so |
| + | <cmath>2 > \tan 2x \Longrightarrow x < \frac 12 \arctan 2.</cmath> | ||
| − | The [[probability]] that <math>x</math> lies in this range is <math>\frac 1{45} \cdot \left(\frac 12 \arctan 2\right) = \frac{\arctan 2}{90}</math> so that <math>m = 2</math>, <math>n = 90</math> and our answer is <math>092</math>. | + | The [[probability]] that <math>x</math> lies in this range is <math>\frac 1{45} \cdot \left(\frac 12 \arctan 2\right) = \frac{\arctan 2}{90}</math> so that <math>m = 2</math>, <math>n = 90</math> and our answer is <math>\boxed{092}</math>. |
== See also == | == See also == | ||
| − | + | {{AIME box|year=2003|n=I|num-b=10|num-a=12}} | |
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[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
Revision as of 15:34, 10 June 2008
Problem
An angle 
is chosen at random from the interval 
Let 
be the probability that the numbers 
and 
are not the lengths of the sides of a triangle. Given that 
where 
is the number of degrees in 
and 
and 
are positive integers with 
find 
Solution
Note that the three expressions are symmetric with respect to interchanging 
and 
, and so the probability is symmetric around 
. Thus, take 
so that 
. Then 
is the largest of the three given expressions and those three lengths not forming a triangle is equivalent to a violation of the triangle inequality
![\[\cos^2 x > \sin^2 x + \sin x \cos x\]](http://latex.artofproblemsolving.com/e/9/5/e9517416a3c31919329a0da7536cd7ea943ff949.png)
This is equivalent to
![\[\cos^2 x - \sin^2 x > \sin x \cos x\]](http://latex.artofproblemsolving.com/d/0/0/d00b48f38038b52b18c0bed05b4995a6adef3920.png)
and, using some of our trigonometric identities, we can re-write this as 
. Since we've chosen 
, 
so
![\[2 > \tan 2x \Longrightarrow x < \frac 12 \arctan 2.\]](http://latex.artofproblemsolving.com/f/1/1/f1190e08567586a123631e30e1c61bdeb88563a8.png)
The probability that lies in this range is 
so that 
, 
and our answer is 
.
See also
| 2003 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
