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Difference between revisions of "2003 AIME I Problems/Problem 11"

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== Problem ==
 
== Problem ==
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An angle <math> x </math> is chosen at random from the interval <math> 0^\circ < x < 90^\circ. </math> Let <math> p </math> be the probability that the numbers <math> \sin^2 x, \cos^2 x, </math> and <math> \sin x \cos x </math> are not the lengths of the sides of a triangle. Given that <math> p = d/n, </math> where <math> d </math> is the number of degrees in <math> \arctan m </math> and <math> m </math> and <math> n </math> are positive integers with <math> m + n < 1000, </math> find <math> m + n. </math>
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An [[angle]] <math> x </math> is chosen at random from the [[interval]] <math> 0^\circ < x < 90^\circ. </math> Let <math> p </math> be the probability that the numbers <math> \sin^2 x, \cos^2 x, </math> and <math> \sin x \cos x </math> are not the lengths of the sides of a triangle. Given that <math> p = d/n, </math> where <math> d </math> is the number of degrees in <math> \arctan m </math> and <math> m </math> and <math> n </math> are [[positive integer]]s with <math> m + n < 1000, </math> find <math> m + n. </math>
  
 
== Solution ==
 
== Solution ==
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<math>\cos^2 x - \sin^2 x > \sin x \cos x</math>
 
<math>\cos^2 x - \sin^2 x > \sin x \cos x</math>
  
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and, using some of our [[trigonometric identities]] we convert this to
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and, using some of our [[trigonometric identities]], we can re-write this as
  
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<math>\cos 2x > \frac 12 \sin 2x</math> and since we've chosen <math>x \in (0, 45)</math> this means
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<math>\cos 2x > \frac 12 \sin 2x</math> and since we've chosen <math>x \in (0, 45)</math> this means <math>\cos 2x > 0</math> so
  
 
<math>2 > \tan 2x</math>  or  <math>x < \frac 12 \arctan 2</math>.
 
<math>2 > \tan 2x</math>  or  <math>x < \frac 12 \arctan 2</math>.
  
  
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The [[probability]] that <math>x</math> lies in this range is <math>\frac 1{45} \cdot \frac 12 \arctan 2 = \frac{\arctan 2}{90}</math> so that <math>m = 2</math>, <math>n = 90</math> and our answer is <math>092</math>.
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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>.
  
 
== See also ==
 
== See also ==
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* [[2003 AIME I Problems/Problem 10 | Previous problem]]
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* [[2003 AIME I Problems/Problem 12 | Next problem]]
 
* [[2003 AIME I Problems]]
 
* [[2003 AIME I Problems]]
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 +
[[Category:Intermediate Trigonometry Problems]]

Revision as of 12:39, 24 October 2006

Problem

An angle $x$ is chosen at random from the interval $0^\circ < x < 90^\circ.$ Let $p$ be the probability that the numbers $\sin^2 x, \cos^2 x,$ and $\sin x \cos x$ are not the lengths of the sides of a triangle. Given that $p = d/n,$ where $d$ is the number of degrees in $\arctan m$ and $m$ and $n$ are positive integers with $m + n < 1000,$ find $m + n.$

Solution

Note that the three expressions are symmetric with respect to interchanging $\sin$ and $\cos$, and so the probability is symmetric around $45^\circ$. Thus, take $0 < x < 45$ so that $\sin x < \cos x$. Then $\cos^2 x$ 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

$\cos^2 x > \sin^2 x + \sin x \cos x$

This is equivalent to

$\cos^2 x - \sin^2 x > \sin x \cos x$

and, using some of our trigonometric identities, we can re-write this as

$\cos 2x > \frac 12 \sin 2x$ and since we've chosen $x \in (0, 45)$ this means $\cos 2x > 0$ so

$2 > \tan 2x$ or $x < \frac 12 \arctan 2$.


The probability that $x$ lies in this range is $\frac 1{45} \cdot \left(\frac 12 \arctan 2\right) = \frac{\arctan 2}{90}$ so that $m = 2$, $n = 90$ and our answer is $092$.

See also

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