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

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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>.
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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>92</math>.
  
 
== See also ==
 
== See also ==

Revision as of 01:53, 6 November 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 $92$.

See also

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