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

 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
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> \text{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>
+
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> \text{arctan}</math> <math>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 ==

Latest revision as of 12:42, 10 October 2020

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 $\text{arctan}$ $m$ 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 $\cos 2x > \frac 12 \sin 2x$. Since we've chosen $x \in (0, 45)$, $\cos 2x > 0$ so

\[2 > \tan 2x \Longrightarrow  x < \frac 12 \arctan 2.\]

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 $\boxed{092}$.

See also

2003 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS