Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
 +
 +
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
 +
 +
<math>\cos^2 x > \sin^2 x + \sin x \cos x</math>
 +
 +
This is equivalent to
 +
 +
<math>\cos^2 x - \sin^2 x > \sin x \cos x</math>
 +
 +
and, using some of our [[trigonometric identities]] we convert this to
 +
 +
<math>\cos 2x > \frac 12 \sin 2x</math> and since we've chosen <math>x \in (0, 45)</math> this means
 +
 +
<math>2 > \tan 2x</math>  or  <math>x < \frac 12 \arctan 2</math>.
 +
 +
 +
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>.
  
 
== See also ==
 
== See also ==
 
* [[2003 AIME I Problems]]
 
* [[2003 AIME I Problems]]

Revision as of 15:31, 11 August 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

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

This is equivalent to

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

and, using some of our trigonometric identities we convert this to

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

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


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

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_11&oldid=9414"