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"

 
 
(10 intermediate revisions by 7 users not shown)
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}</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 ==
 +
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
 +
 +
<cmath>\cos^2 x - \sin^2 x > \sin x \cos x</cmath>
 +
 +
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>\boxed{092}</math>.
 +
 +
== Solution 2 (Complementary Counting) ==
 +
We seek a complementary counting argument, where we look for the probability that <math>\sin^2 x</math>, <math>\cos^2 x</math> and <math>\sin x \cos x</math> form the side lengths of a triangle.
 +
 +
By the triangle inequality, we must have the following three inequalities to be true: <cmath>\sin^2 x + \cos^2 x > \sin x \cos x</cmath> <cmath>\sin^2 x + \sin x \cos x > \cos^2 x</cmath> <cmath>\cos^2 x + \sin x \cos x > \sin^2 x</cmath>
 +
The first inequality will always hold since we have <math>\sin^2 x + \cos^2 x = 1</math>, and <math>1 > \sin x \cos x</math> for all <math>x</math> (The maximum value of <math>\sin x \cos x</math> is <math>\frac{1}{2}</math> when <math>\sin x = \cos x = \frac{\sqrt{2}}{2}</math>).
 +
 +
Now, we examine the second inequality <math>\sin^2 x + \sin x \cos x > \cos^2 x</math>. If we subtract <math>\sin^2 x</math> from both sides, we have <math>\sin x \cos x > \cos^2 x - \sin^2 x</math>. Aha! This resembles our sine and cosine double angle identities. Therefore, our inequality is now <math>\sin 2x > 2\cos 2x</math>. We can divide both sides by <math>\cos 2x</math> and we have <math>\tan 2x > 2</math>.  The solutions to this occur when <math>45 \geq x > \frac{\arctan 2}{2}</math>.
 +
 +
(To understand why it must be <math>x ></math>, we can draw the unit circle, and notice as x moves from <math>\frac{\arctan 2}{2}</math> to <math>90</math>, <math>\tan x</math> approaches <math>\infty</math>. We must cap <math>x</math> at <math>45</math>, since if <math>x > 45</math>, <math>2x > 90</math>, and <math>\tan x</math> will be negative.)
 +
 +
 +
Next, we examine the third inequality, <math>\cos^2 x + \sin x \cos x > \sin^2 x</math>. Once again, we can get our double angle identities for sine and cosine by subtracting <math>\cos^2 2x</math> from both sides. We have, <math>\sin x \cos x > \sin^2 x -\cos^2 x \to \sin 2x > -2\cos 2x</math>.
 +
 +
Next, we again, divide by <math>\cos 2x</math> to produce a <math>\tan 2x</math> (we do this because one trig function is easier to deal with than 2). However, if <math>\cos 2x > 0</math>, we do not need to flip the sign since <math>\sin 2x >0</math>, and so if <math>\cos 2x >0</math>, all values for which that is true satisfy the inequality. So we only consider if <math>\cos 2x < 0</math>, and when we divide by a negative, we must flip the sign. Thus we have <math>\tan 2x < -2</math>.
 +
 +
We can take the <math>\arctan</math> of both sides, and we have <math>\frac{\arctan -2}{2}> x \geq 45</math>. Once again, to better understand this, we can draw the angle <math>x</math> for which <math>\tan 2x = -2</math>, and we notice as <math>2x</math> moves to <math>x=90</math>, <math>\tan 2x</math> approaches <math>- \infty</math>. We must cap <math>x</math> at <math>45</math> since if <math>x<45</math>, we have <math>\tan 2x > 0</math>.
 +
 +
Notice that if we draw the terminal points for <math>\frac{\arctan 2}{2}</math> and <math>\frac{\arctan -2}{2}</math>, they have the same smaller angle with the x and y axis respectively. This means the range of degree measures for which our inequalities hold is <math>90 - \frac{\arctan 2}{2} > x > \frac{\arctan 2}{2}</math> which has an area of <math>90 - \arctan 2</math>. However, we want the complement of this, which has an area of <math>90 - (90 - \arctan 2) = \arctan 2</math>. Therefore, the desired probability is <math>\frac{\arctan2}{90}</math>, and so <math>m+n=2+90=92</math>.
 +
 +
-BossLu99
 +
 +
 +
  
 
== See also ==
 
== See also ==
−
* [[2003 AIME I Problems]]
+
{{AIME box|year=2003|n=I|num-b=10|num-a=12}}
 +
 
 +
[[Category:Intermediate Trigonometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 14:24, 21 June 2024

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 $\boxed{092}$.

Solution 2 (Complementary Counting)

We seek a complementary counting argument, where we look for the probability that $\sin^2 x$, $\cos^2 x$ and $\sin x \cos x$ form the side lengths of a triangle.

By the triangle inequality, we must have the following three inequalities to be true: \[\sin^2 x + \cos^2 x > \sin x \cos x\] \[\sin^2 x + \sin x \cos x > \cos^2 x\] \[\cos^2 x + \sin x \cos x > \sin^2 x\] The first inequality will always hold since we have $\sin^2 x + \cos^2 x = 1$, and $1 > \sin x \cos x$ for all $x$ (The maximum value of $\sin x \cos x$ is $\frac{1}{2}$ when $\sin x = \cos x = \frac{\sqrt{2}}{2}$).

Now, we examine the second inequality $\sin^2 x + \sin x \cos x > \cos^2 x$. If we subtract $\sin^2 x$ from both sides, we have $\sin x \cos x > \cos^2 x - \sin^2 x$. Aha! This resembles our sine and cosine double angle identities. Therefore, our inequality is now $\sin 2x > 2\cos 2x$. We can divide both sides by $\cos 2x$ and we have $\tan 2x > 2$. The solutions to this occur when $45 \geq x > \frac{\arctan 2}{2}$.

(To understand why it must be $x >$, we can draw the unit circle, and notice as x moves from $\frac{\arctan 2}{2}$ to $90$, $\tan x$ approaches $\infty$. We must cap $x$ at $45$, since if $x > 45$, $2x > 90$, and $\tan x$ will be negative.)


Next, we examine the third inequality, $\cos^2 x + \sin x \cos x > \sin^2 x$. Once again, we can get our double angle identities for sine and cosine by subtracting $\cos^2 2x$ from both sides. We have, $\sin x \cos x > \sin^2 x -\cos^2 x \to \sin 2x > -2\cos 2x$.

Next, we again, divide by $\cos 2x$ to produce a $\tan 2x$ (we do this because one trig function is easier to deal with than 2). However, if $\cos 2x > 0$, we do not need to flip the sign since $\sin 2x >0$, and so if $\cos 2x >0$, all values for which that is true satisfy the inequality. So we only consider if $\cos 2x < 0$, and when we divide by a negative, we must flip the sign. Thus we have $\tan 2x < -2$.

We can take the $\arctan$ of both sides, and we have $\frac{\arctan -2}{2}> x \geq 45$. Once again, to better understand this, we can draw the angle $x$ for which $\tan 2x = -2$, and we notice as $2x$ moves to $x=90$, $\tan 2x$ approaches $- \infty$. We must cap $x$ at $45$ since if $x<45$, we have $\tan 2x > 0$.

Notice that if we draw the terminal points for $\frac{\arctan 2}{2}$ and $\frac{\arctan -2}{2}$, they have the same smaller angle with the x and y axis respectively. This means the range of degree measures for which our inequalities hold is $90 - \frac{\arctan 2}{2} > x > \frac{\arctan 2}{2}$ which has an area of $90 - \arctan 2$. However, we want the complement of this, which has an area of $90 - (90 - \arctan 2) = \arctan 2$. Therefore, the desired probability is $\frac{\arctan2}{90}$, and so $m+n=2+90=92$.

-BossLu99



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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_11&oldid=220895"