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2003 AIME I Problems/Problem 11

Revision as of 14:24, 21 June 2024 by Bosslu99 (talk | contribs)
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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

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