Difference between revisions of "2003 AIME I Problems/Problem 11"
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== 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> \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 == | ||
− | |||
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]] | 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 | This is equivalent to | ||
− | <math>\cos^2 x - \sin^2 x > \sin x \cos x</math> | + | <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>. | ||
− | and, | + | 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 == | ||
− | + | {{AIME box|year=2003|n=I|num-b=10|num-a=12}} | |
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[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:24, 21 June 2024
Problem
An angle is chosen at random from the interval Let be the probability that the numbers and are not the lengths of the sides of a triangle. Given that where is the number of degrees in and and are positive integers with find
Solution
Note that the three expressions are symmetric with respect to interchanging and , and so the probability is symmetric around . Thus, take so that . Then 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
This is equivalent to
and, using some of our trigonometric identities, we can re-write this as . Since we've chosen , so
The probability that lies in this range is so that , and our answer is .
Solution 2 (Complementary Counting)
We seek a complementary counting argument, where we look for the probability that , and form the side lengths of a triangle.
By the triangle inequality, we must have the following three inequalities to be true: The first inequality will always hold since we have , and for all (The maximum value of is when ).
Now, we examine the second inequality . If we subtract from both sides, we have . Aha! This resembles our sine and cosine double angle identities. Therefore, our inequality is now . We can divide both sides by and we have . The solutions to this occur when .
(To understand why it must be , we can draw the unit circle, and notice as x moves from to , approaches . We must cap at , since if , , and will be negative.)
Next, we examine the third inequality, . Once again, we can get our double angle identities for sine and cosine by subtracting from both sides. We have, .
Next, we again, divide by to produce a (we do this because one trig function is easier to deal with than 2). However, if , we do not need to flip the sign since , and so if , all values for which that is true satisfy the inequality. So we only consider if , and when we divide by a negative, we must flip the sign. Thus we have .
We can take the of both sides, and we have . Once again, to better understand this, we can draw the angle for which , and we notice as moves to , approaches . We must cap at since if , we have .
Notice that if we draw the terminal points for and , 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 which has an area of . However, we want the complement of this, which has an area of . Therefore, the desired probability is , and so .
-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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