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 | + | 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> | |
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This is equivalent to | This is equivalent to | ||
− | < | + | <cmath>\cos^2 x - \sin^2 x > \sin x \cos x</cmath> |
− | and, using some of our [[trigonometric identities]] we | + | 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> |
− | <math> | + | 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>. |
+ | == See also == | ||
+ | {{AIME box|year=2003|n=I|num-b=10|num-a=12}} | ||
− | + | [[Category:Intermediate Trigonometry Problems]] | |
− | + | {{MAA Notice}} | |
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Revision as of 13:42, 10 October 2020
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 .
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.