Difference between revisions of "2002 AIME II Problems/Problem 10"
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== Solution 2 == | == Solution 2 == | ||
The first case is when the two angles, <math>x</math> and <math>\frac{\pi x}{180}</math>, are coterminal. The second case is when they are reflections of the <math>y</math> axis. | The first case is when the two angles, <math>x</math> and <math>\frac{\pi x}{180}</math>, are coterminal. The second case is when they are reflections of the <math>y</math> axis. | ||
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1. <math>x+2\pi a = \frac{\pi x}{180}</math> for any integer <math>a</math> | 1. <math>x+2\pi a = \frac{\pi x}{180}</math> for any integer <math>a</math> | ||
So <math>x=\frac{360\pi a }{\pi -180}</math> | So <math>x=\frac{360\pi a }{\pi -180}</math> |
Latest revision as of 18:30, 24 June 2020
Contents
Problem
While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of for which the sine of degrees is the same as the sine of radians are and , where , , , and are positive integers. Find .
Solution 1
Note that degrees is equal to radians. Also, for , the two least positive angles such that are , and .
Clearly for positive real values of .
yields: .
yields: .
So, .
Solution 2
The first case is when the two angles, and , are coterminal. The second case is when they are reflections of the axis.
1. for any integer So
2. for any integer So
Choosing carefully such that it's the minimum gives the answer same as above.
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.