Difference between revisions of "2015 AIME I Problems/Problem 13"
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Let <math>p=\sin1\sin3\sin5...\sin89</math> | Let <math>p=\sin1\sin3\sin5...\sin89</math> | ||
− | <cmath>=\sqrt{\sin1\sin3\sin5...\sin177\sin179}</cmath> | + | <cmath>p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}</cmath> |
<cmath>=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}</cmath> | <cmath>=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}</cmath> |
Revision as of 10:39, 21 March 2015
Contents
Problem
With all angles measured in degrees, the product , where and are integers greater than 1. Find .
Solution 1
Let . Then from the identity we deduce that (taking absolute values and noticing ) But because is the reciprocal of and because , if we let our product be then because is positive in the first and second quadrants. Now, notice that are the roots of Hence, we can write , and so It is easy to see that and that our answer is .
Solution 2
Let
because of the identity
we want
Thus the answer is
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.