Difference between revisions of "2013 AIME I Problems/Problem 14"
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Revision as of 18:21, 10 March 2015
Problem 14
For , let
and
so that . Then where and are relatively prime positive integers. Find .
Solution
Solution 1
and
Solving for P, Q we have
Square both side, and use polynomial rational root theorem to solve
The answer is
Solution 2
Use sum to product formulas to rewrite and
Therefore,
Using
Plug in to the previous equation and cancel out the "P" terms to get: .
Then use the pythagorean identity to solve for ,
Solution 3
Note that
Thus, the following identities follow immediately:
Consider, now, the sum . It follows fairly immediately that:
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:
Comparing real and imaginary parts, we find:
Squaring this equation and letting :
Clearing denominators and solving for gives sine as .
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.