Difference between revisions of "2013 AIME I Problems/Problem 14"
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===Solution 2=== | ===Solution 2=== | ||
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− | Use sum to product formulas to rewrite <math>P</math> and <math>Q</math> | + | Use sum to product formulas to rewrite <math>P</math> and <math>Q</math>\ |
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<math>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </math> | <math>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </math> | ||
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Therefore, <math>P \sin \theta - Q \cos \theta = -2P</math> | Therefore, <math>P \sin \theta - Q \cos \theta = -2P</math> | ||
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− | + | Using <math>\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P</math> | |
− | Using <math>\frac{P}{Q} = \frac{2\sqrt2}{7} | ||
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− | + | Plug in to the previous equation and cancel out the "P" terms to get: <math>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</math>. | |
− | Plug in to the previous equation and cancel out the "P" terms to get: <math>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</math> | ||
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Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = -\frac{17}{19} \implies \boxed{036}</math> | Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = -\frac{17}{19} \implies \boxed{036}</math> |
Revision as of 18:20, 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.