Difference between revisions of "2013 AIME I Problems/Problem 14"
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Solving for P, Q we have | Solving for P, Q we have | ||
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<math>\frac{P}{Q} = \frac{\cos\theta }{2 + \sin\theta } = \frac{2\sqrt2}{7}</math> | <math>\frac{P}{Q} = \frac{\cos\theta }{2 + \sin\theta } = \frac{2\sqrt2}{7}</math> | ||
− | Square both side, and | + | Square both side, and convert <math>cos</math> to <math>sin</math>, seeing <math>(19sin\theta+17)(3sin\theta-1)=0</math>, so <math>\sin\theta = -\frac{17}{19} </math> |
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− | <math>\sin\theta = -\frac{17}{19} </math> | ||
The answer is <math>\boxed{036}</math> | The answer is <math>\boxed{036}</math> |
Revision as of 21:37, 24 January 2018
Contents
[hide]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 convert to , seeing , so
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 .
Solution 4
A bit similar to Solution 3. We use because the progression cycles in . So we could rewrite that as .
Similarly, .
Setting complex , we get
.
The important part is the ratio of the imaginary part to the real part. To cancel out the imaginary part from the denominator, we must add to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find a PROPORTION of values. So denominators would cancel out.
.
Setting , we obtain .
Since because , . Adding up, .
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.