Difference between revisions of "2013 AIME I Problems/Problem 14"
(→Solution) |
(→Solution) |
||
| Line 27: | Line 27: | ||
| − | <math>\frac{P}{Q} = \frac{cos\theta\ ( sin\theta + 2)}{8 + 8sin\theta + 2sin^2\theta } | + | <math>\frac{P}{Q} = \frac{cos\theta\ ( sin\theta + 2)}{8 + 8sin\theta + 2sin^2\theta }</math> |
| − | |||
| − | </math> | ||
Square both side, and use polynomial rational root theorem to solve <math>sin\theta</math> | Square both side, and use polynomial rational root theorem to solve <math>sin\theta</math> | ||
Revision as of 15:00, 21 March 2013
Problem 14
14. For 
, let
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots$ (Error compiling LaTeX. Unknown error_msg) $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
and
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots$ (Error compiling LaTeX. Unknown error_msg) $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
so that 
. Then 
where 
and 
are relatively prime positive integers. Find 
.
Solution
(solution)
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
and
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Solve for P, Q we have
Square both side, and use polynomial rational root theorem to solve 
The answer is 036
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 | ||
