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Difference between revisions of "2013 AIME I Problems/Problem 14"

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== Solution ==
 
== Solution ==
 
(solution)
 
(solution)
 +
<math>\begin{align*}</math>
 +
<math>P sin\theta\ + Q cos\theta\ = cos\theta\ - \frac{1}{2}\ P</math>
 +
<math>\end{align*}</math>
 +
and
 +
<math>\begin{align*}</math>
 +
<math>P cos\theta\ + Q sin\theta\ = -2(Q-1)</math>
 +
<math>\end{align*}</math>
 +
 +
Solve for P, Q we have
 +
 +
 +
<math>\frac{P}{Q} = \frac{cos\theta\ ( sin\theta + 2)}{8 + 8sin\theta + 2sin^2\theta }&#036;
 +
 +
</math>
 +
 +
Square both side, and use polynomial rational root theorem to solve <math>sin\theta</math>
 +
 +
<math>sin\theta = -\frac{17}{19} </math>
 +
 +
The answer is 036
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2013|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2013|n=I|num-b=13|num-a=15}}

Revision as of 14:59, 21 March 2013

Problem 14

14. For $\pi \le \theta < 2\pi$, 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 $\frac{P}{Q} = \frac{2\sqrt2}{7}$. Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

(solution) $\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P sin\theta\ + Q cos\theta\ = cos\theta\ - \frac{1}{2}\ P$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg) and $\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) $P cos\theta\ + Q sin\theta\ = -2(Q-1)$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Solve for P, Q we have


$\frac{P}{Q} = \frac{cos\theta\ ( sin\theta + 2)}{8 + 8sin\theta + 2sin^2\theta }&#036;$

Square both side, and use polynomial rational root theorem to solve $sin\theta$

$sin\theta = -\frac{17}{19}$

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
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