Difference between revisions of "2013 AIME I Problems/Problem 14"
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== Solution == | == Solution == | ||
− | + | ===Solution 1=== | |
<math>\begin{align*}</math> | <math>\begin{align*}</math> | ||
<math>P sin\theta\ + Q cos\theta\ = cos\theta\ - \frac{1}{2}\ P</math> | <math>P sin\theta\ + Q cos\theta\ = cos\theta\ - \frac{1}{2}\ P</math> | ||
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The answer is <math>\boxed{036}</math> | The answer is <math>\boxed{036}</math> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | <math>\begin{align*}</math> | ||
+ | Use sum to product formulas to rewrite <math>P</math> and <math>Q</math> | ||
+ | <math>\end{align*}</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> | ||
+ | |||
+ | <math>\begin{align*}</math> | ||
+ | Therefore, <math>P sin\theta\ - Q cos\theta\ = -2P</math> | ||
+ | <math>\end{align*}</math> | ||
+ | |||
+ | <math>\begin{align*}</math> | ||
+ | Using <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>, <math>Q = \frac{7}{2\sqrt2} P</math> | ||
+ | <math>\end{align*}</math> | ||
+ | |||
+ | <math>\begin{align*}</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> | ||
+ | <math>\end{align*}</math> | ||
+ | |||
+ | Then use the pythagorean identity to solve for <math>sin\theta</math>, <math>sin\theta = \boxed{036}</math> | ||
== 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 13:44, 23 March 2013
Contents
[hide]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 1
$\begin{align*}$ (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)
Solving for P, Q we have
Square both side, and use polynomial rational root theorem to solve
The answer is
Solution 2
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
Use sum to product formulas to rewrite and
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
Therefore,
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
Using ,
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg)
Plug in to the previous equation and cancel out the "P" terms to get:
$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Then use the pythagorean identity to solve for ,
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 |