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

Revision as of 13:33, 31 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 1

$\begin{align*}$ (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)

Solving for P, Q we have

$\frac{P}{Q} = \frac{\cos\theta\ ( \sin\theta + 2)}{8 + 8\sin\theta + 2\sin^2\theta } = \frac{2\sqrt2}{7}$

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

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

The answer is $\boxed{036}$

Solution 2

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Use sum to product formulas to rewrite $P$ and $Q$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ...$

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Therefore, $P \sin \theta - Q \cos \theta = -2P$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

$\begin{align*}$ (Error compiling LaTeX. Unknown error_msg) Using $\frac{P}{Q} = \frac{2\sqrt2}{7}$ , $Q = \frac{7}{2\sqrt2} P$ $\end{align*}$ (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: $\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2$ $\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Then use the pythagorean identity to solve for $\sin\theta$ , $\sin\theta = \boxed{036}$

Solution 3

Note that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$

Thus, the following identities follow immediately: $ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)$ $i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)$ $i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)$

Consider, now, the sum $Q+iP$ . It follows fairly immediately that:

$Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}$ $Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}$

This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:

$Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)$ $Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}$

Comparing real and imaginary parts, we find: $\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}$

Squaring this equation and letting $\sin^2(\theta)=x$ :

$\frac{P^2}{Q^2)=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}$ (Error compiling LaTeX. Unknown error_msg)

Solving for $x$ gives the answer.

@@ Line 55: / Line 55: @@
 Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = \boxed{036}</math>
+===Solution 3===
+Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath>
+Thus, the following identities follow immediately:
+<cmath>ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)</cmath>
+<cmath>i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)</cmath>
+<cmath>i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)</cmath>
+Consider, now, the sum <math>Q+iP</math>. It follows fairly immediately that:
+<cmath>Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}</cmath>
+<cmath>Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}</cmath>
+This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:
+<cmath>Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)</cmath>
+<cmath>Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}</cmath>
+Comparing real and imaginary parts, we find:
+<cmath>\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}</cmath>
+Squaring this equation and letting <math>\sin^2(\theta)=x</math>:
+<math>\frac{P^2}{Q^2)=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}</math>
+Solving for <math>x</math> gives the answer.
 == See also ==
 {{AIME box|year=2013|n=I|num-b=13|num-a=15}}

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