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

Revision as of 16:12, 12 February 2016

Problem 14

For $\pi \le \theta < 2\pi$ , let

$\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}$

and

$\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}$

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

$\begin{align*} P \sin\theta\ + Q \cos\theta\ = \cos\theta\ - \frac{1}{2}\ P\\ \end{align*}$ and $\begin{align*} P \cos\theta\ + Q \sin\theta\ = -2(Q-1)\\ \end{align*}$

Solving for P, Q we have

$\frac{P}{Q} = \frac{\cos\theta }{2 + \sin\theta } = \frac{2\sqrt2}{7}$

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

Use sum to product formulas to rewrite $P$ and $Q$

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

Therefore, $P \sin \theta - Q \cos \theta = -2P$

Using $\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P$

Plug in to the previous equation and cancel out the "P" terms to get: $\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2$ .

Then use the pythagorean identity to solve for $\sin\theta$ , $\sin\theta = -\frac{17}{19} \implies \boxed{036}$

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 $\sin^2(\theta)=x$ :

$\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}$

Clearing denominators and solving for $x$ gives sine as $x=-\frac{17}{19}$ .

$017+019=\boxed{036}$

Solution 4

A bit similar to Solution 3. We use $\phi = \theta+90^\circ$ because the progression cycles in $P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)$ . So we could rewrite that as $P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)$ .

Similarly, $Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)$ .

Setting complex $z=q_1+p_1i$ , we get $z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)$

$(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}$ .

The important part is the ratio of the imaginary part $i$ to the real part. To cancel out the imaginary part from the denominator, we must add $0.5i\sin\phi$ 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 $\frac{P}{Q}=\tan\text{arg}(\Sigma)$ a PROPORTION of values. So denominators would cancel out.

$\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}$ .

Setting $\sin\theta=y$ , we obtain $\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}$ $7\sqrt{1-y^2}=2\sqrt{2}(2+y)$ $49-49y^2=8y^2+32y+32$ $57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}$ $y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}$ .

Since $y<0$ because $\pi<\theta<2\pi$ , $y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}$ . Adding up, $17+19=\boxed{036}$ .

@@ Line 83: / Line 83: @@
 A bit similar to Solution 3. We use <math>\phi = \theta+90^\circ</math> because the progression cycles in <math>P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)</math>. So we could rewrite that as <math>P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)</math>.
-Similarly, <math>Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, cos 3\phi\dots)</math>.
+Similarly, <math>Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)</math>.
 Setting complex <math>z=q_1+p_1i</math>, we get <math>z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)</math>

2013 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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