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

m (Solution 1)
(Better (hopefully) explanation)
Line 16: Line 16:
 
== Solution ==
 
== Solution ==
 
===Solution 1===
 
===Solution 1===
 +
 +
Noticing the <math>\sin</math> and <math>\cos</math> in both <math>P</math> and <math>Q,</math> we think of the angle addition identities: <cmath>\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b - \sin a \sin b.</cmath> With this in mind, we multiply <math>P</math> by <math>\sin \theta</math> and <math>Q</math> by <math>\cos \theta</math> to try and use some angle addition identities. Indeed, we get
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
P \sin\theta\ + Q \cos\theta\ = \cos\theta\ - \frac{1}{2}\ P\\
+
P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\
\end{align*}</cmath>
+
&= \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots \\
and
+
&= \cos \theta - \dfrac{1}{2}P
<cmath>\begin{align*}
 
P \cos\theta\ + Q \sin\theta\ = -2(Q-1)\\
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
+
after adding term-by-term. Similar term-by-term adding yields <cmath>P \cos \theta + Q \sin \theta = -2(Q - 1).</cmath>
Solving for P, Q we have
+
This is a system of equations; rearrange and rewrite to get <cmath>P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta</cmath> and <cmath>P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.</cmath> Subtract the two and rearrange to get <cmath>\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.</cmath> Then, square both sides and use Pythagorean Identity to get a quadratic in <math>\sin \theta.</math> Factor that quadratic and solve for <math>\sin \theta = -17/19, 1/3.</math> The answer format tells us it's the negative solution, and our desired answer is <math>17 + 19 = \boxed{036}.</math>
 
 
<math>\frac{P}{Q} = \frac{\cos\theta }{2 + \sin\theta } = \frac{2\sqrt2}{7}</math>
 
 
 
Squaring both sides and converting <math>\cos</math> to <math>\sin</math>, we get <math>(19\sin\theta+17)(3\sin\theta-1)=0</math>, so <math>\sin\theta = -\frac{17}{19} </math>
 
 
 
The answer is <math>\boxed{036}</math>
 
  
 
===Solution 2===
 
===Solution 2===

Revision as of 22:51, 27 November 2019

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

Noticing the $\sin$ and $\cos$ in both $P$ and $Q,$ we think of the angle addition identities: \[\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b - \sin a \sin b.\] With this in mind, we multiply $P$ by $\sin \theta$ and $Q$ by $\cos \theta$ to try and use some angle addition identities. Indeed, we get \begin{align*} P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\ &= \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots \\ &= \cos \theta - \dfrac{1}{2}P \end{align*} after adding term-by-term. Similar term-by-term adding yields \[P \cos \theta + Q \sin \theta = -2(Q - 1).\] This is a system of equations; rearrange and rewrite to get \[P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta\] and \[P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.\] Subtract the two and rearrange to get \[\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.\] Then, square both sides and use Pythagorean Identity to get a quadratic in $\sin \theta.$ Factor that quadratic and solve for $\sin \theta = -17/19, 1/3.$ The answer format tells us it's the negative solution, and our desired answer is $17 + 19 = \boxed{036}.$

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, $17+19=\boxed{036}$.

See also

2013 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png