Difference between revisions of "2013 AIME I Problems/Problem 14"
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− | == Problem | + | == Problem == |
− | + | For <math>\pi \le \theta < 2\pi</math>, let | |
+ | <cmath>\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*}</cmath> | ||
+ | and | ||
+ | <cmath>\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*}</cmath> | ||
+ | so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
+ | ==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*} | ||
+ | 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*}</cmath> | ||
+ | after adding term-by-term. Similar term-by-term adding yields <cmath>P \cos \theta + Q \sin \theta = -2(Q - 1).</cmath> | ||
+ | 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> Since we're given <math>\pi\leq\theta<2\pi,</math> <math>\sin\theta</math> is nonpositive. We therefore use the negative solution, and our desired answer is <math>17 + 19 = \boxed{036}.</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Use sum to product formulas to rewrite <math>P</math> and <math>Q</math> | ||
+ | |||
+ | |||
+ | <cmath>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 + ... </cmath> | ||
+ | |||
+ | Therefore, <cmath>P \sin \theta - Q \cos \theta = -2P</cmath> | ||
+ | |||
+ | Using <cmath>\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P</cmath> | ||
+ | |||
+ | Plug in to the previous equation and cancel out the "P" terms to get: <cmath>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</cmath> | ||
+ | |||
+ | Then use the pythagorean identity to solve for <math>\sin\theta</math>, <cmath>\sin\theta = -\frac{17}{19} \implies \boxed{036}</cmath> | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>. | ||
− | <math> | + | <math>017+019=\boxed{036}</math> |
− | |||
− | |||
− | and | + | ==Solution 4== |
+ | 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>. | ||
+ | |||
+ | Setting complex <math>z=q_1+p_1i</math>, we get <math>z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)</math> | ||
+ | |||
+ | <math>(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}</math>. | ||
+ | |||
+ | The important part is the ratio of the imaginary part <math>i</math> to the real part. To cancel out the imaginary part from the denominator, we must add <math>0.5i\sin\phi</math> 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 <math>\frac{P}{Q}=\tan\text{arg}(\Sigma)</math> a PROPORTION of values. So denominators would cancel out. | ||
+ | |||
+ | <math>\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}</math>. | ||
+ | |||
+ | Setting <math>\sin\theta=y</math>, we obtain | ||
+ | <cmath>\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}</cmath> | ||
+ | <cmath>7\sqrt{1-y^2}=2\sqrt{2}(2+y)</cmath> | ||
+ | <cmath>49-49y^2=8y^2+32y+32</cmath> | ||
+ | <cmath>57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}</cmath> | ||
+ | <cmath>y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}</cmath>. | ||
+ | |||
+ | Since <math>y<0</math> because <math>\pi<\theta<2\pi</math>, <math>y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}</math>. Adding up, <math>17+19=\boxed{036}</math>. | ||
+ | |||
+ | ==Solution 5 (utterly disgusting)== | ||
+ | |||
+ | We notice <math>\sin\theta=-\frac{i}{2}(e^{i\theta}-e^{-i\theta})</math> and <math>\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})</math> | ||
+ | |||
+ | We observe that both <math>P</math> and <math>Q</math> can be split into <math>2</math> parts, namely the terms which contain the <math>\cos</math> and the terms which contain the <math>\sin .</math> | ||
+ | |||
+ | The <math>\cos</math> part of <math>P</math> can be expressed as: | ||
+ | <cmath>\begin{align*}\frac12\cos\theta-\frac18\cos3\theta+\cdots&=\frac14\left(e^{i\theta}\left(1-\frac{e^{i2\theta}}{4}+\cdots\right)+e^{-i\theta}\left(1-\frac{e^{-i2\theta}}{4}+\cdots\right)\right) \\ | ||
+ | &= \frac{1}{4}\left(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}}\right)\\ | ||
+ | &= \frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}.\end{align*}</cmath> | ||
+ | |||
+ | Repeating the above process, we find that the <math>\sin</math> part of <math>P</math> is <cmath>\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}},</cmath> the <math>\cos</math> part of <math>Q</math> is <cmath>\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}},</cmath> and finally, the <math>\sin</math> part of <math>Q</math> is <cmath>\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}.</cmath> | ||
+ | |||
+ | Converting back to trigonometric form, we have <cmath>\begin{align*}\frac{2\sqrt{2}}{7}&=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}\\ | ||
+ | &=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}.\end{align*}</cmath> Using the <math>\sin</math> double identity and simplifying, we have <cmath>\frac{2\sqrt2}{7}=\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}.</cmath> Factoring the denominator, we have <cmath>10-4\sin^2{\theta}-3\sin{\theta}=(5-4\sin\theta)(2+\sin\theta).</cmath> Simplifying <cmath>\begin{align*}\frac{2\sqrt2}{7}&= \frac{\cos{\theta}(5-4\sin{\theta})}{(5-4\sin\theta)(2+\sin\theta)}\\ | ||
+ | &=\frac{\cos\theta}{2+\sin\theta}.\end{align*}</cmath> We set <math>\sin\theta</math> as <math>x</math>, and by the Pythagorean Identity, we have <math>57x^2+32x-17=0</math>. This factors into <math>(19x+17)(3x-1)=0</math>, which yields the 2 solutions <math>x=-\frac{17}{19}, x=\frac{1}{3}</math>. As <math>\pi\leq\theta<2\pi</math>, the latter root is erroneous, and we are left with <math>\sin\theta=-\frac{17}{19}</math>. Thus, our final answer is <math>17+19=\boxed{036}</math>. | ||
+ | |||
+ | ~ASAB | ||
− | + | ==Solution 6== | |
− | < | + | Follow solution 3, up to the point of using the geometric series formula |
− | + | <cmath>Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}</cmath> | |
− | + | Moving everything to the other side, and considering only the imaginary part, we get | |
+ | <cmath>Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0</cmath> | ||
− | == | + | We can then write <math>P = 2 \sqrt{2} k</math>, and <math>Q = 7k</math>, (<math>k \neq 0</math>). Thus, we can substitute and divide out by k. |
− | ( | + | <cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0</cmath> |
− | <math>\ | + | <cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0</cmath> |
− | < | + | <cmath>2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)</cmath> |
− | < | + | <cmath>8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta</cmath> |
− | + | <cmath>\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0</cmath> | |
− | < | + | <cmath>57\sin^{2}\theta+32\sin\theta-17 = 0</cmath> |
− | < | + | <cmath>\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0</cmath> |
− | < | ||
− | + | Since <math>\pi \le \theta < 2\pi</math>, we get <math>\sin \theta < 0</math>, and thus, <math>\sin\theta = \frac{-17}{19} \implies \boxed{036}</math> | |
+ | -Alexlikemath | ||
− | |||
− | + | ==Video Solution== | |
+ | https://youtu.be/036u51CF-EQ?si=SHTrTwSg3LMnE_yH | ||
− | + | ~MathProblemSolvingSkills.com | |
− | |||
== 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:18, 21 August 2024
Contents
Problem
For , let
and
so that
. Then
where
and
are relatively prime positive integers. Find
.
Solution 1
Noticing the and
in both
and
we think of the angle addition identities:
With this in mind, we multiply by
and
by
to try and use some angle addition identities. Indeed, we get
after adding term-by-term. Similar term-by-term adding yields
This is a system of equations; rearrange and rewrite to get
and
Subtract the two and rearrange to get
Then, square both sides and use Pythagorean Identity to get a quadratic in
Factor that quadratic and solve for
Since we're given
is nonpositive. We therefore use the negative solution, and our desired answer is
Solution 2
Use sum to product formulas to rewrite and
Therefore,
Using
Plug in to the previous equation and cancel out the "P" terms to get:
Then use the pythagorean identity to solve for ,
Solution 3
Note that
Thus, the following identities follow immediately:
Consider, now, the sum . It follows fairly immediately that:
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:
Comparing real and imaginary parts, we find:
Squaring this equation and letting :
Clearing denominators and solving for gives sine as
.
Solution 4
A bit similar to Solution 3. We use because the progression cycles in
. So we could rewrite that as
.
Similarly, .
Setting complex , we get
.
The important part is the ratio of the imaginary part to the real part. To cancel out the imaginary part from the denominator, we must add
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
a PROPORTION of values. So denominators would cancel out.
.
Setting , we obtain
.
Since because
,
. Adding up,
.
Solution 5 (utterly disgusting)
We notice and
We observe that both and
can be split into
parts, namely the terms which contain the
and the terms which contain the
The part of
can be expressed as:
Repeating the above process, we find that the part of
is
the
part of
is
and finally, the
part of
is
Converting back to trigonometric form, we have Using the
double identity and simplifying, we have
Factoring the denominator, we have
Simplifying
We set
as
, and by the Pythagorean Identity, we have
. This factors into
, which yields the 2 solutions
. As
, the latter root is erroneous, and we are left with
. Thus, our final answer is
.
~ASAB
Solution 6
Follow solution 3, up to the point of using the geometric series formula
Moving everything to the other side, and considering only the imaginary part, we get
We can then write , and
, (
). Thus, we can substitute and divide out by k.
Since , we get
, and thus,
-Alexlikemath
Video Solution
https://youtu.be/036u51CF-EQ?si=SHTrTwSg3LMnE_yH
~MathProblemSolvingSkills.com
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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