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 | For <math>\pi \le \theta < 2\pi</math>, let | ||
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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>. | 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 | 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 | ||
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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> | 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> | ||
− | + | ==Solution 2== | |
Use sum to product formulas to rewrite <math>P</math> and <math>Q</math> | Use sum to product formulas to rewrite <math>P</math> and <math>Q</math> | ||
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Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = -\frac{17}{19} \implies \boxed{036}</math> | Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = -\frac{17}{19} \implies \boxed{036}</math> | ||
− | + | ==Solution 3== | |
Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath> | Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath> | ||
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Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>. | Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>. | ||
− | <math>017+019=\boxed{036} | + | <math>017+019=\boxed{036} |
− | + | ==Solution 4== | |
− | A bit similar to Solution 3. We use <math>\phi = \theta+90^\circ< | + | 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)< | + | 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< | + | 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>(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< | + | 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>\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< | + | Setting </math>\sin\theta=y<math>, we obtain |
<cmath>\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}</cmath> | <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>7\sqrt{1-y^2}=2\sqrt{2}(2+y)</cmath> | ||
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<cmath>y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}</cmath>. | <cmath>y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}</cmath>. | ||
− | Since <math>y<0< | + | 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 (lots of room for sillies, I wouldn't recommend it)== | |
− | We notice <math>\sin\theta=\frac{-i}{2}(e^{i\theta}-e^{-i\theta})< | + | 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> |
− | With these, we just quickly find the sum of the infinite geometric series' in <math>P< | + | With these, we just quickly find the sum of the infinite geometric series' in </math>P<math> and </math>Q<math>. </math>P<math> has 2 parts, the </math>\cos<math> and the </math>\sin<math> parts. The </math>\cos<math> part is: </math>\frac12\cos\theta-\frac18\cos3\theta+\cdots<math>, which can be turned into: </math>\frac14(e^{i\theta}(1-\frac{e^{i2\theta}}{4}+\cdots)+e^{-i\theta}(1-\frac{e^{-i2\theta}}{4}+\cdots))<math>, which is </math>\frac{1}{4}(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}})<math>. This turns into </math>\frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}<math>. |
− | Following the same process as above, we find that the <math>\sin< | + | Following the same process as above, we find that the </math>\sin<math> part of </math>P<math> is </math>\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}<math>, the </math>\cos<math> part of </math>Q<math> is </math>\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}<math>, and finally, the </math>\sin<math> part of </math>Q<math> is </math>\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}<math>. |
− | We convert all 4 of these equations into trig, and we end up getting <math>\frac{2\sqrt{2}}{7}=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}< | + | We convert all 4 of these equations into trig, and we end up getting </math>\frac{2\sqrt{2}}{7}=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}<math>, we divide by </math>2<math> on both numerator and denominator, and we get </math>\frac{2\sqrt{2}}{7}=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}<math>. We use some trig identities and we get </math>\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}=\frac{2\sqrt2}{7}<math>. We factor the denominator into </math>(5-4\sin\theta)(2+\sin\theta)<math>. We cancel out </math>5-4\sin\theta<math> on both numerator and denominator to get </math>\frac{\cos\theta}{2+\sin\theta}=\frac{2\sqrt2}{7}<math>. We set </math>\sin\theta<math> as </math>x<math>, and we just solve a quadratic in terms of </math>x<math>, </math>\frac{1-x^2}{x^2+4x+4}=\frac{8}{49}<math>, cross multiply and simplify, and we get </math>57x^2+32x-17=0<math>. We can actually factor this to get </math>(19x+17)(3x-1)=0<math>, which yields the 2 solutions </math>x=-\frac{17}{19}, x=\frac{1}{3}<math>. Since </math>\pi\leq\theta<2\pi<math>, the latter solution is deemed invalid, and we are left with </math>\sin\theta=-\frac{17}{19}<math>. Our final answer is </math>17+19=\boxed{036}<math>. |
~ASAB | ~ASAB | ||
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omg this was definitely the hardest problem on the AIME I | omg this was definitely the hardest problem on the AIME I | ||
− | + | ==Solution 6== | |
Follow solution 3, up to the point of using the geometric series formula | 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> | <cmath>Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}</cmath> | ||
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<cmath>Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0</cmath> | <cmath>Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0</cmath> | ||
− | We can then write <math>P = 2 \sqrt{2} k< | + | 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> | <cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0</cmath> | ||
<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}\sqrt{1-\sin^{2}\theta}=\ 0</cmath> | ||
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<cmath>\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0</cmath> | <cmath>\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0</cmath> | ||
− | Since <math>\pi \le \theta < 2\pi< | + | Since </math>\pi \le \theta < 2\pi<math>, we get </math>\sin \theta < 0<math>, and thus, </math>\sin\theta = \frac{-19}{17} \implies \boxed{036}$ |
-Alexlikemath | -Alexlikemath |
Revision as of 20:13, 24 January 2021
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
The answer format tells us it's 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
.
$017+019=\boxed{036}
==Solution 4==
A bit similar to Solution 3. We use$ (Error compiling LaTeX. Unknown error_msg)\phi = \theta+90^\circP\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)
P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)$.
Similarly,$ (Error compiling LaTeX. Unknown error_msg)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)$.
Setting complex$ (Error compiling LaTeX. Unknown error_msg)z=q_1+p_1iz=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)$$ (Error compiling LaTeX. Unknown error_msg)(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}$.
The important part is the ratio of the imaginary part$ (Error compiling LaTeX. Unknown error_msg)i0.5i\sin\phi
\frac{P}{Q}=\tan\text{arg}(\Sigma)
\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}$.
Setting$ (Error compiling LaTeX. Unknown error_msg)\sin\theta=y$, we obtain <cmath>\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}</cmath> <cmath>7\sqrt{1-y^2}=2\sqrt{2}(2+y)</cmath> <cmath>49-49y^2=8y^2+32y+32</cmath> <cmath>57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}</cmath> <cmath>y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}</cmath>.
Since$ (Error compiling LaTeX. Unknown error_msg)y<0\pi<\theta<2\pi
y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}
17+19=\boxed{036}$.
==Solution 5 (lots of room for sillies, I wouldn't recommend it)==
We notice$ (Error compiling LaTeX. Unknown error_msg)\sin\theta=\frac{-i}{2}(e^{i\theta}-e^{-i\theta})\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})
P
Q
P
\cos
\sin
\cos
\frac12\cos\theta-\frac18\cos3\theta+\cdots
\frac14(e^{i\theta}(1-\frac{e^{i2\theta}}{4}+\cdots)+e^{-i\theta}(1-\frac{e^{-i2\theta}}{4}+\cdots))
\frac{1}{4}(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}})
\frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}$.
Following the same process as above, we find that the$ (Error compiling LaTeX. Unknown error_msg)\sinP
\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}
\cos
Q
\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}
\sin
Q
\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}$.
We convert all 4 of these equations into trig, and we end up getting$ (Error compiling LaTeX. Unknown error_msg)\frac{2\sqrt{2}}{7}=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}2
\frac{2\sqrt{2}}{7}=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}
\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}=\frac{2\sqrt2}{7}
(5-4\sin\theta)(2+\sin\theta)
5-4\sin\theta
\frac{\cos\theta}{2+\sin\theta}=\frac{2\sqrt2}{7}
\sin\theta
x
x
\frac{1-x^2}{x^2+4x+4}=\frac{8}{49}
57x^2+32x-17=0
(19x+17)(3x-1)=0
x=-\frac{17}{19}, x=\frac{1}{3}
\pi\leq\theta<2\pi
\sin\theta=-\frac{17}{19}
17+19=\boxed{036}$.
~ASAB
omg this was definitely the hardest problem on the AIME I
==Solution 6== Follow solution 3, up to the point of using the geometric series formula <cmath>Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}</cmath>
Moving everything to the other side, and considering only the imaginary part, we get <cmath>Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0</cmath>
We can then write$ (Error compiling LaTeX. Unknown error_msg)P = 2 \sqrt{2} kQ = 7k
k \neq 0$). Thus, we can substitute and divide out by k.
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0</cmath>
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0</cmath>
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)</cmath>
<cmath>8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta</cmath>
<cmath>\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0</cmath>
<cmath>57\sin^{2}\theta+32\sin\theta-17 = 0</cmath>
<cmath>\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0</cmath>
Since$ (Error compiling LaTeX. Unknown error_msg)\pi \le \theta < 2\pi\sin \theta < 0
\sin\theta = \frac{-19}{17} \implies \boxed{036}$
-Alexlikemath
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.