Difference between revisions of "2013 AIME I Problems/Problem 14"
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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}</math> |
==Solution 4== | ==Solution 4== | ||
− | A bit similar to Solution 3. We use < | + | 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, < | + | 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 < | + | 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 < | + | 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 < | + | 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 < | + | 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)== | ==Solution 5 (lots of room for sillies, I wouldn't recommend it)== | ||
− | We notice < | + | 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 < | + | 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 < | + | 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 < | + | 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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<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 < | + | 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 < | + | 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}</math> |
-Alexlikemath | -Alexlikemath |
Latest revision as of 19:13, 24 January 2021
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 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 .
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 (lots of room for sillies, I wouldn't recommend it)
We notice and
With these, we just quickly find the sum of the infinite geometric series' in and . has 2 parts, the and the parts. The part is: , which can be turned into: , which is . This turns into .
Following the same process as above, we find that the part of is , the part of is , and finally, the part of is .
We convert all 4 of these equations into trig, and we end up getting , we divide by on both numerator and denominator, and we get . We use some trig identities and we get . We factor the denominator into . We cancel out on both numerator and denominator to get . We set as , and we just solve a quadratic in terms of , , cross multiply and simplify, and we get . We can actually factor this to get , which yields the 2 solutions . Since , the latter solution is deemed invalid, and we are left with . Our final answer is .
~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
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
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.