Difference between revisions of "2012 AIME I Problems/Problem 6"
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The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math> | The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math> | ||
| − | ==Solution== | + | ==Solutions== |
| + | |||
| + | ===Solution 1=== | ||
Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{142} = 1.</math> So <math>z</math> must be a <math>142</math>nd root of unity, and thus the imaginary part of <math>z</math> will be <math>\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}</math> for some <math>m</math> with <math>0 \le m < 142.</math> But note that <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math> | Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{142} = 1.</math> So <math>z</math> must be a <math>142</math>nd root of unity, and thus the imaginary part of <math>z</math> will be <math>\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}</math> for some <math>m</math> with <math>0 \le m < 142.</math> But note that <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math> | ||
| + | |||
| + | ===Solution 2=== | ||
| + | Note that <math>w^{143}=w</math> and similar for <math>z</math>, and they are not equal to <math>0</math> because the question implies the imaginary part is positive. Thus <math>w^{142}=z^{142}=1</math>, so each is of the form <math>sin(2*pi*k/142)</math> where <math>k</math> is a positive integer between <math>0</math> and <math>141</math> inclusive. This simplifies to <math>sin(pi*k/71)</math>, and <math>071</math> is prime, so it is the only possible denominator, and thus is the answer. | ||
== See also == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=5|num-a=7}} | {{AIME box|year=2012|n=I|num-b=5|num-a=7}} | ||
Revision as of 14:41, 24 August 2012
Problem 6
The complex numbers 
and 
satisfy 
and the imaginary part of is 
, for relatively prime positive integers 
and 
with 
Find 
Solutions
Solution 1
Substituting the first equation into the second, we find that 
and thus 
So must be a 
nd root of unity, and thus the imaginary part of will be 
for some with 
But note that 
is prime and 
by the conditions of the problem, so the denominator in the argument of this value will always be and thus 
Solution 2
Note that 
and similar for , and they are not equal to 
because the question implies the imaginary part is positive. Thus 
, so each is of the form 
where 
is a positive integer between and 
inclusive. This simplifies to 
, and 
is prime, so it is the only possible denominator, and thus is the answer.
See also
| 2012 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
