Difference between revisions of "2012 AIME I Problems/Problem 6"
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Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{143} = z.</math> We know that <math>z \neq 0,</math> because we are given the imaginary part of <math>z,</math> so we can divide by <math>z</math> to get <math>z^{142} = 1.</math> So, <math>z</math> must be a <math>142</math>nd root of unity, and thus, by De Moivre's theorem, the imaginary part of <math>z</math> will be of the form <math>\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},</math> where <math>k \in \{1, 2, \ldots, 70\}.</math> Note that <math>71</math> is prime and <math>k<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71.</math> 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^{143} = z.</math> We know that <math>z \neq 0,</math> because we are given the imaginary part of <math>z,</math> so we can divide by <math>z</math> to get <math>z^{142} = 1.</math> So, <math>z</math> must be a <math>142</math>nd root of unity, and thus, by De Moivre's theorem, the imaginary part of <math>z</math> will be of the form <math>\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},</math> where <math>k \in \{1, 2, \ldots, 70\}.</math> Note that <math>71</math> is prime and <math>k<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71.</math> Thus, <math>n = \boxed{071}.</math> | ||
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+ | === Video Solution by Richard Rusczyk === | ||
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+ | https://artofproblemsolving.com/videos/amc/2012aimei/342 | ||
+ | |||
+ | ~ dolphin7 | ||
== See also == | == See also == |
Revision as of 12:34, 15 May 2020
Problem 6
The complex numbers and
satisfy
and the imaginary part of
is
, for relatively prime positive integers
and
with
Find
Solution
Substituting the first equation into the second, we find that and thus
We know that
because we are given the imaginary part of
so we can divide by
to get
So,
must be a
nd root of unity, and thus, by De Moivre's theorem, the imaginary part of
will be of the form
where
Note that
is prime and
by the conditions of the problem, so the denominator in the argument of this value will always be
Thus,
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/342
~ dolphin7
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Note: Even Richard Rusczyk said this was a low quality problem. It is a 10 second problem if you know roots of unity, but impossible to to if you don't know them.