Difference between revisions of "2012 AIME I Problems/Problem 6"
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− | 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 the imaginary part of each is of the form <math>\sin\left(\frac{2 \pi k}{142}\right)</math> where <math>k</math> is a positive integer between <math>1</math> and <math>141</math> inclusive. This simplifies to <math>\sin\left(\frac{\pi k}{71}\right)</math>, | + | 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 the imaginary part of each is of the form <math>\sin\left(\frac{2 \pi k}{142}\right)</math> where <math>k</math> is a positive integer between <math>1</math> and <math>141</math> inclusive. This simplifies to <math>\sin\left(\frac{\pi k}{71}\right)</math>. Therefore, the imaginary part of <math>z</math> is <math>\sin\left(\frac{\pi m}{71}\right)</math>, where <math>0 \le m < 71</math>. Since <math>071</math> is prime, it is the only possible denominator, so <math>\boxed{n = 71}</math>. |
== 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 11:29, 24 December 2012
Contents
[hide]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 the imaginary part of each is of the form
where
is a positive integer between
and
inclusive. This simplifies to
. Therefore, the imaginary part of
is
, where
. Since
is prime, it is the only possible denominator, so
.
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 |