Difference between revisions of "2012 AIME I Problems/Problem 6"
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===Solution 2=== | ===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 | + | 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 22:42, 13 October 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 |