2012 AIME I Problems/Problem 6
The complex numbers and satisfy and the imaginary part of is , for relatively prime positive integers and with Find
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
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 sin 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.
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