Difference between revisions of "2012 AIME I Problems/Problem 6"

Revision as of 21:43, 13 October 2012

Problem 6

The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$ , for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$

Solutions

Solution 1

Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{142} = 1.$ So $z$ must be a $142$ nd root of unity, and thus the imaginary part of $z$ will be $\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}$ for some $m$ with $0 \le m < 142.$ But note that $71$ is prime and $m<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71$ and thus $n = \boxed{071.}$

Solution 2

Note that $w^{143}=w$ and similar for $z$ , and they are not equal to $0$ because the question implies the imaginary part is positive. Thus $w^{142}=z^{142}=1$ , so each is of the form $\sin(\frac{2 \pi k}{142})$ where $k$ is a positive integer between $0$ and $141$ inclusive. This simplifies to $sin(pi*k/71)$ , and $071$ is prime, so it is the only possible denominator, and thus is the answer.

@@ Line 8: / Line 8: @@
 ===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.
+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(\frac{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 ==
 {{AIME box|year=2012|n=I|num-b=5|num-a=7}}

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

Page

Toolbox

Search