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Difference between revisions of "2001 AIME II Problems/Problem 14"
m (fixed typo)
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Thus, <math>\theta \equiv 75 \pmod{90}</math>. This yields <math>\theta = 75, 165, 255, 345</math>.  
 
Thus, <math>\theta \equiv 75 \pmod{90}</math>. This yields <math>\theta = 75, 165, 255, 345</math>.  
  
−
Therefore <math>(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\theta_6,\theta_7,\theta_8)=(15,75,105,165,195,255,285,345)</math> and <math>\theta_2+\theta_4+\theta_6+\theta_8+=\boxed{840}</math>
+
Therefore <math>(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\theta_6,\theta_7,\theta_8)=(15,75,105,165,195,255,285,345)</math> and <math>\theta_2+\theta_4+\theta_6+\theta_8=\boxed{840}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=II|num-b=13|num-a=15}}
 
{{AIME box|year=2001|n=II|num-b=13|num-a=15}}

Revision as of 14:40, 26 July 2008

Problem

There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$. These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$, where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$.

Solution

To satisfy $z^{28} - z^{8} - 1 = 0$, $Im(z^{28})=Im(z^{8})$ and $Re(z^{28})=Re(z^{8})+1$.

Since $\mid z \mid = 1$, $z$ is on the unit circle centered at the origin in the complex plane.

Since $Im(z^{28})=Im(z^{8})$, $z^{28}$ and $z^8$ have the same $y$ coordinate.

Since $Re(z^{28})=Re(z^{8})+1$, $z^{28}$ is $1$ unit to the right of $z^{8}$.

It is easy to see that the only possibilities are $(z^{28},z^{8})=(cis(60),cis(120))$ or $(cis{(300)},cis{(240)})$.

For the first possibility:

$z^{28}=cis(28\theta)=cis(60) \Rightarrow 28\theta \equiv 60 \pmod{360} \Rightarrow \theta \equiv 15 \pmod{90}$.

$z^{8}=cis(8\theta)=cis(120) \Rightarrow 8\theta \equiv 120 \pmod{360} \Rightarrow \theta \equiv 15 \pmod{45}$.

Thus, $\theta \equiv 15 \pmod{90}$. This yields $\theta = 15, 105, 195, 285$.

For the second possibility:

$z^{28}=cis(28\theta)=cis(300) \Rightarrow 28\theta \equiv 300 \pmod{360} \Rightarrow \theta \equiv 75 \pmod{90}$.

$z^{8}=cis(8\theta)=cis(240) \Rightarrow 8\theta \equiv 240 \pmod{360} \Rightarrow \theta \equiv 30 \pmod{45}$.

Thus, $\theta \equiv 75 \pmod{90}$. This yields $\theta = 75, 165, 255, 345$.

Therefore $(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\theta_6,\theta_7,\theta_8)=(15,75,105,165,195,255,285,345)$ and $\theta_2+\theta_4+\theta_6+\theta_8=\boxed{840}$.

See also

2001 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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