2001 AIME II Problems/Problem 14

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 $\theta_2+\theta_4+\theta_6+\theta_8=\boxed{840}$ .

2001 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2001 AIME II Problems/Problem 14

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