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2001 AIME II Problems/Problem 14

Revision as of 09:18, 27 July 2008 by Azjps (talk | contribs) (alternative approach)

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

Solution 1

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

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

Since $\text{Im}\,(z^{28})=\text{Im}\,(z^{8})$, $z^{28}$ and $z^8$ have the same $y$ coordinate. Since $\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1$, $z^{28}$ is $1$ unit to the right of $z^{8}$. It is easy to see that the only possibilities are $(z^{28},z^{8})=(\text{cis}\,(60),\text{cis}\,(120))$ or $(\text{cis}\,{(300)},\text{cis}\,{(240)})$.

[asy] pathpen = black+linewidth(0.7); pen l = linewidth(0.6); D(unitcircle); D((-1.5,0)--(1.5,0),l,Arrows(5)); D((0,-1.5)--(0,1.5),l,Arrows(5)); D(D(expi(pi/3))--D(expi(2*pi/3)),EndArrow(3)); D(D(expi(4*pi/3)) -- D(expi(5*pi/3)),BeginArrow(3)); MP("1",(0.5,0));MP("1",(0,3^.5/2),SE);MP("\mathrm{cis}60",expi(1*pi/3),NE);MP("\mathrm{cis}120",expi(2*pi/3),NW);MP("\mathrm{cis}240",expi(4*pi/3),SW);MP("\mathrm{cis}300",expi(5*pi/3),SE); [/asy]

For the first possibility:

\begin{align*} z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(60) \Rightarrow 28\theta \equiv 60 \pmod{360} &\Rightarrow \theta \equiv 15 \pmod{90} \\ z^{8}=\text{cis}\,(8\theta)=\text{cis}\,(120) \Rightarrow 8\theta \equiv 120 \pmod{360} &\Rightarrow \theta \equiv 15 \pmod{45} \end{align*}

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

For the second possibility:

\begin{align*} z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(300) \Rightarrow 28\theta \equiv 300 \pmod{360} &\Rightarrow \theta \equiv 75 \pmod{90} \\ z^{8}=\text{cis}\,(8\theta)=\text{cis}\,(240) \Rightarrow 8\theta \equiv 240 \pmod{360} &\Rightarrow \theta \equiv 30 \pmod{45} \end{align*}

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

Solution 2

Rearrange the given equation as $z^8\left(z^{20}-1\right) = 1$; the magnitudes of both sides must be equal, so

\[\left|z^8\left(z^{20}-1\right)\right| = \left|z^{20}-1\right| = \left| 1 \right| = 1\]

Thus the distance between $z^{20} = \text{cis}\, 20\theta$ and $(1,0)$ on the coordinate plane is $1$. By the distance formula,

\[1 = \sqrt{(\cos 20\theta - 1)^2 + \sin ^2 20\theta} = \sqrt{2 - 2 \cos 20\theta} \Longrightarrow \cos 20\theta = \frac 12\]

And $20\theta = 60, 300 + 360n$, while $z^{20} - 1 = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i - 1 = \text{cis}\,(120,240)$. Thus $z^8 = \frac{1}{z^{20}-1} = \text{cis}^{-1}\, \{120,240\} = \text{cis}\,\{240,120\}$. We thus have $20\theta = 60 + 360n$ and $8\theta = 240 + 360n$ or $20\theta = 300 + 360n$ and $8\theta = 120 + 360n$. From here, follow the above solution.

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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