Difference between revisions of "2001 AIME II Problems/Problem 14"

Revision as of 19:35, 4 July 2013

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.

Revision as of 08:18, 27 July 2008 (view source) Azjps (talk \| contribs) (alternative approach) ← Older edit		Revision as of 19:35, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Algebra Problems]]		[[Category:Intermediate Algebra Problems]]
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2001 AIME II (Problems • Answer Key • Resources)
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