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2001 AIME II Problems/Problem 14: Difference between revisions

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==Solution==
==Solution==


<math>Z</math> can be written in the form <math> \text{cis\,}\theta</math>. Rearranging, we find that  <math> \text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1</math>
<math>z</math> can be written in the form <math> \text{cis\,}\theta</math>. Rearranging, we find that  <math> \text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1</math>


Since the real part of <math>\text{cis\,}{28}\theta</math> is one more than the real part of <math>\text{cis\,} {8}\theta</math> and their imaginary parts are equal, it is clear that either <math>\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta =  -\frac{1}{2}+\frac {\sqrt{3}}{2}i</math>, or <math>\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i</math> and  <math>\text{cis\,} {8}\theta =  -\frac{1}{2}- \frac{\sqrt{3}}{2}i</math>
Since the real part of <math>\text{cis\,}{28}\theta</math> is one more than the real part of <math>\text{cis\,} {8}\theta</math> and their imaginary parts are equal, it is clear that either <math>\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta =  -\frac{1}{2}+\frac {\sqrt{3}}{2}i</math>, or <math>\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i</math> and  <math>\text{cis\,} {8}\theta =  -\frac{1}{2}- \frac{\sqrt{3}}{2}i</math>

Revision as of 17:08, 9 November 2015

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

$z$ can be written in the form $\text{cis\,}\theta$. Rearranging, we find that $\text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1$

Since the real part of $\text{cis\,}{28}\theta$ is one more than the real part of $\text{cis\,} {8}\theta$ and their imaginary parts are equal, it is clear that either $\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}+\frac {\sqrt{3}}{2}i$, or $\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}- \frac{\sqrt{3}}{2}i$

  • Case 1  : $\text{cis\,}{28}\theta = \frac{1}{2}+ \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$

Setting up and solving equations, $Z^{28}= \text{cis\,}{60^\circ}$ and $Z^8= \text{cis\,}{120^\circ}$, we see that the solutions common to both equations have arguments $15^\circ , 105^\circ, 195^\circ,$ and $\ 285^\circ$

  • Case 2  : $\text{cis\,}{28}\theta = \frac{1}{2} -\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac {1}{2} -\frac{\sqrt{3}}{2}i$

Again setting up equations ($Z^{28}= \text{cis\,}{300^\circ}$ and $Z^{8} = \text{cis\,}{240^\circ}$) we see that the common solutions have arguments of $75^\circ, 165^\circ, 255^\circ,$ and $345^\circ$

Listing all of these values, we find that $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\circ$ which is equal to $\boxed {840}$ degrees

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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