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

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Z can be written in the form <math> cis\theta</math>. Rearranging, we see that  <math> cis{28}\theta</math> = <math>cis{8}\theta</math> <math>+1</math>
 
Z can be written in the form <math> cis\theta</math>. Rearranging, we see that  <math> cis{28}\theta</math> = <math>cis{8}\theta</math> <math>+1</math>
  
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Since the real part of <math>cis{28}\theta</math> is one more than the real part of <math>cis {8}\theta</math> and their imaginary parts are the same, it is clear that either <math>cis{28}\theta</math> = <math>{1/2} + \sqrt{3}/{2}</math> and <math>cis {8}\theta</math> =  <math>{-1/2} + \sqrt{3}/{2}</math>, or <math>cis{28}\theta</math> = <math>{1/2} - \sqrt{3}/{2}</math> and <math>cis {8}\theta</math> =  <math>{-1/2} + \sqrt{3}/{2}</math>
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Since the real part of <math>cis{28}\theta</math> is one more than the real part of <math>cis {8}\theta</math> and their imaginary parts are equal, it is clear that either <math>cis{28}\theta</math> = <math>{1/2} + \sqrt{3}/{2}</math> and <math>cis {8}\theta</math> =  <math>{-1/2} + \sqrt{3}/{2}</math>, or <math>cis{28}\theta</math> = <math>{1/2} - \sqrt{3}/{2}</math> and         <math>cis {8}\theta</math> =  <math>{-1/2} + \sqrt{3}/{2}</math>
  
 
*Case One  : <math>cis{28}\theta</math> = <math>{1/2} + \sqrt{3}/{2}</math> and <math>cis {8}\theta</math> =  <math>{-1/2} +\sqrt{3}/{2}</math>
 
*Case One  : <math>cis{28}\theta</math> = <math>{1/2} + \sqrt{3}/{2}</math> and <math>cis {8}\theta</math> =  <math>{-1/2} +\sqrt{3}/{2}</math>

Revision as of 08:18, 22 August 2014

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 $cis\theta$. Rearranging, we see that $cis{28}\theta$ = $cis{8}\theta$ $+1$

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

  • Case One  : $cis{28}\theta$ = ${1/2} + \sqrt{3}/{2}$ and $cis {8}\theta$ = ${-1/2} +\sqrt{3}/{2}$

Setting up and solving equations, $Z^{28}= cis{60^\circ$ (Error compiling LaTeX. Unknown error_msg) and $Z^8= cis{120^\circ$ (Error compiling LaTeX. Unknown error_msg), we see that the only common solutions are $15^\circ , 105^\circ, 195^\circ,$ and $\ 285^\circ$

  • Case 2  : $cis{28}\theta$ = ${1/2} - \sqrt{3}/{2}$ and $cis {8}\theta$ = ${-1/2} -\sqrt{3}/{2}$

Again setting up equations ($Z^{28}= cis{300^\circ$ (Error compiling LaTeX. Unknown error_msg) and $Z^{8} = cis{240^\circ$ (Error compiling LaTeX. Unknown error_msg)) we see that the only common solutions are $75^\circ, 165^\circ, 255^\circ,$ and $345^\circ$

Listing all of these values, it is seen 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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