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

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==Solution==
 
==Solution==
  
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Z can be written in the form <math> cis\theta</math>. Rearranging, we get that  <math> cis{28}\theta</math> = <math>cis{8}\theta</math> <math>+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>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>\frac{1}{2}+\frac {\sqrt{3}}{2}i</math> and <math>cis {8}\theta</math> =  <math>-\frac{1}{2}+\frac {\sqrt{3}}{2}i</math>, or <math>cis{28}\theta</math> = <math>\frac{1}{2} - \frac{\sqrt{3}}{2}i</math> and         <math>cis {8}\theta</math> =  <math>-\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>
  
−
*Case One : <math>cis{28}\theta</math> = <math>\frac{1}{2}+ \frac{\sqrt{3}}{2}i</math> and <math>cis {8}\theta</math> =  <math>-\frac{1}{2}+\frac{\sqrt{3}}{2}i</math>
+
*Case 1 : <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>
−
Setting up and solving equations, <math>Z^{28}= cis{60^\circ</math> and <math>Z^8= cis{120^\circ</math>, we see that the solutions common to both equations have arguments <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math>
+
Setting up and solving equations, <math>Z^{28}= \text{cis\,}{60^\circ}</math> and <math>Z^8= \text{cis\,}{120^\circ}</math>, we see that the solutions common to both equations have arguments <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math>
  
−
*Case 2  : <math>cis{28}\theta</math> = <math>\frac{1}{2} -\frac {\sqrt{3}}{2}i</math> and <math>cis {8}\theta</math> =  <math>-\frac {1}{2} -\frac{\sqrt{3}}{2}i</math>
+
*Case 2  : <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>
  
−
Again setting up equations (<math>Z^{28}= cis{300^\circ</math> and <math>Z^{8} = cis{240^\circ</math>) we see that the common solutions have arguments of  <math>75^\circ, 165^\circ, 255^\circ, </math> and <math>345^\circ</math>
+
Again setting up equations (<math>Z^{28}= \text{cis\,}{300^\circ}</math> and <math>Z^{8} = \text{cis\,}{240^\circ}</math>) we see that the common solutions have arguments of  <math>75^\circ, 165^\circ, 255^\circ, </math> and <math>345^\circ</math>
  
−
Listing all of these values, it is seen that <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math> is equal to <math>(75 + 165 + 255 + 345) ^\circ</math> which is equal to <math>\boxed {840}</math> degrees
+
Listing all of these values, we find that <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math> is equal to <math>(75 + 165 + 255 + 345) ^\circ</math> which is equal to <math>\boxed {840}</math> degrees
  
 
== See also ==
 
== See also ==

Revision as of 17:50, 13 March 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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