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Difference between revisions of "2001 AIME II Problems/Problem 14"
 
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== Problem ==
 
== Problem ==
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There are <math>2n</math> complex numbers that satisfy both <math>z^{28} - z^{8} - 1 = 0</math> and <math>\mid z \mid = 1</math>. These numbers have the form <math>z_{m} = \cos\theta_{m} + i\sin\theta_{m}</math>, where <math>0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360</math> and angles are measured in degrees. Find the value of <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math>.
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There are <math>2n</math> [[complex number]]s that satisfy both <math>z^{28} - z^{8} - 1 = 0</math> and <math>\mid z \mid = 1</math>. These numbers have the form <math>z_{m} = \cos\theta_{m} + i\sin\theta_{m}</math>, where <math>0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360</math> and angles are measured in degrees. Find the value of <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math>.
  
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== Solution ==
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__TOC__
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To satisfy  <math>z^{28} - z^{8} - 1 = 0</math>, <math>Im(z^{28})=Im(z^{8})</math> and <math>Re(z^{28})=Re(z^{8})+1</math>.
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==Solution==
  
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Since <math>\mid z \mid = 1</math>, <math>z</math> is on the unit circle centered at the origin in the complex plane.
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<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>
  
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Since <math>Im(z^{28})=Im(z^{8})</math>, <math>z^{28}</math> and <math>z^8</math> have the same <math>y</math> coordinate.
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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 <math>Re(z^{28})=Re(z^{8})+1</math>, <math>z^{28}</math> is <math>1</math> unit to the right of <math>z^{8}</math>.  
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*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}= \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>. We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.
  
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It is easy to see that the only possibilities are <math>(z^{28},z^{8})=(cis(60),cis(120))</math> or <math>(cis{(300)},cis{(240)})</math>.
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*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>
  
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For the first possibility:
+
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>
  
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<math>z^{28}=cis(28\theta)=cis(60) \Rightarrow 28\theta \equiv 60 \pmod{360} \Rightarrow \theta \equiv 15 \pmod{90}</math>.
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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. We only want the sum of a certain number of theta, not all of it.
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<math>z^{8}=cis(8\theta)=cis(120) \Rightarrow 8\theta \equiv 120 \pmod{360} \Rightarrow \theta \equiv 15 \pmod{45}</math>.
 
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Thus, <math>\theta \equiv 15 \pmod{90}</math>. This yields <math>\theta = 15, 105, 195, 285</math>.
 
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For the second possibility:
 
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<math>z^{28}=cis(28\theta)=cis(300) \Rightarrow 28\theta \equiv 300 \pmod{360} \Rightarrow \theta \equiv 75 \pmod{90}</math>.
 
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<math>z^{8}=cis(8\theta)=cis(240) \Rightarrow 8\theta \equiv 240 \pmod{360} \Rightarrow \theta \equiv 30 \pmod{45}</math>.
 
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Thus, <math>\theta \equiv 75 \pmod{90}</math>. This yields <math>\theta = 75, 165, 255, 345</math>.
 
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Therefore <math>(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\theta_6,\theta_7,\theta_8)=(15,75,105,165,195,255,285,345)</math> and <math>\theta_2+\theta_4+\theta_6+\theta_8+=\boxed{840}</math>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=II|num-b=13|num-a=15}}
 
{{AIME box|year=2001|n=II|num-b=13|num-a=15}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 14:03, 22 July 2018

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 $\ 285^\circ$. We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.

  • 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. We only want the sum of a certain number of theta, not all of it.

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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