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

m (fixed typo)
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== Problem ==
 
== Problem ==
−
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>.
+
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>.
  
 +
__TOC__
 
== Solution ==
 
== Solution ==
−
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>.  
+
=== Solution 1 ===
 +
To satisfy  <math>z^{28} - z^{8} - 1 = 0</math>, <math>\text{Im}\,(z^{28})=\text{Im}\,(z^{8})</math> and <math>\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1</math>.  
  
−
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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Since <math>\mid z \mid = 1</math>, <math>z</math> is on the [[unit circle]] centered at the origin in the [[complex plane]].  
  
−
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.  
+
Since <math>\text{Im}\,(z^{28})=\text{Im}\,(z^{8})</math>, <math>z^{28}</math> and <math>z^8</math> have the same <math>y</math> coordinate. Since <math>\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1</math>, <math>z^{28}</math> is <math>1</math> unit to the right of <math>z^{8}</math>. It is easy to see that the only possibilities are <math>(z^{28},z^{8})=(\text{cis}\,(60),\text{cis}\,(120))</math> or <math>(\text{cis}\,{(300)},\text{cis}\,{(240)})</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>.
+
<center><asy>
−
 
+
pathpen = black+linewidth(0.7); pen l = linewidth(0.6);
−
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>.
+
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></center>
  
 
For the first possibility:  
 
For the first possibility:  
  
−
<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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<cmath>
−
 
+
\begin{align*}
−
<math>z^{8}=cis(8\theta)=cis(120) \Rightarrow 8\theta \equiv 120 \pmod{360} \Rightarrow \theta \equiv 15 \pmod{45}</math>.
+
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*}</cmath>
  
 
Thus, <math>\theta \equiv 15 \pmod{90}</math>. This yields <math>\theta = 15, 105, 195, 285</math>.  
 
Thus, <math>\theta \equiv 15 \pmod{90}</math>. This yields <math>\theta = 15, 105, 195, 285</math>.  
Line 23: Line 29:
 
For the second possibility:  
 
For the second possibility:  
  
−
<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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<cmath> \begin{align*}
−
 
+
z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(300) \Rightarrow 28\theta \equiv 300 \pmod{360} &\Rightarrow \theta \equiv 75 \pmod{90} \\
−
<math>z^{8}=cis(8\theta)=cis(240) \Rightarrow 8\theta \equiv 240 \pmod{360} \Rightarrow \theta \equiv 30 \pmod{45}</math>.
+
z^{8}=\text{cis}\,(8\theta)=\text{cis}\,(240) \Rightarrow 8\theta \equiv 240 \pmod{360} &\Rightarrow \theta \equiv 30 \pmod{45} \end{align*}</cmath>  
  
 
Thus, <math>\theta \equiv 75 \pmod{90}</math>. This yields <math>\theta = 75, 165, 255, 345</math>.  
 
Thus, <math>\theta \equiv 75 \pmod{90}</math>. This yields <math>\theta = 75, 165, 255, 345</math>.  
  
 
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>.  
 
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>.  
 +
 +
=== Solution 2 ===
 +
Rearrange the given equation as <math>z^8\left(z^{20}-1\right) = 1</math>; the magnitudes of both sides must be equal, so
 +
 +
<cmath>\left|z^8\left(z^{20}-1\right)\right| = \left|z^{20}-1\right| = \left| 1 \right| = 1</cmath>
 +
 +
Thus the distance between <math>z^{20} = \text{cis}\, 20\theta </math> and <math>(1,0)</math> on the coordinate plane is <math>1</math>. By the distance formula,
 +
 +
<cmath>1 = \sqrt{(\cos 20\theta - 1)^2 + \sin ^2 20\theta} = \sqrt{2 - 2 \cos 20\theta} \Longrightarrow \cos 20\theta = \frac 12</cmath>
 +
 +
And <math>20\theta = 60, 300 + 360n</math>, while <math>z^{20} - 1 = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i - 1 = \text{cis}\,(120,240)</math>. Thus <math>z^8 = \frac{1}{z^{20}-1} = \text{cis}^{-1}\, \{120,240\} = \text{cis}\,\{240,120\}</math>. We thus have <math>20\theta = 60 + 360n</math> and <math>8\theta = 240 + 360n</math> or <math>20\theta = 300 + 360n</math> and <math>8\theta = 120 + 360n</math>. From here, follow the above solution.
  
 
== 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]]

Revision as of 08:18, 27 July 2008

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.

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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