Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
Line 3: Line 3:
  
 
== Solution ==
 
== 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>.
 +
 
 +
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>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>.
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
<math>z^{8}=cis(8\theta)=cis(120) \Rightarrow 8\theta \equiv 120 \pmod{360} \Rightarrow \theta \equiv 15 \pmod{45}</math>.
 +
 
 +
Thus, <math>\theta \equiv 15 \pmod{90}</math>. This yields <math>\theta = 15, 105, 195, 285</math>.
 +
 
 +
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>.
 +
 
 +
<math>z^{8}=cis(8\theta)=cis(240) \Rightarrow 8\theta \equiv 240 \pmod{360} \Rightarrow \theta \equiv 30 \pmod{45}</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>
  
 
== 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}}

Revision as of 15:36, 26 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

To satisfy $z^{28} - z^{8} - 1 = 0$, $Im(z^{28})=Im(z^{8})$ and $Re(z^{28})=Re(z^{8})+1$.

Since $\mid z \mid = 1$, $z$ is on the unit circle centered at the origin in the complex plane.

Since $Im(z^{28})=Im(z^{8})$, $z^{28}$ and $z^8$ have the same $y$ coordinate.

Since $Re(z^{28})=Re(z^{8})+1$, $z^{28}$ is $1$ unit to the right of $z^{8}$.

It is easy to see that the only possibilities are $(z^{28},z^{8})=(cis(60),cis(120))$ or $(cis{(300)},cis{(240)})$.

For the first possibility:

$z^{28}=cis(28\theta)=cis(60) \Rightarrow 28\theta \equiv 60 \pmod{360} \Rightarrow \theta \equiv 15 \pmod{90}$.

$z^{8}=cis(8\theta)=cis(120) \Rightarrow 8\theta \equiv 120 \pmod{360} \Rightarrow \theta \equiv 15 \pmod{45}$.

Thus, $\theta \equiv 15 \pmod{90}$. This yields $\theta = 15, 105, 195, 285$.

For the second possibility:

$z^{28}=cis(28\theta)=cis(300) \Rightarrow 28\theta \equiv 300 \pmod{360} \Rightarrow \theta \equiv 75 \pmod{90}$.

$z^{8}=cis(8\theta)=cis(240) \Rightarrow 8\theta \equiv 240 \pmod{360} \Rightarrow \theta \equiv 30 \pmod{45}$.

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 $\theta_2+\theta_4+\theta_6+\theta_8+=\boxed{840}$

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_14&oldid=27201"