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>.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
  
 
== See also ==
 
== See also ==
* [[2001 AIME II Problems]]
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{{AIME box|year=2001|n=II|num-b=13|num-a=15}}

Revision as of 23:45, 19 November 2007

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

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See also

2001 AIME II (ProblemsAnswer KeyResources)
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