Difference between revisions of "1997 AIME Problems/Problem 14"
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If <math>\displaystyle \theta=2\pi ik</math>, where k is any constant, the equation reduces to: | If <math>\displaystyle \theta=2\pi ik</math>, where k is any constant, the equation reduces to: | ||
| − | \begin{eqnarray*} | + | <math>\begin{eqnarray*} |
| − | e^{2\pi ik}&=&\cos(2\pi k)+i\sin(2\pi k) | + | e^{2\pi ik}&=&\cos(2\pi k)+i\sin(2\pi k)\\ |
| − | &=&1+0i | + | &=&1+0i\\ |
| − | &=&1+0 | + | &=&1+0\\ |
| − | &=&1 | + | &=&1\\ |
| − | \end{eqnarray*} | + | z^{1997}-1&=&0\\ |
| + | z^{1997}&=&1\\ | ||
| + | z^{1997}&=&e^{2\pi ik}\\ | ||
| + | z&=&e^{\frac{2\pi ik}{1997}} | ||
| + | \end{eqnarray*}<math> | ||
== See also == | == See also == | ||
| − | * [[1997 AIME Problems]] | + | * [[1997 AIME Problems]]</math> |
Revision as of 19:09, 7 March 2007
Problem
Let 
and 
be distinct, randomly chosen roots of the equation 
. Let 
be the probability that 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
The solution requires the use of Euler's formula:
If 
, where k is any constant, the equation reduces to:
$\begin{eqnarray*}
e^{2\pi ik}&=&\cos(2\pi k)+i\sin(2\pi k)\\
&=&1+0i\\
&=&1+0\\
&=&1\\
z^{1997}-1&=&0\\
z^{1997}&=&1\\
z^{1997}&=&e^{2\pi ik}\\
z&=&e^{\frac{2\pi ik}{1997}}
\end{eqnarray*}<math>
== See also ==
- [[1997 AIME Problems]]$ (Error compiling LaTeX. Unknown error_msg)
