Difference between revisions of "1997 AIME Problems/Problem 14"
Ninja glace (talk | contribs) (→Solution) |
Ninja glace (talk | contribs) (→Solution) |
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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: | ||
| − | <math> | + | |
| − | e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)\ | + | <math>\displaystyle e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)</math> |
| − | =1+0i\ | + | |
| − | =1+0\ | + | <math>\displaystyle =1+0i</math> |
| − | =1\ | + | |
| − | z^{1997}-1=0\ | + | <math>\displaystyle =1+0</math> |
| − | z^{1997}=1\ | + | |
| − | z^{1997}=e^{2\pi ik}\ | + | <math>\displaystyle =1</math> |
| − | z=e^{\frac{2\pi ik}{1997}}</math> | + | |
| + | Now, substitute this into the equation: | ||
| + | |||
| + | <math>\displaystyle z^{1997}-1=0</math> | ||
| + | |||
| + | <math>\displaystyle z^{1997}=1</math> | ||
| + | |||
| + | <math>\displaystyle z^{1997}=e^{2\pi ik}</math> | ||
| + | |||
| + | <math>\displaystyle z=e^{\frac{2\pi ik}{1997}}</math> | ||
== See also == | == See also == | ||
* [[1997 AIME Problems]] | * [[1997 AIME Problems]] | ||
Revision as of 19:15, 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:
Now, substitute this into the equation:
