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Difference between revisions of "1997 AIME Problems/Problem 14"

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<math>\displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)</math>
 
<math>\displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)</math>
  
−
If <math>\displaystyle \theta=2\pi ik</math>, where k is any constant, the equation reduces to:
+
If <math>\displaystyle \theta=2\pi k</math>, where k is any constant, the equation reduces to:
  
 
<math>\displaystyle e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)</math>
 
<math>\displaystyle e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)</math>
Line 26: Line 26:
  
 
<math>\displaystyle z=e^{\frac{2\pi ik}{1997}}</math>
 
<math>\displaystyle z=e^{\frac{2\pi ik}{1997}}</math>
 +
 +
<math>\displaystyle z=\cos(\frac{2\pi k}{1997})+i\sin(\frac{2\pi k}{1997})</math>
 +
 +
Now, let <math>\displaystyle v</math> be the root corresponding to <math>\displaystyle \theta=\frac{2\pi m}{1997}</math>, and let <math>\displaystyle w</math> be the root corresponding to <math>\displaystyle \theta=\frac{2\pi n}{1997}</math>
  
 
== See also ==
 
== See also ==
 
* [[1997 AIME Problems]]
 
* [[1997 AIME Problems]]

Revision as of 19:18, 7 March 2007

Problem

Let $\displaystyle v$ and $\displaystyle w$ be distinct, randomly chosen roots of the equation $\displaystyle z^{1997}-1=0$. Let $\displaystyle \frac{m}{n}$ be the probability that $\displaystyle\sqrt{2+\sqrt{3}}\le\left|v+w\right|$, where $\displaystyle m$ and $\displaystyle n$ are relatively prime positive integers. Find $\displaystyle m+n$.

Solution

The solution requires the use of Euler's formula:

$\displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)$

If $\displaystyle \theta=2\pi k$, where k is any constant, the equation reduces to:

$\displaystyle e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)$

$\displaystyle =1+0i$

$\displaystyle =1+0$

$\displaystyle =1$

Now, substitute this into the equation:

$\displaystyle z^{1997}-1=0$

$\displaystyle z^{1997}=1$

$\displaystyle z^{1997}=e^{2\pi ik}$

$\displaystyle z=e^{\frac{2\pi ik}{1997}}$

$\displaystyle z=\cos(\frac{2\pi k}{1997})+i\sin(\frac{2\pi k}{1997})$

Now, let $\displaystyle v$ be the root corresponding to $\displaystyle \theta=\frac{2\pi m}{1997}$, and let $\displaystyle w$ be the root corresponding to $\displaystyle \theta=\frac{2\pi n}{1997}$

See also

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