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

(Solution)
m (I thought that was De Moivre's Theorem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Let <math>\displaystyle v</math> and <math>\displaystyle w</math> be distinct, randomly chosen roots of the equation <math>\displaystyle z^{1997}-1=0</math>.  Let <math>\displaystyle \frac{m}{n}</math> be the probability that <math>\displaystyle\sqrt{2+\sqrt{3}}\le\left|v+w\right|</math>, where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers.  Find <math>\displaystyle m+n</math>.
+
Let <math>\displaystyle v</math> and <math>\displaystyle w</math> be distinct, randomly chosen [[root]]s of the equation <math>\displaystyle z^{1997}-1=0</math>.  Let <math>\displaystyle \frac{m}{n}</math> be the [[probability]] that <math>\displaystyle\sqrt{2+\sqrt{3}}\le\left|v+w\right|</math>, where <math>\displaystyle m</math> and <math>\displaystyle n</math> are [[relatively prime]] [[positive]] [[integer]]s.  Find <math>\displaystyle m+n</math>.
  
 
== Solution ==
 
== Solution ==
The solution requires the use of Euler's formula:
+
:<math>\displaystyle z^{1997}=1</math>
  
<math>\displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)</math>
+
By [[De Moivre's Theorem]], we find that
  
If <math>\displaystyle \theta=2\pi k</math>, where k is any constant, the equation reduces to:
+
:<math>\displaystyle z=\cos\left(\frac{2\pi k}{1997}\right)+i\sin\left(\frac{2\pi k}{1997}\right)</math>
  
<math>\displaystyle e^{2\pi ik}=\cos(2\pi k)+i\sin(2\pi k)</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>. The magnitude of <math>\displaystyle v+w</math> is therefore:
  
<math>\displaystyle =1+0i</math>
+
== See also ==
 
+
{{AIME box|year=1997|num-b=13|num-a=15}}
<math>\displaystyle =1+0</math>
 
 
 
<math>\displaystyle =1</math>
 
 
 
Now, substitute this into the equation:
 
  
<math>\displaystyle z^{1997}-1=0</math>
+
[[Category:Intermediate Complex Numbers Problems]]
 
 
<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>
 
 
 
<math>\displaystyle z=\cos\left(\frac{2\pi k}{1997}\right)+i\sin\left(\frac{2\pi k}{1997}\right)</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>.  The magnitude of <math>\displaystyle v+w</math> is therefore:
 
 
 
== See also ==
 
* [[1997 AIME Problems]]
 

Revision as of 19:30, 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

$\displaystyle z^{1997}=1$

By De Moivre's Theorem, we find that

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

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 $\displaystyle \theta=\frac{2\pi n}{1997}$. The magnitude of $\displaystyle v+w$ is therefore:

See also

1997 AIME (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