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

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== Problem ==
 
== Problem ==
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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>.
+
Let <math>v</math> and <math>w</math> be distinct, randomly chosen [[root]]s of the equation <math>z^{1997}-1=0</math>.  Let <math>\frac{m}{n}</math> be the [[probability]] that <math>\sqrt{2+\sqrt{3}}\le\left|v+w\right|</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] [[positive]] [[integer]]s.  Find <math>m+n</math>.
  
 +
__TOC__
 
== Solution ==
 
== Solution ==
−
:<math>\displaystyle z^{1997}=1=1(\cos 0 + i \sin 0)</math>
+
=== Solution 1 ===
 +
:<math>z^{1997}=1=1(\cos 0 + i \sin 0)</math>
  
 
By [[De Moivre's Theorem]], we find that (<math>k \in \{0,1,\ldots,1996\}</math>)  
 
By [[De Moivre's Theorem]], we find that (<math>k \in \{0,1,\ldots,1996\}</math>)  
  
−
:<math>\displaystyle z=\cos\left(\frac{2\pi k}{1997}\right)+i\sin\left(\frac{2\pi k}{1997}\right)</math>
+
:<math>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:
+
Now, let <math>v</math> be the root corresponding to <math>\theta=\frac{2\pi m}{1997}</math>, and let <math>w</math> be the root corresponding to <math>\theta=\frac{2\pi n}{1997}</math>. The magnitude of <math>v+w</math> is therefore:
 
:<math>\sqrt{\left(\cos\left(\frac{2\pi m}{1997}\right) + \cos\left(\frac{2\pi n}{1997}\right)\right)^2 + \left(\sin\left(\frac{2\pi m}{1997}\right) + \sin\left(\frac{2\pi n}{1997}\right)\right)^2}</math>
 
:<math>\sqrt{\left(\cos\left(\frac{2\pi m}{1997}\right) + \cos\left(\frac{2\pi n}{1997}\right)\right)^2 + \left(\sin\left(\frac{2\pi m}{1997}\right) + \sin\left(\frac{2\pi n}{1997}\right)\right)^2}</math>
 
:<math>=\sqrt{2 + 2\cos\left(\frac{2\pi m}{1997}\right)\cos\left(\frac{2\pi n}{1997}\right) + 2\sin\left(\frac{2\pi m}{1997}\right)\sin\left(\frac{2\pi n}{1997}\right)}</math>
 
:<math>=\sqrt{2 + 2\cos\left(\frac{2\pi m}{1997}\right)\cos\left(\frac{2\pi n}{1997}\right) + 2\sin\left(\frac{2\pi m}{1997}\right)\sin\left(\frac{2\pi n}{1997}\right)}</math>
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We need <math>\cos \left(\frac{2\pi m}{1997}\right)\cos \left(\frac{2\pi n}{1997}\right) + \sin \left(\frac{2\pi m}{1997}\right)\sin \left(\frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}</math>. The [[Trigonometric identities|cosine difference identity]] simplifies that to <math>\cos\left(\frac{2\pi m}{1997} - \frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}</math>. Thus, <math>|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \lfloor \frac{1997}{12} \rfloor =166</math>.
 
We need <math>\cos \left(\frac{2\pi m}{1997}\right)\cos \left(\frac{2\pi n}{1997}\right) + \sin \left(\frac{2\pi m}{1997}\right)\sin \left(\frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}</math>. The [[Trigonometric identities|cosine difference identity]] simplifies that to <math>\cos\left(\frac{2\pi m}{1997} - \frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}</math>. Thus, <math>|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \lfloor \frac{1997}{12} \rfloor =166</math>.
  
−
Therefore, <math>\displaystyle m</math> and <math>\displaystyle n</math> cannot be more than <math>\displaystyle 166</math> away from each other.  This means that for a given value of <math>\displaystyle m</math>, there are <math>\displaystyle 332</math> values for <math>\displaystyle n</math> that satisfy the inequality; <math>\displaystyle 166</math> of them <math>\displaystyle > m</math>, and <math>\displaystyle 166</math> of them <math>\displaystyle < m</math>.  Since <math>\displaystyle m</math> and <math>\displaystyle n</math> must be distinct, <math>\displaystyle n</math> can have <math>\displaystyle 1996</math> possible values.  Therefore, the probability is <math>\displaystyle\frac{332}{1996}=\frac{83}{499}</math>.  The answer is then <math>\displaystyle 499+83=582</math>.
+
Therefore, <math>m</math> and <math>n</math> cannot be more than <math>166</math> away from each other.  This means that for a given value of <math>m</math>, there are <math>332</math> values for <math>n</math> that satisfy the inequality; <math>166</math> of them <math>> m</math>, and <math>166</math> of them <math>< m</math>.  Since <math>m</math> and <math>n</math> must be distinct, <math>n</math> can have <math>1996</math> possible values.  Therefore, the probability is <math>\frac{332}{1996}=\frac{83}{499}</math>.  The answer is then <math>499+83=582</math>.
 +
 
 +
=== Solution 2 ===
 +
The solutions of the equation <math>z^{1997} = 1</math> are the 1997th [[roots of unity]] and are equal to <math>\cos\left(\frac {2\pi k}{1997}\right) + i\sin\left(\frac {2\pi k}{1997}\right)</math> for <math>k = 0,1,\ldots,1996.</math>  They are also located at the vertices of a regular 1997-gon that is centered at the origin in the complex plane.
 +
 
 +
[[Without loss of generality]], let <math>v = 1.</math>  Then
 +
<cmath>
 +
\begin{eqnarray*} |v + w|^2 & = & |\cos\left(\frac {2\pi k}{1997}\right) + i\sin\left(\frac {2\pi k}{1997}\right) + 1|^2 \\
 +
& = & \left|\left[\cos\left(\frac {2\pi k}{1997}\right) + 1\right] + i\sin\left(\frac {2\pi k}{1997}\right)\right|^2 \\
 +
& = & \cos^2\left(\frac {2\pi k}{1997}\right) + 2\cos\left(\frac {2\pi k}{1997}\right) + 1 + \sin^2\left(\frac {2\pi k}{1997}\right) \\
 +
& = & 2 + 2\cos\left(\frac {2\pi k}{1997}\right) \end{eqnarray*}
 +
</cmath>
 +
 
 +
We want <math>|v + w|^2\ge 2 + \sqrt {3}.</math>  From what we just obtained, this is equivalent to <math>\cos\left(\frac {2\pi k}{1997}\right)\ge \frac {\sqrt {3}}2.</math>  This occurs when <math>\frac {\pi}6\ge \frac {2\pi k}{1997}\ge - \frac {\pi}6</math> which is satisfied by <math>k = 166,165,\ldots, - 165, - 166</math> (we don't include 0 because that corresponds to <math>v</math>).  So out of the 1996 possible <math>k</math>, 332 work.  Thus, <math>m/n = 332/1996 = 83/499.</math>  So our answer is <math>83 + 499 = 582.</math>
  
 
== See also ==
 
== See also ==

Revision as of 22:15, 23 November 2007

Problem

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

Solution

Solution 1

$z^{1997}=1=1(\cos 0 + i \sin 0)$

By De Moivre's Theorem, we find that ($k \in \{0,1,\ldots,1996\}$)

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

Now, let $v$ be the root corresponding to $\theta=\frac{2\pi m}{1997}$, and let $w$ be the root corresponding to $\theta=\frac{2\pi n}{1997}$. The magnitude of $v+w$ is therefore:

$\sqrt{\left(\cos\left(\frac{2\pi m}{1997}\right) + \cos\left(\frac{2\pi n}{1997}\right)\right)^2 + \left(\sin\left(\frac{2\pi m}{1997}\right) + \sin\left(\frac{2\pi n}{1997}\right)\right)^2}$
$=\sqrt{2 + 2\cos\left(\frac{2\pi m}{1997}\right)\cos\left(\frac{2\pi n}{1997}\right) + 2\sin\left(\frac{2\pi m}{1997}\right)\sin\left(\frac{2\pi n}{1997}\right)}$

We need $\cos \left(\frac{2\pi m}{1997}\right)\cos \left(\frac{2\pi n}{1997}\right) + \sin \left(\frac{2\pi m}{1997}\right)\sin \left(\frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}$. The cosine difference identity simplifies that to $\cos\left(\frac{2\pi m}{1997} - \frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}$. Thus, $|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \lfloor \frac{1997}{12} \rfloor =166$.

Therefore, $m$ and $n$ cannot be more than $166$ away from each other. This means that for a given value of $m$, there are $332$ values for $n$ that satisfy the inequality; $166$ of them $> m$, and $166$ of them $< m$. Since $m$ and $n$ must be distinct, $n$ can have $1996$ possible values. Therefore, the probability is $\frac{332}{1996}=\frac{83}{499}$. The answer is then $499+83=582$.

Solution 2

The solutions of the equation $z^{1997} = 1$ are the 1997th roots of unity and are equal to $\cos\left(\frac {2\pi k}{1997}\right) + i\sin\left(\frac {2\pi k}{1997}\right)$ for $k = 0,1,\ldots,1996.$ They are also located at the vertices of a regular 1997-gon that is centered at the origin in the complex plane.

Without loss of generality, let $v = 1.$ Then \begin{eqnarray*} |v + w|^2 & = & |\cos\left(\frac {2\pi k}{1997}\right) + i\sin\left(\frac {2\pi k}{1997}\right) + 1|^2 \\ & = & \left|\left[\cos\left(\frac {2\pi k}{1997}\right) + 1\right] + i\sin\left(\frac {2\pi k}{1997}\right)\right|^2 \\ & = & \cos^2\left(\frac {2\pi k}{1997}\right) + 2\cos\left(\frac {2\pi k}{1997}\right) + 1 + \sin^2\left(\frac {2\pi k}{1997}\right) \\ & = & 2 + 2\cos\left(\frac {2\pi k}{1997}\right) \end{eqnarray*}

We want $|v + w|^2\ge 2 + \sqrt {3}.$ From what we just obtained, this is equivalent to $\cos\left(\frac {2\pi k}{1997}\right)\ge \frac {\sqrt {3}}2.$ This occurs when $\frac {\pi}6\ge \frac {2\pi k}{1997}\ge - \frac {\pi}6$ which is satisfied by $k = 166,165,\ldots, - 165, - 166$ (we don't include 0 because that corresponds to $v$). So out of the 1996 possible $k$, 332 work. Thus, $m/n = 332/1996 = 83/499.$ So our answer is $83 + 499 = 582.$

See also

1997 AIME (Problems • Answer Key • Resources)
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
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