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

m (→‎Solution: I screwed up)
m (→‎Solution: minor fixes)
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== Solution ==
 
== Solution ==
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:<math>\displaystyle z^{1997}=1</math>
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:<math>\displaystyle z^{1997}=1=1(\cos 0 + i \sin 0)</math>
  
−
By [[De Moivre's Theorem]], we find that  
+
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>\displaystyle z=\cos\left(\frac{2\pi k}{1997}\right)+i\sin\left(\frac{2\pi k}{1997}\right)</math>
Line 13: Line 13:
 
:<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>
  
−
We need <math>\cos (\frac{2\pi m}{1997})\cos (\frac{2\pi n}{1997}) + \sin (\frac{2\pi m}{1997})\sin (\frac{2\pi n}{1997}) \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>.
  
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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 are greater than <math>\displaystyle m</math>, and <math>\displaystyle 166</math> are less than <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>\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>.
  
 
== See also ==
 
== See also ==

Revision as of 14:22, 1 May 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

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

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

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

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:

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

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

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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