Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1997 AIME Problems/Problem 14: Difference between revisions

← Older editNewer edit →
m I thought that was De Moivre's Theorem
m →Solution: I probably just overcomplicated things ...
Line 10: Line 10:


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>\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>\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>
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 2 \cdot \frac{1997}{2 \pi} = \frac{1997}{6}</math>.


== See also ==
== See also ==

Revision as of 20:37, 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:

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

We need $\cos (\frac{2\pi m}{1997})\cos (\frac{2\pi n}{1997}) + \sin (\frac{2\pi m}{1997})\sin (\frac{2\pi n}{1997}) \ge \frac{\sqrt{3}}{2}$. 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, $|m - n| \le \frac{\pi}{6} \cdot 2 \cdot \frac{1997}{2 \pi} = \frac{1997}{6}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_14&oldid=13772"