1997 AIME Problems/Problem 14

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, $|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \lfloor \frac{1997}{12} \rfloor =166$ .

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 are greater than $\displaystyle m$ , and $\displaystyle 166$ are less than $\displaystyle m$ . 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 $\displaystyle 499+83=582$ .

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

Page

Toolbox

Search

1997 AIME Problems/Problem 14

Problem

Solution

See also