# Probability puzzler involving roots of unity

Problem: Let $v$ and $w$ be roots of $z^{1997} = 1$ chosen at random (uniformly and independently). What is the probability that $|v + w| \ge \sqrt{2 + \sqrt 3}$?

This problem comes from the 1997 AIME.

• Would the downvoter please explain the problem they see with this question? – Potato May 4 '13 at 22:32
• The purpose of this site is to help people, both the askers and future readers. It's not meant to be a contest. – vadim123 May 4 '13 at 22:34
• you are welcome to post whatever you like, and if people think it doesn't belong they will downvote and perhaps close it. You asked why you're getting downvotes, and I explained it to you; I"m not trying to get into an argument. – vadim123 May 4 '13 at 22:42
• @vadim123 Ok, point taken. Thank you for explaining. – Potato May 4 '13 at 22:42
• @vadim123: Whenever one asks a question, there is, on the same page, the option to post an answer simultaneously. Asking questions to which one already knows the answer is expressly permitted. – MJD Jul 6 '13 at 4:06

## 1 Answer

Because of symmetry, we can take $v=1$. Then we let $w=\exp(i \theta)$ We have $$|v+w|=|1+\exp(i \theta)|=|1+\cos \theta + i \sin \theta|=\sqrt{(1+\cos \theta)^2+\sin^2 \theta}=\sqrt {2+2\cos \theta}$$ So we need $\cos \theta \gt \frac {\sqrt 3}2$ or $-\frac \pi 6\lt \theta \lt \frac \pi 6$ This gives $166$ choices for $w$ on each side of $1$, plus $1$ itself, for a chance of $\frac{333}{1997}$