TITLE: Probability puzzler involving roots of unity
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: 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 $\frac{333}{1997}$