1997 AIME Problem 14

Below is the professionally curated solution for Problem 14 of the 1997 AIME, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1997 AIME solutions, or check the answer key.

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Concepts:roots of unitytrigonometric identitybasic probability

Difficulty rating: 2920

14.

Let vv and ww be distinct, randomly chosen roots of the equation z19971=0.z^{1997} - 1 = 0. Let mn\frac{m}{n} be the probability that 2+3v+w,\sqrt{2 + \sqrt{3}} \le |v + w|, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Solution:

By rotational symmetry we may fix vv and let w=ve2πik/1997w = v e^{2\pi i k/1997} with kk uniform in {1,2,,1996}.\{1, 2, \ldots, 1996\}. Then v+w=1+e2πik/1997=2cosπk1997.|v + w| = \left|1 + e^{2\pi i k/1997}\right| = 2\left|\cos\frac{\pi k}{1997}\right|. Also (2cos15)2=2+2cos30=2+3,\left(2\cos 15^\circ\right)^2 = 2 + 2\cos 30^\circ = 2 + \sqrt{3}, so the threshold is 2+3=2cosπ12.\sqrt{2 + \sqrt{3}} = 2\cos\frac{\pi}{12}.

The condition cosπk1997cosπ12\left|\cos\frac{\pi k}{1997}\right| \ge \cos\frac{\pi}{12} holds exactly when πk1997\frac{\pi k}{1997} is within π12\frac{\pi}{12} of 00 or of π,\pi, i.e. k199712=166.41k \le \frac{1997}{12} = 166.41\ldots or k11199712=1830.58.k \ge \frac{11 \cdot 1997}{12} = 1830.58\ldots. That gives 166+166=332166 + 166 = 332 favorable values of k.k.

The probability is 3321996=83499,\frac{332}{1996} = \frac{83}{499}, and 499499 is prime, so m+n=83+499=582.m + n = 83 + 499 = 582.

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