2013 AIME I Problem 14

Below is the professionally curated solution for Problem 14 of the 2013 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AIME I solutions, or check the answer key.

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Concepts:complex numbergeometric sequencetrigonometryquadratic

Difficulty rating: 3270

14.

For πθ<2π,\pi \le \theta \lt 2\pi, let P=12cosθ14sin2θ18cos3θ+116sin4θ+132cos5θ164sin6θ1128cos7θ+P = \frac{1}{2}\cos\theta - \frac{1}{4}\sin 2\theta - \frac{1}{8}\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32}\cos 5\theta - \frac{1}{64}\sin 6\theta - \frac{1}{128}\cos 7\theta + \ldots and Q=112sinθ14cos2θ+18sin3θ+116cos4θ132sin5θ164cos6θ+1128sin7θ+Q = 1 - \frac{1}{2}\sin\theta - \frac{1}{4}\cos 2\theta + \frac{1}{8}\sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta + \frac{1}{128}\sin 7\theta + \ldots so that PQ=227.\frac{P}{Q} = \frac{2\sqrt{2}}{7}. Then sinθ=mn\sin\theta = -\frac{m}{n} where mm and nn are relatively prime positive integers. Find m+n.m + n.

Solution:

The signs and the alternation between sines and cosines suggest powers of i:i: indeed Q+iP=1+12ieiθ+14i2e2iθ+18i3e3iθ+=11ieiθ2=22ieiθ.Q + iP = 1 + \frac{1}{2}ie^{i\theta} + \frac{1}{4}i^2e^{2i\theta} + \frac{1}{8}i^3e^{3i\theta} + \cdots = \frac{1}{1 - \frac{ie^{i\theta}}{2}} = \frac{2}{2 - ie^{i\theta}}. Since 2ieiθ=(2+sinθ)icosθ,2 - ie^{i\theta} = (2 + \sin\theta) - i\cos\theta, multiplying by the conjugate gives Q+iP=2(2+sinθ)+2icosθ5+4sinθ,Q + iP = \frac{2(2 + \sin\theta) + 2i\cos\theta}{5 + 4\sin\theta}, so PQ=cosθ2+sinθ.\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}.

Setting cosθ2+sinθ=227\frac{\cos\theta}{2 + \sin\theta} = \frac{2\sqrt{2}}{7} and squaring, 49(1sin2θ)=8(2+sinθ)2,49(1 - \sin^2\theta) = 8(2 + \sin\theta)^2, which rearranges to 57sin2θ+32sinθ17=(3sinθ1)(19sinθ+17)=0.57\sin^2\theta + 32\sin\theta - 17 = (3\sin\theta - 1)(19\sin\theta + 17) = 0.

Since πθ<2π\pi \le \theta \lt 2\pi forces sinθ0,\sin\theta \le 0, we get sinθ=1719\sin\theta = -\frac{17}{19} (and then cosθ=6219>0,\cos\theta = \frac{6\sqrt{2}}{19} \gt 0, consistent with the positive ratio). Thus m+n=17+19=36.m + n = 17 + 19 = 36.

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