2001 AIME II Problem 14

Attempt Problem 14 of the 2001 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AIME II solutions, or check the answer key.

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14.

There are 2n2n complex numbers that satisfy both z28z81=0z^{28} - z^{8} - 1 = 0 and z=1.|z| = 1. These numbers have the form zm=cosθm+isinθm,z_{m} = \cos\theta_{m} + i\sin\theta_{m}, where 0θ1<θ2<<θ2n<3600 \le \theta_{1} \lt \theta_{2} \lt \ldots \lt \theta_{2n} \lt 360 and angles are measured in degrees. Find the value of θ2+θ4++θ2n.\theta_{2} + \theta_{4} + \cdots + \theta_{2n}.

Answer: 840
Concepts:complex numberDe Moivre’s Theoremmodular arithmetic
Difficulty rating: 3060
Solution:

Write cisθ=cosθ+isinθ.\operatorname{cis}\theta = \cos\theta + i\sin\theta. The equation says z8(z201)=1.z^8(z^{20} - 1) = 1. Taking absolute values and using z=1|z| = 1 gives z201=1,|z^{20} - 1| = 1, so z20z^{20} is at distance 11 from both 00 and 1:1: it is cis(±60).\operatorname{cis}(\pm 60^\circ).

If z20=cis60,z^{20} = \operatorname{cis} 60^\circ, then z201=cis120,z^{20} - 1 = \operatorname{cis} 120^\circ, so z8=cis(120)z^8 = \operatorname{cis}(-120^\circ) and z4=z20(z8)2=cis(60+240)=cis300, \begin{aligned} z^4 &= \frac{z^{20}}{(z^8)^2} \\ &= \operatorname{cis}(60^\circ + 240^\circ) = \operatorname{cis} 300^\circ, \end{aligned} which means 4θ300,4\theta \equiv 300^\circ, i.e. θ75(mod90).\theta \equiv 75^\circ \pmod{90^\circ}. Conversely every such θ\theta works, since then z20=(z4)5=cis60z^{20} = (z^4)^5 = \operatorname{cis} 60^\circ and z8=(z4)2=cis(120).z^8 = (z^4)^2 = \operatorname{cis}(-120^\circ). The case z20=cis(60)z^{20} = \operatorname{cis}(-60^\circ) similarly gives exactly θ15(mod90).\theta \equiv 15^\circ \pmod{90^\circ}.

So the 2n=82n = 8 angles in increasing order are 15,75,105,165,15, 75, 105, 165, 195,255,285,345,195, 255, 285, 345, and θ2+θ4+θ6+θ8\theta_2 + \theta_4 + \theta_6 + \theta_8 =75+165+255+345= 75 + 165 + 255 + 345 =840.= 840.

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