2020 AMC 12A Problem 15

Attempt Problem 15 of the 2020 AMC 12A 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 2020 AMC 12A solutions, or check the answer key.

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15.

In the complex plane, let AA be the set of solutions to z38=0z^3 - 8 = 0 and let BB be the set of solutions to z38z28z+64=0.z^3 - 8z^2 - 8z + 64 = 0. What is the greatest distance between a point of AA and a point of B?B?

232\sqrt{3}

66

99

2212\sqrt{21}

9+39 + \sqrt{3}

Answer: D
Concepts:complex numberfactoringdistance formula
Difficulty rating: 1690
Solution:

The set AA consists of the cube roots of 8:8: 2,2, 1+i3,-1 + i\sqrt3, and 1i3.-1 - i\sqrt3.

Factoring by grouping, z38z28z+64z^3 - 8z^2 - 8z + 64 =z2(z8)8(z8)= z^2(z - 8) - 8(z - 8) =(z8)(z28),= (z - 8)(z^2 - 8), so B={8,22,22},B = \{8, 2\sqrt2, -2\sqrt2\}, all real.

The greatest distance is from 1±i3-1 \pm i\sqrt3 to 8:8: (8(1))2+(3)2\sqrt{(8 - (-1))^2 + (\sqrt3)^2} =81+3= \sqrt{81 + 3} =84=221.= \sqrt{84} = 2\sqrt{21}.

Thus, D is the correct answer.

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