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Difference between revisions of "2020 AMC 12A Problems/Problem 15"

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Plotting the points and looking at the graph will make you realize that <math>1{\pm}i\sqrt{3}</math> and <math>8+0i</math> are the farthest apart and through Pythagorean Theorem, the answer is revealed to be <math>\sqrt{\sqrt{3}^{2}+(8+1)^{2}}=\sqrt{84}=\boxed{\textbf{(D) } 2\sqrt{21}.}</math> ~lopkiloinm
 
Plotting the points and looking at the graph will make you realize that <math>1{\pm}i\sqrt{3}</math> and <math>8+0i</math> are the farthest apart and through Pythagorean Theorem, the answer is revealed to be <math>\sqrt{\sqrt{3}^{2}+(8+1)^{2}}=\sqrt{84}=\boxed{\textbf{(D) } 2\sqrt{21}.}</math> ~lopkiloinm
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==See Also==
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{{AMC12 box|year=2020|ab=A|num-b=1|num-a=3}}
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{{MAA Notice}}

Revision as of 14:38, 1 February 2020

Problem

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

$\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9+\sqrt{3}$

Solution

Realize that $z^{3}-8=0$ will create an equilateral triangle on the complex plane with the first point at $2+0i$ and two other points with equal magnitude at $1{\pm}i\sqrt{3}$.

Also, realize that $z^{3}-8z^{2}-8z+64$ can be factored through grouping: $z^{3}-8z^{2}-8z+64=(z-8)(z^{2}-8).$ $(z-8)(z^{2}-8)$ will create points at $8+0i$ and $\pm2\sqrt{2}+0i.$

Plotting the points and looking at the graph will make you realize that $1{\pm}i\sqrt{3}$ and $8+0i$ are the farthest apart and through Pythagorean Theorem, the answer is revealed to be $\sqrt{\sqrt{3}^{2}+(8+1)^{2}}=\sqrt{84}=\boxed{\textbf{(D) } 2\sqrt{21}.}$ ~lopkiloinm

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
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Problem 3
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All AMC 12 Problems and Solutions

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