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

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[[File:2020 AMC 12A Problem 15.png|center]]
[[File:2020 AMC 12A Problem 15.png|center]]
Graph in Desmos: https://www.desmos.com/calculator/uylcxkffak

Latest revision as of 10:15, 19 April 2021


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 $B?$

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


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


In the graph below, the solutions to $z^{3}-8=0$ are shown in red, and the solutions to $z^{3}-8z^{2}-8z+64=0$ are shown in blue. The greatest distance between one red point and one blue point is shown in a black dashed line segment.

2020 AMC 12A Problem 15.png

Graph in Desmos: https://www.desmos.com/calculator/uylcxkffak


See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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