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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+ | ==Remark== | ||
+ | In the graph below, the solutions to <math>z^{3}-8=0</math> are shown in red, and the solutions to <math>z^{3}-8z^{2}-8z+64=0</math> are shown in blue. The greatest distance between one red point and one blue point is shown in a black dashed line segment. | ||
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+ | <b>File in progress</b> | ||
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+ | ~MRENTHUSIASM | ||
==See Also== | ==See Also== |
Revision as of 10:57, 17 April 2021
Contents
Problem
In the complex plane, let be the set of solutions to and let be the set of solutions to What is the greatest distance between a point of and a point of
Solution
Realize that will create an equilateral triangle on the complex plane with the first point at and two other points with equal magnitude at .
Also, realize that can be factored through grouping: will create points at and
Plotting the points and looking at the graph will make you realize that and are the farthest apart and through Pythagorean Theorem, the answer is revealed to be ~lopkiloinm
Remark
In the graph below, the solutions to are shown in red, and the solutions to are shown in blue. The greatest distance between one red point and one blue point is shown in a black dashed line segment.
File in progress
~MRENTHUSIASM
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.