Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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Simplify left side: | Simplify left side: | ||
− | <cmath>z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}( | + | <cmath>z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)</cmath> |
That is, | That is, | ||
− | <cmath>z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}( | + | <cmath>z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)</cmath> |
We have two roots for both equations, therefore the total number of solution for <math>z</math> is <math>\boxed{\textbf{(D) }4}</math> | We have two roots for both equations, therefore the total number of solution for <math>z</math> is <math>\boxed{\textbf{(D) }4}</math> |
Revision as of 21:11, 15 February 2019
Contents
[hide]Problem
How many nonzero complex numbers have the property that and when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
Solution 1
Convert and into form, giving and . Since the distance from to is , the distance from to must also be , so . Now we must find the requirements for being an equilateral triangle. From , we have and from , we see a monotonic increase of , from to , or equivalently, from to . Hence, there are 2 values that work for . But since the interval also consists of going from to , it also gives us 2 solutions. Our answer is
Here's a graph of how and move as increases- https://www.desmos.com/calculator/xtnpzoqkgs
Someone pls help with LaTeX formatting, thanks -FlatSquare , I did, -Dodgers66
Solution 2
To be equilateral triangle, we should have
Simplify left side:
That is,
We have two roots for both equations, therefore the total number of solution for is
(By Zhen Qin)
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.