Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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Here's a graph of how <math>z</math> and <math>z^3</math> move as <math>\theta</math> increases- https://www.desmos.com/calculator/xtnpzoqkgs | Here's a graph of how <math>z</math> and <math>z^3</math> move as <math>\theta</math> increases- https://www.desmos.com/calculator/xtnpzoqkgs | ||
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==Solution 2== | ==Solution 2== | ||
Revision as of 21:44, 16 February 2019
Contents
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 ![\[r\text{cis}\theta\]](http://latex.artofproblemsolving.com/e/1/1/e11521d98dac9c0fb99d44cb3ecde77cdcd1c815.png)
form, giving ![\[z=r\text{cis}\theta\]](http://latex.artofproblemsolving.com/4/7/1/471451ad083fcd4284c57d60b710b058eb6b505c.png)
and ![\[z^3=r^3\text{cis}(3\theta)\]](http://latex.artofproblemsolving.com/b/f/9/bf90bce6b533810323f7bc856627621b812a98f5.png)
. Since the distance from 
to is 
, the distance from to must also be , so 
. Now we must find ![\[2\theta=\pm\frac{\pi}{3}\]](http://latex.artofproblemsolving.com/1/7/4/1745586dd73018ba5ff984a73b3a7d16b19fbc2e.png)
the requirements for being an equilateral triangle. From 
, we have ![\[\theta=\frac{\pi}{6}\]](http://latex.artofproblemsolving.com/d/3/c/d3c87965c9e9a15318a4a47255b3d45176c72f2a.png)
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
Solution 2
To be equilateral triangle, we should have
![\[\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)\]](http://latex.artofproblemsolving.com/f/1/4/f1449f833c74f8db94418bfdd612abb31763c75d.png)
Simplify left side:
![\[z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)\]](http://latex.artofproblemsolving.com/f/5/8/f5865871947340bf19baff949adc8e4eea388616.png)
That is,
![\[z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)\]](http://latex.artofproblemsolving.com/0/d/f/0df57f9b4125e68466a26437457622e016d3e982.png)
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. 
