Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | Convert z and z^3 into <cmath>r\text{cis}\theta</cmath> form, giving <cmath>z=r\text{cis}\theta</cmath> and <cmath>z^3=r^3\text{cis}(3\theta)</cmath>. Since the distance from 0 to z is r, the distance from 0 to z^3 must also be r, so r=1. Now we must find <cmath>\text{cis}(2\theta)=60</cmath>. From 0 < theta < pi/2, we have <cmath>\theta=\frac{\pi}{2}</cmath> and from pi/2 < theta < pi, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from 180 to 0. Hence, there are 2 values that work for 0 < theta < pi. But since the interval pi < theta < 2pi is identical, because 3theta=theta at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. | + | Convert z and <math>z^3</math> into <cmath>r\text{cis}\theta</cmath> form, giving <cmath>z=r\text{cis}\theta</cmath> and <cmath>z^3=r^3\text{cis}(3\theta)</cmath>. Since the distance from 0 to z is r, the distance from 0 to z^3 must also be r, so r=1. Now we must find <cmath>\text{cis}(2\theta)=60</cmath>. From 0 < theta < pi/2, we have <cmath>\theta=\frac{\pi}{2}</cmath> and from pi/2 < theta < pi, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from 180 to 0. Hence, there are 2 values that work for 0 < theta < pi. But since the interval pi < theta < 2pi is identical, because 3theta=theta at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. <math>\boxed{D}</math> |
-FlatSquare | -FlatSquare | ||
Revision as of 14:45, 14 February 2019
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
Convert z 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 0 to z is r, the distance from 0 to z^3 must also be r, so r=1. Now we must find ![\[\text{cis}(2\theta)=60\]](http://latex.artofproblemsolving.com/6/d/9/6d9fcfef6bf4b46cf437460bae91ac17d60d8696.png)
. From 0 < theta < pi/2, we have ![\[\theta=\frac{\pi}{2}\]](http://latex.artofproblemsolving.com/5/2/8/528f0bc70b7a1eed77297a858eb75d9eee3c27bc.png)
and from pi/2 < theta < pi, we see a monotonic decrease of ![\[\text{cis}(2\theta)\]](http://latex.artofproblemsolving.com/a/6/2/a62485a17672a3a7648a58ba333180def7130c6f.png)
, from 180 to 0. Hence, there are 2 values that work for 0 < theta < pi. But since the interval pi < theta < 2pi is identical, because 3theta=theta at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. 
-FlatSquare
Someone pls help with LaTeX formatting, thanks
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 | |
