2019 AMC 12B Problems/Problem 17
Problem
Solution
Convert z and z^3 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. (D)
-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 | |
