Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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Convert <math>z</math> 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 <math>0</math> to <math>z</math> is <math>r</math>, the distance from <math>0</math> to <math>z^3</math> must also be <math>r</math>, so <math>r=1</math>. Now we must find <cmath>\text{cis}(2\theta)=60</cmath>. From <math>0 < \theta < \pi/2</math>, we have <cmath>\theta=\frac{\pi}{6}</cmath> and from <math>\pi/2 < \theta < \pi</math>, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from <math>180</math> to <math>0</math>. Hence, there are 2 values that work for <math>0 < \theta < \pi</math>. But since the interval <math>\pi < \theta < 2\pi</math> is identical, because <math>3\theta=\theta</math> at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. <math>\boxed{D}</math> | Convert <math>z</math> 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 <math>0</math> to <math>z</math> is <math>r</math>, the distance from <math>0</math> to <math>z^3</math> must also be <math>r</math>, so <math>r=1</math>. Now we must find <cmath>\text{cis}(2\theta)=60</cmath>. From <math>0 < \theta < \pi/2</math>, we have <cmath>\theta=\frac{\pi}{6}</cmath> and from <math>\pi/2 < \theta < \pi</math>, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from <math>180</math> to <math>0</math>. Hence, there are 2 values that work for <math>0 < \theta < \pi</math>. But since the interval <math>\pi < \theta < 2\pi</math> is identical, because <math>3\theta=\theta</math> at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. <math>\boxed{D}</math> | ||
| − | Here's a graph of how the points move as \theta increases: https://www.desmos.com/calculator/xtnpzoqkgs | + | Here's a graph of how the points move as <cmath>\theta</cmath> increases: https://www.desmos.com/calculator/xtnpzoqkgs |
| − | |||
| − | Someone pls help with LaTeX formatting, thanks | + | |
| + | Someone pls help with LaTeX formatting, thanks -FlatSquare | ||
, I did, -Dodgers66 | , I did, -Dodgers66 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=16|num-a=18}} | {{AMC12 box|year=2019|ab=B|num-b=16|num-a=18}} | ||
Revision as of 15:39, 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 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 ![\[\text{cis}(2\theta)=60\]](http://latex.artofproblemsolving.com/6/d/9/6d9fcfef6bf4b46cf437460bae91ac17d60d8696.png)
. From 
, we have ![\[\theta=\frac{\pi}{6}\]](http://latex.artofproblemsolving.com/d/3/c/d3c87965c9e9a15318a4a47255b3d45176c72f2a.png)
and from 
, we see a monotonic decrease of ![\[\text{cis}(2\theta)\]](http://latex.artofproblemsolving.com/a/6/2/a62485a17672a3a7648a58ba333180def7130c6f.png)
, from 
to . Hence, there are 2 values that work for 
. But since the interval 
is identical, because 
at pi, we have 4 solutions. There are not infinitely many solutions since the same four solutions are duplicated. 
Here's a graph of how the points move as
\[\theta\] (Error making remote request. Unexpected URL sent back)
increases: https://www.desmos.com/calculator/xtnpzoqkgs
Someone pls help with LaTeX formatting, thanks -FlatSquare
, I did, -Dodgers66
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 | |
