Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | 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 < | + | 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}{2}</cmath> and from <math>\pi/2 < \theta < \pi</math>, we see a monotonic decrease of <cmath>\text{cis}(2\theta)</cmath>, from <math>180^{\circ}</math> to <math>0^{\circ}</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> |
-FlatSquare | -FlatSquare | ||
Someone pls help with LaTeX formatting, thanks | Someone pls help with LaTeX formatting, thanks | ||
+ | , 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 14:53, 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 form, giving and . Since the distance from to is , the distance from to must also be , so . Now we must find . From , we have and from , we see a monotonic decrease of , 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.
-FlatSquare
Someone pls help with LaTeX formatting, thanks , 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 |