Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}</math> | <math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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>2\theta=\pm\frac{\pi}{3}</cmath>the requirements for being an equilateral triangle. 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 increase of <math>2\theta</math>, from <math>\pi</math> to <math>2\pi</math>, or equivalently, from <math>-\pi</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> also consists of <math>2\theta</math> going from <math>0</math> to <math>2\pi</math>, it also gives us 2 solutions. Our answer is <math>\boxed{\textbf{(D) 4}}</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>2\theta=\pm\frac{\pi}{3}</cmath>the requirements for being an equilateral triangle. 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 increase of <math>2\theta</math>, from <math>\pi</math> to <math>2\pi</math>, or equivalently, from <math>-\pi</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> also consists of <math>2\theta</math> going from <math>0</math> to <math>2\pi</math>, it also gives us 2 solutions. Our answer is <math>\boxed{\textbf{(D) 4}}</math> | ||
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Someone pls help with LaTeX formatting, thanks -FlatSquare | Someone pls help with LaTeX formatting, thanks -FlatSquare | ||
, I did, -Dodgers66 | , I did, -Dodgers66 | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | To be equilateral triangle, we should have | ||
+ | (Z^3-Z)/(Z-0)=cis(pi/3) or cis(2*pi/3) | ||
+ | simplify left side | ||
+ | Z^2-1=cis(pi/3) or cis(2*pi/3) | ||
+ | Z^2=1+cis(pi/3) or 1+cis(2*pi/3) | ||
+ | we have two roots for both equations, therefore the total number of solution for Z is 4 | ||
+ | |||
+ | (By Zhen Qin) | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:49, 15 February 2019
Contents
[hide]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 form, giving and . Since the distance from to is , the distance from to must also be , so . Now we must find the requirements for being an equilateral triangle. From , we have 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 the points move as increases- https://www.desmos.com/calculator/xtnpzoqkgs
Someone pls help with LaTeX formatting, thanks -FlatSquare , I did, -Dodgers66
Solution 2
To be equilateral triangle, we should have (Z^3-Z)/(Z-0)=cis(pi/3) or cis(2*pi/3) simplify left side Z^2-1=cis(pi/3) or cis(2*pi/3) Z^2=1+cis(pi/3) or 1+cis(2*pi/3) we have two roots for both equations, therefore the total number of solution for Z is 4
(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 |
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