Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
− | Convert <math>z</math> and <math>z^3</math> into < | + | Convert <math>z</math> and <math>z^3</math> into modulus-argument (polar) form, giving <math>z=r\text{cis}(\theta)</math> for some <math>r</math> and <math>\theta</math>. Thus, by De Moivre's Theorem, <math>z^3=r^3\text{cis}(3\theta)</math>. Since the distance from <math>0</math> to <math>z</math> is <math>r</math>, and the triangle is equilateral, the distance from <math>0</math> to <math>z^3</math> must also be <math>r</math>, so <math>r^3=r</math>, giving <math>r=1</math>. (We know <math>r \neq 0</math> since the problem statement specifies that <math>z</math> must be nonzero.) |
− | Here's a graph | + | Now, to get from <math>z</math> to <math>z^3</math>, which should be a rotation of <math>120^{\circ}</math> if the triangle is equilateral, we multiply by <math>z^2 = r^2\text{cis}(2\theta)</math>, again using De Moivre's Theorem. Thus we require <math>2\theta=\pm\frac{\pi}{3} + 2\pi k</math> (where <math>k</math> can be any integer). If <math>0 < \theta < \frac{\pi}{2}</math>, we must have <math>\theta=\frac{\pi}{6}</math>, while if <math>\frac{\pi}{2} \leq \theta < \pi</math>, we must have <math>\theta = \frac{5\pi}{6}</math>. Hence there are <math>2</math> values that work for <math>0 < \theta < \pi</math>. By symmetry, the interval <math>\pi \leq \theta < 2\pi</math> will also give <math>2</math> solutions. The answer is thus <math>2 + 2 = \boxed{\textbf{(D) }4}</math>. |
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+ | ''Note'': Here's a graph showing how <math>z</math> and <math>z^3</math> move as <math>\theta</math> increases: https://www.desmos.com/calculator/xtnpzoqkgs. | ||
==Solution 2== | ==Solution 2== | ||
− | + | For the triangle to be equilateral, the vector from <math>z</math> to <math>z^3</math>, i.e <math>z^3 - z</math>, must be a <math>60^{\circ}</math> rotation of the vector from <math>0</math> to <math>z</math>, i.e. just <math>z</math>. Thus we must have | |
<cmath>\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)</cmath> | <cmath>\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)</cmath> | ||
− | + | Simplifying gives | |
− | |||
<cmath>z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)</cmath> | <cmath>z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)</cmath> | ||
− | + | so | |
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− | |||
<cmath>z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)</cmath> | <cmath>z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)</cmath> | ||
− | + | Since any nonzero complex number will have two square roots, each equation gives two solutions. Thus, as before, the total number of possible values of <math>z</math> is <math>\boxed{\textbf{(D) }4}</math>. | |
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==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 23:31, 18 February 2019
Contents
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 modulus-argument (polar) form, giving for some and . Thus, by De Moivre's Theorem, . Since the distance from to is , and the triangle is equilateral, the distance from to must also be , so , giving . (We know since the problem statement specifies that must be nonzero.)
Now, to get from to , which should be a rotation of if the triangle is equilateral, we multiply by , again using De Moivre's Theorem. Thus we require (where can be any integer). If , we must have , while if , we must have . Hence there are values that work for . By symmetry, the interval will also give solutions. The answer is thus .
Note: Here's a graph showing how and move as increases: https://www.desmos.com/calculator/xtnpzoqkgs.
Solution 2
For the triangle to be equilateral, the vector from to , i.e , must be a rotation of the vector from to , i.e. just . Thus we must have
Simplifying gives so
Since any nonzero complex number will have two square roots, each equation gives two solutions. Thus, as before, the total number of possible values of is .
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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