Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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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.) | 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.) | ||
− | Now, to get from <math>z</math> to <math>z^3</math>, which should be a rotation of <math> | + | Now, to get from <math>z</math> to <math>z^3</math>, which should be a rotation of <math>60^{\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>. |
''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. | ''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. | ||
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The first equation in <math>z</math> (meaning <math>z^3 = z \text{cis} 60^\circ{}</math>) gives us: <math>z^2 = cis 60^\circ{}</math>, which gives 2 solutions in <math>z</math>. | The first equation in <math>z</math> (meaning <math>z^3 = z \text{cis} 60^\circ{}</math>) gives us: <math>z^2 = cis 60^\circ{}</math>, which gives 2 solutions in <math>z</math>. | ||
− | The second equation in <math>z</math> (which is <math>z \text{cis} (-60^\circ{}) = z^3</math>) gives us <math>z^2 = | + | The second equation in <math>z</math> (which is <math>z \text{cis} (-60^\circ{}) = z^3</math>) gives us <math>z^2 = \text{cis} (-60^\circ{})</math>, which must give another 2 solutions in <math>z</math>. |
− | Therefore, there are <math>\boxed{(D) 4}</math> solutions for <math>z</math>. (Professor-Mom) | + | Therefore, there are <math>\boxed{(D) 4}</math> solutions for <math>z</math>. (Professor-Mom) |
Note: The motivation for this method came from an older AIME problem, namely https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8. | Note: The motivation for this method came from an older AIME problem, namely https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8. | ||
+ | ==Solution 5== | ||
+ | |||
+ | Let the <math>z = re^{i\theta}</math>, so <math>z^3 = r^3e^{3i\theta}</math>. To have an equilateral triangle, we must have <math>|z^3| = |z|</math>, so <math>r^3 = r</math>, so <math>r= 1</math>. | ||
+ | |||
+ | Note that the angle between <math>z^3</math> and <math>z</math> is <math>3\theta - \theta = 2\theta</math>. Then, by the Law of Cosines, | ||
+ | |||
+ | <cmath> |z^3 - z|^2 = |z|^2 + |z^3|^2 - 2|z||z^3|\cos 2\theta.</cmath> | ||
+ | |||
+ | Since we have an equilateral triangle, it must be that <math>|z^3 - z| =|z| = |z^3| = r = 1</math>. Hence, | ||
+ | <cmath> 1 = 2 - 2\cos 2\theta = 2 - 2(1-2\sin^2\theta) = 4\sin^2\theta,</cmath> | ||
+ | <cmath> \sin^2\theta = \frac{1}{4}</cmath> | ||
+ | <cmath> \sin\theta = \pm\frac{1}{2}</cmath> | ||
+ | <cmath> \theta = \pm\frac{\pi}{6}, \pm\frac{5\pi}{6}.</cmath> | ||
+ | |||
+ | These <math>4</math> values of <math>\theta</math> correspond to <math>\boxed{(D)4}</math> distinct values of <math>z</math>. | ||
+ | ~ brainfertilzer | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 18:19, 12 November 2022
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 (Quick Look)
As before, . Represent in polar form. By De Moivre's Theorem, . To form an equilateral triangle, their difference in angle must be , so From the polar form of , we know that , so cycles in a circle twice. By contrast, represent fixed, distinct points. Thus, intersects these points twice each
Visual: https://www.desmos.com/calculator/rnpxzns0jn
To be more rigorous, you can find the solutions. cycles twice, so , where . Then, , , , . Substitute those values into and check that they are valid.
(Solution by BJHHar)
Solution 3
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 .
Solution 4 (Quick and Easy)
Since the complex numbers and form an equilateral triangle in the complex plane, we note that either is a 60 degrees counterclockwise rotation about the origin from or is a 60 degrees counterclockwise rotation about the origin from .
Therefore, we note that either or
The first equation in (meaning ) gives us: , which gives 2 solutions in .
The second equation in (which is ) gives us , which must give another 2 solutions in .
Therefore, there are solutions for . (Professor-Mom)
Note: The motivation for this method came from an older AIME problem, namely https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8.
Solution 5
Let the , so . To have an equilateral triangle, we must have , so , so .
Note that the angle between and is . Then, by the Law of Cosines,
Since we have an equilateral triangle, it must be that . Hence,
These values of correspond to distinct values of . ~ brainfertilzer
Video Solution
For those who prefer a video: https://www.youtube.com/watch?v=uBL80yd1ihc
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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