Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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Mathislife16 (talk | contribs) (→Solution: fixed a few formatting issues and a few errors regarding the use of cis(theta) to mean the argument of a complex number, switched all degrees to radians, and fixed an incorrect assertion.) |
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==Solution== | ==Solution== | ||
− | 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> | + | 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> |
− | Here's a graph of how the points move as < | + | Here's a graph of how the points move as <math>\theta</math> increases- https://www.desmos.com/calculator/xtnpzoqkgs |
Someone pls help with LaTeX formatting, thanks -FlatSquare | Someone pls help with LaTeX formatting, thanks -FlatSquare |
Revision as of 17:08, 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
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
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 |