Difference between revisions of "2019 AMC 12B Problems/Problem 17"
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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>. | 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>. | ||
+ | ==Video Solution== | ||
+ | |||
+ | For those who prefer a video: https://www.youtube.com/watch?v=uBL80yd1ihc | ||
==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 21:42, 19 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 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
.
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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