2021 Fall AMC 12B Problems/Problem 21
Contents
Problem
For real numbers , let
where
. For how many values of
with
does
Solution 1
Let . Now
.
and
so there is a real root
between
and
. The other
's must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex
's squared is
which is greater than
. If
is real number then
must have magnitude of
, but none of the solutions for
have magnitude of
, so the answer is
~lopkiloinm
Solution 2
For , we get
So either
, i.e.
or
, i.e.
.
For none of these values do we get .
Therefore, the answer is .
Solution 3
We have
Denote . Hence, this problem asks us to find the number of
with
that satisfy
Taking imaginary part of both sides, we have
The sixth equality follows from the property that
.
Therefore, we have either or
or
.
Case 1: .
Because ,
.
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 2: .
Because ,
.
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 3: .
Because ,
.
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
All cases above imply that there is no solution in this problem.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4
Let , so by De Moivre
. The problem essentially asks for the number of real roots of
which lie on the complex unit circle.
Let
be a root of
, and note that we can't have
, else
. Thus, suppose henceforth that
. We then have
, hence the argument of
is either the argument of
, or the argument of
. Since
is real, it follows that
. Now, we can check all of these values and find that none of them work, yielding an answer of
.
-IAmTheHazard
Solution 5 (Geometry)
can be written equivalently as
Thus, we aim to find
such that the sum of the vectors
,
, and
is -1. Notice that
,
,
all lie on the unit circle in the complex plane, and the vector
is collinear with
Since
and we want the three vectors to sum to -1, we either have
and
, or
and
If the first condition is true,
This will imply that
But then
, which violates the condition. Similarly, we can show that the second condition cannot be met either. Thus
does not have any solutions on the interval
Therefore, the answer is
.
-mathy_mathema
Video Solution
~MathProblemSolvingSkills.com
Brute Force Solution
1+cis(x)+cis(3x)=cis(2x) try out all the common values x could be 0,pi/4,pi/3,pi/2,2pi/3,3pi/4,pi,4pi/3,5pi/4,3pi/2,7pi/4,5pi/3,2pi insert them into the equation, and then you find none that work so… let’s move on and pray that our answer is right emilyyunhanq@gmail.com Emily Q
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=vhAc0P09czI
Video Solution by The Power of Logic
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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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