2021 Fall AMC 12B Problems/Problem 21
Problem
For real numbers , let
where
. For how many values of
with
does
Solution 1
Let . Now
.
and
so there is a real number
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 all the solutions for
do not have magnitude of
, so the answer is
~lopkiloinm
Solution 2
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)