2021 Fall AMC 12B Problems/Problem 21
Contents
[hide]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)
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=vhAc0P09czI