2023 AMC 12A Problems/Problem 16
Contents
Problem
Consider the set of complex numbers satisfying . The maximum value of the imaginary part of can be written in the form , where and are relatively prime positive integers. What is ?
Solution 1
First, substitute in .
Let and
We are trying to maximize , so we'll turn the equation into a quadratic to solve for in terms of .
We want to maximize . Since is always negatively contributing to 's value, we want to minimize .
Due to the trivial inequality:
If we plug 's minimum value in, we get that 's maximum value is
Then and
- CherryBerry
Solution 2
We are given that where is some complex number with magnitude . Rearranging the quadratic to standard form and applying the quadratic formula, we have
The imaginary part of is maximized when . (Why? See note below.)
Thus , and so the answer is .
~cantalon
Note: You can cheese/fakesolve by assuming is real. But there is a good reason for it, within the parameters of this problem:
lies on a circle of radius 4, centered at the origin. Thus, is also a circle, centered at . Luckily, the center of this adjusted circle is a negative real number, so the most negative point on that circle () has the largest magnitude (and so its square root also has the largest magnitude of any square root), and its positive square root is the most purely imaginary square root ("vertical"). [We ignore the alternative negative square root, since the problem asks for the most positive value.]
Thus, has both the largest and largest of any square root of , and so the largest product .
[This argument and overall solution would not work if were instead a positive or non-real number, which would happen if the original problem had with .] ~oinava
Solution 3 (Geometry + Logic)
We can write the given condition as Letting , the equation equates to the circle centered at with radius in the complex plane, call it . Thus the locus of is . Let , and since the does not change 's imaginary part, we now need to find with the largest imaginary part such that lies on .
Note that the point on with largest magnitude is and has argument , call it (The leftmost point on ). The value with positive imaginary part such that has an argument of and a magnitude of .
Since across all values of the imaginary part is given by and has the largest possible and the largest possible value of it must have the largest imaginary part.
This can non-rigorously be seen by sketching the oval which is the locus of .
This gives
~AtharvNaphade
Solution 4
To start, we factor to get:
Note that since the magnitude of a product of complex numbers is equal to the product of the magnitudes:
Now, we substitute (Note that we are trying to maximize b):
Since we are trying to maximize , we want the real parts of the components to be as small as possible, which we can do by setting . This leaves us with:
This gives us .
~LTHMath
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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All AMC 12 Problems and Solutions |
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