1983 AIME Problems/Problem 5
Suppose that the sum of the squares of two complex numbers and is and the sum of the cubes is . What is the largest real value that can have?
One way to solve this problem is by substitution. We have
Hence observe that we can write and .
This reduces the equations to and .
Because we want the largest possible , let's find an expression for in terms of .
Substituting, , which factorizes as (the Rational Root Theorem may be used here, along with synthetic division).
The largest possible solution is therefore .
An alternate way to solve this is to let and .
Because we are looking for a value of that is real, we know that , and thus .
Expanding will give two equations, since the real and imaginary parts must match up.
Looking at the imaginary part of that equation, , so , and and are actually complex conjugates.
Looking at the real part of the equation and plugging in , , or .
Now, evaluating the real part of , which equals (ignoring the odd powers of , since they would not result in something in the form of ):
Since we know that , it can be plugged in for in the above equation to yield:
Since the problem is looking for to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, all fail, but does work. Thus, the real part of both numbers is , and their sum is .
Begin by assuming that and are roots of some polynomial of the form , such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), and . Substituting , we deduce that , whose roots are , , and . Since is the sum of the roots and is maximized when , the answer is .
Also, Substituting our above into this, we get . Letting , we have that . Testing , we find that this is a root, to get
Minor edit: Jc426
|1983 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|