1983 AIME Problems/Problem 5
Problem
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?
Solution
Solution 1
One way to solve this problem seems to be by substitution.
and
Because we are only left with 
and 
, substitution won't be too bad. Let 
and 
.
We get 
and
Because we want the largest possible 
, let's find an expression for 
in terms of .
.
Substituting, 
. Factored, 
(the Rational Root Theorem may be used here, along with synthetic division)
The largest possible solution is therefore 
.
Solution 2
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 
See Also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
I started by assuming x and y were roots of some polynomial of the form 
So then 
and 
Substituting 
we arrive at the polynomial 
From rational root theorem we find the roots to be 
Since 
is the sum of the roots and is maximized when b is -4, we the answer is -(-4)=
