Difference between revisions of "1983 AIME Problems/Problem 5"
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== Solution 1== | == Solution 1== | ||
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<math>x^2+y^2=(x+y)^2-2xy=7</math> and | <math>x^2+y^2=(x+y)^2-2xy=7</math> and | ||
Revision as of 10:56, 27 February 2011
Contents
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 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 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 we know that , it can be plugged in for 
in the above equataion 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 | ||
