Difference between revisions of "1983 AIME Problems/Problem 5"
(→Solution 2) |
m |
||
| Line 2: | Line 2: | ||
Suppose that the sum of the squares of two complex numbers <math>x</math> and <math>y</math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value that <math>x + y</math> can have? | Suppose that the sum of the squares of two complex numbers <math>x</math> and <math>y</math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value that <math>x + y</math> can have? | ||
| − | == Solution 1== | + | == Solutions == |
| + | === Solution 1 === | ||
One way to solve this problem seems to be by [[substitution]]. | One way to solve this problem seems to be by [[substitution]]. | ||
| Line 21: | Line 22: | ||
The largest possible solution is therefore <math>x+y=w=\boxed{004}</math>. | The largest possible solution is therefore <math>x+y=w=\boxed{004}</math>. | ||
| − | == Solution 2== | + | ===Solution 2 === |
An alternate way to solve this is to let <math>x=a+bi</math> and <math>y=c+di</math>. | An alternate way to solve this is to let <math>x=a+bi</math> and <math>y=c+di</math>. | ||
| Line 52: | Line 53: | ||
Since the problem is looking for <math>x+y=2a</math> to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, <math>a=10, a=5, a=\frac{5}{2}</math> all fail, but <math>a=2</math> does work. Thus, the real part of both numbers is <math>2</math>, and their sum is <math>\boxed{004}</math> | Since the problem is looking for <math>x+y=2a</math> to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, <math>a=10, a=5, a=\frac{5}{2}</math> all fail, but <math>a=2</math> does work. Thus, the real part of both numbers is <math>2</math>, and their sum is <math>\boxed{004}</math> | ||
| − | == See | + | == See Also == |
{{AIME box|year=1983|num-b=4|num-a=6}} | {{AIME box|year=1983|num-b=4|num-a=6}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 06:00, 16 April 2012
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?
Solutions
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 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 | ||
