Difference between revisions of "1983 AIME Problems/Problem 5"

Revision as of 15:17, 29 June 2012

Problem

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Solution

Solution 1

One way to solve this problem seems to be by substitution.

$x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$

Because we are only left with $x+y$ and $xy$ , substitution won't be too bad. Let $w=x+y$ and $z=xy$ .

We get $w^2-2z=7$ and $w(7-z)=10$

Because we want the largest possible $w$ , let's find an expression for $z$ in terms of $w$ .

$w^2-7=2z \implies z=\frac{w^2-7}{2}$ .

Substituting, $w^3-21w+20=0$ . Factored, $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division)

The largest possible solution is therefore $x+y=w=\boxed{004}$ .

Solution 2

An alternate way to solve this is to let $x=a+bi$ and $y=c+di$ .

Because we are looking for a value of $x+y$ that is real, we know that $d=-b$ , and thus $y=c-bi$ .

Expanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up.

$(a+bi)^2+(c-bi)^2=7+0i$

$(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$

Looking at the imaginary part of that equation, $2ab-2cb=0$ , so $a=c$ , and $x$ and $y$ are actually complex conjugates.

Looking at the real part of the equation and plugging in $a=c$ , $2a^2-2b^2=7$ , or $2b^2=2a^2-7$ .

Now, evaluating the real part of $(a+bi)^3+(a-bi)^3$ , which equals $10$ (ignoring the odd powers of $i$ , since they would not result in something in the form of $10+0i$ ):

$a^3+3a(bi)^2+a^3+3a(-bi)^2=10$

$2a^3-6ab^2=10$

Since we know that $2b^2=2a^2-7$ , it can be plugged in for $b^2$ in the above equation to yield:

$2a^3-3a(2a^2-7)=10$

$-4a^3+21a=10$

$4a^3-21a+10=0$

Since the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, $a=10, a=5, a=\frac{5}{2}$ all fail, but $a=2$ does work. Thus, the real part of both numbers is $2$ , and their sum is $\boxed{004}$

@@ Line 43: / Line 43: @@
 <math>2a^3-6ab^2=10</math>
-Since we know that <math>2b^2=2a^2-7</math>, it can be plugged in for <math>b^2</math> in the above equataion to yield:
+Since we know that <math>2b^2=2a^2-7</math>, it can be plugged in for <math>b^2</math> in the above equation to yield:
 <math>2a^3-3a(2a^2-7)=10</math>

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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