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

(Second solution using less clever subsitutions)
(→‎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
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<math>x^3+y^3=(x+y)(x^2+y^2)-xy(x+y)=(x+y)(7)-xy(x+y)=(7-xy)(x+y)=10</math>
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<math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math>
  
 
Because we are only left with <math>x+y</math> and <math>xy</math>, substitution won't be too bad. Let <math>w=x+y</math> and <math>z=xy</math>.
 
Because we are only left with <math>x+y</math> and <math>xy</math>, substitution won't be too bad. Let <math>w=x+y</math> and <math>z=xy</math>.

Revision as of 13:07, 17 July 2010

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 of $x + y$ can have?

Solution 1

One way to solve this problem seems to be by brute force.

$x^2+y^2=(x+y)^2-2xy=7$ 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 largest possible solution is therefore $x+y=w=4$.

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 $i$):

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

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

Since we know that $2b^2=2a^2-7$, it can be plugged in for $b^2$ in the above equataion 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 $\boxed{004}$

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
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