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

(Solution)
(Solution)
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<math>w(7-z)=10</math>
 
<math>w(7-z)=10</math>
  
Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>.  
+
Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>.
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<math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>.
 
<math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>.
  

Revision as of 23:09, 26 February 2009

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

The best way to solve this problem seems to be by brute force.

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

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 $x+y=w=4$.

See also

1983 AIME (ProblemsAnswer KeyResources)
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