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

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The largest possible solution is therefore <math>4</math>.
 
The largest possible solution is therefore <math>4</math>.
 
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* [[1983 AIME Problems/Problem 4|Previous Problem]]
 
* [[1983 AIME Problems/Problem 6|Next Problem]]
 
* [[1983 AIME Problems|Back to Exam]]
 
  
 
== See also ==
 
== See also ==
* [[AIME Problems and Solutions]]
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{{AIME box|year=1983|num-b=4|num-a=6}}
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Complex Numbers Problems]]
 
[[Category:Intermediate Complex Numbers Problems]]

Revision as of 16:12, 11 March 2007

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.

$w^2+z^2=(w+z)^2-2wz=7$ and $w^3+z^3=(w+z)(w^2+z^2)-wz(w+z)=(w+z)(7)-wz(w+z)=(7-wz)(w+z)=10$

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

We get $x^2-2y=7$ and $x(7-y)=10$

Because we want the largest possible $x$, let's find an expression for $y$ in terms of $x$. $x^2-7=2y \implies y=\frac{x^2-7}{2}$.

Substituting, $x^3-21x+20=0$. Factored, $(x-1)(x+5)(x-4)=0$

The largest possible solution is therefore $4$.

See also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions