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

 
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== Problem ==
 
== Problem ==
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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 of <math>x + y</math> can have?
  
 
== Solution ==
 
== Solution ==
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The best way to solve this problem seems to be by [[brute force]].
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<math>w^2+z^2=(w+z)^2-2wz=7</math> and
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<math>w^3+z^3=(w+z)(w^2+z^2)-wz(w+z)=(w+z)(7)-wz(w+z)=(7-wz)(w+z)=10</math>
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Because we are only left with <math>w+z</math> and <math>wz</math>, [[substitution]] won't be too bad. Let <math>x=w+z</math> and <math>y=wz</math>.
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We get <math>x^2-2y=7</math> and
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<math>x(7-y)=10</math>
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Because we want the largest possible <math>x</math>, let's find an expression for <math>y</math> in terms of <math>x</math>. <math>x^2-7=2y \implies y=\frac{x^2-7}{2}</math>.
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Substituting, <math>x^3-21x+20=0</math>. Factored, <math>(x-1)(x+5)(x-4)=0</math>
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The largest possible solution is therefore <math>4</math>.
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----
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* [[1983 AIME Problems/Problem 4|Previous Problem]]
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* [[1983 AIME Problems/Problem 6|Next Problem]]
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* [[1983 AIME Problems|Back to Exam]]
  
 
== See also ==
 
== See also ==
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* [[1983 AIME Problems]]
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
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[[Category:Intermediate Complex Numbers Problems]]

Revision as of 22:59, 23 July 2006

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

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