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

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The best way to solve this problem seems to be by [[brute force]].
 
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
+
<math>x^2+y^2=(x+y)^2-2xy=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>
+
<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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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>.
+
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>.
  
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We get <math>x^2-2y=7</math> and
+
We get <math>w^2-2z=7</math> and
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<math>x(7-y)=10</math>
+
<math>w(7-z)=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>.
+
Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>. <math>w^2-7=2z \implies z=\frac{w^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>
+
Substituting, <math>w^3-21w+20=0</math>. Factored, <math>(w-1)(w+5)(w-4)=0</math>
  
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The largest possible solution is therefore <math>4</math>.
+
The largest possible solution is therefore <math>x+y=w=\box{4}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:07, 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 $x+y=w=\box{4}$ (Error compiling LaTeX. Unknown error_msg).

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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