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Difference between revisions of "2000 AIME I Problems/Problem 6"

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y - 2\sqrt{xy} + x &=& 4\\
 
y - 2\sqrt{xy} + x &=& 4\\
 
\sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}</cmath>
 
\sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}</cmath>
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Because <math>y > x</math>, we only consider <math>+2</math>.
  
 
For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation.
 
For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation.
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Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>.
 
Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>.
 
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== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2000|n=I|num-b=5|num-a=7}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 10:32, 20 June 2011

Problem

For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^{6}$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$?

Solution

\begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}

Because $y > x$, we only consider $+2$.

For simplicity, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ that satisfy our equation.

The maximum that $\sqrt{y}$ can be is $10^3 - 1 = 999$ because $\sqrt{y}$ must be an integer (this is because $\sqrt{y} - \sqrt{x} = 2$, an integer). Then $\sqrt{x} = 997$, and we continue this downward until $\sqrt{y} = 3$, in which case $\sqrt{x} = 1$. The number of pairs of $(\sqrt{x},\sqrt{y})$, and so $(x,y)$ is then $\boxed{997}$.

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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