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Difference between revisions of "2000 AIME I Problems/Problem 6"
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== Problem ==
 
== Problem ==
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For how many ordered pairs <math>(x,y)</math> of integers is it true that <math>0 < x < y < 10^{6}</math> and that the arithmetic mean of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the geometric mean of <math>x</math> and <math>y</math>?
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For how many [[ordered pair]]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^{6}</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the [[geometric mean]] of <math>x</math> and <math>y</math>?
  
 
== Solution ==
 
== Solution ==
−
From the condition given,
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<cmath>\begin{eqnarray*}
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\frac{x+y}{2} &=& \sqrt{xy} + 2\\
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x+y-4 &=& 2\sqrt{xy}\\
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y - 2\sqrt{xy} + x &=& 4\\
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\sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}</cmath>
  
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<math>\begin{align*}
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For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation.
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\frac{x + y}{2} - 2 &= \sqrt{xy}\\
 
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x + y - 2\sqrt{xy} &= 4\\
 
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(\sqrt{y} - \sqrt{x})^2 &= 4\\
 
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\sqrt{y} - \sqrt{x} &= 2
 
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\end{align*}</math>
 
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The last equation is true because <math>y > x</math>.
 
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Here, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> satisfy our equation, rather than <math>(x,y)</math> directly, because <math>(x,y)</math> can get messy.
 
−
 
 
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The maximum that <math>\sqrt{y}</math> can be is <math>10^3 - 1 = 999</math> because <math>\sqrt{y}</math> must be an integer (this is because  <math>\sqrt{y} - \sqrt{x} = 2</math>, an integer). Then <math>\sqrt{x} = 997</math>, and we continue this downward until <math>\sqrt{y} = 3</math>, in which case <math>\sqrt{x} = 1</math>. The number of pairs of <math>(\sqrt{x},\sqrt{y})</math>, and so <math>(x,y)</math> is, then, <math>999 - 3 + 1 = \boxed{997}</math>.
 
  
 +
The maximum that <math>\sqrt{y}</math> can be is <math>10^3 - 1 = 999</math> because <math>\sqrt{y}</math> must be an integer (this is because <math>\sqrt{y} - \sqrt{x} = 2</math>, an integer). Then <math>\sqrt{x} = 997</math>, and we continue this downward until <math>\sqrt{y} = 3</math>, in which case <math>\sqrt{x} = 1</math>. The number of pairs of <math>(\sqrt{x},\sqrt{y})</math>, and so <math>(x,y)</math> is then <math>\boxed{997}</math>.
 +
<!-- solution lost in edit conflict - azjps -
 +
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>.
 +
-->
 
== 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}}
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 +
[[Category:Intermediate Algebra Problems]]

Revision as of 17:12, 31 December 2007

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*}

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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