Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AIME I Problems/Problem 6"
(→‎Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
 +
=== Solution 1 ===
 
<cmath>\begin{eqnarray*}
 
<cmath>\begin{eqnarray*}
 
\frac{x+y}{2} &=& \sqrt{xy} + 2\\
 
\frac{x+y}{2} &=& \sqrt{xy} + 2\\
Line 17: Line 18:
 
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>.
 
-->
 
-->
 +
 +
=== Solution 2 ===
 +
 +
 +
Let <math>a^2</math> = <math>x</math> and <math>b^2</math> = <math>y</math>
 +
 +
Then <cmath>\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2</cmath>
 +
<cmath>a^2 + b^2 = 2\sqrt{ab} + 4</cmath>
 +
<cmath>(a-b)^2 = 4</cmath>
 +
<cmath>(a-b) = \pm 2</cmath>
 +
 +
This makes counting a lot easier since now we just have to find all pairs <math>(a,b)</math> that differ by 2.
 +
 +
 +
Because <math>\sqrt{10^6} = 10^3</math>, then we can use all positive integers less than 1000 for <math>a</math> and <math>b</math>.
 +
 +
 +
[[Without loss of generality]], let's say <math>a < b</math>.
 +
 +
 +
We can count even and odd pairs separately to make things easier:
 +
 +
 +
Odd: <cmath>(1,3) , (3,5) , (5,7)  .  .  .  (997,999)</cmath>
 +
 +
 +
Even: <cmath>(2,4) , (4,6) , (6,8)  .  .  .  (996,998)</cmath>
 +
 +
 +
This makes 499 odd pairs and 498 even pairs, for a total of <math>\boxed{997}</math> pairs.
  
 
== See also ==
 
== See also ==

Revision as of 23:28, 14 May 2014

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

Solution 1

\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 $\boxed{997}$.

Solution 2

Let $a^2$ = $x$ and $b^2$ = $y$

Then \[\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2\] \[a^2 + b^2 = 2\sqrt{ab} + 4\] \[(a-b)^2 = 4\] \[(a-b) = \pm 2\]

This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2.


Because $\sqrt{10^6} = 10^3$, then we can use all positive integers less than 1000 for $a$ and $b$.


Without loss of generality, let's say $a < b$.


We can count even and odd pairs separately to make things easier:


Odd: \[(1,3) , (3,5) , (5,7) . . . (997,999)\]


Even: \[(2,4) , (4,6) , (6,8) . . . (996,998)\]


This makes 499 odd pairs and 498 even pairs, for a total of $\boxed{997}$ pairs.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_6&oldid=61949"