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

(Solution)
(Solution 3(2-Liner))
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<math>*</math>Note: We are counting the pairs for the values of <math>a</math> and <math>b</math>, which, when squared, translate to the pairs of <math>(x,y)</math> we are trying to find.
 
<math>*</math>Note: We are counting the pairs for the values of <math>a</math> and <math>b</math>, which, when squared, translate to the pairs of <math>(x,y)</math> we are trying to find.
 
===Solution 3(2-Liner)===
 
===Solution 3(2-Liner)===
Our equation is <math>x+y-4=2\sqrt{xy} \implies </math>\sqrt{y}-\sqrt{x}=2<math> since </math>y>x<math>. As a result </math>y<math> must be a perfect square and cannot be </math>10^6, 4, 1<math> so the answer is </math>\boxed{997}$.
+
Our equation is <math>x+y-4=2\sqrt{xy} \implies \sqrt{y}-\sqrt{x}=2</math> since <math>y>x</math>. As a result <math>y</math> must be a perfect square and cannot be <math>10^6, 4, 1</math> so the answer is <math>\boxed{997}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:00, 27 May 2017

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

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.


$*$Note: We are counting the pairs for the values of $a$ and $b$, which, when squared, translate to the pairs of $(x,y)$ we are trying to find.

Solution 3(2-Liner)

Our equation is $x+y-4=2\sqrt{xy} \implies \sqrt{y}-\sqrt{x}=2$ since $y>x$. As a result $y$ must be a perfect square and cannot be $10^6, 4, 1$ so the answer is $\boxed{997}$.

See also

2000 AIME I (ProblemsAnswer KeyResources)
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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