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

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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.
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=== Solution 3 ===
 +
Since the arithmetic mean is 2 more than the geometric mean, <math>\frac{x+y}{2} = 2 + \sqrt{xy}</math>. We can multiply by 2 to get <math>x + y = 4 + 2\sqrt{xy}</math>. Subtracting 4 and squaring gives
 +
<cmath>((x+y)-4)^2 = 4xy</cmath>
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<cmath>((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy</cmath>
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<cmath>x^2 - 2xy + y^2 + 16 - 8x - 8y = 0</cmath>
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Notice that <math>((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y</math>, so the problem asks for solutions of
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<cmath>(x-y-4)^2 = 16y</cmath>
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Since the left hand side is a perfect square, and 16 is a perfect square, <math>y</math> must also be a perfect square. Since <math>0 < y < (1000)^2</math>, <math>y</math> must be from <math>1^2</math> to <math>999^2</math>, giving at most 999 options for <math>y</math>.
 +
 +
However if <math>y = 1^2</math>, you get <math>(x-5)^2 = 16</math>, which has solutions <math>x = 9</math> and <math>x = 1</math>. Both of those solutions are not less than <math>y</math>, so <math>y</math> cannot be equal to 1. If <math>y = 2^2 = 4</math>, you get <math>(x - 8)^2 = 64</math>, which has 2 solutions, <math>x = 16</math>, and <math>x = 0</math>. 16 is not less than 4, and <math>x</math> cannot be 0, so <math>y</math> cannot be 4. However, for all other <math>y</math>, you get exactly 1 solution for <math>x</math>, and that gives a total of <math>999 - 2 = \boxed{997}</math> pairs.
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- asbodke
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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}}

Revision as of 17:52, 19 April 2020

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

Since the arithmetic mean is 2 more than the geometric mean, $\frac{x+y}{2} = 2 + \sqrt{xy}$. We can multiply by 2 to get $x + y = 4 + 2\sqrt{xy}$. Subtracting 4 and squaring gives \[((x+y)-4)^2 = 4xy\] \[((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy\] \[x^2 - 2xy + y^2 + 16 - 8x - 8y = 0\]

Notice that $((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y$, so the problem asks for solutions of \[(x-y-4)^2 = 16y\] Since the left hand side is a perfect square, and 16 is a perfect square, $y$ must also be a perfect square. Since $0 < y < (1000)^2$, $y$ must be from $1^2$ to $999^2$, giving at most 999 options for $y$.

However if $y = 1^2$, you get $(x-5)^2 = 16$, which has solutions $x = 9$ and $x = 1$. Both of those solutions are not less than $y$, so $y$ cannot be equal to 1. If $y = 2^2 = 4$, you get $(x - 8)^2 = 64$, which has 2 solutions, $x = 16$, and $x = 0$. 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$, you get exactly 1 solution for $x$, and that gives a total of $999 - 2 = \boxed{997}$ pairs.

- asbodke

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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