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

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Let <math>a^2</math> = <math>x</math> and <math>b^2</math> = <math>y</math>
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Let <math>a^2</math> = <math>x</math> and <math>b^2</math> = <math>y</math>, where <math>a</math> and <math>b</math> are positive.
  
 
Then <cmath>\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2</cmath>
 
Then <cmath>\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2</cmath>
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[[Without loss of generality]], let's say <math>a < b</math>.
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We know that because <math>x < y</math>, we get <math>a < b</math>.
  
  
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Now, let <math>y = 4k+x.</math> Plugging this back into our expression, we get
 
Now, let <math>y = 4k+x.</math> Plugging this back into our expression, we get
  
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<cmath>(k-1)^2=x-1.</cmath>
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<cmath>(k-1)^2=x.</cmath>
  
 
There, a unique value of <math>x, y</math> is formed for every value of <math>k</math>. However, we must have  
 
There, a unique value of <math>x, y</math> is formed for every value of <math>k</math>. However, we must have  
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<cmath>x=(k-1)^2+1>0.</cmath>
 
<cmath>x=(k-1)^2+1>0.</cmath>
  
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Therefore, there are only <math>997</math> pairs of <math>(x,y).</math>
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Therefore, there are only <math>\boxed{997}</math> pairs of <math>(x,y).</math>
  
 
Solution by Williamgolly
 
Solution by Williamgolly
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=== Solution 5 ===
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First we see that our condition is <math>\frac{x+y}{2} = 2 + \sqrt{xy}</math>. Then we can see that <math>x+y = 4 + 2\sqrt{xy}</math>. From trying a simple example to figure out conditions for <math>x,y</math>, we want to find <math>x-y</math> so we can isolate for <math>x</math>. From doing the example we can note that we can square both sides and subtract <math>4xy</math>: <math>(x-y)^2 = 16 + 16\sqrt{xy} \implies x-y = -2(
 +
\sqrt{1+\sqrt{xy}})</math> (note it is negative because <math>y > x</math>. Clearly the square root must be an integer, so now let <math>\sqrt{xy} = a^2-1</math>. Thus <math>x-y = -2a</math>. Thus <math>x = 2 + \sqrt{xy} - a = 2 + a^2 - 1 -2a</math>. We can then find <math>y</math>, and use the quadratic formula on <math>x,y</math> to ensure they are <math>>0</math> and <math><10^6</math> respectively. Thus we get that <math>y</math> can go up to 999 and <math>x</math> can go down to <math>3</math>, leaving <math>997</math> possibilities for <math>x,y</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 10:31, 24 June 2024

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 $y$?

Solutions

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 $\sqrt{10^6} - 1 = 999$ 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 $b^2$ = $y$, where $a$ and $b$ are positive.

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 $b$.


We know that because $x < y$, we get $a < b$.


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


Solution 4 (Similar to Solution 3)

Rearranging our conditions to

\[x^2-2xy+y^2+16-8x-8y=0 \implies\] \[(y-x)^2=8(x+y-2).\]

Thus, $4|y-x.$

Now, let $y = 4k+x.$ Plugging this back into our expression, we get

\[(k-1)^2=x.\]

There, a unique value of $x, y$ is formed for every value of $k$. However, we must have

\[y<10^6 \implies (k+1)^2< 10^6-1\]

and

\[x=(k-1)^2+1>0.\]

Therefore, there are only $\boxed{997}$ pairs of $(x,y).$

Solution by Williamgolly

Solution 5

First we see that our condition is $\frac{x+y}{2} = 2 + \sqrt{xy}$. Then we can see that $x+y = 4 + 2\sqrt{xy}$. From trying a simple example to figure out conditions for $x,y$, we want to find $x-y$ so we can isolate for $x$. From doing the example we can note that we can square both sides and subtract $4xy$: $(x-y)^2 = 16 + 16\sqrt{xy} \implies x-y = -2( \sqrt{1+\sqrt{xy}})$ (note it is negative because $y > x$. Clearly the square root must be an integer, so now let $\sqrt{xy} = a^2-1$. Thus $x-y = -2a$. Thus $x = 2 + \sqrt{xy} - a = 2 + a^2 - 1 -2a$. We can then find $y$, and use the quadratic formula on $x,y$ to ensure they are $>0$ and $<10^6$ respectively. Thus we get that $y$ can go up to 999 and $x$ can go down to $3$, leaving $997$ possibilities for $x,y$.

See also

2000 AIME I (Problems • Answer Key • Resources)
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