Difference between revisions of "2000 AIME I Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | {{ | + | From the condition given, |
| + | |||
| + | <math>\begin{align*} | ||
| + | \frac{x + y}{2} - 2 &= \sqrt{xy}\\ | ||
| + | x + y - 2\sqrt{xy} &= 4\\ | ||
| + | (\sqrt{y} - \sqrt{x})^2 &= 4\\ | ||
| + | \sqrt{y} - \sqrt{x} &= 2 | ||
| + | \end{align*}</math> | ||
| + | |||
| + | The last equation is true because <math>y > x</math>. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | 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>. | ||
== 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 17:04, 31 December 2007
Problem
For how many ordered pairs 
of integers is it true that 
and that the arithmetic mean of 
and 
is exactly 
more than the geometric mean of and ?
Solution
From the condition given,
$\begin{align*} \frac{x + y}{2} - 2 &= \sqrt{xy}\\ x + y - 2\sqrt{xy} &= 4\\ (\sqrt{y} - \sqrt{x})^2 &= 4\\ \sqrt{y} - \sqrt{x} &= 2 \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
The last equation is true because 
.
Here, we can count how many valid pairs of 
satisfy our equation, rather than directly, because can get messy.
The maximum that 
can be is 
because must be an integer (this is because 
, an integer). Then 
, and we continue this downward until 
, in which case 
. The number of pairs of , and so is, then, 
.
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 | ||
