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> | + | 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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− | + | We know that because <math>x < y</math>, we get <math>a < b</math>. | |
Revision as of 18:25, 7 March 2021
Contents
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
?
Solutions
Solution 1
Because , we only consider
.
For simplicity, we can count how many valid pairs of that satisfy our equation.
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
.
Solution 2
Let =
and
=
, where
and
are positive.
Then
This makes counting a lot easier since now we just have to find all pairs that differ by 2.
Because , then we can use all positive integers less than 1000 for
and
.
We know that because , we get
.
We can count even and odd pairs separately to make things easier*:
Odd:
Even:
This makes 499 odd pairs and 498 even pairs, for a total of pairs.
Note: We are counting the pairs for the values of
and
, which, when squared, translate to the pairs of
we are trying to find.
Solution 3
Since the arithmetic mean is 2 more than the geometric mean, . We can multiply by 2 to get
. Subtracting 4 and squaring gives
Notice that , so the problem asks for solutions of
Since the left hand side is a perfect square, and 16 is a perfect square,
must also be a perfect square. Since
,
must be from
to
, giving at most 999 options for
.
However if , you get
, which has solutions
and
. Both of those solutions are not less than
, so
cannot be equal to 1. If
, you get
, which has 2 solutions,
, and
. 16 is not less than 4, and
cannot be 0, so
cannot be 4. However, for all other
, you get exactly 1 solution for
, and that gives a total of
pairs.
- asbodke
Solution 4 (Similar to Solution 3)
Rearranging our conditions to
Thus,
Now, let Plugging this back into our expression, we get
There, a unique value of is formed for every value of
. However, we must have
and
Therefore, there are only pairs of
Solution by Williamgolly
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.