2000 AIME I Problems/Problem 6
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
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
=
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
.
Without loss of generality, let's say .
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
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 | ||
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