Difference between revisions of "2000 AIME I Problems/Problem 6"
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+ | === Solution 4 (similar to solution 3) === | ||
+ | Rearranging our conditions to | ||
+ | |||
+ | <cmath>x^2-2xy+y^2+16-8x-8y=0 \implies</cmath> | ||
+ | <cmath>(y-)^2=8(x+y-2).</cmath> | ||
+ | |||
+ | Thus, <math>4|y-x.</math> | ||
+ | |||
+ | Now, let <math>y = 4k+x.</math> Plugging this back into our expression, we get | ||
+ | |||
+ | <cmath>(k-1)^2=x-1.</cmath> | ||
+ | |||
+ | There, a unique value of <math>x, y</math> is formed for every value of <math>k</math>. However, we must have | ||
+ | |||
+ | <cmath>y<10^6 \implies (k+1)^2< 10^6-1</cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath>x=(k-1)^2+1>0.</cmath> | ||
+ | |||
+ | Therefore, there are only <math>997</math> pairs of <math>(x,y).</math> | ||
+ | |||
+ | Solution by Williamgolly | ||
== 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 16:12, 6 September 2020
Contents
[hide]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
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.