Difference between revisions of "2000 AIME I Problems/Problem 6"
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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 == |
Revision as of 17:28, 25 December 2021
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 ?
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
Solution 5
First we see that our condition is . Then we can see that . From trying a simple example to figure out conditions for , we want to find so we can isolate for . From doing the example we can note that we can square both sides and subtract : (note it is negative because . Clearly the square root must be an integer, so now let . Thus . Thus . We can then find , and use the quadratic formula on to ensure they are and respectively. Thus we get that can go up to 999 and can go down to , leaving possibilities for .
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.