Difference between revisions of "2000 AIME I Problems/Problem 6"
Tomtomjiji (talk | contribs) (→Solution 1) |
Tomtomjiji (talk | contribs) (→Solution 2) |
||
Line 25: | Line 25: | ||
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> | ||
− | <cmath>a^2 + b^2 = | + | <cmath>a^2 + b^2 = 2ab + 4</cmath> |
<cmath>(a-b)^2 = 4</cmath> | <cmath>(a-b)^2 = 4</cmath> | ||
<cmath>(a-b) = \pm 2</cmath> | <cmath>(a-b) = \pm 2</cmath> |
Revision as of 23:16, 16 June 2014
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.
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.