Difference between revisions of "2000 AIME I Problems/Problem 6"
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== Problem == | == Problem == | ||
− | For how many [[ordered pair]]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^ | + | For how many [[ordered pair]]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the [[geometric mean]] of <math>x</math> and <math>y</math>? |
− | == | + | == Solutions == |
=== Solution 1 === | === Solution 1 === | ||
<cmath>\begin{eqnarray*} | <cmath>\begin{eqnarray*} | ||
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For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation. | For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation. | ||
− | The maximum that <math>\sqrt{y}</math> can be is <math>10^ | + | The maximum that <math>\sqrt{y}</math> can be is <math>\sqrt{10^6} - 1 = 999</math> because <math>\sqrt{y}</math> must be an integer (this is because <math>\sqrt{y} - \sqrt{x} = 2</math>, an integer). Then <math>\sqrt{x} = 997</math>, and we continue this downward until <math>\sqrt{y} = 3</math>, in which case <math>\sqrt{x} = 1</math>. The number of pairs of <math>(\sqrt{x},\sqrt{y})</math>, and so <math>(x,y)</math> is then <math>\boxed{997}</math>. |
<!-- solution lost in edit conflict - azjps - | <!-- solution lost in edit conflict - azjps - | ||
Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>. | Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>. | ||
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=== Solution 2 === | === Solution 2 === | ||
− | 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> | ||
− | <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> | ||
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− | + | We know that because <math>x < y</math>, we get <math>a < b</math>. | |
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− | <math>* | + | <math>*</math>Note: We are counting the pairs for the values of <math>a</math> and <math>b</math>, which, when squared, translate to the pairs of <math>(x,y)</math> we are trying to find. |
+ | |||
+ | === Solution 3 === | ||
+ | Since the arithmetic mean is 2 more than the geometric mean, <math>\frac{x+y}{2} = 2 + \sqrt{xy}</math>. We can multiply by 2 to get <math>x + y = 4 + 2\sqrt{xy}</math>. Subtracting 4 and squaring gives | ||
+ | <cmath>((x+y)-4)^2 = 4xy</cmath> | ||
+ | <cmath>((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy</cmath> | ||
+ | <cmath>x^2 - 2xy + y^2 + 16 - 8x - 8y = 0</cmath> | ||
+ | |||
+ | Notice that <math>((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y</math>, so the problem asks for solutions of | ||
+ | <cmath>(x-y-4)^2 = 16y</cmath> | ||
+ | Since the left hand side is a perfect square, and 16 is a perfect square, <math>y</math> must also be a perfect square. Since <math>0 < y < (1000)^2</math>, <math>y</math> must be from <math>1^2</math> to <math>999^2</math>, giving at most 999 options for <math>y</math>. | ||
+ | |||
+ | However if <math>y = 1^2</math>, you get <math>(x-5)^2 = 16</math>, which has solutions <math>x = 9</math> and <math>x = 1</math>. Both of those solutions are not less than <math>y</math>, so <math>y</math> cannot be equal to 1. If <math>y = 2^2 = 4</math>, you get <math>(x - 8)^2 = 64</math>, which has 2 solutions, <math>x = 16</math>, and <math>x = 0</math>. 16 is not less than 4, and <math>x</math> cannot be 0, so <math>y</math> cannot be 4. However, for all other <math>y</math>, you get exactly 1 solution for <math>x</math>, and that gives a total of <math>999 - 2 = \boxed{997}</math> pairs. | ||
+ | |||
+ | - asbodke | ||
+ | |||
+ | |||
+ | === Solution 4 (Similar to Solution 3) === | ||
+ | Rearranging our conditions to | ||
+ | |||
+ | <cmath>x^2-2xy+y^2+16-8x-8y=0 \implies</cmath> | ||
+ | <cmath>(y-x)^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.</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>\boxed{997}</math> pairs of <math>(x,y).</math> | ||
+ | |||
+ | Solution by Williamgolly | ||
+ | |||
+ | === Solution 5 === | ||
+ | |||
+ | 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 10:31, 24 June 2024
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 |
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