MHB Finding $x$ and $y$ Given $A, G, H \in N$ and $A+G+H=49$

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The discussion revolves around finding the values of x and y given the equations A = (x+y)/2, G = √(xy), and H = (2xy)/(x+y), with the condition that A, G, and H are natural numbers and their sum equals 49. Participants express frustration over the time constraints for providing hints and solutions. The conversation emphasizes the relationships between the variables and the requirement that x must be greater than y, both being positive real numbers. A solution is anticipated but not yet provided, indicating a collaborative effort to solve the problem. The thread highlights the challenge of balancing mathematical reasoning with time management in problem-solving.
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we are given :$x,y\in R ,and \,\, x>y>0, A=\dfrac {x+y}{2},G=\sqrt {xy}, H=\dfrac {2xy}{x+y}$

if $A,G,H\in N, \,\, and \,\,\, A+G+H=49$

please find $,x,y$
 
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Albert said:
we are given :$x,y\in R ,and \,\, x>y>0, A=\dfrac {x+y}{2},G=\sqrt {xy}, H=\dfrac {2xy}{x+y}$

if $A,G,H\in N, \,\, and \,\,\, A+G+H=49$

please find $,x,y$
hint :
$G^2=A\times H$
$A,G,H $ is a geometric sequence
and $A>G>H$
 
Albert said:
hint :
$G^2=A\times H$
$A,G,H $ is a geometric sequence
and $A>G>H$
I hope that you could have given some time before a hint. One day is too short. Here is the solution

Let common ratio be r
so $H + Hr + Hr^2= 49$
so $H(1+r+r^2) = 49$
r need not be integer it could be $\dfrac{p}{q}$ but let us try integer
so we get
H is a factor of 49 so H = 1 or 7
H =1 gives r irrational and H= 7 gives r = 2

so $AM = 28$ or $x+ y = 56$ and $xy =196$

solving these 2 as x is greater $x= 28+14\sqrt{3}$ and $y= 28- 14\sqrt{3}$
For r to be fraction h has to be a product of a sqaure that multyiple has to be factor of 49. So h has to be a square or 7 * square
trying $H = 9$ we get $G = 15$ and $A = 25$ and so $x = 45,y = 5$ as the second solution
 
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kaliprasad said:
I hope that you could have given some time before a hint. One day is too short. Here is the solution

Let common ratio be r
so $H + Hr + Hr^2= 49$
so $H(1+r+r^2) = 49$
r need not be integer it could be $\dfrac{p}{q}$ but let us try integer
so we get
H is a factor of 49 so H = 1 or 7
H =1 gives r irrational and H= 7 gives r = 2

so $AM = 28$ or $x+ y = 56$ and $xy =196$

solving these 2 as x is greater $x= 28+14\sqrt{3}$ and $y= 28- 14\sqrt{3}$
For r to be fraction h has to be a product of a sqaure that multyiple has to be factor of 49. So h has to be a square or 7 * square
trying $H = 9$ we get $G = 15$ and $A = 25$ and so $x = 45,y = 5$ as the second solution
very good !
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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