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   Subtract 2 From Product, Get Sum
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   Author  Topic: Subtract 2 From Product, Get Sum  (Read 383 times)
K Sengupta
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Subtract 2 From Product, Get Sum  
« on: Jul 27th, 2007, 8:04am »
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(A) Analytically determine all possible positive real triplets (x, y, z) satisfying the following system of equations:
 
xy + yz + zx = 12
xyz = 2 + x+ y+ z
 
(B) If the restriction “positive”  as given in  (A) is obviated, and it is known that  x, y and z are distinct real numbers, determine if  the system of equations in terms of (A) still admit of valid solutions.
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Eigenray
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Re: Subtract 2 From Product, Get Sum  
« Reply #1 on: Aug 5th, 2007, 4:01am »
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hidden:
x,y,z are roots of the cubic
 
t3 - st2 + 12t - (2+s) = 0,
 
which has discriminant
 
D = 4s3(2+s) - 122s2 + 4*123 - 18*s*12(2+s) + 27(2+s)2
 = (s-6)2(2s+13)(2s+15).

 
(A) If the roots are all postive and real, then s>0 and D 0.  This is true only for s=6, and x=y=z=2.  (There is also a nice solution using AM-GM similar to this one.)
 
(B) There are 3 distinct real roots when D < 0, which happens for -15/2 < s < -13/2.
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K Sengupta
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Re: Subtract 2 From Product, Get Sum  
« Reply #2 on: Aug 5th, 2007, 7:27am »
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Thanks Eigenray. At the time of posting the problem, the only solution to Part A  I  was aware of was by due utilisation of the AM-GM method.
 
However, I drove myself in circles on Part B before your brilliant methodology, which is simple yet superb,   culminating in  a comprehensive solution to the given problem.
« Last Edit: Aug 5th, 2007, 7:27am by K Sengupta » IP Logged
Eigenray
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Re: Subtract 2 From Product, Get Sum  
« Reply #3 on: Aug 5th, 2007, 9:10am »
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Now, what about all rational solutions?
 
-D must be a perfect square, which means t(t+2) is a perfect square, where t=2s+13, so t = -2u2/(u2+v2) for some integers u,v.  However, we only get 3 rational roots when -s = 187/26, 1177/170, 28033/3770, 41767/6290, 94567/12818, ..., so this doesn't seem too helpful.
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