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Home » Mathematics » High School Pre-Olympiad (Ages 14+)
 
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Elio (n)
Riemann Hypothesis
Riemann Hypothesis

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#1
system
Find the real solutions of the following system :
ax+by=(x-y)^{2}
by+cz=(y-z)^{2}
cz+ax=(z-x)^{2}
where a,b,c are positive reals.
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Elio Nushi - Newton is my avatar and Einstein too
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PostPosted: Sat Nov 07, 2009 12:15 pm  Back to top 
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math154
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer


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#2
Solution
By adding any two equations and subtracting the other, we arrive at
a=\frac{(x-y)(x-z)}x>0\\ b=\frac{(y-x)(y-z)}y>0\\ c=\frac{(z-x)(z-y)}z>0.
Case 1: (x-y)(y-z)(z-x)\ne0. Without loss of generality, assume that x>y>z; then x-y,x-z,y-z>0 so we conclude that x>z>0>y, which is a contradiction.
Case 2: (x-y)(y-z)(z-x)=0. Without loss of generality, assume that x-y=0\implies a,b=0\not>0, which is a contradiction.

PostPosted: Sat Nov 07, 2009 1:14 pm  Back to top 
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Elio (n)
Riemann Hypothesis
Riemann Hypothesis

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#3
HI

How did you get x>z>0>y
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Elio Nushi - Newton is my avatar and Einstein too
"Change is coming in America" Barack Obama

PostPosted: Sun Nov 08, 2009 12:49 am  Back to top 
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math154
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer


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#4
Sorry for not explaining more; I was rushed.

  • x - y,x - z > 0 so for a to be positive, we must have x positive.
  • y - x < 0 and y - z > 0, so for b to be positive, we must have y negative.
  • z - x,z - y < 0 so to for c to be be positive, we must have z positive.

By assumption, x > z, and because z is positive while y is negative, x > z > 0 > y.

PostPosted: Sun Nov 08, 2009 7:16 pm  Back to top 
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