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Thread: Ratio

  1. #1
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    Ratio

    Hello all,

    I need some help on this problem:

    If $\displaystyle \frac{y+z}{pb+qc} = \frac{z+x}{pc+qa} = \frac{x+y}{pa+qb} $, then show that $\displaystyle \frac{2(x+y+z)}{a+b+c} = \frac{(b+c)x + (c+a)y + (a+b)z}{bc+ ca + ab} $
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  2. #2
    Super Member malaygoel's Avatar
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    Hi!
    You can easily get it by using the property
    if a/b = c/d
    then a/b = c/d = (a+c)/(b+d)

    Try it. If you can't get it, tell me.

    Keep Smiling
    Malay
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  3. #3
    Newbie
    Joined
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    I tried ...

    The sum of $\displaystyle \frac{y+z}{pb+qc} = \frac{z+x}{pc+qa} =\frac{x+y}{pa+qb} = \frac{2(x+y+z)}{(p+q)(a+b+c)} $

    It is close to the proof except for $\displaystyle (p+b) $ in the denominator. What should I do next? It can disappear if I assume $\displaystyle (p+b) = 1 $
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  4. #4
    Super Member malaygoel's Avatar
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    $\displaystyle \frac{y+z}{pq+bc}=\frac{ay+az}{pab+qac}$
    $\displaystyle \frac{z+x}{pc+qa}=\frac{bz+bx}{pbc+qab}$
    $\displaystyle \frac{x+y}{pa+qb}=\frac{cx+cy}{pac+qbc}$
    add thse three equations
    $\displaystyle \frac{y+z}{pq+bc}+\frac{z+x}{pc+qa}+\frac{x+y}{pa+ qb}=\frac{ay+az}{pab+qac}+\frac{bz+bx}{pbc+qab}+\f rac{cx+cy}{pac+qbc}$
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