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Math Help - Problemo

  1. #1
    Senior Member OReilly's Avatar
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    Problemo

    I need help to solve this problem.

    If (ay - bx):c = (cx - az):b = (bz - cy):a then prove that
    \frac{x}{a} = \frac{y}{b} = \frac{z}{c}.

    a,b,c \ne 0
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  2. #2
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    Hello, OReilly!

    Here's a start . . .


    (But I sincerely hope there's a better method!)

    If . \frac{ay - bx}{c} \,= \,\frac{cx - az}{b} \,= \,\frac{bz - cy}{a}, then prove that: . \frac{x}{a} = \frac{y}{b} = \frac{z}{c}\;\;\;(a,b,c \ne 0)
    . . . . .[1] . . . . . . . [2] . . . . . . .[3]

    From [1] = [3], we have: . \frac{ay-bx}{c} \:= \:\frac{bz-cy}{a}\quad\Rightarrow\quad a^2y -abx \:=\:bcz - c^2y

    . . Solve for x:\;\;x\:=\:\frac{a^2y + c^2y - bcz}{ab} . (a)


    From [2] = [3], we have: . \frac{cx-az}{b} \:= \:\frac{bz-cy}{a}\quad\Rightarrow\quad acx -a^2z \:= \:b^2z - bcy

    . . Solve for x:\;\;x\:=\:\frac{a^2z + b^2z - bcy}{ac} .(b)


    Equate (a) and (b): . \frac{a^2y + c^2y - bcz}{ab} \:= \:\frac{a^2z + b^2z - bcy}{ac}

    . . and we have: . a^2cy + c^3y - bc^2z \:= \:a^2bz + b^3z - b^2cy

    Rearrange terms: . a^2cy + b^2cy + c^3y \:= \:a^2bz + b^3z + bc^2z

    Factor: . cy(a^2+b^2+c^2) \:=\: bz(a^2+b^2+c^2)

    Divide by (a^2+b^2+c^2):\;\;cy\,= \,bz\quad\Rightarrow\quad\boxed{ \frac{y}{b} \,= \,\frac{z}{c} }

    I'll let you work on the other equality . . .

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  3. #3
    Senior Member OReilly's Avatar
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    Ye, not very ellegant solution, but it works.
    Thanks!
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