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Math Help - Difficult Proof

  1. #1
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    Difficult Proof

    Given that A*B = C*D, prove that \frac{D  B}{\frac{B + D}{2}} = \frac{C  A}{\frac{A + C}{2}}.

    Attempted solution:

    A*B = C*D

    \Rightarrow \frac{B}{D} = \frac{C}{A}

    \Rightarrow 1 - \frac{B}{D} = 1 - \frac{C}{A}

    \Rightarrow \frac{D}{D} - \frac{B}{D} = \frac{A}{A} - \frac{C}{A}

    \Rightarrow \frac{D - B}{D} = \frac{A - C}{A} ....

    Where do I go from here?
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  2. #2
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    Re: Difficult Proof

    Quote Originally Posted by RogueDemon View Post
    Given that A*B = C*D, prove that \frac{D  B}{\frac{B + D}{2}} = \frac{C  A}{\frac{A + C}{2}}.
    This is false, for example, for A = 3, B = 4, C = 2, D = 6. However, the conclusion is true if AD = BC, B + D ≠ 0 and A + C ≠ 0.
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  3. #3
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    Re: Difficult Proof

    Would the conclusion be true if (D - B) and (C - A) were absolute values? If so, would it be possible to prove this?
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  4. #4
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    Re: Difficult Proof

    Interesting. Yes, if AB = CD and neither A = C = 0 nor B = D = 0, then \left|\frac{B-D}{B+D}\right|=\left|\frac{A-C}{A+C}\right|. Under these assumptions, B = 0 iff C = 0 and A = 0 iff D = 0. Check what happens when B = 0 and C = 0, or when A = 0 and D = 0. The remaining case is when neither of the numbers equals 0. Then AB = CD implies A / C = D / B. Let's call this number x. We have

    \left|\frac{B-D}{B+D}\right|=\left|\frac{A-C}{A+C}\right| iff

    \left|\frac{1-D/B}{1+D/B}\right|=\left|\frac{1-C/A}{1+C/A}\right| iff

    \left|\frac{1-x}{1+x}\right|=\left|\frac{1-1/x}{1+1/x}\right| iff

    \left|\frac{1-x}{1+x}\right|=\left|\frac{x-1}{x+1}\right|

    which is true.
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