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Math Help - hard problem

  1. #1
    Senior Member pacman's Avatar
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    hard problem

    In a given triangle ABC, angle B - angle A = 90 degrees.
    Show that 2/c^2 = 1/(b + a)^2 + 1/(b - a)^2
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  2. #2
    MHF Contributor red_dog's Avatar
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    Use the sine law.

    \frac{2}{c^2}=\frac{1}{(b+a)^2}+\frac{1}{(b-a)^2}\Leftrightarrow

    \Leftrightarrow\frac{2}{4R^2\sin^2C}=\frac{1}{4R^2  (\sin B+\sin A)^2}+\frac{1}{4R^2(\sin B-\sin A)^2}\Leftrightarrow

    \Leftrightarrow\frac{2}{\sin^2C}=\frac{1}{4\sin^2\  frac{B+A}{2}\cos^2\frac{B-A}{2}}+\frac{1}{4\sin^2\frac{B-A}{2}\cos^2\frac{B+A}{2}}\Leftrightarrow

    \Leftrightarrow\frac{2}{\sin^2C}=\frac{1}{2\cos^2\  frac{C}{2}}+\frac{1}{2\sin^2\frac{C}{2}}\Leftright  arrow\frac{2}{\sin^2C}=\frac{1}{2\sin^2\frac{C}{2}  \cos^2\frac{C}{2}}\Leftrightarrow

    \Leftrightarrow\sin^2C=4\sin^2\frac{C}{2}\cos^2\fr  ac{C}{2}

    which is true.
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  3. #3
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    Hi,
    I'll write a solution that can definitely be simplified, I'm not happy it uses two other triangle identities, but they are well known and I included links to their proofs below.

    First, note that b-a \neq 0 (otherwise \beta-\alpha=0).

    Here we go:
    \frac{1}{(b+a)^2}+\frac{1}{(b-a)^2}=\frac{1}{(b-a)^2}\left(\left(\frac{b-a}{b+a}\right)^2+1\right) = \frac{1}{(b-a)^2}\left(\left(\frac{\tan(\frac{\beta-\alpha}{2})}{\tan(\frac{\beta+\alpha}{2})}\right)^  2+1\right)<br />

    = \frac{1}{(b-a)^2}\left(\left(\frac{\tan(\frac{\pi/2}{2})}{\tan(\frac{\pi-\gamma}{2})}\right)^2+1\right) = \frac{1}{(b-a)^2}\left(\frac{1}{\tan^2\left(\frac{\pi}{2}-\frac{\gamma}{2}\right)}+1\right)<br />

    <br />
 =\frac{1}{(b-a)^2}\left(\tan^2\left(\frac{\gamma}{2}\right) +1\right)<br />
=\frac{1}{(b-a)^2}\frac{1}{\cos^2(\frac{\gamma}{2})} \left(\sin^2\left(\frac{\gamma}{2}\right)+\cos^2\l  eft(\frac{\gamma}{2}\right)\right)<br />

    <br />
=\frac{1}{(b-a)^2}\frac{1}{\cos^2(\frac{\gamma}{2})} = \frac{2}{(b-a)^2}\left(\frac{\frac{\sqrt2}{2}}{\cos(\frac{\gam  ma}{2})}\right)^2 =  \frac{2}{(b-a)^2}\left(\frac{\sin\left(\frac{\beta-\alpha}{2}\right)}{\cos(\frac{\gamma}{2})}\right)^  2<br />

    <br />
= \frac{2}{(b-a)^2} \left(\frac{b-a}{c}\right)^2 = \frac{2}{c^2}<br />

    comments:
    '2nd =' is "Law of tangents", see e.g. Law of tangents - Wikipedia, the free encyclopedia
    '10th =' is Mollweide's formula, for proof see this thread http://www.mathhelpforum.com/math-he...s-formula.html
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  4. #4
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    Opalg's Avatar
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    Quote Originally Posted by pacman View Post
    In a given triangle ABC, angle B - angle A = 90 degrees.
    Show that 2/c^2 = 1/(b + a)^2 + 1/(b - a)^2
    Since I started to write this, two other solutions have appeared. But here's a third one anyway.

    First, notice that \frac1{(b+a)^2} + \frac1{(b-a)^2} = \frac{(b-a)^2 + (b+a)^2}{(b+a)^2(b-a)^2} = \frac{2(a^2+b^2)}{(b^2-a^2)^2}. So (turning the fractions upside down) what we are trying to prove is equivalent to c^2 = \frac{(b^2-a^2)^2}{(a^2+b^2)}.

    Next, the angles in the triangle are A, B=90º+A and C=90º–2A. Thus \sin B =\sin(90^\circ+A) = \cos A and \sin C = \sin(90^\circ-2A) = \cos (2A). The sine rule in the triangle then tells you that \frac{\sin A}a = \frac{\cos A}b = \frac{\cos(2A)}c = \lambda say. So \sin A = \lambda a, \cos A = \lambda b and \cos(2A) = \lambda c.

    But \sin^2A+\cos^2A = 1, so \lambda^2(a^2+b^2)=1. Also, \cos(2A) = \cos^2A-\sin^2A, so \lambda c = \lambda^2(b^2-a^2). Therefore c^2 = \lambda^2(b^2-a^2)^2 = \frac{(b^2-a^2)^2}{(a^2+b^2)}: just what we wanted!
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  5. #5
    Senior Member pacman's Avatar
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    Thank you GUYS, especially Opalg's solution is very instructive indeed. Now i have 3 perspective of equal value!!!

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