hard problem

**pacman** · August 24th 2009, 03:52 AM

In a given triangle ABC, angle B - angle A = 90 degrees.
Show that 2/c^2 = 1/(b + a)^2 + 1/(b - a)^2

**red_dog** · August 24th 2009, 08:36 AM

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.

**Liwuinan** · August 24th 2009, 08:48 AM

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) $

$= \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) $

$ =\frac{1}{(b-a)^2}\left(\tan^2\left(\frac{\gamma}{2}\right) +1\right) =\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) $

$ =\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 $

$ = \frac{2}{(b-a)^2} \left(\frac{b-a}{c}\right)^2 = \frac{2}{c^2} $

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

**Opalg** · August 24th 2009, 09:00 AM

Originally Posted by pacman

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!

**pacman** · August 25th 2009, 06:53 AM

Thank you GUYS, especially Opalg's solution is very instructive indeed. Now i have 3 perspective of equal value!!!

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