Math Help Forum - View Single Post - In which of the case(s) is the triangle right angled?

red_dog · #2 July 30th, 2008, 11:03 AM

A) $r_a+r_b=r_c-r\Leftrightarrow\frac{S}{p-a}+\frac{S}{p-b}=\frac{S}{p-c}-\frac{S}{p}\Leftrightarrow \frac{2p-a-b}{(p-a)(p-b)}=\frac{c}{p(p-c)}\Leftrightarrow$
$\Leftrightarrow\frac{1}{(p-a)(p-b)}=\frac{1}{p(p-c)}\Leftrightarrow (p-a)(p-b)=p(p-c)\Leftrightarrow$
$\Leftrightarrow p=\frac{ab}{a+b-c}\Leftrightarrow \frac{a+b+c}{2}=\frac{ab}{a+b-c}\Leftrightarrow$
$\Leftrightarrow a^2+b^2=c^2\Leftrightarrow \Delta ABC$ is right angled.

(p is the semiperimeter)

B) $a^2+b^2+c^2=8R^2\Rightarrow 4R^2(\sin^2A+\sin^2B+\sin^2C)=8R^2\Rightarrow$
$\Rightarrow\sin^2A+\sin^2B+\sin^2C=2\Rightarrow$

$\Rightarrow\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos 2C}{2}=2\Rightarrow 1+\cos 2A+\cos 2B+\cos 2C=0\Rightarrow$

$\Rightarrow 2\cos^2A+2\cos(B+C)\cos (B-C)=0\Rightarrow 2\cos A\cos B\cos C=0$

Then, if $\cos A=0\Rightarrow A=\frac{\pi}{2}$
if $\cos B=0\Rightarrow B=\frac{\pi}{2}$
if $\cos C=0\Rightarrow C=\frac{\pi}{2}$

C) $2r_a=2p\Rightarrow \frac{2S}{p-a}=2p\Rightarrow\frac{\sqrt{p(p-a)(p-b)(p-c)}}{p-a}=p\Rightarrow$

$\Rightarrow\sqrt{\frac{p(p-b)(p-c)}{p-a}}=p\Rightarrow\frac{p(p-b)(p-c)}{p-a}=p^2\Rightarrow$

$\Rightarrow(p-b)(p-c)=p(p-a)\Rightarrow b^2+c^2=a^2$ (see A)

D) $2R=r_a-r\Rightarrow 2R=\frac{S}{p-a}-\frac{S}{p}\Rightarrow\frac{abc}{2S}=\frac{aS}{p(p-a)}\Rightarrow$

$\Rightarrow bcp(p-a)=2S^2\Rightarrow bcp(p-a)=2p(p-a)(p-b)(p-c)\Rightarrow$

$\Rightarrow bc=2(p-b)(p-c)\Rightarrow b^2+c^2=a^2$

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