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

  1. #1
    Super Member fardeen_gen's Avatar
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    In which of the case(s) is the triangle right angled?

    In a triangle ABC if following relations hold good, in which of the case(s) the triangle is a right angled triangle?

    A) r2 + r3 = r1 - r
    B) a^2 + b^2 + c^2 = 8R^2
    C) if the diameter of an excircle be equal to the perimeter of the triangle
    D) 2R = r1 - r

    Note: r - inradius
    R - circumradius
    1, 2 ,3 are subscripts denoting different inradiuses
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  2. #2
    MHF Contributor red_dog's Avatar
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    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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