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Thread: One more interesting identity

  1. #1
    Junior Member
    Joined
    Dec 2009
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    One more interesting identity

    Prove that in every triangle holds
    $$a\cot\alpha+b\cot\beta+c\cot\gamma=2(R+r)$$
    where $a$, $b$, $c$, are edges, $\alpha$, $\beta$, $\gamma$ interior angles, and $R$ and $r$ radiuses of circumscribed and inscribed circle in triangle.
    Solution:
    By sinus theorem, we have
    $$a\cot\alpha+b\cot\beta+c\cot\gamma=2R(\cos\alpha +\cos\beta+\cos\gamma=2R(1+4\sin{\frac{\alpha}{2}} \sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$$
    Now, by cosinus theorem

    $$2{\sin{\frac{\alpha}{2}}^2=1-\frac{b^2+c^2-a^2}{2bc}=\dots=\frac{2(s-b)(s-c)}{bc}$$

    So, $\sin\frac{\alpha}{2}=\sqrt\frac{(s-b)(s-c)}{bc}$, and similarly, $\sin\frac{\beta}={2}=sqrt\frac{(s-a)(s-c)}{ac}$ and $\sin\frac{\gamma}{2}=\sqrt\frac{(s-a)(s-b)}{ab}$.
    The rest follow easy from Heron's formula, $P^2=s(s-a)(s-b)(s-c)$, and formulas $abc=4PR$ and $P=rs$.
    Finaly
    $$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac {\gamma}{2}=\frac{r}{4R}$$
    which is desired.
    Last edited by ns1954; Feb 25th 2019 at 09:51 AM.
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