One more interesting identity

Dec 2009
54
0
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.
 
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