Troublesome Triangles

We know that $\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} = 2R$ . Plugging in values we get

$\displaystyle R(\sin{2A} + \sin{2B} + \sin{2C}) \over \displaystyle a \times \frac{b}{2R} + b\times \frac{c}{2R} + c\times \frac{a}{2R}$ $= \displaystyle \frac{a+b+c}{9R}$ .

Rearragging we get ( Using identity $\sin{2A} + \sin{2B} + \sin{2C} = 4\sin{A}\sin{B}\sin{C}$ where $A+B+C = \pi$ )

$\rightarrow 8R^{2}(\sin{A}\sin{B}\sin{C}) = \displaystyle \frac{(a+b+c)(ab+bc+ca)}{9R}$ .

Again using $\displaystyle \frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} = 2R$

$\rightarrow \displaystyle 8R^{2}\times \frac{a}{2R} \times \frac{b}{2R} \times \frac{c}{2R} = \displaystyle \frac{(a+b+c)(ab+bc+ca)}{9R}$

On expanding we get

$\rightarrow (ba^{2} + bc^{2} - 2abc) + (ab^{2} + ac^{2} - 2abc) + (ca^{2} + cb^{2} - 2abc) = 0$

$\rightarrow b(a-c)^{2} + a(b-c)^{2} + c(a-b)^{2} = 0$

This is only possible when $a=b=c$ . So the triangle is $\boxed{equilateral}$

From the symmetrical look of the equation we are tempted to think it to be Equilateral Triangle.
$Let\ S=a=b=c, \ \ A+B+C=60^o,\ \ 2R=\dfrac S {Sin60},\ \ \implies\ R=\dfrac S {\sqrt3}.\\ So\ \ L.H.S.=\dfrac{3*S*\frac 1 2}{3*S*Sin60}=\dfrac 1 {\sqrt3}.\\ R.H.S.=\dfrac{3*S}{ 9* \frac S {\sqrt3} }=\dfrac 1{\frac 3 {\sqrt3}}=\dfrac 1 {\sqrt3}=L.H.S.\\ Thus\ \ it\ \ IS\ \ an \ \ \color{#D61F06}{Equilateral Triangle.}$