Question

In $△ABC$ sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. Then the maximum value of $\frac{a^2+b^2+c^2}{R^2}=$
(where $R$ is circumradius)

Answer 1

Using sine rule :
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\Rightarrow a=2R\sin A,b=2R\sin B,c=2R\sin C$
$\Rightarrow \frac{a^2+b^2+c^2}{R^2}=\frac{4R^2({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)}{R^2}=4({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)$
Now, the value is maximum for equilateral triangle
$\Rightarrow \angle A=\angle B=\angle C={60}^{\circ }$
So, maximum value $=4\times 3\times \frac{3}{4}=9$