A geometrical enigma

s= $\sqrt2$ r and a= $4r \sin60=2\sqrt3 r$ Hence required ratio= $\dfrac{s^2}{\dfrac{\sqrt3 a^2}{4}}$ $= \dfrac{2r^2}{3\sqrt3}=\dfrac{\sqrt{12}}{9}$ $\Large 12+9=\boxed{\color{#D61F06}{\boxed{21}}}$

$\text{Area of regular polygon of n sides in an unit circumcircle } =\frac12*n*(1^2)*Sin\dfrac{2\pi} n.\\ In ~ equilateral ~ \Delta, ~ Circumradius=2*Inradius. ~ But ~ in ~ the ~ \Delta, ~ the ~ circle ~ is ~ Incircle. \\ \therefore~Circumradius=2*Inradius=2*1=2 \text{ is the circumradius for the } \Delta.\\ Required ~ ratio ~\Large =\dfrac{ \frac12*4*(1^2)*Sin\dfrac{2\pi} 4}{\frac12*3*(2^2)*Sin\dfrac{2\pi} 3}=\dfrac{\sqrt{12}}9\\ =\dfrac a b. \therefore ~ a+b=21.$

Let $\text{ r to be the radius of the circle.}$ .

Observe now Rishabh' picture, then $\text{the side of the square inscribed in the circle} = \sqrt{2} \cdot r \Rightarrow \text{Surface of the square} = 2 \cdot r^2$

Let's call $l = \text{side of the equilateral triangle}$ and $s =\text{semiperimeter of the equilateral triangle}$ , then $\text{ Surface of the equilateral triangle} = \frac{1}{2}\cdot l^2\cdot \sin 60º = \frac{\sqrt{3}}{4} \cdot l^2 = r \cdot s = r \cdot \frac{3l}{2} \Rightarrow$ $l = \frac{6r}{ \sqrt{3}} \Rightarrow \text{Surface of the equilateral triangle} = 12r^2 \frac{\sqrt{3}}{4} = \sqrt{3}\cdot 3r^2$ $\frac{\text{Surface of the square}}{\text{Surface of the equilateral triangle}} = \frac{2r^2}{3\sqrt{3}r^2} = \frac{2\sqrt{3}}{9} = \frac{\sqrt{12}}{9}...$