Circle : Triangle

Marvin has given a good solution. I would like to provide an alternate solution that would be time efficient . Let the side length of $\triangle ABC$ = $a$ units

Simply draw a common tangent $DE$ to both circles that touches them at $F$ . We observe that $DE \parallel BC$ .

Though I will not go into the rigour of proving established theorems for what I am about to type, you can do so for your own clarity: we observe that $\triangle ADE$ is also an equilateral triangle and $AGFOI$ is a straight line.

Also observe that the bigger circle is the in-circle of the equilateral $\triangle ABC$ . This makes $O$ the centroid and $AO$ the radius of the circumcircle, if it were drawn.

Hence, radius of bigger circle = $\frac{a}{2\sqrt3} = OF = OI$

$AO = \frac{a}{\sqrt3}$

Now, $AF = AO - OF = \frac{a}{\sqrt3} - \frac{a}{2\sqrt3} = \frac{a}{2\sqrt3}$

Therefore, side length of equilateral $\triangle ADE$ : $AD = DE = EA = \frac{a}{3}$

Observe that the smaller circle is now the in-circle of equilateral $\triangle ADE$ . So, radius of smaller circle = $\frac{a}{6\sqrt3}$

Thus, $\frac{\text{Area of smaller circle}}{\text{Area of equilateral triangle ABC}}$ = $\frac{\pi a^{2}}{(6\sqrt3)^{2}}$ x $\frac{4}{\sqrt3 a^{2}}$

= $\frac{\pi}{27\sqrt3}$

Let the side length of the equilateral triangle be $6$ , then by pythagorean theorem, $AF=\sqrt{6^2-3^2}=\sqrt{27}=3\sqrt{3}$ .

Since $\triangle AJG \sim \triangle AFB$ , we have $\dfrac{JG}{AJ}=\dfrac{FB}{AF}$ $\implies$ $\dfrac{JG}{3}=\dfrac{3}{3\sqrt{3}}$ $\implies$ $JG=\sqrt{3}=GF$ .

$HF=JG+GF=\sqrt{3}+\sqrt{3}=2\sqrt{3}$

$AH=AF-HF=3\sqrt{3}-2\sqrt{3}=\sqrt{3}$

Let $KI=HI=r$ .

$AI=AH-r=\sqrt{3}-r$

Since $\triangle AKI \sim \triangle AJG$ , we have

$\dfrac{r}{JG}=\dfrac{AI}{AG}$ $\implies$ $\dfrac{r}{\sqrt{3}}=\dfrac{\sqrt{3}-r}{2\sqrt{3}}$ $\implies$ $2\sqrt{3}r=3-\sqrt{3}r$ $\implies$ $3\sqrt{3}r=3$ $\implies$ $r=\dfrac{1}{\sqrt{3}}$

So the area of the small circle is $\pi r^2=\pi \left(\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{\pi}{3}$ and the area of the equilateral triangle is $\dfrac{\sqrt{3}}{4}(6^2)=9\sqrt{3}$ .

The desired ratio is $\dfrac{\dfrac{\pi}{3}}{9\sqrt{3}}=\boxed{\dfrac{\pi}{27\sqrt{3}}}$ .