Circles inside an Equilateral Triangle

Let us assign the verticies of the equalateral triangle $A, B$ and $C$ .

$1)$ We can see that the first circle $C_1$ with origin $O_1$ has a radius of $\tan(30) = \frac{\sqrt{3}}{3}$

$2)$ To find the radius $r_1$ of one of the three circles $C_2$ in the corner, we can draw infinitely many circles, each with a common ratio for their radius, toward $C$ . Let us call this common ratio $R$ . So $\frac{\sqrt{3}}{3} + \frac{2\sqrt{3}}{3}R + \frac{2\sqrt{3}}{3}R^2 + ... = O_1 C = \frac{1}{\cos(30)} = \frac{2\sqrt{3}}{3}$

Rearranging and dividing both sides by $\frac{\sqrt{3}}{3}$ we get the infinite sum $R + R^2+R^3+R^4 ... = \frac{1}{2}$

$\frac{R}{1-R} = \frac{1}{2}$ and $R = \frac{1}{3}$

So $r_1 = \frac{\sqrt{3}}{3} * \frac{1}{3} = \frac{\sqrt{3}}{9}$

$3)$ The next question we have is: Where is the other circle. If it was in the corner again (Tangent to two sides of the triangle and to $C_2$ ), then we can see it has a radius of $\frac{\sqrt{3}}{9} *\frac{1}{3} = \frac{\sqrt{3}}{27} \approx 0.0642$ . Now we have to find the radius of the circle $C_3$ and radius $r_3$ , tangent to $C_1, C_2$ and one side of the triangle and compare with $\frac{\sqrt{3}}{27}$ .

For this we can use the equality $\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}$ where $x,y$ are the radii of $C_1, C_2$ and $z$ is the radius of $C_3$ . This can be easily proved geometrically.

$\frac{1}{\sqrt{\frac{\sqrt{3}}{3}}}+\frac{1}{\sqrt{\frac{\sqrt{3}}{9}}} = \frac{1}{\sqrt{r_3}}$

Rearrange to get $r_3 = \frac{-3+2\sqrt{3}}{6} \approx 0.0774$

We can now see that this is the larger of the two circles we were comparing and hence the area is $\pi {r_3}^2 = \pi (\frac{-3+2\sqrt{3}}{6})^2 = \frac{7-4\sqrt{3}}{12}\pi$

Therefore $a+b+c+d = \boxed{26}$