it's only lines and circles

Let the radius of the circles be $r$ . Then we note that the center of the top circle is on the median of equilateral $\triangle ABC$ . Therefore we have:

$\begin{aligned} \frac r{\tan 30^\circ} + r + r & = \frac 12 \\ \sqrt 3 r + 2 r & = \frac 12 \\ \implies r & = \frac 1{2(2+\sqrt 3} = \frac {2-\sqrt 3}2 \\ AD & = \frac 12 - r = \frac 12 - \frac {2-\sqrt 3}2 = \frac {\sqrt 3-1}2 \end{aligned}$

Therefore $a+b+c = 3-1+2 = \boxed 4$ .

Let the radius of each circle be $r$ . Then

$r(\sqrt 3+1+1)=\dfrac 12$ (since the length of each side of the triangle is $1$ )

$\implies r=\dfrac {1}{4+2\sqrt 3}=\dfrac {1}{(\sqrt 3+1)^2}$

So,

$|\overline {AD}|=r(\sqrt 3+1)=\dfrac {1}{\sqrt 3+1}$

$=\dfrac {\sqrt 3-1}{2}$

Hence, $a=3,b=-1,c=2$ and

$a+b+c=3-1+2=\boxed 4$ .