Six congruent semicircles are inscribed in a regular hexagon of side length $1$ so that their diameters are on the sides of the hexagon, one endpoint of each diameter is at a vertex of the hexagon, and adjacent semicircles are tangent to each other. A middle circle centered at the center of the hexagon is tangent to each of the six semicircles, as shown below. The diameter of the middle circle can be written as $\frac{\sqrt{a}-b}c$, where $a$, $b$ and $c$ are integers, $c$ doesn't have square factor. What is $a + b + c $?

1. Let the hexagon be $V_0V_1\cdots V_5$ with side $1$ and let each semicircle have radius $r$. The semicircle on $V_0V_1$ has center $C_0$ on $V_0V_1$ with $V_0C_0=r$, and the one on $V_1V_2$ has center $C_1$ with $V_1C_1=r$. So $C_0V_1=1-r$ and $\angle C_0V_1C_1=120^\circ$.
2. By the law of cosines, $C_0C_1^2=(1-r)^2+r^2+r(1-r)=1-r+r^2$. The semicircles are tangent, so $C_0C_1=2r$, giving $1-r+r^2=4r^2$, i.e.\ $3r^2+r-1=0$ and $r=\dfrac{\sqrt{13}-1}6$. Note that this equation says $1-r=3r^2$.
3. Let $M$ be the midpoint of $V_0V_1$, so $OM\perp V_0V_1$ with $OM=\frac{\sqrt3}2$ and $MC_0=\left|\frac12-r\right|$. In right triangle $OMC_0$,
   $$OC_0^2=\frac34+\left(\frac12-r\right)^2=1-r+r^2=3r^2+r^2=4r^2,$$
   so $OC_0=2r$.
4. The middle circle (radius $\rho$, center $O$) is externally tangent to the semicircle with center $C_0$ and radius $r$, so $OC_0=\rho+r$. Hence $\rho=2r-r=r$.
5. The diameter is $2\rho=2r=\dfrac{\sqrt{13}-1}3$, so $a=13$, $b=1$, $c=3$ ($3$ is square-free), and $a+b+c=\boxed{17}$.