jmc

After the described transformation, we obtain the following diagram.

Drop perpendiculars from $Q$, $R$ and $S$ to $D$, $E$ and $F$ respectively on $BC$.



Since the circles with centers $Q$, $R$ and $S$ are tangent to $BC$, then $D$, $E$ and $F$ are the points of tangency of these circles to $BC$. Thus, $QD=SF=1$ and $RE=r$.

Join $QR$, $RS$, $PS$, $PQ$, and $PR$. Since we are connecting centers of tangent circles, $PQ=PS=2$ and $QR=RS=PR=1+r$.

Join $QS$. By symmetry, $PRE$ is a straight line (that is, $PE$ passes through $R$). Since $QS$ is parallel to $BC$, $QS$ is perpendicular to $PR$, meeting at $Y$.

Since $QD=1$, $YE=1$. Since $RE=r$, $YR=1-r$. Since $QR=1+r$, $YR=1-r$ and $△QYR$ is right-angled at $Y$, then by the Pythagorean Theorem, $$\begin{array}{rl}Q{Y}^2& =Q{R}^2-Y{R}^2\\ & =(1+r{)}^2-(1-r{)}^2\\ & =(1+2r+r^2)-(1-2r+r^2)\\ & =4r.\end{array}$$Since $PR=1+r$ and $YR=1-r$, then $PY=PR-YR=2r$. Since $△PYQ$ is right-angled at $Y$, $$\begin{array}{rl}P{Y}^2+Y{Q}^2& =P{Q}^2\\ (2r{)}^2+4r& =2^2\\ 4r^2+4r& =4\\ r^2+r-1& =0.\end{array}$$By the quadratic formula, $r=\frac{-1\pm \sqrt{1^2-4(1)(-1)}}{2}=\frac{-1\pm \sqrt{5}}{2}$. Since $r>0$, then $r=\frac{-1+\sqrt{5}}{2}$ (which is the reciprocal of the famous ``golden ratio"). Thus, $a+b+c=-1+5+2=\boxed{6}$.