In the diagram, four circles of radius 1 with centres $P$, $Q$, $R$, and $S$ are tangent to one another and to the sides of $\triangle ABC$, as shown. [asy]
size(250);
pair A, B, C, P, Q, R, S;
R=(0,0);
Q=(-2,0);
S=(2,0);
P=(1,1.732);
B=(-5.73,-1);
C=(3.732,-1);
A=(1.366,3.098);
draw(A--B--C--A);
draw(circle(P, 1));
draw(circle(Q, 1));
draw(circle(R, 1));
draw(circle(S, 1));
label("A", A, N);
label("B", B, SW);
label("C", C, SE);
dot(P);
dot(Q);
dot(R);
dot(S);
label("P", P, N);
label("Q", Q, SW);
label("R", R, SW);
label("S", S, SE);
[/asy]

The radius of the circle with center $R$ is decreased so that

$\bullet$ the circle with center $R$ remains tangent to $BC$,

$\bullet$ the circle with center $R$ remains tangent to the other three circles, and

$\bullet$ the circle with center $P$ becomes tangent to the other three circles.

The radii and tangencies of the other three circles stay the same. This changes the size and shape of $\triangle ABC$. $r$ is the new radius of the circle with center $R$. $r$ is of the form $\frac{a+\sqrt{b}}{c}$. Find $a+b+c$.
After the described transformation, we obtain the following diagram. [asy]
size(250);
pair A, B, C, P, Q, R, S;
int x;
x=4;
B=(-.45,0);
Q=(2.414,1);
R=(x,.618);
C=(2x+.45,0);
S=(2x-2.414,1);
P=(x,2.236);
A=(x,3.55);
draw(A--B--C--A);
draw(circle(Q,1));
draw(circle(S,1));
draw(circle(R,.618));
draw(circle(P,1));
label("A", A, N);
label("B", B, SW);
label("C", C, SE);
label("R", R, dir(270));
label("Q", Q, SW);
label("S", S, SE);
label("P", P, N);
dot(Q);
dot(P);
dot(S);
dot(R);
[/asy]

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


[asy]
size(250);
pair P, Q, R, S, B, C, D, E, F, Y;
P=(4,2.236);
Q=(2.414,1);
R=(4,.618);
S=(5.586,1);
B=(0,0);
C=(8,0);
D=(2.414,0);
E=(4,0);
F=(5.586,0);
Y=(4,1);
draw(circle(P,1));
draw(circle(Q,1));
draw(circle(S,1));
draw(circle(R,.618));
draw(B--C);
draw(Q--D);
draw(P--E);
draw(F--S);
draw(Q--S);
draw(Q--P--S--R--Q);
label("D", D, dir(270));
label("E", E, dir(270));
label("F", F, dir(270));
label("P", P, N);
label("Q", Q, NW);
label("S", S, NE);
label("R", R, SW);
label("Y", Y, NW);
dot(P);
dot(Q);
dot(R);
dot(S);
[/asy]

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 $\triangle QYR$ is right-angled at $Y$, then by the Pythagorean Theorem,
\begin{align*}
QY^2 &= QR^2 - YR^2\\
&= (1+r)^2 - (1-r)^2 \\
&= (1+2r+r^2)-(1-2r+r^2) \\
&= 4r.
\end{align*}Since $PR=1+r$ and $YR=1-r$, then $PY = PR-YR=2r$. Since $\triangle PYQ$ is right-angled at $Y$,
\begin{align*}
PY^2 + YQ^2 & = PQ^2 \\
(2r)^2 + 4r & = 2^2 \\
4r^2 + 4r & =4 \\
r^2 + r - 1 & = 0.
\end{align*}By the quadratic formula, $r = \dfrac{-1\pm \sqrt{1^2-4(1)(-1)}}{2} = \dfrac{-1\pm\sqrt{5}}{2}$.
Since $r>0$, then $r = \dfrac{-1+\sqrt{5}}{2}$ (which is the reciprocal of the famous ``golden ratio"). Thus, $a+b+c=-1+5+2=\boxed{6}$.