Circles and triangles

Draw the radii and label the diagram as follows:

Then place the diagram on a coordinate plane so that $C$ is at $(0, 0)$ , $A$ is at $(4, 0)$ , and $B$ is at $(0, 3)$ .

Then $D$ , $E$ , and $F$ are midpoints of each side of the triangle, so $D$ is at $(0, \frac{3}{2})$ , $E$ is at $(2, 0)$ , and $F$ is at $(2, \frac{3}{2})$ .

Let $O$ be at $(p, q)$ , and let the radius of the black circle be $r$ . Then by multiple uses of the distance equation:

$OG = OD + DG = \sqrt{p^2 + (q - \frac{3}{2})^2} + \frac{3}{2} = r$

$OH = OE + EH = \sqrt{(p - 2)^2 + q^2} + 2 = r$

$OI = OF + FI = \sqrt{(p - 2)^2 + (q - \frac{3}{2})^2} + \frac{5}{2} = r$

and these three equations solve to $p = \frac{36}{23}$ , $q = \frac{24}{23}$ , and $r = \frac{72}{23}$ , so $a = 72$ , $b = 23$ , and $a + b = \boxed{95}$ .

Let the vertices of the triangle be $A(0,3)$ , $B(4,0)$ , and $C(0,0)$ . The midpoints of the triangle sides are $D(0,1.5)$ , $E(2,1.5)$ , and $F(2,0)$ . Let the center of the black circle be $O(x,y)$ and its radius be $r$ , and the tangent points of circles be $P$ , $Q$ , and $R$ as shown.

We note that $P$ , $D$ , and $O$ are colinear and that $OP = r$ and $DP = 1.5$ , hence $OD = r-1.5$ . Using the coordinates of $D(0, 1.5)$ and $O(x,y)$ and Pythagorean theorem , we have $x^2 + (y-1.5)^2 = (r-1.5)^2$ . Similarly for $OR$ and $OQ$ , and we get the following three equations.

$\begin{cases} \begin{aligned} x^2 + \left(y-\frac 32 \right)^2 & = \left(r-\frac 32 \right)^2 & ...(1) \\ (x-2)^2 + y^2 & = (r-2)^2 & ...(2) \\ (x-2)^2 + \left(y-\frac 32\right)^2 & = \left(r-\frac 52 \right)^2 & ...(3) \end{aligned} \end{cases}$

From $(1)-(3)$ :

$\begin{aligned} x^2 - (x-2)^2 & = \left(r-\frac 32 \right)^2 - \left(r-\frac 52 \right)^2 \\ 4x - 4 & = 2r - 4 \\ \implies x & = \frac r2 \end{aligned}$

From $(2)-(3)$ :

$\begin{aligned} y^2 - \left(y-\frac 32\right)^2 & = (r-2)^2 - \left(r-\frac 52 \right)^2 \\ 3y - \frac 94 & = r - \frac 94 \\ \implies y & = \frac r3 \end{aligned}$

From $(1)+(2)-(3)$ :

$\begin{aligned} x^2 + y^2 & = r^2 - 2r \\ \frac {r^2}4 + \frac {r^2}9 & = r(r-2) \\ \frac r4 + \frac r9 & = r -2 \\ \left(1-\frac 14 - \frac 19\right) r & = 2 \\ \implies r & = \frac {72}{23} \end{aligned}$

Therefore $a+b = 72 + 23 = \boxed{95}$ .

We put the triangle on a coordinate plane, so that the vertices are $A\left( 0,\ 0 \right)$ , $B\left( 0,\ 3 \right)$ and $C\left( 4,\ 0 \right)$ . Then, the midpoints of $BC$ , $CA$ and $AB$ are $D\left( 2,\ 1.5 \right)$ , $E\left( 2,\ 0 \right)$ and $F\left( 0,\ 1.5 \right)$ respectively (see figure). Denote the center of the black circle and its contact points with the red, blue and green circle by $K$ , $H$ , $I$ and $J$ respectively.

Since the black and the red circles have a common tangent line at $H$ , both $HF$ and $HK$ are perpendicular to this line. Hence, $H$ , $F$ and $K$ are collinear. Likewise, $J$ , $D$ and $K$ are collinear, as well as $I$ , $E$ and $K$ . Let the radius of the black circle be $R$ .

Then,

$KF-KD=\left( KH-FH \right)-\left( KJ-DJ \right)=\left( R-1.5 \right)-\left( R-2.5 \right)=1$ This means that $K$ lies on a hyperbola, who’s foci are $F$ , $D$ and with constant difference of distances $1$ . In fact, $K$ lies on the branch that is closer to focus $D$ .

Any arbitrary point $M\left( x,\ y \right)$ on this branch satisfies the equation

$\begin{aligned} MF-MD=1 & \Leftrightarrow \sqrt{{{\left( x-0 \right)}^{2}}+{{\left( y-1.5 \right)}^{2}}}-\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-1.5 \right)}^{2}}}=1 \\ & \Leftrightarrow 3{{x}^{2}}-{{y}^{2}}-6x+3y=0,\text{ }x>1 \ \ \ \ \ (1)\\ \end{aligned}$ Similarly, $KE-KD=\left( KI-EI \right)-\left( KJ-DJ \right)=\left( R-2 \right)-\left( R-2.5 \right)=0.5$ .

Hence, $K$ lies on a hyperbola, who’s foci are $E$ , $D$ and with constant difference of distances $0.5$ . Again, $K$ lies on the branch that is closer to focus $D$ .

The equation of this branch is

$\begin{aligned} & \sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-0 \right)}^{2}}}-\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-1.5 \right)}^{2}}}=0.5 \\ & \Leftrightarrow {{x}^{2}}-8{{y}^{2}}-4x+12y=0,\text{ }y>0.75 \ \ \ \ \ (2)\\ \end{aligned}$ Solving the system of $(1)$ and $(2)$ , we find the coordinates of $K$ to be $\left( \frac{36}{23},\ \frac{24}{23} \right)$ .

Finally, $R=KI=KE+EI=\sqrt{{{\left( 2-\frac{36}{23} \right)}^{2}}+{{\left( 0-\frac{24}{23} \right)}^{2}}}+2=\dfrac{72}{23}.$ For the answer, $a=72$ , $b=23$ , $a+b=72+23=\boxed{95}.$