Another Circle Problem! (fixed)

Let the radius of the semi-circle be $R = 1$ .

From $OG$ , $FO = OG - GF = 1 - 2r$ .

By the Pythagorean Theorem on $\triangle AFD$ , $AF = \sqrt{AO^2 - FO^2} = \sqrt{1^2 - (1 - 2r)^2} = 2\sqrt{r - r^2}$ .

As an inradius of a right triangle, $r = \frac{1}{2}(AF + FO - AO) = \frac{1}{2}(2\sqrt{r - r^2} + 1 - 2r - 1)$ , which solves to $r = \cfrac{1}{5}$ for $r > 0$ .

Therefore, $\cfrac{r}{R} = \cfrac{\frac{1}{5}}{1} = \cfrac{1}{5}$ , so $a = 1$ , $b = 5$ , and $a + b = \boxed{6}$ .

WLOG, we set $r=1$ . Let $E$ , $D$ be the points of tangency of the lower circle with $OA$ and $OF$ respectively. Denote $\angle JOD$ by $\theta$ .

$OJ$ is the angle bisector of $\angle EOD$ , hence $\angle EOJ=\angle DOJ=\theta$ Since $AC$ is tangent to the upper circle, $OH$ is perpendicular to $AC$ , thus $\triangle AFO$ is a right angled triangle. Hence, $\angle OAF=90{}^\circ -\angle AOF=90{}^\circ -2\theta$ Since $AJ$ is the angle bisector of $\angle OAF$ ,
$\angle OAJ= \dfrac{\angle OAF}{2}=45{}^\circ -\theta$ On right $\triangle ODJ$ ,
$\cot \theta =\dfrac{OD}{JD}=\dfrac{OF-FD}{JD}=\dfrac{\left( R-2 \right)-1}{1}\Rightarrow \cot \theta =R-3 \ \ \ \ \ (1)$ Furthermore, $OE=OD=R-3$ and on right $\triangle JAE$ ,

$\begin{aligned} & \cot \left( 45{}^\circ -\theta \right)=\dfrac{AE}{JE}=\dfrac{AE}{1} \\ & \Rightarrow AE=\cot \left( 45{}^\circ -\theta \right)=\dfrac{\cot \theta \cdot \cot 45{}^\circ +1}{\cot \theta -\cot 45{}^\circ }=\dfrac{\cot \theta +1}{\cot \theta -1} \\ & \overset{\left( 1 \right)}{\mathop{\Rightarrow }}\,AE=\dfrac{\left( R-3 \right)+1}{\left( R-3 \right)-1} \\ & \Rightarrow AE=\dfrac{R-2}{R-4} \ \ \ \ \ (2) \\ \end{aligned}$ Finally, $\begin{aligned} AO=AE+EO & \overset{\left( 2 \right)}{\mathop{\Rightarrow }}\,R=\frac{R-2}{R-4}+R-3 \\ & \Leftrightarrow R\left( R-4 \right)=R-2+\left( R-4 \right)\left( R-3 \right) \\ & \Leftrightarrow \cancel{{{R}^{2}}}-4R=R-2+\cancel{{{R}^{2}}}-7R+12 \\ & \Leftrightarrow R=5 \\ \end{aligned}$

This gives $\frac{r}{R}=\frac{1}{5}$ For the answer, $a=1$ , $b=5$ , thus $a+b=\boxed{6}$ .

Let $R=1$ , the radius of the two congruent circles be $r$ , $JM$ , $JN$ , and $JP$ be perpendicular to $AF$ , $AO$ , and $FO$ respectively, and $\angle FOA = \theta$ . Since radius $OG$ is perpendicular to chord $AC$ , $\angle AFO = 90^\circ$ and $\angle FAO = 90^\circ - \theta$ . Considering the radius $AO$ :

$\begin{aligned} AN + NO & = AO \\ JN \cot \frac {\angle FAO} 2 + JN \cot \frac {\angle FOA}2 & = AO \\ r \cot \left(45^\circ - \frac \theta 2\right) + r \cot \frac \theta 2 & = 1 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \left(\frac {1+t}{1-t} \right) t + \frac rt & = 1 \\ \implies r & = \frac 1{\frac {1+t}{1-t} + \frac 1t} & ...(1) \end{aligned}$

We note that $FO = 1 - 2r$ . Similarly,

$\begin{aligned} FP+PO & = FO \\ r + r \cot \frac \theta 2 & = 1 - 2r \\ \implies r & = \frac 1{3+\frac 1t} & ...(2) \end{aligned}$

From $(1)=(2)$ :

$\begin{aligned} \frac 1{\blue{\frac {1+t}{1-t}}+\frac 1t} & = \frac 1{\blue 3+\frac 1t} \\ \implies \frac {1+t}{1-t} & = 3 \\ 1+t = 3- 3t \\ \implies t & = \frac 12 \\ \implies \frac rR & = r = \frac 1{3+\frac 1t} = \frac 15 \end{aligned}$

Therefore $a+b = 1 + 5 = \boxed 6$ .