Three Congruent Circles

Let the radius of the large circle be $1$ and the radius of the small circle be $r$ . Then the ratio of radii is $r$ . Let the apex angle of the isosceles triangle be $\theta$ . Let the centers of the large and small circles be $O$ and $C$ respectively.

We note that the length $EO = 1$ also equal to $2r + \sin \dfrac \theta 2$ $\implies r = \dfrac {1 - \sin \frac \theta 2}2$ .

Now consider the length of the median $AB$ :

$\begin{aligned} \overbrace{AC + BC}^{=AB} & = \overbrace{AO + BO}^{=AB} \\ \frac r{\sin \frac \theta 2} + r & = 1 + \cos \theta \\ r + r \sin \frac \theta 2 & = \sin \frac \theta 2 + \sin \frac \theta 2 \left(1-2 \sin^2 \frac \theta 2 \right) & \small \blue{\text{Note that }r = \frac {1 - \sin \frac \theta 2 }2} \\ \frac 12 \left(1 -\sin \frac \theta 2 \right)\left(1 + \sin \frac \theta 2 \right) & = 2 \sin \frac \theta 2 -2 \sin^3 \frac \theta 2 & \small \blue{\text{Let }s = \sin \frac \theta 2} \\ 1 - s^2 & = 4s-4s^3 \\ 4s^3 - s^2 - 4s + 1 & = 0 \\ (s - 1)(s+1) \left(s - \frac 14\right) & = 0 & \small \blue{\text{Since }\theta \text{ is acute.}} \\ \implies s & = \frac 14 \\ \implies r & = \frac {1-s}2 = \frac 38 \end{aligned}$

Therefore $a+b = 3+8 = \boxed{11}$ .

Label the diagram as below:

Let the radius of the large circle be $1$ and the radius of the small circle be $r$ , so that the ratio we are looking for will be $r$ .

Also, let $2x = \angle COD$ . Then $\angle AOC = 180 - 2x$ and $\angle OCF = x$ .

By trigonometry on $\triangle OCF$ , $OF = \sin x$ and $CF = \cos x$ .

By trigonometry on $\triangle COD$ , $CD = \sin 2x$ and $OD = \cos 2x$ .

From $OB$ we know that $OB = OF + FB = \sin x + 2r = 1$ .

The small circle is also the incircle of $\triangle ACE$ , so $r = \cfrac{A_{\triangle ACE}}{s_{\triangle ACE}} = \cfrac{\frac{1}{2} \cdot CE \cdot AD}{\frac{1}{2}(2AC + CE)} = \cfrac{\frac{1}{2} \cdot 2 \sin 2x \cdot (1 + \cos 2x)}{\frac{1}{2}(2 \cdot 2 \cos x + 2 \sin 2x)} = 2 \sin x (1 - \sin x)$ .

The equations $\sin x + 2r = 1$ and $r = 2 \sin x (1 - \sin x)$ solve to $r = \frac{3}{8}$ , so $a = 3$ , $b = 8$ , and $a + b = \boxed{11}$ .

Assume radius of large circle as x and small circles as y, the sides of triangle as a,a,b. Now write the following relations from figure.

x^2-(x-2y)^2=(a^2)/4

a(x-2y)=(b/2)x

y=(b/2)[(2a-b)/(2a+b)]^0.5; inscribed circle radius

y=1

Solve using WolframAlpha get x=8/3

Answer = 8+3 =11