Mickey mouse ratios

Note that $EFGH$ is a square. Let the center of square $EFGH$ and the big circle be $O$ , and the side length of square $EFGH$ be $2$ , then the radius of the big circle is $1$ . Let the center of the top right circle be $P$ and the point it is tangent to the center circle be $Q$ . We note that $O$ , $Q$ , $P$ , and $C$ are colinear. We also note that $CQ = OQ = 1$ and that $CQ = r + \sqrt 2 r = 1 \implies r = \dfrac 1{\sqrt 2+1} = \sqrt 2 - 1$ . Then the ratio of the radius of the small circle to that of the big circle is $\dfrac r 1 = \sqrt 2 -1$ . Therefore $a+b = 2+1 = \boxed 3$ .

Circle q tangent circle o at I, then

$\begin{aligned} DI &= IG \\ \sqrt 2 r_q + r_q &= r_o \\ \dfrac {r_q}{r_o} &= \dfrac {1}{\sqrt 2 + 1} \\ &= \sqrt 2 - 1 \end{aligned}$

Without loss of generality, let the length of the side of the square ABCD be 2. Then, referring to the picture, we see DG = DH = 1 and, by the Pythagorean Theorem, HG = Sqrt(2). If r is the radius of the smaller circle, the formula for the in-radius of a circle shows that:

r = 1 / (2+sqrt(2))

By constructing a line from the center of the big circle to the point of tangency with the little circle, we see that (if R is the radius of the big circle)

R = (1/2)*(HG) = sqrt(2) / 2

Therefore,

r/R = (1/(2+sqrt(2))) / (sqrt(2) / 2) = sqrt(2) - 1.

Hence, a = 2 and b = 1, so that a+b = 3