Circle Mania 2.

Note: We just need to find $R$ then $r$ since $r = r'$ using the symmetry about $\overline{BD}$ .

$\overline{BD} = \sqrt{2} = 1 + R + \sqrt{2}R = 1 + (\sqrt{2} + 1)R \implies R = \dfrac{\sqrt{2} - 1}{\sqrt{2} + 1} = 3 - 2\sqrt{2}$

Using $\triangle{O'EO} \implies m^2 = (R + r)^2 - (R - r)^2 = 4Rr \implies m = 2\sqrt{Rr}$

and

Using $\triangle{FO'D} \implies (1 + r)^2 = (1- r)^2 + (1 - R - m)^2 \implies$

$4r = (1 - R)^2 - 2m(1 - r) + m^2 = (1 - R)^2 - 4(1 - R)\sqrt{Rr} + 4Rr \implies$

$4(1 - R)r = (1 - R)^2 - 4\sqrt{Rr}(1 - R) \implies 4r = 1 - R - 4\sqrt{Rr} \implies$

$4r + 4\sqrt{Rr} + R = 1 \implies (\sqrt{R} + 2\sqrt{r})^2 = 1 \implies \sqrt{R} + 2\sqrt{r} = 1 \implies$

$\sqrt{r} = \dfrac{1 - \sqrt{R}}{2} = \dfrac{1 - \sqrt{3 - 2\sqrt{2}}}{2} = \dfrac{1 - \sqrt{(\sqrt{2} - 1})^2}{2} = \dfrac{2 - \sqrt{2}}{2} = \dfrac{\sqrt{2}(\sqrt{2} - 1)}{2}$

$\implies r = \dfrac{2(3 - 2\sqrt{2})}{4} = \dfrac{3 - 2\sqrt{2}}{2} = \dfrac{R}{2} \implies S = \pi R^2 + 2\pi(\dfrac{R}{2})^2 =$

$\dfrac{3}{2}\pi(3 - 2\sqrt{2})^2 =$ $\dfrac{3}{2}\pi(17 - 12\sqrt{2}) = \dfrac{\alpha}{\beta}\pi(\lambda - \omega\sqrt{\beta}) \implies \alpha + \beta + \lambda + \omega = \boxed{34}$ .

Let the radius of the pink circle be $R$ and the radius of the red and blue circles be $r$ . Add $E$ and $F$ as follows:

As a diagonal of a unit square, $BD = \sqrt{2}$ . As a hypotenuse of right $\triangle BFO$ with $OF = R$ , $BO = \sqrt{2}R$ . Therefore, $BD = BO + OE + ED = \sqrt{2}R + R + 1 = \sqrt{2}$ , which solves to $R = 3 - 2\sqrt{2}$ .

By Descartes' Theorem (special case with a straight line), $\cfrac{1}{r} = 1 + \cfrac{1}{R} + 2\sqrt{1 \cdot \cfrac{1}{R}} = 1 + \cfrac{1}{3 - 2\sqrt{2}} + 2\sqrt{1 \cdot \cfrac{1}{3 - 2\sqrt{2}}} = \cfrac{2}{3 - 2\sqrt{2}}$ , so that $r = \frac{1}{2}(3 - 2\sqrt{2})$ .

Therefore, $S = \pi R^2 + 2\pi r^2 = \pi((3 - 2\sqrt{2})^2 + 2(\frac{1}{2}(3 - 2\sqrt{2}))^2) = \frac{3}{2}(17 - 12\sqrt{2})$ , so $\alpha = 3$ , $\beta = 2$ , $\lambda = 17$ , $\omega = 12$ , and $\alpha + \beta + \lambda + \omega = \boxed{34}$ .

$r_{pink} = 3 - 2 \sqrt 2 \\ r_{red} = r_{blue} = \dfrac {3 - 2 \sqrt 2}2 = \dfrac {r_{pink}}2 \\ S = \pi (r_{pink}^2 + r_{red}^2 + r_{blue}^2) \\ = \dfrac 32 \pi r_{pink}^2 \\ = \dfrac 32 \pi (3 - 2 \sqrt 2)^2 \\ = \dfrac 32 \pi (17 - 12 \sqrt 2)$