What's the Radius?

Let the centers of the left and right circles be $O$ and $P$ respectively, $OM$ and $PN$ be perpendicular to $AD$ , $PL$ is perpendicular to $CD$ , and $\angle ECD = \theta$ . Then $\angle CED = 90^\circ - \theta$ and $\angle CEA = 90^\circ + \theta$ . Note that $\angle CAD = 45^\circ$ , and $CP$ , $EP$ , $OE$ , and $AO$ bisect $\angle ECD$ , $\angle CED$ , $\angle CEA$ , and $\angle CAD$ respectively.

We can apply half-angle tangent substitution to solve triangle incircle problems reducing them to solving polynomial equations. From

$\begin{aligned} CL + LD & = CD \\ r \cot \frac \theta 2 + r & = 1 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac rt + r & = 1 \\ \implies t & = \frac r{1-r} \end{aligned}$

From:

$\begin{aligned} AM + ME + EN + ND & = AD \\ r\cot 22.5^\circ + r \cot \left(45^\circ + \frac \theta 2 \right) + r \cot \left(45^\circ - \frac \theta 2 \right) + r & = 1 \\ (\sqrt 2 + 1)r + \frac {1-t}{1+t} \cdot r + \frac {1+t}{1-t} \cdot r + r & = 1 & \small \blue{\text{Note that }t = \frac r{1-r}} \\ (\sqrt 2 + 2)r + (1-2r)r + \frac r{1-2r} & = 1 \\ 4r^3 - (8+2\sqrt 2)r^2 + (6+\sqrt 2)r - 1 & = 0 & \small \blue{\text{for }r \ne \frac 12} \\ \implies r & = \begin{cases} 1 + \dfrac 1{\sqrt 2} \\ \dfrac {1-\sqrt{\sqrt 2 -1}}2 \\ \dfrac {1+\sqrt{\sqrt 2 -1}}2 \end{cases} \end{aligned}$

The acceptable value is $r = \dfrac {1-\sqrt{\sqrt 2-1}}2$ . Therefore $\lambda + \omega = 1+2 = \boxed 3$ .

$A_{\triangle{ECD}} = \dfrac{1}{2}y = \dfrac{1}{2}r(a + y + 1) \implies r = \dfrac{y}{a + y + 1}$

and

$A_{\triangle{ACE}} = \dfrac{1}{2}r(a + \sqrt{2} + 1 - y) = \dfrac{1}{2}(1 - y) \implies$

$r = \dfrac{1 - y}{a + \sqrt{2} + 1 - y} = \dfrac{y}{a + y + 1} \implies (2a + \sqrt{2} + 1)y = a + 1 \implies$

$y = \dfrac{a + 1}{2a + \sqrt{2} + 1}$

In $\triangle{ECD} \implies y^2 + 1 = a^2 \implies (a + 1)^2 + (2a + \sqrt{2} + 1)^2 = a^2(2a + \sqrt{2} + 1)^2$

$\implies (a + 1)(a + 1 - (a - 1)(2a + \sqrt{2} + 1)^2) = 0$ and $a \neq 1 \implies$

$2a^3 + 2\sqrt{2}a^2 - (\sqrt{2} + 1)a - \sqrt{2}(\sqrt{2} + 1) = 0 \implies$

$2a^2(a + \sqrt{2}) - (\sqrt{2} + 1)(a + \sqrt{2}) = 0 \implies (2a^2 - (\sqrt{2} + 1))(a + \sqrt{2}) = 0$

and $a \neq -\sqrt{2} \implies a^2 = \dfrac{\sqrt{2} + 1}{2} \implies \boxed{a = \sqrt{\dfrac{\sqrt{2} + 1}{2}}}$

$\implies y^2 + 1 = \dfrac{\sqrt{2} + 1}{2} \implies \boxed{y = \sqrt{\dfrac{\sqrt{2} - 1}{2}}}$

Let $\alpha = \sqrt{2} + 1$ and $\beta = \sqrt{2} - 1$ $\implies a = \dfrac{\sqrt{\alpha}}{\sqrt{2}}$ and $y = \dfrac{\sqrt{\beta}}{\sqrt{2}} \implies$

$r = \dfrac{y}{a + y + 1} = \dfrac{\sqrt{\beta}}{\sqrt{\alpha} + \sqrt{\beta} + \sqrt{2}} =$ $\dfrac{\sqrt{2}\sqrt{\beta}(\sqrt{\alpha} + \sqrt{\beta} - \sqrt{2})}{4}$

$= \dfrac{\sqrt{2}(1 + \sqrt{2} - 1 - \sqrt{2}\sqrt{\beta})}{4} = \dfrac{2(1 - \sqrt{\beta})}{4} =$

$\dfrac{1 - \sqrt{\sqrt{2} - 1}}{2} = \dfrac{\lambda - \sqrt{\sqrt{\omega} - \lambda}}{\omega} \implies \lambda + \omega = \boxed{3}$ .

By the properties of a unit square, $AD = CD = 1$ and $AC = \sqrt{2}$ . Let $ED = x$ , so that $AE = AD - ED = 1 - x$ , and $EC = \sqrt{ED^2 + CD^2} = \sqrt{x^2 + 1}$ .

Then the areas of $\triangle AEC$ and $\triangle EDC$ are $A_{\triangle AEC} = \frac{1}{2} \cdot AE \cdot CD = \frac{1}{2} \cdot (1 - x) \cdot 1 = \frac{1}{2}(1 - x)$ and $A_{\triangle EDC} = \frac{1}{2} \cdot ED \cdot CD = \frac{1}{2} \cdot x \cdot 1 = \frac{1}{2}x$ .

And the perimeters are $P_{\triangle AEC} = AE + AC + EC = 1 - x + \sqrt{2} + \sqrt{x^2 + 1}$ and $P_{\triangle EDC} = ED + CD + EC = x + 1 + \sqrt{x^2 + 1}$ .

Since the inradius is $r = \cfrac{2A}{P}$ , then $r = \cfrac{2A_{\triangle AEC}}{P_{\triangle AEC}} = \cfrac{2A_{\triangle EDC}}{P_{\triangle EDC}}$ , or $r = \cfrac{1 - x}{1 - x + \sqrt{2} + \sqrt{x^2 + 1}} = \cfrac{x}{x + 1 + \sqrt{x^2 + 1}}$ , which solves to $x = \sqrt{\frac{1}{2}(\sqrt{2} - 1)}$ .

Substituting $x = \sqrt{\frac{1}{2}(\sqrt{2} - 1)}$ into $r = \cfrac{x}{x + 1 + \sqrt{x^2 + 1}}$ and solving gives $r = \cfrac{1 - \sqrt{\sqrt{2} - 1}}{2}$ , so $\lambda = 1$ , $\omega = 2$ , and $\lambda + \omega = \boxed{3}$ .