Traversing Circles

Place the whole diagram on a coordinate plane so that $B$ is at the origin, $A$ is at $A(0, 1)$ , and $C$ is at $C(1, 0)$ , and let $F$ be the intersection of $AE$ and $DC$ at $F(k, k)$ . Also, let $M$ be the midpoint of $AC$ at $M(\frac{1}{2}, \frac{1}{2})$ .

Then $AF$ has the equation $y = \frac{k - 1}{k}x + 1$ and an $x$ -intercept of $E(\frac{k}{1 - k}, 0)$ .

By the distance equation, $AE = \sqrt{1 + \frac{k^2}{(1 - k)^2}}$ , $AF = FC = \sqrt{k^2 + (k - 1)^2}$ , $DF = FE = \sqrt{k^2 + (k - \frac{k}{1 - k})^2}$ , $FM = (\frac{1}{2} - k)\sqrt{2}$ , and $AC = \sqrt{2}$ .

The inradius of right $\triangle ABE$ is $R = \frac{1}{2}(AB + BE - AE) = \frac{1}{2}(1 + \frac{k}{1 - k} - \sqrt{1 + \frac{k^2}{(1 - k)^2}})$ .

The inradius of $\triangle AFC$ is $R = \cfrac{2A_{\triangle AFC}}{P_{\triangle AFC}} = \cfrac{2 \cdot \frac{1}{2} \cdot AC \cdot FM}{AF + FC + AC} = \cfrac{\sqrt{2} \cdot (\frac{1}{2} - k)\sqrt{2}}{2\sqrt{k^2 + (k - 1)^2} + \sqrt{2}}$

Therefore, $R = \frac{1}{2}(1 + \frac{k}{1 - k} - \sqrt{1 + \frac{k^2}{(1 - k)^2}}) = \cfrac{\sqrt{2} \cdot (\frac{1}{2} - k)\sqrt{2}}{2\sqrt{k^2 + (k - 1)^2} + \sqrt{2}}$ , which solves to $k = \frac{1}{4}(2 - \sqrt{2(\sqrt{2} - 1)})$ .

The inradius of $\triangle FEC$ is $r = \cfrac{2A_{\triangle FEC}}{P_{\triangle FEC}} = \cfrac{2 \cdot \frac{1}{2} \cdot EC \cdot k}{EC + FE + FC} = \cfrac{(1 - \frac{k}{1 - k}) \cdot k}{(1 - \frac{k}{1 - k}) + \sqrt{k^2 + (k - \frac{k}{1 - k})^2} + \sqrt{k^2 + (k - 1)^2}}$ , which after substituting $k = \frac{1}{4}(2 - \sqrt{2(\sqrt{2} - 1)})$ solves to $r \approx 0.100644$ , so $\lfloor 10^6r \rfloor = \boxed{100644}$ .

Let the radius of the orange circle be $R$ and $\angle DCB = \theta$ . From the lower orange circle, we have:

$\begin{aligned} R+R \cot \frac \theta 2 & = BE = 1 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ R + \frac Rt & = 1 \\ \implies R & = \frac t{1+t} \end{aligned}$

From the upper orange circle,

$\begin{aligned} \frac R{\frac 12 AC} & = \tan \left(\frac {45^\circ - \theta}2 \right) & \small \blue{\text{Note that }\tan 22.5^\circ = \sqrt 2 -1} \\ \implies \sqrt 2 R & = \frac {\sqrt 2-1-t}{1+(\sqrt 2-1)t} \end{aligned}$

Therefore we have:

$\begin{aligned} \frac {\sqrt 2t}{1+t} & = \frac {\sqrt 2-1-t}{1+(\sqrt 2-1)t} \\ \implies t & = \frac {2\sqrt{\sqrt 2 -1}-1}{3-\sqrt 2} \end{aligned}$

Now consider the lower blue circle,

$\begin{aligned} r \cot \frac {\angle AEC}2 + r \cot \frac {\angle DCB}2 & = CE \\ r \cot \left(\frac {90^\circ + \theta}2 \right) + r \cot \frac \theta 2 & = 1 - \tan \theta \\ r \left(\frac {1-t}{1+t} + \frac 1t \right) & = 1 - \frac {2t}{1-t^2} \\ \frac {1+2t-t^2}{t(1+t)} \cdot r & = \frac {1-2t-t^2}{1-t^2} \\ \implies r & = \frac {t(1-2t-t^2)}{(1-t)(1+2t-t^2)} \\ & = \frac {1371-975\sqrt 2+4164\sqrt{\sqrt 2 -1}-2935\sqrt{2(\sqrt 2-1)}}{9912-7028\sqrt 2+3758\sqrt{\sqrt 2 -1}-2620\sqrt{2(\sqrt 2-1)}} \\ & \approx 0.100644375 \\ \implies \lfloor 10^6 r \rfloor & = \boxed{100644} \end{aligned}$