Tricircle

Let the center of the bottom circle and the top-right circles be $O$ and $P$ respectively, $ON$ be perpendicular to $AB$ and $PM$ be perpendicular to $EF$ . Due to symmetry, $\triangle ADF$ and $\triangle ABE$ are congruent, and $\triangle CEF$ is an isosceles right triangle.

Let $\angle EAB = \theta$ . Note that $AO$ bisects $\angle EAB$ . Using half-angle tangent substitution , we have:

$\begin{aligned} AN+NB & = AB \\ r \cot \frac \theta 2 + r & = 1 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac rt + r & = 1 \\ \frac {1+t}t \cdot r & = 1 & ...(1) \end{aligned}$

Also note that:

$\begin{aligned} \sqrt 2 \cdot CM & = CE \\ \sqrt 2 (1 + \sqrt 2) r & = 1 - \tan \theta = 1 - \frac {2t}{1-t^2} \\ (2+\sqrt 2)r & = \frac {1-2t-t^2}{1-t^2} & ...(2) \end{aligned}$

From $\dfrac {(1)}{(2)}$ :

$\begin{aligned} \frac {1+t}{(2+\sqrt 2)t} & = \frac {1-t^2}{1-2t-t^2} \\ \frac 1{(2+\sqrt 2)t} & = \frac {1-t}{1-2t-t^2} \\ (1+\sqrt 2)t^2 - (4+\sqrt 2)t + 1 & = 0 \\ \implies t & = \frac {4+\sqrt 2-\sqrt{14+4\sqrt 2}}{2+2\sqrt 2} & \small \blue{\text{For }\theta < 90^\circ} \end{aligned}$

Therefore $r = \dfrac t{1+t} = \dfrac {4+\sqrt 2-\sqrt{14+4\sqrt 2}}{6+3\sqrt 2-\sqrt{14+4\sqrt 2}} \approx 0.168807445128$ , $\implies \lfloor 10^6r \rfloor = \boxed{168807}$ .

Let $\overline{DF}=a=\overline{BE}\because \triangle DFA \cong \triangle BEA\Rightarrow \overline{FC}=1-a=\overline{CE}$ $ar(\triangle DFA)=\dfrac{a}{2}=ar(\triangle EBA)$ From Pythagoras Theorem : $\overline{AF}=\sqrt{1+a^2}=\overline{AE}$ From the relation : $rs=\triangle$ where $\triangle=area\space of \space triangle,s=semi-perimeter,r=inradius$

We get that inradius of $\triangle DFA=\dfrac{a}{1+a+\sqrt{1+a^2}}=$ inradius of $\triangle EBA$

Similarly we can find that inradius of $\triangle CFE=\dfrac{1-a}{2+\sqrt{2}}..........[1]$ $\Rightarrow \dfrac{a}{1+a+\sqrt{1+a^2}}=\dfrac{1-a}{2+\sqrt{2}}$ The root of the above equation can be approximated using Newton's method starting from $a_0=\dfrac{1}{2}$

Applying it twice $a\approx 0.423654168508...$

From $[1]$ $r\approx 0.168807785735...\Rightarrow r \times 10^6=168807.785735 \Rightarrow \lfloor r \times 10^6 \rfloor = \boxed{168807}$

Did it similar to Sew Seon-Cheong, found a slightly simpler expression for r: $r=\frac{6+\sqrt{2}-\sqrt{14+4\sqrt{2}}}{12+4\sqrt{2}}$