Three is Four

Construction: Extend line segments $CF$ , $BA$ and let $G$ be the point of intersection. Since $\triangle DCF\sim \triangle AGF$ , $\begin{aligned} \frac{FD}{AF}=\frac{CD}{GA} \implies \frac{x}{1-x}=\frac{1}{GA} \implies GA=\frac{1-x}{x} \end{aligned}$ Using the Pythagorean Theorem in $\triangle DCF$ and $\triangle AGF$ , $\begin{aligned} CF^2&=FD^2+DC^2 \\ CF&=\sqrt{1+x^2} \\ \\ FG^2&=GA^2+AF^2 \\ FG&=\sqrt{\frac{(1-x)^2}{x^2}+(1-x)^2} \\ &=\frac{(1-x)\sqrt{1+x^2}}{x} \end{aligned}$ The circle centered at $O_1$ is the incircle of right $\triangle DCF$ . Therefore, $\begin{aligned} r&=\frac{CD\cdot DF}{CD+DF+FC} \\ &=\frac{x}{1+x+\sqrt{1+x^2}} \end{aligned}$ The circle centered at $O_2$ is the excircle of right $\triangle AGF$ . Therefore, $\begin{aligned} r&=\frac{GA\cdot AF}{GA+FG-AF} \\ &=\frac{\dfrac{1-x}{x}\cdot (1-x)}{\dfrac{1-x}{x}+\dfrac{(1-x)\sqrt{1+x^2}}{x}-(1-x)} \\ &=\frac{1-x}{1+\sqrt{1+x^2}-x} \end{aligned}$ So, we have the following relations, $\begin{cases} r=\dfrac{x}{1+x+\sqrt{1+x^2}} \\ \\ r=\dfrac{1-x}{1+\sqrt{1+x^2}-x} \end{cases}$ Solving the above system of equations, we have, $x\approx 0.6477988...\; ,\; r\approx 0.2281554...\implies \lfloor 10^6 r\rfloor=\boxed{228155}$

The half-angle tangent substitution provide a straight-forward way of solving incircle problems as shown in this solution.

Let $\angle CEB = \theta$ . Consider the bottom-right circle and segment $EB$ .

$\begin{aligned} EH + HB & = EB \\ r \cot \frac {\angle CEB}2 + r & CB \cot \angle CEB \\ r \cot \frac \theta 2 +r & = \cot \theta & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac rt + r & = \frac {1-t^2}{2t} \\ \frac {1+t}t \cdot r & = \frac {1-t^2}{2t} \\ r & = \frac {1-t}2 \\ \implies t & = 1 - 2r \end{aligned}$

Now consider $AB$ :

$\begin{aligned} r + r \cot \left(90^\circ - \frac \theta 2 \right) + r \cot \frac \theta 2 + r & = AB = 1 \\ 2r + r t + \frac rt & = 1 \\ 2rt + rt^2 + r & - t & \small \blue{\text{Note that }t = 1-2r} \\ 2r(1-2r) + r(1-2r)^2 + r & = 1-2r \\ 4r^3 - 8r^2 + 6r - 1 & = 0 & \small \blue{\text{Using Cardano's method}} \end{aligned}$

$\begin{aligned} \implies r & = \frac 23 - \frac 1{3\sqrt[3]{3\sqrt{33}-17}} + \frac {\sqrt[3]{3\sqrt{33}-17}}6 \\ & \approx 0.228155493653962 \\ \implies \lfloor 10^6r \rfloor & = \boxed{228155} \end{aligned}$

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Reference: Cardano's method