Tangential Essentials II

Label the diagram as follows:

By symmetry on the top left circle, $DF$ is the perpendicular bisector of $AC$ , so $\triangle AEF \sim \triangle ACB$ by AA similarity, and since $EF = DF - DE = 1 - 2R$ , that makes $CB = EF \cdot \cfrac{AC}{AE} = (1 - 2R) \cdot 2 = 2 - 4R$ .

By the Pythagorean Theorem on $\triangle ACB$ , $AC = \sqrt{AB^2 - CB^2} = \sqrt{2^2 + (2 - 4R)^2} = 4\sqrt{R - R^2}$ .

Let $k = GC$ . By the Pythagorean Theorem on $\triangle GCB$ , $GB = \sqrt{GC^2 + CB^2} = \sqrt{k^2 + (2 - 4R)^2}$

Also from right $\triangle GCB$ , $R = \frac{1}{2}(GC + CB - GB) = \frac{1}{2}(k + 2 - 4R - \sqrt{k^2 + (2 - 4R)^2})$ , which solves to $k = \cfrac{R(5R - 2)}{3R - 1}$ .

From $\triangle AGB$ , $R = \cfrac{2A_{\triangle AGB}}{P_{\triangle AGB}} = \cfrac{AG \cdot CB}{AG + GB + AB} = \cfrac{(4\sqrt{R - R^2} - k)(2 - 4R)}{(4\sqrt{R - R^2} - k) + \sqrt{k^2 + (2 - 4R)^2} + 2}$ .

Substituting $k = \cfrac{R(5R - 2)}{3R - 1}$ into $R = \cfrac{(4\sqrt{R - R^2} - k)(2 - 4R)}{(4\sqrt{R - R^2} - k) + \sqrt{k^2 + (2 - 4R)^2} + 2}$ and solving numerically gives $R \approx 0.2434772$ , and $\lfloor 10^6 R \rfloor = \boxed{243477}$ .