No 1874 Sangaku Problem

Place the whole diagram so that its center is at the origin and the two large circles are unit circles with centers at $(\pm k, 0)$ .

Then the two large circles have equations of $(x \pm k)^2 + y^2 = 1$ , which intersect at $(0, \pm \sqrt{1 - k^2})$ .

The semi-major axis of the ellipse is $a = k + 1$ , the semi-minor axis of the ellipse is $b = \sqrt{1 - k^2}$ , and the radius of one green circle is $r = 1 - k$ .

Then the ellipse has an equation of $\cfrac{x^2}{(k + 1)^2} + \cfrac{y^2}{1 - k^2} = 1$ .

The three left-most green circles are tangent to a circle also centered at $(k, 0)$ but with a radius of $R = 1 - 2r = 1 - 2(1 - k) = 2k - 1$ , so an equation of $(x - k)^2 + y^2 = (2k - 1)^2$ , and this circle must also be tangent to the ellipse.

Substituting $(x - k)^2 + y^2 = (2k - 1)^2$ into $\cfrac{x^2}{(k + 1)^2} + \cfrac{y^2}{1 - k^2} = 1$ gives $\cfrac{x^2}{(k + 1)^2} + \cfrac{(2k - 1)^2 - (x - k)^2}{1 - k^2} = 1$ , and for that circle to be tangent to the ellipse its discriminant $9k^2 + 2k - 7$ must equal $0$ , and $9k^2 + 2k - 7 = 0$ solves to $k = \cfrac{7}{9}$ .

That makes the semi-major axis $a = k + 1 = \cfrac{7}{9} + 1 = \cfrac{16}{9}$ and the semi-minor axis $b = \sqrt{1 - k^2} = \sqrt{1 - \bigg(\cfrac{7}{9}\bigg)^2} = \cfrac{4\sqrt{2}}{9}$ , for an eccentricity of $e = \cfrac{\sqrt{a^2 - b^2}}{a} = \cfrac{\sqrt{(\frac{16}{9})^2 - (\frac{4\sqrt{2}}{9})^2}}{\frac{16}{9}} = \cfrac{\sqrt{14}}{4} \approx 0.9354143$ , so $\lfloor 10^6 e \rfloor = \boxed{935414}$ .