Elliptical Sangaku

By symmetry, we are looking for circles inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ that are tangent to the ellipse at two points, and whose centres lie on the $x$ -axis. If we consider the point $(u,0)$ , where $|u| < a$ , then the square of the distance from $(u,0)$ to the point $(a\cos t,b\sin t)$ on the ellipse is $D(t) \; = \; (a\cos t - u)^2 + b^2\sin^2t$ and hence $D'(t) \; = \; -2a\sin t(a\cos t - u) + 2b^2\sin t \cos t \; = \; 2\sin t\big(au - (a^2-b^2)\cos t\big) \; = \; 2a\sin t\big(u - ae^2\cos t\big)$ If $|u| \ge ae^2$ , there are only two turning points of $D$ (at $t=0$ and $t=\pi$ ) and hence any circle centred at $(u,0)$ can only have one point of tangency. We must have $|u| < ae^2$ .

Define the angle $0 < \varphi < \tfrac12\pi$ such that $\cos\varphi = e$ .

For any $|u| < ae^2$ we can find a value $w$ such that $u = ae \sin w$ ; since $|\sin w| < e = \cos\varphi$ we deduce that $|w| < \tfrac12\pi - \varphi$ . The minimum distance from $(u,0)$ to the ellipse occurs at the point $(a\cos\theta,b\sin\theta)$ where $\cos\theta = \frac{u}{ae^2} = e^{-1}\sin w$ , and this minimum distance is therefore $r^2$ , where $\begin{aligned} r^2 & = \; \big(ae^{-1}\sin w - ae\sin w\big)^2 + b^2\sin^2\theta \; = \; a^2e^{-2}(1 - e^2)^2\sin^2w + b^2\big(1 - e^{-2}\sin^2w\big) \\ & = \; b^2e^{-2}(1 - e^2)\sin^2w + b^2\big(1 - e^{-2}\sin^2w\big) \; = \; b^2\cos^2w \end{aligned}$ Thus we can draw a circle with centre at the point $(u,0)$ and radius $r$ , where $u = ae\sin w$ and $r = b\cos w$ , which is tangential to the ellipse at precisely two points for any $|w| < \tfrac12\pi - \varphi$ .

Consider two of these circles, given by values $-\tfrac12\pi + \varphi < w_1 < w_2 < \tfrac12\pi - \varphi$ . Then $\begin{aligned} u_2 - u_1 & = \; ae(\sin w_2 - \sin w_1) \; = \; 2a\cos\varphi \cos\tfrac12(w_1+w_2)\,\sin\tfrac12(w_2-w_1) \\ r_1 + r_2 & = \; b(\cos w_1 + \cos w_2) \; = \; 2a\sin\varphi \cos\tfrac12(w_1+w_2)\cos\tfrac12(w_2-w_1) \end{aligned}$ and hence these two circles are tangential to each other provided that $\cos\varphi \sin\tfrac12(w_2-w_1) = \sin\varphi \cos\tfrac12(w_2-w_1)$ , and hence so that $\tan\tfrac12(w_2-w_1) = \tan\varphi$ , and hence provided that $w_2 - w_1 = 2\varphi$ .

Thus we can have seven mutually tangent circles inscribed in the ellipse provided we can choose $w_j = w_0 + 2j\varphi$ for $0 \le j \le 6$ such that $-\tfrac12\pi + \varphi < w_j < \tfrac12\pi - \varphi$ for all $0 \le j \le 6$ . This means that $12\varphi = w_6 - w_0$ must be less than $\pi - 2\varphi$ , and hence that $\varphi < \tfrac{1}{14}\pi$ . Thus we must have $e = \cos\varphi > \cos\tfrac{1}{14}\pi$ , so we deduce that $\alpha = \cos\tfrac{1}{14}\pi$ , and hence $\lfloor 10^6\alpha \rfloor = \boxed{974927}$ .