Three Ellipses and a Circle

Let the circle be centered at the origin and have a radius of $1$ , so that it has an equation of $x^2 + y^2 = 1$ .

Let the each identical ellipse have a major axis of $a$ and a minor axis $b$ . From the segment along the diameter, $2b + 2a + 2b = 2$ , so that $a = 1 - 2b$ .

The minor axis is less than that the major axis, so $b < a$ , or $b < 1 - 2b$ , which means $b < \cfrac{1}{3}$ .

The ellipse on the right is centered at $(1 - b, 0)$ , so its equation is $\cfrac{(x - 1 + b)^2}{b^2} + \cfrac{y^2}{a^2} = 1$ .

Substituting $a = 1 - 2b$ and $y^2 = 1 - x^2$ , the equation becomes $\cfrac{(x - 1 + b)^2}{b^2} + \cfrac{1 - x^2}{(1 - 2b)^2} = 1$ and rearranges to $(x - 1)(x - \cfrac{8b^2 - 5b + 1}{1 - 3b}) = 0$ , which means that the ellipse and the circle intersect at $x = 1$ and $x = \cfrac{8b^2 - 5b + 1}{1 - 3b}$ .

For the ellipse to be inscribed and not intersect the circle, $\cfrac{8b^2 - 5b + 1}{1 - 3b} = x \geq 1$ , which for $b < \cfrac{1}{3}$ solves to $b \geq \cfrac{1}{4}$ .

Since the eccentricity of the ellipse is $e = \cfrac{\sqrt{a^2 - b^2}}{a} = \cfrac{\sqrt{(1 - 2b)^2 - b^2}}{1 - 2b}$ , it is a decreasing function for $\cfrac{1}{4} \leq b < \cfrac{1}{3}$ and has a maximum value at $b = \cfrac{1}{4}$ of $e = \cfrac{\sqrt{(1 - 2 \cdot \frac{1}{4})^2 - (\frac{1}{4})^2}}{1 - 2\cdot \frac{1}{4}} = \cfrac{\sqrt{3}}{2}$ .

Therefore, $\lfloor 10^6 e \rfloor = \lfloor 10^6 \cdot \cfrac{\sqrt{3}}{2} \rfloor = \boxed{866025}$ .