Eccentric Remix!

Consider an ellipse in the form of $\cfrac{x^2}{a^2} + \cfrac{(y - 1 + b)^2}{b^2} = 1$ that is tangent to the unit circle $x^2 + y^2 = 1$ at $(0, 1)$ .

Substituting $x^2 = 1 - y^2$ into $\cfrac{x^2}{a^2} + \cfrac{(y - 1 + b)^2}{b^2} = 1$ and solving for $y$ gives $y = 1$ and $y = \cfrac{-2a^2b + a^2 + b^2}{a^2 - b^2}$ , and if the ellipse is to have the maximum eccentricity then $y = \cfrac{-2a^2b + a^2 + b^2}{a^2 - b^2} = 1$ as well, which solves to $b = a^2$ .

Then $A'C'$ is on the line $y = 1 - 2b = 1 - 2a^2$ , so that $OD' = 1 - 2a^2$ , and by the Pythagorean Theorem on $\triangle OD'C'$ , $D'C' = \sqrt{OC'^2 - OD'^2} = \sqrt{1^2 - (1 - 2a^2)^2} = 2a\sqrt{1 - a^2}$ , which means $A'C' = 2 \cdot D'C' = 2 \cdot 2a \sqrt{1 - a^2} = 4a \sqrt{1 - a^2}$ .

Now rotate the picture to match the diagram in the problem:

By Thales's Theorem $\triangle ABC$ is a right triangle, so by the Pythagorean Theorem on $\triangle ABC$ , $BC = \sqrt{AB^2 - AC^2} = \sqrt{2^2 - (4a \sqrt{1 - a^2})^2} = 2 - 4a^2$ .

From similar triangles, the slope of $AC$ is $m_1 = \cfrac{BC}{AC} = \cfrac{2 - 4a^2}{4a \sqrt{1 - a^2}} = \cfrac{1 - 2a^2}{2a \sqrt{1 - a^2}}$ , so $AC$ has an equation of $y = m_1(x + 1)$ .

Since $BC \perp AC$ , the slope of $BC$ is $m_2 = -\cfrac{1}{m_1} = -\cfrac{2a \sqrt{1 - a^2}}{1 - 2a^2}$ , so $BC$ has an equation of $y = m_2(x - 1)$ .

Let the lower ellipse have a center of $(h, b) = (h, a^2)$ . Then shift that ellipse and $\triangle ABC$ $h$ units to the left and $a^2$ units downwards so that the ellipse has a center at the origin:

Then the ellipse has an equation of $\cfrac{x^2}{a^2} + \cfrac{y^2}{b^2} = 1$ where $b = a^2$ , the line $A''C''$ has an equation of $y = m_1(x + 1 + h) - a^2$ or $y = m_1x + c_1$ where $c_1 = m_1(h + 1) - a^2$ , and the line $B''C''$ has an equation of $y = m_2(x - 1 + h) - a^2$ or $y = m_1x + c_2$ where $c_2 = m_2(h - 1) - a^2$ .

Knowing that y = mx + c is tangent to $\cfrac{x^2}{a^2} + \cfrac{y^2}{b^2} = 1$ if $a^2m^2 + b^2 = c^2$ , we have $a^2m_1^2 + a^4 = c_1^2$ and $a^2m_2^2 + a^4 = c_2^2$ . Substituting $c_1 = m_1(h + 1) - a^2$ , $c_2 = m_2(h - 1) - a^2$ , $m_1 = \cfrac{1 - 2a^2}{2a \sqrt{1 - a^2}}$ , and $m_2 = -\cfrac{2a \sqrt{1 - a^2}}{1 - 2a^2}$ into these two equations and solving for $a$ and $h$ numerically gives $a \approx 0.5259738517...$ .

That means the major axis is $A = 2a \approx 1.0519477034...$ and $\lfloor 10^6 A \rfloor = \boxed{1051947}$ .