Ellipse arrangement in an isosceles triangle

Let the triangle have a base of $2$ and an altitude of $3$ , and place it on the coordinate plane so that its vertices are at $(\pm 1, 0)$ and $(0, 3)$ . Then its congruent sides follow the equations $y = \pm 3x + 3$ .

Let the ellipses have a semi-major axis of $a$ and a semi-minor axis of $b$ . Then the bottom ellipse has a center of $(0, b)$ and an equation of $\cfrac{x^2}{a^2} + \cfrac{(y - b)^2}{b^2} = 1$ , and the top ellipse has a center of $(0, a + 2b)$ and an equation of $\cfrac{x^2}{b^2} + \cfrac{(y - a - 2b)^2}{a^2} = 1$ .

Since the bottom ellipse is inscribed in the triangle, it will touch one its sides in only one place, so after substituting $y = -3x + 3$ into $\cfrac{x^2}{a^2} + \cfrac{(y - b)^2}{b^2} = 1$ , the discriminant of $x$ must be $0$ ; that is, $ab\sqrt{3(3a^2 + 2b - 3)} = 0$ , which solves to $b = \cfrac{1}{2}(3 - 3a^2)$ for positive values of $a$ and $b$ .

Similarly, the top ellipse will also touch one of the triangle's sides in only one place, so after substituting $y = -3x + 3$ and $b = \cfrac{1}{2}(3 - 3a^2)$ into $\cfrac{x^2}{b^2} + \cfrac{(y - a - 2b)^2}{a^2} = 1$ , the discriminant of $x$ must be $0$ ; that is, $a(a^2 - 1)\sqrt{3(3a^2 - 2a - 3)(5a^2 + 6a - 9)} = 0$ , which solves to $a = \cfrac{3\sqrt{6} - 3}{5}$ for $0 < a < 1$ .

If $a = \cfrac{3\sqrt{6} - 3}{5}$ , then $b = \cfrac{1}{2}(3 - 3a^2) = \cfrac{1}{2}\bigg(3 - 3\bigg(\cfrac{3\sqrt{6} - 3}{5}\bigg)^2\bigg) = \cfrac{27\sqrt{6} - 57}{25}$ , and $e = \cfrac{\sqrt{a^2 - b^2}}{a} = \cfrac{\sqrt{(\frac{3\sqrt{6} - 3}{5})^2 - (\frac{27\sqrt{6} - 57}{25})^2}}{\frac{3\sqrt{6} - 3}{5}} = \cfrac{2}{5}\sqrt{7\sqrt{6} - 12} \approx 0.90742$

Therefore, $\lfloor 10^4 e \rfloor = \boxed{9074}$ .