Find the eccentricity! - Part 2

Equation of the asymptote with positive slope to the given hyperbola is $y=\dfrac{m}{n}x$ . From the given conditions of the problem, slope of this line is $\tan {\dfrac{π}{3}}=\sqrt 3$ . So, $\dfrac{m}{n}=\sqrt 3$ and the equation of the line is $y=x\sqrt 3$ . Eccentricity of the hyperbola is $\sqrt {1+(\dfrac{m}{n})^2}=\sqrt {1+3}=2=e_2$ . One of the points of intersection of the asymptotes with the ellipse has coordinates $(\dfrac{ab}{\sqrt {3a^2+b^2}},\dfrac{\sqrt 3ab}{\sqrt {3a^2+b^2}})$ . So $2\times {\dfrac{ab}{\sqrt {3a^2+b^2}}}=\sqrt {a^2-b^2}$ . This yields $e_1^4-8e_1^2+4=0$ , where $e_1=\sqrt {1-(\dfrac{b}{a})^2}$ . Hence $e_1=\sqrt 3-1$ , and $e_1+e_2=\sqrt 3+1=2.7320508075689$ . Therefore the required answer is $\boxed {27320}$

The regular hexagon can be broken up into $6$ equilateral triangles, so the asymptotes of the hyperbola must have slopes of $\pm \tan 60° = \pm \sqrt{3}$ . The asymptotes of a hyperbola also have slopes of $\pm \frac{n}{m}$ . Therefore, $\pm \frac{n}{m} = \pm \sqrt{3}$ . Either way, the eccentricity of the hyperbola is $e_2 = \sqrt{1 + (\frac{n}{m})^2} = \sqrt{1 + (\pm \sqrt{3})^2} = 2$ .

Let $(p, q)$ be the point on the ellipse that intersects with the hyperbola's asymptote in the first quadrant. Since it is on the asymptote that has a slope of $\sqrt{3}$ , $q = \sqrt{3}p$ , and since it is on the ellipse, $\frac{p^2}{a^2} + \frac{q^2}{b^2} = 1$ . Also, since they are points on a regular hexagon, the distance from $(p, q)$ to the origin and the distance from the focus $(\sqrt{a^2 - b^2}, 0)$ to the origin are the same, so that $p^2 + q^2 = a^2 - b^2$ . These equations solve to $\frac{b^2}{a^2} = 2\sqrt{3} - 3$ , which makes the eccentricity of the ellipse $e_1 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - (2\sqrt{3} - 3)} = \sqrt{4 - 2\sqrt{3}} = -1 + \sqrt{3}$ .

Therefore, $\lfloor 10000(e_1 + e_2) \rfloor = \lfloor 10000((-1 + \sqrt{3}) + 2) \rfloor = \lfloor 10000(1 + \sqrt{3}) \rfloor = \boxed{27320}$ .