Packing a Pair of Ellipses

In Optimal packings of two ellipses in a square , Thierry Gensane and Pascal Honvault prove that the critical number ${{e}_{0}}$ occurs when the aspect ratio of the ellipse is

$\dfrac{b}{a}={{E}_{0}}=\sqrt{\dfrac{6\sqrt{3}-3}{11}}\approx 0.8198 \ \ \ \ \ (1)$ where $a$ and $b$ are the semi-major axis and the semi-minor axis of the ellipse.

As seen in the figure, point $A$ lies on the director circle of one of the two ellipses, i.e. on a circle that has the same center ${{O}_{1}}$ as the ellipse and with radius $r=\sqrt{{{a}^{2}}+{{b}^{2}}} \ \ \ \ \ (2)$

Combining relations $(1)$ and $(2)$ we get $r=\sqrt{{{\left( \dfrac{b}{{{E}_{0}}} \right)}^{2}}+{{b}^{2}}}=b\dfrac{\sqrt{1+{{E}_{0}}^{2}}}{{{E}_{0}}}$ Furthermore,

$\begin{aligned} \left[ ABD \right]=\dfrac{1}{2}AK\cdot BD=\dfrac{1}{2}AB\cdot AD & \Rightarrow AK\cdot BD=A{{B}^{2}} \\ & \Rightarrow \left( r+b \right)\sqrt{2}=1 \\ & \Rightarrow \left( b\dfrac{\sqrt{1+{{E}_{0}}^{2}}}{{{E}_{0}}}+b \right)\sqrt{2}=1 \\ & \Rightarrow 2b\left( \dfrac{\sqrt{1+{{E}_{0}}^{2}}}{{{E}_{0}}}+1 \right)=\sqrt{2} \\ & \Rightarrow 2b=\dfrac{\sqrt{2}{{E}_{0}}}{\sqrt{1+{{E}_{0}}^{2}}+{{E}_{0}}} \\ \end{aligned}$ Plugging the value of ${{E}_{0}}$ , we find that the distance between the centers of the two ellipses is $\left| \overline{{{O}_{1}}{{O}_{2}}} \right|=2b=\frac{6\sqrt{2}-\sqrt{6}}{11}$ For the answer, $l=6$ , $m=2$ , $n=11$ , thus, $l+m+n=\boxed{19}$ .

Place the unit square so that its vertices are at $(\pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2})$ , so that the edges are represented by the lines $y = \pm x \pm \frac{\sqrt{2}}{2}$ , and let the ellipses have semi-major axes of $a$ and a semi-major axes of $b$ with centers at $(\pm b, 0)$ so that they are represented by the equations $\cfrac{(x \pm b)^2}{b^2} + \cfrac{y^2}{a^2} = 1$ .

According to the paper in the given hint, the aspect ratio $E_0$ of each ellipse is $E_0 = \sqrt{\cfrac{6\sqrt{3} - 3}{11}} = \cfrac{a}{b}$ , so $a^2 = \cfrac{(6\sqrt{3} - 3)b^2}{11}$ , and substituting this into the equation of the ellipse on the right and rearranging gives $11(x - b)^2 + (6\sqrt{3} - 3)y^2 = 11b^2$ .

To be tangent to the line $y = -x + \frac{\sqrt{2}}{2}$ , the ellipse $11(x - b)^2 + (6\sqrt{3} - 3)y^2 = 11b^2$ will intersect it at only one x-value, so the discriminant of $11(x - b)^2 + (6\sqrt{3} - 3)(-x + \frac{\sqrt{2}}{2})^2 = 11b^2$ will be $0$ . In other words, $22 b^2 + 12 \sqrt{6} b - 6 \sqrt{2} b - 6 \sqrt{3} + 3 = 0$ , which solves to $b = \cfrac{6\sqrt{2} - \sqrt{6}}{22}$ .

The distances between the centers of the two ellipses is then $2b = \cfrac{6\sqrt{2} - \sqrt{6}}{11}$ , so $l = 6$ , $m = 2$ , $n = 11$ , and $l + m + n = \boxed{19}$ .