HMMT February

Solution:
Let $\Gamma$ be an ellipse passing through $A=(2,0)$, $B=(0,3)$, $C=(0,7)$, $D=(6,0)$, and let $P=(0,0)$ be the intersection of $AD$ and $BC$. $\frac{\text{Area of } \Gamma}{\text{Area of } ABCD}$ is unchanged under an affine transformation, so we just have to minimize this quantity over situations where $\Gamma$ is a circle and $\frac{PA}{PD}=\frac{1}{3}$ and $\frac{PB}{BC}=\frac{3}{7}$. In fact, we may assume that $PA=\sqrt{7}$, $PB=3$, $PC=7$, $PD=3\sqrt{7}$. If $\angle P=\theta$, then we can compute lengths to get
$$r=\frac{\text{Area of } \Gamma}{\text{Area of } ABCD}=\pi \frac{32-20\sqrt{7}\cos\theta+21\cos^2\theta}{9\sqrt{7}\cdot\sin^3\theta}$$
Let $x=\cos\theta$. Then if we treat $r$ as a function of $x$,
$$0=\frac{r'}{r}=\frac{3x}{1-x^2}+\frac{42x-20\sqrt{7}}{32-20x\sqrt{7}+21x^2}$$
which means that $21x^3-40x\sqrt{7}+138x-20\sqrt{7}=0$. Letting $y=x\sqrt{7}$ gives
$$0=3y^3-40y^2+138y-140=(y-2)\left(3y^2-34y+70\right)$$
The other quadratic has roots that are greater than $\sqrt{7}$, which means that the minimum ratio is attained when $\cos\theta=x=\frac{y}{\sqrt{7}}=\frac{2}{\sqrt{7}}$. Plugging that back in gives that the optimum $\frac{\text{Area of } \Gamma}{\text{Area of } ABCD}$ is $\frac{28\pi\sqrt{3}}{81}$, so putting this back into the original configuration gives Area of $\Gamma \geq \frac{56\pi\sqrt{3}}{9}$. If you want to check on Geogebra, this minimum occurs when the center of $\Gamma$ is $\left(\frac{8}{3}, \frac{7}{3}\right)$.