HMMT-Feb 2016 guts, Problem 27

Find the smallest possible area of an ellipse passing through $(2,0),(0,3),(0,7)$, and $(6,0)$.

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 $A D$ and $B C$. Area of $\Gamma$ As is inchanged under an affine transformation, so we just have to minimize this quantity over situations where $\Gamma$ is a circle and $\frac{P A}{P D}=\frac{1}{3}$ and $\frac{P B}{B C}=\frac{3}{7}$. In fact, we may assume that $P A=\sqrt{7}, P B=3, P C=7, P D=3 \sqrt{7}$. If $\angle P=\theta$, then we can compute lengths to get \[ r=\frac{\text { Area of } \Gamma}{\text { Area of } A B C D}=\pi \frac{32-20 \sqrt{7} \cos \theta+21 \cos ^{2} \theta}{9 \sqrt{7} \cdot \sin ^{3} \theta} \] Let $x=\cos \theta$. Then if we treat $r$ as a function of $x$, \[ 0=\frac{r^{\prime}}{r}=\frac{3 x}{1-x^{2}}+\frac{42 x-20 \sqrt{7}}{32-20 x \sqrt{7}+21 x^{2}} \] which means that $21 x^{3}-40 x \sqrt{7}+138 x-20 \sqrt{7}=0$. Letting $y=x \sqrt{7}$ gives \[ 0=3 y^{3}-40 y^{2}+138 y-140=(y-2)\left(3 y^{2}-34 y+70\right) \] 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 $\frac{\text { Area of } \Gamma}{\text { Area of } A B C D}$ 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 $\left(\frac{8}{3}, \frac{7}{3}\right)$. $\fbox{\frac{56 \pi \sqrt{3}}{9}}$.

Answer: \frac{56 \pi \sqrt{3}}{9}