Largest ellipsoid section Suppose that we have an ellipsoid in $\mathbb{R}^3$ with centre at $0$. We consider all plane sections that pass through $0$ (which are all ellipses). Of all these ellipses which has the largest perimeter?
What happens in $\mathbb{R}^n$?
 A: For any geometric figure $X$ in a plane, let $\ell_X$ be its perimeter.
Recall

For any convex bodies $A, B$ in a plane, if $A \subseteq B$, then $\ell_A \le \ell_B$.

Let $\hat{z} = (0,0,1)$ and

*

*$\mathcal{E}$ be the solid ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \le 1$ with $a \ge b \ge c$.

*$\mathcal{E}'$ be the solid ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2} + \frac{z^2}{b^2} \le 1$.

For any unit vector $\hat{n} \in S^2$, let  $\Omega(\hat{n})$ and $\Omega'(\hat{n})$ be the intersection of $\mathcal{E}$ and $\mathcal{E}'$ with the plane $\hat{n}\cdot \vec{r} = 0$.
It is clear $\Omega(\hat{n})$ and $\Omega'(\hat{n})$ are solid ellipses, lying in same plane with $\Omega(\hat{n}) \subseteq \Omega'(\hat{n})$. Above result tell us $$\ell_{\Omega(\hat{n})} \le \ell_{\Omega'(\hat{n})}$$
Furthermore, $\Omega'$ is an ellipse with semi-minor axis $b$ and semi-major axis $a' \le a$. By suitable rigid motion, we can "move" and "fit" it inside $\Omega'(\hat{z})$. This means
$$\ell_{\Omega'(\hat{n})} \le \ell_{\Omega'(\hat{z})}$$
Combine these and notice $\Omega(\hat{z}) = \Omega'(\hat{z})$, we obtain
$$\ell_{\Omega(\hat{n})} \le \ell_{\Omega'(\hat{n})} \le \ell_{\Omega'(\hat{z})} = \ell_{\Omega(\hat{z})}$$
So the ellipses with largest parameter is the intersection of $\mathcal{E}$ with $xy$-plane (i.e. the plane containing the two longest axes).
The generalization to $\mathbb{R}^n$ for $n > 3$ is sort of obvious. One just replace

*

*$\mathcal{E}$ by $\sum\limits_{k=1}^n \frac{x_k^2}{a_k^2} \le 1$ with $a_1 \ge a_2 \ge \cdots \ge a_n$.

*$\mathcal{E}'$ by $\frac{1}{a_1^2}x_1^2 + \frac1{a_2^2}\sum\limits_{k=2}^n x_k^2 \le 1$.

Aside from one can no longer label the planes by unit vectors, above arguments continue to work.
