I'm struggling with a problem of Lecture on discrete geometry by Matusek.
Given $C\subset\mathbb{R}^{d}$ a compact and convex set and $p$ an interior point of $C$, it asks to prove the existence of a line $\ell$ passing through $p$ such that the segment $\ell\cap C$ is at least as long as any segment parallel to $\ell$ and contained in $C$.
Actually this was posted before but the solution given was based on unit vector fields on $S^{n}$.
The idea of the proof is to find parallel supporting hyperplanes of $C$ with endpoints $x,y$ such that the segment $[x,y]$ passes through $p$ and this will give the desire line. Nevertheless, I don't know how to start looking for those hyperplanes.
The book post this exercise after the separation theorem and Carathéodory theorem so I guess this theorems are sufficient to prove the above result.
Thank you in advance :)
Here is a proof that I believe works:
First translate the problem so that $p = 0$.
For each direction $v \in S^{d-1}$, consider the line of $v$. Define $f_d(v)$ to be the length of the line segment intersecting $C$ of that line, minus the maximum length of all such line segments parallel to $v$. By compactness of $C, f_d$ is well-defined. By the convexity of $C, f_d$ should be continuous, but I'll leave confirming that up to you. Since $v$ and $-v$ are parallel, $f_d(-v) = f_d(v)$.
Let $\{e_i\}_{i=1}^d$ be the canonical basis for $\Bbb R^d$ and consider the function $$g_d(v) = f_d(v)v\cdot e_d$$ Clearly $g_d$ is also continuous and $g_d(-v) = -g_d(v)$ for all $v$. By the intermediate value theorem, there must be a $v$ for which $g_d(v) = 0$. That means either $f_d(v) = 0$ or $v$ is orthogonal to $e_d$. In the first case, we are done. $v$ is the direction making the line segment through $0$ the longest of all parallel segments.
If $f_d(v) \ne 0$, then $v$ is orthogonal to $e_d$, we intersect $C$ with the hyperplane perpendicular to $e_d$ and apply the canonical isometry of $\Bbb R^{d-1}\times \{0\}$ with $\Bbb R^{d-1}$ to get a compact convex set $C_{d-1}$ with $0$ in its interior, and end up with the same problem one dimension lower. Applying the same logic, we either get a vector for which $f_{d-1}$ is $0$, thus giving us a solution of this new problem, or else $v$ is orthogonal to $e_{d-1}$, and we reduce to that subspace. Note that $f_{d-1}(v) = f_d((v, 0))$ for all $v \in S^{d-2}$.
If this continues all the way down to $C_1$ in $\Bbb R^1$, then $v\in S^0$ is either $1$ or $-1$ and $v\cdot e_1 = \pm 1 \ne 0$. So there $f_1(v) = 0$ is true. In all cases, there is some $1 \le k \le d$ and some $v_k\in S^{k-1}$ with $f_k(v_k) = 0$, and therefore $f_d(v) = 0$ for $v = (v_k,0,\dots,0)$. Thus we have the desired result.