the segment through a fixed interior point of a compact convex set which is at least as long as its parallels The following is an exercise from Lectures on Discrete Geometry by J. Matousek, which I find hard.
Let $C\subset \mathbb{R}^d$ be a compact convex set with a nonempty interior, and let $p\in C$ be an interior point. Show that there exists 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 $C$. 
 A: You can use the following statement: There is no unit vector field $v$ on $S^n$ that is anti-symmetric with respect to the central symmetry of $S^n$. 
This theorem holds because if you would have such a unit field, then you would have a homothopy from $S^n$ to $S^n$ that connects the identity map with a map from $S^n$ to 
itself that factors through a map to $RP^n$. But any such map  $S^n\to RP^n\to S^n$ has and even degree (i.e. the degree can not be $1$).
Now if you accept the statement, you can reason as follows. 
Denote by $C_{\varepsilon}$ the set of points  on distance at most $\varepsilon$ from $C$. It is sufficient to prove the statement for each $C_{\varepsilon}$ and then take the limit $\varepsilon\to 0$. Now, $C_{\varepsilon}$ has $C^1$-smooth boundary, i.e. at each point of $\partial C_{\varepsilon}$ there is a unique supporting hyperplane. 
Suppose that for every segment $[x,y]$ that passes through $p$ and such that $x,y\in \partial C$ the supporting hyperplanes $P_x$ and $P_y$ intersect. Then chose unit vectors in $P_x$ and $P_y$ at $x$ and $y$ correspondingly, that point exactly towards $P_x\cap P_y$. This will give you the desired vector-filed (anti-symmetric with respect to the involution of $\partial C$ that exchanges the ends of segments passing through $p$).
I never saw the book so I don't know if this is the solution that is expected in this book
