Let $S$ be the surface of a convex body, polyhedral or smooth, embedded in $\mathbb{R}^3$. For a point $x \in S$, let $F(x)$ be the set of furthest points from $x$, measured by shortest paths on the surface $S$. Let $f(x)$ be the length of those shortest paths: $|x y|$ for $y \in F(x)$.

It seems natural to hope that

Hypothesis: For any $x \in S$, $f(x) \ge \tfrac{1}{2} \mathrm{diam}(S)$.

Here $\mathrm{diam}(S)$ is the maximum distance between any two points on $S$ (again measured by shortest paths on the surface of $S$). Suppose, for example, that $\rho$ is a diameter-realizing geodesic. Then for any $x \in \rho$, $f(x) \ge \tfrac{1}{2} |\rho|$, just tracking along $\rho$.

A non-comprehensive literature search has failed to uncover a relationship between $f(x)$ and $\mathrm{diam}(S)$.

Itoh, Jin‐ichi and Costin Vǐlcu. "Criteria for farthest points on convex surfaces." Mathematische Nachrichten 282, no. 11 (2009): 1537-1547. Journal link.

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    $\begingroup$ You could rewrite the hypothesis as $\forall X \in S, f(x) \ge \frac12 \max\limits_{y \in S} f(y)$ and avoid mentioning the diameter $\endgroup$ – Henry Sep 8 at 0:31

Denote the diameter by $d$ and distance by $|x-y|$. Then there are $y,z$ such that $d=|y-z|$ and we have by triangle inequality for every $x$: $$d=|y-z|\leq |y-x|+|x-z|\leq 2f(x),$$ so we obtain your inequality. Notice that I did not use convexity, or any other of your assumptions, only the triangle inequality.

  • $\begingroup$ Nice---Thanks! ${}$ $\endgroup$ – Joseph O'Rourke Sep 7 at 0:32
  • $\begingroup$ The key that I missed is that the triangle inequality holds on the surface for the metric of geodesic-shortest paths. $\endgroup$ – Joseph O'Rourke Sep 8 at 0:01

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