# Furthest distance half the diameter?

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.

• 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 Sep 8, 2019 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.
• Nice---Thanks! ${}$ Sep 7, 2019 at 0:32