Let $S \subset \mathbb{R}^n$ be the boundary of a centrally symmetric convex body and provide $S$ with the geodesic metric given by its embedding in Euclidean space (i.e., the distance between two points is the infinimum of the Euclidean lengths of all rectifiable curves on $S$ that join them).

Question. Is the diameter of $S$ realized by a pair of antipodal points?

I am curious about this question when the metric is the induced metric from euclidean space, but I'm mostly inetrested in the case of a polytope provided the graph metric on the 1-skeleton. In the discrete case I do not care too much about constants.

  • $\begingroup$ I edited the question, but the second part about the graph metric is still unclear to me. $\endgroup$ Apr 17, 2014 at 15:04
  • $\begingroup$ By the "graph metric on the 1-skeleton", I take it you mean the metric on the graph formed by the edges. But this metric is not the restriction of the geodesic metric on the whole boundary, nor is the optimum for the full question necessarily realized by vertices (as noted by Noam Elkies for the cube). So the discrete question remains a little unclear to me. $\endgroup$ Apr 21, 2014 at 11:07
  • $\begingroup$ Indeed, these two questions might be very different. Probably it wasn't very wise to write about both in the same MO post. The discrete metric is defined only on the vertex set of the polytope. Paths are restricted to walk on the $1$-skeleton and their length is given by the graph metric, i.e. the length of every edge is prescribed to be $1$ (no matter how far the corresponding vertices with respect to the Euclidean distance). @Dylan Thurston. $\endgroup$ Apr 23, 2014 at 18:33
  • $\begingroup$ Here @alvarezpaiva the previous comment should clarify the confusion. $\endgroup$ Apr 23, 2014 at 18:34

5 Answers 5


For $n=3$ this question was asked in 1996 by James Propp, conjecturing that the answer is Yes. (I got the reference from this page, which concerns the very special case of a rectangular box in ${\bf R}^3$; this is already a nontrivial problem, as Jim Propp noted, and the cuboid page reports calculations claimed to prove it in that case and to locate the diameter, which isn't always a pair of opposite corners.)

David Eppstein's Geometry Junkyard page that quotes Propp's 1996 conjecture also reproduces a 2002 e-mail from Costin Vilcu with a reference to his paper "On Two Conjectures of Steinhaus" (Geom. Ded. 79 (2000), 267-275), whose Proposition 6 (page 273) proves the $n=3$ conjecture.

Perhaps Costin Vilcu knows the current status of the problem for higher $n$. Even the case of rectangular boxes in ${\bf R}^4$ might still be open.


This answer a different questions than was intended. The intended question remains unanswered!

Suppose pair of points $x,y\in K$ attains the diameter. Then $d(x,y)=d(-x,-y)=\operatorname{diam}(K)$. The points $x,-x,y,-y$ are coplanar. Look inside that plane. Consider the parallelogram spanned by these four points. By the parallelogram law, $$d(x,-x)^2+d(y,-y)^2=2d(x,y)^2+2d(x,-y)^2\geq 2d(x,y)^2.$$ Hence, either $d(x,-x)$ or $d(y,-y)$ is least $\operatorname{diam}(K)$.

  • 3
    $\begingroup$ Hi Boris! sorry I was not very clar about what I meant by "induced metric from Euclidean space" I mean the infimal length of all rectifible curves that stay on the boundary of the convex body. Where the length of a curve is measured locally with the Euclidean metric, so if the body is smooth this is a Riemannian metric. $\endgroup$ Apr 14, 2014 at 14:13
  • 1
    $\begingroup$ @Alfredo, you should probably write it as $\operatorname{diam}(\partial K)$. $\endgroup$ Apr 14, 2014 at 14:59
  • $\begingroup$ Aha, the intended question is much more interesting than the one I answered above. $\endgroup$
    – Boris Bukh
    Apr 14, 2014 at 19:09

I am posting this on behalf of Costin Vîlcu (with whom I've had the pleasure of coauthoring). —J.O'Rourke

The posted problem in $\mathbb{R}^3$ was answered in:

Y. G. Nikonorov and Y. V. Nikonorov, The Intrinsic Diameter of the Surface of a Parallelepiped, Discrete Comput. Geom. 40 (2008), 504-527.

I (Costin) know nothing about a higher dimensional analogue.

For doubles of arbitrary dimensional simplices, a similar result is in

J. Itoh and C. Vîlcu, Farthest points and cut loci on some degenerate convex surfaces, J. Geom. 80 (2004), 106-120.

These authors,

K. Grove and P. Petersen, A radius sphere theorem, Inventiones Math. 112 (1993), 577-583,

show that the mapping $F$, associating to each point its set of farthest points, is single-valued and onto for any Alexandrov space with curvature bounded below by $1$ and with $\min_x \mathrm{dist} (x, F(x)) > \pi /2$.


This is more of a followup to Noam's answer. Vilcu had continued to work on the subject, see this paper by him, Rouyer and Itoh: https://www.evernote.com/shard/s24/sh/a5f3c60d-7c69-4a02-81db-e5721253977d/a51556c9763fe199c9981c13d950b97c

However, the most relevant papers seem to be those by Joel Rouyer (unfortunately my university does not subscribe to "Advances in Geometry", so I can't tell with certainty what he does...):

MR2660417 (2011g:52021) Reviewed Rouyer, Joël On antipodes on a convex polyhedron II. (English summary) Adv. Geom. 10 (2010), no. 3, 403–417. 52B10 (51M04 52A15) More links PDF Clipboard Journal Article Make Link

This paper is a sequel to the author's paper [Part I, Adv. Geom. 5 (2005), no. 4, 497–507; MR2174479 (2006h:52005)]. Steinhaus asked about convex surfaces for which every point has precisely one antipode and the antipodal map is an involution. It is open whether the boundary of a convex polyhedron in R3 can satisfy that condition. This paper gives several relevant results about antipodes on convex polyhedra. For example: If a convex polyhedron has a small enough angle at some vertex, then some point has at least two antipodes. For every convex polyhedron, the antipodal map is not a local isometry. If P is a centrally symmetric convex polyhedron, then there is a finite union G⊂P of algebraic arcs such that each point p∈P∖G has a single antipode, which is not −p; hence no centrally symmetric convex polyhedron satisfies Steinhaus's conditions. Reviewed by Margaret M. Bayer


This is the situation in three dimensions for a centrally symmetric surface $S$ with the induced Euclidean metric:

  1. For simply-connected $S$ (convexity is not required), if the diameter of $S$ is attained, it is attained by a pair of antipodal points.
  2. This is not necessarily true if $S$ is not simply connected.

Judging by the answers so far, these might be new results for non-convex $S$. So I'd better prove them.

Outline of Proof of 1 The idea is very simple. Let $X$ and $Y$ be any points in $S$, with antipodal points $X'$ and $Y'$. Let $\gamma_X$ and $\gamma_Y$ be rectifiable paths from $X$ to $X'$ and from $Y$ to $Y'$ respectively. Then we show that there exists a path from $X$ to $Y$ which is no longer than $\max(|\gamma_X|, |\gamma_Y|)$: if $P$ is a point of intersection of $\gamma_X$ and $\gamma_Y$, this will be either the path from $X$ to $P$ to $Y$, or the antipodal image of the path from $X'$ to $P$ to $Y'$. And if $\gamma_X$ doesn't intersect $\gamma_Y$, then the antipodal image of $\gamma_X$ must (because $S$ is simply connected).

Lemma Let $S$ be a simply-connected, centrally symmetric surface in $\mathbb R^3$. Let $X$ and $Y$ be points in $S$, with antipodal points $X'$ and $Y'$ respectively. Let $\gamma_X$ be any rectifiable path from $X$ to $X'$, and $\gamma_Y$ be any rectifiable path from $Y$ to $Y'$. Then there exists a rectifiable path $\gamma_{XY}$ from $X$ to $Y$ with $|\gamma_{XY}| \le \max(|\gamma_X|,|\gamma_Y|)$.

Proof of Lemma Let $\gamma'_X$ denote the antipodal image of $\gamma_X$. Then $\gamma_X+\gamma'_X$, the concatenation of $\gamma_X$ and $\gamma'_X$, is a closed curve; because $S$ is simply connected, this curve divides $S$ into two regions, which are antipodal images of each other. One of these two regions contains $Y$, and the other contains $Y'$; therefore $\gamma_Y$ crosses $\gamma_X+\gamma'_X$, at a point $P$. Without loss of generality, we may assume that $P$ lies on $\gamma_X$.

Now, letting $\gamma_{XP}$ denote the path along $\gamma_X$ from $X$ to $P$ etc., we have $|\gamma_X| = |\gamma_{XP}|+|\gamma_{PX'}|$ and $|\gamma_Y| = |\gamma_{YP}|+|\gamma_{PY'}|$. But then

$$\begin{align} |\gamma_{XPY}|+|\gamma_{X'PY'}| &= (|\gamma_{XP}|+|\gamma_{PY}|) + (|\gamma_{X'P}|+|\gamma_{PY'}|) \\ &= (|\gamma_{XP}|+|\gamma_{PX'}|) + (|\gamma_{YP}|+|\gamma_{PY'}|) \\ &= |\gamma_X|+|\gamma_Y| \\ &\le 2\max(|\gamma_X|,|\gamma_Y|) \end{align}$$

So at least one of $|\gamma_{XPY}|, |\gamma_{X'PY'}|$ is $\le \max(|\gamma_X|,|\gamma_Y|)$. If it is $|\gamma_{XPY}|$, then take $\gamma_{XY} = \gamma_{XPY}$. If it is $|\gamma_{X'PY'}|$, then take $\gamma_{XY} = $ the antipodal image $\gamma'_{X'PY'}$. This proves the Lemma.

Proof of 1 Suppose the diameter $d$ of $S$ is attained by points $X,Y$: that is, all paths from $X$ to $Y$ have length at least $d$. Then $|\gamma_{XY}| \ge d$ for all paths $\gamma_{XY}$ from $X$ to $Y$. Hence, by the Lemma, if $\gamma_X$ and $\gamma_Y$ are paths from $X$ to $X'$ and from $Y$ to $Y'$, then one of $|\gamma_X|$ and $|\gamma_Y|$ is at least $d$. Hence the diameter is attained by antipodal points, either $(X,X')$ or $(Y,Y')$. $\square$

Proof of 2 We describe a centrally symmetric surface $S$, not simply connected, whose diameter is attained, but not by a pair of antipodal points.

Imagine a vertical hollow tube, of radius about 1 and height about 4. Its walls are thin. Half-way up, two long, sharp spikes protrude horizontally from it, one to the left and one to the right. They are conical, with a base radius of about $\frac12$, and length about $10$. (The exact numbers are not important; this is just to give you a picture.) These spikes are also hollow, with thin walls.

The two antipodal points that are furthest apart are the external tips of the spikes. But the diameter is attained by an external spike tip and the opposite, internal spike tip; the distance between these two points is greater, because any path between the two has to climb up over the lip of the tube to get from one to the other.

I hope this is clear enough; if not, I will try to draw some pictures.


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