Consider a convex body $K \subset \mathbb{R}^n$ containing the origin in its interior. Although the body is not necessarily symmetric, let us say that two points in its boundary $\partial K$ are antipodal if the origin lies in the segment that joins them.

Question 1. Assuming that $n$ is odd, does there always exist a pair of parallel affine hyperplanes that support $K$ at a pair of antipodal points?

If the boundary of $K$ is $C^1$ this is true and here is the simple proof: For every point $x \in \partial K$ consider the tangent hyperplane at $x$ and the tangent hyperplane at the point $\bar{x}$ antipodal to it. Translate this second hyperplane so that it passes through $x$ and consider its intersection with $T_x \partial K$. If the hyperplanes are not parallel, we obtain an $(n-2)$-plane tangent to $\partial K$ at $x$. If the tangent hyperplanes at antipodal points are never parallel, we get a continuous field of tangent hyperplanes in the tangent bundle of $\partial K$. Since $\partial K$ is an even-dimensional sphere, this is impossible. Q.E.D.

What happens if the boundary is not smooth and we allow some points to be points of support for multiple support hyperplanes?

Question 2. Assume $n$ is even. Does there exist some convex body $K$ (with a possibly smooth boundary) for which the support planes at antipodal points are never parallel?

These questions came up in a conversation with Constantin Vernicos who convinced me that we didn't really need the answers for what we were doing. Still, I remain curious about the subject.

  • 1
    $\begingroup$ I'm discovering this question now. Interestingly, we faced exactly the same problem in a recent paper about convex cones. In case you are interested, this is Lemma 9 in arxiv.org/abs/1911.09663 (beware that we call antipodal is not the same as you). $\endgroup$ Mar 31, 2021 at 8:59
  • $\begingroup$ @GuillaumeAubrun, thanks for the reference. $\endgroup$ Mar 31, 2021 at 15:06

2 Answers 2


I will try to answer your question 2. The answer is negative. There must be at least two antipodal points with parallel tangent planes. Here is a sketch of a proof. The idea of that proof is that points you are looking for have variational nature. Proof works in any dimension.

Denote the boundary of your body by $\Gamma$. Now we fix a Euclidean structure on $\mathbb R^n$. Let $f\colon S^{n-1}\to \mathbb R$ be a support function of $\Gamma$, here $S^{n-1}$ is a unit sphere and $f(x)=\max_{v\in \Gamma} (x,v)$, $(,)$ is the scalar product. Denote by $v(x) \in \Gamma$ such a point that $f(x)=(x,v(x))$. The tangent plane to $\Gamma$ at the point $v(x)$ is orthogonal to $x$. Then two points $v(x)$ and $v(-x)$ are antipodal if and only if there exists $\lambda>0$ such that $x$ is a critical point of the function $f(y)-\lambda f(-y)$ with zero critical value. The functions $f(y)$ and $f(-y)$ are positive (here we use the fact that the origin belongs to the interior of $\Gamma$), hence for some $\lambda>0$ zero is a critical value of $f(y)-\lambda f(-y)$. That is it.

One can easily generalize that statement - Let two compact convex bodies both contains origin in the interiors. Than there exist at least two rays passing through the origin, such that support planes of their intersections with boundaries are parallel.

  • $\begingroup$ Thanks Petya, this is a very neat proof. You may want to mention (for the non-convex reader) that $df(x)= v(x) \cdot dx$. Since your proof is variational and the support function is just the generating function of the Legendrian submanifold given by the convex body and its tangent spaces, I guess your proof can be generalized to a contact topological statement. $\endgroup$ Dec 14, 2012 at 9:30
  • $\begingroup$ I formulated the precise contact topology question here: mathoverflow.net/questions/116356/… $\endgroup$ Dec 14, 2012 at 9:57

Since Petya answered question 2, let me answer question 1. Set $S=\partial K$ and assume at first that $S$ is smooth. Let $f,g:S^{n-1}\to S^{n-1}$, where $S^{n-1}$ is the unit sphere, be the maps that take an element $x$ to the outward unit normal to $S$ at the points of intersection of $S$ with the half-lines starting at the origin and going in the direction of $-x$ and $x$ respectively. These maps are homeomorphisms and they are homotopic to minus identity and identity respectively. So the map $x\mapsto g^{-1}(-f(x))$ has degree 1, and so it has a fixed point by the Lefschetz theorem (here we use the fact that $n$ is odd, so $n-1$ is even). Any such fixed point gives you a pair of opposite points on $S$ with parallel tangent planes.

upd: the above argument was essentially given in the posting. Let me explain how one can deduce the general case from it. One can approximate (e.g., with respect to the Hausdorff metric) an arbitrary compact convex $K$ with a sequence $(K_i)$ of smooth convex bodies that all have the origin in their interiors. For each $i$ let $a_i,b_i\in\partial K_i$ be opposite points with parallel tangent planes. By choosing a subsequence we may assume that there are $a=\lim a_i,b=\lim b_i$, which will again be opposite points in $\partial K$. By choosing a subsequence again we may assume that $\lim T_{a_i}\partial K_i$ and $\lim T_{b_i}\partial K_i$ also exist. These will be the required support planes.

  • $\begingroup$ Thanks alori. I think your approximation argument can be summarized as follows: when a sequence of smooth convex bodies converges to a given convex body, the sequence of their duals also converges. $\endgroup$ Dec 14, 2012 at 9:33

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