Let $S$ be a bounded convex subset of the plane of unit area. Can we guarantee the existence of a centrally-symmetric subset $C⊆ S$ of area $2/3$?

If $S$ is any triangle, this bound is tight, attained by a hexagon whose vertices are $1/3$ of the way across each side: enter image description here I suspect the triangle is the unique worst case for this problem, but didn't see a great way to prove it - perhaps there is some other convex shape which only has a centrally symmetric subset of area at most $0.6$ or something, though I would be quite surprised.

It's easy to find a $C$ with at least half the area of $S$, because every convex set contains a rectangle of at least half its area (see e.g. this MO thread for references). How much can we improve this lower bound?

In the event of a positive resolution, I am curious whether one can find such a subset only using hexagonal $C$.

I imagine this question is discussed somewhere in the literature; if so, pointers to relevant papers or extensions to higher dimensions would be welcome.


1 Answer 1


Yes, this is the Kovner-Besikovich Theorem.

  • $\begingroup$ Thank you! Do you know if this bound still holds when restricted to hexagonal subsets, or of the lower bounds in higher dimensions? I don't see anything like this mentioned in the linked article. $\endgroup$ Mar 8, 2021 at 18:14
  • $\begingroup$ @RavenclawPrefect Presumably it is better then, since the article points out that the worst are triangles. $\endgroup$
    – Igor Rivin
    Mar 8, 2021 at 18:35
  • 1
    $\begingroup$ Sorry, I think I may have miscommunicated - I am asking about the case when $C$ is restricted to be hexagonal, not when $S$ is. (Since triangles' optimal subset is always a hexagon, any decreased lower bounds would need to be witnessed by a non-triangular set.) $\endgroup$ Mar 8, 2021 at 19:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .