# Does every convex planar set contain a centrally symmetric subset with at least $2/3$ its area?

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: 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.

• 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.) Mar 8, 2021 at 19:04