# Kakeya crossed-needles problem

The Kakeya needle problem asks for the minimum area planar region in which one can completely turn around a line segment through a series of translations and rotations. There is no minimum: There are "Kakeya needle sets" of arbitrarily small area.

I ask the same question but for a rigid plus-sign, two equal-length segments at $$90^\circ$$ sharing their midpoints, forming a $$+$$ shape. Because it seems difficult to achieve $$360^\circ$$ rotation using the type of spikey sets so effective for a single needle, I'm wondering if the answer here might be just a disk?

• Do you know the answer for three lines at equal angles? Jun 23 at 10:46
• @VilleSalo: No, I don't. Nor for a T-shape. I thought perhaps the + is the easiest to settle. Jun 23 at 11:08

I do not know if the minimal area needed to rotate the + is that of the smallest disk containing it. However, I do know that it cannot be done with arbitrarily small area.

This is a special case of a beautiful general theorem by Csörnyei, Héra, and Laczkovich, Theorem 1.2 of Closed sets with the Kakeya property.

I will try to translate the relevant piece of their theorem to have a bit less jargon, but the paper is very clear.

Theorem: Let $$A$$ be a non-empty, closed, connected subset of the plane. Suppose that there exists a non-trivial rigid motion $$m$$ of the plane such that $$A$$ can be continuously moved through rigid motions to $$m(A)$$ inside sets of arbitrarily small area. Then $$A$$ must be a line segment, a half-line, a full line, a circular arc, a circle, or a singleton.

There may be a simple argument for the plus sign that also gives your conjectured bound; I haven't thought about it. Maybe looking at their proof would provide it.

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• Wow, that's an amazing theorem! It still seems possible that there is a set of area smaller than the disk area, but not arbitrarily small. 2 days ago
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• I do not get this. If "full line" means "unbounded straight line", how can it be moved within a set of arbitrarily small area unless it is never rotated? And if it is never rotated, then it is simply not moved at all! 2 days ago
• @user21820 I believe the point is that the motion $m$ can be a nontrivial translation of the plane parallel to the line. Then, of course, the line never changes along $m$, but $m$ is a nontrivial rigid motion of the plane. 2 days ago
• @LukasMiaskiwskyi: I see. That makes sense, thanks! 2 days ago