Generalizing the Borsuk problem: How much can we shrink a planar set of diameter 1 by cutting it into $k$ pieces?

Borsuk's problem asks whether a bounded set in $\mathbb{R}^n$ can be split into $n+1$ sets of strictly smaller diameter. While true when $n=1,2,3$, it fails in dimension $64$ and higher; I believe all other $n$ are open as of this writing.

However, it turns out that at least in the $n=2$ case we can be more precise than "strictly smaller diameter"; if the original set has diameter 1, we can ensure that each piece has diameter at most $\frac{\sqrt{3}}{2}\approx 0.866$, a bound attained by the circle of diameter $1$. To see that this holds, we note that the regular hexagon of width $1$ is a solution to Lebesgue's universal covering problem, and can be split into three sets of diameter $\frac{\sqrt{3}}2$ as well: I am interested in putting bounds on such dissections with more than $3$ pieces: what is the minimum diameter one can ensure when cutting a planar set of unit diameter into $k$ pieces?

Using the same approach as above (finding specific sets with a lower bound, and dissecting a universal cover for sets of diameter 1), I have some bounds for higher $k$ as well, though only for $k=3,4,7$ are they exact:

(Extending this table beyond $k=7$ would be difficult, as working out optimal dissections for the circle would become much more complicated.)

Edit: By taking spokes at $72^\circ$ angles on a regular hexagon (with one spoke meeting the hexagon at the midpoint of a side), I think I can get a slightly better upper bound of around $0.6434$ for the case $k=5$. Optimizing spoke placement further (so that the distances between spoke endpoints are equal) gets me around $0.6223$.

In the limit, I think the diameter of each piece is asymptotic to $\sqrt{\frac{2\pi}{3\sqrt{3}k}}\approx \frac{1.1}{\sqrt{k}}$ by tiling with regular hexagons. Certainly one can do no better than $1/\sqrt{k}$ when dividing the circle, using the isodiametric inequality (if the pieces were any smaller, they would have too little area). Using a trivial dissection of the square, one also has an upper bound of $\frac{\sqrt{2}}{\lceil\sqrt{k}\rceil}$.

Some questions I have about this problem:

• Has this question been investigated before in the literature? If so, what is known?

• Are there any $k$ for which the circle does not present the worst-case scenario for dissection?

• Can the $k=5,6$ upper bounds be substantially improved? I think using Pal's slightly smaller solution to the universal covering problem would allow for a few adjustments when $k=6$, but haven't worked out the details.

what is the minimum diameter one can ensure when cutting a planar set of unit diameter into $k$ pieces?

This problem is considered in 1974 in Problem 102 from [SCY], where the minimum diameter is denoted $\delta_2(k)$. Unfortunately, there are given not much more bounds than in your question. A main tool for the evaluation of $\delta_2(k)$ there is $\delta(k, A)$, the minimum diameter one can ensure when cutting a planar set $A$ of unit diameter into $k$ pieces. Special for $S$ are cases are a disk $D$, a square $S$, and an equilateral triangle $T$. In Problems 103 and table at p. 97 (referenced to paper [Gra] from 1967) bounds $\delta(k, A)$ are shown for $D$ for $k\le 5$, for $T$ and $k\le 10$, and for $S$ and $k\le 4$. Also in [Gra] are evaluated $\delta(k, T)$ for $k\le 15$. When I was a schoolboy, in 1991 I read the article [KK] where were calculated $\delta(2,S)=\tfrac {\sqrt{10}}4$, $\delta(3,S)=\tfrac {\sqrt{130}}{16}=0.712\dots$, and $\delta(5,S)=\tfrac {5\sqrt{34}}{64}=0.455\dots$, found an upper bound $0.4200\dots$ on $\delta(6, S)$, and noted that $\delta(k, D)$ for $k\ge 8$ and $\delta(k,T)$ for $k\ge 16$ are unknown. At pages 96 and 98 are written rather pessimistic thoughts about this approach and in Problem 104 are shown values $\delta_2(2)$, $\delta_2(3)$, $\delta_2(4)$, and $\delta_2(7)$, which you already know. There is noted that no other exact values for $\delta_2(k)$ when $k\ge 2$ are known. Value of $\delta_2(3)$, was, in fact, found by Borsuk [Bor1, Bor2] in 1932–1933 (see also [Gal]). In 1956 a German geometer Lenz [Len1, Len2] thoroughly studied values of $\delta_2(k)$ for small $k$ and calculated $\delta_2(4)$, $\delta_2(5)$ and $\delta_2(7)$. Value of $\delta_2(4)$ was found also by Selfridge [Sel]. In [Gru] is observed that if $G_{11}$ is a regular $11$-gon of diameter $1$ then $\delta_2(6)\ge \delta(6, G_{11})=\frac 1{2\cos (\pi/22)}=0.505141\dots$.

Unfortunately, I don’t speak German, but I guess that in [Len1] at p. 34 are provided bounds $\delta_2(k)\le\tfrac {\sqrt{2}}{\lfloor \sqrt{k}\rfloor}$ for $k\ge 2$ and $\delta_2(k)<\tfrac 1{k-8\pi/\sqrt{27}}\left\lfloor\tfrac {4\pi}{\sqrt{27}}+\sqrt{\tfrac{2\pi k}{\sqrt{27}} }\right\rfloor$ for $k\ge 5$, and at p. 36 a bound $\delta_2(k)\le\tfrac 1{k-1}\left(\tfrac {2}{\sqrt{3}}+\sqrt{\tfrac 43+ \frac{2\pi}{\sqrt{27}}(k-1) }\right)$. Both latter bounds are about $\sqrt{\frac{2\pi}{\sqrt{27}}}k^{-1/2}\approx 1.1 k^{-1/2}$.

But these references are old and some progress could be made from that time.

We should have $\delta_2(k)\approx \sqrt{\frac{2\pi}{\sqrt{27}}}k^{-1/2}$ asymptotically, see below.

A lower bound. Given $k$, Pigeonhole principle implies $\delta_2(k)\ge d(k+1)/2$, where $d(k+1)$ be a maximal possible minimum distance between $k+1$ points of the unit disk, see this thread. This approach should provide an asymptotic bound $\delta_2(k)\ge\approx \sqrt{\tfrac {2\pi}{3\sqrt{3}k}}\approx 1.1 k^{-1/2}$.

An upper bound. Let $C$ a be a (not necessarily convex) subset of the plane that contains a congruent copy of every planar set of unit diameter and $a$ be an area of $S$. The best known bounds for $a$ are about $0.8441$, see a thread about a hard and ungrateful quest for them. If $C$ can be covered by $k$ cells of a hexagonal grid with side $d$ then $\delta_2(k)\le 2d$. This approach should provide an asymptotic bound $\delta_2(k)\le\approx 2\sqrt{\tfrac {2a}{3\sqrt{3}k}}\approx 1.14 k^{-1/2}$.

But Lenz’ bound suggest that we don’t need to use an universal covering set, because at p.11 of [Lit] it is shown that “an area of (greatest) diameter not greater than $1$ is at most $\tfrac{\pi}4$”.

This observation should point to an asymptotically tight upper bound.

References

[Bor1] K. Borsuk, Über die Zerlegung einer euklidischen $n$-dimensionalen Vollkugel in $n$ Mengen, Verhandlungen Intern. Math. Kongr., Zürich 2 (1932) 192.

[Bor2] K. Borsuk, Drei Sätze über die $n$-dimensional Späre, Fundamenta Math. 20 (1933), 177–190.

[Gal] D. Gale, On inscribing $n$-dimensional sets is a regular $n$-simplex, Proc. Amer. Math. Soc. 4 (1953) 222–225.

[Gra] R.L. Graham, On partitions of a equilateral triangle, Canadian Journ. Math. 19 (1967) 394–409.

[Gru] B. Grünbaum, Etudes in combinatorial geometry and the theory of convex bodies, Moskow, Nauka, 1971, in Russian.

[KK] I. Kokorev, L. Kurlyandchik, A big cake on small plates, Kvant 7 (1991) 13–17.

[Len1] H. Lenz, Über die Bedeckung ebener Punktmengen durch solche kleineren Durchmessers, Archiv Math. 7 (1956) 34–40, doi:10.1007/bf01900521.

[Len2] H. Lenz, Zerlegung ebener Bereiche in konvexe Zellen von möglichst kleinem Durchmessers, Jahresber. Deutsch. Math. Vereinigung 58 (1956) 87–97.

[Lit] J.E. Littelwood, A Mathematician’s Miscellany, Methued & Co, London, first published in 1953.

[SCY] D.O. Shklyarskiy, N.N. Chentsov, I.M. Yaglom, Geometrical estimations and combinatorial geometry problems, Moskow, Nauka, 1974, in Russian.

[Sel] J.L. Selfridge, An informal seminar on the coverings of convex sets (Report of the Inst. in the Theory of Numbers), Colorado, 1959. 334.

Here V. P. Filimonov, Covering planar sets, Mat. Sb., 2010, Volume 201, Number 8, 127–160 some more results about this problem, but the article is in Russian. Here link this article on English, But I don't know how to access it.

In any case, you can see pictures of the partitions of the universal covering set to prove the upper bounds of $d_n$(in his article $d_n$ is $\delta_2(n)$). On the last page, you can see the improved results at that time. The first and last columns show improved scores from the bottom and top, respectively