# Distance from constant width bodies

EDIT As @David has observed, my conjecture was clearly wrong for $$\ n:=2.\$$ Let me still give it a chance for $$\ n\ge 3$$.

I'll call a family $$\ F\$$ of bound closed convex subsets of $$\ \mathbb R^n\$$ to be impressive $$\ \Leftarrow:\Rightarrow\$$ each set $$\ B\in F\$$ has constant width greater or equal $$1$$, and each two different members of $$\ F\$$ have disjoint interiors. Family $$\ F\$$ is called assuming $$\ \Leftarrow:\Rightarrow\ \ F\$$ is impressive and the width of each $$\ B\in F\$$ is exactly $$1\$$ (so that all diameters are $$1$$ under the given circumstances).

The following conjecture seems obvious but I don't have a proof:

CONJECTURE   Let $$\ n>2.\$$ There exists real $$\ \delta_n > 0\$$ such that for every impressive $$(n+1)$$-element family $$\ F\$$ in $$\ \mathbb R^n,\$$ and every $$\ x\in\mathbb R^n,\$$ the following inequality holds:

$$\max_{B\in F} d(x\ B)\ \ge\ \delta_n$$

where $$\ d(x\ B) := \inf_{y\in B}\, ||x-y||\$$ (the Euclidean norm is meant).

The harder challenge seem to be the exact computation of the maximal possible $$\ \delta_n.\$$ Now let's still call this maximal constant simply $$\ \delta_n.\$$ Furthermore, I'd like to know also a similar constant $$\gamma_n\$$ restricted to the assuming families, i.e. $$\ \gamma_n\$$ is the maximal constant such that for every $$(n+1)$$-element assuming family $$\ F,\$$ and for every $$\ x\in\mathbb R^n,\$$the following inequality holds:

$$\max_{B\in F} d(x\ B)\ \ge\ \gamma_n$$

Obviously, $$\ \, \delta_n\, \le\ \gamma_n$$.

Needless to say, I apologize if this problem is well-known.

• I imagine that there are higher-dimensional "pyramidlike" analogues of Reuleaux-triangles; is it not possible to let $n+1$ such meet at a point also, as in the $n=2$ case? – Per Alexandersson Sep 14 '14 at 8:41
• @Per, I simply don't know. It'd be interesting to find out one way or another. – Włodzimierz Holsztyński Sep 14 '14 at 9:11
• @PerAlexandersson: I think you are correct. The Reuleaux tetrahedron should do the trick. – Joseph O'Rourke Sep 14 '14 at 18:09
• The Reuleaux tetrahedron is not of constant width. But the Meissner body should work. – Yoav Kallus Sep 14 '14 at 18:32
• @YoavKallus: Thanks for the correction! Changed the image accordingly. – Joseph O'Rourke Sep 15 '14 at 0:24

Not an answer; just an illustration. I had some difficulty understanding the question, so...

Here $n=2$, so the shapes are planar, $\mathbb{R}^2$. I used Reuleaux triangles for the $3=n{+}1$ unit-constant-width bodies $F=\{ B_1, B_2, B_3 \}$ forming an "impressive" and "assuming" family $F$. A particular point $x \in \mathbb{R}^2$ is shown, with segments achieving $d(x,B_i)$. In this case, all three of those min-distances to the bodies are equal, so that is also the max $\gamma_2$. So I think the question is simply asking if there is a lowerbound on the radius of a ball that can nestle in the gap.? I.e., can we ensure that the gap is not arbitrarily small?

Apologies if I am misinterpreting...

Added: To address $d{=}3$ & Per A.'s question, here is an image (from here) of a constant-width Meissner tetrahedron: • But what happens if the Reuleaux triangles are all translates of each other and meet at $x$? Is that a counterexample or am I misunderstanding the question? – David Eppstein Sep 14 '14 at 1:38
• @DavidEppstein: I think that is a counterexample! Let us await Włodz to weigh in... – Joseph O'Rourke Sep 14 '14 at 1:52
• @David--you're right. I am very sorry. I'll soon remove the question. (First, I'll give you and Joseph to read my answer). At least I know that my conjecture was new :-) – Włodzimierz Holsztyński Sep 14 '14 at 5:29
• @Joseph--you may be well right (my eyes trust you). The equivalence still needs to be proved, I think. – Włodzimierz Holsztyński Sep 14 '14 at 5:40
• @David, I decided to leave the conjecture for the remaining case $\ n\ge 3$. Thank you for your counter-exaple (as obvious as it is; the obviousness was my fault :-). – Włodzimierz Holsztyński Sep 14 '14 at 5:43