Zilin Jiang
Joint work with Alexandr Polyanskii
A plank (or slab, strip) of width $w$ is part of $\mathbb{R}^d$ that lies between two parallel hyperplanes at distance $w$.
The width of $C$ is the smallest width of plank covers $C$.
If convex body $C$ is covered by planks, then ...
total width of planks is at least width of $C$.
Alfred Tarski in 1932 proved for disks
Thøger Bang in 1950 proved for convex bodies
Zone of width $\omega$ is part of unit sphere that lies within spherical distance $\omega/2$ of a given great circle.
The total width of zones covering sphere is at least ... $\pi$
Research Problems: Exploring a Planet.
American Mathematical Monthly, 1973.
Appeared in Research Problems in Discrete Geometry
1972 Rosta: 3 zones of equal width;
1974 Linhart: 4 zones of equal width;
2016 Fodor, Vígh and Zarnócz:
if $100$ zones of equal width $w$ cover sphere,
then $100w \ge 2.032$;
2017 J.–Polyanskii: any set of zones, any dimension, charaterize the equality cases.
Proofs of Tarski's plank problem
Proof of Fejes Tóth's zone conjecture
Total width of planks covering unit disk is $\ge 2$.
Planks cover disk$\implies$ Arches cover hemisphere$\implies \sum \pi w_i \ge 2\pi$
Tarski's proof applies to
$2 \times$ inradius of $C$ = width of $C$,
but not to
$\vec{w}_i :=$ "direction" of plank $i$
$L := \{\pm \vec{w}_1 \pm \dots \pm \vec{w}_n\}$
Idea 1: $L$ cannot be covered by the planks.
Idea 2: $L$ can be embeded in $C$ of large width.
Special case: all planks are centered at O.
Claim: $\vec{w}\in L$ achieving max norm is not covered.
$|\vec{w}| \ge |\vec{w} \pm 2\vec{w}_i| \implies w$ not covered by plank $i$.
$\vec{w} = \pm \vec{w}_1 \pm \dots \pm \vec{w}_n \in L$ achieves max norm
$\Leftrightarrow |\epsilon_1 \vec{w}_1 + \dots + \epsilon_n \vec{w}_n|^2$ is maximized on $\{\pm 1\}^n$.
In general, plank $i$: $|\vec{w} \cdot \vec{w}_i$$+ b_i$$| \le |\vec{w}_i|^2$
Optimize quadratic function:
$\sum \epsilon_i\epsilon_j (\vec{w}_i\cdot\vec{w}_j)$ $+\sum b_i\epsilon_i$
Proofs of Tarski's plank problem
Proof of Fejes Tóth's zone conjecture
$\vec{w}_i :=$ direction of zone $i$.
Bang says: some $\vec{w} = $$\epsilon_1$$\vec{w}_1 + \dots +$$\epsilon_n$$\vec{w}_n$ is not covered.
If $|\vec{w}| \le 1$, then $\hat{w}$ is not covered.
Otherwise, $\vec{w} = \vec{w}_1 + \dots + \vec{w}_n$ is of big maginitude ...
Maybe we can merge some zones!
$\angle(\vec{w}_1, \vec{w}_2) \le \alpha_1 + \alpha_2$
$\angle(\vec{w}_1, \vec{w}_2) \le \alpha_1 + \alpha_2$
$\cos \angle(\vec{w}_1, \vec{w}_2) \ge \cos(\alpha_1 + \alpha_2)$
$|\vec{w}_1 + \vec{w}_2|^2 = |\vec{w}_1|^2 + 2|\vec{w}_1||\vec{w}_2|\cos\angle(\vec{w}_1, \vec{w}_2) + |\vec{w}_2|^2$
$\ge \sin^2\alpha_1 + 2\sin\alpha_1\sin\alpha_2\cos(\alpha_1 + \alpha_2) + \sin^2\alpha_2$
$= \dots = \sin(\alpha_1 + \alpha_2)^2$.
Can merge 2 zones when $|\vec{w}_1 + \vec{w}_2| \ge \sin(\alpha_1 + \alpha_2)$.
In general, maybe we can merge some zones when $|\vec{w}_1 + \dots + \vec{w}_n| \ge \sin(\alpha_1 + \dots + \alpha_n)$.
Assume half of total width $\alpha_1 + \dots + \alpha_n < \pi / 2$.
If $|\vec{w}| \leq 1 $, $\hat{w}$ is not covered.
Otherwise $|\vec{w}| > 1 > \sin(\alpha_1 + \dots + \alpha_n)$, merge!
Let disks with radii $r_1, \dots, r_n$ lie in plane.
If no line "seperates" disks,
then they can be coverered by disk of radius $\sum_i r_i$.
Idea: Let $\vec{x}_i$ be center of disk $i$. Consider the disk centered at $\sum_i r_i\vec{x}_i / \sum_i r_i$ with radius $\sum_i r_i$.
Inspiration: Use zone with direction $\vec{w}$ to replace.
Great circle $\leftrightarrow$ Antipodal points
Zone $\leftrightarrow$ Antipodal caps
If every great circle intersects (antipodal) caps...
then total radius of caps is at least $\pi/2$.
The total width of zones covering cap of radius $r$ is ...$\ge 2r$
Conjecture (Fejes Tóth): ... covering spherical convex domain $D$ is at least width of $D$.
Measure relative to $C$ in direction normal to plank.
Conjecture: total relative width of planks covering $C$ is at least $1$.
Keith Ball in 1991 proved for symmetric bodies.
Conjecture: total width of planks covering punctured disk is at least diameter.