On the Circle Covering Theorem by A. W. Goodman and R. E. Goodman

In 1945, A. W. Goodman and R. E. Goodman proved the following conjecture by P. Erd\H{o}s: Given a family of (round) disks of radii $r_1$, $\ldots$, $r_n$ in the plane it is always possible to cover them by a disk of radius $R = \sum r_i$, provided they cannot be separated into two subfamilies by a straight line disjoint from the disks. In this note we show that essentially the same idea may work for different analogues and generalizations of their result. In particular, we prove the following: Given a family of positive homothetic copies of a fixed convex body $K \subset \mathbb{R}^d$ with homothety coefficients $\tau_1, \ldots, \tau_n>0$ it is always possible to cover them by a translate of $\frac{d+1}{2}\left(\sum \tau_i\right)K$, provided they cannot be separated into two subfamilies by a hyperplane disjoint from the homothets.


Introduction
Consider a family K of positive homothetic copies of a fixed convex body K ⊂ R d with homothety coefficients τ 1 , . . . , τ n > 0. Following Hadwiger [6], we call K non-separable if any hyperplane H intersecting conv K intersects a member of K. Answering a question by Erdős, A.W. Goodman and R.E. Goodman [4] proved the following assertion: Theorem 1.1 (A.W. Goodman, R.E. Goodman, 1945) Given a non-separable family K of Euclidean balls of radii r 1 , . . . , r n in R d , it is always possible to cover them by a ball of radius R = r i .
Let us outline here the idea of their proof since we are going to reuse it in different settings.
First, A.W. Goodman and R.E. Goodman prove the following lemma, resembling the 1-dimensional case of the general theorem: (i.e., the center of mass of K if the weights of the balls are chosen to be proportional to the radii). They project the whole family onto d orthogonal directions (chosen arbitrarily) and apply Lemma 1.2 to show that the ball of radius R = r i centered at o indeed covers K.
In [2], K. Bezdek and Z. Lángi show that Theorem 1.1 actually holds not only for balls but also for any centrally-symmetric bodies: Bezdek and Z. Lángi, 2016) Given a non-separable family of homothets of centrally-symmetric convex body K ⊂ R d with homothety coefficients τ 1 , . . . , τ n > 0, it is always possible to cover them by a translate of The idea of their proof is to use Lemma 1.2 to deduce the statement for the case when K is a hypercube, and then deduce the result for sections of the hypercube (which can approximate arbitrary centrally-symmetric bodies).
It is worth noticing that Theorem 1.3 follows from Lemma 1.2 by a more direct argument (however, missed by A.W. Goodman  Another interesting approach to Goodmans' theorem was introduced by K. Bezdek and A. Litvak [3]. They put the problem in the context of studying the packing analogue of Bang's problem through the LP-duality, which gives yet another proof of Goodmans' theorem for the case when K is a Euclidean disk in the plane. One can adapt their argument for the original Bang's problem to get a "dual" counterpart of Goodmans' theorem. We discuss this counterpart and give our proof of a slightly more general statement in Sect. 4.
The paper is organized as follows. In Sect. 2 we prove a strengthening (with factor In Sect. 3 we show that if we weaken the condition of non-separability considering only d+1 directions of separating hyperplanes, then the factor d+1 2 cannot be improved. In Sect. 4 we prove a counterpart of Goodmans' theorem related to the notion somehow opposite to non-separability: Given a positive integer k and a family of Euclidean balls of radii r 1 , . . . , r n in R d , it is always possible to inscribe a ball of radius r = 1 2 r i within their convex hull, provided every hyperplane intersects at most k interiors of the balls.

A Goodmans-Type Result for Non-symmetric Bodies
Let K ⊂ R d be a (not necessarily centrally-symmetric) convex body containing the origin and let K • = {p : p, q ≤ 1 for all q ∈ K } (where ·, · stands for the standard inner product) be its polar body. We define the following parameter of asymmetry: It is an easy exercise in convexity to establish that min{μ > 0 : So an equivalent definition (which is more convenient for our purposes) is The value 1 σ is often referred to as Minkowski's measure of symmetry of body K (see, e.g., [5]). Proof We start by shifting the origin so that

Theorem 2.1 Given a non-separable family of positive homothetic copies of (not
Consider the orthogonal projection π along H onto the direction orthogonal to H . Suppose the segment π(K ) is divided by the projection of the origin in the ratio 1 : s. Since K • ⊂ −σ K • , we may assume that s ∈ [1, σ ]. Identify the image of π with the coordinate line R and denote Fig. 1). Note that the i are proportional to the τ i , and that covers the union I i = π(K). Let us check that I ⊂ I , which would be a contradiction, since π( p) ∈ I , π( p) / ∈ I . First, notice that Fig. 2 Illustration of the proof of Lemma 2.2 Second,

Lemma 2.2 (H. Minkowski and J. Radon) Let K be a convex body in
where σ denotes the parameter of asymmetry of K , defined above.
For the sake of completeness we provide a proof here.
Proof Suppose the origin coincides with the center of mass g = K x dx/ K dx. We show that K • ⊂ −d K • . Consider two parallel support hyperplanes orthogonal to one of the coordinate axes Ox 1 . We use the notation H t = {x = (x 1 , . . . , x d ) : x 1 = t} for hypeplanes orthogonal to this axis. Without loss of generality, these support hyperplanes are H −1 and H s for some s ≥ 1. We need to prove s ≤ d.
Assume that s > d. Consider a cone C defined as follows: its vertex is chosen arbitrarily from K ∩ H s ; its section C ∩ H 0 = K ∩ H 0 ; the cone is truncated by H −1 . Since C is a d-dimensional cone, the x 1 -coordinate of its center of mass divides the segment [−1, s] in ratio 1 : d. Therefore, the center of mass has positive x 1 -coordinate. It follows from convexity of K that C \ K lies (non-strictly) between H −1 and H 0 , hence the center of mass of C \ K has non-positive x 1 -coordinate. Similarly, K \ C lies (non-strictly) between H 0 and H s , hence its center of mass has non-negative x 1coordinate. Thus, the center of mass of K = (C \ (C \ K )) ∪ (K \ C) (see Fig. 2) must have positive x 1 -coordinate, which is a contradiction.

Corollary 2.3 The factor d in Theorem 1.4 can be improved to d+1
2 . Proof The result follows from Theorem 2.1 and Lemma 2.2.
An alternative proof of this corollary that avoids Lemma 2.2 is as follows. We use the notation of Theorem 1.4. Consider the smallest homothet τ K , τ > 0, that can cover K (after a translation to τ K + t, t ∈ R d ). Since it is the smallest, its boundary touches ∂ conv K at some points q 0 , . . ., q m (m ≤ d) such that the corresponding support hyperplanes H 0 , . . ., H m bound a nearly bounded set S, i.e., a set that can be placed between two parallel hyperplanes.
Circumscribe all the bodies from the family K by the smallest homothets of S and apply Theorem 2.1 for them (note that if m < d then S is unbounded, but that does not ruin our argument). Since S is a cylinder based on an m-dimensional simplex, its parameter of asymmetry equals m ≤ d, and we are done. Remark 2.4 Up to this moment the best possible factor for non-symmetric case is unknown. Bezdek and Lángi [2] give a sequence of examples in R d showing that it is impossible to obtain a factor less than 2 3 + 2

A Sharp Goodmans-Type Result for Simplices
Consider the case when K ⊂ R d is a simplex. In this section we are only interested in separating hyperplanes parallel to a facet of K . Proof A proof of possibility to cover follows the same lines as (and is even simpler than) the proof of Theorem 2.1. Let K have its center of mass at the origin. For a family K = {o i + τ i K }, consider a homothet d+1 Assuming d+1 2 τ i K + o does not cover K, we find a hyperplane H (strictly) separating a point p ∈ conv K \ d+1 Note that H can be chosen among the hyperplanes spanned by the facets of d+1 2 τ i K + o , so H is parallel to one of them.
After projecting everything along H onto the direction orthogonal to H , we repeat the same argument as before and show that (in the notation from Theorem 2.1) which contradicts our assumption. Next, we construct an example showing that the factor d+1 2 cannot be improved. Consider a simplex where N is an arbitrary large integer. Section it with all hyperplanes of the form {x i = t} or of the form d i=1 x i = t (for t ∈ Z). Consider all the smallest simplices generated by these cuts and positively homothetic to K . We use coordinates to denote the simplex lying in the hypercube For d = 2 (see Fig. 3) we compose K of the simplices with the following coordinates: For d = 3: For general d: It is rather straightforward to check that each b i ranges over the set {0, 1, . . . , d N }, and their sum is not greater than d(d+1) 2 N . Therefore, the chosen family K is indeed non-separable by hyperplanes parallel to the facets of K . Moreover, the chosen simplices touch all the facets of K , so K is the smallest simplex covering K. Finally, we note that any one-dimensional parameter of K (say, its diameter) is d(d+1)N 2(d N+1) times greater than the sum of the corresponding parameters of the elements of K, and this ratio tends to d+1 2 as N → ∞. Proof Mark all the segment endpoints and subdivide all the segments by the marked points. Next, put the origin at the leftmost marked point and numerate the segments between the marked points from left to right. We say that the i-th segment is of multiplicity 0 ≤ k i ≤ k if it is covered k i times. We keep the notation I i for the new segments with multiplicities, c i for their midpoints, and i for their lengths. Note that the value i c i i is preserved after this change of notation: it is the coordinate of the center of mass of the segments regarded as solid one-dimensional bodies of uniform density.

Remark 4.3
The estimate in Theorem 4.2 is sharp for any k, as can be seen from the example of k translates of K lying along the line so that consecutive translates touch.