On the Circle Covering Theorem by A.W. Goodman and R.E. Goodman
Arseniy Akopyan 0 1 2 3
Alexey Balitskiy 0 1 2 3
Mikhail Grigorev 0 1 2 3
Editor in Charge: János Pach
0 Moscow Institute of Physics and Technology , Institutskiy per. 9, 141700 Dolgoprudny , Russia
1 Institute of Science and Technology Austria (IST Austria) , Am Campus 1, 3400 Klosterneuburg , Austria
2 Department of Mathematics, Massachusetts Institute of Technology , 182 Memorial Dr., Cambridge, MA 02142 , USA
3 Institute for Information Transmission Problems RAS , Bolshoy Karetny per. 19, 127994 Moscow , Russia
In 1945, A.W. Goodman and R.E. Goodman proved the following conjecture by P. Erd o˝s: Given a family of (round) disks of radii r1, . . ., rn in the plane, it is always possible to cover them by a disk of radius R = ri , 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 ⊂ Rd with homothety coefficients τ1, . . . , τn > 0, it is always possible to cover them by a translate of d +21 τi K , provided they cannot be separated into two subfamilies by a hyperplane disjoint from the homothets.
GoodmanGoodman theorem; Nonseparable family; Positive homothets

Mathematics Subject Classification 52C10 · 52C17
1 Introduction
Consider a family K of positive homothetic copies of a fixed convex body K ⊂ Rd
with homothety coefficients τ1, . . . , τn > 0. Following Hadwiger [6], we call K
nonseparable if any hyperplane H intersecting conv K intersects a member of K.
Answering a question by Erdo˝s, A.W. Goodman and R.E. Goodman [4] proved the
following assertion:
Theorem 1.1 (A.W. Goodman, R.E. Goodman, 1945) Given a nonseparable family
K of Euclidean balls of radii r1, . . . , rn in Rd , it is always possible to cover them by
a ball of radius R = ri .
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 1dimensional case of the general theorem:
Lemma 1.2 Let I1, . . . , In ⊂ R be segments of lengths 1, . . . , n with midpoints
c1, . . . , cn. Assume the union Ii is a segment (i.e. the family of segments is
nonseparable). Then the segment I of length i with midpoint at the center of mass
c = i cii covers Ii .
Next, for a family K = {oi + ri B} (B denotes the unit ball centered at the origin
of Rd ), A.W. Goodman and R.E. Goodman consider the point o = riroii (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 = ri 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 centrallysymmetric bodies:
Theorem 1.3 (K. Bezdek and Z. Lángi, 2016) Given a nonseparable family of
homothets of centrallysymmetric convex body K ⊂ Rd with homothety coefficients
τ1, . . . , τn > 0, it is always possible to cover them by a translate of τi K .
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 centrallysymmetric 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 and R.E. Goodman). In 2001, F. Petrov
proposed a particular case of the problem (when K is a Euclidean ball) to Open
Mathematical Contest of Saint Petersburg Lyceum N−o 239 [1]. He assumed the following
solution (working for any symmetric K as well): For a family K = {oi +τi K }, consider
a homothet τi K +o with center o = τiτoii . If τi K +o does not cover K, then
there exists a hyperplane H separating a point p ∈ conv K \ τi K + o from
τi K + o . Projection onto the direction orthogonal to H reveals a contradiction
with Lemma 1.2.
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 LPduality, 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
d +21 instead of d) of the following result of K. Bezdek and Z. Lángi:
Theorem 1.4 (K. Bezdek and Z. Lángi, 2016) Given a nonseparable family of
positive homothetic copies of a (not necessarily centrallysymmetric) convex body K ⊂ Rd
with homothety coefficients τ1, . . . , τn > 0, it is always possible to cover them by a
translate of d τi K .
2 A GoodmansType Result for Nonsymmetric Bodies
Let K ⊂ Rd be a (not necessarily centrallysymmetric) 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:
min min {μ > 0 : (K − q) ⊂ −μ(K − q)}.
σ = q∈int K
It is an easy exercise in convexity to establish that min{μ > 0 : (K − q) ⊂ −μ
(K − q)} = min {μ > 0 : (K − q)◦ ⊂ −μ(K − q)◦}. So an equivalent definition
(which is more convenient for our purposes) is
min min {μ > 0 : (K − q)◦ ⊂ −μ(K − q)◦}.
σ = q∈int K
The value σ1 is often referred to as Minkowski’s measure of symmetry of body K (see,
e.g., [5]).
Theorem 2.1 Given a nonseparable family of positive homothetic copies of (not
necessarily centrallysymmetric) convex body K ⊂ Rd with homothety coefficients
Fig. 1 Illustration of the proof of Theorem 2.1
Proof We start by shifting the origin so that K ◦ ⊂ −σ K ◦.
For a family K = {oi + τi K }, consider the homothet σ +21 τi K + o with center
o = τiτoii . Assume that σ +21 τi K + o does not cover K, hence there exists a
hyperplane H (strictly) separating a point p ∈ conv K \ σ +21 τi K + o from
σ +21 τi K + o . 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 Ii = [ai , bi ] = π oi + τi K , ci = π(oi ),
i = bi − ai , L = i (see Fig. 1). Note that the i are proportional to the τi , and that
s(ci −ai ) = bi −ci . Denote c = π(o) = Li ci and I = [a, b] = π σ +21 τi K +o
the segment of length σ +21 L divided by c in the ratio 1 : s.
Also consider the midpoints ci = ai +2bi . By Lemma 1.2, the segment I = [a , b ]
of length L with midpoint at c = Li ci covers the union Ii = π(K). Let us check
that I ⊂ I , which would be a contradiction, since π( p) ∈ I , π( p) ∈/ I .
First, notice that
ci =
= ci ,
1 1 1
a = c − 2 L ≥ c − 2 L ≥ c − 1 + s
L = a.
Fig. 2 Illustration of the proof of Lemma 2.2
ci − ci =
1 1 s − 1
b = c + 2 L = c + (c − c) + 2 L = c + 2(s + 1)
s − 1 1 s σ + 1
≤ c + 2(s + 1) L + 2 L ≤ c + 1 + s 2
L = b.
Lemma 2.2 (H. Minkowski and J. Radon) Let K be a convex body in Rd . Then σ ≤ d,
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 d x / K d x . We
show that K ◦ ⊂ −d K ◦. Consider two parallel support hyperplanes orthogonal to one
of the coordinate axes O x1. We use the notation Ht = {x = (x1, . . . , xd ) : x1 = t }
for hypeplanes orthogonal to this axis. Without loss of generality, these support
hyperplanes are H−1 and Hs 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 ∩ Hs ; its section C ∩ H0 = K ∩ H0; the cone is truncated by H−1.
Since C is a ddimensional cone, the x1coordinate of its center of mass divides the
segment [−1, s] in ratio 1 : d. Therefore, the center of mass has positive x1coordinate.
It follows from convexity of K that C \ K lies (nonstrictly) between H−1 and H0,
hence the center of mass of C \ K has nonpositive x1coordinate. Similarly, K \ C
lies (nonstrictly) between H0 and Hs , hence its center of mass has nonnegative
x1coordinate. Thus, the center of mass of K = (C \ (C \ K )) ∪ (K \ C ) (see Fig. 2)
must have positive x1coordinate, which is a contradiction.
Corollary 2.3 The factor d in Theorem 1.4 can be improved to d +21 .
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 ∈ Rd ). Since it is the smallest, its boundary
touches ∂ conv K at some points q0, . . ., qm (m ≤ d) such that the corresponding
support hyperplanes H0, . . ., Hm 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 mdimensional simplex, its
parameter of asymmetry equals m ≤ d, and we are done.
Remark 2.4 Up to this moment the best possible factor for nonsymmetric case is
unknown. Bezdek and Lángi [2] give a sequence of examples in Rd showing that it is
impossible to obtain a factor less than 23 + 3 √23 (> 1) for any d ≥ 2.
3 A Sharp GoodmansType Result for Simplices
Consider the case when K ⊂ Rd is a simplex. In this section we are only interested
in separating hyperplanes parallel to a facet of K .
Theorem 3.1 Let K be a family of positive homothetic copies of a simplex K ⊂ Rd
with homothety coefficients τ1, . . . , τn > 0. Suppose any hyperplane H (parallel to
taofcaocveetrof KK) ibnytearstercatninsglatceonovf d +2K1 inteτriseKct.s Ma omreeomvbeer,r tohfe Kfa.cTtohrend +2it1 iscapnonsostibblee
improved.
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 = {oi + τi K }, consider a homothet d +21
Assuming d +21 τi K + o does not cover K, we find a hyperplane H (strictly)
separating a point p ∈ conv K \ d +21 τi K + o from d +21 τi K + o . Note that
H can be chosen among the hyperplanes spanned by the facets of d +21 τi 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 +21 cannot be improved.
Consider a simplex
K =
x = (x1, . . . , xd ) ∈ Rd : xi ≥ 0,
i=1
xi ≤
where N is an arbitrary large integer. Section it with all hyperplanes of the form
{xi = t } or of the form id=1 xi = t (for t ∈ Z). Consider all the smallest simplices
generated by these cuts and positively homothetic to K . We use coordinates
Fig. 3 Example for d = 2 and N = 5
i=1
bi ≤
d(d + 1)
N ,
2
to denote the simplex lying in the hypercube {bi ≤ xi ≤ bi + 1, i = 1, . . . , d}.
For d = 2 (see Fig. 3) we compose K of the simplices with the following
coordinates:
For d = 3:
It is rather straightforward to check that each bi ranges over the set {0, 1, . . . , d N },
and their sum is not greater than d(d2+1) N . Therefore, the chosen family K is indeed
nonseparable 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 onedimensional parameter of K (say, its diameter) is d2((dd+N1+)1N) times
greater than the sum of the corresponding parameters of the elements of K, and this
ratio tends to d+1 as N → ∞.
2
4 A “Dual” Version of Goodmans’ Theorem
Lemma 4.1 Let I1, . . . , In ⊂ R be segments of lengths 1, . . . , n with midpoints
c1, . . . , cn. Assume every point on the line belongs to at most k of the interiors of the
Ii . Then the segment I of length k1 i with midpoint at the center of mass c = i cii
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 ≤ ki ≤ k if it is covered ki times. We keep the notation Ii for the new
segments with multiplicities, ci for their midpoints, and i for their lengths. Note that
the value i ci is preserved after this change of notation: it is the coordinate of the
i
center of mass of the segments regarded as solid onedimensional bodies of uniform
density.
Note that ci = 1 + · · · + i−1 + 21 i . We prove that
c =
(this would mean that the left endpoint of I is contained in conv
endpoint everything is similar).
The inequality in question
Ii ; for the right
2
,
which is true, since k ≥ ki .
Theorem 4.2 Let k be a positive integer, and K be a family of positive homothetic
copies (with homothety coefficients τ1, . . . , τn > 0) of a centrallysymmetric convex
body K ⊂ Rd . Suppose any hyperplane intersects at most k interiors of the homothets.
Then it is possible to put a translate of k1 τi K into their convex hull.
Proof As usual, for a family K = {oi + τi K }, consider a homothet k1
with center o = τiτoii . Assume k1 τi K + o does not fit into conv K, then
there exists a hyperplane H separating a point p ∈ k1 τi K + o from conv K.
After projecting onto the direction orthogonal to H , we use Lemma 4.1 to obtain a
contradiction.
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.
Acknowledgements Open access funding provided by Institute of Science and Technology (IST
Austria). The authors are grateful to Rom Pinchasi and Alexandr Polyanskii for fruitful discussions. Also the
authors thank Roman Karasev, Kevin Kaczorowski, and the anonymous referees for careful reading and
suggested revisions. The research of the first author is supported by People Programme (Marie Curie Actions)
of the European Union’s Seventh Framework Programme (FP7/20072013) under REA grant agreement
n◦[291734]. The research of the second author is supported by the Russian Foundation for Basic Research
Grant 150199563 A and Grant 153120403 (mol_a_ved).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0
International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution,
and reproduction in any medium, provided you give appropriate credit to the original author(s) and the
source, provide a link to the Creative Commons license, and indicate if changes were made.
1. Berlov , S. , Ivanov , S. , Karpov , D. , Kokhas' , K. , Petrov , F. , Khrabrov , A. : Problems from St . Petersburg School Olympiad on Mathematics 2000  2002 . Nevskiy Dialekt, St. Petersburg ( 2006 )
2. Bezdek , K. , Lángi , Z. : On nonseparable families of positive homothetic convex bodies . Discrete Comput. Geom . 56 ( 3 ), 802  813 ( 2016 ). doi:10.1007/s00454 016  9815 1
3. Bezdek , K. , Litvak , A.E. : Packing convex bodies by cylinders . Discrete Comput. Geom . 55 ( 3 ), 725  738 ( 2016 ). doi:10.1007/s00454 016  9760 z
4. Goodman , A.W. , Goodman , R.E. : A circle covering theorem . Am. Math. Monthly 52 ( 9 ), 494  498 ( 1945 ). doi:10.2307/2304537
5. Grünbaum , B. : Measures of symmetry for convex sets . In: Proceedings of Symposia in Pure Mathematics, vol. 7 , pp. 233  270 . American Mathematical Society, Providence ( 1963 )
6. Hadwiger , H. : Nonseparable convex systems . Am. Math. Monthly 54(10) , 583  585 ( 1947 ). doi:10. 2307/2304497