ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 23 (2023) #P1.09
https://doi.org/10.26493/1855-3974.2550.d96
(Also available at http://amc-journal.eu)
On generalized Minkowski arrangements*
Máté Kadlicskó
Department of Geometry, Budapest University of Technology,
Egry József utca 1., Budapest, Hungary, 1111
Zsolt Lángi †
MTA-BME Morphodynamics Research Group and Department of Geometry,
Budapest University of Technology,
Egry József utca 1., Budapest, Hungary, 1111
Received 9 Februar 2021, accepted 14 December 2021, published online 28 October 2022
Abstract
The concept of a Minkowski arrangement was introduced by Fejes Tóth in 1965 as a
family of centrally symmetric convex bodies with the property that no member of the family
contains the center of any other member in its interior. This notion was generalized by Fejes
Tóth in 1967, who called a family of centrally symmetric convex bodies a generalized
Minkowski arrangement of order µ for some 0 < µ < 1 if no member K of the family
overlaps the homothetic copy of any other member K ′ with ratio µ and with the same
center as K ′. In this note we prove a sharp upper bound on the total area of the elements
of a generalized Minkowski arrangement of order µ of finitely many circular disks in the
Euclidean plane. This result is a common generalization of a similar result of Fejes Tóth
for Minkowski arrangements of circular disks, and a result of Böröczky and Szabó about
the maximum density of a generalized Minkowski arrangement of circular disks in the
plane. In addition, we give a sharp upper bound on the density of a generalized Minkowski
arrangement of homothetic copies of a centrally symmetric convex body.
Keywords: Arrangement, Minkowski arrangement, density, homothetic copy.
Math. Subj. Class. (2020): 52C15, 52C26, 52A10
*The authors express their gratitude to K. Bezdek for directing their attention to this interesting problem, and
to two anonymous referees for many helpful suggestions. The second named author is supported by the National
Research, Development and Innovation Office, NKFI, K-119670, the János Bolyai Research Scholarship of the
Hungarian Academy of Sciences, and the BME IE-VIZ TKP2020 and ÚNKP-20-5 New National Excellence
Programs by the Ministry of Innovation and Technology.
†Corresponsing author.
E-mail addresses: kadlicsko.mate@gmail.com (Máté Kadlicskó), zlangi@math.bme.hu (Zsolt Lángi)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 23 (2023) #P1.09
1 Introduction
The notion of a Minkowski arrangement of convex bodies was introduced by L. Fejes
Tóth in [6], who defined it as a family F of centrally symmetric convex bodies in the
d-dimensional Euclidean space Rd, with the property that no member of F contains the
center of any other member of F in its interior. He used this concept to show, in particular,
that the density of a Minkowski arrangement of homothets of any given plane convex body
with positive homogeneity is at most four. Here an arrangement is meant to have positive
homogeneity if the set of the homothety ratios is bounded from both directions by positive
constants. It is worth mentioning that the above result is a generalization of the planar case
of the famous Minkowski Theorem from lattice geometry [11]. Furthermore, Fejes Tóth
proved in [6] that the density of a Minkowski arrangement of circular disks in R2 with
positive homogeneity is maximal for a Minkowski arrangement of congruent circular disks
whose centers are the points of a hexagonal lattice and each disk contains the centers of six
other members on its boundary.
In [7], extending the investigation to finite Minkowski arrangements, Fejes Tóth gave a
sharp upper bound on the total area of the members of a Minkowski arrangement of finitely
many circular disks, and showed that this result immediately implies the density estimate
in [6] for infinite Minkowski circle-arrangements. Following a different direction, in [8]
for any 0 < µ < 1 Fejes Tóth defined a generalized Minkowski arrangements of order µ as
a family F of centrally symmetric convex bodies with the property that for any two distinct
members K,K ′ of F , K does not overlap the µ-core of K ′, defined as the homothetic
copy of K ′ of ratio µ and concentric with K ′. In this paper he made the conjecture that
for any 0 < µ ≤
√
3 − 1, the density of a generalized Minkowski arrangement of circular
disks with positive homogeneity is maximal for a generalized Minkowski arrangement of
congruent disks whose centers are the points of a hexagonal lattice and each disk touches
the µ-core of six other members of the family. According to [8], this conjecture was verified
by Böröczky and Szabó in a seminar talk in 1965, though the first written proof seems to be
published only in [4] in 2002. It was observed both in [8] and [4] that if
√
3− 1 < µ < 1,
then, since the above hexagonal arrangement does not cover the plane, that arrangement
has no maximal density.
In this paper we prove a sharp estimate on the total area of a generalized Minkowski
arrangement of finitely many circular disks, with a characterization of the equality case.
Our result includes the result in [7] as a special case, and immediately implies the one in
[4]. The proof of our statement relies on tools from both [4, 7], but uses also some new
ideas. In addition, we also generalize a result from Fejes Tóth [6] to find a sharp upper
bound on the density of a generalized Minkowski arrangement of homothetic copies of a
centrally symmetric convex body.
For completeness, we mention that similar statements for (generalized) Minkowski ar-
rangements in other geometries and in higher dimensional spaces were examined, e.g. in
[5, 9, 13]. Minkowski arrangements consisting of congruent convex bodies were consid-
ered in [3]. Estimates for the maximum cardinality of mutually intersecting members in a
(generalized) Minkowski arrangement can be found in [10, 14, 15, 17]. The problem inves-
tigated in this paper is similar in nature to those dealing with the volume of the convex hull
of a family of convex bodies, which has a rich literature. This includes a result of Oler [16]
(see also [2]), which is also of lattice geometric origin [20], and the notion of parametric
density of Betke, Henk and Wills [1]. In particular, our problem is closely related to the
notion of density with respect to outer parallel domains defined in [2]. Applications of
M. Kadlicskó and Z. Lángi: On generalized Minkowski arrangements 3
(generalized) Minkowski arrangements in other branches of mathematics can be found in
[18, 19].
As a preliminary observation, we start with the following generalization of Remark 2
of [6], stating the same property for (not generalized) Minkowski arrangements of plane
convex bodies. In Proposition 1.1, by vold(·) we denote d-dimensional volume, and by Bd
we denote the closed Euclidean unit ball centered at the origin.
Proposition 1.1. Let 0 < µ < 1, let K ⊂ Rd be an origin-symmetric convex body and
let F = {x1 + λ1K,x2 + λ2K, . . . } be a generalized Minkowski arrangement of order µ,
where xi ∈ Rd, λi > 0 for each i = 1, 2, . . . . Assume that F is of positive homogeneity,
that is, there are constants 0 < C1 < C2 satisfying C1 ≤ λi ≤ C2 for all values of i, and
define the (upper) density δ(F) of F in the usual way as
δ(F) = lim sup
R→∞
∑
xi∈RBd vold(xi + λiK)
vold(RBd)
,
if it exists. Then
δ(F) ≤ 2
d
(1 + µ)d
, (1.1)
where equality is attained, e.g. if {x1, x2, . . .} is a lattice with K as its fundamental region,
and λi = 2/(1 + µ) for all values of i.
Proof. Note that the equality part of Proposition 1.1 clearly holds, and thus, we prove only
the inequality in (1.1). Let || · ||K : Rd → [0,∞) denote the norm with K as its unit ball.
Then, by the definition of a generalized Minkowski arrangement, we have
||xi − xj ||K ≥ max{λi + µλj , λj + µλi}
≥ 1
2
((λi + µλj) + (λj + µλi))
=
1 + µ
2
(λi + λj),
implying that the homothets xi+(λi/2) ·(1 + µ)K are pairwise non-overlapping. In other
words, the family F ′ = {xi + (λi/2) · (1 + µ)K : i = 1, 2, . . .} is a packing. Thus, the
density of F ′ is at most one, from which (1.1) readily follows. Furthermore, if K is the
fundamental region of a lattice formed by the xi’s and λi = 2/(1 + µ) for all values of i,
then F ′ is a tiling, implying the equality case.
Following the terminology of Fejes Tóth in [7] and to permit a simpler formulation
of our main result, in the remaining part of the paper we consider generalized Minkowski
arrangements of open circular disks, where we note that generalized Minkowski arrange-
ments can be defined for families of open circular disks in the same way as for families of
closed circular disks.
To state our main result, we need some preparation, where we denote the boundary of
a set by bd(·). Consider some generalized Minkowski arrangement F = {Bi = xi +
ρi int(B
2) : i = 1, 2, . . . , n} of open circular disks in R2 of order µ, where 0 < µ < 1. Set
U(F) =
⋃n
i=1 Bi =
⋃
F . Then each circular arc Γ in bd(U(F)) corresponds to a circular
sector, which can be obtained as the union of the segments connecting a point of Γ to the
center of the disk in F whose boundary contains Γ. We call the union of these circular
4 Ars Math. Contemp. 23 (2023) #P1.09
sectors the outer shell of F . Now consider a point p ∈ bd(U(F)) belonging to at least
two members of F , say Bi and Bj , such that xi, xj and p are not collinear. Assume that
the convex angular region bounded by the two closed half lines starting at p and passing
through xi and xj , respectively, do not contain the center of another element of F in its
interior which contains p on its boundary. We call the union of the triangles conv{p, xi, xj}
satisfying these conditions the inner shell of F . We denote the inner and the outer shell of F
by I(F) and O(F), respectively. Finally, we call the set C(F) = U(F)\(I(F)∪O(F)) the
core of F (cf. Figure 1). Clearly, the outer shell of any generalized Minkowski arrangement
of open circular disks is nonempty, but there are arrangements for which I(F) = ∅ or
C(F) = ∅.
Figure 1: The outer and inner shell, and the core of an arrangement, shown in white, light
grey and dark grey, respectively.
If the intersection of two members of F is nonempty, then we call this intersection a
digon. If a digon touches the µ-cores of both disks defining it, we call the digon thick.
A digon which is not contained in a third member of F is called a free digon. Our main
theorem is as follows, where area(X) denotes the area of the set X .
Theorem 1.2. Let 0 < µ ≤
√
3−1, and let F = {Bi = xi+ρi int(B2) : i = 1, 2, . . . , n}
be a generalized Minkowski arrangement of finitely many open circular disks of order µ.
Then
T = π
n∑
i=1
ρ2i ≤
2π√
3(1 + µ)2
area(C(F))+
+
4 · arccos( 1+µ2 )
(1 + µ) ·
√
(3 + µ)(1− µ)
area(I(F)) + area(O(F)),
M. Kadlicskó and Z. Lángi: On generalized Minkowski arrangements 5
where T is the total area of the circles, with equality if and only if each free digon in F is
thick.
In the paper, for any points x, y, z ∈ R2, we denote by [x, y] the closed segment with
endpoints x, y, by [x, y, z] the triangle conv{x, y, z}, by |x| the Euclidean norm of x, and if
x and z are distinct from y, by ∠xyz we denote the measure of the angle between the closed
half lines starting at y and passing through x and z. Note that according to our definition,
∠xyz is at most π for any x, z ̸= y. For brevity we call an open circular disk a disk, and
a generalized Minkowski arrangement of disks of order µ a µ-arrangement. Throughout
Sections 2 and 3 we assume that 0 < µ ≤
√
3− 1.
In Section 2, we prove some preliminary lemmas. In Section 3, we prove Theorem 1.2.
Finally, in Section 4, we collect additional remarks and questions.
2 Preliminaries
For any Bi, Bj ∈ F , if Bi ∩Bj ̸= ∅, we call the two intersection points of bd(Bi) and
bd(Bj) the vertices of the digon Bi ∩Bj .
First, we recall the following lemma of Fejes Tóth [7, Lemma 2]. To prove it, we ob-
serve that for any µ > 0, a generalized Minkowski arrangement of order µ is a Minkowski
arrangement as well.
Lemma 2.1. Let Bi, Bj , Bk ∈ F such that the digon Bi ∩ Bj is contained in Bk. Then
the digon Bi ∩Bk is free (with respect to F).
From now on, we call the maximal subfamilies F ′ of F (with respect to containment)
with the property that
⋃
Bi∈F ′ Bi is connected the connected components of F . Our next
lemma has been proved by Fejes Tóth in [7] for Minkowski arrangements of order µ = 0.
His argument can be applied to prove Lemma 2.2 for an arbitrary value of µ. Here we
include this proof for completeness.
Lemma 2.2. If F ′ is a connected component of F in which each free digon is thick, then
the elements of F ′ are congruent.
Proof. We need to show that for any Bi, Bj ∈ F ′, Bi and Bj are congruent. Observe that
by connectedness, we may assume that Bi ∩ Bj is a digon. If Bi ∩ Bj is free, then it is
thick, which implies that Bi and Bj are congruent. If Bi ∩ Bj is not free, then there is a
disk Bk ∈ F ′ containing it. By Lemma 2.1, the digons Bi ∩ Bk and Bj ∩ Bk are free.
Thus Bk is congruent to both Bi and Bj .
In the remaining part of Section 2, we examine densities of some circular sectors in
certain triangles. The computations in the proofs of these lemmas were carried out by a
Maple 18.00 software.
Lemma 2.3. Let 0 < γ < π and A,B > 0 be arbitrary. Let T = [x, y, z] be a triangle such
that ∠xzy = γ, and |x− z| = A and |y − z| = B. Let ∆ = ∆(γ,A,B), α = α(γ,A,B)
and β = β(γ,A,B) denote the functions with variables γ,A,B whose values are the area
and the angles of T at x and y, respectively, and set fA,B(γ) =
(
αA2 + βB2
)
/∆. Then,
for any A,B > 0, the function fA,B(γ) is strictly decreasing on the interval γ ∈ (0, π).
6 Ars Math. Contemp. 23 (2023) #P1.09
B
A
γ
z
β
y
α
x
Figure 2: Notation in Lemma 2.3.
Proof. Without loss of generality, assume that A ≤ B, and let g = αA2 + βB2. Then, by
an elementary computation, we have that
g = A2 arccot
A−B cos γ
B sin γ
+B2 arccot
B −A cos γ
A sin γ
, and ∆ =
1
2
AB sin γ.
We regard g and ∆ as functions of γ. We intend to show that g′∆− g∆′ is negative on the
interval (0, π) for all A,B > 0. Let h = g′ · ∆/∆′ − g, and note that this expression is
continuous on (0, π/2) and (π/2, π) for all A,B > 0. By differentiating and simplifying,
we obtain
h′ =
−2
(
A2(1 + cos2(γ)) +B2(1 + cos2(γ))− 4AB cos(γ)
)
A2B2 sin2(γ)
cos2 (γ)(A2 +B2 − 2AB cos(γ))2
,
which is negative on its domain. This implies that g′∆ − g∆′ is strictly decreasing on
(0, π/2) and strictly increasing on (π/2, π). On the other hand, we have limγ→0+(g′∆ −
g∆′) = −A3Bπ, and limγ→π− (g′∆− g∆′) = 0. This yields the assertion.
Lemma 2.4. Consider two disks Bi, Bj ∈ F such that |xi − xj | < ρi + ρj , and let v be a
vertex of the digon Bi ∩ Bj . Let T = [xi, xj , v], ∆ = area(T ), and let αi = ∠vxixj and
αj = ∠vxjxi. Then
1
2
αiρ
2
i +
1
2
αjρ
2
j ≤
4 arccos 1+µ2
(1 + µ)
√
(1− µ)(3 + µ)
∆, (2.1)
with equality if and only if ρi = ρj and |xi − xj | = ρi(1 + µ).
Proof. First, an elementary computation shows that if ρi = ρj and |xi − xj | = ρi(1 + µ),
then there is equality in (2.1).
Without loss of generality, let ρi = 1, and 0 < ρj = ρ ≤ 1. By Lemma 2.3, we
may assume that |xi − xj | = 1 + µρ. Thus, the side lengths of T are 1, ρ, 1 + µρ.
M. Kadlicskó and Z. Lángi: On generalized Minkowski arrangements 7
v
αi
xi
αj
xj
ρi ρj
Figure 3: Notation in Lemma 2.4.
Applying the Law of Cosines and Heron’s formula to T we obtain that
1
2αiρ
2
i +
1
2αjρ
2
j
∆
=
f(ρ, µ)
g(ρ, µ)
,
where
f(ρ, µ) =
1
2
arccos
1 + (1 + µρ)2 − r2
2(1 + µρ)
+
1
2
ρ2 arccos
ρ2 + (1 + µρ)2 − 1
2ρ(1 + µρ)
,
and
g(ρ, µ) = ρ
√
2 + ρ+ µρ)(2− ρ+ µρ)(1− µ2).
In the remaining part we show that
f(ρ, µ)
g(ρ, µ)
<
4 arccos 1+µ2
(1 + µ)
√
(1− µ)(3 + µ)
if 0 < ρ < 1 and 0 ≤ µ ≤
√
3− 1. To do it we distinguish two separate cases.
Case 1: 0 < ρ ≤ 1/5. In this case we estimate f(ρ, µ)/g(ρ, µ) as follows. Let the
part of [xi, xj ] covered by both disks Bi and Bj be denoted by S. Then S is a segment of
length (1−µ)ρ. On the other hand, if Ai denotes the convex circular sector of Bi bounded
by the radii [xi, v] and [xi, xj ] ∩ Bi, and we define Aj analogously, then the sets Ai ∩ Aj
and (Ai ∪ Aj) \ T are covered by the rectangle with S as a side which contains v on the
side parallel to S. The area of this rectangle is twice the area of the triangle conv(S∪{v}),
implying that
f(ρ, µ)
g(ρ, µ)
≤ 1 + 2(1− µ)ρ
1 + µρ
.
We show that if 0 < ρ ≤ 1/5, then the right-hand side quantity in this inequality is strictly
less than the right-hand side quantity in (2.1). By differentiating with respect to ρ, we
see that as a function of ρ, 1 + (2(1− µ)ρ) /(1 + µρ) is strictly increasing on its domain
8 Ars Math. Contemp. 23 (2023) #P1.09
and attains its maximum at ρ = 1/5. Thus, using the fact that this maximum is equal to
(7− µ)/(5 + µ), we need to show that
4 arccos 1+µ2
(1 + µ)
√
(1− µ)(3 + µ)
− 7− µ
5 + µ
> 0.
Clearly, the function
µ 7→
arccos 1+µ2
1+µ
2
is strictly decreasing on the interval [0,
√
3 − 1]. By differentiation one can easily check
that the function
µ 7→ 7− µ
5 + µ
√
(1− µ)(3 + µ)
is also strictly increasing on the same interval. Thus, we obtain that the above expression
is minimal if µ =
√
3− 1, implying that it is at least 0.11570 . . ..
Case 2: 1/5 < ρ ≤ 1.
We show that in this case the partial derivative ∂ρ (f(ρ, µ)/g(ρ, µ)), or equivalently, the
quantity h(ρ, µ) = f ′ρ(ρ, µ)g(ρ, µ) − g′ρ(ρ, µ)f(ρ, µ) is strictly positive. By plotting the
latter quantity on the rectangle 0 ≤ µ ≤
√
3 − 1, 1/5 ≤ ρ ≤ 1, its minimum seems to be
approximately 0.00146046085. To use this fact, we upper bound the two partial derivatives
of this function, and compute its values on a grid. In particular, using the monotonicity
properties of the functions f, g, we obtain that under our conditions |f(ρ, µ)| < 1.25 and
|g(ρ, µ)| ≤ 0.5. Furthermore, using the inequalities 0 ≤ µ ≤
√
3 − 1, 1/5 ≤ ρ ≤ 1 and
also the triangle inequality to estimate the derivatives of f and g, we obtain that
|f ′ρ(ρ, µ)| < 1.95, |f ′µ(ρ, µ)| < 2.8, |f ′′ρρ(ρ, µ)| < 2.95, |f ′′ρµ(ρ, µ)| < 9.8,
and
|g′ρ(ρ, µ)| < 0.93, |g′µ(ρ, µ)| < 1.08, |g′′ρρ(ρ, µ)| < 2.64, |g′′ρµ(ρ, µ)| < 15.1.
These inequalities imply that |h′ρ(ρ, µ)| < 4.78 and |h′µ(ρ, µ)| < 28.49, and hence, for any
∆ρ and ∆µ, we have h(ρ + ∆ρ, µ + ∆µ) > h(ρ, µ) − 4.78|∆ρ| − 28.49|∆µ|. Thus, we
divided the rectangle [0.2, 1] × [0,
√
3 − 1] into a 8691 × 8691 grid, and by numerically
computing the value of h(ρ, µ) at the gridpoints, we showed that at any such point the value
of h (up to 12 digits) is at least 0.00144. According to our estimates above, this implies
that h(ρ, µ) ≥ 0.00002 for all values of ρ and µ.
Before our next lemma, recall that B2 denotes the closed unit disk centered at the
origin.
Lemma 2.5. For some 0 < ν < 1, let x, y, z ∈ R2 be non-collinear points, and let
{Bu = u + ρuB2 : u ∈ {x, y, z}} be a ν-arrangement of disks; that is, assume that for
any {u, v} ⊂ {x, y, z}, we have |u − v| ≥ max{ρu, ρv} + νmin{ρu, ρv}. Assume that
for any {u, v} ⊂ {x, y, z}, Bu ∩ Bv ̸= ∅, and that the union of the three disks covers the
triangle [x, y, z]. Then ν ≤
√
3− 1.
M. Kadlicskó and Z. Lángi: On generalized Minkowski arrangements 9
x y
z
ρj
ρx
ρy
ρz
T
Figure 4: Notation in Lemma 2.5. The circles drawn with dotted lines represent the µ-cores
of the disks.
Proof. Without loss of generality, assume that 0 < ρz ≤ ρy ≤ ρx. Since the disks
are compact sets, by the Knaster-Kuratowski-Mazurkiewicz lemma [12], there is a point
q of T belonging to all the disks, or in other words, there is some point q ∈ T such
that |q − u| ≤ ρu for any u ∈ {x, y, z}. Recalling the notation T = [x, y, z] from the
introduction, let T ′ = [x′, y′, z′] be a triangle with edge lengths |y′ − x′| = ρx + νρy ,
|z′ − x′| = ρx + νρz and |z′ − y′| = ρy + νρz , and note that these lengths satisfy the
triangle inequality.
We show that the disks x′+ρxB2, y′+ρyB2 and z′+ρzB2 and T ′ satisfy the conditions
in the lemma. To do this, we show the following, more general statement, which, together
with the trivial observation that any edge of T ′ is covered by the two disks centered at
its endpoints, clearly implies what we want: For any triangles T = [x, y, z] and T ′ =
[x′, y′, z′] satisfying |u′ − v′| ≤ |u − v| for any u, v ∈ {x, y, z}, and for any point q ∈ T
there is a point q′ ∈ T ′ such that |q′ − u′| ≤ |q − u| for any u ∈ {x, y, z}. The main
tool in the proof of this statement is the following straightforward consequence of the Law
of Cosines, stating that if the side lengths of a triangle are A,B,C, and the angle of the
triangle opposite of the side of length C is γ, then for any fixed values of A and B, C is a
strictly increasing function of γ on the interval (0, π).
To apply it, observe that if we fix x, y and q, and rotate [x, z] around x towards [x, q],
we strictly decrease |z − y| and |z − q| and do not change |y − x|, |z − x|, |x − q| and
|y−q|. Thus, we may replace z by a point z∗ satisfying |z∗−y| = |z′−y′|, or the property
that z∗, q, x are collinear. Repeating this transformation by x or y playing the role of z
we obtain either a triangle congruent to T ′ in which q satisfies the required conditions, or
a triangle in which q is a boundary point. In other words, without loss of generality we
may assume that q ∈ bd(T ). If q ∈ {x, y, z}, then the statement is trivial, and so we
assume that q is a relative interior point of, say, [x, y]. In this case, if |z − x| > |z′ − x′| or
|z − y| > |z′ − y′|, then we may rotate [y, z] or [x, z] around y or x, respectively. Finally,
if |y − x| > |y′ − x′|, then one of the angles ∠yxz or ∠xyz, say ∠xyz, is acute, and then
we may rotate [z, y] around z towards [z, q]. This implies the statement.
10 Ars Math. Contemp. 23 (2023) #P1.09
By our argument, it is sufficient to prove Lemma 2.5 under the assumption that |y−x| =
ρx+νρy , |z−x| = ρx+νρz and |z−y| = ρy+νρz . Consider the case that ρx > ρy . Let q be
a point of T belonging to each disk, implying that |q−u| ≤ ρu for all u ∈ {x, y, z}. Clearly,
from our conditions it follows that |x−q| > ρx−ρy . Let us define a 1-parameter family of
configurations, with the parameter t ∈ [0, ρx−ρy], by setting x(t) = x−tw, where w is the
unit vector in the direction of x− q, ρx(t) = ρx − t, and keeping q, y, z, ρy, ρz fixed. Note
that in this family q ∈ Bx(t) = x(t)+ρx(t)B2, which implies that |x(t)−u| ≤ ρx(t)+ρu
for u ∈ {y, z}. Thus, for any {u, v} ⊂ {x(t), y, z}, there is a point of [u, v] belonging to
both Bu and Bv . This, together with the property that q belongs to all three disks and using
the convexity of the disks, yields that the triangle [x(t), y, z] is covered by Bx(t)∪By∪Bz .
Let the angle between u − x(t) and w be denoted by φ. Then, using the linearity of
directional derivatives, we have that for f(t) = |x(t) − u|, f ′(t) = − cosφ ≥ −1 for
u ∈ {y, z}, implying |x(t) − u| ≥ |x − u| − t = ρx(t) + νρu for u ∈ {y, z}, and also
that the configuration is a ν-arrangement for all values of t. Hence, all configurations in
this family, and in particular, the configuration with t = ρx − ρy satisfies the conditions
in the lemma. Thus, repeating again the argument in the first part of the proof, we may
assume that ρx = ρy ≥ ρz , |y − x| = (1 + µ)ρx and |z − x| = |z − y| = ρx + νρz .
Finally, if ρx = ρy > ρz , then we may assume that q lies on the symmetry axis of T
and satisfies |x − q| = |y − q| > ρx − ρz . In this case we apply a similar argument
by moving x and y towards q at unit speed and decreasing ρx = ρy simultaneously till
they reach ρz , and, again repeating the argument in the first part of the proof, obtain that
the family {ū + ρzB2 : ū ∈ {x̄, ȳ, z̄}}, where T̄ = [x̄, ȳ, z̄] is a regular triangle of side
lengths (1 + ν)ρz , covers T̄ . Thus, the inequality ν ≤
√
3 − 1 follows by an elementary
computation.
In our next lemma, for any disk Bi ∈ F we denote by B̄i the closure xi + ρiB2 of Bi.
Lemma 2.6. Let Bi, Bj , Bk ∈ F such that B̄u ∩ B̄v ̸⊆ Bw for any {u, v, w} = {i, j, k}.
Let T = [xi, xj , xk], ∆ = area(T ), and αu = ∠xvxuxw. If T ⊂ B̄i ∪ B̄j ∪ B̄k, then
1
2
∑
u∈{i,j,k}
αuρ
2
u ≤
2π√
3(1 + µ)2
∆, (2.2)
with equality if and only if ρi = ρj = ρk, and T is a regular triangle of side length
(1 + µ)ρi.
Proof. In the proof we call
δ =
∑
u∈{i,j,k} αuρ
2
u
2∆
the density of the configuration.
Consider the 1-parameter families of disks Bu(ν) = xu+(1 + µ) / (1 + ν) ρu int(B2),
where u ∈ {i, j, k} and ν ∈ [µ, 1]. Observe that the three disks Bu(ν), where u ∈ {i, j, k},
form a ν-arrangement for any ν ≥ µ. Indeed, in this case for any {u, v} ⊂ {i, j, k}, if
ρu ≤ ρv , we have
1 + µ
1 + ν
ρv + ν
(
1 + µ
1 + ν
ρu
)
= ρv + µρu −
ν − µ
1 + ν
(ρv − ρu) ≤ ρv + µρu ≤ |xu − xv|.
M. Kadlicskó and Z. Lángi: On generalized Minkowski arrangements 11
Furthermore, for any ν ≥ µ, we have
(1 + µ)2
∑
u∈{i,j,k}
αuρ
2
u = (1 + ν)
2
∑
u∈{i,j,k}
αu
(
1 + µ
1 + ν
)2
ρ2u.
Thus, it is sufficient to prove the assertion for the maximal value ν̄ of ν such that the
conditions T ⊂ B̄i(ν)∪B̄j(ν)∪B̄k(ν) and B̄u∩B̄v ̸⊆ Bw are satisfied for any {u, v, w} =
{i, j, k}. Since the relation B̄u ∩ B̄v ̸⊆ Bw implies, in particular, that B̄u ∩ B̄v ̸= ∅, in this
case the conditions of Lemma 2.5 are satisfied, yielding ν̄ ≤
√
3 − 1. Hence, with a little
abuse of notation, we may assume that ν̄ = µ. Then one of the following holds:
(i) The intersection of the disks B̄u is a single point.
(ii) For some {u, v, w} = {i, j, k}, B̄u ∩ B̄v ⊂ B̄w and B̄u ∩ B̄v ̸⊂ Bw.
Before investigating (i) and (ii), we remark that during this process, which we refer to
as µ-increasing process, even though there might be non-maximal values of ν for which the
modified configuration satisfies the conditions of the lemma and also (i) or (ii), we always
choose the maximal value. This value is determined by the centers of the original disks and
the ratios of their radii.
First, consider (i). Then, clearly, the unique intersection point q of the disks lies in T ,
and note that either q lies in the boundary of all three disks, or two disks touch at q. We
describe the proof only in the first case, as in the second one we may apply a straightforward
modification of our argument. Thus, in this case we may decompose T into three triangles
[xi, xj , q], [xi, xk, q] and [xj , xk, q] satisfying the conditions in Lemma 2.4, and obtain
1
2
∑
u∈{i,j,k}
αuρ
2
u ≤
4 arccos 1+µ2
(1 + µ)
√
(1− µ)(3 + µ)
∆ ≤ 2π√
3(1 + µ)2
∆,
where the second inequality follows from the fact that the two expressions are equal if
µ =
√
3− 1, and (
2 arccos
1 + µ
2
−
π
√
(1− µ)(3 + µ)√
3(1 + µ)
)′
> 0
if µ ∈ [0,
√
3 − 1]. Here, by Lemma 2.4, equality holds only if ρi = ρj = ρk, and T is a
regular triangle of side length (1+µ)ρi. On the other hand, under these conditions in (2.2)
we have equality. This implies Lemma 2.6 for (i).
In the remaining part of the proof, we show that if (ii) is satisfied, the density of the
configuration is strictly less than 2π/
(√
3(1 + µ)2
)
. Let q be a common point of bd(B̄w)
and, say, B̄u. If q is a relative interior point of an arc in bd(B̄u ∩ B̄v), then one of the disks
is contained in another one, which contradicts the fact that the disks Bu, Bv, Bw form a
µ-arrangement. Thus, we have that either B̄u ∩ B̄v = {q}, or that q is a vertex of the digon
Bu ∩ Bv . If B̄u ∩ B̄v = {q}, then the conditions of (i) are satisfied, and thus, we assume
that q is a vertex of the digon Bu ∩ Bv . By choosing a suitable coordinate system and
rescaling and relabeling, if necessary, we may assume that Bu = int(B2), xv lies on the
positive half of the x-axis, and xw is written in the form xw = (ζw, ηw), where ηw > 0,
12 Ars Math. Contemp. 23 (2023) #P1.09
B̄u
B̄w
B̄v
Figure 5: An illustration for the proof of Lemma 2.6.
and the radical line of Bu and Bv separates xv and xw (cf. Figure 5). Set ρ = ρw. We
show that ηw > (1 + µ)ρ/2.
Case 1: if ρ ≥ 1. Then we have |xw| ≥ ρ+ µ.
Let the radical line of Bu and Bv be the line {x = t} for some 0 < t ≤ 1. Then, as
this line separates xv and xw, we have ζw ≤ t, and by (ii) we have q = (t,−
√
1− t2).
This implies that |xw − q| ≤ |xw − xu|, |xw − xv|, from which we have 0 ≤ ζw. Let
S denote the half-infinite strip S = {(ζ, η) ∈ R2 : 0 ≤ ζ ≤ t, η ≥ 0}, and set s =
(t,−
√
1− t2+ρ). Note that by our considerations, xw ∈ S and |xw− q| = ρ, which yield
ηw ≤ −
√
1− t2 + ρ. From this it follows that ρ + µ ≤ |xw| ≤ |s|, or in other words, we
have t2 + (ρ −
√
1− t2)2 ≥ (ρ + µ)2. By solving this inequality for t with parameters ρ
and µ, we obtain that t ≥ t0, 1 ≤ ρ ≤
(
1− µ2
)
/ (2µ) and 0 ≤ µ ≤
√
2− 1, where
t0 =
√
1−
(
1− 2µρ− µ2
2ρ
)2
.
Let p = (ζp, ηp) be the unique point in S with |p| = ρ + µ and |p − q| = ρ, and
observe that ηw ≥ ηp. Now we find the minimal value of ηp if t is permitted to change
and ρ is fixed. Set p′ = (ζp,−
√
1− ζ2p). Since the bisector of [p′, q] separates p′ and
p, it follows that |p − p′| ≥ |p − q| = ρ with equality only if p′ = q and p = s, or in
other words, if t = t0. This yields that ζp is maximal if t = t0. On the other hand, since
|p| = ρ + µ and p lies in the first quadrant, ηp is minimal if ζp is maximal. Thus, for a
fixed value of ρ, ηp is minimal if t = t0 and p = s = (t0,−
√
1− t20 + ρ), implying that
ηw ≥ −
√
1− t20 + ρ =
(
2ρ2 + µ2 + 2µρ− 1
)
/(2ρ). Now, ρ ≥ 1 and µ < 1 yields that
2ρ2 + µ2 + 2µρ− 1
2ρ
− (1 + µ)ρ
2
=
ρ2 − µρ2 + 2µρ− 1
2ρ
≥ µ
2ρ
> 0,
implying the statement.
M. Kadlicskó and Z. Lángi: On generalized Minkowski arrangements 13
Case 2: if 0 < ρ ≤ 1. In this case the inequality ηw > (1 + µ)ρ/2 follows by a similar
consideration.
In the remaining part of the proof, let
σ(µ) =
2π√
3(1 + µ)2
.
Now we prove the lemma for (ii). Suppose for contradiction that for some configuration
{Bu, Bv, Bw} satisfying (ii) the density is at least σ(µ); here we label the disks as in the
previous part of the proof. Let B′w = x
′
w + ρw int(B
2) denote the reflection of Bw to
the line through [xu, xv]. By the inequality ηw > (1 + µ)ρ/2 proved in the two previous
cases, we have that {Bu, Bv, Bw, B′w} is a µ-arrangement, where we observe that by the
strict inequality, Bw and B′w do not touch each others cores. Furthermore, each triangle
[xu, xw, x
′
w] and [xv, xw, x
′
w] is covered by the three disks from this family centered at
the vertices of the triangle, and the intersection of no two disks from one of these triples
is contained in the third one. Thus, the conditions of Lemma 2.6 are satisfied for both
{Bu, Bw, B′w} and {Bv, Bw, B′w}. Observe that as by our assumption the density in T is
σ(µ), it follows that the density in at least one of the triangles [xu, xw, x′w] and [xv, xw, x
′
w],
say in T ′ = [xu, xw, x′w], is at least σ(µ). In other words, under our condition there
is an axially symmetric arrangement with density at least σ(µ). Now we apply the µ-
increasing process as in the first part of the proof and obtain a µ′-arrangement {B̂u =
xu + (1 + µ)/(1 + µ
′)ρu int(B
2), B̂w = xw + (1 + µ)/(1 + µ
′)ρw int(B
2), B̂′w = x
′
w +
(1 + µ)/(1 + µ′)ρw int(B
2)} with density σ(µ′) and µ′ ≥ µ that satisfies either (i) or (ii).
If it satisfies (i), we have that the density of this configuration is at most σ(µ′) with equality
if only if T ′ is a regular triangle of side length (1 + µ′)ρ, where ρ is the common radius of
the three disks. On the other hand, this implies that in case of equality, the disks centered at
xw and x′w touch each others’ cores which, by the properties of the µ-increasing process,
contradicts the fact that Bw and B′w do not touch each others’ µ-cores. Thus, we have that
the configuration satisfies (ii).
From Lemma 2.1 it follows that B̂w ∩ B̂′w ⊂ B̂u. Thus, applying the previous con-
sideration with B̂u playing the role of Bw, we obtain that the distance of xu from the
line through [xw, x′w] is greater than (1 + µ
′)/2ρu. Thus, defining B̂′u = x
′
u + (1 + µ)/
(1 + µ′)ρu int(B
2) as the reflection of B′u about the line through [xw, x
′
w], we have that
{B̂u, B̂w, B̂′w, B̂′u} is a µ′-arrangement such that {B̂u, B̂′u, B̂w} and {B̂u, B̂′u, B̂′w} satisfy
the conditions of Lemma 2.6. Without loss of generality, we may assume that the density
of {B̂u, B̂′u, B̂w} is at least σ(µ′). Again applying the µ-increasing procedure described
in the beginning of the proof, we obtain a µ′′-arrangement of three disks, with µ′′ ≥ µ′,
concentric with the original ones that satisfy the conditions of the lemma and also (i) or
(ii). Like in the previous paragraph, (i) leads to a contradiction, and we have that it satisfies
(ii). Now, again repeating the argument we obtain a µ′′′ -arrangement{
y +
1 + µ
1 + µ′′′
ρu int(B
2), xw +
1 + µ
1 + µ′′′′
ρw int(B
2), x′w +
1 + µ
1 + µ′′′
ρw int(B
2)
}
,
with density at least σ(µ′′′) and µ′′′ ≥ µ′′, that satisfies the conditions of the lemma,
where either y = xu or y = x′u. On the other hand, since in the µ-increasing process we
choose the maximal value of the parameter satisfying the required conditions, this yields
that µ′ = µ′′ = µ′′′. But in this case the property that {B̂u, B̂′u, B̂w} satisfies (ii) yields
that {B̂u, B̂′u, B̂w} does not; a contradiction.
14 Ars Math. Contemp. 23 (2023) #P1.09
3 Proof of Theorem 1.2
The idea of the proof follows that in [7] with suitable modifications. In the proof we de-
compose U(F) =
⋃n
i=1 Bi, by associating a polygon to each vertex of certain free digons
formed by two disks. Before doing it, we first prove some properties of µ-arrangements.
Let q be a vertex of a free digon, say, D = B1 ∩ B2. We show that the convex
angular region R bounded by the closed half lines starting at q and passing through x1
and x2, respectively, does not contain the center of any element of F different from B1
and B2 containing q on its boundary. Indeed, suppose for contradiction that there is a disk
B3 = x3+ρ3 int(B
2) ∈ F with q ∈ bd(B3) and x3 ∈ R. Since [q, x1, x2]\{q} ⊂ B1∪B2,
from this and the fact that F is a Minkowski-arrangement, it follows that the line through
[x1, x2] strictly separates x3 from q. As this line is the bisector of the segment [q, q′], where
q′ is the vertex of D different from q, from this it also follows that |x3 − q| > |x3 − q′|.
Thus, q′ ∈ B3.
Observe that in a Minkowski arrangement any disk intersects the boundary of another
one in an arc shorter than a semicircle. This implies, in particular, that B3 ∩ bd(B1) and
B3 ∩ bd(B2) are arcs shorter than a semicircle. On the other hand, from this the fact that
q, q′ ∈ B3 yields that bd(D) ⊂ B3, implying, by the properties of convexity, that D ⊂ B3,
which contradicts our assumption that D is a free digon.
Note that, in particular, we have shown that if a member of F contains both vertices of
a digon, then it contains the digon.
B1 B2
B(t)
Figure 6: The 1-parameter family of disks inscribed in B1 ∩B2.
Observe that the disks inscribed in D can be written as a 1-parameter family of disks
B(t) continuous with respect to Hausdorff distance, where t ∈ (0, 1) and B(t) tends to
{q} as t → 0+ (cf. Figure 6); here the term ‘inscribed’ means that the disk is contained
in Bi ∩ Bj and touches both disks from inside. We show that if some member Bk of F ,
different from B1 and B2, contains B(t) for some value of t, then Bk contains exactly
one vertex of D. Indeed, assume that some Bk contains some B(t) but it does not contain
any vertex of D. Then for i ∈ {1, 2}, Bk ∩ bd(Bi) is a circular arc Γi in bd(D). Let
Li be the half line starting at the midpoint of Γi, and pointing in the direction of the outer
normal vector of Bi at this point. Note that as D is a plane convex body, L1 ∩ L2 = ∅. On
the other hand, since B1, B2, Bk are a Minkowski arrangement, from this it follows that
xk ∈ L1∩L2; a contradiction. The property that no Bk contains both vertices of D follows
from the fact that D is a free digon. Thus, if q ∈ Bk for an element Bk ∈ F , then there is
M. Kadlicskó and Z. Lángi: On generalized Minkowski arrangements 15
some value t0 ∈ (0, 1) such that B(t) ⊆ Bk if and only if t ∈ (0, t0].
In the proof, we call the disks Bi, Bj adjacent, if Bi ∩ Bj is a digon, and there is a
member of the family B(t) defined in the previous paragraph that is not contained in any
element of F different from Bi and Bj . Here, we remark that any two adjacent disks define
a free digon, and if a vertex of a free digon is a boundary point of U(F), then the digon is
defined by a pair of adjacent disks.
Consider a pair of adjacent disks, say B1 and B2, and let q be a vertex of D = B1∩B2.
If q is a boundary point of the union U(F), then we call the triangle [x1, x2, q] a shell
triangle, and observe that by the consideration in the previous paragraph, the union of shell
triangles coincides with the inner shell of F .
If q is not a boundary point of U(F), then there is a maximal value t0 ∈ (0, 1) such
that B(t0) = x+ ρB2 is contained in an element Bi of F satisfying q ∈ Bi. Then, clearly,
B(t0) touches any such Bi from inside, and since B1 and B2 are adjacent, there is no
element of F containing B(t0) and the vertex of D different from q. Without loss of gen-
erality, assume that the elements of F touched by B(t0) from inside are B1, B2, . . . , Bk.
Since B1 and B2 are adjacent and there is no element of F containing both B(t0) and the
vertex of D different from q, we have that the tangent points of B1 and B2 on bd(B(t0)) are
consecutive points among the tangent points of all the disks Bi, where 1 ≤ i ≤ k. Thus,
we may assume that the tangent points of B1, B2, . . . , Bk on B(t0) are in this counter-
clockwise order on bd(B(t0)). Let x denote the center of B(t0). Since F is a Minkowski
arrangement, for any 1 ≤ i < j ≤ k, the triangle [x, xi, xj ] contains the center of no ele-
ment of F apart from Bi and Bj , which yields that the points x1, x2, . . . , xk are in convex
position, and their convex hull Pq contains x in its interior but it does not contain the center
of any element of F different from x1, x2, . . . , xk (cf. also [7]). We call Pq a core polygon.
We remark that since F is a µ-arrangement, the longest side of the triangle [x, xi, xi+1],
for i = 1, 2 . . . , k, is [xi, xi+1]. This implies that ∠xixxi+1 > π/3, and also that k < 6.
Furthermore, it is easy to see that for any i = 1, 2, . . . , k, the disks Bi and Bi+1 are
adjacent. Thus, any edge of a core polygon is an edge of another core polygon or a shell
triangle. This property, combined with the observation that no core polygon or shell triangle
contains any center of an element of F other than their vertices, implies that core polygons
cover the core of F without interstices and overlap (see also [7]).
Let us decompose all core polygons of F into triangles, which we call core triangles, by
drawing all diagonals in the polygon starting at a fixed vertex, and note that the conditions
in Lemma 2.6 are satisfied for all core triangles. Now, the inequality part of Theorem 1.2
follows from Lemmas 2.4 and 2.6, with equality if and only if each core triangle is a regular
triangle [xi, xj , xk] of side length (1 + µ)ρ, where ρ = ρi = ρj = ρk, and each shell
triangle [xi, xj , q], where q is a vertex of the digon Bi ∩ Bj is an isosceles triangle whose
base is of length (1 + µ)ρ, and ρ = ρi = ρj . Furthermore, since to decompose a core
polygon into core triangles we can draw diagonals starting at any vertex of the polygon, we
have that in case of equality in the inequality in Theorem 1.2, all sides and all diagonals of
any core polygon are of equal length. From this we have that all core polygons are regular
triangles, implying that all free digons in F are thick.
On the other hand, assume that all free digons in F are thick. Then, from Lemma 2.2
it follows that any connected component of F contains congruent disks. Since an adjacent
pair of disks defines a free digon, from this we have that, in a component consisting of disks
of radius ρ > 0, the distance between the centers of two disks defining a shell triangle, and
the edge-lengths of any core polygon, are equal to (1 + µ)ρ. Furthermore, since all disks
16 Ars Math. Contemp. 23 (2023) #P1.09
centered at the vertices of a core polygon are touched by the same disk from inside, we also
have that all core polygons in the component are regular k-gons of edge-length (1 + µ)ρ,
where 3 ≤ k ≤ 5. This and the fact that any edge of a core polygon connects the vertices
of an adjacent pair of disks yield that if the intersection of any two disks centered at two
different vertices of a core polygon is more than one point, then it is a free digon. Thus,
any diagonal of a core polygon in this component is of length (1 + µ)ρ, implying that any
core polygon is a regular triangle, from which the equality in Theorem 1.2 readily follows.
4 Remarks and open questions
Remark 4.1. If
√
3 − 1 < µ < 1, then by Lemma 2.5, C(F) = ∅ for any µ-arrangement
F of order µ.
Remark 4.2. Observe that the proof of Theorem 1.2 can be extended to some value µ >√
3 − 1 if and only if Lemma 2.4 can be extended to this value µ. Nevertheless, from
the continuity of the functions in the proof of Lemma 2.4, it follows that there is some
µ0 >
√
3−1 such that the lemma holds for any µ ∈ (
√
3−1, µ0]. Nevertheless, we cannot
extend the proof for all µ < 1 due to numeric problems.
Remark 4.2 readily implies Remark 4.3.
Remark 4.3. There is some µ0 >
√
3 − 1 such that if µ ∈ (
√
3 − 1, µ0], and F is a
µ-arrangment of finitely many disks, then the total area of the disks is
T ≤
4 · arccos( 1+µ2 )
(1 + µ) ·
√
(3 + µ)(1− µ)
area(I(F)) + area(O(F)),
with equality if and only if every free digon in F is thick.
Conjecture 4.4. The statement in Remark 4.3 holds for any µ-arrangement of finitely many
disks with
√
3− 1 < µ < 1.
Let 0 < µ < 1 and let F = {Ki : i = 1, 2, . . .} be a generalized Minkowski arrange-
ment of order µ of homothets of an origin-symmetric convex body in Rd with positive
homogeneity. Then we define the (upper) density of F with respect to U(F) as
δU (F) = lim sup
R→∞
∑
Bi⊂RB2 area (Bi)
area
(⋃
Bi⊂RB2 Bi
) .
Clearly, we have δ(F) ≤ δU (F) for any arrangement F .
Our next statement is an immediate consequence of Theorem 1.2 and Remark 4.3.
Corollary 4.5. There is some value
√
3− 1 < µ0 < 1 such that for any µ-arrangement F
of Euclidean disks in R2, we have
δU (F) ≤

2π√
3(1+µ)2
, if 0 ≤ µ ≤
√
3− 1, and
4·arccos( 1+µ2 )
(1+µ)·
√
(3+µ)(1−µ)
, if
√
3− 1 < µ ≤ µ0.
For any 0 ≤ µ < 1, let u, v ∈ R2 be two unit vectors whose angle is π3 , and let
Fhex(µ) denote the family of disks of radius (1 + µ) whose set of centers is the lattice
M. Kadlicskó and Z. Lángi: On generalized Minkowski arrangements 17
{ku + mv : k,m ∈ Z}. Then Fhex(µ) is a µ-arrangement, and by Corollary 4.5, for
any µ ∈ [0,
√
3 − 1], it has maximal density on the family of µ-arrangements of positive
homogeneity. Nevertheless, as Fejes Tóth observed in [8] (see also [4] or Section 1), the
same does not hold if µ >
√
3 − 1. Indeed, an elementary computation shows that in
this case Fhex(µ) does not cover the plane, and thus, by adding disks to it that lie in the
uncovered part of the plane we can obtain a µ-arrangement with greater density.
Fejes Tóth suggested the following construction to obtain µ-arrangements with large
densities. Let τ > 0 be sufficiently small, and, with a little abuse of notation, let τFhex(µ)
denote the family of the homothetic copies of the disks in Fhex(µ) of homothety ratio τ
and the origin as the center of homothety. Let F1hex(µ) denote the µ-arrangement obtained
by adding those elements of τFhex(µ) to Fhex(µ) that do not overlap any element of it.
Iteratively, if for some positive integer k, Fkhex(µ) is defined, then let F
k+1
hex (µ) denote the
union of Fkhex(µ) and the subfamily of those elements of τk+1Fhex(µ) that do not overlap
any element of it. Then, as was observed also in [8], choosing suitable values for τ and
k, the value of δU (Fhex(µ)) can be approximated arbitrarily well by δ(Fkhex(µ)). We note
that the same idea immediately leads to the following observation.
Remark 4.6. The supremums of δ(F) and δU (F) coincide on the family of the µ-arrange-
ments F in R2 of positive homogeneity.
We finish the paper with the following conjecture.
Conjecture 4.7. For any µ ∈ (
√
3 − 1, 1) and any µ-arrangement F in R2, we have
δ(F) ≤ δU (Fhex(µ)).
ORCID iDs
Máté Kadlicskó https://orcid.org/0000-0003-2948-8823
Zsolt Lángi https://orcid.org/0000-0002-5999-5343
References
[1] U. Betke, M. Henk and J. M. Wills, Finite and infinite packings, J. Reine Angew. Math. 453
(1994), 165–191, doi:10.1515/crll.1994.453.165, https://doi.org/10.1515/crll.
1994.453.165.
[2] K. Bezdek and Z. Lángi, Bounds for totally separable translative packings in the plane, Discrete
Comput. Geom. 63 (2020), 49–72, doi:10.1007/s00454-018-0029-6, https://doi.org/
10.1007/s00454-018-0029-6.
[3] M. Bleicher and J. Osborn, Minkowskian distribution of convex discs, Acta Math. Acad. Sci.
Hungar. 18 (1967), 5–17.
[4] K. Böröczky and L. Szabó, Minkowski arrangements of circles in the plane, 70, part I, pp.
87–92, 2002.
[5] K. Böröczky and L. Szabó, Minkowski arrangements of spheres, Monatsh. Math.
141 (2004), 11–19, doi:10.1007/s00605-002-0002-5, https://doi.org/10.1007/
s00605-002-0002-5.
[6] L. Fejes Tóth, Minkowskian distribution of discs, Proc. Am. Math. Soc. 16 (1965), 999–1004,
doi:10.2307/2035601, https://doi.org/10.2307/2035601.
[7] L. Fejes Tóth, Minkowskian circle-aggregates, Math. Ann. 171 (1967), 97–103, doi:10.1007/
bf01351644, https://doi.org/10.1007/bf01351644.
18 Ars Math. Contemp. 23 (2023) #P1.09
[8] L. Fejes Tóth, Packings and coverings in the plane, in: Proc. Colloquium on Convexity (Copen-
hagen, 1965), Kobenhavns Univ. Mat. Inst., Copenhagen, 1967 pp. 78–87.
[9] L. Fejes Tóth, Minkowski circle packings on the sphere, Discrete Comput. Geom. 22 (1999),
161–166, doi:10.1007/pl00009451, https://doi.org/10.1007/pl00009451.
[10] V. Földvári, Bounds on convex bodies in pairwise intersecting Minkowski arrangement of or-
der µ, J. Geom. 111 (2020), Paper No. 27, 11, doi:10.1007/s00022-020-00538-3, https:
//doi.org/10.1007/s00022-020-00538-3.
[11] P. M. Gruber and C. G. Lekkerkerker, Geometry of Numbers, volume 37 of North-Holland
Mathematical Library, North-Holland Publishing Co., Amsterdam, 2nd edition, 1987.
[12] B. Knaster, C. Kuratowski and S. Mazurkiewicz, Ein beweis des fixpunktsatzes für n-
dimensionale simplexe, Fundam. Math. 14 (1929), 132–137, http://eudml.org/doc/
212127.
[13] J. Molnár, Aggregati di cerchi di Minkowski, Ann. Mat. Pura Appl. (4) 71 (1966), 101–108,
doi:10.1007/bf02413737, https://doi.org/10.1007/bf02413737.
[14] M. Naszódi, J. Pach and K. Swanepoel, Arrangements of homothets of a convex body, Math-
ematika 63 (2017), 696–710, doi:10.1112/s0025579317000122, https://doi.org/10.
1112/s0025579317000122.
[15] M. Naszódi and K. J. Swanepoel, Arrangements of homothets of a convex body II, Contrib.
Discrete Math. 13 (2018), 116–123, doi:10.11575/cdm.v13i2.62732, https://doi.org/
10.11575/cdm.v13i2.62732.
[16] N. Oler, An inequality in the geometry of numbers, Acta Math. 105 (1961), 19–48, doi:10.
1007/bf02559533, https://doi.org/10.1007/bf02559533.
[17] A. Polyanskii, Pairwise intersecting homothets of a convex body, Discrete Math. 340
(2017), 1950–1956, doi:10.1016/j.disc.2017.04.002, https://doi.org/10.1016/j.
disc.2017.04.002.
[18] J. Solymosi and C. Wong, An application of kissing number in sum-product estimates, Acta
Math. Hungar. 155 (2018), 47–60, doi:10.1007/s10474-018-0831-x, https://doi.org/
10.1007/s10474-018-0831-x.
[19] G. Wolansky, A discrete approximation of the entropy functional for probability measure on
the plane, in: Nonlinear analysis and optimization, Amer. Math. Soc., Providence, RI, volume
659 of Contemp. Math., pp. 283–288, 2016, doi:10.1090/conm/659/13155, https://doi.
org/10.1090/conm/659/13155.
[20] H. Zassenhaus, Modern developments in the geometry of numbers, Bull. Am. Math. Soc.
67 (1961), 427–439, doi:10.1090/s0002-9904-1961-10614-9, https://doi.org/10.
1090/s0002-9904-1961-10614-9.