OON GENERALIZED MINKOWSKI ARRANGEMENTS
MÁTÉ KADLICSKÓ AND ZSOLT LÁNGI
Abstract.
The concept of a Minkowski arrangement was introduced by FejesTóth in 1965 as a family of centrally symmetric convex bodies with the prop-erty that no member of the family contains the center of any other member inits interior. This notion was generalized by Fejes Tóth in 1967, who called afamily of centrally symmetric convex bodies a generalized Minkowski arrange-ment of order µ for some < µ < if no member K of the family overlaps thehomothetic copy of any other member K (cid:48) with ratio µ and concentric with K (cid:48) .In this note we prove a sharp upper bound on the total area of the elements ofa generalized Minkowski arrangement of order µ of finitely many circular disksin the Euclidean plane. This result is a common generalization of a similar re-sult of Fejes Tóth for Minkowski arrangements of circular disks, and a result ofBöröczky and Szabó about the maximum density of a generalized Minkowskiarrangement of circular disks in the plane. In addition, we give a sharp upperbound on the density of a generalized Minkowski arrangement of homotheticcopies of a centrally symmetric convex body. Introduction
The notion of a
Minkowski arrangement of convex bodies was introduced byL. Fejes Tóth in [7], who defined it as a family F of centrally symmetric convexbodies in the d -dimensional Euclidean space R d , with the property that no memberof F contains the center of any other member of F in its interior. He used thisconcept to show, in particular, that the density of a Minkowski arrangement ofhomothets of any given plane convex body with positive homogeneity is at mostfour. Here an arrangement is meant to have positive homogeneity if the set ofthe homothety ratios is bounded from both directions by positive constants. It isworth mentioning that the above result is a generalization of the planar case ofthe famous Minkowski Theorem from lattice geometry. Furthermore, Fejes Tóthproved in [7] that the density of a Minkowski arrangement of circular disks in R with positive homogeneity is maximal for a Minkowski arrangement of congruentcircular disks whose centers are the points of a hexagonal lattice and each diskcontains the centers of six other members on its boundary.In [9], extending the investigation to finite Minkowski arrangements, Fejes Tóthgave a sharp upper bound on the total area of the members of a Minkowski ar-rangement of finitely many circular disks, and showed that this result immediately Mathematics Subject Classification.
Key words and phrases. arrangement, Minkowski arrangement, density, homothetic copy.The second named author is supported by the National Research, Development and InnovationOffice, NKFI, K-119670, the János Bolyai Research Scholarship of the Hungarian Academy ofSciences, and the BME IE-VIZ TKP2020 and ÚNKP-20-5 New National Excellence Programs bythe Ministry of Innovation and Technology. a r X i v : . [ m a t h . M G ] F e b M. KADLICSKÓ AND Z. LÁNGI implies the density estimate in [7] for infinite Minkowski circle-arrangements. Fol-lowing a different direction, in [8] for any < µ < Fejes Tóth defined a generalizedMinkowski arrangements of order µ as a family F of centrally symmetric convexbodies with the property that for any two distinct members K, K (cid:48) of F , K does notoverlap the µ -core of K (cid:48) , defined as the homothetic copy of K (cid:48) of ratio µ and con-centric with K (cid:48) . In this paper he made the conjecture that for any < µ ≤ √ − ,the density of a generalized Minkowski arrangement of circular disks with positivehomogeneity is maximal for a generalized Minkowski arrangement of congruentdisks 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 wasverified by Böröczky and Szabó in a seminar talk in 1965, though the first writtenproof seems to be published only in [5] in 2002. It was observed both in [8] and[5] that if √ − < µ < , then, since the above hexagonal arrangement does notcover the plane, that arrangement has no maximal density.In this paper we prove a sharp estimate on the total area of a generalizedMinkowski arrangement of finitely many circular disks, with a characterizationof the equality case. Our result includes the result in [9] as a special case, andimmediately implies the one in [5]. The proof of our statement relies on tools fromboth [5, 9], but uses also some new ideas. In addition, we also generalize a resultfrom Fejes Tóth [7] to find a sharp upper bound on the density of a generalizedMinkowski arrangement of homothetic copies of a centrally symmetric convex body.For completeness, we mention that similar statements for (generalized) Minkowskiarrangements in other geometries and in higher dimensional spaces were examined,e.g. in [6, 10, 13]. Minkowski arrangements consisting of congruent convex bod-ies were considered in [4]. Estimates for the maximum cardinality of mutuallyintersecting members in a (generalized) Minkowski arrangement can be found in[11, 14, 15, 17]. The problem investigated in this paper is similar in nature to thosedealing with the volume of the convex hull of a family of convex bodies, which hasa rich literature. This includes a result of Oler [16] (see also [3]), which is also oflattice geometric origin [20], and the notion of parametric density of Betke, Henkand Wills [1]. In particular, our problem is closely related to the notion of densitywith respect to outer parallel domains defined in [3]. Applications of (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 Re-mark 2 of [7], stating the same property for (not generalized) Minkowski arrange-ments of plane convex bodies. In Proposition 1, by vol d ( · ) we denote d -dimensionalvolume, and by B d we denote the closed Euclidean unit ball centered at the origin. Proposition 1.
Let < µ < , let K ⊂ R d be an origin-symmetric convex bodyand let F = { x + λ K, x + λ K, . . . } be a generalized Minkowski arrangement oforder µ , where x i ∈ R d , λ i > for each i = 1 , , . . . . Assume that F is of positivehomogeneity, that is, there are constants < C < C satisfying C ≤ λ i ≤ C forall values of i , and define the (upper) density δ ( F ) of F in the usual way as δ ( F ) = lim sup R →∞ (cid:80) x i ∈ R B d vol d ( x i + λ i K )vol d ( R B d ) , if it exists. Then (1) δ ( F ) ≤ d (1 + µ ) d , INKOWSKI ARRANGEMENTS 3 where equality is attained, e.g. if { x , x , . . . } is a lattice with K as its fundamentalregion, and λ i = 1 + µ for all values of i .Proof. Note that the equality part of Proposition 1 clearly holds, and thus, weprove only the inequality in (1). Let || · || K : R d → [0 , ∞ ) denote the norm with K as its unit ball. Then, by the definition of a generalized Minkowski arrangement,we have || x i − x j || K ≥ max { λ i + µλ j , λ j + µλ i } ≥≥
12 (( λ i + µλ j ) + ( λ j + µλ i )) = 1 + µ λ i + λ j ) , implying that the homothets x i + λ i (1+ µ )2 K are pairwise non-overlapping. In otherwords, the family F (cid:48) = (cid:110) x i + λ i (1+ µ )2 K : i = 1 , , . . . (cid:111) is a packing. Thus, thedensity of F (cid:48) is at most one, from which (1) readily follows. (cid:3) Following the terminology of Fejes Tóth in [9] and to permit a simpler formulationof our main result, in the remaining part of the paper we consider generalizedMinkowski arrangements of open circular disks, where we note that generalizedMinkowski arrangements can be defined for families of open circular disks in thesame way as for families of closed circular disks.To state our main result, we need some preparation. Consider some generalizedMinkowski arrangement F = { B i = x i + ρ i B : i = 1 , , . . . , n } of open circulardisks in R of order µ , where < µ < . Set U ( F ) = (cid:83) ni =1 B i . Then each circulararc Γ in the boundary of U ( F ) corresponds to a circular sector, which can beobtained as the convex hull of Γ and the center of the disk in F whose boundarycontains Γ . We call the union of these circular sectors the outer shell of F . Nowconsider a point p ∈ bd( U ( F )) belonging to at least two members of F , say B i and B j , such that x i , x j and p are not collinear. We choose the notation in such a waythat the convex angular region bounded by the two closed half lines starting at p and passing through x i and x j , respectively, do not contain the center of anotherelement of F which contains p on its boundary. We call the union of the triangles conv { p, x i , x j } satisfying these conditions the inner shell of F . We denote the innerand 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 shellof any generalized Minkowski arrangement is nonempty, but there are arrangementsfor which I ( F ) = ∅ or C ( F ) = ∅ .If the intersection of two members of F is nonempty, then we call this intersectiona 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.
Let < µ ≤ √ − , and let F = { B i = x i + ρ i int( B ) : i =1 , , . . . , n } be a generalized Minkowski arrangement of finitely many open circulardisks of order µ . Then T = π n (cid:88) i =1 ρ i ≤ π √ µ ) area( C ( F ))++ 4 · arccos( µ )(1 + µ ) · (cid:112) (3 + µ )(1 − µ ) area( I ( F )) + area( O ( F )) , M. KADLICSKÓ AND Z. LÁNGI
Figure 1.
The outer and inner shell, and the core of an arrange-ment. where T is the total area of the circles, with equality if and only if each free digonin F is thick. In the paper, for any points x, y, z ∈ R , we denote by [ x, y ] the closed segmentwith endpoints x, y , by [ x, y, z ] the triangle conv { x, y, z } , by | x | the Euclidean normof x , and if x and z are distinct from y , by ∠ xyz we denote the measure of the anglebetween the closed half lines starting at y and passing through x and z . Note thataccording to our definition, ∠ xyz is at most π for any x, z (cid:54) = y . By bd( · ) we denotethe boundary of a set. Furthermore, 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 < µ ≤ √ − .In Section 2, we prove some preliminary lemmas. In Section 3 we prove Theo-rem 1. Finally, in Section 4 we collect additional remarks and questions.2. Preliminaries
For any B i , B j ∈ F , if B i ∩ B j (cid:54) = ∅ , we call the two intersection points of bd( B i ) and bd( B j ) the vertices of the digon B i ∩ B j .First, we recall a lemma from [9]. To prove it we observe that for any µ > , ageneralized Minkowski arrangement of order µ is a Minkowski arrangement as well. Lemma 1.
Let B i , B j , B k ∈ F such that the digon B i ∩ B j is contained in B k .Then the digon B i ∩ B k is free. From now on, we call the maximal subfamilies F (cid:48) of F (with respect to contain-ment) with the property that (cid:83) B i ∈F (cid:48) B i is connected the connected components of F . INKOWSKI ARRANGEMENTS 5
Lemma 2. If F (cid:48) is a connected component of F in which each free digon is thick,then the elements of F (cid:48) are congruent.Proof. We need to show that for any B i , B j ∈ F (cid:48) , B i and B j are congruent. Observethat by connectedness, we may assume that B i ∩ B j is a digon. If B i ∩ B j is free,then it is thick, which implies that B i and B j are congruent. If B i ∩ B j is not free,then there is a disk B k ∈ F (cid:48) containing it. By Lemma 1, the digons B i ∩ B k and B j ∩ B k are free. Thus B k is congruent to both B i and B j . (cid:3) In the remaining part of Section 2, we examine densities of some circular sectorsin certain triangles. The computations in the proofs of these lemmas were carriedout by a Maple 18.00 software.
Lemma 3.
Let < γ < π and A, B > be arbitrary. Let T = [ x, y, z ] be a trianglesuch 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 area and the angles of T at x and y , respectively, and set f A,B ( γ ) = αA + βB ∆ . Then, for any A, B > , the function f A,B ( γ ) is strictly decreasing on the interval γ ∈ (0 , π ) .Proof. Without loss of generality, assume that A ≤ B , and let g = αA + βB .Then, by an elementary computation, we have that g = A arccot A − B cos γB sin γ + A arccot B − A cos γA sin γ , and ∆ = 12 AB sin γ. We regard g and ∆ as functions of γ . We intend to show that g (cid:48) ∆ − g ∆ (cid:48) is negativeon the interval (0 , π ) for all A, B > . Let h = g (cid:48) ∆∆ (cid:48) − g , and note that thisexpression is continuous on (cid:0) , π (cid:1) and (cid:0) π , π (cid:1) for all A, B > . By differentiatingand simplifying, we obtain h (cid:48) = − (cid:0) A (1 + cos ( γ )) + B (1 + cos ( γ )) − AB cos( γ ) (cid:1) A B sin ( γ )cos ( γ )( A + B − AB cos( γ )) , which is negative on its domain. This implies that g (cid:48) ∆ − g ∆ (cid:48) is strictly decreas-ing on (cid:0) , π (cid:1) and strictly increasing on (cid:0) π , π (cid:1) . On the other hand, we have lim γ → + ( g (cid:48) ∆ − g ∆ (cid:48) ) = − A Bπ , and lim γ → π − ( g (cid:48) ∆ − g ∆ (cid:48) ) = 0 . This yields theassertion. (cid:3) Lemma 4.
Consider two disks B i , B j ∈ F such that | x i − x j | < ρ i + ρ j , and let v bea vertex of the digon B i ∩ B j . Let T = [ x i , x j , v ] , ∆ = area( T ) , and let α i = ∠ vx i x j and α j = ∠ vx j x i . Then (2) α i ρ i + 12 α j ρ j ≤ µ (1 + µ ) (cid:112) (1 − µ )(3 + µ ) ∆ , with equality if and only if ρ i = ρ j and | x i − x j | = ρ i (1 + µ ) .Proof. Without loss of generality, let ρ i = 1 , and < ρ j = ρ ≤ . By Lemma 3, wemay assume that | x i − x j | = 1 + µρ . Then, by the Law of Cosines and by Heron’sformula, we may write α i ρ i + α j ρ j ∆ as a C ∞ -class function f ( r, µ ) , where ( r, µ ) isa point of the rectangle R = (0 , × (0 , √ − . Numeric computations show that ∂ µ ∂ r f is negative on R , and f (cid:48) r ( r, √ − is positive on (0 , (see Figure 2). Thisimplies that ∂ r f ( r, µ ) > for all ( r, µ ) ∈ R , from which the lemma readily follows. (cid:3) M. KADLICSKÓ AND Z. LÁNGI . . . . r . . . . . Figure 2.
The graph of the function f (cid:48) r ( r, √ − on the interval [0 , . Lemma 5.
For some < ν < , let x, y, z ∈ R be non-collinear points, and let { B u = u + ρ u B : u ∈ { x, y, z }} be a ν -arrangement of disks; that is, assume thatfor any { u, v } ⊂ { x, y, z } , we have | u − v | ≥ max { ρ u , ρ v } + ν min { ρ u , ρ v } . Assumethat for any u, v ∈ { x, y, z } , B u ∩ B v (cid:54) = ∅ , and that the union of the three diskscovers the triangle [ x, y, z ] . Then ν ≤ √ − .Proof. Without loss of generality, assume that < ρ z ≤ ρ y ≤ ρ x . Since the disksare compact sets, by the Knaster-Kuratowski-Mazurkiewicz lemma [12], there is apoint q of T belonging to each disk, or in other words, there is some point q ∈ T such that | q − u | ≤ ρ u for any u ∈ { x, y, z } . Let T (cid:48) = [ x (cid:48) , y (cid:48) , z (cid:48) ] be a triangle withedge lengths | y (cid:48) − x (cid:48) | = ρ x + νρ y , | z (cid:48) − x (cid:48) | = ρ x + νρ z and | z (cid:48) − y (cid:48) | = ρ y + νρ z , andnote that these lengths satisfy the triangle inequality.We show that the disks x (cid:48) + ρ x B , y (cid:48) + ρ y B and z (cid:48) + ρ z B cover T (cid:48) . To dothis, we show the following, more general statement, which clearly implies what wewant: For any triangles T = [ x, y, z ] and T (cid:48) = [ x (cid:48) , y (cid:48) , z (cid:48) ] satisfying | u (cid:48) − v (cid:48) | ≤ | u − v | for any u, v ∈ { x, y, z } , and for any point q ∈ T there is a point q (cid:48) ∈ T (cid:48) such that | q (cid:48) − u (cid:48) | ≤ | q − u | for any u ∈ { x, y, z } . The main tool in the proof of this statementis Cauchy’s Arm Lemma, which states, in particular, that if the side lengths of atriangle 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 (cid:48) − y (cid:48) | ,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 (cid:48) in which q satisfiesthe required conditions, or a triangle in which q is a boundary point. In otherwords, 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 pointof, say, [ x, y ] . In this case, if | z − x | > | z (cid:48) − x (cid:48) | or | z − y | > | z (cid:48) − x (cid:48) | , then we mayrotate [ y, z ] or [ x, z ] around y or x , respectively. Finally, if | y − x | > | y (cid:48) − x (cid:48) | , then INKOWSKI ARRANGEMENTS 7 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.By our argument, it is sufficient to prove Lemma 5 under the assumption that | y − x | = ρ x + νρ y , | z − x | = ρ x + νρ z and | z − y | = ρ y + νρ z . Consider thecase that ρ x > ρ y . Let q be a point of T belonging to each disk, implying that | q − u | ≤ ρ u for all u ∈ { x, y, z } . Clearly, from our conditions it follows that | x − q | > ρ x − ρ y . Let us define a -parameter family of configurations, with theparameter t ∈ [0 , ρ x − ρ y ] , by setting x ( t ) = x − tv , where v is the unit vector in thedirection of x − q , ρ x ( t ) = ρ x − t , and keeping q, y, z, ρ y , ρ z fixed. Note that duringthis motion, q ∈ x ( t ) + ρ x ( t ) B implies that | x ( t ) − u | ≤ ρ x ( t ) + ρ u for u ∈ { y, z } .Furthermore, since d | x ( t ) − u | d t ≥ − for u ∈ { y, z } , we have that the configuration isa ν -arrangement for all values of t . On the other hand, for any value of t , the factthat q ∈ x ( t ) + ρ x ( t ) B yields [ x ( t ) , u ] is covered by ( x ( t ) + ρ x ( t ) B ) ∪ ( u + ρ u B ) for u ∈ { y, z } , which implies that the conditions of Lemma 5 are satisfied for any t .In particular, the configuration with t = ρ x − ρ y also satisfies the conditions in thelemma; thus, applying the argument in the first part of the proof, we may assumethat ρ x = ρ y . Finally, if ρ x = ρ y > ρ z , then we have | z − x | = | z − y | = ρ x + νρ z , andwe may assume that q lies on the symmetry axis of T and satisfies | z − q | > ρ x − ρ z .In this case we apply a similar argument by moving x and y towards q at unitspeed and decreasing ρ x = ρ y simultaneously till they reach ρ z , and obtain thatthe family { ¯ u + ρ z B : ¯ u ∈ { ¯ x, ¯ y, ¯ z }} , where ¯ T = [¯ x, ¯ y, ¯ z ] is a regular triangle ofside lengths (1 + ν ) ρ z , covers ¯ T . Thus, the inequality ν ≤ √ − follows by anelementary computation. (cid:3) In our last lemma, for any disk B i ∈ F we denote by ¯ B i the closure x i + ρ i B of B i . Lemma 6.
Let B i , B j , B k ∈ F such that ¯ B u ∩ ¯ B v (cid:54)⊆ B w for any { u, v, w } = { i, j, k } .Let T = [ x i , x j , x k ] , ∆ = area( T ) , and α u = ∠ x v x u x w for any { u, v, w } = { i, j, k } .If T ⊂ ¯ B i ∪ ¯ B j ∪ ¯ B k , then (3) (cid:88) u ∈{ i,j,k } α u ρ u ≤ π √ µ ) ∆ , with equality if and only if ρ i = ρ j = ρ k , and T is an equilateral triangle of sidelength (1 + µ ) ρ i .Proof. In the proof we call δ = (cid:80) u ∈{ i,j,k } α u ρ u the density of the configuration.Consider the -parameter families of disks B u ( ν ) = x u + µ ν ρ u int( B ) , where u ∈ { i, j, k } and ν ∈ [ µ, . Observe that the three disks B u ( 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 ρ u ≤ ρ v , we have µ ν ρ v + ν ( µ ν ρ u ) = ρ v + µρ u − ν − µ ν ( ρ v − ρ u ) ≤ ρ v + µρ u ≤| x u − x v | . Furthermore, for any ν ≥ µ , we have (1 + µ ) (cid:80) u ∈{ i,j,k } α u ρ u = (1 + ν ) (cid:80) u ∈{ i,j,k } α u (cid:16) µ ν (cid:17) ρ u . Thus, it is sufficient to prove the assertion for themaximal value ¯ ν of ν such that the conditions T ⊂ ¯ B i ( ν ) ∪ ¯ B j ( ν ) ∪ ¯ B k ( ν ) and ¯ B u ∩ ¯ B v (cid:54)⊆ B w are satisfied for any { u, v, w } = { i, j, k } . Since the relation ( ¯ B u ∩ ¯ B v ) (cid:54)⊆ B w implies, in particular, that ¯ B u ∩ ¯ B v (cid:54) = ∅ , in this case the conditions ofLemma 5 are satisfied, yielding ¯ ν ≤ √ − . Hence, with a little abuse of notation,we may assume that ¯ ν = µ . Then one of the following holds: M. KADLICSKÓ AND Z. LÁNGI ¯ B u ¯ B w ¯ B v Figure 3.
An illustration for the proof of Lemma 6.(i) The intersection of the disks ¯ B u is a single point.(ii) For some { u, v, w } = { i, j, k } , ¯ B u ∩ ¯ B v ⊂ ¯ B w and ¯ B u ∩ ¯ B v (cid:54)⊂ B w .First, consider (i). Then, clearly, the unique intersection point q of the diskslies in T . Thus, in this case we may decompose T into three triangles [ x i , x j , q ] , [ x i , x k , q ] and [ x j , x k , q ] satisfying the conditions in Lemma 4, and obtain (cid:88) u ∈{ i,j,k } α u ρ u ≤ µ (1 + µ ) (cid:112) (1 − µ )(3 + µ ) ∆ ≤ π √ µ ) ∆ , where the second inequality follows by an elementary computation. Here, byLemma 4, equality holds only if ρ i = ρ j = ρ k , and T is an equilateral triangleof side length (1 + µ ) ρ i . On the other hand, under these conditions in (3) we haveequality. This implies Lemma 6 for (i).In the remaining part of the proof, we deal with (ii). Let q be a common pointof bd( ¯ B w ) and, say, ¯ B u . If q is a relative interior point of an arc in bd( ¯ B u ∩ ¯ B v ) ,then ¯ B u ⊆ ¯ B w , which contradicts the fact that the disks B u , B v , B w form a µ -arrangement. Thus, we have that either ¯ B u ∩ ¯ B v = { q } , or that q is a vertex ofthe digon B u ∩ B v . By choosing a suitable coordinate system and rescaling andrelabeling, if necessary, we may assume that B u = int( B ) , x v lies on the positivehalf of the x -axis, and x w is written in the form x w = ( ζ w , η w ) , where η w > , andthe radical line of B u and B v separates x v and x w (cf. Figure 3). Set ρ = ρ w . Weshow that η w ≥ (1+ µ ) ρ . Case 1 , if ρ ≥ . Then we have | x w | ≥ ρ + µ .Let the radical line of B u and B v be the line { x = t } for some < t ≤ .Then, by (ii), the point q = ( t, −√ − t ) is a point of ¯ B w . This implies that | x w − q | ≤ | x w − x u | , | x w − x v | , from which we have ≤ ζ w ≤ t . Let S denote thehalf-infinite strip S = { ( ζ, η ) ∈ R : 0 ≤ ζ ≤ t, η ≥ } , and set s = ( t, −√ − t + ρ ) .Observe that if | s | < ρ + µ , then S ∩ (cid:0) q + ρ B (cid:1) ⊂ ( ρ + µ ) B , which contradicts our INKOWSKI ARRANGEMENTS 9 conditions that | x w | ≥ ρ + µ and | x w − q | ≤ ρ . Thus, | s | ≥ ρ + µ , which impliesthat t ≥ (cid:114) − (cid:16) − µρ − µ ρ (cid:17) , ≤ ρ ≤ − µ µ and ≤ µ ≤ √ − .Let p = ( ζ p , η p ) be the unique point in S with | p | = ρ + µ and | p − q | = ρ , andobserve that η w ≥ η p . Furthermore, for a fixed value of ρ , by decreasing t also η w is decreased. Indeed, decreasing t corresponds to rotating the coordinate systemin counterclockwise direction while q is fixed, and in this case η w clearly decreases.Thus, we may assume that t = (cid:114) − (cid:16) − µρ − µ ρ (cid:17) . Define the function f ( ρ, µ ) = ( µ + ρ ) − (cid:18) − µρ − µ ρ (cid:19) − (cid:18) (1 + µ ) ρ (cid:19) , where ( ρ, µ ) ∈ D = { ( ρ, µ ) ∈ R : 1 ≤ ρ ≤ − µ µ , ≤ µ ≤ √ − } , and note that toshow the inequality η w ≥ (1+ µ ) ρ it is sufficient to show that f ( ρ, µ ) ≥ on D . Toshow it we observe that ∂ ρ ( ρ f ) is negative on D , and f (1 , µ ) = 2 µ + 4 µ ≥ . Case 2 , if < ρ ≤ . In this case the inequality η w ≥ (1+ µ ) ρ follows by a similarconsideration.Now we prove the lemma for (ii). Assume that for some configuration { B u , B v , B w } satisfying (ii) the density is at least π √ µ ) . Let B (cid:48) w = x (cid:48) w + ρ w int( B ) denotethe reflection of B w to the line through [ x u , x w ] . Then the conditions of Lemma 6are satisfied for { B u , B w , B (cid:48) w } and { B v , B w , B (cid:48) w } . Thus, by (i), we have that thereis an axially symmetric configuration satisfying (ii) and density at least π √ µ ) .Without loss of generality, we may assume that this configuration is { B u , B v , B (cid:48) v } ,where B v and B (cid:48) v are congruent, and | x v − x u | = | x (cid:48) v − x u | . Then, by Lemma 1, ¯ B v ∩ ¯ B (cid:48) v ⊂ B u . Let B (cid:48) u be the reflected copy of B u to the line through [ x v , x (cid:48) v ] .By our consideration, the conditions of Lemma 1 are satisfied for { B u , B (cid:48) u , B v } and { B u , B (cid:48) u , B (cid:48) v } . Thus, applying the resizing argument in the second paragraph of theproof, it follows that for some value of µ , { B u , B (cid:48) u , B v } or { B u , B v , B (cid:48) v } satisfies (i),which implies the statement. (cid:3) Proof of Theorem 1
The idea of the proof follows that in [9] with suitable modifications. In the proofwe decompose U ( F ) = (cid:83) ni =1 B i , by associating a polygon to each vertex of certainfree 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 = B ∩ B . We show that the convexangular region R bounded by the closed half lines starting at q and passing through x and x , respectively, does not contain the center of any element of F differentfrom B and B containing q on its boundary. Indeed, suppose for contradictionthat there is a disk B = x + ρ int( B ) ∈ F with q ∈ bd( B ) and x ∈ R . Since [ q, x , x ] \{ q } ⊂ B ∪ B , from this it follows that the line through [ x , x ] strictlyseparates x from q . As this line is the bisector of the segment [ q, q (cid:48) ] , where q (cid:48) isthe vertex of D different from q , from this it also follows that | x − q | > | x − q (cid:48) | .Thus, q (cid:48) ∈ B .Observe that in a Minkowski arrangement any disk intersects the boundary ofanother one in an arc shorter than a semicircle. This implies, in particular, that B ∩ bd( B ) and B ∩ bd( B ) are arcs shorter than a semicircle. On the other hand, from this the fact that q, q (cid:48) ∈ B yields that bd( D ) ⊂ B , implying, by theproperties of convexity, that D ⊂ B , which contradicts our assumption that D isa free digon.Note that, in particular, we have shown that if a member of F contains bothvertices of a digon, then it contains the digon. B B B ( t ) Figure 4.
The -parameter family of disks inscribed in B ∩ B .Observe that the disks inscribed in D can be written as a -parameter family ofdisks B ( t ) continuous with respect to Hausdorff distance, where t ∈ (0 , and B ( t ) tends to { q } as t → + (cf. Figure 4); here the term ‘inscribed’ means that the diskis contained in B i ∩ B j and touches both disks from inside. We show that if somemember B k of F , different from B and B , contains B ( t ) for some value of t , then B k contains exactly one vertex of D . Indeed, assume that some B k contains some B ( t ) but it does not contain any vertex of D . Then for i ∈ { , } , B k ∩ bd( B i ) is a circular arc Γ i in bd( B ) ∩ bd( B ) . Let L i be the half line starting at themidpoint of Γ i , and pointing in the direction of the outer normal vector of B i atthis point. Since B , B , B k are a Minkowski arrangement, from this it follows that x k ∈ L ∩ L , which contradicts the fact that L ∩ L = ∅ . The property that no B k contains both vertices of D follows from the fact that D is a free digon. Basedon these observations, we have that if q ∈ B k for an element B k ∈ F , then there issome value t ∈ (0 , such that B ( t ) ⊆ B k if and only if t ∈ (0 , t ] .In the proof, we call the disks B i , B j adjacent , if B i ∩ B j is a digon, and there is amember of the family B ( t ) defined in the previous paragraph that is not containedin any element of F different from B i and B j . Here, we remark that any twoadjacent disks define a free digon, and if a vertex of a free digon is a boundarypoint of U ( F ) , then the digon is defined by a pair of adjacent disks.Consider a pair of adjacent disks, say B and B , and let q be a vertex of D = B ∩ B . If q is a boundary point of the union U ( F ) , then we call the triangle [ x , x , q ] a shell triangle , and observe that by the consideration in the previousparagraph, 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 t ∈ (0 , such that B ( t ) = x + ρ B is contained in an element B i of F satisfying q ∈ B i .Then, clearly, B ( t ) touches any such B i from inside, and there is no element of F containing B ( t ) and the vertex of D different from q . Without loss of generality,assume that the elements of F touched by B ( t ) from inside are B , B , . . . , B k .Since B and B are adjacent, we have that the tangent points of B and B INKOWSKI ARRANGEMENTS 11 are consecutive points among the tangent points of all B i s, where ≤ i ≤ k .Thus, we may assume that the tangent points of B , B , . . . , B k on B ( t ) are inthis counterclockwise order on bd( B ( t )) . Since F is a Minkowski arrangement,we have that the points x , x , . . . , x k are in convex position, their convex hull P q contains the center x of B ( t ) in its interior but it does not contain the center ofany element of F different from x , x , . . . , x k . We call P q a core polygon .We remark that since F is a µ -arrangement, the longest side of the triangle [ x, x i , x i +1 ] , for i = 1 , . . . , k , is [ x i , x i +1 ] . This implies that ∠ x i xx i +1 > π , andalso that k < . Furthermore, it is easy to see that for any i = 1 , , . . . , k , the disks B i and B i +1 are adjacent. Thus, any edge of a core polygon is an edge of anothercore polygon or a shell triangle. This property, combined with the observation thatno core polygon or shell triangle contains any center of an element of F other thantheir vertices, implies that core polygons cover the core of F without intersticesand overlap (see also [9]).Let us decompose all core polygons of F into triangles, which we call core tri-angles , by drawing all diagonals in the polygon starting at a fixed vertex, and notethat the conditions in Lemma 6 are satisfied for all core triangles. Now, the in-equality part of Theorem 1 follows from Lemmas 4 and 6, with equality if and onlyif each core triangle is a regular triangle [ x i , x j , x k ] of side length (1 + µ ) ρ , where ρ = ρ i = ρ j = ρ k , and each shell triangle [ x i , x j , q ] , where q is a vertex of the digon B i ∩ B j is an isosceles triangle whose base is of length (1 + µ ) ρ , and ρ = ρ i = ρ j .Furthermore, since to decompose a core polygon into core triangles we can drawdiagonals starting at any vertex of the polygon, we have that in case of equalityin the inequality in Theorem 1, all sides and all diagonals of any core polygon areof equal length, from which 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 Lemma2 it follows that any connected component of F contains congruent disks. Since anadjacent pair of disks defines a free digon, from this we have that, in a componentconsisting of disks of radius ρ > , the distance between the centers of two disksdefining a shell triangle, and the edge-lengths of any core polygon, are equal to (1 + µ ) ρ . Furthermore, since all disks centered at the vertices of a core polygonare touched by the same disk from inside, we also have that all core polygons inthe component are regular k -gons of edge-length (1 + µ ) ρ , where ≤ k ≤ . Thisand the fact that any edge of a core polygon connects the vertices of an adjacentpair of disks yield that if the intersection of any two disks centered at two differentvertices 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 , implyingthat any core polygon is a regular triangle, from which the equality in Theorem 1readily follows. 4. Remarks and open questions
Remark 1. If √ − < µ < , then by Lemma 5, C ( F ) = ∅ for any µ -arrangement F of order µ . Remark 2.
Observe that the proof of Theorem 1 can be extended to some value µ > √ − if and only if Lemma 4 can be extended to this value µ . Nevertheless,from the continuity of the functions in the proof of Lemma 4, it follows that there is some µ > √ − such that the lemma holds for any µ ∈ ( √ − , µ ] . Nevertheless,we cannot extend the proof for all µ < due to numeric problems.Remark 2 readily implies Remark 3. Remark 3.
There is some µ > √ − such that if µ ∈ ( √ − , µ ] , and F is a µ -arrangment of finitely many disks, then the total area of the disks is T ≤ · arccos( µ )(1 + µ ) · (cid:112) (3 + µ )(1 − µ ) area( I ( F )) + area( O ( F )) , with equality if and only if every free digon in F is thick. Conjecture 1.
The statement in Remark 3 holds for any µ -arrangement of finitelymany disks with √ − < µ < .Let < µ < and let F = { K i : i = 1 , , . . . } be a generalized Minkowskiarrangement of order µ of homothets of an origin-symmetric convex body in R d with positive homogeneity. Then we define the (upper) density of F with respectto (cid:83) F as δ U ( F ) = lim sup R →∞ (cid:80) B i ⊂ R B area ( B i )area (cid:0)(cid:83) B i ⊂ R B B i (cid:1) . Clearly, we have δ ( F ) ≤ δ U ( F ) for any arrangement F .Our next statement is an immediate result of Theorem 1 and Remark 3. Corollary 1.
There is some value √ − < µ < such that for any µ -arrangement F of disks in R , we have δ U ( F ) ≤ π √ µ ) , if ≤ µ ≤ √ − , and · arccos( µ )(1+ µ ) · √ (3+ µ )(1 − µ ) , if √ − < µ ≤ µ . For any ≤ µ < , let u, v ∈ R be two unit vectors whose angle is π , and let F hex ( µ ) denote the family of disks of radius (1+ µ ) whose set of centers is the lattice { ku + mv : k, m ∈ Z } . Then F hex ( µ ) is a µ -arrangement, and by Corollary 1, for any µ ∈ [0 , √ − , it has maximal density on the family of µ -arrangements of positivehomogeneity. Nevertheless, as Fejes Tóth observed in [8] (see also [5] or Section 1),the same does not hold if µ > √ − . Indeed, an elementary computation showsthat in this case F hex ( µ ) does not cover the plane, and thus, by adding disks toit that lie in the uncovered part of the plane we can obtain a µ -arrangement withgreater density.Fejes Tóth suggested the following construction to obtain µ -arrangements withlarge densities. Let τ > be sufficiently small, and, with a little abuse of notation,let τ F hex ( µ ) denote the family of the homothetic copies of the disks in F hex ( µ ) ofhomothety ratio τ and the origin as the center of homothety. Let F ( µ ) denotethe µ -arrangement obtained by adding those elements of τ F hex ( µ ) to F hex ( µ ) thatdo not overlap any element of it. Iteratively, if, for some positive integer k , F k hex ( µ ) is defined, then let F k +1hex ( µ ) denote the union of F k hex ( µ ) and the subfamily ofthose elements of τ k +1 F hex ( µ ) that do not overlap any element of it. Then, as wasobserved also in [8], choosing suitable values for τ and k , the value of δ U ( F hex ( µ )) can be approximated arbitrarily well by δ ( F k hex ( µ )) . We note that the same ideaimmediately leads to the following observation. INKOWSKI ARRANGEMENTS 13
Remark 4.
The supremums of δ ( F ) and δ U ( F ) coincide on the family of the µ -arrangements F in R of positive homogeneity.We finish the paper with the following conjecture. Conjecture 2.
For any µ ∈ ( √ − , and any µ -arrangement F in R , we have δ ( F ) ≤ δ U ( F hex ( µ )) . Acknowledgments.
The authors express their gratitude to K. Bezdek to direct their attention to thisinteresting problem.
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