The smallest convex K-gon containing N congruent disks
aa r X i v : . [ m a t h . O C ] F e b THE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS
ORGIL-ERDENE ERDENEBAATAR and UUGANBAATAR NINJBAT
Department of MathematicsThe National University of Mongolia, Ikh Surguuliin gudamj-1, 14201Ulaanbaatar, Mongolia.
Abstract.
Consider the problem of finding the smallest area convex k -goncontaining n ∈ N congruent disks without an overlap. By using Wegner in-equality in sphere packing theory we give a lower bound for the area of suchpolygons. For several cases where this bound is tight we construct correspond-ing optimal polygons. We also discuss its solution for some cases where thisbound is not tight, e.g. n = 2 and k is odd, and n = 3 , k = 4. On the wayto prove our results we prove a result on geometric invariants between twopolygons whose sides are pairwise parallel, and give a new characterisation forthe trisectrix of Maclaurin. Introduction
Most finite sphere packing problems fall into two types [11]. • Free packing : Locate a finite set of congruent balls in the space so thatthe volume of their convex hull is minimal. In the two dimensional space,notable results on this problem are the Thue-Groemer and Oler inequalities(see Chap. 4.3 in [4], [9]) and the Wegner inequality (see Theorem 2).In higher dimensions, L.F. T´oth’s sausage conjecture is a partially solvedmajor open problem [2]. • Bin packing : Locate a finite set of congruent balls in the smallest volumecontainer of a specific kind. In the two dimensional space, the containeris usually a circle [8], an equilateral triangle [14] or a square [15]. In suchcases, the smallest containers and the corresponding optimal packings areknown when the number of disks is not so big, e.g. up to 20 [10].We study the following finite packing problem which contains elements of bothtypes.
Problem 1.
For k ≥ find the smallest area convex k -gon containing n ∈ N unitradius (’unit’ in brief ) disks without an overlap. E-mail address : [email protected] and [email protected] .2010 Mathematics Subject Classification.
Key words and phrases.
Finite packing, Smallest polygon, Wegner inequality, Maclaurin tri-sectrix, Parallel polygons, Erd˝os-Oler conjecture.To appear in Elemente der Mathematik. K -GON CONTAINING N CONGRUENT DISKS
The solution to this problem for n = 1 is given by the following well-known result(see Chap. 2 in [1]). Theorem 1.
When n = 1 the regular k -gon circumscribing a unit disk is the onlysolution to Problem 1 for all k ≥ . In Theorem 4 of this paper we give an extension of this result as an inequalitybounding area of the containing polygon from below. This inequality is tight inmany cases including • n = 1 and k ≥ • n = 2 and k = 2 k ′ for k ′ ≥ • n ∈ { , } and k = 3 k ′ and • n is a centered hexagonal number and k = 6 k ′ .The solution of Problem 1 for these tight cases is obtained in Theorem 5. Then wediscuss its solution for other cases where this bound is not binding; for n = 2 , k =2 k ′ + 1 in the remark following Theorem 5, and for n = 3 , k = 4 in Theorem 6.The latter case is essential as it demonstrates the possibility of disks being packednon-efficiently inside the minimal polygon.Along the way to prove our main results, we prove two intermediate results whichare interesting on their own. The first one gives geometric invariants between twopolygons whose sides are pairwise parallel; see Proposition 1. The second onegives a simple geometric characterisation for a well known curve, the trisectrix ofMaclaurin; see Proposition 2. In the final section we discuss some open problems.2. Preliminaries
In addition to the usual ones, we use the following definitions. A region is asubset of the plane with finite area and when X is a region, k X k denotes its area.The line segment connecting points A, B is denoted as AB , | AB | is its length and( AB ) is its interior. Let a finite set of unit disks be located in R without an overlap,i.e. each pair has disjoint interior. Their joint tangent is a line which is tangent toat least two of the disks and supports their convex hull. Each joint tangent boundsa half plane which contains the disks. Intersection of these half planes is called as tangent polygon of the disks (see Fig. 1 ( a ), ( b )). Clearly, every tangent polygonis convex and it is a convex polygon as long as the centres of the disks are not allcollinear.The regular k -gon circumscribing a unit disk is called as unit k -gon . Thus, unittriangle has sides of length 2 √
3, while unit square has sides of length 2. m ′ th ofa regular mk -gon is a polygon obtained after cutting the original polygon by twoapothems intersecting at πm angle; see Fig. 2.Let C ⊂ R be a convex disk and P be a convex polygon containing C . Using aterminology in Pach and Agarwal [17], a cap of P w.r.t. C is the region enclosedby the boundary of C and two consecutive sides of P which are tangent to C ; seeFig. 1 ( c ). k Cap C ( P ) k denotes the sum of areas of all caps of P w.r.t. C .A set of n unit disks constitute a Groemer packing if each pair has disjoint interiorand the convex hull of their centres is either a line segment of length 2( n −
1) orcan be triangulated into equilateral triangles of edge length two using the n centresas vertices [12]. If in addition, perimeter of the hull is 2 (cid:6) √ n − − (cid:7) , where ⌈ x ⌉ = min { z ∈ Z : z ≥ x } , then the Groemer packing is a Wegner packing [5]; seeFig. 3. By using these geometric properties, it is easy to show that the convex hull
HE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS 3
Figure 1: Tangent polygons (a quadrilateral and an infinite strip) and a cap.Figure 2: Half, one-third, one-sixth of a regular hexagon.of the centres of the disks in a Wegner packing has at most six sides; see Chap.4.3 in [4]. n ∈ N is called exceptional if there is no Wegner packing of n unit disks.The smallest exceptional number is 121 and they constitute less than 5% of all N [5]. The following result known as Wegner inequality; see Chap. 4.3 in [4], [5]. Theorem 2. (Wegner inequality) If H is the convex hull of n ∈ N non-overlappingunit disks then k H k ≥ √ · ( n −
1) + (2 − √ · (cid:6) √ n − − (cid:7) + π. Equality holds if and only if the disks are packed in a Wegner packing.
Figure 3: Wegner packing of n = 10 disks. THE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS
The following result is a reliable tool when one studies the smallest circumscrib-ing polygons of convex figures [6, 20].
Theorem 3.
Let C ⊂ R be a convex disk and P be the smallest area convexpolygon containing it. Then, midpoints of sides of P lie on the boundary of C . If an internal angle at a vertex of a polygon is greater than π , then the vertexis reflex . If P = A ...A n and Q = B ...B n are two simple polygons with the sameorientation and A i A i +1 k B i B i +1 for all 1 ≤ i ≤ n , then they are called parallelpolygons . Finally, the Maclaurin trisectrix is a cubic plane curve defined as thelocus of the point of intersection of two lines, each rotating at a uniform rate aboutseparate points, so that the ratio of the rates of rotation is 1:3 and the lines initiallycoincide with the line passing the two points. Its polar equation is r = a sec θ andCartesian equation is y = x ( x +3 a ) a − x [19].3. The main results
We shall prove two intermediate results.
Proposition 1.
Let P and Q be two simple parallel polygons. Then (a) They have the same number of reflex vertices, and (b)
If one is convex so is the other and their corresponding internal angles areequal.Proof.
Let P = A ...A n and Q = B ...B n and we denote their internal angles as ∡ A i = α i , ∡ B i = β i for 1 ≤ i ≤ n . Let I = { i ∈ N : 1 ≤ i ≤ n, α i = β i } , I = { i ∈ N : 1 ≤ i ≤ n, α i < β i } and I = { i ∈ N : 1 ≤ i ≤ n, α i > β i } . Since A i A i +1 k B i B i +1 for 1 ≤ i ≤ n , i ∈ I implies α i + π = β i and i ∈ I implies α i = β i + π .Note that X i ∈ I α i + X i ∈ I α i + X i ∈ I α i = X i ∈ I β i + X i ∈ I β i + X i ∈ I β i = ( n − π. Since P i ∈ I α i = P i ∈ I β i , P i ∈ I ( α i + π ) = P i ∈ I β i and P i ∈ I α i = P i ∈ I ( β i + π ),we get X i ∈ I α i + X i ∈ I β i + | I | π = X i ∈ I α i + | I | π + X i ∈ I β i which simplifies to | I | = | I | . This proves Proposition 1 (a) after noting that | I | is the number of reflex vertices in Q whose corresponding vertex in P is normal,i.e. non-reflex, and | I | is the number of reflex vertices in P whose correspondingvertex in Q is normal.Assume P is convex. Then, by part (a) both polygons have 0 reflex vertices.Hence, Q is convex. Since n = | I | + | I | + | I | and | I | = | I | = 0, we have n = | I | ,i.e. α i = β i for all 1 ≤ i ≤ n . (cid:3) Remarks : There seem to be a slight confusion in the computational geometry lit-erature regarding to geometric invariants between parallel polygons. For example,on p.2 [13] it is (mistakenly) claimed that “two polygons are parallel iff they havethe same sequence of angles”. The above result clarifies the situation.Let P be a convex k -gon containing n unit disks without an overlap such thateach side of P is tangent to at least one of the disks. Let us pick one of the disksand for each side of P there are two tangents to the disk which are parallel to HE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS 5 it. Choose the one which is closer to the side and the polygon whose sides arecontained in these tangents is called as shrink of P for the picked disk; see Fig. 5.By construction, P and its shrink are parallel and by Proposition 1 (b) they havethe same internal angles.Our second intermediate result is as follows. Proposition 2.
Let ω be a circle with centre O and radius r ω , l be a line tangentto ω at T and X be an arbitrary point on l . Let m be the other tangent from X to ω , R = m ∩ ω and X ′ be reflection of X on m w.r.t. R . Then t ( ω, l ) , the locus of X ′ , is a Maclaurin trisectrix. Conversely, if t ( ω, l ) is a Maclaurin trisectrix thenthere exist circle ω and line l which generates it as described.Proof. Let us introduce a polar coordinate system with pole at O and axis on T O -ray. The angular coordinates are measured in the counterclockwise direction;see Fig. 4. Let the azimuth of X be φ ∈ [0 , π ) ∪ ( π , π ]. If φ ∈ [0 , π ), thenFigure 4: A characterisation of Maclaurin trisectrix. | OX | = r ω sec φ . By construction △ OX ′ X is isosceles with | OX ′ | = | OX | . Since △ OT X = △ ORX = △ ORX ′ we have ∡ T OX ′ = 3 φ . Thus, X ′ has polar coordi-nates ( r ω sec φ, φ ) which gives the polar equation r = r ω sec θ . If φ ∈ ( π , π ] weget the same equation after replacing φ in our analysis by φ ′ = 2 π − φ . Note that t ( ω, l ) has an asymptote which is perpendicular to the polar axis and passes overthe point (2 π, r ω ).Conversely, suppose t ( ω, l ) is a curve with polar equation r = a sec θ . Then ω is chosen as the circle with centre at the pole and radius a , l is tangent to ω atpoint T ( a, t ( ω, l ). (cid:3) Remarks : Another derivation of this curve and some motivations for finding alter-native derivations of classical curves are given in [18].Let us now prove our first main result.
Theorem 4. If P is a convex k -gon containing n ∈ N non-overlapping unit disksthen k P k ≥ √ · ( n −
1) + (2 − √ · (cid:6) √ n − − (cid:7) + k · tan πk . THE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS
Equality holds if and only if the disks are located in a Wegner packing, P is equian-gular, each side of P is tangent to at least one of the disks and each cap of P w.r.t.the convex hull of the disks is a cap w.r.t. a unit disk.Proof. By Theorem 3 we may assume that each side of P is tangent to the convexhull of the disks, which we denote as H . This is equivalent to assume that each sideis tangent to at least one of the disks. Since k P k = k H k + k P \ H k , by Theorem 2it suffices to prove that(1) k P \ H k ≥ k · tan πk − π where the right hand side is the sum of cap areas of a unit k -gon w.r.t. the circum-scribed unit disk, while the left hand side is k Cap H ( P ) k .Let ω be one of the disks and P ′ be the shrink of P for ω ; see Fig. 5 (a). SinceFigure 5: Shrinking P to P ′ in (a); adding an auxiliary disk in (b). P ′ is a circumscribing convex k -gon of ω , by Theorem 1(2) k Cap ω ( P ′ ) k ≥ k · tan πk − π and equality holds iff P ′ is regular.We claim that each cap of P ′ w.r.t. ω is smaller (in area) than the correspondingcap of P w.r.t. H . To see this, let A , ..., A k be the vertices of P and S be the setof n disks that it contains. Choose one of the vertices, A i . If A i − A i and A i A i +1 are tangent to the same unit disk in S , then the cap of P w.r.t. H with a vertexat A i is a cap w.r.t. a unit disk. Since shrinking preserves the internal angle at A i ,the corresponding cap of P ′ is a translation of this cap; see Fig. 5 (b). Thus, theyare congruent.If otherwise, A i − A i and A i A i +1 are tangent to two different unit disks in S .Let they be τ with centre T and κ with centre K , respectively. Draw a line passingthrough T which is parallel to A i − A i and another line passing through K whichis parallel to A i A i +1 as in Fig. 5 (b). Let A ∗ i be their intersection. Note that thispoint is well defined, on the bisector of ∠ A i and because of convexity it is inside P and closer to A i than both T and K . Let ( τ, κ ) be the unit disk centred at A ∗ i .By construction ( τ, κ ) is tangent to A i − A i and A i A i +1 . If H ′ is the convex hullof S ∪ { ( τ, κ ) } , then the cap of P w.r.t. H ′ with a vertex at A i is the same as the HE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS 7 corresponding cap of P ′ w.r.t. ω by the above argument. But the former cap issmaller than the cap of P w.r.t. H with a vertex at A i by the amount k H ′ k − k H k = k△ A ∗ i T K k + | T A ∗ i | + | A ∗ i K | − | T K | > k Cap H ( P ) k ≥ k Cap ω ( P ′ ) k and equality holds iff each cap of P w.r.t. H is a cap w.r.t. a unit disk. Inequalities(2) and (3) imply (1).From the proof above it should be clear that equality holds iff • Each side of P is tangent to at least one of the disks, and • k H k = √ · ( n −
1) + (2 − √ · (cid:6) √ n − − (cid:7) + π , which happens iff thedisks constitute a Wegner packing by Theorem 2, and • Each cap of P w.r.t. the convex hull of the disks is a cap w.r.t. a unit disk,and • P ′ is regular, which happens when P is equiangular (recall that P ′ and P have the same internal angles). (cid:3) Remarks : In above we used Theorem 2 to prove Theorem 4. One should noticethat the reverse implication is also possible and very much the same.The following result shows that the above inequality is tight in many cases.
Theorem 5. If (a) n = 2 and k = 2 k ′ with k ′ ≥ , or (b) n ∈ { , } and k = 3 k ′ , or (c) n ∈ N \ { } is not exceptional and k = 6 k ′ then the inequality in Theorem 4 is tight and the solution of Problem 1 can beconstructed. In particular, when n = 3 m ( m −
1) + 1 and k = 6 the solution is theregular hexagon with sides of length m −
1) + √ .Proof. Let us prove (a). Two unit disks are Wegner packed iff they are tangent.Let O , O be the centres of the disks and let us construct the solution of Problem1 as follows. First, draw the tangent polygon of the disks; recall that in this caseit is an infinite strip. Cut out a rectangle whose two opposite sides are containedin the joint tangents of the disks, and the other two sides pass through O and O .For each of these latter two sides, take half of unit 2 k ′ -gon and paste (i.e. glue) itover its side of length 2 with the rectangle. The resulting 2 k ′ -gon contains the twodisks and satisfies all the equality requirements in Theorem 4; see Fig. 6.Let us prove (b). Let n = 3 and it is easy to see that centres of three Wegnerpacked disks constitute vertices of an equilateral triangle with sides of length 2 asin Fig. 7 ( a ). Then, their tangent polygon T satisfies all the equality conditions inTheorem 4. Thus, T is the solution when k ′ = 1. For k ′ = 2, we can cut from T three equilateral triangles, each has a vertex common with T and a side tangent toconvex hull of the circles. The remaining hexagon which we denote by T satisfiesthe necessary conditions; see Fig. 7 ( b ). In general, T k ′ is constructed as follows.Remove all caps of T w.r.t. the convex hull of the disks and replace each by theunion of caps of one-third of unit 3 k ′ -gon w.r.t. the circumscribed unit disk. Itis easy to check that this construction is well defined and satisfies the necessaryconditions. THE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS
Figure 6: Construction of optimal polygons for n = 2 and k = 4 in (a), k = 6 in(b).If n = 6, a similar argument proves the statement after noting that centres ofsix Wegner packed disks constitute vertices of an equilateral triangle with sides oflength 4 as in Fig. 7 (c).Figure 7: Wegner packing and optimal polygons when n = 3 , n is not exceptional, there is a Wegner packing of n disks. Because of the perimeter condition, linear packing of n disks is never aWegner packing for n = 2. Thus, the convex hull of the centres of Wegner packed n = 2 disks which we denote as T c has at least three sides. Let T be the tangentialpolygon of the packed disks. Since T c has at least three and at most six sides andis triangulated into equilateral triangles, and T c and T are parallel, T has three tosix sides and its internal angles are either π or π (recall Proposition 1 (b)). Thisgives us the following possibilities: • T is an equilateral triangle, or • T is a quadrilateral with two internal angles of π and two of π , or • T is a pentagon with four internal angles of π and one of π , or • T is a hexagon with internal angles of π .In each case, take a vertex with angle π and cut out an equilateral triangle from T which shares this vertex and the side of it which does not contain this vertex istangent to the disk closest to the vertex. Let the resulting polygon be T . Then,by construction, T is a convex hexagon containing all the disks, each of its internalangle is π , and each side is tangent to at least one of the disks and each cap is acap w.r.t. a unit disk. Thus, T satisfies all equality conditions in Theorem 4 andis a solution to Problem 1 when n is not exceptional and k = 6. HE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS 9
In general, T k ′ is constructed as follows. Remove all six caps of T w.r.t. theconvex hull of the disks and replace each by the union of caps of a one sixth of theunit 6 k ′ -gon w.r.t. the unit disk which it circumscribes. It is easy to check thatthis construction is well defined and satisfies the equality conditions.If n = 3 m ( m −
1) + 1, i.e. the centred hexagonal number, it is easy to verifythat the Wegner packing of n disks is so that the convex hull of their centres is aregular hexagon with sides of length 2 m . Then, their tangent polygon is the regularhexagon of sides of length 2( m −
1) + √ and satisfies all the equality conditions inTheorem 4. Thus, the tangent hexagon is the solution. (cid:3) Remarks : With some effort one can show that a construction similar to that inTheorem 5 (a) works for n = 2 and k = 2 k ′ + 1. This time, we need to paste half ofunit 2 k ′ -gon and half of unit 2( k ′ + 1)-gon to the central rectangle. Moreover, thisreasoning can be applied to finding the smallest convex k -gon contianing n linearlypacked disks.So far our solutions for Problem 1 relied on the cases where the disks are Wegnerpacked, i.e. efficiently packed. The following result shows that this is not alwaysthe case. Theorem 6.
Let
M N KL be the smallest area convex quadrilateral containing threeefficiently packed unit disks and P ∗ be the × rectangle in which the disks arepacked linearly. Then k M N KL k > k P ∗ k .Proof. Let ω i , i = 1 , , O i be their centres. We know that | O O | = | O O | = | O O | = 2. Let △ ABC be the tangent polygon of the diskssuch that
A, B are on the joint tangent of ω , ω , B, C are on the joint tangent of ω , ω and C, A are on the joint tangent of ω , ω . Let ω ∩ AB = T , ω ∩ AB = T , ω ∩ BC = T , ω ∩ BC = T , ω ∩ AC = T and ω ∩ AC = T . Further let △ AA ′ A ′′ be such that A ′ ∈ AB , A ′′ ∈ AC and ω is an excircle tangent to A ′ A ′′ on its midpoint. Let △ BB ′ B ′′ and △ CC ′ C ′′ be defined analogously; see Fig. 8.Let us draw the section of Maclaurin trisectrix for ω and AB -line ranging be-tween A ′′ and the reflection A w.r.t. T . We call this curve as the trisectrixfor ( ω , AA ′ ). Draw similarly trisectrices for ( ω , AA ′′ ), ( ω , BB ′ ), ( ω , BB ′′ ),( ω , CC ′ ) and ( ω , CC ′′ ). Let I , I , I denote pairwise intersections of these sixcurves. Let R ∈ ( AA ′ ) be such that R I is tangent to ω by its midpoint. Wedefine other points R i , 2 ≤ i ≤ r ij with i, j ∈ { , , } and i < j denote the radical axis of ω i and ω j and O be their common intersection. Becauseof symmetry in our configuration, I ∈ r , I ∈ r , I ∈ r . Let r ∩ AB = P and r ∩ BC = P . Claim:
M N KL ∈ { A ′ BCA ′′ , B ′ CAB ′′ , C ′ ABC ′′ , I R CR , I R AR , I R BR } .We assume that M N KL is clockwise oriented. Let H be the convex hull of ω , ω , ω . Then(4) k H k = π + 6 + √ . By Theorem 3 we know that each side of
M N KL is tangent to H on its midpoint.This implies that each side is tangent to at least one of the disks. Since there arefour sides and three disks, one disk must be tangent to at least two consecutivesides. Let these be M N and
M L . This implies M ∈ △ ABC . Let us first assumethat M is located in the A -vertex cap of △ ABC , i.e.
M N , M L are tangent to ω .We shall analyse the following cases. K -GON CONTAINING N CONGRUENT DISKS
Figure 8: Tangent polygon of the three disks.(a) Assume M = A ′ . Then (by Theorem 3) L = A ′′ , N is on A ′ B -ray while K is on A ′′ C -ray. Since N K must be tangent to H , this implies N = B and K = C . So we end up with A ′ BCA ′′ . The case of M = A ′′ is treatedanalogously.(b) Assume M is located in the A ′ -vertex cap of A ′ BCA ′′ w.r.t. ω . Then L islocated either in the same cap or in △ AA ′ A ′′ while N is either in the samecap or in the region bounded by trisectrix for ( ω , AA ′′ ) and A ′ T , or on T B -ray. Location of L implies that K must be either in the A -vertex capor in the region bounded by trisectrix for ( ω , AA ′ ) and AT . In all cases, N K necessarily cuts at least one of the disks. Thus, M can not be in the A ′ -vertex cap of A ′ BCA ′′ . Similarly, it can not be in the A ′′ -vertex cap of A ′ BCA ′′ .(c) Assume M ∈ int ( △ AA ′ A ′′ ). Then N is located in one of A ′ -vertex capor the interior of the region bounded by trisectrix for ( ω , AA ′′ ) and A ′ T .The former is not possible by (b). L is located in one of A ′′ -vertex capor the interior of the region bounded by the trisectrix for ( ω , AA ′ ) and A ′′ T . Again the former is not possible. Position of N implies K is eitherin B ′′ -vertex cap, or above B ′′ B ′ -line. The former case is not possible by(b). Similarly, position of L implies K is located below C ′ C ′′ -line. Let Q be the intersection of B ′ B ′′ -line and C ′ C ′′ -line. Our last two conclusionsimply that K -vertex cap of M N KL contains △ B ′ QC ′′ , whose area is(5) k△ B ′ QC ′′ k = 4 + √ √ . Notice that k M N KL k > k H k + k△ B ′ QC ′′ k . On the other hand(6) k A ′ BCA ′′ k = 11 + 6 √ √ . HE SMALLEST CONVEX K -GON CONTAINING N CONGRUENT DISKS 11
By using (4), (5), (6), we know k M N KL k−k A ′ BCA ′′ k > k H k + k△ B ′ QC ′′ k−k A ′ BCA ′′ k >
0. Thus, MNKL is not the smallest.(d) Assume M ∈ ( R A ′ ). Then L is on the section strictly between A ′′ and I of the trisectrix for ( ω , AA ′ ), while N is on the A ′ B -ray. The formerconclusion implies K is located above BC -line and below I R -line. Then N K can only be tangent to ω . This in turn implies N ∈ ( T B ). Thendistance from N to a point on T T -arc of ω is much smaller than distancefrom K to the same point. In particular the former distance is at most | BT | = √
3, while the latter distance is greater than | T T | = 4 (to seethis just project K to BC ). Thus, such M N KL can not be the smallest.Similarly, M ∈ ( R A ′′ ) gives a non-optimal solution.(e) Assume M ∈ ( AR ). Then L is in the interior the region bounded by thetrisectrix for ( ω , CC ′′ ). This implies K ∈ int ( △ CC ′ C ′′ ). We can repeatthe argument in (c) to reach a contradiction. Similarly, M ∈ ( AR ), thenwe reach to a contradiction.(f) Assume M = R . Then L = I , which implies K = R . These imply that M N KL = R BR I . Similarly, M = R implies M N KL = R I R C .(g) Assume M = A . Then N is on T B -ray and L is on T C -ray. Since both N K and KL are tangent to H , we must have either L ∈ T C or N ∈ T B .First assume L ∈ T C . If L ∈ ( T C ), then by arguments above we knowthat either L = C ′ or L = R . In the first case we end up with ABC ′′ C ′ and in the second case with AR I R . If L = C , then M N KL = AB ′′ B ′ C .Similarly N ∈ ( T B ) implies M N KL ∈ { B ′ CAB ′′ , I R AR , ABC ′′ C ′ } .Thus, we conclude that when M is located in the A -vertex cap of △ ABC , M N KL ∈{ A ′ BCA ′′ , B ′ CAB ′′ , C ′ ABC ′′ , I R CR , I R AR , I R BR } . A similar argumentshows that M N KL is one of these six polygons when M is located in the C -vertexcap or B -vertex cap of △ ABC . This proves our claim.Symmetries in our configuration imply k A ′ BCA ′′ k = k B ′ CAB ′′ k = k C ′ ABC ′′ k and k I R CR k = k I R AR k = k I R BR k . Then k A ′ BCA ′′ k = √ √ >
12 = k P ∗ k . Notice that k I R BR k = k R P OI k + k P BP O k + k P CI O k . Since eachof these three quadrilaterals contains a unit disk and is non-regular, each has areagreater than 4 by Theorem 1. Then, k I R BR k >
12 = k P ∗ k . (cid:3) Remarks : By now we know that n = 3 , k = 4 is the first case where the disks arepacked non-efficiently inside the smallest containing polygon, when we order ( n, k )lexicographically. 4. Final discussions
Let us discuss some open problems. Because of Theorem 6, packing of n disksin the smallest convex k -gon is not always the most efficient. However by Dowkerinequality [7], at any fixed location of the disks, the area of the smallest containing k -gon tends rather fast to the area of their convex hull as k → ∞ . This is asupportive fact for the efficient packing to realise inside the smallest containing k -gon when k is large. It is believed that efficient packing of n ∈ N unit disks isa Groemer packing [5]. We can then ask whether the packing of n disks in thesmallest convex k -gon is always a Groemer packing?The following assertion seems very plausible. K -GON CONTAINING N CONGRUENT DISKS
Conjecture 1. If n = m ( m +1)2 , i.e. the m ′ th triangular number, then the smallesttriangle containing n unit disks is the equilateral triangle of side m −
1) + 2 √ . In Theorem 1 and Theorem 5 (b) we proved it for m = 1 , ,
3, but our approachstopped at m = 4 where the Wegner packing of 10 disks is not triangular; seeFig. 3. One can show that the smallest triangle containing two unit disks is theisosceles right triangle with hypothenuses of length 6+4 √
2, i.e. it is not equilateral.However, it is plausible that the smallest triangle containing m ( m +1)2 − m > Conjecture 2. If n > is a triangular number, then the smallest triangle con-taining n disks is the same as that containing n − disks. In Theorem 5, we showed that if n is a centred hexagonal number, then thesmallest containing hexagon is regular. Then, we can propose the following analogyof this conjecture. Conjecture 3. If n ∈ N is a centred hexagonal number, then the smallest hexagoncontaining n disks is the same as that containing n − disks. The following assertion would simplify some of our proofs. Let P be the smallestconvex k -gon containing a convex disk C . Let P be the convex polygon obtainedafter cutting the largest (area) triangles from each cap of P so that P still contains C . Is it true that minimality of P implies minimality of P ? The answer is positiveif C is a unit disk, negative if C is the convex hull of two tangent unit disks and k = 3, but positive in the latter case when k = 4 (see Theorem 5 (a)). What if C is an ellipse? Moreover, does minimality of both P and P imply minimality ofmembers of the sequence of obtained by the ’greedy cut’ procedure described? Acknowledgements : We are grateful to participants of 2018 Fall Meeting of theMongolian Mathematical Society for useful discussions and to Bilguun Bolortuya forresearch assitance. Financial support from the Asia Research Centre in Mongolia(P2018-3567) is acknowledged.
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