Optimal Packings of 4 Equal Circles on a Square Flat...

Optimal Packings of 4 Equal Circles ona Square Flat Torus

Madeline Brandt and Hanson Smith

Summer 2013

1.0 Introduction

Suppose that it is a nice, Friday afternoon, and you decide that you want to makesome cookies to top off your long week of studying mathematics. You make your dough,grease your cookie sheet, and prepare to arrange your equally sized balls of dough. Howshould you arrange them so that the interiors of the cookies are mutually disjoint whenthey come out of the oven? How can you arrange the cookies so that you maximizetheir radius, and thus the total amount of cookie you will eat? What if your cookiesheet is a flat torus? Our aim in this report is to find all locally and globally maximallydense arrangements of four equal circles on any flat torus. Broadly, we accomplishedthis by considering all potential packing graphs for a circle packing, embedding thecombinatorial multigraphs onto tori in all possible ways, and then finding which torithe equilateral packing graphs would embed on.

We may associate a combinatorial graph to a packing of n equal circles by allowingthe centers of the circles to be represented by vertices, and allowing edges to connectvertices corresponding to tangent circles. We can then use the resulting graph as atensegrity framework in which all edges are struts. The results of Connelly [1] restrictthe number of edges on these graphs to be greater than 2n − 1 and using Euclideangeometry, the number of edges can be restricted to less than 3n. Thus, the search foroptimally dense packings begins by considering a finite number of combinatorial graphsas candidates for the packing graphs of optimally dense packings. Using restrictionson vertex degree, the number of combinatorial graphs to consider can be reduced.Using techniques from Topological Graph theory, the combinatorial graphs may beembedded onto tori in all possible ways. These can then be constructed as equilateralpacking graphs, and results from Rigidity Theory may be used in order to determineif the resulting tensegrity frameworks are rigid (and thus locally maximally dense).Finally, we determine which set of tori the equilateral embeddings embedded on. Inthe case where a torus had multiple locally maximally dense packings, we determinedthe optimal packing by comparing the densities of the packings.

Packings of small numbers of equal circles on flat tori have been considered byothers. Heppes considered optimal packings of one through four equal circles on therectangular flat torus in [5]. We show that our results match his results in Appendix 2.Dickinson, Guillot, Keaton, and Xhumari consider optimal packings of up to five equalcircles on a square flat torus and packings of up to six equal circles on a triangular flattorus in [2] and [3].

1

2.0 Definitions and Background

We will begin with a discussion of some of the basic definitions and theorems that willbe used to prove the result. An equal circle packing is an arrangement of equallysized circles with disjoint interiors in some container. The container considered in thispaper is the flat torus. A flat torus is the quotient of the plane with respect to alattice (figure 2.1). By the classification of un-oriented two dimensional lattices, it isalways possible to find a basis for a lattice in which one vector is given by ~v1 = 〈1, 0〉and the other is given by ~v2 = 〈x, y〉, where x2 +y2 ≥ 1 and 0 ≤ x ≤ 1

2. For this reason,

the moduli space of flat tori is defined as the set {(x, y)|x2 + y2 ≥ 1 ∧ 0 ≤ x ≤ 12}

(figure 2.2). Every torus is represented by a point in the moduli space, and no twopoints in the moduli space correspond to the same torus.

Figure 2.1: A flat torus.

In order to discuss the relative density of packings, more terms are needed. Twopackings P and P ′ are defined to be ε-close if there exists a one-to-one correspondencebetween circles in P and circles in P ′ such that for all corresponding circle centers cin P and c′ in P ′, the distance between c and c′ is less than ε. A packing P is locallymaximally dense if there exists an ε > 0 such that all ε-close packings of equal circleshave a packing density no greater than that of P . A packing P is globally maximallydense if it is the densest possible packing.

In order to determine all locally and globally maximally dense packings, we will belooking at the graphs associated to a packing. Let P be a packing of circles on a flattorus. Then the equilateral packing graph, GP , is a graph with vertices at circlecenters and edges as the line segments connecting tangent circles (figure 2.3). Note that,in an equilateral packing graph, vertices have location and edges have length. Thus,every packing of equal circles is associated to an embedding of an equilateral graphon a flat torus. A combinatorial multigraph is a collection of points and (possibly

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Figure 2.2: The moduli space of flat tori.

multiple) lines connecting points. The combinatorial multigraph whose equilateralembedding on a torus is GP will be very important in determining all locally maximallydense packings.

Figure 2.3: An equilateral packing graph on an equal circle packing of 4 circleson a flat torus.

We will now discuss the necessary definitions and theorems from rigidity theoryas they place several useful restrictions on the equilateral packing graphs and combi-natorial multigraphs. A curious reader may consult [2] or [1] for deeper and broaderexplanations of these terms. A strut tensegrity framework is an embedded graphin which all edges are restrained to only increase in length, and all vertices are vari-able. These vertices can be moved using a flex of the framework, where a flex is amovement of the vertices which respects the restrictions of the framework. Thus, anequilateral packing graph can naturally be thought of as a strut tensegrity framework.Note that any flex will either increase or preserve the radius of the circles and thereforethe density of the packing. A flex is called a trivial flex if it moves each vertex of theframework in the same way. A strut tensegrity framework is rigid if the only flex ofthe framework is a trivial one.

GVSU REU 2013 3 Brandt & Smith

We say that p′ is an infinitesimal flex of the graph if

(pi − pj) · (p′i − p′j) ≥ 0 for (i, j) as a strut of our graph.

Furthermore, p′ is a trivial infinitesimal flex of our graph if it is the derivative att = 0 of a rigid Euclidean motion Rt : Ed → Ed

p′i = (d

dtRtpi)t=0, i = 1, ..., n.

If the graph admits only trivial infinitesimal flexes, we say that is is infinitesimallyrigid. All of the above definitions are directly from [1].

The following are results from [2] and [1] which we will use in order to prove ourresult.

Proposition 1. If the strut tensegrity framework associated to a circle packing P isinfinitesimally rigid or rigid, then the circle packing P is locally maximally dense. [2]

Proposition 2 (Graph Restrictions). Let P be a locally maximally dense packing ofn ≥ 4 circles on a flat torus with no free circles. Then the packing graph associated toP satisfies the following:

1. It contains at least 2n− 1 and at most 3n edges,

2. Every vertex is connected to at least 3 and at most 6 others. [2]

Instead of starting with tori and trying to find the most dense packings on them,we will start with a list of combinatorial graphs and find the ones that correspond tolocally maximally dense packings. Proposition 2 allows us to do this, because it ensuresthat we can make a finite list of these combinatorial graphs. By proposition 2, the listof all combinatorial multigraphs with four vertices and between seven and twelve edgescontains every combinatorial multigraph that corresponds to a locally maximally densepacking of four equal circles on a flat torus (there were 825 such graphs).[6] In orderto remove some of these graphs, there are restrictions that can be made.


3.0 Theorems About Potential Packing Graphs

Restrictions on Combinatorial Packing Graphs

Now, we will prove some results which were found and used in order show that someof the combinatorial graphs did not correspond to circle packings.

Proposition 3. Let P be a packing of equal circles, and GP be the combinatorial graphassociated to P . If one of the vertices of GP has degree 6, then GP has at least 12 edges.

Proof. LetGa be the vertex inGP of degree 6, and let A be the circle in P correspondingto Ga, and let a be the center of A. The angles between two adjacent edges connectedto Ga must be 60◦, because if any one of them had degree less than 60◦ then the twocircles tangent to A corresponding to these edges in GP would overlap. Furthermore,if this angle where greater than 60◦, then the angle between two other edges connectedto Ga would be less than 60◦ since the sum of the angles around a point is 360◦. Thus,the circles in P which are tangent to A and correspond to edges which are separatedby 60◦ in GP must be tangent to one another. Therefore, there are 6 edges connectingthe vertices in GP corresponding to the circles tangent to A, and 6 edges connected toGa, so GP has at least 12 total edges.

Students in the 2011 REU also noted and proved this result.[4]

We will prove that in an equal circle packing of 4 or more circles on a flat torus, notwo circles can share 3 or more tangencies, but first, we need a lemma.

Lemma 1 (Pick’s Theorem on any Lattice). Let P be a lattice polygon on any lattice.Let A be the area of a fundamental domain of the lattice, i be the number of latticepoints contained in P , and b be the number of lattice points in the boundary of P . Thenthe area of P is equal to A(i+ b

2− 1).

Proof. Let (a, b) and (c, d) be a basis for a lattice Λ, and let P be any polygon withvertices in that lattice. Let φ : R2 → R2 be the linear transformation which changesfrom coordinates in Λ to coordinates in the lattice generated by (1, 0) and (0, 1). Thus,φ(x, y) = (ax + cy, bx + dy). We can apply the Change of Variable theorem to P andφ because P is a compact and connected subset of R2 with a boundary of volume zero,

5

and φ is an injective, invertible map on R2 with a continuous derivative. Then,

Area of P =

∫P

1

=

∫φ−1(P )

1 · |φ′| By the Change of Variable Theorem.[7]

=

∣∣∣∣a cb d

∣∣∣∣ · Area of φ−1(P )

= A(i+ b/2− 1) By Pick’s Theorem on φ−1(P ) and the volume of

the parallelogram generated by (a, b) and (c, d).[7]

Proposition 4. In an equal circle packing on a flat torus with four (or more) circlesno two circles can share three or more tangencies.

Proof. Suppose two circles in a packing of four or more equal circles on a flat torus aremutually tangent at three points. Call the circles sharing three tangencies A and B,and lift the packing on the torus to a packing of the plane. Pick an arbitrary lift ofcircle A and call its center a. By assumption this lift of circle A will have three lifts ofcircle B tangent to it. Let b1, b2, and b3 be the center points of these lifts of circle B,as in Figure 3.1. Note that the vectors between the centers any two lifts of B will belattice vectors. Let ~v1 be the vector between b1 and b2 and let ~v2 be vector between b1

and b3. Note that ~v1 and ~v2 must be linearly independent. Let the point at b1 + ~v1 + ~v2

be called b4. Note that this point will also be the center of a lift of B, and it will bedistinct from b1, b2, and b3.

Let P be the parallelogram defined by vertices b1, b2, b3, and b4. Let the area ofthe fundamental domain of the lattice be AFD, and let the area of P be AP . ApplyingPick’s Theorem (and the generalization above) to P and using the minimum num-ber of lattice points on the interior and on the boundary we lower bound AP . Thus,AP ≥ (0 + 4

2− 1)AFD = AFD.

We now find a lower bound the density, ρ, by using the lower bound on AFD. Letd be the diameter of the circles. We compute AP in terms of d using triangles 4b1ab3,


Figure 3.1: Diagram for Proposition 4

4b1ab2, and 4b3ab2:

AP = 2(area(4b1b2a) + area(4b1b3a) + area(4b2b3a))

= 2((d sin (θ))(d cos (θ)) + (d sin (ϕ))(d cos (ϕ)) + (d sin (π − θ − ϕ))(d cos (π − θ − ϕ)))

= 2d2(sin (2θ)

2+

sin (2ϕ)

2+

sin (2(π − θ − ϕ))

2

)= d2(sin (2θ) + sin (2ϕ) + sin (−2(θ + ϕ)))

= d2(sin (2θ) + sin (2ϕ)− sin (2(θ + ϕ)))

≤ 3d2 because of the upper bound of 1 on each of the sine functions.

Taking into consideration all 4 circles of this packing, the total area covered by the4 circles would be πd2. We can then see that

ρ =Area covered by 4 circles

AFD≥ Area covered by 4 circles

AP≥ πd2

3d2=π

3> 1,

which is a contradiction, since the packing density must be less then 1. Hence, twocircles cannot share three or more tangencies. The case of two circles sharing four or


more tangencies is eliminated by the above density argument since it is a subcase oftwo circles sharing three tangencies.

By applying Propositions 2.3, 3 and 4, we were able to reduce the number ofcombinatorial graphs we were considering from 825 to 14. The next step of the processwas to embed these graphs onto tori in all possible ways, and eliminate ones withgeometric problems preventing them from actually corresponding to circle packings.

Embedding the Graphs onto Tori

After eliminating all the combinatorial multigraphs we could, we turned to previouslyknown techniques from topological graph theory in order to find all 2-cell embeddingsof a combinatorial graph onto the topological torus. More details on this process canbe found in [2]. We used an algorithm written by Professor Dickinson in order toimplement Edmond’s permutation technique. This returned all possible embeddings ofour 14 combinatorial graphs. It also eliminated ones which had immediate geometricproblems or forbidden face patterns [2]. This left us with 31 toroidal embeddings ofgraphs to examine. Each of these is described in Appendix 1. The naming scheme forthe graphs and their embeddings goes as follows: “v” stands for vertex number, so allof the embeddings begin with “v4,” “e” stands for edge number, “n” gives the numbera given combinatorial graph is listed as in [6], “t” denotes the toroidal embedding. Thenumbers following the “t” give the face pattern and the type of the embedding.

Theorems About Equilateral Graph Embeddings

that are not Packing Graphs

Many of these toroidal embeddings did not correspond to equilateral packing graphson tori. The following lemmas eliminate many toroidal embeddings; however, othersare dealt with case-by-case in Appendix 1.

Lemma 2. All quadrilaterals in packing graphs are rhombi.

Proof. Since all edges in a packing graph have the same length, all quadrilaterals arerhombi.

Lemma 3 (Pentagon Lemma). If two edges of a pentagon (formed from edges in apotential packing graph) in a two-cell embedding of a graph on a flat torus are parallel,then the potential packing graph is not the packing graph associated to an equal circlepacking on a flat torus.

Proof. Let ABCDE be an edge pentagon in a potential packing graph. Notice that allsegments have equal length. Let AB be parallel to DC. Since AB, AC, and DC are


all the same length, and AB is parallel to DC, we have that ABCD is a rhombus.Thus, the distance between points B and D is the same as the length of AC, and sothe circles with centers at B and D should be tangent, but the packing graph doesnot indicate this. Therefore, the potential packing graph is not associated to an equalcircle packing.

After eliminating toroidal embeddings which could not become equilateral packinggraphs using lemma 3 and other ad-hoc methods, 14 equilateral packing graphs thatcould correspond to locally maximally dense circle packings remained.

Locally Maximally Dense Packings and the Mod-

uli Space

By looking at the equilateral packing graphs as tensegrity frameworks we found whetheror not they had proper stress, defined in [1]. Proposition 1 eliminates graphs that neverhave proper stress and are therefore not rigid frameworks. After this process, thereremain only 10 locally maximally dense packings on various tori.

Packings of three equal circles on tori which have room for a fourth circle also needto be examined, since the methods used do not reveal packings which contain a freecircle. A free circle is a circle which is not held fixed by any other circles. Uponexamining these packings, no packings of three circles which could admit a rattler werefound(we did find packings of 3 circles which could admit a fourth, fixed circle). Amore detailed discussion can be found in Appendix 3.

Each equilateral packing graph has a lot of structure and anywhere from zero to twofree parameters. Thus each packing graph corresponds to some subset of the modulispace which it embeds on.


4.0 Globally Maximally Dense Packings of 4 Equal

Circles

After finding the locally maximally dense circle packings, further investigation wasneed to find the globally maximally dense packings on regions in the moduli spacewith two or more locally maximally dense packings. Using Mathematica, two locallymaximally dense packings on the same region with large differences in density couldbe separated. The Plot3D function proved particularly useful for separating the lessdense locally maximally dense packings from the globally maximally dense packings.The packing v4e07n036t12 and a portion of the packing v4e08n061t11 were shown notto be globally maximally dense packings in this manner.

The Plot3D function lacked the precision to distinguish between two locally densepackings occupying one region of the moduli space, so further analysis was needed. Ulti-mately, it was shown that v4e09n095t11 is more dense than the portion of v4e08n061t11occupying the same region. For the full argument please see the Appendix.

Theorem 1. The globally maximally dense packings of four equal circles with no self-tangencies on flat tori are given by the packing graphs which occupy the followingregions in the moduli space (see the figure below).

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We can see the density of the globally maximally dense packing on a given torus inthe moduli space in the figure below. Pink areas are more dense, while blue areas areless dense.


Appendix 1: Details on Each Combinatorial Em-

bedding

v4e07n034T31

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Figure 1: v4e07n034T31

In v4e07n034T31, a pentagon is formed by 12413, and a quadrilateral is formedby 1342, with 13 and 24 parallel, so this packing graph does not correspond to a circlepacking by the Pentagon Lemma.

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v4e07n035t11

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Figure 2: v4e07n035t11

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Β4


This embedding did correspond to a locally maximally dense packing. Figure 3shows the equilateral packing graph, where π/3 < α < 2π/3 and π/3 < β < π, withthe restriction that α/2 + β > π, because of symmetry. It occupies the region of themoduli space displayed below. It is the globally maximally dense packing in this region.The least dense point is displayed below, and occurs when α ≈ 1.78 and β ≈ 2.25. Thedensity at this point is approximately 0.7.


Figure 4: Moduli space for v4e07n035T11

v4e07n035t12

Since 1342 is a parallelogram, 13 and 42 are parallel. Thus in the hexagon 131242 ver-tices 1 and 2 are the same distance apart as vertices 3 and 4. If this graph represents acircle packing, then the circles centered at 4 and 3 will be tangent inside the hexagon.Since this edge doesn’t appear on the graph it fails.



v4e07n036t11

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Figure 6: v4e07n036t11Α

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This packing did not correspond to a locally maximally dense packing. It wasobserved in Mathematica that it never had proper stresses.


v4e07n036t12

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Β

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This corresponded to a locally maximally dense packing in some regions. We allowedπ/3 < α < π and 2π/3 < β < π, but the packing did not have proper stress in thisentire region. It had proper stress in two areas, but these were found to correspond tothe same packings, by a symmetry. Ultimately, the packing was restricted to π/3 < α <π and 2π/3 < β < 5π/6, and α < 1/6(π−6(− tan−1(1/3 cot(π/6+β))−π)). It occupiesthe following region of the moduli space and is not globally maximally dense (we had tocompare it to the packings v4e07n035T11, v4e09n094T11, and v4e08n061T11). Amongall locally maximally dense packings without a free circle, however, the numerical leastdensity is approached by a sequence of packings in this domain. It approaches thedensity 2

3Π√12

.


Figure 10: Moduli space for v4e07n036t12

v4e07n036t31

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Proposition 5. The potential packing graph of v4e07n036t31 does not correspond toa locally maximally dense circle packing of four equal circles on a flat torus.

Proof. Suppose v4e07n036T31 corresponds to a a locally maximally dense circle pack-ing. Consider the hexagon 121423. Notice that opposite sides 13 and 41 form two sidesof the triangle 4134. Let the radii of the circles in this packing be 1

2and note that the

two lifts of point 1 must be less than 2 units apart by the triangle inequality on 4121.


Figure 12: The Construction

Now, consider a line through the segment 31 and another line through the segment 14where both segments form sides of the hexagon. Because the two triangles 4134 arecongruent equilateral triangles and are translates of one another the lines will intersectwith a 60◦ angle.

Figure 13: The Regions

Consider the two regions partitioned by the intersecting lines as candidates for thelocation of point 2 where point 2 has distance one from vertex points 3 and 4 of thehexagon. Furthermore, define Region 1 to be open along its boundaries with Region2. In other words, Region 2 contains all the points on the two intersecting lines. Now,point 2 cannot be in Region 2 as this would imply that the interior angle of the hexagonat points 3 or 4 respectively would be greater than or equal to 180◦. Thus, if this pack-ing is locally maximally dense, point 2 must be in Region 1.


Consider the triangle formed by the previously mentioned lines through 13 and 41and the lattice vector between the two lifts of point 1. Recall that the lattice vectorbetween the two lifts of 1 is strictly less than two units in length. Now, because theangle at the intersection of the lines 13 and 41 is 60◦, the Law of Sines shows that oneof the other sides of the triangle must have length less than 2. To see this note thatthe sum of the two angles at lifts of point 1 in the triangle is 120◦ and thus one angleat point 1 must be less than or equal to 60◦. Designate this smaller angle as α. Thus,for l being the side between lifts of point 1 with l < 2 and s as the side opposite theangle α, sin(60◦)

l= sin(α)

s. Therefore s = sin(α)

sin(60◦)l and since sine is an increasing function

on the interval [0◦, 90◦] one sees that s ≤ 2. Note that if s = 2, then α = 60◦ and thetriangle is an equilateral triangle. This, however, is a contradiction as it would implythat l = 2. Thus s < 2.

Figure 14: Triangles Argument

Call the intersection of the two lines c. Now we will show that circles of radius 1centered at points 3 and 4 cover 443c and therefore there is no possible location forpoint 2. Call the distance between c and point 1 through the hexagon t. We note thatt > 1. Furthermore, let m be the length from point 4 to point 3 through the hexagon.Using the Law of Cosines and noting that the angle at c is 60◦ we have

m2 = (s− 1)2 + (t− 1)2 − (s− 1)(t− 1)

l2 = s2 + t2 − st

Combining the above we have

s+ t = l2 −m2 + 1 < 22 − 12 + 1 = 4


Thus

s+ t < 4 or (s− 1) + (t− 1) < 2

Note that c4 has length t− 1 and 3c has length s− 1. Call these lengths t1 and s1

respectively. This implies

s1 + t1 < 2

In particulart12< 1 (1)

From s < 2 we haves1 < 1 (2)

Now, since s+ t = l2−m2 + 1, l < 2, s > 1 and t > 1, we see m2 = l2− s− t+ 1 <22 − 1− 1 + 1 = 3 or m <

√3 or

m

2< 1 (3)

Define a as the midpoint of the segment c4 and define b as the midpoint of thesegment 43. Take the triangle 4ca3 and let n be the length of a3. Applying the Lawof Cosines, we have n2 = (s1)2 + ( t1

2)2− s1

t12

. On the region 0 < s1 < 1 and 0 < t12< 1,

n2 has a maximum when s1 = 1 and t12

= 1. Therefore

n < 1 (4)

Equations 1 and 3 imply that the circle of radius 1 centered at point 4 contains4ba4. Equations 4 and 2 imply that the circle of radius 1 centered at point 3 contains43ca. Equations 4 and 3 imply that the circle of radius 1 centered at point 3 contains43ab. Therefore the union of the unit radius circles centered at points 3 and 4 cover443c This is a contradiction and we conclude that v4e07n036T31 cannot be the packinggraph of a circle packing.


v4e07n036t41

Since 1342 is a parallelogram, 13 and 42 are parallel. Thus in the hexagon 124123 ver-tices 1 and 2 are the same distance apart as vertices 3 and 4. If this graph represents acircle packing, then the circles centered at 4 and 3 will be tangent inside the hexagon.Since this edge doesn’t appear on the graph it fails.



v4e07n036t51

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Proposition 6. The potential packing graph v4e07n036t51 fails to correspond to alocally maximally dense circle packing on any flat torus.

Proof. Suppose v4e07n036t51 corresponds to a locally maximally dense circle pack-ing on a flat torus. Furthermore, suppose that all edges on the corresponding equilateralpacking graph are length 1. Note, because they are opposite sides of the parallelogram1342, all lifts of the segments 41 and 32 are parallel. Likewise, all lifts of the segments13 and 24 are parallel. Furthermore, call ∠324 inside the parallelogram angle α andcall ∠132 inside the parallelogram angle β. Now take pentagons 34212 and 43121 and,by using parallel lines and examining the octagon 12142124, note they have 180◦ rota-tional symmetry about the midpoint of edge 43. Call pentagon 34212 P and pentagon43121 Q. In pentagon P , label ∠234 as γ, ∠342 as δ, ∠421 as ε, ∠212 as ζ and ∠123as η. Note that, respectively, these angle are ∠143, ∠431, ∠312, ∠121 and ∠214 in Q.

We will find a contradiction by continuously deforming α from 60◦ to 120◦ andshowing that η or ε will always be less than 60◦ for α in this range. First, it is nec-essary to define some lengths. Let 22 through P be a and let its length be la. Thesymmetric line from 1 to 1 through Q will be denoted a′. Designate 13 through P as band let its length be lb. Denote the symmetric line 24 through Q as b′. Let the diagonalthrough the parallelogram be d and denote its length ld.

Suppose α = 60◦. Though this value of α is not represented by the equilateral pack-ing graph, we begin with this hypothesis in order to show that no value of α greaterthan 60◦ can be represented. Drawing a line perpendicular to d through 2 and forminga right triangle, we see that sin(30◦) = 1

2and thus ld = 1. Notice that segment 12 is

also length 1 and forms a diagonal between parallel segments 41 and 23. We want to


Figure 17: Labels

show that lb = 1 and b and b′ are parallel, so that 3421 is a parallelogram.

Note that the equilateral packing graph for this packing can be constructed withonly the free parameters α and γ. Furthermore, note that η and ε have symmetryabout γ = 120◦. Thus 120◦ < γ < 180◦. Let us fix γ. Now, if α = 60◦ and point 4 inthe parallelogram is at the origin, then point 3 has coordinates (1

2,√

32

) and point 4 has

coordinates (12− cos(δ),

√3

2+ sin(δ)). So the coordinates of point 2 between points 4

and 1 are (32− cos(δ),

√3

2+ sin(δ)) and the coordinates of point 2 between 1 and 3 are

(1, 0). To intersect the circles of radius 1 centered at the two lifts of point 2 in P weuse Mathematica.

Solve[{(x - 1)^2 + y^2 ==

1, (x - (3/2 - Cos[\[Delta]]))^2 + (y - (Sqrt[3]/2 +

Sin[\[Delta]]))^2 == 1}, {x, y}];

sols = FullSimplify[%]

We examine the first solution.

Plot[x /. sols[[1]], {\[Beta], Pi/3, 2*Pi/3}, Frame -> True,

FrameStyle -> Black]

Plot[y /. sols[[1]], {\[Beta], Pi/3, 2*Pi/3}, Frame -> True,

FrameStyle -> Black]

However, Mathematica can’t simplify this. The second solution has the coordinates ofpoint 1 at (3

2,√

32

). Since this is distance 1 away from point 3 we conclude that angle


η = 60◦. Thus when α = 60◦, then 3421 is a parallelogram.

Note that because the sum of angles around a point is 360◦, β + γ + δ = 360◦. ByProfessor Dickinson’s equation we can write l1 is terms of γ and δ.

l2a = 3− 2(cos(γ) + cos(δ)− cos(γ + δ)) (5)

Now, since α + β = 180◦, if α is increased, then β will decrease. Thus, sinceβ + γ + δ = 360◦ and since γ is fixed as a free parameter, increasing α will increaseδ. Note that because circles are disjoint, 60◦ < δ < 180◦. Furthermore, since thecircles must be disjoint and since 1423 is a parallelogram, 60◦ < β < 120◦. Thus240◦ < γ + δ < 300◦. Note that cosine is strictly decreasing on the interval (60◦, 180◦),but strictly increasing on the interval (240◦, 300◦). Thus, by (1), increasing δ increasesla.

Take the isosceles triangle formed by a and the segments 21 and 12 of P and callit R. Since the lengths of the segments 21 and 12 are 1, increasing la increases angleζ. Trigonometrically, dropping a line perpendicular to a through point 1 forms a righttriangle. Because sin( ζ

2) = la

1and because sine is strictly increasing on the interval

(30◦, 90◦), increasing la increases ζ. Furthermore, call the portions of the angles ε andη partitioned by R εR and ηR respectively, and note that as ζ increases εR and ηR willdecrease.

Figure 18: Triangle R

Now take the quadrilateral 3422 inside P and call it S. Define the isosceles triangle234 inside S as T and define the side opposite γ as t with length lt. Drop a line per-


pendicular to a through 4 and call the line h and the intersection of a and h, i. Callthe right triangle i24 with hypotenuse t, U and call the right triangle i42, V . Call theportion of angle δ in T , δT , the portion of δ in U , δU , the portion of δ in V , δV , theportion of ε in V , εV , the portion of η in T , ηT , and the portion of angle η in U , ηU . Itis important to remember here that the packing is parameterized in terms of γ so thatas α and therefore δ are changed, γ remains constant.

Figure 19: Quadrilateral S

Now, we increase α and therefore δ and la; however, γ and t remain constant. ThusδT and ηT are constant. So δU and/or δV have increased. Assume, without loss ofgenerality for it can be shown that this assumption is symmetric, that δU increased.Call the original value of δU , δU [original] and the original length of h, lh[original].

Thus cos(δU) = lhlt

and cos(δU [original]) = lh[original]lt

. However, since cosine is strictlydecreasing on the interval (0◦, 90◦), we see that lh < lh[original]. We have shownthat increasing δ, decreases lh. However sin(εV ) = lh

1and sin(ηU) = lh

lt. Since sine is

strictly increasing on the interval (0◦, 90◦), we see that increasing δ decreases εV and ηU .

We have shown that increasing α increases δ and la. Furthermore, we have shownthat increasing la decreases εR and ηR and that increasing δ decreases εV and ηU . Now,since ε = εV + εR and η = ηT + ηU + ηR where ηT is constant, increasing α decreasesε and η. However, it was shown in the hypothesis that η = 60◦ when α = 60◦. Thusη is never greater than 60◦. Note that when 60◦ < γ < 120◦ then ε is the angle thatis never greater than 60◦. We conclude that v4e07n036t15 cannot represent a locallymaximally dense packing of four circles on a flat torus. �


v4e07n037t11

The same lattice vector connects points 1 and 1 and points 3 and 3. By side-side-sidethe triangles 4121 and 4343, with 11 and 33 formed by lattice vectors, are equivalent.Thus 34 of 4343 is parallel to 12 or 21 of 4121. Therefore, considering the pentagon14312, Lemma 2 shows that this potential packing graph doesn’t represent a circlepacking.



v4e07n037t21

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This packing is not locally maximally dense. This packing has 3 degrees of freedom,α, β, and γ and it is never properly stressed.


v4e08n057t21

v4e08n057T21 is pictured in figure 23. Since C ′C and A′A are the same length(because they are lifts by the same vector), and all edges in the graph are the samelength, we have that 4ADA′ is congruent to 4CBC ′, by SSS. We also note that C ′Cand A′A are parallel. This implies that either BC is parallel to AD, or BC is parallelto A′D. If BC is parallel to AD, then this implies that ABCD is a rhombus, and soCD is the same length as AB. Thus, the circles with centers at D and C are tangent,and so this potential packing graph is not associated to an equal circle packing. If BCis parallel to A′D, then this implies that A′BCD is a rhombus, and so CA′ is the samelength as DB. Thus, the circles with centers at A′ and C are tangent, and so thispotential packing graph is not associated to an equal circle packing.



v4e08n057t22

v4e08n057T22 is pictured in figure 24. Since C ′C and A′A are the same length(because they are lifts by the same vector), and all edges in the graph are the samelength, we have that 4ADA′ is congruent to 4CBC ′ by SSS. We also note that C ′Cand A′A are parallel. This implies that either C ′B is parallel to DA′, or BC is parallelto DA′. If C ′B is parallel to DA′, then ∠BC ′A is the same as ∠CA′D = 60◦, andso 4AC ′B is equilateral, implying that AB is the same length as CB. Thus, thecircles with centers at A and B are tangent, and so this potential packing graph is notassociated to an equal circle packing. If BC is parallel to DA′, then ∠BCD is thesame as ∠CDA′ = 60◦, and so BD is the same length as CA′. Thus, the circles withcenters at B and D are tangent, and so this potential packing graph is not associatedto an equal circle packing.



v4e08n058t31

A pentagon is formed by 14312, and a quadrilateral (formed by two triangles together)is formed by 4213, with parallel sides 12 and 34, so this packing graph does not corre-spond to a circle packing by the Pentagon Lemma.

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v4e08n058t41

In v4e08n058T41, a pentagon is formed by 12134 and a quadrilateral is formed by3421, with parallel sides 43 and 12, so this packing graph does not correspond to acircle packing by the Pentagon Lemma.

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v4e08n059t11

v4e08n059T11 is pictured in figure 28. Since CB is parallel to AC ′, and AC ′ is paral-lel to A′C, we have that A′C is parallel to CB. We also note that A′C is parallel to DA,so DA is parallel to CB. This implies that ∠DAC is equal to ∠ACB, which is 60◦, sothat4DAC is an equilateral triangle. Hence, the circles at D and C should be tangent.

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v4e08n059t21

In v4e08n059T21, a pentagon is formed by 12423, and a quadrilateral is formed by1243, with parallel sides 13 and 24, so this packing graph does not correspond to acircle packing by the Pentagon Lemma.

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v4e08n060t11

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This packing is locally maximally dense. It has one free parameter, α, which canrange from π/3 to 2π/3. All of these arrangements have proper stress. It occupies theregions of the moduli space shown (in red), and does not include either endpoint. Eachendpoint is a triangular close packing, one containing self tangencies. This packing isglobally maximally dense for the region of the moduli space it occupies.

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Figure 31: v4e08n060t11pic



v4e08n060t12

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This packing has 3 free parameters (labeled in the picture below), and is not locallymaximally dense. By observing the stresses, it was found that either a pair of stresseshas opposite signs or the packing graph as a bar framework is not rigid.


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v4e08n061t11

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This packing is locally maximally dense for all 2π/3 < α < π, and all π/3 < β <2π/3 (with angles as labeled below). It is required that α + β < 5π/3, because thisrestriction ensures that the circles do not overlap. This packing leads to overlappingregions in the moduli space: the blue region (pictured below) occupies a region noother packings occupy, the yellow region overlaps both the blue region in this packing,the red region in this packing, and packings v4e09n095T11 and v4e10n141T12, and thered region overlaps with packing v4e09n095T11. The yellow region is less dense thanthe other two regions. The packing on the blue region is the globally maximally densepacking for the the region it occupies. We prove in Appendix 4 that v4e09n095T11 is


the more dense packing.

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Figure 37: A component of the moduli space for v4e08n061t11


v4e08n061t21

v4e08n061T21 is pictured in figure 40. We begin by noting that4ABA′ is congruentto 4C ′D′C, because they have all equal side lengths. We also have that AA′ is parallelto CC ′. Then we have two cases. In case 1, AB is parallel to CD′. In this case,∠D′CB equals ∠ABC, which equals 60◦, so D′ and B should connect. In the othercase (pictured), we have that AB is parallel to C ′D′. Since AB is parallel to A′B′′,D′C ′ is parallel to A′B′′, implying that ∠D′C ′A′ equals ∠C ′A′B′′, which equals 60◦.This means that A′ and D′ should connect, and so this potential packing graph is notassociated to an equal circle packing.



v4e08n061t22

v4e08n061T22 is pictured in figure 41. Since all side lengths are equal, 4BCB′ iscongruent to 4DA′′D′. Also, BB′ is parallel to DD′. This implies that either BC isparallel to DA′′, or BC is parallel to A′′D′. In the case where BC is parallel to DA′′,we find that CBDA′′ is a rhombus, implying that C and A′′ should be connected. Inthe case where BC is parallel to D′A′′, we also have that CB′ is parallel to DA′′, andso B′CDG is a rhombus, implying that B′ and A′′ should be connected.



v4e08n061t31

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This packing is not locally maximally dense because when observing the stresses,it was found that this packing is always either not properly stressed or not rigid as abar framework.

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v4e09n094t11

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This packing has one free parameter, α (as labeled below), which is restricted fromπ/3 to π originally, but because of symmetry, the upper bound changed to 2π/3. Thispacking is stressed everywhere. The moduli space of this packing does not include theα = π/3 endpoint (corresponds to triangular close packing on a triangular flat torus),but it does include the α = 2π/3 endpoint. This packing is the globally maximallydense packing for the region of the moduli space it occupies.

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v4e09n094t21

In v4e09n094T21, a pentagon is formed by 12413, and a quadrilateral is formedby 1243 (formed by two triangles), with parallel sides 42 and 13, so this does notcorrespond to a circle packing by the Pentagon Lemma.

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v4e09n094t22

In v4e09n094T22, a pentagon is formed by 12413, and a quadrilateral is formed by1242, with parallel sides 42 and 13, and so this does not correspond to a circle packingby the Pentagon Lemma.

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v4e09n094t31

v4e09n094T31 is pictured in figure 50. We can use lemma 3 to show that this poten-tial packing graph is not associated to an equal circle packing after noticing that BCis parallel to A′D, and that A′D is parallel to D′A, and that D′ABCA′′ is a pentagon.

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v4e09n095t11

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This packing has two free parameters, α and β, which both range from π/3 to 2π/3and are labeled in the figure below. By observation of symmetry in the packing, theparameters can be further restricted to α > 2β − π/3. There is proper stress in thisentire region. In addition, the bottom edge (without the endpoints) in the moduli spaceis included (none of the other boundaries are). This packing is the globally maximallydense packing for the region of the moduli space it occupies.

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v4e10n141t11

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This packing has one free parameter, α, as in the packing below. We restricted0 < α < π/6, with symmetry at π/12. It has proper stress everywhere in the interior.The top point in the moduli space (α = 0) is not included, but the bottom endpoint(α = π/12) is included. This packing is the globally maximally dense packing for theregion of the moduli space it occupies.


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v4e10n141t12

This packing has one free parameter, α, which ranges from π/3 to π/2. This packinghas proper stress everywhere, except at the endpoints, which are two triangular closepackings. This packing is the globally maximally dense packing for the region of themoduli space it occupies.


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v4e12n288t11

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This packing corresponds to the triangular close packing on the triangular flat torus.With no free parameters, it is an endpoint for many other packings. It is the globallymaximally dense packing for the region of the moduli space it occupies and it is themost dense packing of four circles on any flat torus.

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v4e12n288t12

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This corresponds to the triangular close packing on the 2/√

3 : 1 rectangular torus.With no free parameters, it is an endpoint for many other packings. It is the globallymaximally dense packing for the region of the moduli space it occupies and it is themost dense packing of four circles on any flat torus.


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Appendix 2: Our Results Match Those of Heppes

Heppes looks at what the common radius of 4 equal circles packed on a rectangulartorus is, where the torus is the quotient of the lattice generated by (0, b) and (0, 1),where b ≥ 1. [5]

Case 1: if b = cos(π/12−ε)π/12+ε , where 0 < ε < π/12.

Heppes claims that the radius of the circles, r4(b), in this region is 14 cos(π/12+ε)

. This

should correspond to our packing v4e10n141t11 (where circles have radius 1).First, we need to check that Heppes’ tori in this region are the same as our tori

for v4e10n141t11. b = 1 corresponds to the square flat torus, and this v4e10n141t11also shares this as an endpoint. If we make a picture of the other endpoint of packingv4e10n141t11, and solve for what b is given that the side length of the smaller side ofthe torus is 1, we find that b = 2/

√3 (you can find this by drawing a triangle, as in

the figure), and this matches Heppes’ result.

The next step is to calculate the radius of the circles in our packing. We had onefree parameter, α, which is equal to ε + π/12. Using triangles in the packing (seefigure), we find that 2r(cos(ε + π/12)) = 1/2, and so r = 1

4 cos(π/12+ε), which matches

Heppes’ result.

54

Case 2: 2√3 ≤ b ≤ 2

√3

Here, Heppes claims that r4(b) =√

(1/2)2 + (b/4)2/2. This should correspond to ourpacking v4e08n060t11.

We already know that the lower boundary matches, from the previous case. Theother bound comes from the following picture of the upper point in the packing ofv4e08n060t11. Since the radii of the circles here are 1/2 by the Dickinson Self-TangencyTheorem, we have that b = 2

√3, which matches Heppes’ result.

Below we have our packing v4e08n060t11 displayed, and we will check that weget the same value for r assuming a torus side length of 1 and b. Using the trianglepictured, we find that (4r)2 = 12 + (b/2)2, and after some simplification, find thatr =

√(1/2)2 + (b/4)2/2.


Case 3: b ≥ 2√3

By the bounds on the previous example and the Dickinson Self-Tangency Theorem,our results match with Heppes in this case.


Appendix 3: Packings of 3 Circles With Room for

a 4th

By Professor Dickinson’s proofs, we only need to consider packings with equilateralpacking graphs containing regular hexagons and (not necessarily regular) heptagons.Observing all of the toroidal embeddings of graphs associated with packings of 3 circleson tori, we find that only 4 contain hexagons and heptagons. These were v3e05v1t11(7-gon), v3e05v1t21 (6-gon), v3e06v3t11 (6-gon), and v3e06v3t12 (6-gon).

v3e05v1t11

Observation of this packing shows that it looks like another circle will fit at the coor-dinates (3/2,

√3/2) (this 4th circle is in blue in the picture below). It looks like this

circle is tangent to circle 3 twice and circle 2 twice. A quick check reveals that theseare real tangencies (each of the distances is 1). We note that this implies that circle 4is fixed, and so this solution for where to place a 4th circle is unique.

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Figure 66: v3e05v1t11

The original packing of 3 circles allowed the angles α and β to range from 2π/3to π. However, for some combinations of these angles, there is not room for the 4thcircle. Some algebra reveals that in order for there to be a 4th circle, we must add theadditional restriction of α + β < 5π/3.

57

Adding this 4th circle yields the same packing as v4e095t11. Adding the restrictionthat α + β < 5π/3 to the moduli space on v3e05v1t11 shows that this packing is onthe same tori as v4e095t11, so we have already accounted for this case (of adding a 4thcircle to a packing of 3 circles) in the packings we found.

v3e05v1t21

This packing has room for a 4th circle only when there is a regular hexagon, so α, β =2π/3. This leaves room for one circle which will be tangent to all circles already inthe packing twice, so we are left with the triangular close packing on a rectangular2/√

3 : 1 torus (packing v4e12n288t12 for us). Hence, we have already handled thiscase.

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v3e06v3t11

This packing has room for a 4th circle only when there is a regular hexagon, so α =2π/3. This leaves room for one circle which will be tangent to all circles already inthe packing twice, so we are left with the triangular close packing on a rectangular2/√

3 : 1 torus (packing v4e12n288t12 for us). Hence, we have already handled thiscase.


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v3e06v3t12

This packing has room for a 4th circle only when there is a regular hexagon, so α =2π/3. This leaves room for one circle which will be tangent to all circles already inthe packing twice, so we are left with the triangular close packing on a triangular flattorus (packing v4e12n288t11 for us). Hence, we have already handled this case.

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Appendix 4: Packing v4e09n095t11 is More Dense

than Packing v4e08n061t11

Proposition 7. The density of packing v4e09n095T11 is greater than the density ofpacking v4e08n061T11 on all tori which both of them fit on.

Proof. Let α be the angle ∠C1AF in packing v4e09n095T11, and let β be the angle∠C2AF in packing v4e08n061T11 (pictured in figure 71). Let d be the diameter of thecircles. Let both packings be on a torus given by ~v1 = 〈1, 0〉, and ~v2, where ~v2 is inthe area of the moduli space displayed below (because this the region where packingv4e08n061T11 and packing v4e09n095T11 embed on the same tori). Let this region becalled M .

Figure 70: The portion of the moduli space occupied by both v4e08n061T11and v4e09n095T11, M . The lower, curved boundary (including endpoints) is notcontained in M , while the upper boundary and the left boundary are containedin M . We choose not to preform inversion because in the diagrams used in thisproof, ~v2 must land in this region.

First, we need to show that there are no tori in M in which the densities of thesepackings are equal. A situation like this is pictured in figure 71. If the packings areon the same torus and they have the same density, then the radii of their circles mustbe the same. This implies that α = β, because in both of these packings, these anglesdetermine the radii of the circles in the packings since 1 = 2d cos(α). We also restrictα, β > 0 because this corresponds to the lower boundary of M , and so this boundaryis not included in M . The purple circles in this figure represent the possible locationsof the end of ~v2 for each packing. We are looking for points which are equidistant fromA′ and B′.

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A simplified picture is given in figure 72. Our goal now is to prove that the densitiesof the two packings are only equal when they are on a rectangular torus (the reason forthis will be explained later). First, we note that ∠A′CB′ = 2π−π/3−2π/3−(π−2α) =2α. Then, we construct the perpendicular bisector of segment A′B′, and this tells usthat ∠ECB′ = α and that ∠B′EC = π/2. Thus, 4CFA is congruent to 4BECby AAS (since segments AC and CB′ are both d). These congruent triangles implythat |CE| = 1/2. Now, we want to construct point C ′, which is where the end of ~v2

is, we want it to be d away from both B′ and A′, and so it is on the perpendicularbisector of B′A′. Then, 4C ′B′E is congruent to 4CB′E because they are both righttriangles, share side B′E and have the same length of hypotenuse, d. Hence, |C ′C| = 1.The location of C ′, then, is at C + (1/2 − 2 cos(∠GCE), d sin(α) + sin(∠GCE)), but∠GCE = π/3 +α−∠ACG = π/3 +α−α = π/3, so the x-coordinate of C ′ is 0. Thisimplies that ~v2 is along the y-axis.

If ~v2 is on the y-axis, however, we are looking at packings of rectangular flat tori.Looking back to region M , we see that there is only one rectangular torus which Mcould possibly contain. However, having v4e08n061T11 and v4e09n095T11 on thistorus requires that α = β = 0 (the lower boundary of M), so this torus is not actuallyan element of M . We conclude that there are no tori in M for which these packingshave the same density.

Figure 71: The case where α = β. Here, the red packing corresponds to packingv4e08n061T11 and the green packing corresponds to packing v4e09n095T11.

Now, we note that the density of a packing associated to a given equilateral packinggraph is a continuous function of some parameters in that equilateral packing graph.


Figure 72: The case where α = β

Let D1 be the density of v4e09n095T11, and D2 be the density of v4e08n061T11. Wejust showed that D1 −D2 is never equal to 0. Let f = D1 −D2. We can see that f iseither always positive or always negative as follows. Suppose that this is not the case,so for some sets of parameters a and b, f(a) is positive and f(b) is negative. Then theIntermediate Value Theorem implies that for some set of parameters c, f(c) = 0, butwe just showed that this is not the case. Hence, D1 −D2 is either always positive oralways negative. In other words, one packing is always more dense than the other one.

Now we will test a point in the moduli space to show that the density of packingv4e09n095t11 is always greater than v4e08n061t11. We will examine the density of thepackings at (x, y) = (0.17, 1.01).

In the Mathematica files, the packings are parameterized in terms of the angles[Alpha] and [Beta] for v4e08n061t11 and [Gamma] and [Delta] for v4e09n095t11.This parameterization is particularly convenient because the radii of the circles can becomputed directly from [Alpha] and [Gamma]. This is because [Alpha] and [Gamma]

correspond to the angle opposite the basis vector (1, 0) in the isosceles triangle 4212in v4e08n061t11 and the isosceles triangle 4121 in v4e09n095t11.

Starting with v4e09n095t11 we define a function in Mathematica for the diameterof the circles

d95=1/(2(Sin[\[Gamma]/2]))

Then we parameterize the lattice vectors as


LVec95={{1,0},

{d95*(Cos[(\[Pi]-\[Gamma])/2]-Cos[\[Pi]/3-((\[Pi]-\[Gamma])/2)]

+Cos[\[Delta]+(\[Pi]-\[Gamma])/2]),d95*(Sin[(\[Pi]-\[Gamma])/2]

+Sin[\[Pi]/3-((\[Pi]-\[Gamma])/2)]

+Sin[\[Delta]+(\[Pi]-\[Gamma])/2])}}

Now we use the Solve function

Solve[LVec95[[2]]=={.17,1.01},{\[Gamma],\[Delta]}]

The output is

{{\[Gamma]->2.65232,\[Delta]->1.27161}}

However, in Mathematica [Gamma] and [Delta] are stored to 15 digits of precision.Thus we can copy [Gamma] and use Solve to find the radius of the circles, r.

Solve[r==1/(4*Sin[(2.652317130555382)/2]),{r}]

The output is

{{r->0.257672}}

Now we move onto the packing v4e08n061t11. This packing is parameterized sothat one must use the negative of inversion to find the correct region in the modulispace. To simplify matters we will instead perform the negative of inversion on ourcoordinates.

z = 0.17 + 1.01i⇒ −1

0.17 + 1.01i=

−1

0.17 + 1.01i· 0.17− 1.01i

0.17− 1.01i=−.17 + 1.01i

1.049

Thus the new x coordinate is −0.171.049

and the new y coordinate is 1.011.049

. Note thatbecause reflection doesn’t change density we can change the sign of our x coordinate.For convenience we define a function for the diameter of the circles in the packing.

d61=1/(2Sin[\[Alpha]/2])

We proceed as before by defining a function for the lattice vectors.

LVec61={{1,0},

{d61*(Cos[(\[Pi]-\[Alpha])/2]-Cos[5\[Pi]/6

-(\[Alpha]/2)]-Cos[\[Beta]-((\[Pi]-\[Alpha])/2)])

,d61*(Sin[(\[Pi]-\[Alpha])/2]+Sin[5\[Pi]/6

-(\[Alpha]/2)]+Sin[\[Beta]-((\[Pi]-\[Alpha])/2)])}}

Then we use the Solve function


Solve[LVec61[[2]]=={(.17/1.049),(1.01/1.049)},{\[Alpha],\[Beta]}]

The output is

{{\[Alpha]->1.25313,\[Beta]->0.308525},{\[Alpha]->3.03043,\[Beta]->1.40075}}

Checking this against the constructions in Geogebra, we see that the second valuefor [Alpha] is the correct one. We proceed by solving for the radii of the circles R with[Alpha].

Solve[R==1/(4*Sin[3.0304310486873725/2]),{R}]

The output is

{{R->0.250387}}

Comparing this to our earlier results, 0.257672 > 0.250387 and we conclude thev4e09n095t11 is the more dense packing on the torus given by ~v2 = 〈0.17, 1.01〉. Sincewe have found a case in which D1 > D2, and we know that one density is always greaterthan the other density, we know that the density of packing v4e09n095T11 is greaterthan the density of packing v4e08n061T11 on all tori which both of them fit on.


Bibliography

[1] Connelly, Robert. Rigid Circle and Sphere Packings, Structural Topology (1988),no. 14, 43-60, Dual French-English text.

[2] Dickinson, Guillot, Keaton, Xhumari. Optimal Packings of Up to 5 Equal Circleson a Square Flat Torus.

[3] Dickinson, Guillot, Keaton, Xhumari. Oprimal Packings of Up to Six Equal Circleson a Triangular Flat Torus.

[4] Ellsworth, Anna Victoria, and Kenkel, Jenny. Packings of 3 Equal Circles on FlatTori. Final Report from the 2011 REU.

[5] Heppes, Aladar. Densest Circle Packing on the Flat Torus. Periodica MathematicaHungarica: Vol. 39 (1-3), 1999, pp 129-134.

[6] Royal, Gordon. Small Multigraphs. Found at http://school.maths.uwa.edu.au/ gor-don/remote/graphs/multigraphs/index.html.

[7] Shurman, Jerry. Multivariable Calculus. Found at http://people.reed.edu/ jer-ry/211/vcalc.pdf. 111-119, 316.

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