Sphere packings, I

Discrete & Computational Geometry, Jan 1997

We describe a program to prove the Kepler conjecture on sphere packings. We then carry out the first step of this program. Each packing determines a decomposition of space into Delaunay simplices, which are grouped together into finite configurations called Delaunay stars. A score, which is related to the density of packings, is assigned to each Delaunay star. We conjecture that the score of every Delaunay star is at most the score of the stars in the face-centered cubic and hexagonal close packings. This conjecture implies the Kepler conjecture. To complete the first step of the program, we show that every Delaunay star that satisfies a certain regularity condition satisfies the conjecture.

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Sphere packings, I

Discrete Comput Geom Geometry Discrete& Computational Sphere Packings 0 0 Department of Mathematics, University of Michigan , Ann Arbor, M148109 , USA We describe a program to prove the Kepler conjecture on sphere packings. We then carry out the first step of this program. Each packing determines a decomposition of space into Delaunay simplices, which are grouped together into finite configurations called Delaunay stars. A score, which is related to the density of packings, is assigned to each Delaunay star.We conjecture that the score of every Delaunay star is at most the score of the stars in the face-centered cubic and hexagonal close packings. This conjecture implies the Kepler conjecture. To complete the first step of the program, we show that every Delaunay star that satisfies a certain regularity condition satisfies the conjecture. T. C. Hales 1. Introduction The Kepler conjecture asserts that no packing of spheres in three dimensions has density exceeding that of the face-centered cubic lattice packing. This density is z r / V r ~ 0.74048. In an earlier paper [H2] we showed how to reduce the Kepler conjecture to a finite calculation. That paper also gave numerical evidence in support of the method and conjecture. This finite calculation is a series of optimization problems involving up to 53 spheres in an explicit compact region of Euclidean space. Computers have little difficulty in treating problems of this size numerically, but a naive attempt to make a thorough study of the possible arrangements of these spheres would quickly exhaust the w o r d ' s computer resources. The first purpose of this paper is to describe a program designed to give a rigorous proof of the Kepler conjecture. A sketch of a related program appears in [H2]. Although the approach of [H2] is based on substantial numerical evidence, some of the constructions of that paper are needlessly complicated. This paper streamlines some of those constructions and replaces others with constructions that are more amenable to rigorous methods. For this program to succeed, the original optimization problem must be partitioned into a series of much smaller problems that may be treated by current computer technology or hand calculation. The second purpose of this paper is to carry out the first step of the proposed program. A statement of the result is contained in Theorem 1 below. Background to another approach to this problem is found in [H3]. To add more detail to the proposed program, we recall some constructions from earlier papers [HI], [H2]. Begin with a packing of nonoverlapping spheres of radius I in Euclidean three-space. The density of a packing is defined in [HI]. It is defined as a limit of the ratio of the volume of the unit balls in a large region of space to the volume of the large region. The density of the packing may be improved by adding spheres until there is no further room to do so. The resulting packing is said to be saturated. It has the property that no point in space has distance greater than 2 from the center of some sphere. Every saturated packing givesrise to a decomposition of space into simplices called the Delaunay decomposition. The vertices of each Delaunay simplex are centers of spheres of the packing. None of the centers of the spheres of the packing lie in the interior of the circumscribing sphere of any Delaunay simplex. In fact, this property is enough to determine the Delaunay decomposition completely except for certain degenerate packings. A degeneracy occurs, for instance, when two Delaunay simplices have the same circumscribing sphere. In practice, these degeneracies are important, because they occur in the face-centered cubic and hexagonal close packings. The paper [H2] shows how to resolve the degeneracies by taking a small perturbation of the packing. In general, the Delaunay decomposition will depend on this perturbation. We refer to the centers of the packing as vertices, since the structure of the simplicial decomposition of space is our primary concern. For a proof that the Delaunay decomposition is a dissection of space into simplices, we refer the reader to [R]. The Delaunay decomposition is dual to the well-known Voronoi decomposition. If the vertices of the Delaunay simplices are in nondegenerate position, two vertices are joined by an edge exactly when the two corresponding Voronoi cells share a face, three vertices form a face exactly when the three Voronoi cells share an edge, and four vertices form a simplex exactly when the four corresponding Voronoi cells share a vertex. In other words, two vertices are joined by an edge if they lie on a sphere that does not contain any other of the vertices, and so forth (again assuming the vertices to be in nondegenerate position). The collection of all simplices that share a given vertex is called a Delaunay star. (This is a provisional definition: it is refined below.) Every Delaunay simplex has edges between 2 and 4 in length and, because of the saturation of the packing, a circumradius of at most 2. We assume that every simplex S in this paper comes with a fixed order on its edges, 1. . . . . 6. The order on the edges is to be arranged so that the first, second, and third edges meet at a vertex. We may also assume that the edges numbered i and i + 3 are opposite edges for i = 1, 2, 3. We define S(yl . . . . . Yt) to be the (ordered) simplex whose i th edge has length yi. If S is a Delaunay simplex in a fixed Delaunay star, then it has a distinguished vertex, the vertex common to all simplices in the star. In this situation we assume that the edges are numbered so that the first, second, and third edges meet at the distinguished vertex. A function, known as the compression I'(S), is defined on the space of all Delaunay simplices. Let 8oct = (-3Jr + 12arccos(1/vc3))/~/8 ~ 0.720903 be the density of a regular octahedron with edges of length 2. That is, place a unit ball at each vertex of the octahedron, and let/~octbe the ratio of the volume of the part of the balls in the octahedron to the volume of the octahedron. Let S be a Delaunay simplex. Let B be the union of four unit balls placed at each of the vertices of S. Define the compression as We extend the definition of compression to Delaunay stars D* by setting F(D*) = F(S), with the sum running over all the Delaunay simplices in the star. In this and subsequent work, we single out for special treatment the edges of length between 2 and 2.51. The constant 2.51 was determined experimentally to have a number of desirable properties. This constant appears throughout the paper. We call vertices that come within 2.51 of each other close neighbors. We say that the convex hull of four vertices is a quasi-regular tetrahedron (or simply a tetrahedron) if all four vertices are close neighbors of one another. Suppose that we have a configuration of six vertices in bijection with the vertices of an octahedron with the property that two vertices are close neighbors if and only if the corresponding vertices of the octahedron are adjacent. Suppose further that exactly one of the three diagonals has length at most 2~/2. In this case we call the convex hull of the six vertices a quasi-regular octahedron (or simply an octahedron). The compatibility of quasi-regular tetrahedra and octahedra with the Delaunay decomposition is established in Section 3. We think of Euclidean space as the union of quasi-regular tetrahedra, octahedra, and various less-interesting Delaunay simplices. From now on, a Delaunay star is to be the collection of all quasi-regular tetrahedra, octahedra, and Delaunay simplices that share a common vertex v. This collection of Delaunay simplices and quasi-regular solids is often, but not always, the same as the objects called Delaunay stars in [H2]. We warn the reader of this shift in terminology. It is convenient to measure the compression in multiples of the compression of the regular simplex of edge length 2. We define apoint (abbreviated pt) to be 1TM(S (2,2, 2, 2, 2, 2)). We have pt = 1lzr/3 - 12 arccos(1/V~) ~ 0.0553736. One of the main purposes of this paper and its sequel is to replace the compression by a function (called the score) that has better properties than the compression. Further details on the definition of score appear in Section 2. We are now able to state the main theorem of this paper. Theorem 1. If a Delaunay star is composed entirely of quasi-regular tetrahedra, then its score is less than 8 pt. The idea of the proof is the following. Consider the unit sphere whose center is the center of the Delaunay star D*. The intersection of a simplex in D* with this unit sphere is a spherical triangle. For example, a regular tetrahedron with edges of length 2 gives a triangle on the unit sphere of arc length zr/3. The star D* gives a triangulation of the unit sphere. The restriction on the lengths of the edges of a quasi-regular tetrahedron constrains the triangles in the triangulation. We classify all triangulations that potentially come from a star scoring more than 8 pt. Section 5 develops a long list of properties that must be satisfied by the triangulation of a high-scoring star. It is then necessary to classify all the triangulations that possess the properties on this list. The original classification was carded out by a computer program, which generated all potential triangulations and checked them against the list. D. J. Muder has made a significant improvement in the argument by giving a direct, computer-free classification. His result appears in the Appendix. As it turns out, there is only one triangulation that satisfies all of the properties on the list. Section 7 proves that Delaunay stars with this triangulation score less than 8 pt. This will complete the main thread of the argument. There are a number of estimates in this paper that are established by computer. These estimates are used throughout the paper, even though their proofs are not discussed until Sections 8 and 9. These sections may be viewed as a series of technical appendices giving explicit formulas for the compression, dihedral angles, solid angles, volumes, and other quantities that must be estimated. The final section states the inequalities and gix/es details about the computerized verification. There is no vicious circle here: the results of Sections 8 and 9 do not rely on anything from Sections 2-7. There are several functions of a Delaunay simplex that are used throughout this paper. The compression F(S) has been defined above. The dihedral angle dih(S) is defined to be the dihedral angle of the simplex S along the first edge (with respect to the fixed order on the edges of S). Set dihmin = dih(S(2, 2, 2.51, 2, 2, 2.51)) ~ 0.8639; dihmax = dih(S(2.51, 2, 2, 2.51,2, 2)) = arccos(-29003/96999) ~ 1.874A.n/..We will see that dihmin and dihmaxare lower and upper bounds on the dihedral angles of quasiregular tetrahedra. The solid angle (measured in steradians) at the vertex joining the first, second, and third edges is denoted sol(S). The intersection of S with the ball of unit radius centered at this vertex has volume sol(S)/3 (see [HI, 2.1]). For example, sol(S(2, 2, 2, 2, 2, 2)) ~ 0.55. Let rad(S) be the circumradius of the simplex S. In Section 2 we define two other functions: vor(S), which is related to the volume of Voronoi cells, and the score a(S). Finally, let r/(a, b, c) denote the circumradius of a triangle with edges a, b, c. Explicit formulas for all these functions appear in Section 8. 2. The Program By proving Theorem 1, the main purpose of this paper will be achieved. Nevertheless, it might be helpful to give a series of comments about how Theorem 1 may be viewed as the solution to the first of a handful of optimization problems that would collectively provide a solution to the Kepler conjecture. We begin with some notation and terminology. We fix a Delaunay star D* about a vertex v0, which we take to be the origin, and we consider the unit sphere at v0. Let vl and v2 be vertices of D* such that v0, vl, and Va are all close neighbors of one another. We take the radial projections Pi of vi to the unit sphere with center at the origin and connect the points Pl and Pa by a geodesic arc on the sphere. We mark all such arcs on the unit sphere. Lemma 3.10 shows that the arcs meet only at their endpoints. The closures of the connected components of the complement of these arcs are regions on the unit sphere, called the standard regions. We may remove the arcs that do not bound one of the regions. The resulting system of edges and regions is referred to as the standard decomposition of the unit sphere. Let C be the cone with vertex vo over one of the standard regions. The collection of the Delaunay simplices, quasi-regular tetrahedra, and quasi-regular octahedra of D* in C (together with the distinguished vertex v0) is called a standard cluster. Each Delaunay simplex in D* belongs to a unique standard cluster. Each triangle in the standard decomposition of the unit sphere is associated with a unique quasi-regular tetrahedron, and each tetrahedron determines a triangle in the standard decomposition (Lemma 3.7). We may identify quasi-regular tetrahedra with clusters over triangular regions. We assign a score to each standard cluster in [H4, 3]. In this section we define the score of a quasi-regular tetrahedron and describe the properties that the score should have in general. Let S be a quasi-regular tetrahedron. It is a standard cluster in a Delaunay star with center v0. If the circumradius of S is at most 1.41, then we define the score to be I'(S). If the circumradius is greater than 1.41, then embed the simplex S in Euclidean threespace. Partition Euclidean space into four infinite regions (infinite Voronoi cells) by associating with each vertex of S the points of space closest to that vertex. By intersecting S with each of the four regions, we partition S into four pieces So, $1, $2, and $3, corresponding to its four vertices v0, v~, v2, and v3. Let sol/be the solid angle at the vertex vi of the simplex. The expression -48o~t vol(S0) + 4 sol0/3 is an analytic function of the lengths of the edges for simplices S that contain their circumcenters. (Explicit formulas appear in Section 8.) This function may be analytically continued to a function of the lengths of the edges for simplices S that do not necessarily contain their circumcenters. Let vor(S) be defined as the analytic continuation of -4(~oa vol(S0) + 4 sol0/3. In this case define the score of S to be vor(S). Write (r(S) for the score of a quasiregular tetrahedron. In summary, the score is Ir(s), a(S) = [vor(S) if the circumradius of S is at most 1.41, otherwise. The quasi-regular tetrahedron S appears in four Delaunay stars. In the other three Delaunay stars, the distinguished vertices will be vl, v2, and v3, so that S, viewed as a standard cluster in the other Delaunay stars, will have scores -8o~t vol(Si) + soli/3 (or their analytic continuations), for i = 1, 2, 3. By definition, r ( s ) = 3 i=0 (-8oc, vol(L) + soli/3). The sum of the scores of S, for each of the four vertices of S, is 4F(S). This is the same total that is obtained by summing the compression of S at each of its vertices. This is the property we need to relate the score to the density of the packing. It means that although the score reapportions the compression among neighboring Delaunay stars, the average of the compression over a large region of space equals the average of the score over the same region, up to a negligible boundary term. The analytic continuation in the definition of vor(S) has the following geometric interpretation. If the circumcenter of a quasi-regular tetrahedron S is not contained in S, a small tip of the infinite Voronoi cell at v0 (or some other vertex) will protrude through the opposite face of the Delaunay simplex. The volume of this small protruding tip is not counted in -48o~t vol(S0) + 4 sol0/3, but it is counted in the analytic continuation. The analytic continuations of the scores of S for each of the other three vertices acquire a term representing the negative volume of a part of the tip. The three parts together constitute the entire tip, so that the negative volumes exactly offset the volume of the tip, and the sum of the four scores of S is 4F(S). Details appear in [H4]. The general definition of the score will have similar properties. To each standard cluster of a Delaunay star D* a score will be assigned. The rough idea is to let the score of a simplex in a cluster be the compression F(S) is the circumradius of every face of S is small, and otherwise to let the score be defined by Voronoi cells (in a way that generalizes the definition for quasi-regular tetrahedra). The score cr(D*) of a Delaunay star is defined as the sum of the scores of its standard clusters. The score has the following properties [H4, 3.1 and 3.5]: 1. The score of a standard cluster depends only on the cluster, and not on the way it sits in a Delaunay star or in the Delaunay decomposition of space. 2. The Delaunay stars of the face-centered cubic and hexagonal close packings score exactly 8 pt. 3. The score is asymptotic to the compression over large regions of space. We make this more precise. Let A denote the vertices of a saturated packing. Let A s denote the vertices inside the ball of radius N. (Fix a center.) Let D*(v) denote the Delaunay star at v ~ A. Then the score satisfies (in Landau's notation) = r(D*(v)) + O ( N b . AN AN L e m m a 2.1. If the score of every Delaunay star in a saturated packing is at most s < 16zr/3, then the density ofthepacking is at most 16ZrSoct/(16~r - 3s). If the score of every Delaunay star in a packing is at most 8 pt, then the density of the packing is at most / Proof The second claim is the special case s = 8 pt. The proof relies on property 3. The number of vertices such that D* (v) meets the boundary of the ball B s of radius N has order O (N2). Since the Delaunay stars give a fourfold cover of R3, we have ( 4 -8oct vol(Bs) + IANI ~ ) = Z ( - 8 o c t v o l ( D * ( v ) ) + ^N 4 sol(D*(v)) 3 + O ( N 2) C(D*(v)) + O ( N a) = Z AN = Z a ( D * ( v ) ) + O(N 2) AN < s l A s l + O(N2). Rearranging, we get 4re [A N [ < 8oc, 3 vol(Bs) - (1 - 3s/16zr) + 0~(N 2) vol(BN) In the limit the left-hand side is the density and the right-hand side is the bound. Similar arguments can be found in [HI] and [H2]. [] The following conjecture is fundamental. By the lemma, this conjecture implies the Kepler conjecture. The lemma also shows that weaker bounds than 8 pt on the score might be used to give new upper bounds on the density of sphere packings. Conjecture 2.2. The score of every Delaunay star is at most 8 pt. The basic philosophy behind the approach of this paper is that quasi-regular tetrahedra are the only clusters that give a positive score, standard clusters over quadrilateral regions should be the only other clusters that may give a score of zero, and every other standard cluster should give a negative score. Moreover, we will prove that no quasi-regular tetrahedron gives more than 1pt. Thus, heuristically, we try to obtain a high score by including as many triangular regions as possible. If we allow any other shape, preference should be given to quadrilaterals. Any other shape of region should be avoided if possible. If these other regions occur, they should be accompanied by additional triangular regions to compensate for the negative score of the region. We will see later in this paper that even triangular regions tend to give a low score unless they are arranged to give five triangles around each vertex. The main steps in a proof of the Kepler conjecture are 1. A proof that even if all regions are triangular the total score is less than 8 pt. 2. A proof that standard clusters in regions of more than three sides score at most Opt. 3. A proof that if aU of the standard regions are triangles or quadrilaterals, then the total score is less than 8 pt (excluding the case of pentagonal prisms). 4. A proof that if some standard region has more than four sides, then the star scores less than 8 pt. 5. A proof that pentagonal prisms score less than 8 pt. The division of the problem into these steps is quite arbitrary. They were originally intended to be steps of roughly equal magnitude, although is has turned out that a construction in [H4] has made the second step substantially easier than the long calculations of the third step. This paper carries out the first step. The second step of the program is also complete [H4]. Partial results are known for the third step [H5]. In the fourth step it will be necessary to argue that these regions take up too much space, give too little in return, and have such strongly incompatible shapes that they cannot be part of a winning strategy. To make step 5 precise, we definepentagonalprisms to be Delaunay stars whose standard decomposition has ten triangles and five quadrilaterals, with the five quadrilaterals in a band around the equator, capped on both ends by five triangles (Diagram 2.3). The conjecture in this section asserts, in particular, that pentagonal prisms, which created such difficulties in [H2], score less than 8 pt. The final step has been separated from the third step, because the estimates are expected to be more delicate for pentagonal prisms than for a general Delaunay star in the third step. One of the main shortcomings of the compression is that pentagonal prisms have compression greater than 8 pt (see [H2]). Numerical evidence suggests that the upper bound on the compression is attained by a pentagonal prism, denoted Dppap in [H2], at Diagram 2.3 about 8.156 pt, and this means that the link between the compression and the Kepler conjecture is indirect. The score appears to correct this shortcoming. What evidence is there for the conjecture and program? They come as the result of extensive computer experimentation. I have checked the conjecture against much of the data obtained in the numerical studies of [H2, 9.3]. The data suggest that the score tends to give a dramatic improvement over the compression, often improving the bound by more than a point. The score of the particularly troublesome pentagonal prism Dppdp drops safely under 8 pt. I have checked a broad assortment of other pentagonal prisms and have found them all to score less than 8 pt. The second step shows that no serious pathologies can arise. The only way to form a Delaunay star with a positive score is by arranging a number of quasi-regular tetrahedra around a vertex (together with other standard clusters than can only lower the score). There must be at least eight tetrahedra to score 8 pt, and if there are any distortions in these tetrahedra, there must be at least nine. However, as this paper shows, too many quasi-regular tetrahedra in any star are also harmful. Future papers will impose additional limits on the structure of the optimal Delaunay star. This section studies the compatibility of the Delaunay simplices and the quasi-regular solids. Fix three vertices Vl, v2, and v3 that are close rieighbors to one another. Let T be the triangle with vertices vi. It does not follow that T is the face of a Delaunay simplex. However, as we will see, when T is not the face of a simplex, the arrangement of the surrounding simplices is almost completely determined. If T is not the face of a Delaunay simplex, then we will show that there are two additional vertices Voand V'o, where vo and v0' are close neighbors to vl,/)2, and v3. This means that there are two quasi-regular tetrahedra Sl and $2 with vertices (vo, vl, v2, v3) and (v~,/)1, v2, v3), respectively, that have the common face T (see Diagram 3.1 (a)). We see that $1 tO $2 is the union Of three Delaunay simplices with vertices (/)0, v~,/)l, v2), (/)0,/)~, v2, v3), and (Vo, v~, v3, vl) in Diagram 3.1(b). This section establishes that this is the only situation in which quasi-regular tetrahedra are not Delaunay simplices: they must come in pairs and their union must be three Delaunay simplices joined along a common edge. The decomposition of this paper is obtained by taking each such triple of Delaunay simplices (3.1 (b)) and replacing the triple by a pair of quasi-regular tetrahedra (3.1(a)). Diagram 3.1 From the dual perspective of Voronoi cells, the Voronoi cell at v0(or v~) would have a small tip protruding from Sl through T, if the vertex v~ were not present. The vertex v~ slices off this protruding tip so that the Voronoi ceils at o0 and v~ have a small face in common. L e m m a 3.2. Suppose that the circumradius o f the triangle T is less that .v/2. Then T is the face o f a Delaunay simplex. Proof. Let r < ~/2 be the radius of the circle that circumscribes T, and let c be the center of the circle. The sphere of radius r at c does not contain any vertices of D* other than vl, v2, and v3. By the definition of the Delaunay decomposition (as described in the Introduction), this implies that T is the face of a simplex. [] R e m a r k 3.3. We have several constraints on the edge lengths, if T is not a face of a Delaunay simplex. Consider the circumradius 17(a, b, c ) o f a triangle whose edges have lengths a, b, c between 2 and 2.51. Since 2.512 < 22+ 22, we see that the triangle is acute, so that r/(a, b, c, ) is monotonically increasing in a, b, and c. This gives simple estimates relating the circumradius to a, b, and c. The circumradius is at most r/(2.5 I, 2.51, 2.51 ) = 2.51/~/~ ~ 1.449. If the circumradius is at least ~/2 ~ 1.41421, then a, b, and c are greater than 2.3(I/(2.3, 2.51,2.51) ~ 1.41191 < ~/'2). Under the same hypothesis, two of a, b, and c are greater than 2.41 (r/(2.41, 2.41, 2.51) < Vr2). Finally, at least one edge has length greater than 2.44(r/(2.44, 2.44, 2.44) < V~). L e m m a 3.4. Let T be the triangle with vertices 13i . Assume that the vertices vi are close neighbors o f one another. Suppose there is a vertex vo that lies closer to the circumcenter o f T than the vertices o f T do. Then the vertex 1)o satisfies 2 < I1)0- vii < 2.15,for i = 1, 2, 3. In particular, the convex hull ofvo . . . . . 1)3 is a quasi-regular tetrahedron S with face T. Another way of stating the hypothesis on the circumcenter is to say that the plane of T separates 1)0 from the circumcenter of S. Because of the constraints on the edge lengths in Remark 3.3, the three other faces of S are faces of Delaunay simplices. Proof. We defer the proof to Section 8.2.5. [] Lemma 3.5. Let 1), 1)1, 1J2, 1)3,and 1)4 be distinct vertices with pairwise distances at least 2. Suppose that the pairs (vi, vj) are close neighbors for {i, j} ~ {1, 4}. Then v does not lie in the convex hull of (vl, v2, v3, v4). Proof. For a contradiction, suppose v lies in the convex hull. Since Iv - vi I > 2 and for {i, j} r {1, 4}, vi and vj are close neighbors, the angle formed by vi and vj at the vertex v is at most 00 = arccos(1 - 2.512/8) -~ 1.357. Each such pair of vertices gives a geodesic arc of length at most 00 radians on the unit sphere centered at v. We obtain in this way a triangulation of the unit sphere by four triangles, two with edges of length at most 00 = 2 arcsin(2.51/4) radians, and two others that fit together to form a quadrilateral with edges of at most 00 radians. By the spherical law of cosines, the area formula for a spherical triangle, and [H2, 6.1], each of the first two triangles has area at most 3 arccos(cos 00/(1 + cos00)) - rr ~ 1.04. By the same lemma, the quadrilateral has area at most the area of a regular quadrilateral of side 00, or about 2.8. Since the combined area of the two triangles and quadrilateral is less than 4~r, they cannot give the desired triangulation. To see that v cannot lie on the boundary, it is enough to check that a triangle having two edges of lengths between 2 and 2.51 cannot contain a point that has distance 2 or more from each vertex. We leave this as an exercise. [] Corollary 3.5.1, No vertex of the packing is ever an interior point of a quasi-regular tetrahedron or octahedron. Proof. The corollary is immediate for a tetrahedron. For an octahedron, draw the distinguished diagonal and apply the lemma to each of the four resulting simplices. [] Let T be a triangle of circumradius between ~ and 2.51/~/3, with edges of length between 2 and 2.51. Consider the line through the circumcenter of T, perpendicular to the plane of T. Let s be the finite segment in this line whose endpoints are the circumcenters of the two simplices with face T formed by placing an additional vertex at distance 2 from the three vertices of T on either side of the plane through T. Lemma 3.6. Let S be a simplex formed by the vertices of T and a fourth vertex v~. Suppose that the circumcenter of S lies on the segment s. Assume that vo' has distance at least 2from each of the vertices o f T . Then vo' has distance less than 2.3from each of the vertices of T. Proof. Let 1)1, 1)2, 1)3 be the vertices of T. For a contradiction, assume that 1)~ has distance at least 2.3 from a vertex 1)1 of T. Lemma 3.4 shows that the plane through T does not separate v~ from the circumcenter of S. If the circumcenter lies in s (and is not separated from 1)~ by the plane through T), then by moving 1)~ to decrease the circumradius, the circumcenter remains in s. Let S be the simplex with vertices 1)~,1)1, 1)2,and v3. The circumcenter of S lies in the interior of S. We omit the proof, because it is established by methods similar to (but longer than) the proof of Lemma 3.4. Thus, the circumradius is increasing in the lengths of I1)~- vi I, for i = l, 2, 3 (see paragraph 8.2.4). Let R be the circumradius of a simplex with face T and center an endpoint ofs. We will prove that the circumradius of S is greater than R, contrary to our hypothesis. Moving v~ to decrease the circumradius, we may take the distances to vi to be precisely 2.3, 2, and 2. We may move 1)i, v2, and 1)3 along their fixed circumscribing circle until lyE- 1)31= 2.51 and I1)1- 1)21 = I1)1- 1)31in a way that does not decrease any of the distances from v~ to 1)i. Repeating the previous step, we may retain our assumption that 1)~has distances exactly 2.3, 2, and 2 from the vertices 1)i as before. We have reduced the problem to a one-dimensional family of tetrahedra parametrized by the radius r of the circumscribing circle of T. Set x(r) = I1)1 - 1)31 = 11)2 - 1)31. To obtain our desired contradiction, we must show that the circumradius R'(r) of the simplex S(2.3, 2, 2, 2.51, x(r), x(r)) satisfies R'(r) > R(r). Since both R' and R are increasing in r, for r ~ [V"2, 2.51/x/3], the desired inequality follows if we evaluate the 200 constants g ' ( r i ) - g ( r i + l ) for ri = ~/2 ( 2 . 5 1 _ Vr~) i \ ,,/'] 200' for i = 0 . . . . . 199, and check that they are all positive. (The smallest is about 0.00005799, which occurs for i = 199.) [] Let T be a triangle made up of three close neighbors. Suppose that T is not the face of a Delaunay simplex. There exists a vertex v0 whose distance to the circumcenter of T is less than the circumradius of T. Let S be the quasi-regular tetrahedron formed by v0 and the vertices of T. It is not a Delaunay simplex, so there exists a vertex v6 that is less than the circumradius of S from the circumcenter of S. Let S' be the simplex formed by v6 and the vertices of T. It is not a Delaunay simplex either. The circumcenter of S lies in the segment s of Lemma 3.4, so the circumcenter of S' does too. The lengths of the edges of S and S' other than T are constrained by Lemmas 3.4 and 3.6. In particular, in light of Remark 3.3, the faces other than T of S and S' are faces of Delaunay simplices. If vo and vo' lie on the same side of the plane through T, then either vo' lies in S, v0 lies in S', or the faces of S and S' intersect nonsimplicially. None of these situations can occur because a nondegenerate Delaunay decomposition is a Euclidean simplicial complex and because of Lemma 3.5. We conclude that v0 and v6 lie on opposite sides of the plane through T. S U S' is bounded by Delaunay faces, so S U S' is a union of Delaunay simplices. The fourth vertex of the Delaunay simplex in S US' with face (v0, v l, v2) cannot be v3 (S is not a Delaunay simplex), so it must be v0/. Similarly, (vo, vo/, v2, v3) and (v0, vo!, v3, vl) are Delaunay simplices. These three Delaunay simplices cannot be quasi-regular tetrahedra by Lemma 8.3.2. The assumption that T is not a face has completely determined the surrounding geometry: there are two quasi-regular tetrahedra S and S' along T such that S U S' is a union of three Delaunay simplices. Lemma 3.7. Let L be a union of standard regions. Suppose that the boundary of L consists of three edges. Then either L or its complement is a single triangle. For example, the interior of L cannot have the form of Diagram 3.8. Lemma 3.5 (proof) shows that if all regions are triangles, then there are at least 12 triangles, so that the exterior of L cannot have the form of Diagram 3.8 either. Diagram3.8 Proof. Replacing L by its complement if necessary, we may assume that the area of L is less than its complement. The triangular boundary corresponds to four vertices vo (the origin), vl, v2, and v3. The close-neighbor constraints on the lengths show that the convex hull of v0. . . . . v3 is a quasi-regular tetrahedron. By construction each quasiregular tetrahedron is a single cluster. [] We say that a point v ~ ]]~3 is enclosed by a region on the unit sphere is the interior of the cone (with vertex v0) over that region contains v. For example, in Diagram 3.9, the point v is enclosed by the given spherical triangle. vo Diagram3.9 The following lemma was used in Section 2 to define the standard decomposition. Lemma 3.10. Fix a Delaunay star D* with center vo. Draw geodesic arcs on the unit sphere at vofor every triple of close neighbors vo, vl, v2 (as in Section 2). The resulting system of arcs do not meet except at endpoints. Proof. Our proof is based on the fact that a nondegenerate Delaunay decomposition is a Euclidean simplicial complex. Let Tl and/'2 be two triangles made from two such triples of close neighbors. We have T/ C Si U S~, where Si and Si are the Delaunay simplices with face T/if T,. is the face of a Delaunay simplex, and they are the two quasi-regular tetrahedra with face T/constructed above, otherwise. Since a nondegenerate Delaunay decomposition is a Euclidean simplicial complex, $1 U S'l meets $2 t_JS~ simplicially. By the restrictions on the lengths of the edges in Lemmas 3.4 and 3.6, this forces/'1 to intersect T2 simplicially. The result follows. [] Quadrilaterals Fix a Delaunay star composed entirely of quasi-regular tetrahedra and consider the associated triangulation of the unit sphere. Let L be a region of the sphere bounded by four edges of the triangulation. L will be the union of two or more triangles. Replacing L by its complement in the unit sphere if necessary, we assume that the area of L is less than that of its complement. We claim that, in this context, L is the union of either two or four triangles, as illustrated in Diagram 4.1. In particular, L encloses at most one vertex. If a diagonal to the quadrilateral L is an edge of the triangulation, the region L is divided into two triangles each associated with a quasi-regular tetrahedron. In particular, there is no enclosed vertex. (Lemma 3.7 precludes any subtriangulation of a triangular region.) If, however, there is a single enclosed vertex and neither diagonal is an edge of the triangulation, then the only possible triangulation of L is the one of the diagram. Proposition 4.2 completes the proof of the claim. Diagram4.1 Proposition 4.2. A union of regions (of area less than 2~r) bounded by exactly four edges cannot enclose two vertices of distance at most 2.51 from the origin. This argument is somewhat delicate: if our parameter 2.51 had been set at 2.541, for instance, such an arrangement would exist. First we prove a useful reduction. A s s u m e a figure exists with vectors vl . . . . . 1)4, v, a n d v' subject to the L e m m a 4.3. constraints 2 <_ [vii < 2.51, 2 <_ Ivi - vi+ll < ki, 2 < I v i - Vi+2[, 2 < IV-- V'I, hi < I w - v i i , 2 < Iwl _< e f o r w = v , v ' and i = 1. . . . . 4 ( m o d 4 ) , where s hi, and ki are f i x e d constants that satisfy s ~ [2.51, 2~/2], hi E [2, 2~r ki E [2, 2.51]. Let L be the quadrilateral on the unit sphere with vertices vi/lvi[ and edges running between consecutive vertices. A s s u m e that v and v' lie in the cone at the origin obtained by scaling L. Then another figure exists m a d e o f a (new) collection o f vectors vl . . . . . v4, v, a n d v' subject to the constraints above together with the additional constraints [i)i - Vi+l[ = ki, Ivil = 2 f o r i = l . . . . . 4, Ivl = Iv'l = e . Moreover, the quadrilateral L may be a s s u m e d to be convex. Proof. B y rescaling v and v', we m a y assume that ]vl = Io'l = e. ( M o v i n g v or v' a w a y from the origin increases its distance from the other vertices o f the configuration.) The diagonals satisfy Iv2 - v4[ > 2.1 and Iv1 - v31 > 2.1. Otherwise, if say Iv1 - v31 < 2.1, then the faces with vertices (Vl, v2, v3) and (Vl, v4, v3) have circumradius less than ~,/2. The edge f r o m 0 to v has length at most v"2, so this edge cannot intersect these faces by the Euclidean simplicial c o m p l e x argument used before L e m m a 3.7 and in L e m m a 3.10. B y L e m m a 3.5, v cannot lie in the convex hull o f (0, vl, v3, vi). This leaves v nowhere to go, and a figure with IVl - w31 < 2. I does not exist. Next, we claim that we m a y a s s u m e that the quadrilateral L is convex (in the sense that it contains the geodesic arcs between any two points in the region). T o see this, suppose the vertex vi lies in the cone over the convex hull o f the other three vertices vj. Consider the plane P through the origin, vi-1, and vi+l. T h e reflection v~ o f vi through P is no closer to v, v', or vi+2 and has the s a m e distance to the origin, v i - l , and vi+l. Thus, replacing vi with v; if necessary, we m a y assume that L is convex. Most o f the remaining deformations are described as pivots. We fix an axis and rotate a vertex around a circle centered along and perpendicular to the given axis. If, for example, Iv3 - v4l < k3, we pivot the vertex 04 around the axis through 0 and Vl until Iv3 - v41 ----k3. It follows f r o m the choice o f axes that the distances f r o m v4 to the origin and vl are left unchanged, and it follows from the convexity o f L that Iw - Va[ increases for w = v2, 03, v, and v'. Similarly, we m a y pivot vertices vi along the axis through the origin and vi+l until [vi - Vi-l[ = ki, for all i. Fix an axis through two opposite vertices (say Vl and v3) and pivot another vertex (say v2) around the axis toward the origin. We wish to continue by picking different axes and pivoting until [vii = 2, for i = 1, 2, 3, 4. However, this process appears to break down in the event that a vertex 1)i has distance hi from one o f the enclosed vertices and the pivot toward the origin moves vi closer to that enclosed vector. We must check that this situation can be avoided. Interchanging the roles o f (1)1, v3) with (v2, 1)4) as necessary, we continue to pivot until Ioll = 11)31 = 2 or 11)21 = 11)41= 2 (say the former). We claim that either v2 has distance greater than h2 from v or that pivoting 1)2 around the axis through 1)1 and v3 moves v2 away from v. If not, we find that Iv1[ = I031 = 2, Iv - w21 = h2 and that 1)lies in the cone C = C(1)2) with vertex v2 spanned by the vectors from v2 to the origin, Vl, and v3. (This relies on the convexity o f the region L.) To complete the proof, we show that this figure made from (0, vl, v2, v3, v) cannot exist. Contract the edge (v, 02) as much as possible keeping the triangle (0, vl, v3) fixed, subject to the constraints that 1) ~ C(1)2) and Iw - w'l > 2, for w = v, 1)2 and w' = 0, vl, 1)3.This contraction gives Iv - 1)21< h2 < 2Vr2. Case 1: v lies in the plane o f ( 0, 02, 1 )3). This gives an impossibility: crossing edges (1), 1)2), (0, v3) o f length less than 2~/'2. Similarly, v cannot lie in the plane o f (0, '1)1, 1)2). Case 2: v lies in the interior o f the cone C(v2). The contraction gives Iv - w'l = 2 for w = v, v2 and w' = 0, 1)1, 1)3.The edge (1), v2)divides the convex hull o f (0, v l, v2, 1)3, v) into three simplices. Consider the dihedral angles o f these simplices along this edge. The dihedral angle o f the simplex (1)1, v2, 1)3, 1)) is less than Jr. The dihedral angles o f the other two are less than dih(S(2q"2, 2, 2, 2, 2, 2)) = zr/2. Hence, the dihedral angles along the diagonal cannot sum to 2zr and the figure does not exist. Case 3: v lies in t h e p l a n e o f ( v 2 , 1:1, 1)3). Let r be the radius o f the circle in this plane passing through 1)1 and 1)3 obtained by intersecting the plane with a sphere o f radius 2 at the origin. We have 2r > [vl - 1)31 > 2 V ~ because otherwise we have the impossible situation o f crossing edges (1)1, 1)3) and (v2, 1)) o f length less than 2~/~. Let H be the perpendicular bisector o f the segment (1)1, 1)3). By reflecting v through H if necessary we may assume that- v and v2 lie on the same side o f H , Say the side o f v3. Furthermore, by contracting (1), v2), we may assume without loss o f generality that 11)3- v[ = 11)21)31 = 2. Let f ( r ) = Iv - v21, as a function o f r . f is increasing in r. The inequalities 2~/'2 > h2 > I v - v2l > f ( ~ ) = 2~/~ give the desired contradiction. [] Proof o f Proposition 4.2. Assume for a contradiction that o and v' are vertices enclosed by L. Let the center o f the Delaunay star be at the origin, and let vl . . . . . v4, indexed consecutively, be the four vertices o f the Delaunay star that determine the extreme points o f L. We describe a sequence o f deformations o f the configuration (formed by the vertices vl . . . . . v4, v, v') that transform the original configuration o f vertices into particular rigid arrangements below. We show that these rigid arrangements cannot exist, and from this it follows that the original configuration does not exist either. These deformations preserve the constraints o f the problem. To be explicit, we assume that 2 _< Iwl _< 2.51, that 2 <_ [w - vii if w ~ vi, that 2 < Iv - v'[, and that [vi - vi+l [ _< 2.51, for i = 1, 2, 3, 4 and w = vl . . . . . v4, v, v'. Here and elsewhere we take our subscripts modulo 4, so that vl = v5, and so forth. The deformations also keep v and v' in the cone at the origin that is determined b y the vertices vi. We consider s o m e deformations that increase Iv - v'l. B y L e m m a 4.3, we m a y assume that Ivil = 2, lol = Iv'l = Ivi - vi+ll = 2.51, for i = 1 , 2 , 3 , 4 . If, for s o m e i, we have [vi - v[ > 2 and Ivi - v'l > 2, then we fix v i - i and vi+2 and pivot vi+l around the axis through the origin and vi+2 away from v and v'. T h e constraints Ivi+l - vii = Ivi - Vi-l[ = 2.51 force us to drag vi to a new position on the sphere o f radius 2. B y m a k i n g this pivot sufficiently small, we m a y a s s u m e that Ivi - w l and [Vi+l - w[, for w = v, v', are greater than 2. The vertices v and v' cannot both have distance 2 from both vi+2 and v i - l , for then we would have v = v'. So one o f them, say v, has distance exactly 2 f r o m at most one o f vi+2 and v i - i (say vi+2). Thus, v m a y be pivoted around the axis through the origin and vi+2 away from v'. In this way, we increase Iv - v'l until [vi - vl = 2 or Ivi - v'l = 2, f o r / = 1 , 2 , 3 , 4 . Suppose one o f v, v' (say v) has distance 2 from vi, vi+1, and 0i+2. The configuration is c o m p l e t e l y rigid. B y symmetry, the vertices v i - i and v' must be the reflections o f vi+l and v, respectively, through the plane through 0, vi, and 1)i+2. In particular, v' has distance 2 f r o m vl, v i - l , and vi+2. We pick coordinates and evaluate the length Iv - v'[. We find that Iv - v'[ ~ 1.746, contrary to the hypothesis that the centers o f the spheres o f our packing are separated by distances o f at least 2. Thus, the hypothesis that v has distance 2 f r o m three other vertices is incorrect. Diagram4.4 We are left with one o f the configurations of D i a g r a m 4.4. An edge is drawn in the diagram, w h e n the distance between the two endpoints has the smallest possible value (that is, 2.51 for the four edges o f the quadrilateral, and 2 for the remaining edges). D e f o r m the figure o f case (a) along the one remaining degree o f freedom until Iv - v'l = 2. In case (a), referring to the notation established by D i a g r a m 4.5, we have a quadrilateral on the unit sphere o f edges tl = 2 a r c s i n ( 2 . 5 1 / 4 ) ~ 1.357 radians, t2 = a r c c o s ( 2 . 5 1 / 4 ) radians, and t3 = 2 arcsin(1/2.51) radians. The f o r m o f this quadrilateral is determined by the angle a , and it is clear that the angle/~(c~) is decreasing in or. We have 0 = dihmax = dih(S(2.51, 2, 2, 2.51, 2, 2)) ~ 1.87A.A.AA. The figure exists if and only if there exists ot such that 3(t~) +/~(2zr - 0 - or) = 2Jr - 0. By symmetry, we m a y assume that ot < J r - 0 / 2 -~ 2.20437. The condition Iv - vl[ > 2 implies that However, by monotonicity, ~(ot) + ~(2:r - 0 - u) < ~(oti) -4-/~(2zr - 0 - ffi -- 0.1) < 2Zr -- 0, for ui < u < 0.1 + oti, with oti = 1.21 + 0.1 i, and i = 0, 1. . . . . 9, as a direct calculation o f the constants/~(ui) + fl(2zr - 0 - o~i - 0.1) will reveal. (The largest constant, which is about 27r - 0 - 0.113., occurs for i = 0.) Hence, the figure o f D i a g r a m 4.4(a) does not exist. tj VI Diagram4.5 To rule out Diagram 4.4(b), we reflect v, if necessary, to its image through the plane P through (0, vl, v3), so that P separates v and v'. The vertex v can then be pivoted a w a y from v' along the axis through vl and v3. This decreases Ivl, but we m a y rescale so that Ivl = 2.51. Eventually Iv - v21 = 2 or Iv - v41 = 2. This is the previously considered case in which v has distance 2 from three o f the vertices vi. This completes the p r o o f that the original arrangement o f two enclosed vertices does not exist. [] Restrictions I f a Delaunay star D* is c o m p o s e d entirely o f tetrahedra, then we obtain a triangulation o f the unit sphere. As explained in Section 2, we wish to prove that no matter what the triangulation is, we always obtain a score less than 8 pt. In this section we m a k e a long list o f properties that a configuration must have if it is to have a score o f 8 pt or more. The next section and the Appendix show that only one triangulation satisfies all o f these properties. Additional arguments will show that this triangulation scores less than 8 pt. This will complete the p r o o f o f T h e o r e m 1. In this section the term vertices refers to the vertices o f the triangulation o f the unit sphere. The edges o f the triangulation give a planar graph. We adopt the standard terminology o f graph theory to describe the triangulation. We speak o f the degree o f a vertex, adjacent vertices, and so forth. The n triangles around a vertex are referred to as an n-gon. We also refer to the corresponding n tetrahedra that give the triangles of the polygon. We say that a triangulation contains a pattern (al . . . . . a,), for ai E N, if there are distinct vertices vi of degrees ai that are pairwise nonadjacent, for i = 1. . . . . n. Let N be the number of vertices in the triangulation, and let Ni be the number of vertices of degree i. We have N = ~ Ni. In this section we start to use our various inequalities from Section 9 related to the score. Since the score tr(S) may be either F(S) or vor(S) depending on the circumradius of S, there are two cases to consider for every inequality. In general, the inequalities for F(S) are more difficult to establish. In the following sections we only cite the inequalities pertaining to F(S). Section 9 shows how all the same inequalities hold for vor(S). Proposition 5.1. Consider a Delaunay star D* that is composed entirely of quasiregular tetrahedra. Suppose that o (D*) > 8 pt. Then the following restrictions hold on the triangulation of the unit sphere given by the standard decomposition: 1. 1 3 < N < 15. 2. N = N 4 + N s + N 6 . 3. A region bounded by three edges is either a single triangle or the complement of a single triangle. 4. Two degree 4 vertices cannot be adjacent. 5. N4 < 2 . 6. Patterns (6, 6, 6) and (6, 6, 4) do not exist. 7. The pattern (6, 6) or (6, 4, 4) implies that N > 14. 8. If there are two adjacent degree 6 vertices, and a third degree 6 vertex adjacent to neither of the first two, then N = 15 and all other vertices are adjacent to at least one of these three. 9. The triangulation is made of geodesic arcs on the sphere whose radian lengths are between 0.8 and 1.36. L e m m a 5.2. Consider a vertex of degree n,for some 4 < n < 7. Let Sl . . . . . S, be the tetrahedra that give the n triangles. Then ~'~i~=lo (Si) is less than zn, where z4 = 0.33 pt, z5 = 4.52 pt, z6 = - 1.52 pt, and z7 = - 8 . 9 pt. Suppose that n > 6. Let Si . . . . . $4 be any four of the n tetrahedra around the vertex. Then ~'~.4i=l tr(Si) < 1.5 pt. Proof. When n = 4, this is Lemma 9.5. When n = 5, this is Lemma 9.6. When n = 6 or n = 7, we have, by Calculation 9.4, /I n ~-"~cr(Si) < ~'~(0.378979dih(Si) - 0.410894) i=1 i=1 = 2rr(f).~7R979) -- 0.410894n. The right-hand side evaluates to about -1.520014 pt and -8.940403 pt, respectively, when n = 6 and n = 7. Assume that n > 6, and select any four $I . . . . . $4 of the n tetrahedra around the vertex. Let $5 and $6 be two other tetrahedra around the vertex. The dihedral angles of $5 and $6 are at least dihmin, by Calculation 9.3. Each of the four triangles (associated with SI . . . . . $4) must then, on average, have an angle at most (2rr - 2 dihmin)/4 at v. By Calculation 9.4, 4 4 E a(Si) < Z ( 0 . 3 7 8 9 7 9 dih(Si) - 0.410894) i=1 i=l < (2Jr - 2 dihmin)0.378979 + 4(-0.410894) < 1.5 pt. [] Proof of Proposition 5.1. Lett = 2 ( N - 2 ) be the number oftriangles, an even number. By Euler's theorem on polyhedra, 3/73 + 2 N 4 + N 5 + 0 N 6 - N7 . . . . Suppose that t < 18. By Calculation 9.8, Suppose t < 24. For each vertex v, set (v = Y'~vtri, the sum running over the tetrahedra around v. Clearly, ~r cri = ( ~ v (v)/3. Pick a vertex v that is not a vertex of any of the seven triangles of the heptagon. By Lemma 5.2, we see that (~ < 0.33 pt if v has degree 4, (v < 4.52 pt if v has degree 5, (v < - 1.52 pt if v has degree 6, and (v < - 8 . 9 pt if v has degree 7. In particular, if v has degree n, then (~ falls short of n points by at least 0.48 pt. Thus, ~_, tri < ~ (t) (7) cri + ~_, pt - 0.48 pt < ( - 8 . 9 + (24 - 7) - 0.48) pt < 8 pt. (t-7) Finally, we assume that t = 26 and N = 15. If N6 = 0 and N7 = 1, then Euler's theorem gives the incompatible conditions 2N4 + N5 - 1 = 12 and N4 + N5 + 1 = 15. Thus, N6 > 0 or N7 > 1. This gives a second k-gon (k = 6 or 7) around a vertex v. This second polygon shares at most two triangles with the original heptagon. This leaves at least four triangles of a second polygon exterior to the first. Thus, by Lemma 5.2, E cri <_~_~ cri + E cri + E (26) (7) (4) ( t - 11) pt < - 8 . 9 pt + 1.5 pt + (26 - 11) pt < 8 pt. (Proof of 5.1.1, continued) Assume N = 12. We define three classes of quasi-regular tetrahedra. In the first class all the edges have lengths between 2 and 2.1. In the second class the fourth, fifth, and sixth edges have lengths greater than 2.1. The third class is everything else. Set e = 0.001, a = -0.419351, b = 0.2856354. The following are established by Calculations 9.10-9.12: or(S) < a sol(S) + b + e for S in the first class, a ( S ) _< a sol(S) + b for S in the second class, or(S) < a sol(S) + b - 5e for S in the third class. Consider a vertex v of degree n = 4, 5, or 6 and the surrounding tetrahedra Sl . . . . . Sn. We claim that ( This follows directly from the stated inequalities if none of these tetrahedra are in the first class. It is also obvious if at least one of these tetrahedra is in the third class, because then the inequality is violated by at most - 5 e + (n - 1)e _< 0. So assume that all of the tetrahedra are in the first two classes with at least one in the first class. By the restrictions on the lengths of the edges, a tetrahedron in the first class cannot be adjacent to one in the second class. We conclude that the tetrahedra are all in the first class. By Lemma 5.2, zn <_n(O.904) pt. If )-'~.(n) sol(S/) _ n(0.56176), then (n) SO we may assume that ~(n) sol(S/) > n(0.56176). By Calculation 9.13, sol(S) = 2arccot ( 2 - ~ / 2 ) . Proof. (See [H2, p. 64].) We use the branch of arccot taking values in [0, zr]. [] The function a is increasing in yl, Y2, and y3 on [21 4] and is decreasing in the variables y4, Ys, and Y6 on the same interval. We claim that a > 0 for any Delaunay simplex. We prove this in the case A > 0, leaving the degenerate case A = 0 to the reader. The area of the face T opposite the origin is at most 4~/3 (since its edges are all at most 4). There exists a plane that is tangent to the unit" sphere between this face and the unit sphere [ H I , 2.1]. The area of the radial projection of T to this plane is less than the area of a disk D of radius 1.5 on the plane centered at the point of tangency. The solid angle (that is, the area of the radial projection of T to the unit sphere) is less than the area of the radial projection of D to the unit sphere). This area is 2~r(1 - cos(arctan(1.5))) < zr. L e m m a 8.4.2 now gives the result. This allows us to use L e m m a 8.4.2 in the form sol(S) = 2 arctan(A1/2/(2a)). 8.4.3. The solid angle is the area of a spherical triangle. Let x, y, and z be the cosines of the radian lengths of the edges of the triangle. By the spherical law of cosines, the solid angle, expressed as a function of x, y, and z is c ( x , y , z ) + c ( y , z , x ) + c ( z , x , y ) - T r , c ( x , y , z ) := arccos The partial derivative with r e s p e c t to x of this expression for the solid angle is ( - 1 - x + y + z ) / ( ( x + 1)~/t), where t = 1 - x 2 - y2 _ z 2 + 2xyz. The second derivative of this expression evaluated at - 1 - x + y + z = 0 is - 2 ( 1 - y)(1 - z)t -3/2 < 0. So the unique critical point is always a local maximum. If.neither edge at a vertex of a spherical triangle is constrained, then the triangle can be contracted or expanded by moving the vertex. If the lengths of the edges are constrained to lie in a product of intervals, then the minimum area occurs when two of the edges are as short as possible and the third is at one of the extremes. The m a x i m u m area is attained when two of the edges are as long as possible and the third is at one of the extremes or at a critical point: x = y + z - 1 (or it symmetries, y = x + z - 1, z = x + y - 1). ( x y z ) ~/(1 - y T ) ( T - z 2 ) " The compression F(S) of a Delaunay simplex S is defined as (8.5.1) where (8.5.2) F(S) = -6oct vol(S) + )-'~4=1 soli (S) We let ao = a(yl, Y2, Y3, Y4, Y5, Y6), where a is the function defined in (8.4.1). We also let aj, a2, and a3 be the functions a for the vertices denoted vl, rE, and 03 in Diagram 8.1.2. For example, al = a(yl, Y5, Y6, Y4, Y2, Y3). Set t = -v/A/2. By the results of Sections 8.1 and 8.4, we find that F(S) = -8o~rt/6 + ~ 3o 2 arctan(t/ai). Recall that ai > 0 on a cell C. We wish to give an elementary upper bound of F on C. Set, for t >__0 and a = (a0, al, a2, a3) ~ Ii4+, - 3 ~ 6 y(t, a) := i=0 \ a i / " The partial derivative of y with respect to ai is - 2 t / ( 3 ( t 2 + a2)) < O, so y(t, a) < y(t, a-), where a - = (a0 , a I , a 2 , a 3) is a lower bound of a on C, as determined by Section 8.4. We study y(t) = y(t, a - ) as a function of t to obtain an upper bound on F (S). We note that y ' ( t ) = - - 8oct 6 Upper and lower bounds on t are known from Section 8.1. An upper bound on y for t in [train,tmax] is the maximum of s and s where e is the tangent to y at any point in [tn~n, tmax] Voronoi cells We assume in this section that the simplex S has the property that the circumcenter of each of the three faces with vertex at the origin lies in the cone at the origin over the face. This condition is automatically satisfied if these three faces are acute Mangles. In particular, it is satisfied for a quasi-regular tetrahedron. When the cone over a quasi-regular tetrahedron S contains the circumcenter of S, Section 2 sets vor(S) = -48o~t vol(S0) + 4 sol(S)/3, where S0 is the intersection of S with a Voronoi cell at the origin. Otherwise, vor(S) is defined as an analytic continuation. This section gives formulas for vol(S0). As usual, set S = S(yl . . . . . Y6) and xi = y2. Suppose at first that the circumcenter of S is contained in the cone over S. The polyhedron S0 breaks into six pieces, called the Rogers simplices. A Rogers simplex is the convex hull of the origin, the midpoint of an edge (the first, second, or third edge), the circumcenter of a face along the given edge, and the circumcenter of S. Each Rogers simplex has the form R = R(a, b, c) := S(a, b, c, (c2 - b2) I/2, (c 2 - a2) 1/2, (b2 -- a2) 1/2) for some 1 < a < b < c. Here a is the half-length of an edge, b is the circumradius of a face, and c is the circumradius of the Delaunay simplex. The volume is vol(R(a, b, c)) = a(b 2 - a 2 ) 1 / 2 ( c 2 - b2)1/2/6. The density 8(a, b, c) of R = R(a, b, c) is defined as the ratio of the volume of the intersection of R with a unit ball at the origin to the volume of R. It follows from the definitions that (8.6. l) vor(S) = E 4 vol(R(a, b, c))(-8oct + 8 (a, b, c)), where c is the circumradius rad(S), and (a, b) runs over the six pairs ( ~ , r/(yl, Y2, Y6)), ( 2 ' r/(y2, Y3, Y4)), Y3, -~ rl(y3, Yt, Ys)), ( 2 , rl(Yl, Y2, Y6)) , ( ~ , r/(y2, Y3, Y4)), ( 2 ' r/(y3, Yl, Ys')Upper and lower bounds on vol(R (a, b, c)) follow without difficult from upper and lower bounds on a, b, and c. Thus, an upper bound on 8 (a, b, c) leads to an upper bound on vor(S). The next lemma, which is due to Rogers, gives a good upper bound [R]. Lemma 8.6.2. l < a < b < c . The density 8(a, b, c) is monotonically decreasing in each variable for Proof. Let 1 _< al _< bi _< Cl, 1 _< a2 < b2 _< c2, a2 < al, b2 < bl, C2 ~ Cl. The points of R(al, bl, Cl) are realized geometrically by linear combinations sl = )`1 (al, 0, 0) + )`2(al, (b~ - al2)1/2, 0) + )`3(al, (b~ - a2) 1/2, (c2 - b2)1/2), where )`~, )`2, )`3 ~ 0 and )`l + ),2 + )`3 __5 1. The points of R(a2, b2, c2) are realized geometrically by linear combinations s2 = kl (a2, O, O) + Z2(a2, (b2 - a21)/2,2 O) + X3(a2, (b2 - a21)/2,2 (c2 - b2)1/2), with the same restrictions on )`i. Then )Sl 12 -- Isal 2 = )`~()`t + 2)`2 + 2)`3)(a~ - a 2) + )`2(L2 + 2)`3)(b~ - b~) + )`23(c2l - c~). So lsll 2 >__ Is212. This means that the linear transformation sl ~-~ s2 that carries the simplex $1 to $2 moves points of the simplex Sl closer to the origin. In particular, the linear transformation carries the part in $1 of the unit ball at the origin into the unit ball. This means that the density of R(al, bl, Cl) is at most that of R(a2, b2, c2). [] If the circumcenter of S is not in the cone over S, then the analytic continuation gives vor(S) = y ~ 4eR vol(R(a, b, c))(-8oa + 6(a, b, c)), R where eR = 1 if the face of the Delaunay simplex S corresponding to R has positive orientation, and eR = - I otherwise. (The face of S "corresponding" to R(a, b, c) is the one used to compute the circumradius b.) 8.6.3. A calculation based on the explicit coordinates of S and its circumcenter given in Lemma 8.1.4 shows that ( eR vol R (Yl ~-, r/(yl, Y2, Y6), rad(S) ) ) = XI(X2+ X6-- XI)X(X4, X5, X3,XI,X2, X6) 48U(Xl, X2, x6)A(Xl . . . . , X6) 1/2 If the circumcenter of S is not contained in S, then the same formula holds by analytic continuation. By definition, eR = - 1 exactly when the function X is negative. Although this formula is more explicit than the earlier formula, it tends to give weaker estimates of vor(S) and was not used in the calculations in Section 9. 8.6.4. There is another approximation to vor(S) that will be useful. Set Sy = S(2, 2, 2, y, y, y). (We hope there is no confusion with the previous notation Si.) For 1 _< a < b < c, let vol(R(a, b, c)) = a((b 2 - a2)(c 2 - b2))1/2/6 be as above. Set r(a) = vol(R(a, r/(2, 2, 2a), 1.41)). L e m m a 8.6.5. Assume that the circumradius of a quasi-regular tetrahedron S is a t least 1.41, and that 6 < Yl + Y2 + Y3 < 6.3. Set a = (Yl + Y2 + Y3 - 4)/2. Pick y to satisfy sol(Sy) = sol(S). Then Proof. If S is any quasi-regular tetrahedron, let Sm be the simplex defining the "tangent" Voronoi cell, that is, Sm is the simplex with the same origin that cuts out the same spherical triangle as S on the unit sphere, but that satisfies Yl = Y2 = Y3 = 2. The lengths of the fourth, fifth, and sixth edges of St~ are between ~/8 2(2.3) = ~ . 4 and 2.51. The faces of Sm are acute triangles. A calculation similar to the proof in paragraph 8.2.5, based on X (3.4, 3.4, 4, 4, 4, 2.512) > 0, shows that the circumcenter of St~ is contained in the cone over S ~ , Since S ~ is obtained by "truncating" S, we observe that vor(Stan) - 4~oct vol(S\Stan), where S and St~ are the pieces of Voronoi cells denoted $0 in Section 2 for S and Stan, respectively. By a convexity result of Fejes Ttth, vor(Stan) < vor(Sy) IF-F,p. 125]. (This inequality relies on the fact that the circumcenter of St~ is contained in the cone over S~..) Let al be the half-length of the first, second, or third edge. Since vol(R(al, b, c)) is increasing in c, we obtain a lower bound on vol(R (al, b, c)) for c = 1.41, the lower bound on the circumradius of S. The function vol(R(at, b, 1.41)), considered as a function of b, has at most one critical point in [aj, 1.41] and it is always a local maximum (by a second derivative test). Thus, (8.6.6) vol(R(aj, b, c)) > min(vol(R(al, brain, 1.41)), vol(R(al, bmax, 1.41))), where brain and bronxare upper and lower bounds on b ~ [al, 1.41]. A lower bound on b is 17(2,2, 2al), which means that vol(R(al, brain, 1.41)) may be replaced with r(al) in inequality (8.6.6). By Heron's formula, the function r/(a, b, c), for acute triangles, is convex in pairs of variables: l~aa?]bb -- l~2b = Thus, an upper bound on bmaxis 17(2,2.51,2a), where a = (Yl + Y2 + Y3 - 4)/2. This means that vol(R((al, bmax, 1.41))) may be replaced with the function vol(R(al, 17(2,2.51, 2a), 1.41)) in inequality (8.6.6). Now 1 < al < a, and vol(R(al, b, c)) is decreasing in the first variable (for 1 < at and b < ~f2), so we may use the lower bound vol(R(a, r/(2, 2.51, 2a), 1.41)) instead. By Calculations 9.20.2 and 9.20.3, we conclude that vol(R(al, b, c)) > r(a). This lower bound is valid for each Rogers simplex. The volume of S\Stan is then at least ( ( 1 - y ~ ) + (1 - ~23) + (1 - y ~ ) ) 2r(a)" The concavity of 1 - 8 / y 3 gives ~ - - l ( 1 - 8/y3i) > 1 - 1/a 3. We have established that vol(S\Smn) > 1 - ~ 2r(a). The result follows. [] 8.6.7. We conclude our discussion of Voronoi cells with a few additional comments about the case in which analytic continuation is used to define vor(S), with S a quasiregular tetrahedron. Assume the circumcenter c of S lies outside S and that the face T of S with negative orientation is the one bounded by the first, second, and sixth edges. It follows from Section 3 that Yl, Y2, Y6 ~ [2.3, 2.51] and Y3, y4, Y5 E [2, 2.15]. Let Pl (resp. P2) be the point on T equidistant from the origin, v3, and 1)1 (resp. i)2), iPll 2 xi = "-~ -I XlU(Xl, X2, X6)(--3Cl Jr- X3 Jr XS)2 4 ( a A / O X 4 ) 2 Let p0 be the circumcenter of T (see Diagram 8.6.8). V! V2 Pj 1/o Diagram 8.6.8 en = - I for the Rogers simplex with vertices the origin, P0, c, and vl/2. It lies outside S. The other Rogers simplex along the first edge has eR = l, so that the part common to both of these Rogers simplices cancels in the definition of vor(S). This means that vor(S) becomes the sum of the usual contributions from the two Rogers simplices along the third edge, 4 vol(R(a, b, c))(-8oct + 8(a, b, c)) for (a, b, c) -- (yl/2, JT(Yl,Y3, Y5), IPl]) and (y2/2, r/(y2, Y3, Y4), IP2I), and v(S') :-----48octvol(S") < (r/(yl, 2.51,2.51) 2 - Xl'~!/2__ Qpl(Y1, 2.51, 2, 2, 2, 2.51)12 _ X I ] 1/2 - 4/ 4 1 _< (~(2.51, 2.51, 2.51) 2 2"512) 1/2 / 2"15/1224) ~lpl (2.51, 2.51, 2, 2, 2, 2.51)12 - - < 0.1381. The inequality that replaces Yl with 2.51 results from Calculation 9.21. 8.7. A Final Reduction Let S be a Delaunay simplex. Suppose that the lengths of the edges Yl, Y5, and Y6 are greater than 2. Let S' be a simplex formed by contracting the vertex joining edges 1, 5, and 6 along the first edge by a small amount. We assume that the lengths Y'l, Ys, and y~ of the new edges are still at least 2 and that the circumradius of S' is at most 2, so that S' is a Delaunay simplex. Proposition 8.7.1. F(S') > r'(s). We write sol/, for i = 1, 2, 3, for the solid angles at the three vertices Pl, P2, and P3 of S terminating the edges 1, 2, and 3. Let P'l be the vertex terminating edge 1 of S'. Similarly, we write solI, for i = 1, 2, 3, for the solid angles at the corresponding vertices of S'. We set vol(V) = vol(S) - vol(S') and wi = s o l / - solI. It follows directly from the construction of S' that w2 and w3 are positive. The dihedral angle ot along the first edge is the same for S and S'. The angle ~i of the triangle (0, pl, Pi) at Pl is less than the angle fl~ of the triangle (0, P'l, Pi) at P'l, for i = 2, 3. It follows that Wl is negative, since - w j is the area of the quadrilateral region of Diagram 8.7.2 on the unit sphere. s' v, Lemma 8.7.3. 8octvol(V) > lo2/3 + w3/3. The lemma immediately implies the proposition because wl < 0 and r(s') - p(s) W2 3 W3 3 + 8o~tvol(V). Proof. Let T' be the face of S' with vertices p'p P2, and P3- We consider S' as a function of t, where t is the distance from Pl to the plane containing T'. (See Diagram 8.7.2.) It is enough to establish the lemma for t infinitesimal. As shown in Diagram 8.7.2, let Vl be the pyramid formed by intersecting V with the plane through Pl that meets the fifth edge at distance to = 1.15 from P3 and the sixth edge at distance to from P2- Also, let the intersection of V with a ball of radius to centered at Pi be denoted Vi, for i = 2, 3. For t sufficiently small, the region 111 is (essentially) disjoint from I12 and V3. We claim that vol(Vl) > vol(V2 N V3). Let 0 be the angle of T' subtended by the fifth and sixth edges of S'. Then vol(Vl) = Bt/3, where B is the area of the intersection of Vl and T': (8.7.4) B = sin O(y~ - to)(Y'6 - to) 2 Since S' is a Delaunay simplex, the estimates ~r/6 < 0 < 2zr/3 from [HI, 2.3] hold. In particular, sin0 > 0.5, so B > 0.25(2 - 1.15) 2, and vol(Vl) > 0.06t. If vol(V2 ~ V3) is nonempty, the fourth edge of S must have length less than 2t0. The dihedral angle a' of V along the fourth edge is then less than the tangent of the angle, which is at most t + O(t2), in Landau's notation. As in [HI, 5], we obtain the estimate vol(V2 f3 V3) < et' f This establishes the claim. Thus, for t sufficiently small ro t2 - t 2 dt = < 0.025t + O(t2). ~oct vol(V) >_ 8oct(vol(Vl) -]- vol(V2) '1- voI(V3) - voI(V2 f3 V3)) > ('oct/03' ( ? + - ~ ) > 1 " 0 9 ( 3 + 3 ) " 9. Floating-Point Calculations This section describes various inequalities that have been established by the method of subdivision on SUN workstations. The full source code (in C++) for these calculations is available [H6]. Floating-point operations on computers are subject to round-off errors, making many machine computations unreliable. Methods of interval arithmetic give users control over round-offerrors [AH]. These methods may be reliably implemented on machines that allow arithmetic with directed rounding, for example, those conforming to the IEEE/ANSI standard 754 [W], [IEEE], [P]. Interval arithmetic produces an interval in the real line that is guaranteed to contain the result of an arithmetic operation. As the round-off errors accumulate, the interval grows wider, and the correct answer remains trapped in the interval. Apart from the risk of compiler errors and defective hardware, a bound established by interval arithmetic is as reliable as a result established by integer arithmetic on a computer. We have used interval arithmetic wherever computer precision is a potential issue (Calculations 9.1-9.19, in particular). Every inequality of this section has been reduced to a finite number of inequalities of the form r(xo) < 0, where r is a rational function o f x ~ ]~n and x0 is a given element in ]Rn. To evaluate each rational expression, interval arithmetic is used to obtain an interval Y containing r(xo). T h e stronger inequality, y < 0 for all y e Y, which may be verified by computer, implies that r (x0) < 0. To reduce the calculations to rational expressions r(xo), rational approximations to the functions Vff, arctan(x), and arccos(x) with explicit error bounds are required. These were obtained from [H7]. Reliable approximations to various constants (such as rr, ~r and 8oct) with explicit error bounds are also required. These were obtained in Mathematica and were double checked against Maple. Let S = S(yl . . . . . Y6) be a quasi-regular tetrahedron. We label the indices as in Diagram 8.1.2. Let I" = F(S(yl . . . . . Y6)) be the compression. We also let dih = dih(S(yl . . . . . Y6)) be the dihedral angle along the first edge, and let sol = sol(S(yl . . . . . Y6)) be the solid angle at the origin, that is, the solid angle formed by the first, second, and third edges (yl, y2, Y3) of S. All of the following inequalities are to be considered as inequalities of analytic functions of Yl. . . . . Y6. Although each of the calculations is expressed as an inequality between functions of six variables, Proposition 8.7.1 has been invoked repeatedly to reduce the number of variables to three or four. For instance, suppose that we wish to establish/(sol, F) < 0, w h e r e / ( s o l , F) is an expression in F(S(yl . . . . . Y6)) and sol(S(yl . . . . . y6)). Invoking Proposition 8.7.1 three times, we may assume that the vertices marked vl, v2, and v3 in Diagram 8.1.2 each terminate an edge of minimal length. It is then sufficient to establish the inequality in seven situations of smaller dimension; that is, we may assume the edges i 6 I have minimal length, where I is one of {1, 4}, {2, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}, {1, 2, 3}. Similarly, for an inequality in F and dihedral angle, we reduce to the seven cases I = {1,4}, {2, 5}, {3, 6}, {1, 2, 3}, {1, 5, 6}, {3, 4, 5}, {2, 4, 6}. The first two calculations are inequalities of the compression F of a quasi-regular tetrahedron. C a l c u l a t i o n 9.1. P ~ 1 pt. This first inequality and Calculation 9.3 are the only ones that are not strict inequalities. Set So = S(2, 2, 2, 2, 2, 2). By definition, F(S0) = 1 pt. We must give a direct proof that So gives the maximum in an explicit neighborhood. Then we use the method of subdivision to bound F away from 1pt outside the given neighborhood. An infinitesimal version of the following result is proved in [HI]. L e m m a g . l . 1 . If y i ~ [ 2 , 2 . 0 6 ] , f o r i = 1 . . . . . 6, then F(S(yl . . . . . Y6)) < F ( S 0 ) , with equality if and only if S(yl . . . . . Y6) = So. Set aoo = a(2, 2, 2, 2, 2, 2) = 2 0 , Ao = A(22, 22, 22, 22, 22, 22), to = ~ / ~ o / 2 = 4V/-2, and bo = 2(1 + to2/ao2o) -1 = 50/81. Set f = maxi(yi - 2) _< 0.06 and a - = a(2, 2, 2, 2 + f , 2 + f , 2 + f ) = 20 - 1 2 f - 3 f 2. As in Section 8, we set t = ~r We have arctan(x) < arctan(xo) + (x - x o ) / ( 1 + x2), if x, xo > 0. We verify below that A >_ A o under the restrictions given above. Then (9.1.2) F ( S ) 8o, t - - g - - + 2 Z a r c t a n i=O t \ a i / < r ( s 0 ) = r ( S o ) + b o Z i=O ai _<F(So)+ (At-+t0A~( 8o~t24 bo a Set co = - 8 o a / 2 4 + b o / a - > 0. Write Yi = 2 + 35, with 0 < 35 < f . Set xi = 4 + ei = y/2 = 4 + 435 + f/2. Then 35 < el~4, for i = 1. . . . . 6. Set e = 4 f + f 2 . Set/~(el . . . . . er) = A(Xl . . . . . xr). We find that a2Zx/ae2 0et = 4 + e4 + e5 - e6 > 0 and similarly that 02A/OeiOel > 0, for i = 3, 5, 6. So 07~/Oej is minimized by taking e2 = e3 = e5 = e6 = 0, and this partial derivative is at least 0e~ (el, 0, 0, e4, 0, 0) = 16 - 8el - 2ele4 - e2 > 16 - 8e - 3e 2 > 0. Thus A > Ao. The partial derivative OA/Oel is at most ~ e l (el, e, e, e4, e, e) = 16 + 16e - 8el + nee4 - 2ele4 - e2 < 16 + 16e + 4ee4 - e42 < 1 6 + 1 6 e + 3 e 2. So A - Ao < (16 + 16e + 3e2)(el + . . . + e6). We expand )~-~=o(ai - a o o ) = 5(el + . . . +e6) + h i + h 2 + h 3 as a sum of homogeneous polynomials hi of degree i in f l . . . . . f6. A calculation shows that hi = 0. Also h2 is quadratic in each variable f / w i t h negative leading coefficient, so h2 attains its minimum at an extreme point of the cube [0, f]6. A calculation then shows that h2 > - 1 4 f 2 on [0, f]6. By discarding all the positive terms of h3, we find that h3 > -ft2f4- flf~- f/f5 - f2f/- f/f6- f3f~ > - 6 f 3. Thus, ~i=3o(ai - aoo) > 5(el +... + e6) - 14f 2 - 6 f 3. Since e = max(e/), we find thatf < e/4 _5<(el +--" + e6)/4.This gives + e6) --5 + . Similarly, we expand ~i=3o(aoo - ai) 2 as a polynomial in fl . . . . . f r . To obtain an upper bound, we discard all the negative terms of the polynomial and evaluate all the positive terms at (fl . . . . . f r ) = (f, . . . . f ) . This gives 3 E ( a o o - ai) 2 < 4 9 4 4 f 2 + 7 2 9 6 f 3 + 3 6 8 4 f 4 + 8 2 8 f 5 + 7 3 f 6 i----0 < 88 + . . . + e6)(4944f + 7 2 9 6 f 2 + 3 6 8 4 f 3 + 8 2 8 f 4 + 7 3 f 5 ) . F(S) (9.1.3) (el + F(So) ~-~6e6) -< We now insert these estimates back into inequality (9.1.2). This gives ~ - ( 1 6 + 16e + 3e2) + b0t0 ._5 + 3 . 5 f + 1 . 5 f 2) 0 a~0 ( The right-hand side of this inequality is a rational function of f . (Both a - and e depend on f . ) Each of the three terms on the right-hand side is increasing in f . Therefore, the right-hand side reaches its maximum at f = 0.06. Direct evaluation at f = 0.06 gives F(S) - F(So) < -0.00156(el + . . - + e6). To verify Calculation 9.14, the computer examined only 7 cells. Calculation 9.1 required 1899 ceils. Calculation 9.6.1 required over 2 million cells. The number of cells required in the verification of the other inequalities falls between these extremes. Calculationg.2. F < 0.5pt, i f y l ~ [2.2, 2.51]. The next several calculations are concerned with the relationship between the dihedral angle and the compression. Calculation 9.3. dih(S(yl . . . . . Y6)) > dihmin := dih(S(2, 2.51, 2, 2, 2.51, 2)) 0.8639. Since this bound is realized by a simplex, we must carry out the appropriate local analysis in a neighborhood of S(2, 2.51, 2, 2, 2.51, 2). L e m m a 9.3.1. Suppose that2 < Yi < 2.2,fori = 1,3,6, andthat2 < Yi = 2.51,for i = 2, 4, 5. Then dih(S(yl . . . . . Y6)) > dihmin. Proof This is an application of Lemma 8.3.1. By that lemma, dih(S) is increasing in x4, so we fix x4 = 4. The sign of Ocosdih/Ox2 is the sign of OA/Ox3, and a simple estimate based on the explicit formulas of Section 8 shows that OA/Ox3 > 0 under the given constraints. Thus, we minimize dih(S) by setting x2 = 2.512. By symmetry, we setx5 = 2.512. Now consider dih(S) as a function of xl, x3, and x6. The sign of Ocos dih/Ox3 is the sign of OA(xl, 2.512, x3, 22, 2.512, x6)/Ox2. The maximum of this partial (about -2.39593) is attained when Xl, x3, and x6 are as large as possible: Xl = x3 = x6 = 2.22. So dih(S) is increasing in x3. We take x3 = 4, and by symmetry x6 = 4. We have 0 cos dih(S) OXl 2q (x I) U(Xl, X2, X6)3/2U(Xl,X3, X5)3/2' where tl(Xl) ,~, 464.622 - 1865.14xl + 326.954x~ - 92.9817x~ + 4x 4. An estimate of the derivative of q (x) shows that ti (x) attains its maximum at xl = 4, and tl (4) < - 6 6 9 1 < 0. Thus dih(S) is minimized when Xl = 4. [] Calculation 9.4. 1-" < 0.378979 dih - 0.410894. Lemma 9.5. If $1, $2, $3, and $4 are any four tetrahedra such that dih(S]) + dih(S2) + dih(S3) + dih(S4) > 2zr, then F(S1) + . . . + F(S4) < 0.33 pt. This lemma is a consequence of the following three calculations, each established by interval arithmetic and the method of subdivision. Caleulation 9.5.1. provided dih ~ [dih(S(2, 2, 2, 2, 2, 2)), 1.42068]. Calculation 9.5.2. F < -0.19145 dih + 0.2910494, F < -0.0965385 dih + 0.1562106, provided dih ~ [1.42068, dih(S(2, 2, 2, 2.51, 2, 2))]. Calculation 9.5.3. F < -O.19145dih+0.31004, provided dih > dih(S(2, 2, 2, 2.51,2, 2)). To deduce Lemma 9.5, we consider the piecewise linear bound s on F obtained from these estimates. The linear pieces are s < 1 pt, s C3, and Ca on [dihmin,d]], [dl, d2], [d2, d3], and [d3, dihmax],where dl = dih(S(2, 2, 2, 2, 2, 2)), d2 = 1.42068, and d3 = dih(S(2, 2, 2, 2.51, 2, 2)). (See Calculation 9.1 and Lemma 8.3.2.) Diagram 9.5.4 illustrates these linear bounds. (There are small discontinuities at dl, d2, and d3 that may be eliminated by replacing s by s + ei, for some ei > 0, for i = 1, 2, and 4.) We then ask for the maximum of s + s + s + s under the constraint (h +t2+ta+t4)/4 > zr/2 ~ [dE, d3]. Since s is constant on (dihmin,dl) and decreasing on [dl, dihmax], we may assume that ti > d], for all i. Since the slope of s is equal to the slope of s we may assume that ti > dE, for all i, or that ti < d3, for all i. If ti < d3, for all i, then we find that ti > 2rr - 3d3 > dE. So in either case, ti > dE, for all i. By convexity, an upper bound is 4e3 (rr/2) < 0.33 pt. 11 d.1~2 l13 ~14 lpt 0pt Diagram 9.5.4 Lemma 9.6. /fSl . . . . . $5 are anyfive tetrahedra such that dih(S1) + . - . + dih(Ss) >_ 2rr, then F(S]) + . . . + F(Ss) < 4.52 pt. This is a consequence of two other calculations. C a l c u l a t i o n 9.6.1. F < - 0 . 2 0 7 0 4 5 dih + 0.31023815, provided dih ~ [do, 27r - 4 d o ] , where do = dih(S(2, 2, 2, 2, 2, 2)). C a l c u l a t i o n 9.6.2. F < 0.028792018, if dih > 2rr - 4 d 0 . 1 pt 0 pt Diagram 9.6.3 C a l c u l a t i o n 9.7. F < 0.389195 d i h - 0.435643, if yl > 2.05. The next inequalities relate the solid angles to the compression. C a l c u l a t i o n 9.8. F < - 0 . 3 7 6 4 2 1 0 1 sol + 0.287389. C a l c u l a t i o n 9.9. F < 0.446634 sol - 0.190249. C a l c u l a t i o n 9.10. 1. . . . . 6. F < - 0 . 4 1 9 3 5 1 sol + 0.2856354 + 0.001, if yi ~ [2, 2.1], for i = The following calculation involves the circumradius. We leave it to the reader to check that the dimension-reduction techniques o f L e m m a 8.7.1 m a y still be applied. C a l c u l a t i o n 9.11. I" < - 0 . 4 1 9 3 5 1 sol + 0.2856354, provided that Y4, Ys, Y6 > 2.1 and that the circumradius o f S is at most 1.41. C a l c u l a t i o n 9 . 1 2 . i, and y4 ~ [2, 21 ]. F < - 0 . 4 1 9 3 5 1 sol + 0 . 2 8 5 6 3 5 4 - 5(0.001), i f y i > 2.1 for s o m e C a l c u l a t i o n 9.13. F < - 0 . 6 5 5 5 7 sol + 0.418, i f y l ~ [2, 2.1], for i = 1. . . . . 6. Calculation 9.14. sol(S) > 0.21. Calculation 9.15. F + (K - sol)/3 < 0.564978 dih - 0.614725, where K (47r - 6.48)/12, provided Yl ~ [2, 2.05], and Y2, Y3 E [2, 2.2]. Calculation 9.16. dih > 0.98 and sol > 0.45, provided Yl e [2, 2.05], and Y2, Y3 [2, 2.2]. Let F(S) be replaced by vor(S) in each of the Calculations 9.1-9.16 to obtain a new list of inequalities 9.1'-9.16'. (In 9.11' we drop the constraint on the circumradius of S.) We claim that all of the inequalities 9.*' hold whenever S is a quasi-regulartetrahedron of circumradius at least 1.41. In fact, inequalities 9. !', 9.2', 9.4', 9.5.1 ', 9.5.2', 9.5.3', 9.6.1 ', 9.6.2', and 9.7' follow directly from 9.17 and the inequalities dihmin < dih < dihmax. Calculations 9.3, 9.14, and 9.16 are independent of F, and so do not require modification. Inequalities 9.8', 9.11', and 9.12' also rely on 9.17, 9.18, and 9.19, inequality 9.9' on 9.14, and inequality 9.15' on 9.16. Inequalities 9.10' and 9.13' are vacuous by the comments of paragraph 8.2.4. Write S = S(yl, y2, Y3, y4, ys, y6), vor = vor(S), and let rad = rad(S) be the circumradius of S. L e m m a 9.17. If the circumradius is at least 1.41, then vor < - 1 . 8 pt. ProoJ~ If sol > 0.91882, then the lemma is a consequence of Lemma 9.18. (The proof of Lemma 9.18, under the restriction sol > 0.91882, is independent of the proof of this lemma.) Assume that sol < 0.91882. If S' and S and Delaunay stars, related as in Section 8.7, then vor(S') > vor(S) because S~ is obtained by slicing a slab from 30. This means that we may apply the dimension-reduction techniques described at the beginning of this section, unless the deformation decreases the circumradius to 1.41. An interval calculation establishes the result when yl + Y2 + Y3 > 6.3 and the circumradius constraint is met (Calculation 9.17.1). When the dimension-reduction techniques apply, the dimension of the search space may be reduced to four, and an interval calculation similar to the others in this section gives the result (Calculation 9.17.2). We lacked the computer resources to perform the interval analysis directly when the dimension-reduction techniques fail and found it necessary to break the problem up into smaller pieces when Yl + Y + 2 + Y3 < 6.3. We make use of the following calculations. Calculation 9.17.1. vor(S) < - 1.8 pt provided yl + y2 Jr Y3 > 6.3, rad(S) = 1.41, and sol(S) < 0.91882. Calculation 9.17.2. vor(S) < -1.8ptprovidedthatrad(S) > 1.41,yl+y2-I-y3 >_6.3, sol(S) _< 0.912882, and S lies in one of the seven subspaces of smaller dimension associated with I (sol, F) at the beginning of the section. Assume that yl + y2 + Y3 < 6.192 and that rad > 1.41. Then Assume that yl + Y2 + Y3 < 6.106 and that rad > 1.41. Then Assume that Yl + Y2 + Y3 < 6.064 and that tad > 1.41. Then Assume that Yl + Y2 -k Y3 < 6.032 and that tad > 1.41. Then In the interval arithmetic verification of Calculations 9.17.3, we may assume that rad = 1.41 and that Yl + Y2 + Y3 is equal to the given upper bound 6.3, 6.192, etc. To see this we note that the circumradius constraint is preserved by a deformation of S that increases Yl, Y2, or Y3 while keeping fixed the spherical triangle on the unit sphere at the origin cut out by S. We increase Yl, y2, and Y3 in this way until the sum equals the given upper bound. Then fixing Yl, y2, and Y3, and one of y4, ys, and Y6, we decrease the other two edges in such a way as to decrease the solid angle and circumradius until rad = 1.41. This deformation argument would break down if we encountered a configuration in which two of Y4, Ys, and Y6 equal 2, but this cannot happen when rad(S) > 1.41 because this constraint on the edges would lead to the contradiction 1.41 < rad(S) < rad(S(2.3, 2.3, 2.3, 2, 2, 2.51)) < 1.39 It is possible to reduce Calculations 9.17.3 further to the four-dimensional situation where two of y4, Y5 and Y6 are either 2 or 2.51. Consider a simplex S with vertices 0, 01, 02, and 03. Let pi be the corresponding vertices of the spherical triangle cut out by S on the unit sphere at the origin. Fix the origin, v2, and 03, and vary the vertex 01. The locus on the unit sphere described as pl traces out spherical triangles of fixed area is an arc of a Lexell circle C. Define the "interior" of C to be the points on the side of C corresponding to spherical triangles of smaller area. The locus traced by Vl on the circumsphere (of S) with Iv~l constant is a circle. Let C', also a circle, be the radial projection of this locus to the unit sphere. Define the "interior" of C' to be the points coming from larger Ivl I. The two circles C and C' meet at Pl, either tangentially or transversely. The interior of C cannot be contained in the interior of C' because the Lexell arc contains p*2, the point on the unit sphere antipodal to P2 [FT, p. 23], but the interior of of C' does not. Furthermore, if Iv~l = 2 and I021 + I031 > 4, then 02 or 03 lies in the interior of C and C', so that the circles have interior points in common. This means that 01 can always be moved in such a way that the solid angle is decreasing, the circumradius is constant, and the length I0~1 is decreasing (or constant if loll = 2). If any two of y4, Ys, and Y6 are not at an extreme point, this argument can be applied to Vl, v2, or v3 to decrease the solid angle. This proves the reduction. We are now in a position to prove the lemma for simplices satisfying yl +Y2 +Y3 < 6.3. Calculation allows us to assume sol > 0.767. As in Section 8.6, let Sy = S(2, 2, 2, y, y, y). We rely on the fact that vor(Sy)is decreasing in y, for y 6 [2.26, 2.41 ]. In fact, the results of Section 8.6 specialize to the formula (9.17.4) and the sign of its derivative is determined by a routine Mathematica calculation. It is clear that sol(Sy) is continuous and increasing in y. Since so1($2.26) < 0.767 and sol(S2.41) > 0.91882, our conditions imply that sol(S) = sol(Sy) for some y ~ [2.26, 2.41]. Let r(a) and vol(R(a, b, c)) be the functions introduced in Section 8.6.4. This suggests the following procedure. Pick y so that sol(S) > sol(Sy). Calculate the smallest (or at least a reasonably small) a for which 2r(a) Proof. We adopt the notation and techniques of Lemma 9.17. If sol < 0.918819, then the result follows from Lemma 9.17. (The proof of Lemma 9.17 is independent of the argument that follows under that restriction on solid angles.) Let f ( S ) = -0.419351 sol(S) + 0.2856354 - vor(S). We show that f ( S ) is positive. We use the inequality f ( S ) > f(Stan) "+"4oct vol(S\St~) of Lemma 8.6.4. A routine calculation based on formula (9.17.4) shows that f ( S y ) is increasing, for y ~ [2.4085, 2.51]. If sol > 0.951385, then we appeal to Fejes T6th's convexity argument described in Section 8.6. To justify the use of his argument, we must verify that the cone over Stan contains the circumcenter of S. The first, second, and third edges have length 2, and the fourth, fifth, and sixth edges are between 2.21 and 2.51 (Calculation 9.18.2). These are is less than - 1 . 8 pt. Monotonicity (Calculation 9.20.1) and Lemma 8.6.5 imply that vor(S) < - 1 . 8 p t , ifyl +Y2+Y3 > 2 ( 2 + a ) . To treat the case that remains (Yl + y 2 + y 3 < 2(2 + a)), use Calculations 9.17.3 to obtain a new lower bound on sol(S), and hence a new value for y. The procedure is repeated until a = 0.016. Calculation completes the argument by covering the case Yl + Y2 + Y3 < 6.032. We leave it to the reader to check that ~(2.2626, 1.096), ~(2.326, 1.053), ~(2.364, 1.032), ~(2.391, 1.016) are less than - 1 . 8 pt and that so1($2.2626) < 0.767, so1($2.326) < 0.83, so1($2.364) < 0.87, so1($2.391) < 0.9. This completes the proof of Lemma 9.17. L e m m a 9.18. Ifrad(S) > 1.41, then vor < -0.419351 sol +0.2856354. stronger restrictions on the edges than in the proof of Lemma 8.6.5, so the justification there applies here as well. We observe that SO1($2.4366) < 0.951385 so that f ( S ) > f(Stan) > f(Sy) > f($2.4366) ~ 0.000024 > 0, where y satisfies sol(Sy) = sol(Stan). We may assume that 0.918819 < sol < 0.951385. L e m m a 9.18.1. The combined volume of the two Rogers simplices along a common edge of a quasi-regular tetrahedron S is at least 0.01. Proof. The combined volume is at least that of the right-circular cone of height a and base a wedge of radius v ' ~ - a 2 and dihedral angle dihmin, where a is the half-length of an edge and b is a lower bound on the circumradius of a face with an edge 2a. This gives the lower bound of We minimize b by setting b2 = )?(2, 2 , 2 a ) 2 = 4 / ( 4 - a 2 ) . Then b 2 - a 2 = (a 2 - 2)2/(4 - a2), which is decreasing in a ~ [1, 2.51/2]. Thus, we obtain a lower bound on b 2 - a 2 by setting a = 2.51/2, and this gives the estimate of the lemma. [] We remark that so1($2.4085) < 0.918819. If 1.15 < (Yl + y2 + Y3- 4)/2, then Lemma 9.18.1 and Section 8.6 give f(S)>f(S2.4o85)+48oct(l_ 1.1531) 0 . 0 1 > 0 . (Analytic continuation is not required here, because of the constraints on the edges in Calculation 9.18.2.) We may now assume that Yl + Y2+ Y3 < 6.3. To continue, we need a few more calculations. Calculation 9.18.2. If sol > 0.918, then y4, Ys, y6 > 2.21. Calculation If yl + y2 + Y3 < 6.02 and rad > 1.41, then sol > 0.928. Calculation I f y l + Y2 + Y3 < 6.0084 and rad > 1.41, then sol > 0.933. Calculation If yl + Y2 + y3 < 6.00644 and rad > 1.41, then sol > 0.942. In the verification of these calculations, we make the same reductions as in Calculations 9.17.3. Adapting the procedure of Lemmas 9.17 and 8.6.5, we fine that a lower bound on the solid angle leads to an estimate of a constant a with the property that f (S) > 0 whenever Yl +Y2 +Y3 > 4 + 2 a . That is, we p i c k a so that ~l(y, a) := f(Sy)+88oa(1 - 1/a2)r(a) is positive, where y is chosen so that sol(Sy) is a lower bound on the solid angle. The values are all positive. This yields the bound Yl + Y2 + Y3 < 6.0034. Assume that S satisfies Yl + y2 + Y3 < 6.0034 and rad(S) > 1.41. Then by Calculation, sol(S) = sol(Stun) _> 0.942. Also rad(St~) > rad(S)/1.0017 > 1.41/1.0017 > 1.4076, because rescaling St~ by a factor of 1.0017 gives a simplex containing S. This means that St~ satisfies the hypotheses o f the following calculation. Calculation 9.18.4 completes the proof of Lemma 9.18. [] Calculation9.18.4. f (Stan) > O. If rad(St~ > 1.4076 and 0.933 < sol(St~) < 0.951385, then In this verification we may assume that the fourth, fifth, and sixth edges of St~ are at least 2.27, for otherwise the circumradius is at most rad(S(2, 2, 2, 2.27, 2.51, 2.51)) < 1.4076. In the verification of Calculation 9.18.4, we also rely on the fact that fl(X4, X5,X6) :----f ( S ( 2 , 2, 2, ~ / ~ , ~r ~ ) ) is increasing in (x4, x5, x6) ~ [2.272, 2.512]. Here is a sketch justifying this fact. The details were carried out in Mathematica with high-precision arithmetic. The explicit formulas of Section 8.6 lead to an expression for Ofl/Ox4 as W(x4, X5, X6)(-X4 "1-X5 "~ X6) ( - 1 6 + Xa)2A 3/2 where W is a polynomial in x4, xs, and x6 (with 13 terms). To show that W is positive, expand it in a Taylor polynomial about (x4, xs, x6) = (2.272, 2.272, 2.272) and check that the lower bound of inequality (7.1) is positive. Calculation 9.19. If y4 ~ [2, 2.1], then sol.< 0.906. For a _< b ___ c, we have vol(R(a, b, c)) = a((b 2 - a2)(c 2 - b2))1/2/6. The final four calculations are particularly simple (to the extent that any of these calculations are simple), since they involve a single variable. They were verified in Mathematica with rational arithmetic. Calculation 9.20.1. on [1, 1.15]. Calculation 9.20.2. [1, 1.151. The function (1 - 1/a 2) vol(R(a, 77(2, 2, 2a), 1.41)) is increasing The function vol(R(a, 17(2,2, 2a), 1.41)) is decreasing for a Calculation 9.20.3. For a E [1, 1.15] we have vol(R(a, 17(2,2, 2a), 1.41)) < vol(R(a, 17(2, 2.51, 2a), 1.41)). Calculation 9.21. Let pl -- Pl (Yl. . . . . Y6) be the point in Euclidean space introduced in Section 8.6.7. For y e [2.3, 2.51], d y 0(2.51,2.51, y ) > - 0 . 7 5 > ~ y y Ip~(y, 2.51,2,2,2,2.51)l 2 Acknowledgments I would like to thank D. J. Muder for the Appendix and the referees for suggesting other substantial improvements. Appendix. Proof of Theorem 6.1 D. J. Muder Notation and Observations. Let P be a point of degree d. If we consider P to be the center of the configuration, then thefirst rim of points will be the set of d points adjacent to P, and the second rim will be those at distance 2 from P. Let 8(P) be the sum of the degrees of the first rim points. If d = 6, it is easy to see that the number of points on the second rim is s = 8(P) - 24, and the total number of points of distance at most 2 from P is 8(P) - 17. The second rim is thus an s-gon. This s-gon and the triangulation of the P-side of the s-gon are referred to as the inner graph. The inner-graph degree of a second rim point is called its inner degree. The number of second-rim points with inner degree 5 is equal to the number of degree 4 points in the first rim, and the number of second-rim points with inner degree 3 is equal to the number of degree 6 points on the first rim. All other second-rim points have inner degree 4. Points beyond the second rim are called extra points. If there are no extra points, then there are at least two second-rim points whose degrees are equal to their inner degrees. Let N~ (P) be the number of points of degree A in rim i around P. Notice that if d = 6, then a(P) = 30 - N~(P) + N~(P). Further, Euler's formula gives N - 12 = N6 - N4. Putting this together with 5.1.1 and 5.1.5, we see 13_< 1 2 + N 6 - N 4 _ < 15, l <_ l + N4 <_N6 <_3+ N4 <_5. L e m m a 1. There is no triangle whose vertices all have degree 6. L e m m a 2. Suppose N6 > 3. Let G6 be the subgraph of points of degree 6. Either G6 is three points in two components, or it is a single (open or closed) path, with no other edges. With these lemmas and this notation, we consider the different possible values of N6. N6 = 1. This forces N4 = 0. Let P be the degree 6 point. Then 8(P) = 30, all thirteen points are in the inner graph, and all six second-rim points have inner degree 4. However, at least two of these second-rim points have inner degree equal to degree. Contradiction. N6 = 2. If the two degree 6 points are adjacent, let P be either one. Now 8 (P) = 31 - N4~, and we see that there are no extra points, and all second-rim points have degree 5. All but N41 second-rim points have inner degree less than 5, and at least two of them will have inner degree equal to degree. So N4 > N4l = 2, and N < 12. Contradiction. If the degree 6 points are not adjacent, then the (6, 6) forces N > 14, so N4 = 0. If 8(P) is either of the two degree 6 points, 8(P) = 30, so the second rim has six points, all of inner degree 4, and there is one extra point. If the extra point has degree 5, it is adjacent to all but one of the second-rim points. This unique second-rim point is then part of a quadrilateral, which can only be triangulated by the diagonal edge that does not include the extra point. This creates two second-rim points of degree 6, in violation of our assumptions. Therefore the extra point has degree 6, and we have Diagram 6.2. N6 = 3. Either G6 is an open path or it has two components. In either case there is a (6, 6), so N > 14 and Na = 0 or 1. In the first case let P be the center of the path. Now 8 (P) = 32 - N4l . We see that there are no extra points and any degree 4 point is in the first rim. The second rim has at most one point of inner degree greater than 4, and two points whose inner degrees are equal to their degrees. So there must be a point of degree 4 on the second rim. Contradiction. In the second case 5.1.8 forces N = 15, hence N4 = 0. If P is one of the points in the two-point component of G6, then 8(P) = 31, so there is one extra point and seven second-rim points. Either one or two of the second rim points is not connected to the extra point. In either case at least one of these two has degree = inner degree, which is at most 4. Contradiction. N6 = 4. Now G6 is either an open or closed path of length 4. If open, then, by 5.1.8, N = 15 and N4 = 1. If P is either of the interior degree 6 points, then 8 (P) = 32 - N41, the second rim is an (8 - Nl)-gon with N~ points of inner degree at least 5 and N41extra points. We cannot have N4l = 0, or else there would be at least two second-rim points of degree 4, so N~ = 1. This holds no matter which interior degree 6 point we started with, so the degree 4 point must be adjacent to both. It must also be adjacent to one of the other two, or else there is a (6, 6, 4). Therefore there is a degree 6 point whose firstrim degree sequence is 6 / 4 / 6 / 5 / 5 / 5 , producing a second-rim inner degree sequence of 4/3/5/3/4/4/4. The point of inner degree 5 is the fourth neighbor of the degree 4 point, and cannot be degree 6 without making G6 closed. Therefore there is an edge connecting the two points of inner degree 3. Adding this edge to the inner graph creates a hexagon with all points of "inner" degree 4, and the extra point in its interior. The extra point is connected to five of these points, and the sixth has degree 4. Contradiction. If G6 is a closed path, then a degree 4 point Q must triangulate this quadrilateral. Using Q as the center, there are four first-rim points and eight second-rim points. The second rim can be drawn in a square with the four points of inner degree 3 on the comers and the four points of inner degree 4 at the midpoints. There is at most one additional degree 4 point (other than Q). If it is anywhere but at one of the second-rim midpoints, a (6, 6, 4) exists. The four midpoints cannot be adjacent to any additional second-rim points without creating an illegal triangle or quadrilateral. Since there can be at most one more degree 4 point, at least three of the four second-rim midpoints must be joined to the extra point(s). Two midpoints of consecutive sides cannot be joined to the same extra point without forcing the corresponding comer to be degree 4, which is impossible. So there must be two extra points, and thus no degree 4 points other than Q. One extra point must be joined to the midpoints of each pair of parallel sides of the square. However, all these edges cannot be drawn without intersecting. N6 = 5. Now G6 is a path of length 5, closed so that a (6, 6, 6) does not exist. Any point in this pentagon is nonadjacent to two others, so the situation of 5.1.8 applies in five different ways. Let P, Q, R ~ G6 with P and Q adjacent, but neither adjacent to R. Then there are eight points adjacent to either P or Q, and six points adjacent to R, so there are exactly two common points in these two sets. In our case these two points are precisely the other two points of degree 6. This means that no point inside or outside the pentagon can be connected to more than two points of the pentagon. Thus the only possible configuration for the 15 points is to form three concentric pentagons, with G6 as the middle one. Prior to triangulating the inner and outer pentagons, all of their points have degree 4. Triangulating either creates more degree 6 points. Contradiction. Proofof Lemma 1. Let Pl, P2, and P3 be the vertices of such a triangle. If Pj has no neighbors of degree 4, then the only possibility is 8(Pj) = 32, which forces Ng = 4, N~ --- 2, and N = 15. All second-rim points have inner degree at most 4, and there are no extra points. Therefore two second-rim points have degree and inner degree 4. Call them QI and Q2. Therefore, N4 > 2, N6 _> 5, and N2 _> 2. Let Rl and R2 be second-rim points of degree 6. Each Qk and Re must be adjacent, or else (Pj, Re, Qk) is a (6, 6, 4). However, all edges from Qj are inner-graph edges. So QI R1Q2R2 must be a second-rim quadrilateral, which is impossible. Thus N4~(Pj) _> 1 for each Pj. This forces N4 = 2. At least one of the degree 4 points is adjacent to two of the Pj. One of these two, say P1, is not adjacent to the other degree 4 point. Since 8(P) _< 32, we are left with four possible first-rim degree sequences for/'1: (1) 6 / 6 / 4 / 5 / 5 / 5 ; (2) 6 / 6 / 4 / 6 / 5 / 5 ; ( 3 ) 6 / 6 / 4 / 5 / 6 / 5 : or (4) 6 / 6 / 4 / 5 / 5 / 6 . The last three possibilities are easily dealt with. Sequence ( 3 ) contains a (6, 6, 4). Sequence (4) creates two 6-6-6 triangles joined at an edge. Let PI and P2 be the vertices of the edge, and let P3 and P4 be the other vertices of the triangles. The conditions N41(Pj) _> 1 and N4 = 2 force/'3 and P4 to have sequence (1). In (2) we can invoke 5.1.8 with the three degree 6 points in the first rim. Call them Rl, R2, and T, with RI adjacent to RE. In order for the numbers to work out, there can be precisely two points which are adjacent to T as well as one of the Rj. However, in (2) the center, the first-rim degree 4 point, and the second-rim neighbor of the first-rim degree 4 point must all fit this description. Sequence (1) is m o r e difficult to eliminate. Let Rl . . . . . R6 be the first-rim points listed in the order o f (1). Let $1 . . . $7 be the second-rim points, with $I being adjacent to both Rj and R2, and $7 having only Rl as a first-rim neighbor. There is at least one degree 6 point in the second r i m - - i f it is at Sl, then 8(Rl) ----32; if it is anywhere else we can invoke 5.1.8. In either case, N = 15 and there are two degree 6 points outside the first rim. These must lie in {$6, $7, Sl, $3} to avoid making a (6, 6, 4) with R1 and R3. At m o s t one o f these points can be adjacent to Rl (since 8 ( R l ) < 33), so $3 m u s t have degree 6. Now $3 must connect to both the degree 6 and degree 4 points in {$6, $7, $1 } to avoid making a (6, 6, 6) or (6, 6, 4) with the center point. However, $3 has inner degree 5 and so can connect to at most one o f those points. Contradiction. 8o. (A - Ao ) 24 (t + to) 3 t - t o Calculation 9.17.3 .1. sol > 0 . 767 . Calculation 9.17.3 .2. sol > 0 . 83 . Calculation 9.17.3 .3. sol > 0 . 87 . Calculation 9.17.3 .4. sol > 0 .9. Calculation 9.17.3 .5. sol > 0 . 91882 . Assume that yl + y2 -I- Y3 < 6.3 and that rad > 1 . 41 . Then [AH] G. Alefeld and J. Herzeberger ,Introduction to Interval Computations, Academic Press, New York, 1983 . [FT] L. Fejes T6th , Lagerungen in der Ebene, auf der Kugel und im Raum , Springer-Verlag,Berlin, 1953 . [H1] T. C. Hales , Remarks on the density of sphere packings in three dimensions , Combinatorica , 13 ( 2 ) ( 1993 ), 181 - 197 . [H2] T. C. Hales , The sphere packing problem , J. Comput. Appl . Math, 44 ( 1992 ), 41 - 76 . [H3] T. C. Hales, The status of the Kepler conjecture , Math. lntelligencer, 16 ( 3 ) ( 1994 ), 47 - 58 . [H4] T. C. Hales , Sphere packings, II, Discrete Comput . Geom., to appear. [H5] T. C. Hales , Sphere packings, Ilia, in preparation. [H6] T. C. Hales , Packings, http://www-personal.math.lsa.umich.edu/~hales/packings.html [H7] J. E Hart et al., Computer Approximations , Wiley,New York, 1968 . [IEEEI IEEE Standard for Binary Floating-PointArithmetic , ANSI/IEEEStd. 754- 1985 , IEEE, New York. [PI W. H. Press et al., Numerical recipes in C, Less-Numerical Algorithms , second edition, Cambridge UniversityPress, Cambridge, 1992 . Chapter 20. [R] C. A. Rogers, The packing of equal spheres , Proc. London Math. Soc. (3) 8 ( 1958 ), 609 -- 620 . [W] What every computer scientist should know about floating-pointarithmetic , Comput. Surveys , 23 ( 1 ) ( 1991 ), 5 - 48 .

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T. C. Hales. Sphere packings, I, Discrete & Computational Geometry, 1997, 1-51, DOI: 10.1007/BF02770863