The strong thirteen spheres problem
Abstract
The thirteen spheres problem is asking if 13 equal size nonoverlapping spheres in three dimensions can touch another sphere of the same size. This problem was the subject of the famous discussion between Isaac Newton and David Gregory in 1694. The problem was solved by Schütte and van der Waerden only in 1953.
A natural extension of this problem is the strong thirteen spheres problem (or the Tammes problem for 13 points) which asks to find an arrangement and the maximum radius of 13 equal size nonoverlapping spheres touching the unit sphere. In the paper we give a solution of this longstanding open problem in geometry. Our computerassisted proof is based on a enumeration of the socalled irreducible graphs.
1 Introduction
1.1 The thirteen spheres problem
The kissing number is the highest number of equal nonoverlapping spheres in that touch another sphere of the same size. In three dimensions the kissing number problem is asking how many white billiard balls can kiss (touch) a black ball.
The most symmetrical configuration, 12 balls around another, is achieved if the 12 balls are placed at positions corresponding to the vertices of a regular icosahedron concentric with the central ball. However, these 12 outer balls do not kiss each other and may all be moved freely. So perhaps if you moved all of them to one side, a 13th ball would possibly fit in?
This problem was the subject of the famous discussion between Isaac Newton and David Gregory in 1694 (May 4, 1694; see [29] for details of this discussion). Most reports say that Newton believed the answer was 12 balls, while Gregory thought that 13 might be possible. However, Casselman [10] found some puzzling features in this story.
This problem is often called the thirteen spheres problem. Hoppe [15] thought he had solved the problem (1874). But there was a mistake  an analysis of this mistake was published by Hales in 1994 [14] (see also [29]). Finally this problem was solved by Schütte and van der Waerden in 1953 [28]. A subsequent twopage sketch of an elegant proof was given by Leech [17] in 1956. Leech’s proof was presented in the first edition of the wellknown book by Aigner and Ziegler [1], the authors removed this chapter from the second edition because a complete proof would have to include so much spherical trigonometry.
1.2 The Tammes problem
If unit spheres kiss the unit sphere in , then the set of kissing points is an arrangement on the central sphere such that the (Euclidean) distance between any two points is at least 1. So the kissing number problem can be stated in other way: How many points can be placed on the surface of so that the angular separation between any two points be at least ?
This leads to an important generalization: a finite subset of is called a spherical code if for every pair of with its angular distance is at least .
Let be a finite subset of . Denote
Then is a spherical code.
Denote by the largest angular separation with that can be attained in , i.e.
In other words, how are congruent, not overlapping circles distributed on the sphere when their common radius of the circles has to be as large as possible?
This question, also known as the problem of the ‘‘inimical dictators’’: Where should dictators build their palaces on a planet so as to be as far away from each other as possible? The problem was first asked by the Dutch botanist Tammes [30] (see [8, Section 1.6: Problem 6]), who was led to this problem by examining the distribution of openings on the pollen grains of different flowers.
1.3 The Tammes problem for
The first unsolved case of the Tammes problem is , which is particularly interesting because of its relation to the kissing problem and the Kepler conjecture [6, 13, 29].
Actually this problem is equivalent to the strong thirteen spheres problem, which asks to find an arrangement and the maximum radius of 13 equal size nonoverlapping spheres in touching the unit sphere.
It is clear that the equality implies . Böröczky and Szabó [6] proved that . Recently Bachoc and Vallentin [3] have shown that .
We note that there is an arrangement of 13 points on such that the distance between any two points of the arrangement is at least (see [13, Ch. VI, Sec. 4]). This arrangement is shown in Fig. 1.
Remark. Denote the constant by . The value can be found analytically. Indeed, we have (see for notations and functions Fig. 9 and Section 3): , where . This yields:
Thus, we have and .
2 Main theorem
In this paper we present a solution of the Tammes problem for .
Theorem 1.
The arrangement of 13 points in which is shown in Fig. 1 is the best possible, the maximal arrangement is unique up to isometry, and .
2.1 Basic definitions
Contact graphs. Let be a finite set in . The contact graph is the graph with vertices in and edges such that .
Shift of a single vertex. Let be a finite set in . Let be a vertex of with , i.e. there is such that . We say that there exists a shift of if can be slightly shifted to such that .
Danzer’s flip. Danzer [11, Sec. 1] defined the following flip. Let be vertices of with . We say that is flipped over if is replaced by its mirror image relative to the great circle (see Fig. 2). We say that this flip is Danzer’s flip if .
Irreducible graphs. We say that the graph is irreducible^{1}^{1}1This terminology was used by Schütte  van der Waerden [27, 28], Fejes Tóth [13], and Danzer [11]. (or jammed) if there are neither Danzer’s flips nor shifts of vertices.
and . Denote by the arrangement of 13 points in Fig. 1. Let . It is not hard to see that the graph is irreducible.
Maximal graphs . Let be a subset of with and . Denote by the graph . Actually, this definition does not assume that is unique. We use this designation for some with .
Graphs . Let us define four planar graphs (see Fig. 3), where , and . Note that , is obtained from by removing certain edges.
2.2 Main lemmas
Lemma 1.
is isomorphic to with or .
Lemma 2.
is isomorphic to and .
It is clear that Lemma 2 yields Theorem 1. Now our goal is to prove these lemmas.
3 Properties of
3.1 Combinatorial properties of
Proposition 3.1.
Let be a finite set in . Then is a planar graph.
Proof.
Let with . Then the shortest arcs and don’t intersect. Otherwise, the length of at least one of the arcs has to be less than . This yields the planarity of . ∎
Proposition 3.2.
Let be a subset of with and . Then for the graph is irreducible.
Proposition 3.3.
Let . If the graph is irreducible, then degrees of its vertices can take only the values (isolated vertices), , , or .
Proposition 3.4.
Let with . If the graph is irreducible, then its faces are polygons with at most vertices.
Böröczky and Szabó [6, Lemma 8 and Lemma 9(iii)] considered isolated vertices in irreducible graphs with 13 vertices.
Proposition 3.5.
Let with . Let the graph be irreducible. If contains an isolated vertex, then it lies in the interior of a hexagon of and this hexagon cannot contain other vertices of .
Combining these propositions, we obtain the following combinatorial properties of .
Corollary 3.1.

is a planar graph;

Any vertex of is of degree or ;

Any face of is a polygon with or vertices;

If contains an isolated vertex , then lies in a hexagonal face. Moreover, a hexagonal face of cannot contain two or more isolated vertices.
3.2 Geometric properties of
Let with . Let the graph be irreducible. Note that all faces of are convex polygons. (Otherwise, a ‘‘concave’’ vertex of a polygon can be shifted to the interior of .) Then the faces of the graph in are regular triangles, rhombi, convex equilateral pentagons, and convex equilateral hexagons. Polygons with more than six vertices cannot occur. Note that the triangles, rhombi, or pentagons of cannot contain isolated vertices in their interiors. The lengths of all edges of equal .
Consider as parameters (variables) of in the set of all angles of its faces and . Clearly, the graph , , and the set uniquely (up to isometry) determine embedding isolated vertices in .
We obviously have the following constraints for these parameters.
Proposition 3.6.

for all ;

for all , where
is the angle of a regular triangle in with sides of length ;

for all vertices of , where is the set of subscripts of angles that are adjacent to ;
Let be a face of . Then is a polygon with vertices, where , or . Consider all possible cases.
1. : triangle. In this case, is a regular triangle.
Proposition 3.7.
Let be a triangular face of with angles . Then .
2. : quadrilateral. In this case, is a rhombus. Then we have and . Using the spherical Pythagorean theorem, one can show that
Then
Since , we have (Fig. 4).
Proposition 3.8.
Let be a quadrilateral of with angles . Then and for all .
3. : pentagon. In this case, is a convex equilateral pentagon . Let be its angles. Then is uniquely determined by and any pair of these angles, for instance, by (Fig. 5).
It is not hard for given parameters and to find as functions of , i.e. , where . Let and . Then we have for all . We have that all .
Denote by the image of after Danzer’s flip. Let denote the minimum distance between and , where . If is a face of and is irreducible, then does not admit Danzer’s flips. Therefore, for all . Thus we have the following proposition.
Proposition 3.9.
Let be a pentagonal face of with angles . Then and for all .
4. : hexagon. In this case, is a convex equilateral hexagon with angles . Clearly, is uniquely defined by any three angles and .
Let for . Let for . Then we have for all .
In fact, for the case we have two subcases: (a) has no isolated vertices, and (b) has an isolated vertex.
It is easy to see that for case 4(a) there exists an analog of Proposition 3.9. Let denote the minimum distance between and , where .
Proposition 3.10.
Let be a hexagonal face of with angles . Suppose the face has no isolated vertices in its interior. Then and for all (Fig. 7).
Now consider case 4(b). Denote by the set of all points in the interior of such that there is a pair with . Clearly, .
Let be defined by a pair . Denote by the set of all such that and . Let
Since contains an isolated vertex, we have .
Proposition 3.11.
Let be a hexagonal face of with angles . Suppose the face has an isolated vertex in its interior.
Then for all and .
4 Proof of Lemma 1
Here we give a sketch of our computer proof. For more details see http://dcs.isa.ru/taras/tammes13/ .
The proof consists of two parts:
(I) Create the list of all graphs with 13 vertices that
satisfy Corollary 3.1;
(II) Using linear approximations and linear programming remove
from the list all graphs that do not
satisfy the geometric properties of (see Propositions
3.63.11).
(I). To create we use the program plantri (see [24]).^{2}^{2}2The authors of this program are Gunnar Brinkmann and Brendan McKay. This program is the isomorphfree generator of planar graphs, including triangulations, quadrangulations, and convex polytopes. (The paper [9] describes plantri’s principles of operation, the basis for its efficiency, and recursive algorithms behind many of its capabilities.)
The program plantri generates 94,754,965 graphs in , i.e. graphs that satisfy Corollary 3.1. Namely, contains 30,829,972 graphs with triangular and quadrilateral faces; 49,665,852 with at least one pentagonal face and with triangular and quadrilaterals; 13,489,261 with at least one hexagonal face which do not contain isolated vertices; 769,375^{3}^{3}3Perhaps contains isomorphic graphs. graphs with one isolated vertex, ^{3}^{3}footnotemark: 3 with two isolated vertices, and no graphs with three or more isolated vertices.
(II). Let us consider a graph from . We start from the level of approximation . Now using Propositions 3.63.11 we write linear equalities and inequalities for the parameters (angles) of this graph.
For we use the following linear equalities and
inequalities:
(i) 13 linear equalities in
Proposition 3.6(3);
(ii) Since ,
we have
;
(iii) For a quadrilateral
from Proposition 3.8 we have equalities ,
and inequalities ;
(iv) For a quadrilateral, (ii) and
yield ;
(v) Let be a pentagonal face. Consider all vectors
that satisfy Proposition 3.9 (see Fig. 6). We use
a convex polytope in which contains .
Actually, is defined by certain linear inequalities. For
instance,
,
, etc;
(vi) For a hexagonal face that contains no isolated
vertices, using
Proposition 3.10, we find a set of three polytopes
, which are defined
by the inequalities
and ;
(vii) For a hexagonal face with an isolated vertex, Proposition
3.11 yields .
Using this set of linear inequalities, we find minimal and maximal value of each variable by linear programming. This gives us a convex region in the space of possible solutions that contains all possible solutions for given graph (if they exist). If the region becomes empty, this means that we can eliminate the graph considered. This step ‘‘kills’’ almost all graphs. After this step, there remain graphs without hexagons, graphs with hexagons and without isolated vertices, graphs with one isolated vertex, and graphs with two isolated vertices.
For we use the following idea. This region is smaller than the original region, so we can adjust linear estimates for nonlinear equalities and inequalities. For quadrilaterals we adjust inequalities using (iv). For pentagons we are using an additional set of inequalities. Namely, using functions , , and bounds for can be obtained minimal and maximal linear bounds for .
Repeating this procedure , we obtain a chain of nested convex regions, which contain all possible solutions. This chain converges to empty or nonempty region. If this result is empty, the graph is eliminated. After this step, only graphs remain in the main group, graphs remain in the second group, graphs remain in the third group, and no graphs remain in the fourth group.
For the level of approximation , we split the region into two smaller regions and repeat the same procedure as for independently. For graphs with empty hexagons, we make a specific split by taking different values of from item (vi) (see above).
Repeating the splitting procedure, we ‘‘kill’’ all graphs except .
This result gives us two surprises. We expected that subgraphs were to remain, because they can be infinitesimally close to , and so they cannot be eliminated by computer program. But we didn’t expect that all other graphs would be killed. Also manually, we found two subgraphs which could be contact graphs: and . But we missed the graph with one isolated vertex, which was found by computer program.
Remark. In Fig. 8 are presented examples of graphs which are not isomorphic to and have been eliminated only after many iterations. The most ‘‘surviving’’ graph is . This graph is also a subgraph of . After eliminating four edges, the graph contains four pentagons. The reason why it was eliminated because there are angles which are slightly bigger than , so that the pentagons are not convex. Therefore, this graph is not irreducible. Other most surviving graphs were ‘‘strong’’ because they have several pentagons and hexagons. Note that here we use weak bounds for pentagons and hexagons given by (v),(vi),(vii). Our elimination procedure works very fast when we have sufficiently many triangles and quadrilaterals, and it works worse (slowly) when we have several pentagons and hexagons.
5 Proof of Lemma 2
Proof.
This proof is based on geometric properties of . In Section 4 we substitute all nonlinear equations by certain linear inequalities. Note that a statement is a byproduct of this approximation. Here we prove that based on original equations.
Lemma 1 says that , where , or 3. We are going to prove that if with , then .
5.0. Angles of . Let . For we have (see Fig. 9):
Therefore, for the value are functions in the variables . Since we have also an additional equation for the vertex (see Fig. 1 and Fig. 9):
the value is a function in , as well as is a function in the variables and . Thus, all and are functions in or in .
Now we consider three cases , where .
5.1. The case . In this case . Then for the vertex we have the equation:
From this it follows that and therefore all are functions in . Note that
Thus, is a function in (see Fig. 10).
If , then . Since the function is monotone decreasing, we have only if . Thus, .
5.2. The case . It is already shown that and all are functions in . Let
We can see from Fig. 11 that the intersection consists of one point with It is not hard to prove this fact. Indeed, conversely, and there is a point on the boundary of such that or . Therefore, we have the same case as in 5.1, a contradiction. Thus, .
5.3. The case . This case can be considered by the same method as the case . Actually, for given all angles can be found by the same formulas as in 5.0. On the other hand,
Then all are functions in the variables and . Since (or equivalently ), we have the equation:
It yields that all depend on two parameters.
The vertex is isolated. In fact, we can shift this point in such a way that at least two edges , where , have lengths . Then for two other edges we have inequalities: and .
Arguing as in 5.2, we can show that there are parameters such that and at least one of the inequalities becomes equality. It is not hard to see that there are exactly two geometrically nonequivalent cases with exactly one edge or , such that . These cases are shown in Fig. 12.
Actually, the first subcase is case 5.1. For the second subcase consider the pentagon . All angles of can be found as functions in . Since and any two angles of define all other angles we can use one of these equations to find as a function in . Then (see Fig. 12) is a function in . In fact, the graph of the function is very similar to the graph in Fig. 10, and is a monotone decreasing function. Thus, cannot be greater than , and .
We see that if , then is isomorphic to . Moreover, is uniquely defined up to isometry and . This completes the proof. ∎
Acknowledgment. We wish to thank Robert Connelly, Alexey Glazyrin, Nikita Netsvetaev, and Günter Ziegler for helpful discussions and comments.
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O. R. Musin, Department of Mathematics, University of Texas at Brownsville, 80 Fort Brown, Brownsville, TX, 78520, USA.
Email address:
A. S. Tarasov, Institute for System Analysis, Russian Academy of Science, 9 Pr. 60letiya Oktyabrya, Moscow, Russia.
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