Separating circles on the sphere by polygonal tilings

Cell partitions of the plane play an important role in solving optimization problems in the area of discrete geometry (see Fejes Toth FejesToth and Pach Pach . Suppose we have a locally finite family of points or circular discs or general convex discs in the plane. If no two of the discs have an interior point in common, then the family is said to form a packing. The phrase circle packing will refer to a packing of circluar discs. The Voronoi diagram associated with a packing is a partition of a plane into cells, where each cell consists of all points of the plane closer to a particular member of the packing, than to any other. Voronoi diagrams are also called nearest point partitions. In the simplest case, when a point set is given, the cells are convex polygons. Note that some of the cells could be unbounded (half planes, strips, angular sectors), thus for the purpose of a this paper, we say that convex polygons are convex regions bounded by a finite number of segments, lines or half lines.

If the circles in the packing are congruent, then the Voronoi diagram of the circles is the same as the Voronoi diagram of the centers of the circles, therefore it is a separating polygonal tiling (Figure 1a). If the circles are not necessarily congruent, then the Voronoi diagrams might have cells with curved boundaries (Figure 1b), thus it is not necessarily a separating polygonal tiling. Moreover, the Voronoi diagram determined by the centers of the circles might have cells, which only partially contain circles.

In this paper, we study the existence of separating polygonal tilings for various packings. Our work was inspired by the following problem which appeared in a Russian math olympiad problem collection skljarszkij ,

Section 2 contains an outline of the solution of Exersice 1. The proof presented in Section 2 shows also that spheres of any locally finite sphere packing in $R^{n}$ can be separated by a polyhedral tiling. Figure 1d is also included in skljarszkij to let readers realise that they are packings of convex discs, whose members cannot be separated by polygonal tilings. This paper starts with two additional generalizations of Exercise 1.

The first real generalizations concerns the separability of circle packings on the sphere. Arcs of the great circles are the geodesics on the sphere, thus we want to partition the sphere into convex regions bounded by arcs of great circles (Figure 2a). Such tilings on the sphere will be called polygonal tilings. This automatically means that we consider only packings of circular discs which are smaller than a hemisphere. A very natural idea would be to use a stereographic projection, say from the north pole to a plane tangent to the sphere at the south pole. This projection maps circles of the sphere to planar circles. In view of Exercise 1, the planar circles can be separated by a polygonal tiling. Pulling back the boundary segments of this tiling to the sphere, one would get a separating tiling on the sphere. The problem is that although the boundary of such tiles consists of arcs of circles, they are not necessarily arcs of great circles. In Section 3 a different approach is presented to prove

The second generalization concerns circle packings in the hyperbolic plane. In the hyperbolic plane all the needed concepts (lines, circles, convexity, polygons) exist, thus one can ask the polygonal separation problem exactly in the same way as in the Euclidean plane. A natural approach would be to get a better understanding of the properties of the polygonal tiles either in the Poincare disc model or in the Klein disc model and use Euclidean arguments to show the existence of separating polygonal tilings. An advantage of the Poincare disc model is that it is conformal (circles and angles are not distorted), a disadvantage is that lines of the geometry are circular arcs orthogonal to the boundary circle of the disc (Figure 2b). In the Klein disc model lines of the hyperbolic plane are chords of the disc. Circles in the Klein disc model become ellipses increasingly flattened as they are nearer to the edge of the model (Figure 2c). Thus in the Klein disc model while lines are simple to study for separation, circles are not. In Section 4 a different approach is presented to prove

Finally, we turn to the question of characterizing circles with the separating property. This part of the paper focuses to convex discs the Euclidean plane,

We define ’caps’ of convex discs, in particular ’isosceles caps’ as they play an important role in such discussion.

Section 5 contains two theorems concerning isosceles caps of a convex discs in the Euclidean plane. These theorems have analogous versions on the sphere and in the hyperbolic plane, but we do not include them in Section 5, as they are not going to be applied in Section 6.

The above two theorems will be applied in Section 6, where the following theorem will be proved,

Note that similarity does not exists neither on the sphere nor in the hyperbolic plane, so one would need to restrict the question of separation for packings of congruent copies of discs in those two spaces. Even under this restriction, Theorem 5 is not true on the sphere. Just take any convex disc inside of a hemisphere, with an area greater than $\frac{2}{3}$ of the surface area of the sphere. Obviously, there is room only for at most two copies of $C$ on the sphere, and two copies can always be separated by a great circle. In the hyperbolic plane it is not known if circles are characterised with the separation property.

In this section, we include the outline of a known solution of Exercise 1, so that the reader could see how the generalizations are related to this original solution.

By this definition, points inside the circle have negative power, points outside of the circle have positive power, and points on the circle have zero power. For external points, the power equals the square of the length of a tangent from the point to the circle.

Similarly to the Voronoi partition, Lemma 1 implies that the least power partition is a separating polygonal tiling.

In this section we prove Theorem 1, which is formally stated in the introduction. In plain terms, we are going to construct a separating polygonal tiling for every finite circle packing on the sphere.

For the purpose of the construction we introduce the spherical potential of a point with respect to a given circle. First we review some standard notations: distance $\widehat{PQ}$ between two points $P,Q$ on the sphere is measured by the central angle of the shorter great circular arc connecting the two points. Similarly, radius of a circle is measured by the central angle corresponding to the arc representing the radius of the circle.

The spherical potential was introduced specifically for this proof. The potential $p(A,C)$ depends on the distance $\widehat{AO}$ through cos(.). Further the point is on the sphere from the center, less potential it has.

We prove the following two lemmas.

According to the spherical law of cosine, in any right triangle with hypotenuse $c$ and legs $a,b$ we have $\cos a\cos b=cosc$. Thus, the numerators in the previous equation can be replaced with appropriate products of cos’s:

After cancelling with $\cos\widehat{AP}$ we get that P is also an equipotential point. ∎

.

Moving a point P from M to N along the arc MN, the inequality changes from $p(X,C_{1})>P(X,C_{2})$ to $p(X,C_{1})<P(X,C_{2})$. A simple continuity argument implies that there is at least one equipotential point on the arc MN.

which is the same as

Solving this for x, we have that

Since this constant is not equal to $0$, we have a unique solution for $x$ on the half circle, i.e. on the interval $[0,\pi)$. Note that the antipodal point of this unique $P$ is also an equipotential point with respect to the circles $C_{1}$ and $C_{2}$. ∎

Uniqueness of the perpendicular projections mean that all equipotential points lie on a great circle perpendicular to the great circle $C$. Moreover, by monotonicity of the potential it turns out that this great circle is separating $C_{1}$ and $C_{2}$. Similarly to the Voronoi partition, we have now that the least potential partition is a separating polygonal tiling, which completes the proof of Theorem 1.

It turns out that after appropriately redefining the potential of a point $A$ with respect to a circle $C$ of center $O$ and radius $r$, the entire proof of the spherical separation problem goes through word by word. In fact, the potential $p(A,C)=\frac{\cosh\widehat{AO}}{\cosh r}$ does the job.

Returning to $C$, for comparison put a circle $c$ next to $C$ (Figure 5). Let $p$ and $P$ be the lowest points of $c$ and $C$. Consider a random circumscribed triangle $t$ with contact points $p,q,r$ of circle $c$. Let $T$ be the circumscribed triangle of $C$, whose sides are parallel to the sides of triangle $t$. Denote the corresponding tangency points of $T$ and $C$ by $P,Q,R$. In view of the above elementary exercise, the triangles $pqr$ and $PQR$ are not only similar, but they are in parallel positions. Let us now fix two of the contact points $\{p,q,r\}$ (note that there are three ways to do this) and change triangle $t$ by running the third contact point along the arc of circle $c$. The later observation implies that $R$ runs along a circular arc. Moreover, the three arcs corresponding to the three choices of the two contact points of $\{p,q,r\}$, form a complete circle. ∎

Note that the above proof uses similarity, thus it does not work on the sphere and in the hyperbolic plane. Although we are not going to use the spherical and hyperbolic versions of Theorem 3, we present here a second Euclidean proof, which can be modified for those two spaces.

In this section we prove Theorem 5, which is formally stated in the introduction. With other words, we will give instructions how to arrange finite many similar copies of a given non circular discs, which already form a nonseparable packing. Assume the convex disc $C$ is not a circle. In view of Theorem 3, $C$ has a non isosceles cap $EPF$ with $EP>FP$. For future reference chose $\epsilon>0$ to be smaller than $EP-FP$. Let $\alpha$ be the angle of the cap EFP. Let $n$ be the largest integer so that $n\alpha\leq 2\pi$ and let $\beta=2\pi-n\alpha$. For our construction we need to start with an $(n+1)$ sided polygon $R$, whose outside angles are all equal to $\alpha$ except one which is equal to $\beta$ and whose sides are all shorter than $\epsilon$. Figure 8 shows an arrangement of $n$ congruent copies of $C$ so that the attached $\alpha$ caps together with an empty $\beta$ angular sector tile the neighbourhood of polygon $R$. According to Theorem 4 disc $C$ has an isosceles $\beta$ cap. Finally, we place a scaled copy of $C$ with the isosceles $\beta$ cap fitting in the $\beta$ angular sector. By scaling we can ensure that the side lengths of the $\beta$ cap are between $EP-\epsilon$ and $FP$. The choice of $\epsilon$ and the clockwise rotational condition guaranties that no separating line between consecutive copies of $C$ can not intersect the polygon $R$, thus separating tiling cannot exists. ∎