Tilings with the neighborhood property

The neighborhood N(T) of a tile T is the set of all tiles which meet T in at least one point. If for each tile T there is a different tile T1 such that N(T)=N(T1) then we say the tiling has the neighborhood property (NEBP). Grünbaum and Shepard conjecture that it is impossible to have a monohedral tiling of the plane such that every tile T has two different tiles T1, T2 with N(T)=N(T1)=N(T2). If all tiles are convex we show this conjecture is true by characterizing the convex plane tilings with NEBP. More precisely we prove that a convex plane tiling with NEBP has only triangular tiles and each tile has a 3-valent vertex. Removing 3-valent vertices and the incident edges from such a tiling yields an edge-to-edge planar triangulation. Conversely, given any edge-to-edge planar triangulation followed by insertion of a vertex and three edges that triangulate each triangle yields a convex plane tiling with NEBP. We exhibit an infinite family of nonconvex monohedral plane tilings with NEBP. We briefly discuss tilings of R3 with NEBP and exhibit a monohedral tetrahedral tiling of R3 with NEBP.


INTRODUCTION.
A plane tiling denoted by 7" is a countable family of closed sets which cover the plane without gaps or overlaps. Tiles can intersect along their boundaries but interiors are disjoint. We assume all of our tiles are (closed) topological disks. A tiling T is convex if all tiles are convex, and monohedral if every tile in T is congruent to a fixed tile T, which is called the prototile of 7". The neighborhood N(T), in a gi,;,en tiling, is the set of tiles consisting of T and all tiles which meet T in at least one point. If for each tile T1 there is a different tile T2 such that N(T1) N(T2), then we call this the neighborhood property (NEBP). If our tiles are polygons then a tiling is edge-to-edge if the intersection of every pair of nondisjoint tiles is either a vertex or an edge. Cm:inbaum and Shephard [5] posed the following problem: Show that there is no monohedral tiling, in which the prototile is a polygonal disk, such that for tile T there exist two other tiles T and T. such that N(T) N(T N(T2). We show this is true if the tiles are convex by characterizing the convex plane tilings with the NEBP. 2. CONVEX PLANE TILINGS.
In R a cone of support to a closed set A at a point p on its boundary is the intersection of all closed half spaces which contain A and which contain p in their boundary hyperplanes. Let S be a convex set. A point z in S is called an extreme point of S if there exists no nondegenerate line segment in S that contains z in its relative interior. The set of extreme points of S is called the profile of S. It is a theorem that if S is a compact convex subset of R', then S is the convex hull of its profile. If 7 9 is a set of points then we let cony79 denote the convex hull of 79.
Here we assume that our tiles are convex and (as stated previously) topological disks. Let T and T2 be different tiles for which N(T) N(T). They clearly meet and since they are convex they can be separated by a straight line L. We assume that L is oriented to be the z-axis, and T is above the z-axis.
Let # be a point of T1 that is at maximum distance from the z-axis. Such aft exists since T1 is compact. If a nondegenerate line segment L' in T1 contained ft in its relative interior or if the supporting cone at ft were a half-space, then there would be a tile Ts touching T1 at ft where T3 N(T2). Thus the point ft is an extreme point of T1 and the supporting cone at ft is not a half-space. Let L1 and L2 be two rays frrming the support cone at ft. It is clear that these two rays L and L 2 must intersect L. Also, the convexity ofour tiles implies that at least three tiles meet at ft. Suppose Ts and T4 meet T1 at ft and are separated from T1 by lines Ls and /-,4, respectively.
Arguing symmetrically for T, there would be a and a which would meet T2 at an extreme point ft', and which are separated from T byL and L. It is then elementary to establish that the lines L, L3, L intersect in a common point P3, and P3 E T1 N T f T3 N .A nalogously, L, L4, L intersect in a common point P4, and P4 5 T T N T4 N . Now {p, P4ft} c T, T is convex, and T is bounded by L3, L4, L, hence T1 is the triangle with vertices ft, P3 and P4. Analogously, T2 is a triangle with vertices ft', P3 and P4. Clearly T and T' share an edge. Suppose another tile T5 met T at ft. Then T5 could meet T only at/9 3 or P4. But this is dearly impossible because ofthe location of T, T4 and the fact that Ts is convex. We have proven: PROPOSITION 2.1. Let 7" be a convex plane tiling with NEBP. Then the tiles of 7" are triangles.
If T : T' and N(T) N(T'), then T and T' share an edge. In addition, exactly three tiles meet at vertex ft(ft') ofT(T') which is not on the shared edge.
PROOF. Let ea be the edge of T3 and el be the edge of T1 which are both contained in the line Ls that separates T3 and T. The proof of Proposition 2.1 shows that e C ca. If N(T:;) N(T), then es e by Proposition 2.1. IfN(Tz) -N(T), let T3 play the role of T in Proposition 2.1 to get that es C e. Thus es el. A symmetric argument shows that the edge of T and that of T4 that lie along L4 must coincide. The result follows since T is an arbitrary tile. PROPOSITION 2.3. Let 7" be a convex plane tiling with NEBP. Then each tile (which is a triangle) has exactly one 3-valent vertex and the other two vertices are at least 6-valent. Further, the edge opposite the 3-valent vertex in a tile is shared by two tiles whose neighborhoods are equal.
PROOF. By Proposition 2.1 the valency of vertex # in T is 3 and since tiles are convex, the third vertex of T3, besides P3 and ft, must be at a strictly greater distance from L than ft. Let this vertex be P5 and let the analogous additional vertex of , besides P3 and ft', be p. Note that T3 conv{ft, P3, P5 } and conv{ft', P3, I }, T4 conv{ft, P4, } and conv{ft', P4, P }. But there is no tile with vertices P3, P, 1 since such a tile would have no vertex of valency three. Hence the valency of P3, and analogously P4, must be at least 6. The final statement of our result is then immediate.
Define a plane triangulatton refinements as follows Begin with any edge-to-edge tiling of the plane by triangles. Insert a point PT in the interior of each triangle T and draw an edge from PT to each of the vertices of T. Do this for each of the tiles in the original plane triangulation. TItEOREM 2.1. A convex plane tiling with NEBP is equivalent to a plane triangulation refinement.
PROOF. A plane triangulation refinement is easily checked to be a convex plane tiling with NEBP. Conversely, let T be a convex plane tiling with NEBP. Delete all vertices that are initially 3-valent and the incident edges. By Proposition 2.3 every tile in 7" is removed by this process. What remains is an edge-to-edge triangulation of the plane Using the discarded 3-valent vertices as the pr's gives a refinement that recreates 7". Our result is clear. PROPOSITION 2.4. For any fixed positive integer k there is a convex plane tiling with NEBP using exactly k different tile shapes PROOF. Begin with an edge-to-edge tiling of the plane with equilateral triangles. For k 1 we produce a refinement by choosing Pr to be the centroid of each triangle. For k 2 we produce a refinement by choosing Pr on the line through the centroid and perpendicular to the base of T, but Pr is not set at the centroid. For k 3 choose PT so three distinct triangles tile T. Clearly, by choosing the placement of the pr's we can produce a refinement with any given positive number of k distinct tile shapes. 3

. CONVEX MONOHEDRAL PLANE TILINGS WITH NEBP.
We are now in position to completely characterize the monohedral convex plane tilings with NEBP. THEOREM 3.1. There is a unique monohedral convex plane tiling 7" with NEBP.
PROOF. We need a simple fact that is easy to verify. Let T be a triangle and let T1, T2 and T3 be three triangles that tile T edge-to-edge. If T1 is congruent to Tg then T3 is an isosceles triangle. If all three tiles are equal, then each is the triangle with angles 30 , 30 and 120. It is now clear that removal of all 3-valent vertices from 7" along with incident edges produces an edge-to-edge tiling of the plane with equilateral triangles. Such a tiling is unique, so our tiling T is obviously unique. This tiling is exhibited in Figure 3  The automorphism group of a tiling 7is the set of isometrics that preserve the tiling Such a group partitions the tiles into transitivity classes where any two tiles within a class can be mapped to each other via some automorphism. A tiling 7" is isohedral if there is only one transitivity class. The plane isohedral tilings are completely known (see [3] or [4]) and among the isohedral plane tilings only the tiling of Figure 3.1 has NEBP. Thus we have: There is a unique isohedral plane tiling with NEBP Note that Theorem 3.1 and Proposition 3.1 prove the Grnbaum and Shephard conjecture (stated in the introduction) for plane tilings when the tilings are either convex or isohedral. 4. NONCONVEX PLANE TILINGS WITH NEBP.
One simple example of a nonconvex plane tiling with NEBP is given as follows: With center C on a straight line L draw circles with varying radii that become arbitrarily large The tiles are the simplyconnected regions. The pair of tiles sharing the same two edges lying along L will have the same neighborhood.
We now produce monohedral nonconvex plane tiling with NEBP Figure 4 shows how to "dent in" the edges of each triangle in the tiling of Figure 3.1 so that the tiling is still monohedral. The following lemma is an easy exercise and the subsequent proposition is immediate. LEMMA 4.1. If the denting process is applied to Figure 3.1 to produce a monohedral polygonal nonconvex plane tiling with NEBP, then each tile has an odd number of sides. PROPOSITION 4.1. For each odd integer k > 3, there is a monohedral plane tiling with NEBP using tiles with k sides.
One has to wonder here whether all monohedral plane tilings with NEBP arise from Figure 3.1 by a suitable denting process. From the proof ofProposition 2.4 and the "denting" procedure we have: THEOREM 4.1. For any fixed positive integer k there is a nonconvex plane tiling with NEBP using exactly k different tile shapes. 5. SOME TILINGS OF R 3 WITH NEBP.
We can extend a tiling of the plane to a tiling of/3 by a natural lifting process (mentioned to us by B. Cjrnbaum). Let T be a plane tiling (taken to be in the xy-coordinate plane) and d a fixed positive distance. Define L('T, d) as follows: for each T in 7", construct a prism with base T and height d. This produces a slab and now we stack such slabs (face-to-face) to fill/3. Our next result is easy to verify. We generalize the refinement process described prior to Theorem 2.1. Begin with a face-to-face tiling "T or R 3 using tetrahedral and/or hexahedra with triangular faces. Insert a point Pr in the interior of each tile T. Clearly T can be subdivided into tetrahedra that are determined by Pr and each face ofT. Do this for each T in 7" to get a refinement of the original face-to-face tiling. The following theorems is easy to verify: THEOREM 5.1. The refinement of a face-to-face tiling of R 3 using tetrahedra and/or hexadehra with triangular faces, is a tetrahedral tiling of 3 with NEBP.
A tetrahedron is isosceles if opposite edges are equal. An isosceles tetrahedron clearly has congruent faces. Also, a tetrahedron is isosceles iff the medians are equal, e., iff the centroid is equidistant from the vertices of the tetrahedron. THEOREM 5.2. If there is a monohedral face-to-face tiling of R 3 with isosceles tetrahedra, then there is a monohedral face-to-face tiling with tetrahedra of R 3 with NEBP.
PROOF. The medians of the isosceles tetrahedra are equal and the faces are congruent. If we choose PT to be the centroid of a given tetrahedron, then the resulting tetrahedra in the refinement will be congruent. Our result follows by Theorem 5.1.
The isosceles tetrahedron To with side lengths 2, 2, V/', V/-, v/', and v/, tiles E 3 face-to-face (see [2]). In Figure 5.1 three copies ofT0 are shown to tile a prism. This prism can easily be stacked and translated to produce a face-to-face tiling of R s using To.
Using Theorem 5.2 the following is then clear: COROLLARY 5.2. There is a monohedral tetrahedral tiling of R s with NEBP. 6. FINAL REMARKS.
One question posed earlier is whether all monohedral plane tilings with NEBP arise from Figure 3.1 by a suitable denting process. Another natural question is whether there is a succinct characterization of convex tilings of R 3 with NEBP as there was for convex plane tilings with NEBP.