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Properties of Nodes in Pentagonal Tilings

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Properties of Nodes in Pentagonal Tilings Original Paper Forma, 24, 117–121, 2009 Properties of Nodes in Pentagonal Tilings Teruhisa Sugimoto1∗ and Tohru Ogawa2.3 1The Institute of Statistical Mathematics, 10-3 Midori-cho, Tachikawa, Tokyo 190-8562, Japan 2(Emeritus Professor of University of Tsukuba), 1-1-1 Tennodai, Tsukuba-shi, Ibaraki 305-8577, Japan 3The Interdisciplinary Institute of Science, Technology and Art, Suzukidaini-building 211, 2-5-28 Kitahara, Asaka-shi, Saitama 351-0036, Japan ∗E-mail address: [email protected] (Received November 17, 2008; Accepted March 13, 2010) A node of valence k in an edge-to-edge tiling is a point that is the common vertex of k tiles. We show that an edge-to-edge tiling of plane by pentagons each of which has m nodes of valence 3 and 5 − m nodes of valence k has properties of (m, k) = (3, 4) or (m, k) = (4, 6) if it is normal. Then we discuss tilings by congruent convex pentagons using the properties. Key words: Convex Pentagon, Tiling, Tile, Pentagonal Tiling, Node 1. Introduction the edge-to-edge tilings are topologically pentagonal, non- Tiling by polygons is to cover the plane with polygons edge-to-edge tilings are not. Thus our interest lies more without gaps or overlaps. In this paper, the “plane” means in edge-to-edge tiling since it is more essential than non- the familiar Euclidean plane of elementary geometry. If all edge-to-edge tiling (Sugimoto and Ogawa, 2005). Then, tiles in a tiling are same size and shape we say that the the convex pentagonal tiles that belong to type 4, type 5, tiling is monohedral. If a tiling by polygons is monohe- type 6, type 7, type 8, or type 9 form edge-to-edge tilings dral, the polygon is called the prototile of monohedral tiling (see Fig. 1). In addition, although the tilings of type 1 and (Grunbaum¨ and Shephard, 1987). Then the polygon which type 2, among 14 types, are generally non-edge-to-edge is the prototile of monohedral tiling is called particularly the (see Fig. 1), the tilings by convex pentagonal tiles, which polygonal tile. belong to type 1 or type 2, can be edge-to-edge in special As to polygonal tiles, it is known as following. All trian- cases (see Fig. 3). gles and quadrilaterals tile the plane. The convex hexagon Then, in our study, a point that is the common vertex that is tileable can be categorized into three types. In the of k polygons (tiles) in an edge-to-edge tiling is called a cases of convex polygons with seven or more edges, no node of valence k (Sugimoto and Ogawa, 2006b). The polygonal tiles exist (Grunbaum¨ and Shephard, 1987). valence k of a node in edge-to-edge tiling is at least three. The classification problem of convex pentagons that can There exist positive numbers r and R such that any tile tile the plane is still open and is generally termed the convex contains a certain disk of radius r and is contained in a pentagonal tiling problem. At present, for the convex pen- certain disk of radius R in which case we say the tiles tagonal tiles, 14 types are known (see Fig. 1) (Grunbaum¨ in tiling are uniformly bounded (Grunbaum¨ and Shephard, and Shephard, 1987; Wells, 1991; Sugimoto and Ogawa, 1987; Bagina, 2004). A tiling by polygons is called normal 2006b). As shown in Fig. 1, each convex pentagonal tile is if it is edge-to-edge and all the tiles in tiling are uniformly defined by the lengths of its edges and the magnitudes of its bounded. angles, but some degrees of freedom remain. Then, unless In this paper, it is the main purpose to show the following a convex pentagon is a new prototile of monohedral tiling, a proposition. convex pentagonal tile belongs to one or more of 14 types. PROPOSITION 1 Let J be an edge-to-edge tiling of plane For example, although the tiling in Fig. 2 is new, the pentag- by pentagons each of which has m nodes of valence 3 and onal tiles in the figure are not new (Sugimoto and Ogawa, 5 − m nodes of valence k (0 < m < 5, k ≥ 4) .IfJ is a 2006a, 2009). This is because these pentagonal tiles belong normal tiling, (m, k) = (3, 4) or (m, k) = (4, 6). to both type 1 and type 7. The tilings in Fig. 1 themselves can be distinguished Although the edge-to-edge tiling J by pentagons is the into two kinds: edge-to-edge tilings, and non-edge-to-edge normal, it is not necessary to be monohedral. It is clear that tilings. The tiling by polygon is called edge-to-edge if any Proposition 1 has no trouble if all the pentagons in J are the two polygons either do intersect or share only one vertex or same size and shape. Therefore, after proving Proposition only one edge and vertices of the polygons coincide with 1, we consider the tilings by congruent convex pentagons vertices of the tiling. On the other hand, in non-edge-to- through Proposition 1. edge tilings, the vertices of polygons may contact the edges We are interested in the convex pentagonal tiling problem of adjoining polygons. As to the pentagonal tilings, while and study it. The properties of edge-to-edge tiling by pen- 117 118 T. Sugimoto and T. Ogawa type 1 type 2 type 3 A ++=360 B C . A ++=360 B D , A ===120, C D a =. d a =, b d =+. c e d c a D C d C A c b e B E D b B B e E A e c C D a E A b a d type 4 type 5 type 6 C ==90 E , A = 120 , A + B + D = 360 , a =, e c =. d C = 60 , A = 2,C a =, b c =. d abe= = , cd = . b e B a E e A D D c a a E d A d E A C D e b b B C d B C c c type 7 type 8 type 9 2+B C = 360 , 2+A B = 360 , 2+E B = 360 , 2+=360,D A 2+D C =360, 2+D C =360 , a = b = c = d. a ===. b c d a = b = c = d. e e e D d E D d E d D C a E a C a A C c c B c A B A B b b b type 10 type 11 type 12 A =90, A = 90 , A = 90 , B +=180, E C + E = 180 , C + E = 180 , 2+=D E 360 , 2+B C = 360 , 2B + C = 360 , 2+C B =360 , d = e = 2+a c. 2=a d = c + e. a ==+. b c e e E e D D d D d d e E a C E C a C A B A B c a c A B c b b b type 13 type 14 A = C = 90 , A = 90 , B 145.34 , 2=B 2=E 360 D, CD 69.32 , 124.66 , 2=c 2=d e. E 110.68 , d 2acd = 2 = = e. e e D D d C E E a c a C A B A B c b b Fig. 1. Convex pentagonal tiles of type 1–type 14. tagons introduced in Proposition 1 in this paper may not be of the tiling in F, respectively. The tiling J0 is balanced sufficient for solving the convex pentagonal tiling problem. if it is normal and satisfies the following condition: the However, by accumulating such properties one by one, we limits limρ→∞ N(F)/P(F) and limρ→∞ E(F)/P(F) exist are steadily approaching the complete solution of the prob- and are finite (see Section 3.3 in Grunbaum¨ and Shephard lem. (1987)). Proof. Given a closed disk W of radius ρ(>0) on normal 2. Proof of Proposition 1 tiling J ,letP(F1), P(F2), and P(F) denote the number First, the definition required for proof is introduced. of pentagons in F1, the number of pentagons in F2,and Given a normal tiling J0 by polygons, let W be a closed the number of pentagons in F, respectively. Since J is disk of radius ρ(> 0) on the plane. Then, let F1 and F2 normal, the following relations hold (see Statement 3.2.2 in denote the set of the polygons contained in W and the set Grunbaum¨ and Shephard (1987)): of polygons meeting the boundary of W but not contained P(F ) P(F ) in W, respectively. Here, we define F := F ∪ F and de- lim 1 = 1 and lim 2 = 0. (1) 1 2 ρ→∞ ( ) ρ→∞ ( ) note by P(F) the number of polygons in F. In addition, P F P F let E(F) and N(F) denote the numbers of edges and nodes Two tiles are called adjacent if they have an edge in Properties of Nodes in Pentagonal Tilings 119 c Condition B b C A a AB = = 135 , d C = 90 , D E D = 112.5 , e E = 67.5 , abcd = = = . Nodes’ combinations 2 EBC + + = 360 , 2 EAC + + = 360 , 2 DA + = 360 , 2 BC + = 360 , A + B + C = 360 . Fig. 2. Tiling by convex pentagons that belong to both type 1 and type 7. (a) c b (b) e d B D C A E C a d a c D E A B e b ABC+ + = 360 , ABD+ + = 360 , ad=. acde===. Fig. 3.
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