正４面体と他の正多面体との共通の辺展開図に関する研究

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(1)Vol.2010-AL-132 No.1 2010/11/19. 情報処理学会研究報告 IPSJ SIG Technical Report 1. 1. 1. 1. 1. 2. 3. 4. 5. 1/4 1/4. 正４面体と他の正多面体との共通の辺展開図に関する研究. 5. 3 -1/2. 堀. 山. 貴. 史†1. 上原. 隆. 平†2. 1/4 1/4. a. Nonexistence of Common Edge Developments of Regular Tetrahedron and Other Platonic Solids Takashi Horiyama. †1. 6. 7. 6. Figure 1 A polygon that folds to a regular tetrahedron and a cube of size 1 : 1 : √ 3 − 1/2(= 1.232).. ２個（以上）のプラトン立体を折ることのできる一つの展開図が存在するかどうかは，未解決問題である．本稿ではこの問題に対する部分的な解を与える．具体的には，正４面体を除くプラトン立体の辺展開図の中に，正４面体の展開図は存在しないことを示す．. 4. 3. 2. 1. a 8. 7. 8. a. 10. 9. 9. 10. Figure 2 A polygon that folds to a regular icosahedron and a tetramonohedron of size 1 : 1.145 : 1.25.. a. Figure 3 How to make a tetrahedron from a short pipe.. and Ryuhei Uehara†2. 0.9965 : 0.99658) . It is an open problem that whether there is a polygon that can fold to two (or more) Platonic solids. We give a partial answer to this problem. There is no other Platonic solid but regular tetrahedra that can be edge-cut to a polygon that can fold to a regular tetrahedron.. • A polygon folding to a regular icosahedron and a tetramonohedron of size 1 : 1.145 : 1.25 (Figure 2; bold lines with label “a” are glued, dotted lines are the outlines of the tetramonohedron, and each pair of numbers represents the same point that will be contained in a flat surface of the tetramonohedron). Observing these interesting polygons, it is natural to ask that whether there is a poly-. 1. Introduction. gon that can fold to two (or more) different Platonic solids. This question has arisen. There are some polygons that can fold to two different polyhedra:. several times independently (see 4) [Section 25.8.3]), and still open.. • A polygon folding to a regular octahedron and a tetramonohedron 4) [Figure 25.50]. √ • A polygon folding to a regular tetrahedron and a box of size 1×1× 3−1/2 = 1.232.. We give a partial answer to this problem. In this paper, we investigate a polygon P that folds to two different polyhedra X and Y where Y is fixed to a regular tetrahedron.. (Figure 1; this is found by Hirata in 2000. See also 4) [Figures 25.51 and 22.52]). √ • A polygon folding to a regular cube and a tetramonohedron of size 1 : 34/6 : √ 34/6 = 0.9718, and a regular octahedron and a tetramonohedron of size 1.0072 :. One reason is that this is the simplest polyhedron. In fact, in the above polygons, we can see that one of two polyhedra is always a tetrahedron. Moreover, we can make a tetrahedron by gluing two open sides of a pipe (Figure 3), and this idea with some imagination leads us to the other polygons that make a tetrahedron and another polyhedron (Figures 2 and 4). But another important reason comes from the following theorem:. †1 埼玉大学, Saitama University †2 北陸先端科学技術大学院大学, Japan Advanced Institute of Science and Technology. Theorem 1 (See 1)) Let Y be a regular tetrahedron and P be any development. 1. c 2010 Information Processing Society of Japan.

(2) Vol.2010-AL-132 No.1 2010/11/19. 情報処理学会研究報告 IPSJ SIG Technical Report. Proof. If some points of X remains out of any edge of a development P of X, P is not. (1). flat. This contradicts that P is a development of X. Let P be a polygon on the plane, and R be the set of three points (called rotation. (2). centers) on the boundary of P . Then P has a tiling called symmetry group p2 if P. (3). fills the plane by the repetition of 2-fold rotations around the points in R. The filling should contain no gaps nor overlaps. The rotation defines an equivalence relation on the points in the plane. Two points p1 , p2 are mutually equivalent if p1 can be moved. (1) A polygon folds to a regular octahedron and a tetramonohedron. (2) and (3) Polygons fold to a regular cube and a tetramonohedron.. to p2 by the 2-fold rotations. More on the properties of p2 tiling can be seen, e.g., in 7).. Figure 4 Polygons that fold to a tetramonohedron and another Platonic solid.. Based on the notion of p2 tiling, we can refine Theorem 1 as follows: Theorem 6 (See 1)，2)) P is a development of a regular tetrahedron if and only. of Y . Then P is a tiling. That is, P fills a plane.. if (1) P has a p2 tiling, (2) four of the rotation centers define the triangular lattice so. That is, if another polyhedron X has a development that folds to a regular tetrahedron,. that the four are the lattice points, and (3) no two of the four rotation centers belong. it should be a tiling by Theorem 1. This is a very useful property.. to the same equivalent class on the tiling.. We also assume that the development P of X is obtained by edge-cutting. That is,. The proof is a straight-forward extension of Akiyama’s one 1) (c.f. 2)). We note that P. we obtain P by just cutting edges of X. The main theorem in this paper is as follows:. is any development and is not limited to a development by edge-cutting; you can cut surfaces with any curve.. Theorem 2 There exists no polygon P such that (1) P is a development of a regular tetrahedron, and (2) P is a development by edge-cutting of X, where X is a Platonic. 3. Regular polyhedra versus regular tetrahedron. solid except a regular tetrahedron.. Let X be a Platonic solid except regular tetrahedron. That is, X is one of regular. 2. Preliminaries. cube, regular octahedron, regular dodecahedron, and regular icosahedron. In this sec-. We first show some basic lemmas about developments of a convex polyhedron ob-. tion, we show that each regular polyhedra cannot share its development with a regular. tained by an edge-cutting.. tetrahedron.. Lemma 3 4) [Lemma 22.1.1] Any edge-cutting of a polyhedron produces a tree.. 3.1 Regular dodecahedron. Lemma 4 Let T be a tree produced by an edge-cutting of a convex polyhedron X.. The easiest case is that X is a regular dodecahedron.. Then, each leaf and branch corresponds to a point of X.. Lemma 7 Let P be any development of a regular dodecahedron X by edge-cutting.. Proof. Each leaf has to have a curvature less than 2π; otherwise, the development P. Then P cannot fold to a regular tetrahedron Y .. of X produced by T has redundant cutting (see 6) [Lemma 1] for the details of this. Proof. The polygon P consists of 12 regular pentagons. It is well known that regular. argument). Since only points have curvatures less than 2π, the lemma follows for leaves.. pentagons cannot make a tiling; only regular triangles, squares, or regular hexagons can. For each branch, the lemma is clear since each edge is adjacent to other edges at a point.. make a tiling. Thus, by Theorem 1, P cannot fold to Y . 3.2 Regular cube and regular octahedron. Lemma 5 All points of a polyhedron X are on the edges of any development of X.. Figure 5 (a) is a development of a regular cube by edge-cutting, and has two p2. 2. c 2010 Information Processing Society of Japan.

(3) Vol.2010-AL-132 No.1 2010/11/19. 情報処理学会研究報告 IPSJ SIG Technical Report q2 q1. b a. Figure 6 Unmatching lengths. (a). (b) either (1) Y can be obtained by edge-to-edge gluing of P 0 , or (2) Y can be obtained by edge-to-edge gluing of P 0 with some gluings of rolling belts on P 0 . If P 0 contains rolling belts, each of them has length (`/2) for some positive integer ` (remind that each edge of X has a unit length). Proof.(Sketch.) This is obtained by slight extension of Theorem 25.3.1 in 4). Hirata. (c). showed the case (1). The case (2) is a simple extension of (1)5) .. Figure 5 A development of a cube and its tilings.. We now show the main theorem for a regular icosahedron. tilings in Figure 5 (b) and (c). In both situations, the development has six rotation. Theorem 9 Let P be any development by edge-cutting of X. Then P cannot fold. centers (black circles in the figure), while no four of the rotation centers consist the. to a regular tetrahedron Y .. triangular lattice. We enumerate all possible tilings of each one of eleven edge-cutting. Proof. Let P 0 = (v0 , u0 , v1 , u1 , . . . , v21 , u21 , v0 ) be the refined polygon of P described. developments of a regular cube. As a result, there is no development that satisfies the. above. Let U = {v0 , u0 , v1 , u1 , . . . , v21 , u21 }. We suppose that P 0 can fold to a regu-. required property of a development of a regular tetrahedron in Theorem 6. For a regular. lar tetrahedron Y to derive contradictions. Since Y has the same area of the regular √ icosahedron, each edge of Y is of length 5.. octahedron, we have the same result.. By Lemma 3, P 0 is glued to a tree T on Y . Since T has at least two leaves, there are. 3.3 Regular icosahedron It is well known that a regular icosahedron has 43380 different edge-cutting devel-. two points q1 and q2 in U that correspond to two points on the regular tetrahedron by. 3). opments . Hence case-analysis like a regular cube and a regular octahedron is not. 4. Then, by Lemma 8, we have two cases.. √ In case (1), q1 and q2 are some grid points on the triangular lattice of size 5. Hence √ the distance between q1 and q2 on the triangular lattice of unit length is 5c for some. reasonable. We let X be a regular icosahedron with unit length edges and P a development by edge-cutting of X. Let T be a set of edges on X such that cutting them yields P . Then, by Lemma 3, T induces a tree with 11 edges on X. Thus P consists. positive integer c. Thus, there exist two nonnegative integers a, b with a + b > 0 such √ that ( 5c)2 = a2 + b2 − 2ab cos(5π/6) (see Figure 6). When a > 0 and b > 0, this √ equation implies that 5c2 = a2 + b2 + 2 3ab. But there are no three positive integers. of 22 edges and 22 vertices. (We note that some vertices may be on a straight line of length 2 or more. But we consider that each edge has a unit length and each longer line. a, b and c that satisfy this equation. Hence we have a contradiction. If a = 0 or b = 0, √ √ 5c = b or 5c = a, that is also a contradiction.. consists of unit edges.) We denote these 22 vertices by v0 , v1 , . . . , v21 consecutively in clockwise. For each i with 0 ≤ i < 21, let ui be the center point on the edge (vi , vi+1 ),. we have. and u21 be the center point on the edge (v21 , v0 ). We let P 0 be the refined polygon. In case (2), we first observe that there are at most two rolling belts on Y ; otherwise, some belts are not independent. Using the similar manner, only considerable case is √ that q1 and q2 are on the different rolling belts, and the distance between them are 5c. v0 , u0 , v1 , u1 , . . . , v21 , u21 obtained from P . Then, we have the following lemma: Lemma 8 Let Y be any convex polyhedron obtained by any folding of P 0 . Then. 3. c 2010 Information Processing Society of Japan.

(4) Vol.2010-AL-132 No.1 2010/11/19. 情報処理学会研究報告 IPSJ SIG Technical Report. for some positive integer c. However, in the case, each belt produces an integer length √ edge on Y . On the other hand, each edge on Y has length 5. That is a contradiction.. References 1) Akiyama, J.: Tile-Makers and Semi-Tile-Makers, The Mathematical Association of Amerika, Vol.Monthly 114, pp.602–609 (2007). 2) Akiyama, J. and Nara, C.: Developments of Polyhedra Using Oblique Coordinates, J.Indonesia.Math.Soc., Vol.13, No.1, pp.99–114 (2007). 3) Buekenhout, F. and Parker, M.: The number of nets of the regular convex polytopes in dimension ≤ 4, Discrete Mathematics, Vol.186, No.1-3, pp.69–94 (1998). 4) Demaine, E. D. and O’Rourke, J.: Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Cambridge University Press (2007). 5) Hirata, K.: Personal communication (2010). 6) Mitani, J. and Uehara, R.: Polygons Folding to Plural Incongruent Orthogonal Boxes, Canadian Conference on Computational Geometry (CCCG 2008), pp.39–42 (2008). 7) Schattschneider, D.: The plane symmetry groups: their recognition and notation, American Mathematical Monthly, Vol.85, pp.439–450 (1978). 8) Shirakawa, T.: Personal communication (2010).. 4. c 2010 Information Processing Society of Japan.

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