US20100225057A1 - Three dimensional puzzle - Google Patents
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- US20100225057A1 US20100225057A1 US12/727,390 US72739010A US2010225057A1 US 20100225057 A1 US20100225057 A1 US 20100225057A1 US 72739010 A US72739010 A US 72739010A US 2010225057 A1 US2010225057 A1 US 2010225057A1
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- A—HUMAN NECESSITIES
- A63—SPORTS; GAMES; AMUSEMENTS
- A63F—CARD, BOARD, OR ROULETTE GAMES; INDOOR GAMES USING SMALL MOVING PLAYING BODIES; VIDEO GAMES; GAMES NOT OTHERWISE PROVIDED FOR
- A63F9/00—Games not otherwise provided for
- A63F9/06—Patience; Other games for self-amusement
- A63F9/12—Three-dimensional jig-saw puzzles
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- G—PHYSICS
- G09—EDUCATION; CRYPTOGRAPHY; DISPLAY; ADVERTISING; SEALS
- G09B—EDUCATIONAL OR DEMONSTRATION APPLIANCES; APPLIANCES FOR TEACHING, OR COMMUNICATING WITH, THE BLIND, DEAF OR MUTE; MODELS; PLANETARIA; GLOBES; MAPS; DIAGRAMS
- G09B23/00—Models for scientific, medical, or mathematical purposes, e.g. full-sized devices for demonstration purposes
- G09B23/02—Models for scientific, medical, or mathematical purposes, e.g. full-sized devices for demonstration purposes for mathematics
- G09B23/04—Models for scientific, medical, or mathematical purposes, e.g. full-sized devices for demonstration purposes for mathematics for geometry, trigonometry, projection or perspective
Abstract
A three-dimensional puzzle which forms a regular polyhedron and has not conventionally existed is realized. In addition, a three-dimensional puzzle which realizes a Fedrov space filling solid and has not conventionally existed is realized. According to the invention, a three-dimensional puzzle is provided having a regular polyhedron consisting of a plurality of convex polyhedrons which fill an interior of the regular polyhedron comprising the plurality of convex polyhedrons having a plurality of a pair of convex polyhedrons in a mirroring image relationship, wherein the plurality of convex polyhedrons are indivisible into two or more congruent shaped polyhedrons. In addition, the plurality of convex polyhedrons may be four convex polyhedrons and include three pairs of convex polyhedrons in a mirroring image relationship. Further the plurality of convex polyhedrons may be five convex polyhedrons and include four pairs of convex polyhedrons in a mirroring image relationship.
Description
- This application is based upon and claims the benefit of priority from the prior Japanese Patent Application No. 2008-156057, filed on Jun. 14, 2008, Japanese Patent Application No. 2008-268221, filed on Oct. 17, 2008, Japanese Patent Application No. 2008-277198, filed on Oct. 28, 2008, and PCT Application No. PCT/JP2009/060763, filed on Jun. 12, 2009, the entire contents of which are incorporated herein by reference.
- 1. Technical Field
- The present invention is related to a three-dimensional puzzle. In particular, the present invention is related to a regular polyhedron puzzle which includes a plurality of convex polyhedrons which fill an interior, and a three-dimensional puzzle which can realize a Fedorov space-filling three dimensional solid.
- 2. Description of the Related Art
- A regular polyhedron puzzle is known in which the regular polyhedron is divided into several convex polyhedrons. The convex polyhedrons which form this conventional regular polyhedron puzzle do not have particular regularity or characteristics, and are appropriately divided in order to adjust the difficulty and complexity of the puzzle (e.g. Japan Laid Open Utility Model Publication No. S63-200867, Japan Laid Open Patent Publication No. S58-049168, Japan Registered Utility Model Application No. 3107739, International Laid Open Pamphlet No. 2006/075666 and Japan Laid Open Patent Publication No. H03-155891).
- The present invention forms a regular polyhedron and realizes a three-dimensional puzzle which has not conventionally existed. In addition, it is possible to realize a Fedrov space filling solid and realize a three-dimensional puzzle which has not conventionally existed.
- According to one embodiment of the present invention, a three-dimensional puzzle is provided including four types of convex polyhedrons from which a regular tetrahedron, a cube, a regular octahedron, a regular dodecahedron or a regular icosahedron are formed, wherein three types among the four types of convex polyhedrons each having a pair of convex polyhedrons in a mirroring image relationship, the four types of convex polyhedrons are indivisible into two or more congruent shaped polyhedrons, and the regular tetrahedron, the cube, the regular octahedron, the regular dodecahedron and the regular icosahedron are formed using only the four types of convex polyhedrons so that the interior of the regular tetrahedron, the cube, the regular octahedron, the regular dodecahedron and the regular icosahedron are filled.
- In addition, according to one embodiment of the present invention a three-dimensional puzzle is provided including five types of convex polyhedrons from which a regular tetrahedron, a cube, a regular octahedron, a regular dodecahedron or a regular icosahedron are formed, wherein four types among the five types of convex polyhedrons each having a pair of convex polyhedrons in a mirroring image relationship, the five types of convex polyhedrons are indivisible into two or more congruent shaped polyhedrons, and the regular tetrahedron, the cube, the regular octahedron, the regular dodecahedron and the regular icosahedron are formed using only the five types of convex polyhedrons so that the interior of the regular tetrahedron, the cube, the regular octahedron, the regular dodecahedron and the regular icosahedron are filled.
- According to one embodiment of the present invention, a Fedorov space-filling three-dimensional puzzle is provided including a plurality of first convex polyhedrons and a plurality of second convex polyhedrons which are in a mirroring relationship with the first convex polyhedrons from which the three-dimensional puzzle is formed, wherein the plurality of first convex polyhedrons and the plurality of second convex polyhedrons each are indivisible into two or more congruent shaped polyhedrons, and the plurality of first convex polyhedrons and the plurality of second convex polyhedrons form all the Fedorov space-filling three dimensional solids by filling the interior of all the Fedorov space-filling three dimensional solids.
- The Fedorov space-filling three-dimensional puzzle may also include an elongated rhombic dodecahedron which is formed using the plurality of first convex polyhedrons and the plurality of second convex polyhedrons includes a truncated octahedron, a parallel hexahedron, a skewed hexagonal prism and a rhombic dodecahedron.
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FIG. 1 is a diagram which shows a three-dimensional puzzle (regular tetrahedron) 100 related to one embodiment of the present invention; -
FIG. 2( a) is a diagram which shows acube 200,FIG. 2( b) is a diagram which shows atriangular pyramid 201 including a peak B which has been cut away; -
FIG. 3( a) is a diagram which shows the appearance of forming aquadrangular pyramid 301 from fourtriangular pyramids 201,FIG. 3( b) is a diagram which shows the appearance of forming aregular octahedron 300 from twoquadrangular pyramids 301; -
FIG. 4( a) is a diagram which shows three points Q, T, U respectively placed on the three edges of thetriangular pyramid 201,FIG. 4( b) is a diagram which shows atriangular pyramid 203 including a peak A which has been cut away from thetriangular pyramid 201; -
FIG. 5( a) and (b) are diagrams which show thetriangular pyramid 203 including the peaks C, F which has been cut away,FIG. 5( c) is a diagram which shows aregular icosahedron 400 formed from four pairs of aheptahedron 207 and a mirror imagesymmetrical heptahedron 209; -
FIG. 6( a) is a diagram which shows aregular dodecahedron 500,FIG. 6( b) is a diagram which shows a cut away through a plane which passes through the peaks A, B, C, D and a plane which passes through the peaks A, B, E, F,FIG. 6( c) shows the appearance of thecube 200 which is left at the center of the cut away; -
FIG. 7( a) is a diagram which shows a view of theregular tetrahedron 100 again,FIG. 7( b) is a diagram which shows theregular tetrahedron 100 cut away through a plane which passes through the center points I, J, K, L on the four edges AF, FC, CH, and HA,FIG. 7( c) is a diagram of an extractedprism 101,FIG. 7( d) is a diagram of a cut away through a plane which passes through three points J, L, M when M is the center point of the edge FH,FIG. 7( e) is a diagram in which apentahedron 103 is divided into twocongruent tetrahedrons -
FIG. 8 is a diagram which shows a development of apiece 110 which is the shape of an atom α which forms the three-dimensional puzzle related to one embodiment of the present invention; -
FIG. 9 is a diagram which shows a view of anequihepta 207 again,FIG. 9( b) is a diagram which shows the appearance of the equihepta divided into three solids; -
FIG. 10 is a development of apiece 210 which is the shape of an atom γ which forms the three-dimensional puzzle related to one embodiment of the present invention; -
FIG. 11 is a development of apiece 203 which is the shape of an atom γ which forms the three-dimensional puzzle related to one embodiment of the present invention; -
FIG. 12( a) is a diagram which shows aroof 501 related to one embodiment of the present invention again,FIG. 12( b) shows apiece 510 which is the shape of an atom δ which forms the three-dimensional puzzle related to one embodiment of the present invention; -
FIG. 13 shows thepiece 510 which is the shape of an atom δ which forms the three-dimensional puzzle related to one embodiment of the present invention; -
FIG. 14 is a diagram which shows a cube three-dimensional puzzle 200 related to one embodiment of the present invention; -
FIG. 15 is a diagram ofparts 201 of a triangular pyramid related to one embodiment of the present invention which is formed usingparts 207 comprised of apiece 210 which is the shape of the atom β and apiece 203 which is the shape of the atom γ; -
FIG. 16 is a diagram ofparts 207 related to one embodiment of the present invention which is formed using apiece 210 which is the shape of three atoms β; -
FIG. 17 is a diagram which shows a regular octahedron three-dimensional puzzle 300 related to one embodiment of the present invention; -
FIG. 18 is a diagram which shows a regular dodecahedron three-dimensional puzzle 500 related to one embodiment of the present invention; -
FIG. 19 is a diagram which shows a regular icosahedron three-dimensional puzzle 400 related to one embodiment of the present invention; -
FIG. 20( a) is a diagram which shows the appearance of theroof 501 which is reversed so as to fill the interior of thecube 200,FIG. 20( b) is a diagram which shows ahexahedron 601 related to one embodiment of the present invention,FIG. 20( c) shows a development of thehexahedron 601,FIG. 20( d) is a diagram which shows the appearance when the interior of thecube 200 is filled with the two mirror imagesymmetrical hexahedrons -
FIG. 21 is a development of a piece 610 which is the shape of an atom ε related to one embodiment of the present invention; -
FIG. 22( a) is a view of thepiece 110 which is the shape of the atom α again,FIG. 22( b) is a diagram which shows thepiece 110 divided into atetrahedron 121 and atetrahedron 123,FIG. 22( c) is a diagram which shows apiece 151 which is the shape of an atom ξ and apiece 152 which is the shape of an atom ξ′ related to one embodiment of the present invention,FIG. 22( d) is a diagram which shows a quadrangular pyramid formed by gluing atetrahedron 123 and atetrahedron 124; -
FIG. 23 is a development of thepiece 151 which is the shape of the atom ξ related to one embodiment of the present invention; -
FIG. 24 is a diagram which shows the appearance of an end of thepiece 203 which is the shape of the atom γ which is divided into apiece 800 which is the shape of the atom θ and apiece 900 which is the shape of the atom η; -
FIG. 25 is a development of thepiece 800 which is the shape of the atom θ related to one embodiment of the present invention; -
FIG. 26 is a development of thepiece 900 which is the shape of the atom η related to one embodiment of the present invention; -
FIG. 27 is a diagram which shows apiece 700 which is the shape of a polyhedron corresponding to two atoms α related to an embodiment of the present invention; -
FIG. 28 is a diagram which shows a regular tetrahedron three-dimensional puzzle related to one embodiment of the present invention; -
FIGS. 29( a) to (e) are diagrams of five types of Fedorov space-filling three-dimensional puzzles related to one embodiment of the present invention.FIG. 29( a) shows a parallel hexahedron,FIG. 29( b) shows a rhombic dodecahedron,FIG. 29( c) shows a skewed hexagonal prism,FIG. 29( d) shows an elongated rhombic dodecahedron,FIG. 29( e) shows a truncated octahedron; -
FIGS. 30( a) to (e) are exemplary diagrams which show the appearance where space is filled by stacking of translates of five types of Fedorov space-filling three-dimensional puzzles.FIG. 30( a) shows the appearance of space filling by a parallel hexahedron,FIG. 30( b) shows the appearance of space filling by a rhombic dodecahedron,FIG. 30( c) shows the appearance of space filling by a skewed hexagonal prism,FIG. 30( d) shows the appearance of space filling by a elongated rhombic dodecahedron,FIG. 30( e) shows the appearance of space filling by a truncated octahedron; -
FIG. 31( a) is a diagram which shows acube 1200,FIG. 31( b) shows the cube divided into six congruentquadrangular pyramids 1201; -
FIG. 32( a) is a diagram which shows one of thequadrangular pyramids 1201FIG. 32( b) shows one of thequadrangular pyramids 1201 divided intoright tetrahedrons 1203 by two perpendicular planes,FIG. 32( c) is a diagram which shows aright tetrahedron 1203; -
FIG. 33( a) shows an exemplary view where thequadrangular pyramids 1201 are glued to each side of thecube 1200,FIG. 33( b) is a diagram which shows arhombic dodecahedron 1300; -
FIG. 34( a) shows asphenoid 1100 in which tworight tetrahedrons 1203 are glued to the bottom side of right angled isosceles triangles,FIG. 34( b) shows atriangular prism 1401 in which threesphenoids 1100 are combined together,FIG. 34( c) shows a skewedhexagonal prism 1400; -
FIG. 35( a) shows arhombic dodecahedron 1300,FIG. 35( b) is an exemplary diagram which shows the formation of an elongatedrhombic dodecahedron 1500 by therhombic dodecahedron 1300 and a helmet-shaped polyhedron 1501; -
FIG. 36( a) is a diagram which shows thesphenoid 1100 related to one embodiment of the present invention,FIG. 36( b) shows thesphenoid 1100 divided into c-squadrons 1105; -
FIG. 37( a) is an enneahedron of adiamond 1601 in which four c-squadrons 1105 are combined together,FIG. 37( b) is an exemplary diagram of a truncated octahedron (1600) formed by arranging six of theenneahedrons 1601; -
FIG. 38( a) is a diagram which shows a c-squadron 1105 related to one embodiment of the present invention,FIG. 38( b) is an exemplary diagram which shows the c-squadron 1105 divided intoσ 1101 and σ′ 1103,FIG. 38( c) is an exemplary diagram which shows the formation of aright tetrahedron 1203 using two each ofσ 1101 and σ′ 1103; -
FIG. 39 is a development ofσ 1101 and σ′ 1103 related to one embodiment of the present invention; -
FIG. 40 is an exemplary diagram which shows the formation of ahexahedron 1701 usingσ 1101 and σ′ 1103 related to one embodiment of the present invention; -
FIG. 41( a) is an upper side diagram which shows the formation of apolyhedron 1703 using threehexahedrons 1701,FIG. 41( b) is a bottom side diagram of thepolyhedron 1703; -
FIG. 42( a) is an exemplary diagram which shows the formation of acube 1705 with an open hole using thepolyhedron 1703,FIG. 42( b) is a diagram which shows a combination of thecube 1705 with open hole and half of thetruncated octahedron 1603,FIG. 42( c) is a diagram which shows half of a cube formed by thecube 1705 with open hole and half of thetruncated octahedron 1603,FIG. 42( d) is a diagram which shows thecube 1700 which is formed; -
FIG. 43( a) is a diagram which shows half of arhombic dodecahedron 1301,FIG. 43( b) is a diagram which shows a helmet-shapedpolyhedron 1501,FIG. 43( c) is a diagram which shows apolyhedron 1503 cut away from the helmet-shapedpolyhedron 1501,FIG. 43( d) is a diagram which shows the formation of a skewedhexagonal prism 1420 from thepolyhedron 1503 and half of therhombic dodecahedron 1301,FIG. 43( e) is a diagram which shows the skewedhexagonal prism 1420; -
FIG. 44 shows an arrangement of apuzzle 1651 in which a truncated octahedron three-dimensional puzzle related to one embodiment of the present invention is halved,FIG. 44( a) shows a side face view,FIG. 44( b) shows a side face view,FIG. 44( c) shows an upper face view,FIG. 44( d) shows a bottom face view; -
FIG. 45 shows an arrangement of apuzzle 1650 in which a truncated octahedron three-dimensional puzzle related to one embodiment of the present invention is halved,FIG. 45( a) shows a side face view,FIG. 45( b) shows a side face view,FIG. 45( c) shows a bottom face view; -
FIG. 46 is a diagram which shows an arrangement of a cube three-dimensional puzzle related to one embodiment of the present invention,FIG. 46( a) shows a bottom face view of apuzzle 1651 in which a truncated octahedron three-dimensional puzzle is halved,FIG. 46( b) shows apuzzle 1751 formed by cutting a cube with an open hole,FIG. 46( c) shows half of acube puzzle 1753 in which thepuzzle 1651 which is half of a truncated octahedron three-dimensional puzzle, is glued to thepuzzle 1751 formed by cutting a cube with an open hole,FIG. 46( d) shows a cube three-dimensional puzzle 1750; -
FIG. 47 is a diagram which shows an arrangement of a rhombic dodecahedron three-dimensional puzzle related to one embodiment of the present invention,FIG. 47( a) shows half of acube puzzle 1753,FIG. 47( b) shows apuzzle 1353 where half of a cube is removed from apuzzle 1351 which is half of a rhombic dodecahedron,FIG. 47( c) shows thepuzzle 1351 which is half of a rhombic dodecahedron,FIG. 47( d) shows a side face view of arhombic dodecahedron puzzle 1350, FIG. 47.(e) shows an upper side view of therhombic dodecahedron puzzle 1350; -
FIG. 48 is a diagram which shows an arrangement of a skewed hexagonal prism three-dimensional puzzle related to one embodiment of the present invention,FIG. 48( a) shows a side face view of apuzzle 1553 which has been cut away from a helmet-shapedpolyhedron puzzle 1551,FIG. 48( b) shows a front face view ofFIG. 48( a),FIG. 48( c) shows a back face view ofFIG. 48( a),FIG. 48( d) shows an exemplary diagram of the formation of a skewedhexagonal prism puzzle 1450 by thepuzzle 1553 which is cut away from the helmet-shapedpolyhedron 1551, and thepuzzle 1351 which is half of therhombic dodecahedron 1350,FIG. 48( e) shows an upper face view of the skewedhexagonal prism puzzle 1450,FIG. 48( f) shows a side face view of the skewedhexagonal prism puzzle 1450; and -
FIG. 49 is a diagram which shows an arrangement of an elongated rhombic dodecahedron three-dimensional puzzle related to one embodiment of the present invention,FIG. 49( a) shows a side face view of the helmet-shapedpolyhedron puzzle 1551,FIG. 49( b) shows a front face view of the helmet-shapedpolyhedron puzzle 1551,FIG. 49( c) shows a back face view of the helmet-shapedpolyhedron puzzle 1551,FIG. 49( d) shows an exemplary diagram of the formation of an elongatedrhombic dodecahedron puzzle 1550 by the helmet-shapedpolyhedron puzzle 1551 and therhombic dodecahedron puzzle 1350,FIG. 49( e) shows an upper face view of the elongatedrhombic dodecahedron puzzle 1550,FIG. 49( f) shows a side face view of the elongatedrhombic dodecahedron puzzle 1550. -
- 100 regular tetrahedron
- 101 prism
- 103 pentahedron
- 110 atom α
- 111 atom α′
- 121 tetrahedron LJKW
- 123 tetrahedron LHKW
- 124 mirror image of
tetrahedron 123 - 151 atom ξ
- 152 atom ξ
- 200 cube
- 201 regular triangular pyramid
- 203 golden tetra (atom γ)
- 204 atom γ′
- 205 polyhedron except golden tetra from regular triangular pyramid
- 207 equihepta
- 209 mirror image of equihepta
- 210 atom β
- 211 atom β′
- 300 regular octahedron
- 301 quadrangular pyramid
- 400 regular icosahedron
- 500 regular dodecahedron
- 501 roof
- 510 atom δ
- 601 hexahedron
- 602 mirror image of
hexahedron 601 - 610 atom ε
- 611 atom ε′
- 700 α2
- 701 quadrangular pyramid LHKWK′
- 800 atom θ
- 801 atom θ′
- 900 atom η
- 901 atom η′
- 1100 sphenoid
- 1101 σ
- 1103 σ′
- 1105 c-squadron
- 1200 cube
- 1201 quadrangular pyramid
- 1203 triangular pyramid (right tetrahedron)
- 1300 rhombic dodecahedron
- 1301 polyhedron which is a rhombic dodecahedron cut in half
- 1350 rhombic dodecahedron puzzle
- 1351 puzzle which is a rhombic dodecahedron cut in half
- 1353 a puzzle except half a cube from a puzzle which is a rhombic dodecahedron cut in half
- 1400 skewed hexagonal prism
- 1401 triangular prism
- 1403 mirror image symmetrical triangular prism
- 1420 skewed hexagonal prism
- 1450 skewed hexagonal prism puzzle
- 1500 elongated rhombic dodecahedron
- 1501 helmet-shaped polyhedron
- 1503 polyhedron cut away from helmet-shaped
polyhedron 501 - 1505 skewed triangular prism
- 1507 mirror image of skewed triangular prism 505
- 1550 elongated rhombic dodecahedron puzzle
- 1551 helmet-shaped polyhedron puzzle
- 1553 puzzle cut away from helmet-shaped polyhedron puzzle 551
- 1600 truncated octahedron
- 1601 enneahedron (diamond)
- 1603 half of truncated octahedron
- 1650 truncated octahedron puzzle
- 1651 half of truncated octahedron puzzle
- 1700 cube
- 1701 hexahedron
- 1703 polyhedron
- 1705 cube with open hole
- 1707 polyhedron formed by cutting through cube with open hole
- 1709 polyhedron formed by cutting through half of cube
- 1750 cube puzzle
- 1751 puzzle formed by cutting through cube with open hole
- 1753 half of cube puzzle
- When the interior of a regular polyhedron is filled perfectly by a plurality of convex polyhedrons, these convex polyhedrons can be called the component elements (here called “atoms”) of a regular polyhedron. In other words, this means the regular polyhedron can be divided into several atoms. Only a regular tetrahedron, a cube, a regular octahedron, a regular icosahedron and a regular dodecahedron exist as a regular polyhedron.
- However, there are limitless methods for dividing a regular polyhedron into several convex polyhedrons. That is, there are limitless atoms for forming a regular polyhedron.
- Thus, the inventors of the present invention keenly examined which atoms should be adopted in order to be able to fill all the regular polyhedrons with their atoms while reducing the number of different atoms. As a result, a shape of the minimum number of atoms for filling all the regular polyhedrons was discovered. The circumstances in which the inventors of the present invention discovered a shape of the minimum number of atoms for filling all the regular polyhedrons is explained below.
- First, a step which excludes a self-evident atom is explained. For example, a regular tetrahedron is examined as a polyhedron. Let the center of the regular tetrahedron be the peak, and four triangles on the surface of the regular tetrahedron as the bottom surface and the regular tetrahedron is divided into four congruent triangular pyramids. Because the bottom surface of this triangular pyramid is a regular triangle, this triangular pyramid has threefold rotation symmetry and mirror image symmetry and these are further divided into six congruent triangular pyramids. Here, when the mirror image symmetrical convex parts are defined as the same atom, it is possible to perfectly fill the interior of regular tetrahedron using 24 atoms of single type.
- It is possible to apply the same division method as the division method of the regular tetrahedron stated above to a regular polyhedron. That is, a cube and regular octahedron are filled by 48 atoms and a regular dodecahedron and regular icosahedron are filled by 120 atoms. However, each of these atoms is obvious and effective in filling a specific regular polyhedron. However, when each of these atoms is diverted to an atom of another regular polyhedron, it instantly becomes an atom with bad efficiency.
- The condition A below is attached in order to remove this obvious atom.
- It is possible to use any atom as at least two types of regular polyhedron.
- Then, when an efficient atom is adopted for a specific polyhedron, it is necessary to create a method which can use this atom effectively in the filling of another regular polyhedron. Here, it is preferred that as many atoms as possible should be used to fill any one of the regular polyhedrons. However, a supplement is also required. This is because a cube is self-expanding, and when one filling method is discovered, it is possible to easily increase the number of atoms by 23, 32 etc. simply by lining it up down, left right. Consequently, a cube is examined restricted to a basic filling method which does not use the self expanding properties. In this way, filling all of the regular polyhedrons using as many atoms of as few varieties as possible while satisfying [condition A] becomes a problem.
- Furthermore, defining a mirror image symmetrical convex polyhedron as the same variety of atom.
- (1) because a regular polyhedron has various symmetries it is predicted that a lot of mirror image symmetrical polyhedrons will be used, (2) because it is common sense to identify a pair of reverse diagrams in a tiling problem very similar to this filling problem
- Five types of regular polyhedron are known, a regular tetrahedron, a regular hexahedron (cube), a regular octahedron, a regular dodecahedron and a regular icosahedron. In order to fill these regular polyhedrons with various atoms so as to satisfy [condition A], first, it is necessary to investigate the mutual relationships between the five types of regular polyhedron.
- In the
cube 200 as inFIG. 2( a), the upper face square is determined by ABCD and the bottom face square by EFGH, the cube is cut by a plane which passes through the three peaks A, C, F and atriangular pyramid 201 which includes the peak B is cut away as inFIG. 2( b). Because this cross section is a regular triangle, if a triangular pyramid which includes the peaks D, E, G is cut away by the same method, the remaining solid ACFH becomes aregular tetrahedron 100 as shown inFIG. 2 (a). The six dash lines inFIG. 2 (a) show theregular tetrahedron 100 obtained by this cutting away. - Next, as is shown in
FIG. 2( b), fourtriangular pyramids 201 which are cut away form a bottom surface of a regular triangle and a side face of a right isosceles triangle. The peaks B, C, D, E, G of thesetriangular pyramids 201 are gathered at one point, and when arranged so that the bottom face forms a square, aquadrangular prism 301 which has four side faces being all regular triangles is formed, shown inFIG. 3( a). When the bottom face squares as shown inFIG. 3( b) of two of thesequadrangular prisms 301 are attached to each other (when a pair of bottom face squares are glued together) aregular octahedron 300 is formed. One relationship between thecube 200, theregular tetrahedron 100 and theregular octahedron 300 is obtained using this cutting method. - Next, focusing on the
triangular pyramid 201 shown inFIG. 2( b), as is shown inFIG. 4( a) three points Q, T, U are each placed at proportions on three edges AB, AC, AF which satisfy the following formula: -
[formula 1] -
AQ: QB=AT:TC=AU:UF=1:τ (1) - Here, τ is a golden proportion, and as is common knowledge can be obtained by the formula (2):
-
[formula 2] -
- Then, as is shown in
FIG. 4( b) atriangular pyramid 203 including the peak A is cut away from thetriangular pyramid 201. As is shown inFIG. 5( a), (b), when thetriangular pyramid 203 including the peaks C, F is also cut away, aheptahedron 207 is left in the center. Here, the three points R, S, V are each placed on the three edges CB, FB, FC at a proportion which satisfies the formula (3) below: -
[formula 3] -
CR:RB=FS:SB=FV:VC=1:τ (3 - When a
heptahedron 207 is created in the center ofFIG. 5 b) using this method, the back face triangle TUV is a regular triangle and the three said face right triangles TQU, USV, VRT are congruent with the left hand side right triangle when the back face regular triangle TUV is divided in half left and right. - Next, the relationship established in the formula (4) below will be focused on.
-
[Formula 4] -
QBS=RBQ=SBR=90° (4) - Four pairs of this
heptahedron 207 and its mirror imagesymmetrical heptahedron 209 are created. By attaching (gluing) these together, aregular icosahedron 400 is formed as shown inFIG. 5( c). As a result, one relationship is established between thetriangular pyramid 201 and theregular icosahedron 400. - Next, a
regular dodecahedron 500 is shown inFIG. 6( a). As is shown inFIG. 6( b), the upper part is cut away by a plane which passes through four peaks A, B, C, D, and the right part is cut away by a plane which passes through four peaks A, B, E, F. Then, two cross sections become congruent squares and the two cut away solids becomecongruent pentahedrons 501. As is shown by the dash line inFIG. 6( c), when further four planes are cut away, acube 200 is left at the center. One relationship between the regular dodecahedron and the cube is obtained. - The
regular tetrahedron 100,regular octahedron 300 andregular icosahedron 400 are obtained from dividing thecube 200 as a result of the above stated examination. In addition, thecube 200 is obtained from dividing theregular dodecahedron 500. This dividing satisfies the condition [condition A]. - Next, it is examined how it is possible to fill the entire regular polyhedron while satisfying condition [condition A] with as few types of atoms as possible. The case is considered where the
regular tetrahedron 100 is formed from the dash line regular tetrahedron itself shown inFIG. 2( a). Next, thecube 200 is formed by three types of convex polyhedron (five types if mirror images are included), namely oneregular tetrahedron 100, twoheptahedrons 207 and two mirror image symmetrical heptahedrons 208, and sixtetrahedrons 203 and six mirror image symmetrical tetrahedrons. Next, theregular octahedron 300 is formed by two types of convex polyhedron (four types if mirror images are included), namely fourheptahedrons 207 and four mirror image symmetrical heptahedrons 208, and twelvetetrahedrons 203 and twelve mirror image symmetrical tetrahedrons. Next, the regular dodecahedron as shown inFIG. 6( b) is formed by four types of convex polyhedron (six type if mirror images are included), namely sixpentahedrons 501, oneregular tetrahedron 100, twoheptahedrons 207 and two mirror image symmetrical heptahedrons 208, sixtetrahedrons 203 and six mirror image symmetrical tetrahedrons. Lastly, a regular icosahedron is formed by one type of convex polyhedron consisting of fourheptahedrons 207 and four mirror image symmetrical heptahedrons 208. - From the examination above, it is clear that any of four types of convex polyhedron (six types if mirror images are included) are used in forming a regular polyhedron, namely
- [1] regular tetrahedron 100 (used for
regular tetrahedron 100,cube 200 and regular dodecahedron 500), - [2] heptahedron (equihepta) 207 (and its mirror image 208) (used in
cube 100,regular octahedron 300 and regular icosahedron 400), - [3] tetrahedron (golden tetra) 203 (and its mirror image 204) (used in
cube 100 and regular octahedron 300), and - [4] pentahedron (roof) 501 (used in regular dodecahedron 500).
- If the four types of convex polyhedron are each divided into several congruent convex polyhedrons, the number of atoms which form the regular polyhedron increases. Obvious dividing is removed and examined. First, in the
regular tetrahedron 100 in [1], theregular tetrahedron 100 which is extracted fromFIG. 2( a) is shown again inFIG. 7( a) and theregular tetrahedron 100 is cut by a plane which passes through the midpoints I, J, K, L on the four edges AF, FC, CH, HA. Then, as is shown inFIG. 7( b), this cross section becomes a square (petri figure) and two solids which are divided become congruent prisms (triangular prisms) 101.FIG. 7( c) shows a view from a different angle with one of theprisms 101 extracted. When theprism 101 is cut by a plane which passes through the three points J, L, M where M is at the center of the edge FH as shown inFIG. 7( d), twocongruent pentahedrons 103 are formed. Moreover, because eachpentahedron 103 is symmetrical, when theleft side pentahedron 103 is taken, it is possible to further divide the pentahedron into twocongruent tetrahedrons FIG. 7( e).FIG. 8 shows a development of the right side tetrahedron 110 (KJLH) and the dimensions are described as ratios. One atom which forms a regular polyhedron with thetetrahedron 110 as α is defined below. Then, as is shown inFIG. 1 , because theregular tetrahedron 100 is formed by eight atoms, this can be written as α8, as in a molecular formula. - Here, as is shown in
FIG. 7( e) atom α is the tetrahedron (convex polyhedron) 110 with peaks KJLH, and itsmirror image tetrahedron 111 with peaks MJLH. The atom α satisfies the conditions shown in formula (5) below. -
[formula 5] -
JL:LK:KL:LH:HL:LJ:JK:KH:JH=2:√2:√2:√2:√2:2:√2:√2:√6 (5) - Next, when the
equihepta 207 in [2] is examined, theequihepta 207 shown again inFIG. 9( a) has three rotational symmetry when the regular triangle TUV is the bottom surface and the point B is the peak. As a result, when the center of the regular triangle is O, and the equihepta is cut in a perpendicular direction by three faces divided into 120° pieces through O, the equihepta is always divided into threecongruent solids 210. This division is shown inFIG. 9( b) and a development is shown inFIG. 10 . - One atom which forms a regular polyhedron with the
heptahedron 210 as β is defined below. Theatom β 210, as shown inFIG. 10 , is a heptahedron (convex polyhedron) having edges BX, XY, YO, OA1, A1Z, ZT, TZ, ZA1, A1O, OY, YX, XR, RX, XB, BO, BZ, BR, RT, TY and TA1. - Here, a, k, k′, b, c, d satisfy the following formulas (6) to (11)
-
(formula 6) -
a=3−√5=2(2−τ) (6) -
(formula 7) -
k=1/(τ+2)=(5−√5)/10 (7) -
(formula 8) -
k′=1−k=(5+√5)/10 (8) -
(formula 9) -
b=2/√3 (9) -
(formula 10) -
C=(2/τ2)√(τ2−2k′+3k′ 2) (10) -
(formula 11) -
d=(2/τ2)√(4/3−4k′+4k′ 2) (11) - At this time,
atom β 210 satisfies the conditions shown in formula (12) below. -
- Because a regular icosahedron is formed by twenty-four atoms β 210, it is possible to describe this as β24.
- Next, when the
golden tetra 203 in [3] is examined, it is impossible to divide thegolden tetra 203 into a plurality of congruent convex polyhedrons. Thus, one atom which forms a regular polyhedron with thegolden tetra 203 as γ is defined.FIG. 11 is a development of theatom γ 203 and dimensions are shown according to ratios. The atom γ, as is shown inFIG. 11 , is a tetrahedron (convex polyhedron) having edges CR, RC, CV, VC, CT, TC, RT, RV and VT. The atom γ satisfies the formula (13) below. -
(formula 13) -
CR:RC:CV:VC:CT:TC:RT:RV:VT =a:a:√3a:√3a:a:a:√2a:2a: √2(√5−1) (13) - Then, because a regular octahedron is formed by twenty-four β and γ, it is possible to describe this as β24γ24. In addition, because a cube in
FIG. 1( a) is formed by eight α, twelve β and twelve γ, it is possible to describe this as α8β12γ12. - Next, when the
roof 501 in [4] is examined, while considering that theroof 501 has up down left and right symmetry, it is divided into twocongruent pentahedrons 501 as is shown inFIG. 12( a). One atom which forms a regular polyhedron with thepentahedron 510 as δ is defined.FIG. 13 is a development of theatom δ 510 and dimensions are shown according to ratios. Theatom δ 510, as is shown inFIG. 13 , is a tetrahedron (convex polyhedron) having edges BE, ED1, D1A, AD1, D1E, EB, BC1, C1E, C1D1, BA and AE. Theatom δ 510 satisfies the formula (14) below. -
(formula 14) -
BE:ED 1 :D 1 A:AD 1 :D 1 E:EB:BC 1 :C 1B: C1 E:C 1 D:BA:AE=2:√(4−τ):√(4−τ):√(4−τ): √(4−τ):2:2(τ−1):2(τ−1):2(τ−1): τ−1:2:2√2 (14) - A solution for forming a regular polyhedron by a minimum number of atoms α, β, γ and δ is obtained as described above. The number of each regular polyhedron using the atoms α, β, γ, δ can be expressed as:
-
Regular tetrahedron α8 Cube α8β12γ12 Regular octahedron β24γ24 Regular dodecahedron α8β12γ12δ12 Regular icosahedron β24 - However, a mirror image symmetrical atom is also shown by α, β, γ, δ. If a prime symbol (′) is attached to distinguish a mirror image symmetrical atom, then they can be expressed as:
-
Regular tetrahedron α4α′4 Cube α4α′4β6β′6γ6γ′6 Regular octahedron β12β′12γ12γ′12 Regular dodecahedron α4α′4β6β′6γ6τ′γδ12 Regular icosahedrons β12β′12 - From the above, it was discovered that the minimum number of atoms for forming a regular polyhedron is four types α, β, γ, δ.
- Here, it is clear that the minimum number of atoms for forming a regular polyhedron can be expressed as
theorem 1 under the definitions (1)-(4) below. - (1) polyhedrons P and Q are “congruent” means either that P and Q are identical or that P and Q have a mirror image relationship.
- (2) a polyhedron is “indecomposable” means that P is indivisible into two or more congruent polyhedrons.
- (3) Let II be a group of polyhedrons P1, P2, . . . Pn.
- i.e. II={P1, P2, . . . Pn}
- Let E be a group of indecomposable polyhedrons e1, e2, . . . em,
- i.e. E={c1, c2, . . . cm}, ∀e1 is indecomposable, ci, and cj are non-congruent (i≠j)
- At this time, E is a group E (II) of the element (atom) of II means satisfying formula (15) below.
-
(formula 15) -
(∀i<1≦i≦n), ai is a nonnegative integer. (15) - That is, whichever polyhedron Pi belongs to II, it is possible to divide into an indecomposable polyhedron belonging to E.
- (4) The element number (atom number) e (II) of II means the minimum number of an order among various element groups with respect to II.
- That is e (II)=min|E (II)|
-
- Let II1={regular polyhedron group}
- e (II1)≦4, e (II1)={α, β, γ, δ}
- Here, a three-dimensional puzzle of the present invention related to the present embodiment which uses the minimum number of atoms α, β, γ, δ for forming all the regular polyhedrons will be explained in detail. The three-dimensional puzzle of the present invention related to the present embodiment creates a regular polyhedron using the minimum number of atoms α, β, γ, δ, and has the excellent effects of being able to visually explain the above stated
theorem 1. Consequently, the three-dimensional puzzle of the present invention related to the present embodiment is excellent as an educational material for explaining the above statedtheorem 1. -
FIG. 1 is referenced.FIG. 1 is a three-dimensional puzzle of the present invention related to the present embodiment and shows a regular tetrahedron three-dimensional puzzle 100. As is shown inFIG. 1 , the three-dimensional puzzle (regular tetrahedron three-dimensional puzzle 100) can be formed by four pieces which have the shape ofatom α 110 and four pieces which have a shape of a mirror image α′ 111 ofatom α 110 as explained in detail above. Furthermore, the regular tetrahedron three-dimensional puzzle 100 of the present invention related to the present embodiment is one example and there are also examples where the arrangement of theatom α 110 and the atom α′ 111 can be changed while maintaining the relative positional relationship between them. -
FIG. 14 is referenced.FIG. 14 is a three-dimensional puzzle of the present invention related to the present embodiment and shows a cube three-dimensional puzzle 200. The cube includes theregular tetrahedron 100 which is comprised of the atom α (and α′) as explained in detail above, and is formed by gluing atriangular pyramid 201 which has side faces which are right isosceles triangles to each of the four faces of the cube. In addition, as is shown inFIG. 15 , thetriangular pyramid 201 is a part arranged with threepieces 203 which has the shape of the atom γ with parts 207 (FIG. 16 ) comprised of threepieces 210 which have the shape of atom β at the center. Therefore, the three-dimensional puzzle of the present invention related to the present embodiment (cube puzzle 200) can be formed by using fourpieces 110 which have the shape of atom α and fourpieces 111 which have the shape of the atom α mirror image atom α′, sixpieces 210 which have the shape of atom β and six pieces 211 which have the shape of the atoms β mirror image atom β′, and sixpieces 203 which have the shape of atom γ and six pieces 204 which have the shape of the atom γ mirror image atom γ′. Furthermore, the cube three-dimensional puzzle 200 of the present invention related to the present embodiment is one example and there are also examples where the arrangement of the atoms α and α′, β and β′, and γ and γ′ can be changed while maintaining the relative positional relationship between them. -
FIG. 17 is referenced.FIG. 14 is a three-dimensional puzzle of the present invention related to the present embodiment and shows a regular octahedron three-dimensional puzzle 300. The regular octahedron three-dimensional puzzle 300 is formed by gluing eight regular threesided pyramids 201 which have sides which are right isosceles triangles. Therefore, the three-dimensional puzzle of the present invention related to the present embodiment (regular octahedron puzzle 300) can be formed by using twelvepieces 210 which have the shape of atom β and twelve pieces 211 which have the shape of the atom β mirror image atom β′, and twelvepieces 203 which have the shape of atom γ and twelve pieces 204 which have the shape of the atom γ mirror image atom γ′. Furthermore, the regular octahedron three-dimensional puzzle 300 of the present invention related to the present embodiment is one example and there are also examples where the arrangement of the atoms β and β′, and γ and γ′ can be changed while maintaining the relative positional relationship between them. -
FIG. 18 is referenced.FIG. 18 is a three-dimensional puzzle of the present invention related to the present embodiment and shows a regular dodecahedron three-dimensional puzzle 500. The regular dodecahedron three-dimensional puzzle 500 includes thecube 200 at the center as shown inFIG. 6( c), and is formed by arrangingroof parts 501 comprised of twopieces 510 which have a shape of the atom δ on six sides ofcubes 200. Therefore, the three-dimensional puzzle of the present invention related to the present embodiment (regular dodecahedron puzzle 500) can be formed by using fourpieces 110 which have the shape of atom α and fourpieces 111 which have the shape of the atom α mirror image atom α′, sixpieces 210 which have the shape of atom β and six pieces 211 which have the shape of the atoms β mirror image atom β′, sixpieces 203 which have the shape of atom γ and six pieces 204 which have the shape of the atom γ mirror image atom γ′, and twelvepieces 510 which have the shape of the atom δ. Furthermore, the regular dodecahedron three-dimensional puzzle 500 of the present invention related to the present embodiment shown inFIG. 18 is one example and there are also examples where the arrangement of the atoms α 110 and α′111,β 210 and β 211′,γ 203 and γ′ 203, can be changed while maintaining the relative positional relationship between them. -
FIG. 19 is referenced.FIG. 19 is a three-dimensional puzzle of the present invention related to the present embodiment and shows a regular icosahedron three-dimensional puzzle 400. Four pairs ofparts 207 which have the shape of an equihepta comprised from threepieces 210 which have the shape of the atom β shown inFIG. 16 , and theparts 209 which have the equihepta mirror image symmetrical shape are glued together to form the regular icosahedron. Therefore, the three-dimensional puzzle of the present invention related to the present embodiment (regular icosahedron puzzle 400) can be formed by using twelvepieces 210 which have the shape of atom β and twelve pieces 211 which have the shape of the atoms β mirror image atom β′. Furthermore, the regular icosahedron three-dimensional puzzle 400 of the present invention related to the present embodiment shown inFIG. 19 is one example and there are also examples where the arrangement of the atoms β 210 and β′ 211, can be changed while maintaining the relative positional relationship between them. - As explained above, because the three-dimensional puzzle of the present invention related to the present embodiment can be formed by a minimum number of atoms (convex polyhedrons) which fill all the regular polyhedrons, the present invention demonstrated excellent effects of being able to form all the regular polyhedrons (regular tetrahedron, cube, regular octahedron, regular dodecahedron, regular icosahedron) even with few parts. In addition, it is possible to visually prove the theorem discovered by the inventors and the present invention can also be used as an excellent educational material for explaining the above stated
theorem 1. - In the second embodiment, an example of a three-dimensional puzzle which can form a cube and a regular dodecahedron without using the atom α of embodiment one is explained.
-
FIG. 20 is referenced.FIG. 20( a) shows the appearance when six roofs 501 (δ2) comprised from two atoms δ 510 are reversed so as to fill the interior of acube 200. InFIG. 20( a) only theroof 501 of the back face, side face and bottom face is shown to facilitate visualization. Each face of the pentahedron 501 (δ2) which is reversed overlaps so that no gaps exist as is clear from the dihedral angle of each face. However, a long thin gap is produced in the eight corners of the cube and four mirror imagesymmetrical hexahedrons 601 and fourhexahedrons 602 are required to fill these gaps.FIG. 20( b) shows ahexahedron 601 for filling the forefront left side lower gap inFIG. 20( a), andFIG. 20( c) shows a development of this. In addition,FIG. 20( d) shows the appearance when two lower left gaps of two of the mirror imagesymmetrical hexahedrons regular dodecahedron 500 andcube 200 is obtained which is different to the relationship in the first embodiment. - A cube which is different to the cube explained in the first embodiment is comprised of two types of convex polyhedron, namely, six roofs 501 (δ2), four
hexahedrons 601 and four mirror imagesymmetrical hexahedrons 602. In addition, theregular dodecahedron 500 is comprised of two types of convex polyhedron, namely, twelve roofs 501 (δ2), fourhexahedrons 601 and four mirror imagesymmetrical hexahedrons 602. - From the above investigation it is clear that the five types of convex polyhedrons are used in the structure of any two or more regular polyhedrons:
- [1] regular tetrahedron 100 (used in a
regular tetrahedron 100 and cube 200). - [2] equihepta 207 and its mirror image 209 (used in a
cube 100,regular octahedron 300 and regular icosahedron 500). - [3]
golden tetra 203 and its mirror image 204 (used in acube 200 and regular octahedron 300). - [4] roof 501 (used in a regular dodecahedron 500).
- [5]
hexahedron 601 and its mirror image 602 (used in acube 200 and regular dodecahedron 500). - Here, because the convex polyhedrons [1] to [4] were explained in the first embodiment, an explanation is omitted here.
- When the
hexahedron 601 in [5] is examined, it has three rotational symmetry as is clear from the hexahedron inFIG. 20( b) and is the development inFIG. 20( c). As a result, by cutting in a perpendicular direction by three faces of 120° each, the hexahedron is divided into three congruent polyhedrons. A development of three congruent quadrangular pyramids 610 which have been cut along the edges of thehexahedron 601 is shown inFIG. 21 . Furthermore, dimensions are described using ratios in the development. A quadrangular pyramid 610 shown inFIG. 21 is defined as atom ε. - Atom ε 610 is a pentahedron (convex polyhedron) which has edges D1E1, E1F1, F1G1, G1H1, H1I1, I1J1, J1K1, K1D1, D1F1, D1I1, D1J1 and F1I1 Atom ε satisfies the following formula (16)
-
(formula sixteen) -
D 1 E 1 :E i F 1 :F 1 G 1 :G 1 H 1 :H 1 I 1 :I 1 J 1 : J 1 K 1 :K 1 D 1 :D 1 F 1 :D 1 I 1 :D 1 J 1 :F 1 I 1=√3:2−τ:2−τ:2−τ:√(18−11τ): √(18−11τ):2−τ:√3:√(4−τ):2(τ−1): √(4−τ):τ−1 (16) - Because the
cube 200 is comprised of twelve atoms δ 510 and twenty-four atoms ε (twelve atoms ε and twelve mirror image atoms ε′), it is expressed as δ12ε24 (δ12ε12ε′12) and because theregular dodecahedron 500 is comprised of twenty-four δ and twenty-four ε, it is expressed as δ24ε24 (δ24ε12ε′12). - By using the atoms α, β, γ, δ, ε, each individual regular polyhedron becomes:
-
Regular tetrahedron α8 Cube α8β12γ12, δ12ε24 Regular octahedron β24γ24 Regular dodecahedron δ24ε24
However, the same symbols α, β, γ, δ, ε, also indicate mirror image symmetrical atoms. If the mirror image symmetrical atoms are attached with a prime symbol (′) then: -
Regular tetrahedron α4α′4 Cube α4α′4β6β′6γ6γ′6, δ12ε12ε′12 Regular octahedron β12β′12γ12γ′12 Regular dodecahedron δ24ε12ε′12 Regular icosahedron β12β′12 - In the three-dimensional puzzle of the present invention related to the present embodiment, it is possible to form a cube and a regular dodecahedron without using atom α by using ε in addition to the minimum number of atoms α, β, γ, δ which form all the regular polyhedrons. Consequently, the three-dimensional puzzle of the present invention related to the present embodiment is created from regular polyhedrons using the atom ε in addition to the minimum number of atoms α, β, γ, δ, it is possible to visually explain the above stated
theorem 1 and demonstrate excellent effects which can compare both by a transformation example using ε which is the fifth atom. Therefore, the three-dimensional of the present invention related to the present embodiment is also excellent as an educational material. - In the present embodiment, an example of a three-dimensional puzzle which can form a
regular tetrahedron 100,cube 200 andregular octahedron 300 without using the atom α of the first embodiment is explained. -
FIG. 22 is referenced.FIG. 22( a) shows theatom α 110 shown inFIG. 7( e) seen from a different direction. Next, theatom α 110 is divided in two as inFIG. 22( b) from the peak L perpendicular to the edge JH with the foot being W. Then, theleft side tetrahedron 121 becomes three portions by dividing theregular tetrahedron 100 in a perpendicular direction from the peak and this can be divided into two mirror imagesymmetrical tetrahedrons FIG. 22( c). Thetetrahedron 151 shown inFIG. 22( c) is an atom and defined as ξ. A development of theatom ξ 151 is shown inFIG. 23 . Each dimension is shown by ratios. As is shown inFIG. 22( c), theatom ξ 151 is a tetrahedron (convex polyhedron) which has edges LM1, M1L1, L1M1, M1J, JM1, M1L, LL1, LJ and L1J. The atom ξ satisfies the following formula (17). Furthermore, as is shown inFIG. 22( c) the atom ξ has a mirror symmetrical atom ξ′ 152. -
(formula 17) -
LK 1 :M 1 L 1 :L 1 M 1 :M 1 J:JM 1 :M 1 L:LL 1 ,:LJ:L 1 J=2/√3:1 /√6:1/√6:√(2/3):√(2/3):√(2/3):2√3: √(3/2):√2:1/√2 (17) - In addition, with respect to the
right side tetrahedron 123 inFIG. 22( b), a mirror imagesymmetrical tetrahedron 124 is created and is combined (glued) by the face of triangle HLW.FIG. 22( d) is a quadrangular pyramid obtained from this method and is attached together by an unforeseen shape with the atom β 210 andatom γ 203. - In order to show this, first the
atom γ 203 is focused on. This is a slightly long and triangular pyramid, however, as is shown inFIG. 24 , theatom θ 800 is cut from the end of the atom γ. Then, a development of each of the polyhedrons is shown inFIG. 25 andFIG. 26 . The polyhedron in which theatom θ 800 is cut from the end of the atom γ is defined asatom η 900. In each ofFIG. 25 andFIG. 26 which are developments, the dimension of each edge is shown as a ratio. Furthermore, the atom η 900 and theatom θ 800 each has a mirror symmetrical atom η′ 901 and θ′ 801. -
Atom θ 800 is a tetrahedron (convex polyhedron) which has edges VO1, O1P1, P1O1, O1N1, N1O1, O1V, VP1, VN1 and N1P1 as is shown inFIG. 25 . Furthermore, the atom θ has a mirror symmetrical atom θ′ 801. Theatom θ 800 satisfies the formulas (18) and (19) below. -
(formula 18) -
e=√(2/5(9−4√5)) (18) -
(formula 19) -
VO 1 :O 1 P 1 :P 1 O 1:O1 N 1 :N 1O1 :O 1 V: VP 1 :VN 1 :N 1 P 1=√3ak:e:e:ak:ak:√3ak√2(√5−2):2ak:√3e (19) -
Atom η 900 is a pentahedron (convex polyhedron) which has edges O1N1, N1O1, O1P1, P1O1, O1C, CT, TC, CR, RC, CO1, N1P1, P1T, TR and N1R, as is shown inFIG. 25 . Furthermore, theatom η 900 has a mirror symmetrical atom η′ 901. -
FIG. 26 is a development of the atom η 900 and satisfies the formula (20) below. -
(formula 20) -
O 1 N 1 :N 1 O 1 :O 1 P 1 :P 1 O 1 :O 1 C:CT:TC:CR:RC: CO 1 :N 1 P 1 :P 1 T:TR:N 1 R=ak:ak:e:e;√3ak′:a:a:a:a:√3ak′: √3e:√2:2a:2ak′ (20) - By combining one of three types of atom,
β 210,η 900 andθ 800, it is possible to make aquadrangular pyramid 701 shown inFIG. 22( d). - By adding four atoms ξ 151 to the
quadrangular pyramid 701, two atoms α 110, that is, thepolyhedron 700 shown inFIG. 27 is obtained. As stated above, the regular tetrahedron is formed by sixteenξ 151, fourβ 210, four η 900 and fourθ 800, in total β4ξ16η4θ4 (β2β′2ξ8ξ′8η2η′2θ2θ′2). Furthermore, as already indicated, theregular tetrahedron 200 is also formed by ξ6. - When a regular tetrahedron is formed by this method, the entire structure using the
atom α 110 is replaced with this, and theregular tetrahedron 100, thecube 200, theregular octahedron 300 which are formed by the atoms β 210,ξ 151,η 900 and θ 801 each become: -
Regular tetrahedron β4ξ16η4θ4ξ6 (β2β′2ξ8ξ′8η2η′2θ2θ′2ξ3ξ′3) Cube β16ξ16η16θ16 (β8β′8ξ8ξ′8η8η′8θ8θ′8) Regular octahedron β24η24θ24 (β12β′12η12η′12θ12θ′12)
In this way, the number of atoms which are used in the structure increases significantly, however, the type of atom only increases by one. - In the three-dimensional puzzle of the present invention related to
embodiment 1, an example in which it is possible to make a structure with a minimum number of atoms α, β, γ, δ which form all the regular polyhedrons is explained. However, the three-dimensional puzzle of the present invention related to the present embodiment can form a regular tetrahedron, cube and regular dodecahedron without using the atom α. Consequently, because the three-dimensional puzzle of the present invention related to the present embodiment can form the atom α by combining different atoms and form a regular tetrahedron, cube and regular dodecahedron without using the atom α, it is possible to visually explain the above statedtheorem 1 and demonstrate excellent effects which can compare both by a transformation example using ξ, θ, η which are the sixth to eighth atoms. Therefore, the three-dimensional puzzle of the present invention related to the present embodiment is also excellent as an educational material. - In the fourth embodiment, an example is explained in which the three-dimensional puzzle which has the atoms explained in the first to third embodiments as structural components is applied to a Fedorov polyhedron or a Fedorov space filling solid. As stated previously, in a convex polyhedron, convex polyhedrons P and Q are congruent if either the two convex polyhedrons P and Q are the same or when they have a relationship where one of the convex polyhedrons is a mirror image of the other.
- When a non-congruent convex polyhedron Pi is 1≦i≦n, then II={P1, P2, . . . , Pn}. If at least one element P which is included in Pi, is formed by face to face joining of congruent convex polyhedrons of convex polyhedron σ, then convex polyhedron σ is called an atom of II.
- In addition, a parellelohedron is a convex polyhedron which tiles three-dimensional space using a face to face joining of its translates. Minkowski obtained the following results for a general d-dimensional parallelohedron.
-
- If P is a d-dimensional parallelohedron, then
- 1) P is centrally symmetric,
- 2) All faces of P are centrally symmetric,
- 3) The projection of P along any of its (d-2) faces onto the complementary 2-plane is either a parallelogram or a centrally symmetric hexagon.
- The number fd-1 (P) of faces of a d-parallelohedron P does not exceed 2(2d−1) and there is a parallelohedron P with fd-1=2(2d−1).
- Dolbilin extended Minkowski's theorems for non-face to face tilings of space. There are also numerous studies on parallelohedra discussed by Alexandrov and Gruber.
- In 1890, a Russian crystallographer, Evgraf Fedorov, established that there are exactly five types of parallelohedra, namely, parallelopiped (cube), rhombic dodecahedron, skewed hexagonal prism (parallel hexagonal prism), elongated rhombic dodecahedron and truncated octahedron shown in
FIG. 29( a) toFIG. 29( e). These parallelohedra are called “Fedrov polyhedrons” or “Fedrov space filling solids”.FIG. 30( a) toFIG. 30( e) shows the space filling characteristics of each type of parallelohedron. - The characteristics which are common to these space filling solids is not only that any one type of solid can fill space without any gaps, but also that any two solids can be overlapped by only translation of it as filling the space of each solid.
- Let II1 be a group consisting of a cube, skewed hexagonal prism, rhombic dodecahedron, elongated rhombic dodecahedron and truncated octahedron. The inventors keenly examined which atom should be adopted for filling all the space filling solids included in II1 by the minimum number of atoms. As a result, the shape of the minimum number of atoms for filling all the space filling solids was discovered. Next, the circumstances in which the inventors discovered the shape of the minimum number of atoms for filling all the space filling solids is explained. In addition, in the present embodiment, a three-dimensional puzzle having convex polyhedrons which form five types of Fedrov space filling solids is explained.
- The
cube 1200 inFIG. 31( a) is decomposed into six congruentquadrangular pyramids 1201 shown inFIG. 31( b) using six planes, each of which passes through the center of the cube and contains two opposite edges. Thequadrangular pyramid 1201, as shown inFIG. 31( a), has a square base and the four sides are congruent isosceles triangles. - The
quadrangular pyramid 1201 is further divided into four congruenttriangular pyramids 1203 as shown inFIG. 32( b) by two perpendicular planes passing through the peak and through each of the two diagonals of the base. When thesetriangular pyramids 1203 are shown again inFIG. 32( c), because three faces are orthogonal at one peak, thetriangular pyramid 1203 is called aright tetra 1203. Thus, thecube 1200 can be divided into twenty-four congruentright tetra 1203. If the process is reversed, then it is possible to form thecube 1200 by twenty-four congruentright tetra 1203. - In addition, as is shown in
FIG. 33( a) andFIG. 33( b), by attaching aquadrangular pyramid 1201 to each of the six congruent faces of thecube 1200, arhombic dodecahedron 1300 is formed. This shows that it is possible to form arhombic dodecahedron 1300 from forty-eight congruentright tetra 1203. In this way, theright tetra 1203 is a common constituent element of thecube 1200 and therhombic dodecahedron 1300. - As is shown in
FIG. 32( c), the base of theright tetra 1203 is a right isosceles triangle. Tworight tetras 1203 which are glued at the base of these right isosceles triangles form anothertetrahedron 1100 as is shown inFIG. 34( a). This tetrahedron is called aSphenoid 1100. InFIG. 34( a) the dash line represents the gluing surface of two triangular pyramids. - It is known that the sphenoid 1100 has noteworthy characteristics which are called self-expanding. This is because when eight sphenoid 1100 are combined, the length of each edge expands to exactly twice that of a similar triangular pyramid which makes it easy to understand that the sphenoid is a space filling solid.
- Moreover, when three sphenoid 1100 are combined it is possible to form a
triangular prism 1401 with a regular triangle as a perpendicular cross section shown inFIG. 34( b). Then, the skewedhexagonal prism 1400 of Fedorov shown inFIG. 34( c) is formed using three pairs of thetriangular prism 1401 and its mirror image symmetricaltriangular prism 1403 so that the upper and lower faces are adjusted to form parallel hexagonals. Therefore, a skewedhexagonal prism 1400 can be formed by eighteensphenoids 1100. - Next,
FIG. 35( a) shows anelongated rhombic dodecahedron 1500. This elongatedrhombic dodecahedron 1500 is formed from two parts. One part is arhombic dodecahedron 1300 and the other part is aconcave polyhedron 1501 which can be obtained from the rhombic dodecahedron. In order to obtain the secondconvex polyhedron 1501, we consider four peaks of therhombic dodecahedron 1300. The elongatedrhombic dodecahedron 1500 is cut along the edges containing these peaks shown inFIG. 35 , and the elongatedrhombic dodecahedron 1500 is opened at this peak. The obtainedpolyhedron 1501 is formed in the shape of a helmet. As stated previously, therhombic dodecahedron 1300 consists of forty-eightright tetras 1203. Therefore, the elongatedrhombic dodecahedron 1500 can be formed using ninety-sixright tetras 1203. - The sphenoid 1100 has another important property. As is shown in
FIG. 36( a), it can be divided into fourcongruent hexahedrons 1105 as is shown inFIG. 36( b) by using six planes wherein each plane passes through the midpoint of an edge of the sphenoid 1100 and must also be perpendicular to the edge. Each of thesehexahedrons 1105 is called a c-squadron. - By arranging these four c-squadrons in an appropriate way, it is possible to form an enneahedron (diamond) 1601 as shown in
FIG. 37( a). Then, while adjusting so that a square and a regular hexahedron appear on the surfaces, atruncated octahedron 1600 is formed as shown inFIG. 37( b) by combining (glued) together sixdiamonds 1601. In other words, the c-squadron 1105 is a constituent element of thetruncated octahedron 1600 and thetruncated octahedron 1600 can be formed using twenty-four c-squadrons 1105. - From the above, it can be seen that the cube (parallel hexagram) 1200 can be formed by twenty-four
right tetras 1203, therhombic dodecahedron 1300 can be formed by forty-eightright tetras 1203, the skewedhexagonal prism 1400 can be formed by eighteensphenoids 1100, the elongatedrhombic dodecahedron 1500 can be formed by ninety-six tetras and the truncated octahedron can be formed by twenty-four c-squadrons 1105. - Next, an examination is made whether common constituent elements exist between the
right tetra 1203, sphenoid 1100 and c-squadron 1105. - Here, one among the four c-
squadrons 1105 shown inFIG. 36( b) is shown again inFIG. 38( a). The c-squadron 1105 is a solid with left and right symmetry and is divided into twocongruent pentahedrons FIG. 38( b). Then, as is shown inFIG. 38( c), by using two each of thesenew pentahedrons right tetra 1203 shown inFIG. 32( c). - In addition, as is shown in
FIG. 1 , it is possible to form the sphenoid 1100 with eight of these pentahedrons (fourpentahedrons 1101 and four pentahedrons 1103). That is, thepentahedrons - It is not possible to divide the
pentahedrons pentahedrons pentahedron 1101 is defined as a and itsmirror image 1103 is defined as σ′. A development including dimensions of theatom σ 1101 and atom σ′ 1103 is shown inFIG. 39 . As is shown inFIG. 39 , the atom σ is a pentahedron (convex polyhedron) which has edges ab, bc, ca, cf, fc, ca, ad, da, ab, be, eb, de, ef, and fd, and the atom σ′ is a pentahedron (convex polyhedron) which has edges a′b′, b′c′, c′a′, c′f′, f′ c′ c′a′, a′d′, d′ a′ a′b′, b′e′, e′ b′ d′e′, e′ f′ and f′d′. The atom σ satisfies the following formula (21). Furthermore, the formula of the atom σ′ which is the mirror image of the atom σ is omitted. -
(formula 21) -
ab, bc, ca, cf, fc, ca, ad, da, ab, be, eb, de, ef, fd=2:√2:√2:√6:√6:√2:√2:√2:2:4:4:3√2: 2√3:√6 (21) - By using the
atom σ 1101 and atom σ′ 1103, it is clear that the cube (parallel hexahedron) 1200 can be formed by ninety-six atoms σ (forty-eightatoms σ 1101 and forty-eight atoms σ′ 1103), therhombic dodecahedron 1300 can be formed by one hundred and ninety-two atoms σ (ninety-sixatoms σ 1101 and ninety-six atoms σ′ 1103), thetruncated octahedron 1600 can be formed by forty-eight atoms σ (twenty-fouratoms σ 1101 and twenty-four atoms σ′ 1103), the skewedhexagonal prism 1400 can be formed by one hundred and forty-four atoms σ (seventy-twoatoms σ 1101 and seventy-two atoms σ′ 1103) and the elongatedrhombic dodecahedron 1500 can be formed by three-hundred and eighty-four atoms σ (one-hundred and ninety-twoatoms σ 1101 one-hundred and ninety-two atoms σ′ 1103). - In addition, a cube is formed by ninety-six
atoms σ 1101 and atoms σ′ 1103, however, it is possible to form a cube with as little as twenty-four atoms σ (twelve atoms δ 1101 and twelve atoms σ′ 1103). - From the above, it was discovered that the minimum number of atoms for forming a Fedrov space filling solid is one type of atom σ (and its mirror image σ′).
- Here, it is clear that it is possible to express the minimum number of atoms which form the Fedrov space filling solids as
theorem 2 under the definitions (1) to (4) below. - (1) The polyhedrons P and Q are “congruent” means that P and Q are either exactly the same shape or are in a mirror image relationship.
- (2) A polyhedron is “indecomposable” means P are indivisible into two or more congruent polyhedrons.
- (3) Let II be a group of polyhedrons P1, P2, . . . Pn.
- i.e. II={P1, P2, . . . Pn}
- Let E be a group of indecomposable polyhedrons e1, e2, . . . em,
- i.e. E={e1, e2, . . . em}, ∀e1 is indecomposable, ei and ej are non-congruent (i≠j)
- At this time, E is a group E (II) of the element (atom) of II means satisfying formula (15) below.
-
- That is, whichever polyhedron Pi belongs to II, it is possible to divide into an indecomposable polyhedron belonging to E.
- (4) The element number (atom number) e (II) of II means the minimum number of an order among various element groups with respect to II.
- That is e (II)=min|E (II)|
-
- Let II2={a group of Fedrov space filling solids}
- e (II2)≦1, E (II2)={σ}
- Here, the three-dimensional puzzle of the present invention related to the present embodiment which uses a minimum number of atoms σ to form all the Fedrov space filling solids will be explained in detail. The three-dimensional puzzle of the present invention related to the present embodiment creates Fedrov space filling solids using a minimum number of atoms σ and demonstrates excellent effects of being able to visually explain the above stated
theorem 2. Consequently, the three-dimensional puzzle of the present invention related to the present embodiment is excellent as an educational material for explaining the above statedtheorem 2. -
FIG. 28 is referenced.FIG. 28 is a three-dimensional puzzle of the present invention related to the present embodiment and shows a sphenoid three-dimensional puzzle. As is shown inFIG. 28 , the three-dimensional puzzle (sphenoid) of the present invention related to the present embodiment can be formed using four pieces which have a shape of anatom σ 1101 and four pieces which have a shape of theatom σ 1101 mirror image σ′ 1103 as explained above in detail. - The parallel hexagram three-dimensional puzzle of the present invention related to the present embodiment can be formed using forty-eight pieces which have a shape of an
atom σ 1101 and forty-eight pieces which have a shape of theatom σ 1101 mirror image σ′ 1103 as explained above in detail. - The skewed hexagonal three-dimensional puzzle of the present invention related to the present embodiment can be formed using seventy-two pieces which have a shape of an
atom σ 1101 and seventy-two which have a shape of theatom σ 1101 mirror image σ′ 1103 as explained above in detail. - The truncated octahedron three-dimensional puzzle of the present invention related to the present embodiment can be formed using twenty-four pieces which have a shape of an
atom σ 1101 and twenty-four which have a shape of theatom σ 1101 mirror image σ′ 1103 as explained above in detail. - The rhombic dodecahedron three-dimensional puzzle of the present invention related to the present embodiment can be formed using ninety-six pieces which have a shape of an
atom σ 1101 and ninety-six which have a shape of theatom σ 1101 mirror image σ′ 1103 as explained above in detail. - The elongated rhombic dodecahedron three-dimensional puzzle of the present invention related to the present embodiment can be formed using one hundred and ninety-two pieces which have a shape of an
atom σ 1101 and one hundred and ninety-two which have a shape of theatom σ 1101 mirror image σ′ 1103 as explained above in detail. - The three-dimensional puzzle of the present invention related to the present embodiment which has the shape of a Fedrov space filling solid comprised of five types of polyhedron, namely a parallel hexagram, skewed hexagonal prism, truncated octahedron, rhombic dodecahedron and elongated rhombic dodecahedron, can be assembled from a piece which has a shape of one
atom σ 1101 and a piece which has the shape of one atom σ′ 1103 mirror image. As a result, it is possible to visually explain thetheorem 2 stated above and the three-dimensional puzzle of the present invention related to the present embodiment is excellent as an educational material. - The inventors discovered that the five types of Fedrov space filling solids each require a multiple of twenty-four c-squadrons, that is, a multiple of twenty-four
atoms σ 1101 and σ′ 1103. - From the proof of the elongated rhombic dodecahedron in
theorem 2 andFIG. 35 , it is clear that the elongatedrhombic dodecahedron 1500 includes arhombic dodecahedron 1300. In addition, from the proof of thecube 1200 and therhombic dodecahedron 1300 intheorem 1, it was shown that therhombic dodecahedron 1300 includes thecube 1200. Therhombic dodecahedron 1300 is obtained by gluing the square bottom of thequadrangular pyramid 1201 to each face of thecube 1200. Therefore, the elongatedrhombic dodecahedron 1500 includes therhombic dodecahedron 1300 and therhombic dodecahedron 1300 includes thecube 1200. - The
truncated octahedron 1600 includes thecube 1200 by a special method. Whenatom σ 1101 and atom σ′ 1103 which are congruent pentahedrons shown inFIG. 38( b) are considered, ahexahedron 1701 including seven vertices and eleven edges shown inFIG. 40 by gluing the atoms. - This
hexahedron 1701 has a face comprised of three triangles (one right isosceles triangle and two congruent right triangles), two quadrilaterals and one pentagon. This polyhedron is called a tripenquadron. - The
polyhedron 1703 show inFIG. 41( a) is obtained when threetripenquadrons 1701 are glued along their right triangular faces so that they all meet at the vertex. The bottom face of thispolyhedron 1703 is a regular hexahedron shown inFIG. 41( b). - When eight of these
polyhedrons 1703 are glued along the right isosceles triangular faces, acube 1705 is obtained having six holes at the center of each face as shown inFIG. 42( a). In order to examine the interior of thecube 1705 with open hole, the cube is opened by a plane through the diagonals of two opposite faces which intersect these faces perpendicularly. When a diagonal cross section shown inFIG. 42( b) is viewed, it reveals that the cross section in the interior of thepolyhedron 1707 made by cutting thecube 1705 with open holes consist of regular hexagons and that the hole is obtained by carving out half of atruncated octahedron 1603.FIG. 42( c) andFIG. 42( d) show a cube which is the result of assembling the truncated octahedron and cube with open holes. It is clear that thetruncated octahedron 1600 is included in thecube 1700 in a special way. - Finally, it is shown that a skewed hexagonal prism is included in an elongated rhombic dodecahedron in a particular way. The elongated
rhombic dodecahedron 1500 comprised of therhombic dodecahedron 1300 and helmet shapedpolyhedron 1501 is again considered. - First, as is shown in
FIG. 43( a), therhombic dodecahedron 1300 is divided into twocongruent polyhedrons 1301 and only one of these is considered. Next, a part of the helmet shapedpolyhedron 1501 shown inFIG. 43( b) is cut off as shown inFIG. 43( c). Theright side polyhedron 1503 inFIG. 43( c) is V-shaped and consists of two skewedtriangular prisms triangle prism 1507 is a mirror image of the skewedtriangular prism 1505. Each skewed triangular prism is formed by gluing three sphenoids as shown inFIG. 36( a). - If the V-shaped
polyhedron 1503 is glued to the lower half of therhombic dodecahedron 1301, the skewedhexagonal prism 1420 shown inFIG. 43( d) andFIG. 43( e) is obtained. In this way, it is shown that the skewedhexagonal prism 1420 is included within the elongatedrhombic dodecahedron 1500 in a special way. - The three-dimensional puzzle of the present invention related to the present embodiment which has the shape of a Fedrov space filling solid comprised of five types of polyhedron, namely a cube, skewed hexagonal prism, rhombic dodecahedron, truncated octahedron, and elongated rhombic dodecahedron, can be assembled from a piece which has a shape of one
atom σ 1101 and a piece which has the shape of one atom σ′ 1103 of its mirror image. In addition, it is possible to visually explain the inclusion of the cube, skewed hexagonal prism, rhombic dodecahedron and truncated octahedron in the elongated rhombic dodecahedron and the three-dimensional puzzle of the present invention related to the present embodiment is excellent as an educational material. -
FIG. 44 shows the arrangement of apuzzle 1651 which is half of a truncated octahedron three-dimensional puzzle of the present invention related to the present embodiment. (a) shows a side face view, (b) shows a side face view, (c) shows an upper face view and (d) shows a bottom face view. In addition,FIG. 45 shows the arrangement of a truncated octahedron three-dimensional puzzle 1650. (a) shows a side face view, (b) shows a side view and (c) shows a bottom face view. The truncated octahedron three-dimensional puzzle is formed using twenty-four pieces which have the shape of theatom σ 1101 and twenty-four pieces which have the shape of the mirror image σ′ 1103 of theatom σ 1101 as explained in detail above. -
FIG. 46 shows an arrangement of a cube three-dimensional puzzle of the present invention related to the present embodiment. (a) shows the bottom face of apuzzle 1651 which is half of the truncated octahedron three-dimensional puzzle, (b) shows apuzzle 1751 by a cross section of a cube with open holes, (c) shows apuzzle 1753 of half of a cube in which thepuzzle 1651 which is half of the truncated octahedron three-dimensional puzzle is glued to puzzle 1751 formed by cutting the cube with open holes, (d) shows a cube threedimensional puzzle 1750. The cube three-dimensional puzzle can be formed using forty-eight pieces which have the shape of theatom σ 1101 and forty-eight pieces which have the shape of the mirror image σ′ 1103 of the atom δ 1101 as explained in detail above. -
FIG. 47 shows an arrangement of a rhombic dodecahedron three-dimensional puzzle of the present invention related to the present embodiment. (a) showspuzzle 1753 of half of a cube, (b) showspuzzle 1353 without the half of the cube from thepuzzle 1351 which is half of a rhombic dodecahedron, (c) shows thepuzzle 1351 which is half of a rhombic dodecahedron, (d) shows a side face view ofrhombic dodecahedron puzzle 1350, (e) shows an upper face view of therhombic dodecahedron puzzle 1350. The rhombic dodecahedron three-dimensional puzzle can be formed using ninety-six pieces which have the shape of theatom σ 1101 and ninety-six pieces which have the shape of the mirror image σ′ 1103 of theatom σ60 1101 as explained in detail above. -
FIG. 48 shows an arrangement of a skewed hexagonal prism three-dimensional puzzle of the present invention related to the present embodiment. (a) shows a side face view ofpuzzle 1553 cut from a helmet shapedpolyhedron puzzle 1551, (b) shows a front face view of thepuzzle 1553, (c) shows a back face view of thepuzzle 1553, (d) shows an exemplary view of the formation of a skewedhexagonal prism 1450 by thepuzzle 1553 which is cut away from the helmet shapedpolyhedron puzzle 1551 and apuzzle 1351 which is half of therhombic dodecahedron 1350, (e) shows an upper face view of the skewedhexagonal prism 1450, (f) shows a side face view of the skewedhexagonal prism 1450. The skewedhexagonal prism 1450 three-dimensional puzzle can be formed using seventy-two pieces which have the shape of theatom σ 1101 and seventy-two pieces which have the shape of the mirror image σ′ 1103 of theatom σ 1101 as explained in detail above. -
FIG. 49 shows an arrangement of an elongated rhombic dodecahedron three-dimensional puzzle of the present invention related to the present embodiment. (a) shows a side face view of the helmet shapedpolyhedron puzzle 1551, (b) shows a front face view of the helmet shapedpolyhedron puzzle 1551, (c) shows a back view of the helmet shapedpolyhedron puzzle 1551, (d) shows an exemplary view of the formation of an elongated rhombic dodecahedron three-dimensional puzzle 1550 by the helmet shapedpolyhedron puzzle 1551 and therhombic dodecahedron puzzle 1350, (e) shows an upper face view of the elongated rhombic dodecahedron three-dimensional puzzle 1550, (f) shows a side face view of the elongated rhombic dodecahedron three-dimensional puzzle 1550. The elongated rhombic dodecahedron three-dimensional puzzle 1550 can be formed using one hundred and ninety-two which have the shape of theatom σ 1101 and one hundred and ninety-two which have the shape of the mirror image σ′ 1103 of theatom σ 1101 as explained in detail above. - The three-dimensional puzzle of the present invention related to the present embodiment which has the shape of a Fedrov space filling solid comprised of five types of polyhedron, namely a cube, skewed hexagonal prism, rhombic dodecahedron, truncated octahedron and elongated rhombic dodecahedron, can be assembled from a piece which has a shape of one
atom σ 1101 and a piece which has the shape of one atom σ′ 1103 of its mirror image. In addition, it is possible to visually explain thetheorem 2 stated above and the three-dimensional puzzle of the present invention related to the present embodiment is excellent as an educational material. - Because the three-dimensional puzzle of the present invention is formed with the minimum number of atoms (convex polyhedrons) for filing all of the regular polyhedrons, it is possible to form all of the regular polyhedrons (regular tetrahedron, cube, regular octahedron, regular dodecahedron and regular icosahedron) even if the number of parts is small. In addition, it is possible to visually prove the novel theorem discovered by the inventors and the present invention can be used as an excellent educational material for explaining this theorem.
- According to the three-dimensional puzzle of the present invention, by using a different atom in addition to the minimum number of atoms for constructing all of the regular polyhedrons, it is possible to form a cube and regular dodecahedron without using one of the minimum number of atoms. In addition, it is possible to visually explain the novel theorem discovered by the inventors, compare both puzzles by a transformation example which uses a different atom, and the present invention can be used as an excellent educational material for explaining this theorem.
- According to the three-dimensional puzzle of the present invention, even if one of the minimum number of atoms for constructing all of the regular polyhedrons is not used, it is possible to form a regular polyhedron, cube and regular dodecahedron by using a different atom. In addition, it is possible to visually explain the novel theorem discovered by the inventors, compare both puzzles by a transformation example which uses a different atom, and the present invention can be used as an excellent educational material for explaining this theorem.
- According to the present invention, it is possible to realize a Fedorov space-filling solid and provide a three-dimensional puzzle which has not existed until now.
Claims (2)
1. A three-dimensional puzzle comprising:
four types of convex polyhedrons from which a regular tetrahedron, a cube, a regular octahedron, a regular dodecahedron or a regular icosahedron are formed;
wherein
three types among the four types of convex polyhedrons each having a pair of convex polyhedrons in a mirroring image relationship;
the four type of convex polyhedrons are indivisible into two or more congruent shaped polyhedrons; and
the regular tetrahedron, the cube, the regular octahedron, the regular dodecahedron and the regular icosahedron are formed using only the four types of convex polyhedrons so that the interior of the regular tetrahedron, the cube, the regular octahedron, the regular dodecahedron and the regular icosahedron are filled.
2. A three-dimensional puzzle comprising:
five types of convex polyhedrons from which a regular tetrahedron, a cube, a regular octahedron, a regular dodecahedron or a regular icosahedron are formed;
wherein
four types among the five types of convex polyhedrons each having a pair of convex polyhedrons in a mirroring image relationship;
the five type of convex polyhedrons are indivisible into two or more congruent shaped polyhedrons; and
the regular tetrahedron, the cube, the regular octahedron, the regular dodecahedron and the regular icosahedron are formed using only the five types of convex polyhedrons so that the interior of the regular tetrahedron, the cube, the regular octahedron, the regular dodecahedron and the regular icosahedron are filled.
Applications Claiming Priority (7)
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JP2008-277198 | 2008-10-28 | ||
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USD845401S1 (en) * | 2017-11-04 | 2019-04-09 | Octarine Investments Limited | Pyramid |
US10569185B2 (en) | 2014-09-16 | 2020-02-25 | Andreas Hoenigschmid | Three-dimensional geometric art toy |
US11318369B2 (en) * | 2018-06-07 | 2022-05-03 | National Tsing Hua University | Multiple rhombic dodecahedron puzzle |
US11697058B1 (en) | 2022-08-21 | 2023-07-11 | Andreas Hoenigschmid | Triple inversion geometric transformations |
US11878255B2 (en) | 2022-01-12 | 2024-01-23 | Kevin Schlapi | Puzzle kits |
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US5660387A (en) * | 1996-01-23 | 1997-08-26 | Stokes; William T. | Polyhedron puzzle |
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Also Published As
Publication number | Publication date |
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JP2010017517A (en) | 2010-01-28 |
JP4310419B1 (en) | 2009-08-12 |
JP2010017518A (en) | 2010-01-28 |
EP2204224A1 (en) | 2010-07-07 |
JP4310418B1 (en) | 2009-08-12 |
WO2009151119A1 (en) | 2009-12-17 |
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