U.S. patent application number 12/727390 was filed with the patent office on 2010-09-09 for three dimensional puzzle.
This patent application is currently assigned to Tokai University Educational System. Invention is credited to Hitoshi Akiyama, Gisaku Nakamura, Ikuro Sato, Norihasa Tamiya.
Application Number | 20100225057 12/727390 |
Document ID | / |
Family ID | 41036692 |
Filed Date | 2010-09-09 |
United States Patent
Application |
20100225057 |
Kind Code |
A1 |
Akiyama; Hitoshi ; et
al. |
September 9, 2010 |
THREE DIMENSIONAL PUZZLE
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.
Inventors: |
Akiyama; Hitoshi; (Tokyo,
JP) ; Nakamura; Gisaku; ( Tokyo, JP) ; Sato;
Ikuro; (Sendai-shi, JP) ; Tamiya; Norihasa;
(Houfu-shi, JP) |
Correspondence
Address: |
PEARNE & GORDON LLP
1801 EAST 9TH STREET, SUITE 1200
CLEVELAND
OH
44114-3108
US
|
Assignee: |
Tokai University Educational
System
Tokyo
JP
|
Family ID: |
41036692 |
Appl. No.: |
12/727390 |
Filed: |
March 19, 2010 |
Related U.S. Patent Documents
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Application
Number |
Filing Date |
Patent Number |
|
|
PCT/JP2009/060763 |
Jun 12, 2009 |
|
|
|
12727390 |
|
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Current U.S.
Class: |
273/153S |
Current CPC
Class: |
G09B 23/04 20130101;
A63F 9/12 20130101 |
Class at
Publication: |
273/153.S |
International
Class: |
A63F 9/12 20060101
A63F009/12 |
Foreign Application Data
Date |
Code |
Application Number |
Jun 14, 2008 |
JP |
2008-156057 |
Oct 17, 2008 |
JP |
2008-268221 |
Oct 28, 2008 |
JP |
2008-277198 |
Claims
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.
Description
CROSS REFERENCE TO RELATED APPLICATIONS
[0001] 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.
BACKGROUND OF THE INVENTION
[0002] 1. Technical Field
[0003] 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.
[0004] 2. Description of the Related Art
[0005] 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).
BRIEF SUMMARY OF THE INVENTION
[0006] 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.
[0007] 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.
[0008] 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.
[0009] 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.
[0010] 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.
BRIEF DESCRIPTION OF THE DRAWINGS
[0011] FIG. 1 is a diagram which shows a three-dimensional puzzle
(regular tetrahedron) 100 related to one embodiment of the present
invention;
[0012] FIG. 2(a) is a diagram which shows a cube 200, FIG. 2(b) is
a diagram which shows a triangular pyramid 201 including a peak B
which has been cut away;
[0013] FIG. 3(a) is a diagram which shows the appearance of forming
a quadrangular pyramid 301 from four triangular pyramids 201, FIG.
3(b) is a diagram which shows the appearance of forming a regular
octahedron 300 from two quadrangular pyramids 301;
[0014] FIG. 4(a) is a diagram which shows three points Q, T, U
respectively placed on the three edges of the triangular pyramid
201, FIG. 4(b) is a diagram which shows a triangular pyramid 203
including a peak A which has been cut away from the triangular
pyramid 201;
[0015] FIG. 5(a) and (b) are diagrams which show the triangular
pyramid 203 including the peaks C, F which has been cut away, FIG.
5(c) is a diagram which shows a regular icosahedron 400 formed from
four pairs of a heptahedron 207 and a mirror image symmetrical
heptahedron 209;
[0016] FIG. 6(a) is a diagram which shows a regular 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 the
cube 200 which is left at the center of the cut away;
[0017] FIG. 7(a) is a diagram which shows a view of the regular
tetrahedron 100 again, FIG. 7(b) is a diagram which shows the
regular 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 extracted prism 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 a pentahedron 103 is divided into
two congruent tetrahedrons 110 and 111;
[0018] FIG. 8 is a diagram which shows a development of a piece 110
which is the shape of an atom .alpha. which forms the
three-dimensional puzzle related to one embodiment of the present
invention;
[0019] FIG. 9 is a diagram which shows a view of an equihepta 207
again, FIG. 9(b) is a diagram which shows the appearance of the
equihepta divided into three solids;
[0020] FIG. 10 is a development of a piece 210 which is the shape
of an atom .gamma. which forms the three-dimensional puzzle related
to one embodiment of the present invention;
[0021] FIG. 11 is a development of a piece 203 which is the shape
of an atom .gamma. which forms the three-dimensional puzzle related
to one embodiment of the present invention;
[0022] FIG. 12(a) is a diagram which shows a roof 501 related to
one embodiment of the present invention again, FIG. 12(b) shows a
piece 510 which is the shape of an atom .delta. which forms the
three-dimensional puzzle related to one embodiment of the present
invention;
[0023] FIG. 13 shows the piece 510 which is the shape of an atom
.delta. which forms the three-dimensional puzzle related to one
embodiment of the present invention;
[0024] FIG. 14 is a diagram which shows a cube three-dimensional
puzzle 200 related to one embodiment of the present invention;
[0025] FIG. 15 is a diagram of parts 201 of a triangular pyramid
related to one embodiment of the present invention which is formed
using parts 207 comprised of a piece 210 which is the shape of the
atom .beta. and a piece 203 which is the shape of the atom
.gamma.;
[0026] FIG. 16 is a diagram of parts 207 related to one embodiment
of the present invention which is formed using a piece 210 which is
the shape of three atoms .beta.;
[0027] FIG. 17 is a diagram which shows a regular octahedron
three-dimensional puzzle 300 related to one embodiment of the
present invention;
[0028] FIG. 18 is a diagram which shows a regular dodecahedron
three-dimensional puzzle 500 related to one embodiment of the
present invention;
[0029] FIG. 19 is a diagram which shows a regular icosahedron
three-dimensional puzzle 400 related to one embodiment of the
present invention;
[0030] FIG. 20(a) is a diagram which shows the appearance of the
roof 501 which is reversed so as to fill the interior of the cube
200, FIG. 20(b) is a diagram which shows a hexahedron 601 related
to one embodiment of the present invention, FIG. 20(c) shows a
development of the hexahedron 601, FIG. 20(d) is a diagram which
shows the appearance when the interior of the cube 200 is filled
with the two mirror image symmetrical hexahedrons 601 and 602;
[0031] FIG. 21 is a development of a piece 610 which is the shape
of an atom .epsilon. related to one embodiment of the present
invention;
[0032] FIG. 22(a) is a view of the piece 110 which is the shape of
the atom .alpha. again, FIG. 22(b) is a diagram which shows the
piece 110 divided into a tetrahedron 121 and a tetrahedron 123,
FIG. 22(c) is a diagram which shows a piece 151 which is the shape
of an atom .xi. and a piece 152 which is the shape of an atom .xi.'
related to one embodiment of the present invention, FIG. 22(d) is a
diagram which shows a quadrangular pyramid formed by gluing a
tetrahedron 123 and a tetrahedron 124;
[0033] FIG. 23 is a development of the piece 151 which is the shape
of the atom .xi. related to one embodiment of the present
invention;
[0034] FIG. 24 is a diagram which shows the appearance of an end of
the piece 203 which is the shape of the atom .gamma. which is
divided into a piece 800 which is the shape of the atom .theta. and
a piece 900 which is the shape of the atom .eta.;
[0035] FIG. 25 is a development of the piece 800 which is the shape
of the atom .theta. related to one embodiment of the present
invention;
[0036] FIG. 26 is a development of the piece 900 which is the shape
of the atom .eta. related to one embodiment of the present
invention;
[0037] FIG. 27 is a diagram which shows a piece 700 which is the
shape of a polyhedron corresponding to two atoms .alpha. related to
an embodiment of the present invention;
[0038] FIG. 28 is a diagram which shows a regular tetrahedron
three-dimensional puzzle related to one embodiment of the present
invention;
[0039] 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;
[0040] 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;
[0041] FIG. 31(a) is a diagram which shows a cube 1200, FIG. 31(b)
shows the cube divided into six congruent quadrangular pyramids
1201;
[0042] FIG. 32(a) is a diagram which shows one of the quadrangular
pyramids 1201 FIG. 32(b) shows one of the quadrangular pyramids
1201 divided into right tetrahedrons 1203 by two perpendicular
planes, FIG. 32(c) is a diagram which shows a right tetrahedron
1203;
[0043] FIG. 33(a) shows an exemplary view where the quadrangular
pyramids 1201 are glued to each side of the cube 1200, FIG. 33(b)
is a diagram which shows a rhombic dodecahedron 1300;
[0044] FIG. 34(a) shows a sphenoid 1100 in which two right
tetrahedrons 1203 are glued to the bottom side of right angled
isosceles triangles, FIG. 34(b) shows a triangular prism 1401 in
which three sphenoids 1100 are combined together, FIG. 34(c) shows
a skewed hexagonal prism 1400;
[0045] FIG. 35(a) shows a rhombic dodecahedron 1300, FIG. 35(b) is
an exemplary diagram which shows the formation of an elongated
rhombic dodecahedron 1500 by the rhombic dodecahedron 1300 and a
helmet-shaped polyhedron 1501;
[0046] FIG. 36(a) is a diagram which shows the sphenoid 1100
related to one embodiment of the present invention, FIG. 36(b)
shows the sphenoid 1100 divided into c-squadrons 1105;
[0047] FIG. 37(a) is an enneahedron of a diamond 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
the enneahedrons 1601;
[0048] 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 into
.sigma. 1101 and .sigma.' 1103, FIG. 38(c) is an exemplary diagram
which shows the formation of a right tetrahedron 1203 using two
each of .sigma. 1101 and .sigma.' 1103;
[0049] FIG. 39 is a development of .sigma. 1101 and .sigma.' 1103
related to one embodiment of the present invention;
[0050] FIG. 40 is an exemplary diagram which shows the formation of
a hexahedron 1701 using .sigma. 1101 and .sigma.' 1103 related to
one embodiment of the present invention;
[0051] FIG. 41(a) is an upper side diagram which shows the
formation of a polyhedron 1703 using three hexahedrons 1701, FIG.
41(b) is a bottom side diagram of the polyhedron 1703;
[0052] FIG. 42(a) is an exemplary diagram which shows the formation
of a cube 1705 with an open hole using the polyhedron 1703, FIG.
42(b) is a diagram which shows a combination of the cube 1705 with
open hole and half of the truncated octahedron 1603, FIG. 42(c) is
a diagram which shows half of a cube formed by the cube 1705 with
open hole and half of the truncated octahedron 1603, FIG. 42(d) is
a diagram which shows the cube 1700 which is formed;
[0053] FIG. 43(a) is a diagram which shows half of a rhombic
dodecahedron 1301, FIG. 43(b) is a diagram which shows a
helmet-shaped polyhedron 1501, FIG. 43(c) is a diagram which shows
a polyhedron 1503 cut away from the helmet-shaped polyhedron 1501,
FIG. 43(d) is a diagram which shows the formation of a skewed
hexagonal prism 1420 from the polyhedron 1503 and half of the
rhombic dodecahedron 1301, FIG. 43(e) is a diagram which shows the
skewed hexagonal prism 1420;
[0054] FIG. 44 shows an arrangement of a puzzle 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;
[0055] FIG. 45 shows an arrangement of a puzzle 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;
[0056] 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 a puzzle 1651 in
which a truncated octahedron three-dimensional puzzle is halved,
FIG. 46(b) shows a puzzle 1751 formed by cutting a cube with an
open hole, FIG. 46(c) shows half of a cube puzzle 1753 in which the
puzzle 1651 which is half of a truncated octahedron
three-dimensional puzzle, is glued to the puzzle 1751 formed by
cutting a cube with an open hole, FIG. 46(d) shows a cube
three-dimensional puzzle 1750;
[0057] 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 a cube puzzle 1753,
FIG. 47(b) shows a puzzle 1353 where half of a cube is removed from
a puzzle 1351 which is half of a rhombic dodecahedron, FIG. 47(c)
shows the puzzle 1351 which is half of a rhombic dodecahedron, FIG.
47(d) shows a side face view of a rhombic dodecahedron puzzle 1350,
FIG. 47.(e) shows an upper side view of the rhombic dodecahedron
puzzle 1350;
[0058] 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 a
puzzle 1553 which has been cut away from a helmet-shaped polyhedron
puzzle 1551, FIG. 48(b) shows a front face view of FIG. 48(a), FIG.
48(c) shows a back face view of FIG. 48(a), FIG. 48(d) shows an
exemplary diagram of the formation of a skewed hexagonal prism
puzzle 1450 by the puzzle 1553 which is cut away from the
helmet-shaped polyhedron 1551, and the puzzle 1351 which is half of
the rhombic dodecahedron 1350, FIG. 48(e) shows an upper face view
of the skewed hexagonal prism puzzle 1450, FIG. 48(f) shows a side
face view of the skewed hexagonal prism puzzle 1450; and
[0059] 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-shaped polyhedron puzzle 1551, FIG. 49(b)
shows a front face view of the helmet-shaped polyhedron puzzle
1551, FIG. 49(c) shows a back face view of the helmet-shaped
polyhedron puzzle 1551, FIG. 49(d) shows an exemplary diagram of
the formation of an elongated rhombic dodecahedron puzzle 1550 by
the helmet-shaped polyhedron puzzle 1551 and the rhombic
dodecahedron puzzle 1350, FIG. 49(e) shows an upper face view of
the elongated rhombic dodecahedron puzzle 1550, FIG. 49(f) shows a
side face view of the elongated rhombic dodecahedron puzzle
1550.
EXPLANATION OF THE REFERENCE SYMBOLS
[0060] 100 regular tetrahedron [0061] 101 prism [0062] 103
pentahedron [0063] 110 atom .alpha. [0064] 111 atom .alpha.' [0065]
121 tetrahedron LJKW [0066] 123 tetrahedron LHKW [0067] 124 mirror
image of tetrahedron 123 [0068] 151 atom .xi. [0069] 152 atom .xi.
[0070] 200 cube [0071] 201 regular triangular pyramid [0072] 203
golden tetra (atom .gamma.) [0073] 204 atom .gamma.' [0074] 205
polyhedron except golden tetra from regular triangular pyramid
[0075] 207 equihepta [0076] 209 mirror image of equihepta [0077]
210 atom .beta. [0078] 211 atom .beta.' [0079] 300 regular
octahedron [0080] 301 quadrangular pyramid [0081] 400 regular
icosahedron [0082] 500 regular dodecahedron [0083] 501 roof [0084]
510 atom .delta. [0085] 601 hexahedron [0086] 602 mirror image of
hexahedron 601 [0087] 610 atom .epsilon. [0088] 611 atom .epsilon.'
[0089] 700 .alpha.2 [0090] 701 quadrangular pyramid LHKWK' [0091]
800 atom .theta. [0092] 801 atom .theta.' [0093] 900 atom .eta.
[0094] 901 atom .eta.' [0095] 1100 sphenoid [0096] 1101 .sigma.
[0097] 1103 .sigma.' [0098] 1105 c-squadron [0099] 1200 cube [0100]
1201 quadrangular pyramid [0101] 1203 triangular pyramid (right
tetrahedron) [0102] 1300 rhombic dodecahedron [0103] 1301
polyhedron which is a rhombic dodecahedron cut in half [0104] 1350
rhombic dodecahedron puzzle [0105] 1351 puzzle which is a rhombic
dodecahedron cut in half [0106] 1353 a puzzle except half a cube
from a puzzle which is a rhombic dodecahedron cut in half [0107]
1400 skewed hexagonal prism [0108] 1401 triangular prism [0109]
1403 mirror image symmetrical triangular prism [0110] 1420 skewed
hexagonal prism [0111] 1450 skewed hexagonal prism puzzle [0112]
1500 elongated rhombic dodecahedron [0113] 1501 helmet-shaped
polyhedron [0114] 1503 polyhedron cut away from helmet-shaped
polyhedron 501 [0115] 1505 skewed triangular prism [0116] 1507
mirror image of skewed triangular prism 505 [0117] 1550 elongated
rhombic dodecahedron puzzle [0118] 1551 helmet-shaped polyhedron
puzzle [0119] 1553 puzzle cut away from helmet-shaped polyhedron
puzzle 551 [0120] 1600 truncated octahedron [0121] 1601 enneahedron
(diamond) [0122] 1603 half of truncated octahedron [0123] 1650
truncated octahedron puzzle [0124] 1651 half of truncated
octahedron puzzle [0125] 1700 cube [0126] 1701 hexahedron [0127]
1703 polyhedron [0128] 1705 cube with open hole [0129] 1707
polyhedron formed by cutting through cube with open hole [0130]
1709 polyhedron formed by cutting through half of cube [0131] 1750
cube puzzle [0132] 1751 puzzle formed by cutting through cube with
open hole [0133] 1753 half of cube puzzle
DETAILED DESCRIPTION OF THE INVENTION
First Embodiment
[0134] 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.
[0135] 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.
[0136] 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.
[0137] 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.
[0138] 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.
[0139] The condition A below is attached in order to remove this
obvious atom.
[Condition A]
[0140] It is possible to use any atom as at least two types of
regular polyhedron.
[0141] 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 2.sup.3, 3.sup.2 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.
[0142] Furthermore, defining a mirror image symmetrical convex
polyhedron as the same variety of atom. [0143] (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
[0144] 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.
[0145] In the cube 200 as in FIG. 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 a
triangular pyramid 201 which includes the peak B is cut away as in
FIG. 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 a regular
tetrahedron 100 as shown in FIG. 2 (a). The six dash lines in FIG.
2 (a) show the regular tetrahedron 100 obtained by this cutting
away.
[0146] Next, as is shown in FIG. 2(b), four triangular 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 these triangular pyramids 201 are gathered at one point, and
when arranged so that the bottom face forms a square, a
quadrangular prism 301 which has four side faces being all regular
triangles is formed, shown in FIG. 3(a). When the bottom face
squares as shown in FIG. 3(b) of two of these quadrangular prisms
301 are attached to each other (when a pair of bottom face squares
are glued together) a regular octahedron 300 is formed. One
relationship between the cube 200, the regular tetrahedron 100 and
the regular octahedron 300 is obtained using this cutting
method.
[0147] Next, focusing on the triangular pyramid 201 shown in FIG.
2(b), as is shown in FIG. 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:.tau. (1)
Here, .tau. is a golden proportion, and as is common knowledge can
be obtained by the formula (2):
[formula 2]
.tau. = 5 + 1 2 .apprxeq. 1.618033989 ( 2 ) ##EQU00001##
[0148] Then, as is shown in FIG. 4(b) a triangular pyramid 203
including the peak A is cut away from the triangular pyramid 201.
As is shown in FIG. 5(a), (b), when the triangular pyramid 203
including the peaks C, F is also cut away, a heptahedron 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:.tau. (3
[0149] When a heptahedron 207 is created in the center of FIG. 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.
[0150] Next, the relationship established in the formula (4) below
will be focused on.
[Formula 4]
QBS=RBQ=SBR=90.degree. (4)
[0151] Four pairs of this heptahedron 207 and its mirror image
symmetrical heptahedron 209 are created. By attaching (gluing)
these together, a regular icosahedron 400 is formed as shown in
FIG. 5(c). As a result, one relationship is established between the
triangular pyramid 201 and the regular icosahedron 400.
[0152] Next, a regular dodecahedron 500 is shown in FIG. 6(a). As
is shown in FIG. 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 become congruent pentahedrons 501. As is shown by the dash
line in FIG. 6(c), when further four planes are cut away, a cube
200 is left at the center. One relationship between the regular
dodecahedron and the cube is obtained.
[0153] The regular tetrahedron 100, regular octahedron 300 and
regular icosahedron 400 are obtained from dividing the cube 200 as
a result of the above stated examination. In addition, the cube 200
is obtained from dividing the regular dodecahedron 500. This
dividing satisfies the condition [condition A].
[0154] 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 in FIG. 2(a). Next, the cube 200 is formed
by three types of convex polyhedron (five types if mirror images
are included), namely one regular tetrahedron 100, two heptahedrons
207 and two mirror image symmetrical heptahedrons 208, and six
tetrahedrons 203 and six mirror image symmetrical tetrahedrons.
Next, the regular octahedron 300 is formed by two types of convex
polyhedron (four types if mirror images are included), namely four
heptahedrons 207 and four mirror image symmetrical heptahedrons
208, and twelve tetrahedrons 203 and twelve mirror image
symmetrical tetrahedrons. Next, the regular dodecahedron as shown
in FIG. 6(b) is formed by four types of convex polyhedron (six type
if mirror images are included), namely six pentahedrons 501, one
regular tetrahedron 100, two heptahedrons 207 and two mirror image
symmetrical heptahedrons 208, six tetrahedrons 203 and six mirror
image symmetrical tetrahedrons. Lastly, a regular icosahedron is
formed by one type of convex polyhedron consisting of four
heptahedrons 207 and four mirror image symmetrical heptahedrons
208.
[0155] 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
[0156] [1] regular tetrahedron 100 (used for regular tetrahedron
100, cube 200 and regular dodecahedron 500),
[0157] [2] heptahedron (equihepta) 207 (and its mirror image 208)
(used in cube 100, regular octahedron 300 and regular icosahedron
400),
[0158] [3] tetrahedron (golden tetra) 203 (and its mirror image
204) (used in cube 100 and regular octahedron 300), and
[0159] [4] pentahedron (roof) 501 (used in regular dodecahedron
500).
[0160] 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], the
regular tetrahedron 100 which is extracted from FIG. 2(a) is shown
again in FIG. 7(a) and the regular 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 in FIG. 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 the prisms 101
extracted. When the prism 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 in FIG. 7(d), two congruent pentahedrons 103 are
formed. Moreover, because each pentahedron 103 is symmetrical, when
the left side pentahedron 103 is taken, it is possible to further
divide the pentahedron into two congruent tetrahedrons 110 and 111
as is shown in 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 the
tetrahedron 110 as .alpha. is defined below. Then, as is shown in
FIG. 1, because the regular tetrahedron 100 is formed by eight
atoms, this can be written as .alpha..sub.8, as in a molecular
formula.
[0161] Here, as is shown in FIG. 7(e) atom .alpha. is the
tetrahedron (convex polyhedron) 110 with peaks KJLH, and its mirror
image tetrahedron 111 with peaks MJLH. The atom .alpha. 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)
[0162] Next, when the equihepta 207 in [2] is examined, the
equihepta 207 shown again in FIG. 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.degree. pieces through O,
the equihepta is always divided into three congruent solids 210.
This division is shown in FIG. 9(b) and a development is shown in
FIG. 10.
[0163] One atom which forms a regular polyhedron with the
heptahedron 210 as .beta. is defined below. The atom .beta. 210, as
shown in FIG. 10, is a heptahedron (convex polyhedron) having edges
BX, XY, YO, OA.sub.1, A.sub.1Z, ZT, TZ, ZA.sub.1, A.sub.1O, OY, YX,
XR, RX, XB, BO, BZ, BR, RT, TY and TA.sub.1.
[0164] Here, a, k, k', b, c, d satisfy the following formulas (6)
to (11)
(formula 6)
a=3- 5=2(2-.tau.) (6)
(formula 7)
k=1/(.tau.+2)=(5- 5)/10 (7)
(formula 8)
k'=1-k=(5+ 5)/10 (8)
(formula 9)
b=2/ 3 (9)
(formula 10)
C=(2/.tau..sup.2) (.tau..sup.2-2k'+3k'.sup.2) (10)
(formula 11)
d=(2/.tau..sup.2) (4/3-4k'+4k'.sup.2) (11)
[0165] At this time, atom .beta. 210 satisfies the conditions shown
in formula (12) below.
( formula 12 ) BX : XY : YO : OA 1 : A 1 Z : ZT : TZ : ZA 1 : A 1 O
: OY : YX : XR : RX : XB : BO : BZ : BR : RT : TY : TA 1 = d : ak :
c : c : ak : 3 ak : 3 ak : ak : c : c : ak : 3 ak ' : 3 ak ' : d :
b : d : 2 - a : a : 2 ak ' : 2 ak ( 12 ) ##EQU00002##
[0166] Because a regular icosahedron is formed by twenty-four atoms
.beta. 210, it is possible to describe this as .beta..sub.24.
[0167] Next, when the golden tetra 203 in [3] is examined, it is
impossible to divide the golden tetra 203 into a plurality of
congruent convex polyhedrons. Thus, one atom which forms a regular
polyhedron with the golden tetra 203 as .gamma. is defined. FIG. 11
is a development of the atom .gamma. 203 and dimensions are shown
according to ratios. The atom .gamma., as is shown in FIG. 11, is a
tetrahedron (convex polyhedron) having edges CR, RC, CV, VC, CT,
TC, RT, RV and VT. The atom .gamma. 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)
[0168] Then, because a regular octahedron is formed by twenty-four
.beta. and .gamma., it is possible to describe this as
.beta..sub.24.gamma..sub.24. In addition, because a cube in FIG.
1(a) is formed by eight .alpha., twelve .beta. and twelve .gamma.,
it is possible to describe this as
.alpha..sub.8.beta..sub.12.gamma..sub.12.
[0169] Next, when the roof 501 in [4] is examined, while
considering that the roof 501 has up down left and right symmetry,
it is divided into two congruent pentahedrons 501 as is shown in
FIG. 12(a). One atom which forms a regular polyhedron with the
pentahedron 510 as .delta. is defined. FIG. 13 is a development of
the atom .delta. 510 and dimensions are shown according to ratios.
The atom .delta. 510, as is shown in FIG. 13, is a tetrahedron
(convex polyhedron) having edges BE, ED.sub.1, D.sub.1A, AD.sub.1,
D.sub.1E, EB, BC.sub.1, C.sub.1E, C.sub.1D.sub.1, BA and AE. The
atom .delta. 510 satisfies the formula (14) below.
(formula 14)
BE:ED.sub.1:D.sub.1A:AD.sub.1:D.sub.1E:EB:BC.sub.1:C.sub.1B:
C.sub.1E:C.sub.1D:BA:AE=2: (4-.tau.): (4-.tau.): (4-.tau.):
(4-.tau.):2:2(.tau.-1):2(.tau.-1):2(.tau.-1): .tau.-1:2:2 2
(14)
[0170] A solution for forming a regular polyhedron by a minimum
number of atoms .alpha., .beta., .gamma. and .delta. is obtained as
described above. The number of each regular polyhedron using the
atoms .alpha., .beta., .gamma., .delta. can be expressed as:
TABLE-US-00001 Regular tetrahedron .alpha..sub.8 Cube
.alpha..sub.8.beta..sub.12.gamma..sub.12 Regular octahedron
.beta..sub.24.gamma..sub.24 Regular dodecahedron
.alpha..sub.8.beta..sub.12.gamma..sub.12.delta..sub.12 Regular
icosahedron .beta..sub.24
[0171] However, a mirror image symmetrical atom is also shown by
.alpha., .beta., .gamma., .delta.. If a prime symbol (') is
attached to distinguish a mirror image symmetrical atom, then they
can be expressed as:
TABLE-US-00002 Regular tetrahedron .alpha..sub.4.alpha.'.sub.4 Cube
.alpha..sub.4.alpha.'.sub.4.beta..sub.6.beta.'.sub.6.gamma..sub.6.ga-
mma.'.sub.6 Regular octahedron
.beta..sub.12.beta.'.sub.12.gamma..sub.12.gamma.'.sub.12 Regular
dodecahedron
.alpha..sub.4.alpha.'.sub.4.beta..sub.6.beta.'.sub.6.gamma..sub.6.tau.'.s-
ub..gamma..delta..sub.12 Regular icosahedrons
.beta..sub.12.beta.'.sub.12
[0172] From the above, it was discovered that the minimum number of
atoms for forming a regular polyhedron is four types .alpha.,
.beta., .gamma., .delta..
[0173] 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. [0174] (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. [0175] (2) a polyhedron is
"indecomposable" means that P is indivisible into two or more
congruent polyhedrons. [0176] (3) Let II be a group of polyhedrons
P.sub.1, P.sub.2, . . . P.sub.n. [0177] i.e. II={P.sub.1, P.sub.2,
. . . P.sub.n} [0178] Let E be a group of indecomposable
polyhedrons e.sub.1, e.sub.2, . . . e.sub.m, [0179] i.e.
E={c.sub.1, c.sub.2, . . . c.sub.m}, .A-inverted.e.sub.1 is
indecomposable, c.sub.i, and c.sub.j are non-congruent (i.noteq.j)
[0180] At this time, E is a group E (II) of the element (atom) of
II means satisfying formula (15) below.
[0180] (formula 15)
(.A-inverted..sub.i<1.ltoreq.i.ltoreq.n), a.sub.i is a
nonnegative integer. (15)
That is, whichever polyhedron P.sub.i belongs to II, it is possible
to divide into an indecomposable polyhedron belonging to E. [0181]
(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.
[0182] That is e (II)=min|E (II)|
(Theorem 1)
[0182] [0183] Let II.sub.1={regular polyhedron group} [0184] e
(II.sub.1).ltoreq.4, e (II.sub.1)={.alpha., .beta., .gamma.,
.delta.}
[0185] Here, a three-dimensional puzzle of the present invention
related to the present embodiment which uses the minimum number of
atoms .alpha., .beta., .gamma., .delta. 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
.alpha., .beta., .gamma., .delta., 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 stated theorem 1.
(Regular Tetrahedron Three-Dimensional Puzzle)
[0186] 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 in FIG. 1, the three-dimensional puzzle (regular tetrahedron
three-dimensional puzzle 100) can be formed by four pieces which
have the shape of atom .alpha. 110 and four pieces which have a
shape of a mirror image .alpha.' 111 of atom .alpha. 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 the atom .alpha. 110 and the atom .alpha.'
111 can be changed while maintaining the relative positional
relationship between them.
(Cube Three-Dimensional Puzzle)
[0187] 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 the
regular tetrahedron 100 which is comprised of the atom .alpha. (and
.alpha.') as explained in detail above, and is formed by gluing a
triangular 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 in FIG. 15, the triangular pyramid 201 is a
part arranged with three pieces 203 which has the shape of the atom
.gamma. with parts 207 (FIG. 16) comprised of three pieces 210
which have the shape of atom .beta. 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 four
pieces 110 which have the shape of atom .alpha. and four pieces 111
which have the shape of the atom .alpha. mirror image atom
.alpha.', six pieces 210 which have the shape of atom .beta. and
six pieces 211 which have the shape of the atoms .beta. mirror
image atom .beta.', and six pieces 203 which have the shape of atom
.gamma. and six pieces 204 which have the shape of the atom .gamma.
mirror image atom .gamma.'. 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 .alpha. and .alpha.', .beta. and .beta.',
and .gamma. and .gamma.' can be changed while maintaining the
relative positional relationship between them.
(Regular Octahedron Three-Dimensional Puzzle)
[0188] 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 three sided 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 twelve pieces 210
which have the shape of atom .beta. and twelve pieces 211 which
have the shape of the atom .beta. mirror image atom .beta.', and
twelve pieces 203 which have the shape of atom .gamma. and twelve
pieces 204 which have the shape of the atom .gamma. mirror image
atom .gamma.'. 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 .beta. and .beta.', and .gamma.
and .gamma.' can be changed while maintaining the relative
positional relationship between them.
(Regular Dodecahedron Three-Dimensional Puzzle)
[0189] 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 the cube
200 at the center as shown in FIG. 6(c), and is formed by arranging
roof parts 501 comprised of two pieces 510 which have a shape of
the atom .delta. on six sides of cubes 200. Therefore, the
three-dimensional puzzle of the present invention related to the
present embodiment (regular dodecahedron puzzle 500) can be formed
by using four pieces 110 which have the shape of atom .alpha. and
four pieces 111 which have the shape of the atom .alpha. mirror
image atom .alpha.', six pieces 210 which have the shape of atom
.beta. and six pieces 211 which have the shape of the atoms .beta.
mirror image atom .beta.', six pieces 203 which have the shape of
atom .gamma. and six pieces 204 which have the shape of the atom
.gamma. mirror image atom .gamma.', and twelve pieces 510 which
have the shape of the atom .delta.. Furthermore, the regular
dodecahedron three-dimensional puzzle 500 of the present invention
related to the present embodiment shown in FIG. 18 is one example
and there are also examples where the arrangement of the atoms
.alpha. 110 and .alpha.'111, .beta. 210 and .beta. 211', .gamma.
203 and .gamma.' 203, can be changed while maintaining the relative
positional relationship between them.
(Regular Icosahedron Three-Dimensional Puzzle)
[0190] 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 of parts 207 which have the shape of an equihepta comprised
from three pieces 210 which have the shape of the atom .beta. shown
in FIG. 16, and the parts 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 twelve pieces 210 which have the
shape of atom .beta. and twelve pieces 211 which have the shape of
the atoms .beta. mirror image atom .beta.'. Furthermore, the
regular icosahedron three-dimensional puzzle 400 of the present
invention related to the present embodiment shown in FIG. 19 is one
example and there are also examples where the arrangement of the
atoms .beta. 210 and .beta.' 211, can be changed while maintaining
the relative positional relationship between them.
[0191] 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.
Second Embodiment
[0192] In the second embodiment, an example of a three-dimensional
puzzle which can form a cube and a regular dodecahedron without
using the atom .alpha. of embodiment one is explained.
[0193] FIG. 20 is referenced. FIG. 20(a) shows the appearance when
six roofs 501 (.delta..sup.2) comprised from two atoms .delta. 510
are reversed so as to fill the interior of a cube 200. In FIG.
20(a) only the roof 501 of the back face, side face and bottom face
is shown to facilitate visualization. Each face of the pentahedron
501 (.delta..sub.2) 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 image symmetrical hexahedrons 601 and four hexahedrons 602
are required to fill these gaps. FIG. 20(b) shows a hexahedron 601
for filling the forefront left side lower gap in FIG. 20(a), and
FIG. 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
image symmetrical hexahedrons 601 and 602 are filled. In this way,
a relationship between a regular dodecahedron 500 and cube 200 is
obtained which is different to the relationship in the first
embodiment.
[0194] 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 (.delta..sub.2), four hexahedrons 601 and four mirror
image symmetrical hexahedrons 602. In addition, the regular
dodecahedron 500 is comprised of two types of convex polyhedron,
namely, twelve roofs 501 (.delta..sub.2), four hexahedrons 601 and
four mirror image symmetrical hexahedrons 602.
[0195] 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: [0196] [1] regular tetrahedron 100 (used in a
regular tetrahedron 100 and cube 200). [0197] [2] equihepta 207 and
its mirror image 209 (used in a cube 100, regular octahedron 300
and regular icosahedron 500). [0198] [3] golden tetra 203 and its
mirror image 204 (used in a cube 200 and regular octahedron 300).
[0199] [4] roof 501 (used in a regular dodecahedron 500). [0200]
[5] hexahedron 601 and its mirror image 602 (used in a cube 200 and
regular dodecahedron 500).
[0201] Here, because the convex polyhedrons [1] to [4] were
explained in the first embodiment, an explanation is omitted
here.
[0202] When the hexahedron 601 in [5] is examined, it has three
rotational symmetry as is clear from the hexahedron in FIG. 20(b)
and is the development in FIG. 20(c). As a result, by cutting in a
perpendicular direction by three faces of 120.degree. 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 the hexahedron 601 is shown in FIG. 21.
Furthermore, dimensions are described using ratios in the
development. A quadrangular pyramid 610 shown in FIG. 21 is defined
as atom .epsilon..
[0203] Atom .epsilon. 610 is a pentahedron (convex polyhedron)
which has edges D.sub.1E.sub.1, E.sub.1F.sub.1, F.sub.1G.sub.1,
G.sub.1H.sub.1, H.sub.1I.sub.1, I.sub.1J.sub.1, J.sub.1K.sub.1,
K.sub.1D.sub.1, D.sub.1F.sub.1, D.sub.1I.sub.1, D.sub.1J.sub.1 and
F.sub.1I.sub.1 Atom .epsilon. satisfies the following formula
(16)
(formula sixteen)
D.sub.1E.sub.1:E.sub.iF.sub.1:F.sub.1G.sub.1:G.sub.1H.sub.1:H.sub.1I.sub-
.1:I.sub.1J.sub.1:
J.sub.1K.sub.1:K.sub.1D.sub.1:D.sub.1F.sub.1:D.sub.1I.sub.1:D.sub.1J.sub.-
1:F.sub.1I.sub.1= 3:2-.tau.:2-.tau.:2-.tau.: (18-11.tau.):
(18-11.tau.):2-.tau.: 3: (4-.tau.):2(.tau.-1): (4-.tau.):.tau.-1
(16)
[0204] Because the cube 200 is comprised of twelve atoms .delta.
510 and twenty-four atoms .epsilon. (twelve atoms .epsilon. and
twelve mirror image atoms .epsilon.'), it is expressed as
.delta..sub.12.epsilon..sub.24
(.delta..sub.12.epsilon..sub.12.epsilon.'.sub.12) and because the
regular dodecahedron 500 is comprised of twenty-four .delta. and
twenty-four .epsilon., it is expressed as
.delta..sub.24.epsilon..sub.24
(.delta..sub.24.epsilon..sub.12.epsilon.'.sub.12).
[0205] By using the atoms .alpha., .beta., .gamma., .delta.,
.epsilon., each individual regular polyhedron becomes:
TABLE-US-00003 Regular tetrahedron .alpha..sub.8 Cube
.alpha..sub.8.beta..sub.12.gamma..sub.12,
.delta..sub.12.epsilon..sub.24 Regular octahedron
.beta..sub.24.gamma..sub.24 Regular dodecahedron
.delta..sub.24.epsilon..sub.24
However, the same symbols .alpha., .beta., .gamma., .delta.,
.epsilon., also indicate mirror image symmetrical atoms. If the
mirror image symmetrical atoms are attached with a prime symbol (')
then:
TABLE-US-00004 Regular tetrahedron .alpha..sub.4.alpha.'.sub.4 Cube
.alpha..sub.4.alpha.'.sub.4.beta..sub.6.beta.'.sub.6.gamma..sub.6.ga-
mma.'.sub.6, .delta..sub.12.epsilon..sub.12.epsilon.'.sub.12
Regular octahedron
.beta..sub.12.beta.'.sub.12.gamma..sub.12.gamma.'.sub.12 Regular
dodecahedron .delta..sub.24.epsilon..sub.12.epsilon.'.sub.12
Regular icosahedron .beta..sub.12.beta.'.sub.12
[0206] 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 .alpha. by using
.epsilon. in addition to the minimum number of atoms .alpha.,
.beta., .gamma., .delta. 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 .epsilon. in addition to the minimum
number of atoms .alpha., .beta., .gamma., .delta., it is possible
to visually explain the above stated theorem 1 and demonstrate
excellent effects which can compare both by a transformation
example using .epsilon. 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.
Third Embodiment
[0207] In the present embodiment, an example of a three-dimensional
puzzle which can form a regular tetrahedron 100, cube 200 and
regular octahedron 300 without using the atom .alpha. of the first
embodiment is explained.
[0208] FIG. 22 is referenced. FIG. 22(a) shows the atom .alpha. 110
shown in FIG. 7(e) seen from a different direction. Next, the atom
.alpha. 110 is divided in two as in FIG. 22(b) from the peak L
perpendicular to the edge JH with the foot being W. Then, the left
side tetrahedron 121 becomes three portions by dividing the regular
tetrahedron 100 in a perpendicular direction from the peak and this
can be divided into two mirror image symmetrical tetrahedrons 151
and 152 as shown in FIG. 22(c). The tetrahedron 151 shown in FIG.
22(c) is an atom and defined as .xi.. A development of the atom
.xi. 151 is shown in FIG. 23. Each dimension is shown by ratios. As
is shown in FIG. 22(c), the atom .xi. 151 is a tetrahedron (convex
polyhedron) which has edges LM.sub.1, M.sub.1L.sub.1,
L.sub.1M.sub.1, M.sub.1J, JM.sub.1, M.sub.1L, LL.sub.1, LJ and
L.sub.1J. The atom .xi. satisfies the following formula (17).
Furthermore, as is shown in FIG. 22(c) the atom .xi. has a mirror
symmetrical atom .xi.' 152.
(formula 17)
LK.sub.1:M.sub.1L.sub.1:L.sub.1M.sub.1:M.sub.1J:JM.sub.1:M.sub.1L:LL.sub-
.1,:LJ:L.sub.1J=2/ 3:1 / 6:1/ 6: (2/3): (2/3): (2/3):2 3: (3/2):
2:1/ 2 (17)
[0209] In addition, with respect to the right side tetrahedron 123
in FIG. 22(b), a mirror image symmetrical 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 .beta. 210
and atom .gamma. 203.
[0210] In order to show this, first the atom .gamma. 203 is focused
on. This is a slightly long and triangular pyramid, however, as is
shown in FIG. 24, the atom .theta. 800 is cut from the end of the
atom .gamma.. Then, a development of each of the polyhedrons is
shown in FIG. 25 and FIG. 26. The polyhedron in which the atom
.theta. 800 is cut from the end of the atom .gamma. is defined as
atom .eta. 900. In each of FIG. 25 and FIG. 26 which are
developments, the dimension of each edge is shown as a ratio.
Furthermore, the atom .eta. 900 and the atom .theta. 800 each has a
mirror symmetrical atom .eta.' 901 and .theta.' 801.
[0211] Atom .theta. 800 is a tetrahedron (convex polyhedron) which
has edges VO.sub.1, O.sub.1P.sub.1, P.sub.1O.sub.1, O.sub.1N.sub.1,
N.sub.1O.sub.1, O.sub.1V, VP.sub.1, VN.sub.1 and N.sub.1P.sub.1 as
is shown in FIG. 25. Furthermore, the atom .theta. has a mirror
symmetrical atom .theta.' 801. The atom .theta. 800 satisfies the
formulas (18) and (19) below.
(formula 18)
e= (2/5(9-4 5)) (18)
(formula 19)
VO.sub.1:O.sub.1P.sub.1:P.sub.1O.sub.1:O.sub.1N.sub.1:N.sub.1O.sub.1:O.s-
ub.1V: VP.sub.1:VN.sub.1:N.sub.1P.sub.1= 3ak:e:e:ak:ak: 3ak 2(
5-2):2ak: 3e (19)
[0212] Atom .eta. 900 is a pentahedron (convex polyhedron) which
has edges O.sub.1N.sub.1, N.sub.1O.sub.1, O.sub.1P.sub.1,
P.sub.1O.sub.1, O.sub.1C, CT, TC, CR, RC, CO.sub.1, N.sub.1P.sub.1,
P.sub.1T, TR and N.sub.1R, as is shown in FIG. 25. Furthermore, the
atom .eta. 900 has a mirror symmetrical atom .eta.' 901.
[0213] FIG. 26 is a development of the atom .eta. 900 and satisfies
the formula (20) below.
(formula 20)
O.sub.1N.sub.1:N.sub.1O.sub.1:O.sub.1P.sub.1:P.sub.1O.sub.1:O.sub.1C:CT:-
TC:CR:RC: CO.sub.1:N.sub.1P.sub.1:P.sub.1T:TR:N.sub.1R=ak:ak:e:e;
3ak':a:a:a:a: 3ak': 3e: 2:2a:2ak' (20)
[0214] By combining one of three types of atom, .beta. 210, .eta.
900 and .theta. 800, it is possible to make a quadrangular pyramid
701 shown in FIG. 22(d).
[0215] By adding four atoms .xi. 151 to the quadrangular pyramid
701, two atoms .alpha. 110, that is, the polyhedron 700 shown in
FIG. 27 is obtained. As stated above, the regular tetrahedron is
formed by sixteen .xi. 151, four .beta. 210, four .eta. 900 and
four .theta. 800, in total
.beta..sub.4.xi..sub.16.eta..sub.4.theta..sub.4
(.beta..sub.2.beta.'.sub.2.xi..sub.8.xi.'.sub.8.eta..sub.2.eta.'.sub.2.th-
eta..sub.2.theta.'.sub.2). Furthermore, as already indicated, the
regular tetrahedron 200 is also formed by .xi..sub.6.
[0216] When a regular tetrahedron is formed by this method, the
entire structure using the atom .alpha. 110 is replaced with this,
and the regular tetrahedron 100, the cube 200, the regular
octahedron 300 which are formed by the atoms .beta. 210, .xi. 151,
.eta. 900 and .theta. 801 each become:
TABLE-US-00005 Regular tetrahedron
.beta..sub.4.xi..sub.16.eta..sub.4.theta..sub.4.xi..sub.6
(.beta..sub.2.beta.'.sub.2.xi..sub.8.xi.'.sub.8.eta..sub.2.eta.'.sub.2.th-
eta..sub.2.theta.'.sub.2.xi..sub.3.xi.'.sub.3) Cube
.beta..sub.16.xi..sub.16.eta..sub.16.theta..sub.16
(.beta..sub.8.beta.'.sub.8.xi..sub.8.xi.'.sub.8.eta..sub.8.eta.'.sub.8.th-
eta..sub.8.theta.'.sub.8) Regular octahedron
.beta..sub.24.eta..sub.24.theta..sub.24
(.beta..sub.12.beta.'.sub.12.eta..sub.12.eta.'.sub.12.theta..sub.12.theta-
.'.sub.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.
[0217] 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 .alpha., .beta.,
.gamma., .delta. 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
.alpha.. Consequently, because the three-dimensional puzzle of the
present invention related to the present embodiment can form the
atom .alpha. by combining different atoms and form a regular
tetrahedron, cube and regular dodecahedron without using the atom
.alpha., it is possible to visually explain the above stated
theorem 1 and demonstrate excellent effects which can compare both
by a transformation example using .xi., .theta., .eta. 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.
Fourth Embodiment
[0218] 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.
[0219] When a non-congruent convex polyhedron P.sub.i is
1.ltoreq.i.ltoreq.n, then II={P.sub.1, P.sub.2, . . . , P.sub.n}.
If at least one element P which is included in P.sub.i, is formed
by face to face joining of congruent convex polyhedrons of convex
polyhedron .sigma., then convex polyhedron .sigma. is called an
atom of II.
[0220] 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.
(Minkowski Theorem A)
[0221] If P is a d-dimensional parallelohedron, then [0222] 1) P is
centrally symmetric, [0223] 2) All faces of P are centrally
symmetric, [0224] 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.
(Minkowski Theorem B)
[0225] The number f.sub.d-1 (P) of faces of a d-parallelohedron P
does not exceed 2(2.sup.d-1) and there is a parallelohedron P with
f.sub.d-1=2(2.sup.d-1).
[0226] 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.
[0227] 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) to FIG.
29(e). These parallelohedra are called "Fedrov polyhedrons" or
"Fedrov space filling solids". FIG. 30(a) to FIG. 30(e) shows the
space filling characteristics of each type of parallelohedron.
[0228] 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.
[0229] Let II.sub.1 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 II.sub.1 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.
(Cube and Rhombic Dodecahedron)
[0230] The cube 1200 in FIG. 31(a) is decomposed into six congruent
quadrangular pyramids 1201 shown in FIG. 31(b) using six planes,
each of which passes through the center of the cube and contains
two opposite edges. The quadrangular pyramid 1201, as shown in FIG.
31(a), has a square base and the four sides are congruent isosceles
triangles.
[0231] The quadrangular pyramid 1201 is further divided into four
congruent triangular pyramids 1203 as shown in FIG. 32(b) by two
perpendicular planes passing through the peak and through each of
the two diagonals of the base. When these triangular pyramids 1203
are shown again in FIG. 32(c), because three faces are orthogonal
at one peak, the triangular pyramid 1203 is called a right tetra
1203. Thus, the cube 1200 can be divided into twenty-four congruent
right tetra 1203. If the process is reversed, then it is possible
to form the cube 1200 by twenty-four congruent right tetra
1203.
[0232] In addition, as is shown in FIG. 33(a) and FIG. 33(b), by
attaching a quadrangular pyramid 1201 to each of the six congruent
faces of the cube 1200, a rhombic dodecahedron 1300 is formed. This
shows that it is possible to form a rhombic dodecahedron 1300 from
forty-eight congruent right tetra 1203. In this way, the right
tetra 1203 is a common constituent element of the cube 1200 and the
rhombic dodecahedron 1300.
(Skewed Hexagonal Prism)
[0233] As is shown in FIG. 32(c), the base of the right tetra 1203
is a right isosceles triangle. Two right tetras 1203 which are
glued at the base of these right isosceles triangles form another
tetrahedron 1100 as is shown in FIG. 34(a). This tetrahedron is
called a Sphenoid 1100. In FIG. 34(a) the dash line represents the
gluing surface of two triangular pyramids.
[0234] 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.
[0235] 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 in FIG. 34(b). Then, the skewed
hexagonal prism 1400 of Fedorov shown in FIG. 34(c) is formed using
three pairs of the triangular prism 1401 and its mirror image
symmetrical triangular prism 1403 so that the upper and lower faces
are adjusted to form parallel hexagonals. Therefore, a skewed
hexagonal prism 1400 can be formed by eighteen sphenoids 1100.
(Elongated Rhombic Dodecahedron)
[0236] Next, FIG. 35(a) shows an elongated rhombic dodecahedron
1500. This elongated rhombic dodecahedron 1500 is formed from two
parts. One part is a rhombic dodecahedron 1300 and the other part
is a concave polyhedron 1501 which can be obtained from the rhombic
dodecahedron. In order to obtain the second convex polyhedron 1501,
we consider four peaks of the rhombic dodecahedron 1300. The
elongated rhombic dodecahedron 1500 is cut along the edges
containing these peaks shown in FIG. 35, and the elongated rhombic
dodecahedron 1500 is opened at this peak. The obtained polyhedron
1501 is formed in the shape of a helmet. As stated previously, the
rhombic dodecahedron 1300 consists of forty-eight right tetras
1203. Therefore, the elongated rhombic dodecahedron 1500 can be
formed using ninety-six right tetras 1203.
(Truncated Octahedron)
[0237] The sphenoid 1100 has another important property. As is
shown in FIG. 36(a), it can be divided into four congruent
hexahedrons 1105 as is shown in FIG. 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
these hexahedrons 1105 is called a c-squadron.
[0238] 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, a truncated octahedron 1600 is
formed as shown in FIG. 37(b) by combining (glued) together six
diamonds 1601. In other words, the c-squadron 1105 is a constituent
element of the truncated octahedron 1600 and the truncated
octahedron 1600 can be formed using twenty-four c-squadrons
1105.
[0239] From the above, it can be seen that the cube (parallel
hexagram) 1200 can be formed by twenty-four right tetras 1203, the
rhombic dodecahedron 1300 can be formed by forty-eight right tetras
1203, the skewed hexagonal prism 1400 can be formed by eighteen
sphenoids 1100, the elongated rhombic dodecahedron 1500 can be
formed by ninety-six tetras and the truncated octahedron can be
formed by twenty-four c-squadrons 1105.
[0240] Next, an examination is made whether common constituent
elements exist between the right tetra 1203, sphenoid 1100 and
c-squadron 1105.
[0241] Here, one among the four c-squadrons 1105 shown in FIG.
36(b) is shown again in FIG. 38(a). The c-squadron 1105 is a solid
with left and right symmetry and is divided into two congruent
pentahedrons 1101 and 1103 as is shown in FIG. 38(b). Then, as is
shown in FIG. 38(c), by using two each of these new pentahedrons
1101 and 1103, it is possible to form the right tetra 1203 shown in
FIG. 32(c).
[0242] In addition, as is shown in FIG. 1, it is possible to form
the sphenoid 1100 with eight of these pentahedrons (four
pentahedrons 1101 and four pentahedrons 1103). That is, the
pentahedrons 1101 and 1103 are a common constituent element of the
five types of Fedrov space filling solid.
[0243] It is not possible to divide the pentahedrons 1101 and 1103
into further congruent solids. The pentahedrons 1101 and 1103 are
atoms of the five types of Fedrov space filling solids, and the
pentahedron 1101 is defined as a and its mirror image 1103 is
defined as .sigma.'. A development including dimensions of the atom
.sigma. 1101 and atom .sigma.' 1103 is shown in FIG. 39. As is
shown in FIG. 39, the atom .sigma. 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 .sigma.' 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
.sigma. satisfies the following formula (21). Furthermore, the
formula of the atom .sigma.' which is the mirror image of the atom
.sigma. 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)
[0244] By using the atom .sigma. 1101 and atom .sigma.' 1103, it is
clear that the cube (parallel hexahedron) 1200 can be formed by
ninety-six atoms .sigma. (forty-eight atoms .sigma. 1101 and
forty-eight atoms .sigma.' 1103), the rhombic dodecahedron 1300 can
be formed by one hundred and ninety-two atoms .sigma. (ninety-six
atoms .sigma. 1101 and ninety-six atoms .sigma.' 1103), the
truncated octahedron 1600 can be formed by forty-eight atoms
.sigma. (twenty-four atoms .sigma. 1101 and twenty-four atoms
.sigma.' 1103), the skewed hexagonal prism 1400 can be formed by
one hundred and forty-four atoms .sigma. (seventy-two atoms .sigma.
1101 and seventy-two atoms .sigma.' 1103) and the elongated rhombic
dodecahedron 1500 can be formed by three-hundred and eighty-four
atoms .sigma. (one-hundred and ninety-two atoms .sigma. 1101
one-hundred and ninety-two atoms .sigma.' 1103).
[0245] In addition, a cube is formed by ninety-six atoms .sigma.
1101 and atoms .sigma.' 1103, however, it is possible to form a
cube with as little as twenty-four atoms .sigma. (twelve atoms
.delta. 1101 and twelve atoms .sigma.' 1103).
[0246] From the above, it was discovered that the minimum number of
atoms for forming a Fedrov space filling solid is one type of atom
.sigma. (and its mirror image .sigma.').
[0247] 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. [0248] (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.
[0249] (2) A polyhedron is "indecomposable" means P are indivisible
into two or more congruent polyhedrons. [0250] (3) Let II be a
group of polyhedrons P.sub.1, P.sub.2, . . . P.sub.n. [0251] i.e.
II={P.sub.1, P.sub.2, . . . P.sub.n} [0252] Let E be a group of
indecomposable polyhedrons e.sub.1, e.sub.2, . . . e.sub.m, [0253]
i.e. E={e.sub.1, e.sub.2, . . . e.sub.m}, .A-inverted.e.sub.1 is
indecomposable, e.sub.i and e.sub.j are non-congruent (i.noteq.j)
[0254] At this time, E is a group E (II) of the element (atom) of
II means satisfying formula (15) below.
[0254] ( formula 22 ) P i = i = 1 m a i e i ( .A-inverted. i < 1
.ltoreq. i .ltoreq. n ) , a i is a nonnegative integer . ( 22 )
##EQU00003##
[0255] That is, whichever polyhedron P.sub.i belongs to II, it is
possible to divide into an indecomposable polyhedron belonging to
E. [0256] (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. [0257] That is e (II)=min|E (II)|
(Theorem 2)
[0257] [0258] Let II.sub.2={a group of Fedrov space filling solids}
[0259] e (II.sub.2).ltoreq.1, E (II.sub.2)={.sigma.}
[0260] Here, the three-dimensional puzzle of the present invention
related to the present embodiment which uses a minimum number of
atoms .sigma. 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 .sigma. 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
stated theorem 2.
(Regular Tetrahedron Three-Dimensional Puzzle)
[0261] 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 in FIG. 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 an atom .sigma. 1101 and four pieces which
have a shape of the atom .sigma. 1101 mirror image .sigma.' 1103 as
explained above in detail.
(Parallel Hexagram Three-Dimensional Puzzle)
[0262] 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 .sigma. 1101
and forty-eight pieces which have a shape of the atom .sigma. 1101
mirror image .sigma.' 1103 as explained above in detail.
(Skewed Hexagonal Prism Three-Dimensional Puzzle)
[0263] 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 .sigma. 1101 and
seventy-two which have a shape of the atom .sigma. 1101 mirror
image .sigma.' 1103 as explained above in detail.
(Truncated Octahedron Three-Dimensional Puzzle)
[0264] 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 .sigma. 1101
and twenty-four which have a shape of the atom .sigma. 1101 mirror
image .sigma.' 1103 as explained above in detail.
(Rhombic Dodecahedron Three-Dimensional Puzzle)
[0265] 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 .sigma. 1101
and ninety-six which have a shape of the atom .sigma. 1101 mirror
image .sigma.' 1103 as explained above in detail.
(Elongated Rhombic Dodecahedron Three-Dimensional Puzzle)
[0266] 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 .sigma. 1101 and one hundred and ninety-two which have a
shape of the atom .sigma. 1101 mirror image .sigma.' 1103 as
explained above in detail.
[0267] 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 .sigma. 1101
and a piece which has the shape of one atom .sigma.' 1103 mirror
image. As a result, it is possible to visually explain the theorem
2 stated above and the three-dimensional puzzle of the present
invention related to the present embodiment is excellent as an
educational material.
Fifth Embodiment
(Elongated Rhombic Dodecahedron Including a Space Filling
Solid)
[0268] 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 .sigma. 1101 and .sigma.'
1103.
[0269] From the proof of the elongated rhombic dodecahedron in
theorem 2 and FIG. 35, it is clear that the elongated rhombic
dodecahedron 1500 includes a rhombic dodecahedron 1300. In
addition, from the proof of the cube 1200 and the rhombic
dodecahedron 1300 in theorem 1, it was shown that the rhombic
dodecahedron 1300 includes the cube 1200. The rhombic dodecahedron
1300 is obtained by gluing the square bottom of the quadrangular
pyramid 1201 to each face of the cube 1200. Therefore, the
elongated rhombic dodecahedron 1500 includes the rhombic
dodecahedron 1300 and the rhombic dodecahedron 1300 includes the
cube 1200.
[0270] The truncated octahedron 1600 includes the cube 1200 by a
special method. When atom .sigma. 1101 and atom .sigma.' 1103 which
are congruent pentahedrons shown in FIG. 38(b) are considered, a
hexahedron 1701 including seven vertices and eleven edges shown in
FIG. 40 by gluing the atoms.
[0271] 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.
[0272] The polyhedron 1703 show in FIG. 41(a) is obtained when
three tripenquadrons 1701 are glued along their right triangular
faces so that they all meet at the vertex. The bottom face of this
polyhedron 1703 is a regular hexahedron shown in FIG. 41(b).
[0273] When eight of these polyhedrons 1703 are glued along the
right isosceles triangular faces, a cube 1705 is obtained having
six holes at the center of each face as shown in FIG. 42(a). In
order to examine the interior of the cube 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 in FIG. 42(b) is viewed, it reveals that the
cross section in the interior of the polyhedron 1707 made by
cutting the cube 1705 with open holes consist of regular hexagons
and that the hole is obtained by carving out half of a truncated
octahedron 1603. FIG. 42(c) and FIG. 42(d) show a cube which is the
result of assembling the truncated octahedron and cube with open
holes. It is clear that the truncated octahedron 1600 is included
in the cube 1700 in a special way.
[0274] 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 the rhombic
dodecahedron 1300 and helmet shaped polyhedron 1501 is again
considered.
[0275] First, as is shown in FIG. 43(a), the rhombic dodecahedron
1300 is divided into two congruent polyhedrons 1301 and only one of
these is considered. Next, a part of the helmet shaped polyhedron
1501 shown in FIG. 43(b) is cut off as shown in FIG. 43(c). The
right side polyhedron 1503 in FIG. 43(c) is V-shaped and consists
of two skewed triangular prisms 1505 and 1507 and the skewed
triangle prism 1507 is a mirror image of the skewed triangular
prism 1505. Each skewed triangular prism is formed by gluing three
sphenoids as shown in FIG. 36(a).
[0276] If the V-shaped polyhedron 1503 is glued to the lower half
of the rhombic dodecahedron 1301, the skewed hexagonal prism 1420
shown in FIG. 43(d) and FIG. 43(e) is obtained. In this way, it is
shown that the skewed hexagonal prism 1420 is included within the
elongated rhombic dodecahedron 1500 in a special way.
[0277] 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 .sigma. 1101 and a piece
which has the shape of one atom .sigma.' 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.
(Truncated Octahedron Three-Dimensional Puzzle)
[0278] FIG. 44 shows the arrangement of a puzzle 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 the atom .sigma. 1101 and twenty-four pieces
which have the shape of the mirror image .sigma.' 1103 of the atom
.sigma. 1101 as explained in detail above.
(Cube Three-Dimensional Puzzle)
[0279] 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 a puzzle 1651 which is half of the
truncated octahedron three-dimensional puzzle, (b) shows a puzzle
1751 by a cross section of a cube with open holes, (c) shows a
puzzle 1753 of half of a cube in which the puzzle 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 three dimensional puzzle 1750. The cube
three-dimensional puzzle can be formed using forty-eight pieces
which have the shape of the atom .sigma. 1101 and forty-eight
pieces which have the shape of the mirror image .sigma.' 1103 of
the atom .delta. 1101 as explained in detail above.
(Rhombic Dodecahedron Three-Dimensional Puzzle)
[0280] FIG. 47 shows an arrangement of a rhombic dodecahedron
three-dimensional puzzle of the present invention related to the
present embodiment. (a) shows puzzle 1753 of half of a cube, (b)
shows puzzle 1353 without the half of the cube from the puzzle 1351
which is half of a rhombic dodecahedron, (c) shows the puzzle 1351
which is half of a rhombic dodecahedron, (d) shows a side face view
of rhombic dodecahedron puzzle 1350, (e) shows an upper face view
of the rhombic dodecahedron puzzle 1350. The rhombic dodecahedron
three-dimensional puzzle can be formed using ninety-six pieces
which have the shape of the atom .sigma. 1101 and ninety-six pieces
which have the shape of the mirror image .sigma.' 1103 of the atom
.sigma.60 1101 as explained in detail above.
(Skewed Hexagonal Prism Three-Dimensional Puzzle)
[0281] 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 of puzzle 1553 cut
from a helmet shaped polyhedron puzzle 1551, (b) shows a front face
view of the puzzle 1553, (c) shows a back face view of the puzzle
1553, (d) shows an exemplary view of the formation of a skewed
hexagonal prism 1450 by the puzzle 1553 which is cut away from the
helmet shaped polyhedron puzzle 1551 and a puzzle 1351 which is
half of the rhombic dodecahedron 1350, (e) shows an upper face view
of the skewed hexagonal prism 1450, (f) shows a side face view of
the skewed hexagonal prism 1450. The skewed hexagonal prism 1450
three-dimensional puzzle can be formed using seventy-two pieces
which have the shape of the atom .sigma. 1101 and seventy-two
pieces which have the shape of the mirror image .sigma.' 1103 of
the atom .sigma. 1101 as explained in detail above.
(Elongated Rhombic Dodecahedron Three-Dimensional Puzzle)
[0282] 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 shaped polyhedron puzzle 1551, (b) shows a front face
view of the helmet shaped polyhedron puzzle 1551, (c) shows a back
view of the helmet shaped polyhedron puzzle 1551, (d) shows an
exemplary view of the formation of an elongated rhombic
dodecahedron three-dimensional puzzle 1550 by the helmet shaped
polyhedron puzzle 1551 and the rhombic 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 the
atom .sigma. 1101 and one hundred and ninety-two which have the
shape of the mirror image .sigma.' 1103 of the atom .sigma. 1101 as
explained in detail above.
[0283] 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 .sigma. 1101 and a piece
which has the shape of one atom .sigma.' 1103 of its mirror image.
In addition, it is possible to visually explain the theorem 2
stated above and the three-dimensional puzzle of the present
invention related to the present embodiment is excellent as an
educational material.
[0284] 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.
[0285] 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.
[0286] 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.
[0287] 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.
* * * * *