FOREWORD
P O L Y H E D R A

THE REALM OF GEOMETRIC BEAUTY

By

P R E F A C E

The total lack of a single book dealing, in a succinct manner, with the origin, development and natural geometric construction of the most important duals of the thirteen semiregular Archimedean polyhedra was the inspiration for this unique work. Furthermore, I wished to present these polyhedra in such a form as to emphasize their supreme eternal beauty.
It was my desire to develop this topic with such care, order and accuracy that both the student and the scholar could methodically proceed to a clear understanding of the principle underlying this ancient but perennial discipline.
U. A. G.
University of San Francisco [1962]

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Ars sine scientia nihil.

THE ORIGIN OF THE THIRTEEN DUALS
OF THE SEMI-REGULAR ARCHIMEDEAN POLYHEDRA
The Graziotti Approach

In space, every figure having height, width and depth, and formed of planes, lines and points, has associated with it another figure composed of planes, lines and points--its so-called dual configuration. In this association lies a far reaching principle of mathematics--the principle of duality. Each of the points of the first body is transformed into a plane of the second body, and a plane is transformed into a point. The new body thus formed is called the dual of the first. The construction of the dual elements must be carried out in a prescribed manner. A demonstration of the formation of a well known regular body--the octahedron--will serve to illustrate the construction. According to Diagram 1, construct at each vertex of the octahedron a plane, perpendicular to the line joining the vertex to the center of symmetry of the octahedron. To each vertex, a unique plane always corresponds. Each of the planes is the dual of the vertex. The intersection of these planes form a hexahedron. Hence, the hexahedron is the dual of the octahedron.

 Diagram 1

When the principle is applied to regular, semi-regular, or some of the starred polyhedra, it is a direct conjugation with respect to the surface of the sphere in which they can be inscribed. The duals obtained from the group of the five regular polyhedra (Platonic bodies) are also five regular polyhedra, because they repeat alternately their own shapes.1 In fact, the tetrahedron always produces another tetrahedron similar to its prototype.2 The octahedron and the icosahedron produce the hexahedron and the dodecahedron respectively. The duals obtained from the thirteen principal semi regular or Archimedean polyhedra3 have all the centroids of their faces equidistant from the center of the symmetry of the polyhedron, but their faces are not regular polygons. regularity lies only between the surface and the center of symmetry of the solid. Therefore, the Archimedean duals can also be defined as perpendicularly-regular, in contrast to the thirteen Archimedean bodies which are distinguished by being facially-regular, because their faces are all equiangular and equilateral, though not similar.

All the mathematicians and geometricians who work in the field of polyhedra are aware of the origin of the Archimedean bodies. They know that these are derived in a more or less direct way from the five Platonic polyhedra, but few of them know the origin of the duals of Archimedes. The principal reason that these duals are unknown or only obscurely known is that no one, until the present, undertook their study -- searching for their geometric origins, and devulging the results. Another reason is that they involve a rather complex construction if solved by synthetic geometry, and highly involved calculations in spherical trigonometry, if solved analytically.

Mining engineers may be the only people thus far who know the morphological properties of this type of polyhedron, since nearly all these figures appear in crystal and skeletal forms found in nature. This group--identified here as b-type--includes thirteen principal bodies which are not inscribable in a sphere. Only an in-sphere is tangent to the centroids of their faces. They differ from the first group, the a-type, mainly because they are isohedral, which means that faces of a single polyhedron are alike, although the perimeter of their faces forms irregular polygons. These polygons are of three kinds: triangular, quadrilateral, and pentagonal. Their anguloids—the solid corners at the vertices—are trihedrals, tetrahedrals, pentahedrals, hexahedrals, octahedrals, and decahedrals. Four of them — the rhombic dodecahedron, the trapezoidal icositetrahedron, the rhombic triacontahedron, and the pentagonal icositetrahedron — were known long ago. Three of them are found in nature. The first two are the crystalline forms of the mineral garnet, and the fourth is the crystalline form of the mineral cuprite, CU2O, with the crystals taking the form of 24 hedra or faces, bounded by congruent irregular pentagons. The third is not found in nature, but it has a simple and well known solution (see Plate 7). The design is harmonious and rhythmical, consisting of twelve pentagonal stars, each composed of five rhombs totally overlapping each other, in such a way that the twelve pentagonal stars wrap exactly two times the surface of the rhombic triacontahedron. The first of these four was already investigated by Kepler. He shows how the figure is constructed from the hexahedron.4

The other nine of the thirteen b-type figures were unknown to mathematicians and geometers until the last century when I. H. T. Muller5 mentioned them for the first time in 1852 in his work, Trigometrie. The real credit for finding the solution belongs to a brilliant French mathematician named Catalan, who discovered all thirteen of them 1862. For his discovery, he was awarded the first prize bestowed by the French Academy of Sciences.6 the problem given to the contestants for the Grand Prix in Mathematics was "To Perfect the Geometric Theory of Polyhedra." Catalan's winning paper was called "Mémoire sur la Théorie des Polyèdres". He approached the subject purely from the trigonometrical point of view, however, and did not discover either the natural or the simplest geometric origins. It is a matter of experience that people usually do things in the most difficult manner before finding the simplest approach to accomplish the same result.

D'Arcy Thompson7 also believed that these polyhedra could never be constructed by geometrical means such as Archimedes used in constructing the thirteen a-type bodies by the inductive process. Thompson stated flatly that "it [the solution] is a matter of spherical trigonometry rather than simple geometry, and the problem for that reason, remained long unsolved." It was unsolved until the work of Catalan appeared in 1862. Thompson's statement must be refuted for the benefit of science.

In the following pages the author shows for the first time in history the construction of the main thirteen Archimedean duals and their solution by geometrical means. The beautiful and dynamic forms are illustrated, and comments are made upon them. They are given, along with their prototypes, in sequence form, to the point of their final derivative stage.
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PROOF BASED ON THE TRIAKIS OCTAHEDRON

In order to prove the geometric construction of the triakis octahedron, we must show that three basic properties exist in the completed polyhedron. These three basic properties are: 1. Symmetry and congruency of the faces. 2. Perpendicular and regularity of the faces relative to the center of symmetry of the polyhedron. 3. 'Convexity' of the surface of the polyhedron. The following demonstration of the existence of the above properties in the triakis octahedron uses the same procedure that would be applied to any of the thirteen duals of Archimedes. Because of the symmetry of the processes of constructing the polyhedra and the similarities which exist among them, the demonstration below for this one polyhedron proves, in a general sense, the processes for all the 13 figures. If a rigorous proof of each solid is desired, the reader can apply on his own the basic method outlined here. The proof consists of cutting the completed polyhedron by suitable planes, chosen to show the triangular relationships of the intersections of the faces.

SYMMETRY AND CONGRUENCY OF THE FACES

If the faces of the polyhedron are counted, we find there are the required 24 faces. By cutting the solid figure in half, and passing a plane through four vertices, showing on this plane the circumference of the finished polyhedron as well as the lines of intersection of this plane and the various planes of the original hexahedron, we will have Diagram 2.
The fact that the four vertices lie on the same plane and that this plane does in fact 'cut the solid in half', can be proven if we remember that the vertices of the diagram are also the vertices of the pyramids constructed on the face of the original hexahedron, and hence must lie directly above the centers of the faces of the hexahedron. In Diagram 2, a face of the polyhedron has been shown rabbeted so as to be seen in plane view. Diagram 3 shows one of the pyramids constructed on the face of the hexahedron, and this will help the reader to visualize the triangle relationship in Diagram 2.

 Diagram 2-3

•• A1, A2, A3, etc. = Vertices of the starred hexahedron and of the triakis tetrahedron. •• C = Center of symmetry of the three polyhedra. •• M = Mid points of lines joining vertices. •• m = Mid points of edges of the hexahedron. •• O = Third apex of triangular face (at the corner of original hexahedron). •• S¯¯S = Axis of rotation of plane A1¯¯O¯¯A4¯¯M¯¯A1 •• A1¯¯O¯¯A4¯¯M¯¯A1 = Rabbeted face

From an inspection of the triangular relationship of Diagram 2, the reader will see that the following are true:

1. Each triangular face is typically bounded by the line A1¯¯A4, A1¯¯O, and A4¯¯O.
A1¯¯A4 = A1¯¯A2 = A2¯¯A3 = A3¯¯A4.
A1¯¯O = A4¯¯O = A2¯¯O = etc. = S.

2. From (1) above, all faces are congruent since their three respective sides are equal.

3. Since C¯¯A1 = C¯¯A4, the line C¯¯m extended is the perpendicular bisector of A1¯¯A4. Similarly O M is the perpendicular bisector of A1¯¯A4.

An analysis of this plane through the four vertices shows the symmetry of the polyhedron, and, in (2) above, the faces are shown to be congruent, thus demonstrating the first basic property.

PERPENDICULAR REGULARITY

In Diagram 2 it has been seen that the midpoint of A1¯¯A4 (M), the midpoint of the edge of the hexahedron (m) and the center of symmetry of the hexahedron (C), lie on a straight line. This line, extended, wil also pass through the midpoint of the opposite edge of the hexahedron. If we draw the plane determined by this line and by the edge of the original hexahedron and show on this plane the circumference of the solid figure and of the original hexahedron, we will have Diagram 4.

 Diagram 4 Orthogonal projection of the starred hexahedron.

Rectangle O1 O2 O4 O5 O1 given by rotating the diagram at left.
O1 O2 A3 O4 O5 A6 O1 = Section of the starred hexahedron.

Since in Diagram 2 the line C¯¯m is perpendicular to A1¯¯A4, and since in Diagram 4 we have taken a plane which has to be perpendicular to that of Diagram 2, then the plane of Diagram 4 is perpendicular to A1¯¯A4. The lines M¯¯O then represent the plane of adjacent faces of the solid figure perpendicular to the plane of Diagram 4, and, if we can prove the perpendicular distances from C to M¯¯O1 and to M¯¯O2 are the same, we have proven that the perpendicular distances from two adjacent faces, and hence from all faces, to the center of symmetry of the polyhedron are the same. That the perpendicular distances of the lines M¯¯O1 and M¯¯O2 to C are the same can be seen from the fact that Δ C¯¯M¯¯O1 and Δ C¯¯M¯¯O2 are congruent.

From Diagram 4 then, we can show that all faces of the polyhedron have the same perpendicular distance to the center, and hence the second basic property required of the polyhedron has been demonstrated.

CONVEXITY

To demonstrate the third required property of the polyhedron, we must show that wherever two faces join, they form a convex joint in the surface of the polyhedron. The faces join in only two ways: along the short side, A1¯¯O, or along the long side, A1¯¯A2. That the two faces meeting along the line A1¯¯A2, the long side, form a convex juncture can be seen from an analysis of Diagram 2 and 4. It requires another view of the figure to demonstrate convexity along the lines A1¯¯O.

 Diagram 5

In diagram 5 the polyhedron has been cut by the plane determined by three, points O, M, and M1. From Diagram 2 we know that

A1¯¯O = S
A1¯¯M = ½(A1¯¯A2) = S {(2 + Ö2) /4}

Therefore A1¯¯O > A1¯¯M, and M¯¯O¯¯M¹¯¯M is a triangle with its vertex O farthest from the line A1¯¯C. From this, two faces joining along the lines A1¯¯O must form a convex surface. Hence, two adjacent faces must always form a convex surface. And the third basic property is demonstrated.

POLYHEDRA IMAGES

(Click on an image to view a larger size)

 Prospectus page 9 1 Plate 1 2 Plate 2 3 Plate 3 4 Plate 4 5 Plate 5 6 Plate 6 7 Plate 7 8 Plate 8 9 Plate 9 10 Plate 10 11 Plate 11 12 Plate 12 13 Plate 13

PLATES WITH COMMENTARIES

1
vertical
Plate1

Morphological Characteristics of the Triakis Tetrahedron

 Faces 12 congruent obtuse angles Edges 18 Vertices 8 Anguloids formed by 4 trihedrals and 4 hexahedrals Facial angles 112o 53' 8", 33o 33' 26", 33o 33' 26" Dihedral Angle θ = 129o 31' 16" Symbol E.3.62. (for each face, 3 edges at one vertex and 6 edges at each of the other two vertices

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1
horizontal

Plate 1

2
vertical
Plate 2

Morphological Characteristics of the Triakis Octahedron

 Faces 24 congruent abtuse-angled triangles Edges 36 Vertices 14 Anguloids formed by 8 trihedrals and 6 octahedrals Facial angles 117o 15' 17", 31o 22' 21", 31o 22' 21" Dihedral Angle θ = 147o 21' 0" Symbol E. 3. 82.
2
horizontal
Plate 2

3
vertical
Plate 3

Morphological Characteristics of the Triakis Hexahedron

 Faces 24 congruent, acute-angled triangle Edges 36 Vertices 14 Anguloids formed by 6 tetrahedrals and 8 hexahedrals Facial angles 83° 37' 14", 48° 11½', 11½' Dihedral Angle θ = 143° 7' 48" Symbol E. 4. 6².
3
horizontal
Plate 3

4
vertical
Plate 4

Morphological Characteristics of the Triakis Icosahedron

 Faces 60 congruent obtuse-angled triangles Edges 90 Vertices 32 Anguloids formed by 20 trihedrals and 12 decahedrals Facial angles 119° 2' 21", 30° 28' 49", 30° 28' 49" Dihedral Angle θ = 160° 36' 45" Symbol E. 3. 12²
4
horizontal
Plate 4

5
vertical
Plate 5

Morphological Characteristics of the Trapezoidal Icosistetrahedron

 Faces 24 congruent trapezoidals Edges 48 Vertices 26 Anguloids 8 trihedrals and 18 tetrhedrals Facial angles 115° 15' 48", 81° 34' 44", 138° 6' 34" Dihedral Angle θ = 138° 6' 34" Symbol E. 3. 4³
5
horizontal
Plate 5

6
vertical
Plate 6

Morphological Characteristics of the Rhombic Dodecahedron

 Faces 12 congruent Edges 24 Vertices 14 Anguloids 8 trihedrals and 6 tetrahedrals Facial angles 109° 28', 70° 32' Dihedral Angle θ = 120° Symbol E. (3.4)²

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6
horizontal
Plate 6

7
vertical
Plate 7

Morphological Characteristics of the Rhombic Triacontahedron

 Faces 30 congruent rhombs Edges 60 Vertices 32 Anguloids 20 trihedrals and 12 pentahedrals Facial angles 116° 34', 63° 26' Dihedral Angle θ = 144° Symbol E. (3.5)².
7
horizontal
Plate 7

8
vertical
Plate 8

Morphological Characteristics of the Hexakis Octahedron

 Faces 48 equal scalene triangles Edges 72 Vertices 26 Anguloids formed by 12 tetrahedrals, 8 hexahedrals, and 6 octahedrals Facial angles 87° 12', 55° 1½', 37° 46½' Dihedral Angle θ = 155° 5' 0" Symbol E. 4. 6. 8.
8
horizontal
Plate 8

9
vertical
Plate 9

Morphological Characteristics of the Pentakis Dodecahedron

 Faces 60 congruent, isosceles triangles Edges 90 Vertices 32 Anguloids 12 pentahedrals and 20 hexahedrals Facial angles 68° 37' 8", 55° 41' 26", 55° 41' 26" Dihedral Angle θ = 156° 43' 7" Symbol E. 5. 6².

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9
horizontal
Plate 9

10
vertical
Plate 10

Morphological Characteristics of the Trapezoidal Hexecontahedron

 Faces 60 congruent trapezoids Edges 120 Vertices 62 Anguloids 20 trihedrals, 30 tetrahedrals and 12 pentahedrals Facial angles 118° 16' 7", 86° 58' 27", 67° 46' 59" Dihedral Angle θ = 154° 8' Symbol E. 3. 4. 5. 4.
10
horizontal
Plate 10

11
vertical
Plate 11

Morphological Characteristics of the Hexakis Icosahedron

 Faces 120 congruent, scalene traingles Edges 150 Vertices 62 Anguloids formed by 30 tetrahedrals, 20 hexahedrals and 12 decahedrals Facial angles 88° 59' 31", 58° 14' 17", 32° 46' 12" Dihedral Angle θ = 164° 53' 17" Symbol E. 4. 6. 10.
11
horizontal
Plate poli/ 11

12
vertical
Plate 12

Morphological Characteristics of the Pentagonal Icositetrahedron

 Faces 24 equal but irregular pentagons Edges 60 Vertices 38 Anguloids formed by 32 trihedrals and 6 tetrahedrals Facial angles 114° 48½', 80° 46' Dihedral Angle θ = 136° 20' Symbol E. 34. 4.

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12
horizontal
Plate 12

13
vertical
Plate 13

Morphological Characteristics of the Pentagonal Hexecontrahedron

 Faces 60 congruent,irregular pentagons Edges 150 Vertices 92 Anguloids 80 thrihedrals and 12 pentahedral Facial angles 118° 8' 12", 67deg; 27' 12", 153° 10' 24" Dihedral Angle θ = 153° 10' 24" Symbol E. 34. 5.

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13
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Plate 13