P O L Y H E D R A
THE REALM OF GEOMETRIC BEAUTY
By
Ugo Adriano Graziotti
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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 1 
Ars sine scientia nihil.
THE ORIGIN OF THE THIRTEEN DUALS
OF THE SEMIREGULAR 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 pointsits socalled dual configuration. In this association lies a far reaching principle of mathematicsthe 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 bodythe octahedronwill 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

This illustration shows the hexahedron circumscribed by the octahedron, and the octahedron
circumscribing a smaller hexahedron, etc., etc. Since the octahedron is obtained in direct
conjunction with the hexahedron and vice versa, the operation of inscribing them within
each other may be repeated to infinity.
When the principle is applied to regular, semiregular, 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 polyhedra^{3} 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 perpendicularlyregular, in contrast to the thirteen Archimedean bodies which are distinguished by being faciallyregular, 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 groupidentified here as btypeincludes thirteen principal bodies which are not inscribable in a sphere. Only an insphere is tangent to the centroids of their faces. They differ from the first group, the atype, 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, CU_{2}O, 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 btype figures were unknown to mathematicians and geometers until the last century when I. H. T. Muller^{5} 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 Thompson^{7} also believed that these polyhedra could never be constructed by geometrical means such as Archimedes used in constructing the thirteen atype 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.
____________________
^{1}The five regular polyhedra are also defined as Platonic solids, Platonic bodies, or Platonic figures, since Plato (428347 B.C.) was the first to make an extensive description of their construction and their mystical properties in relation to the firmament in one of his works, The Timaeus.
In the writing of Proclus (410485), his Commentary to Timaeus shows that Plato wrote a special book on the five polyhedra. Unfortunately, the work has been lost.
^{2}The tetrahedron is capable of producing 2 more regular polyhedra: octahedron and icosahedron.
^{3}Archimedes (287?212 B.C.) wrote numerous works, including one on Investigation relating to Polyhedra. This is known from a scholium of a manuscript written by Pappus (third century A.D.), at present preserved in the Archives of the Vatican Library. The work was lost shorly after the times of Pappus and the amanuensis of the Middle Ages never made copies to be transmitted to posterity.
^{4}Johannes Kepler (15711630), Great Books of the Western World, 1952, Vol 16, p. 866.
^{5}Enciclopedia Italiana di Scienze, Lettere, ed Arti, Roma 193543, Vol. 27, p. 641 ff.
^{6}Journal de l'École Impériale Polytechnique, Paris 1865, Vol. 24, pp. 171.
^{7}D'Arcy Thompson, On Growth and Form , Cambridge University Press, Sec. Ed. Reprinted, Vol II, 1959.
 2 
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 23

•• A_{1}, A_{2}, A_{3}, 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 A_{1}¯¯O¯¯A_{4}¯¯M¯¯A_{1}
•• A_{1}¯¯O¯¯A_{4}¯¯M¯¯A_{1} = 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 A_{1}¯¯A_{4}, A_{1}¯¯O, and A_{4}¯¯O.
A_{1}¯¯A_{4} = A_{1}¯¯A_{2} = A_{2}¯¯A_{3} = A_{3}¯¯A_{4}.
A_{1}¯¯O = A_{4}¯¯O = A_{2}¯¯O = etc. = S.
2. From (1) above, all faces are congruent since their three respective sides are equal.
3. Since C¯¯A_{1} = C¯¯A_{4}, the line C¯¯m extended is the perpendicular bisector of A_{1}¯¯A_{4}. Similarly O M is the perpendicular bisector of A_{1}¯¯A_{4}.
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 A_{1}¯¯A_{4} (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.
Rectangle O_{1} O_{2} O_{4} O_{5} O_{1} given by rotating the diagram at left.
O_{1} O_{2} A_{3} O_{4} O_{5} A_{6} O_{1} = Section of the starred hexahedron.
Since in Diagram 2 the line C¯¯m is perpendicular to A_{1}¯¯A_{4}, 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 A_{1}¯¯A_{4}. 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¯¯O_{1} and to M¯¯O_{2} 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¯¯O_{1} and M¯¯O_{2} to C are the same can be seen from the fact that Δ C¯¯M¯¯O_{1} and Δ C¯¯M¯¯O_{2} 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, A_{1}¯¯O, or along the long side, A_{1}¯¯A_{2}. That the two faces meeting along the line A_{1}¯¯A_{2}, 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 A_{1}¯¯O.
Diagram 5

In diagram 5 the polyhedron has been cut by the plane determined by three, points O, M, and M^{1}. From Diagram 2 we know that
A_{1}¯¯O = S
A_{1}¯¯M = ½(A_{1}¯¯A_{2}) = S {(2 + Ö2) /4}
Therefore A_{1}¯¯O > A_{1}¯¯M, and M¯¯O¯¯M¹¯¯M is a triangle with its vertex O farthest from the line A_{1}¯¯C. From this, two faces joining along the lines A_{1}¯¯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)
PLATES WITH COMMENTARIES
1
vertical
Plate1
PLATE 1: Solid Triakis Tetrahedron
The natural geometric solution to the problem illustrated in Plate 1 is in three steps. We must start with the archetype, the regular tetrahedron. Take four more tetrahedra equal in size to the first, and construct them on the original tetrahedron in such a way that each one adheres to one of the four original faces. Now the tetrahedron has become the hidden core of the new polyhedron, a starred tetrahedron(central figure in the Plate)^{1}, composed of 12 equilateral triangles, 8 vertices and 18 edges. The vertices of this figure form the faces of the new polyhedron. By properly joining the four vertices of the starred tetrahedron with 6 straight lines of equal length, we will transform the original 12 equilateral triangular faces into 12 obtuseangled triangles, thus producing the last link in the chain and the final solution of the problem. The last figure is known as triakis tetrahedron(top figure in the Plate), etymologically meaning that each of the four faces of the regular tetrahedron is now increased by three supplementary faces. The triakis tetrahedron is the dual of the truncated tetrahedron.
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 
112^{o} 53' 8", 33^{o} 33' 26", 33^{o} 33' 26" 
Dihedral Angle 
θ = 129^{o} 31' 16" 
Symbol 
E.3.6^{2}. (for each face, 3 edges at one vertex and 6 edges at each of the other two vertices 
_________________
^{1}This solid is a "Uniform polyhedron" in the broad sense since all its faces are regular polygons of the same size and all its vertices are congruent. The regular starred hexahedron of Plate 2 and the regular starred dodecahedron of Plate 4 are also "Uniform polyhedra". In addition, the author has constructed four more uniform polyhedra, including the regular starred icosahedron. These last four will not be considered in this book. Thus, the total number of uniform polyhedra is not "at least 75", as asserted by U. S. M. Coxeter and others, but at least 82.

1
horizontal
Plate 1 
2
vertical
Plate 2
PLATE 2: Solid Triakis Octahedron
The Triakis Octahedron is the result of a chain composed of three links, the first of which is the hexahedron (the lowest figure in the Plate). It is the archetype of the triakis octahedron. The central figure in this Plate represents the starred hexahedron which is formed of 6 tetragonal pyramids constructed one on each face of the original solid, the length of the base and of the lateral edges of the pyramids being equal. The invisible core of the star polyhedron is the hexahedron; the surface of this solid is composed of 24 equilateral triangles, 14 vertices and 36 edges. It is a uniform polyhedron. By properly joining the 6 vertices of the stellated hexahedron by means of straight lines, we transform the figure into a new one made by 24 obtuseangled congruent triangles and known by the name of triakis octahedron (top figure). This name literally means that each of the 8 faces of the octahedron contains 3 other faces, The dual of the triakis octahedron is the truncated hexahedron, and in either way they cannot be conjugated. Consequently the geometric solution of this plate is the only possible one.
Morphological Characteristics of the Triakis Octahedron
Faces 
24 congruent abtuseangled triangles 
Edges 
36 
Vertices 
14 
Anguloids 
formed by 8 trihedrals and 6 octahedrals 
Facial angles 
117^{o} 15' 17", 31^{o} 22' 21", 31^{o} 22' 21" 
Dihedral Angle 
θ = 147^{o} 21' 0" 
Symbol 
E. 3. 8^{2}. 

2
horizontal
Plate 2 
3
vertical
Plate 3
PLATE 3: Solid Tetrakis Hexahedron
For this geometrical operation, the archetype is the cuboctahedron (lower figure). The central figure is called a stellated cuboctahedron composed of 6 quadrangular and 8 triangular pyramids, constructed on the faces of the original cuboctahedron. The tetragonal pyramids alone (without the trigonal pyramids) form the regular octahedron, since the cuboctahedron is a solid obtained by truncating it. The trigonal pyramids are all regular tetrahedrons, whose bases correspond to the midpoints of the edges of the octahedron. By properly joining, with segments of straight lines, the 14 vertices of the starred cuboctahedron, we produce a new solid having 24 isosceles triangles, known as the tetrakis hexahedron, which literally means that each of the six faces of the hexahedron is now composed of 4 supplementary faces (top figure). The dual of the tetrakis hexahedron is the truncated octahedron from the tetrakis hexahedron. This construction is directly obtained by means of one single step. However, the simplicity obtained above does not occur in the operation of deriving the tetrakis hexahedron from the truncated octahedron.
Morphological Characteristics of the Triakis Hexahedron
Faces 
24 congruent, acuteangled 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
PLATE 4: Solid Triakis Icosahedron
For this geometrical operation the archetype is the regular pentagonal dodecahedron (lowest figure). The central figure is a stellated dodecahedron, composed of 12 pentagonal pyramids, constructed on the faces of the original dodecahedron. The length of the lateral edges of the pyramids, from the base to the vertex, is equal to that of the base. The result will be a starred dodecahedron, having 12 blunted vertices (see central figure). It is a uniform polyhedron. Its shape is between that of the pentakis dodecahedron and that of the great stellated dodecahedron (the small stellated dodecahedron of Poinsot). It has 60 equilateral triangles, 32 vertices and 90 edges. By properly joining the vertices with segments of straight lines, we produce a new solid having 60 isosceles triangular faces, which is known as the triakis icosahedronliterally meaning that each of the 20 faces of the icosahedron is composed of three supplementary faces (see top figure represented in perspective). The dual of the triakis icosahedron is the truncated dodecahedron but neither the centroids of the former nor the centroids of the latter are capable of generating their duals. Consequently, the method presented in Plate 4 is the only possible way for obtaining the desired solution.
Morphological Characteristics of the Triakis Icosahedron
Faces 
60 congruent obtuseangled 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
PLATE 5: Solid Trapezoidal Icositetrahedron
The solid trapezoidal icositetrahedron is obtained by applying several consecutive geometrical operations to two interlinked, reciprocal polyhedra. The two archetypal solids are the cuboctahedron, to the left, and the rhombic dodecahedron to the right (lowest figures). The central figure is composed of the two lower polyhedra, interlinked, which assume the appearance of a starred polyhedron, the core of which is the small rhombicuboctahedron. But notice that the real stellated small rhombicuboctahedron (see Plate 8, central figure) has a quite different aspect from this one, for its vertices are more expanded, and all the edges of its trigonal and tetragonal pyramids are of equal length. In this normal interpenetration the edges of the cuboctahedron are intersected at their mid points by the edges of the rhombic dodecahedron, but the edges of the latter are not divided into two equal parts. Consequently the resulting trigonal pyramids are lower than the tetragonal ones.
By properly joining the vertices of the compound polyhedron with straight lines of equal length we obtain another solid which is known as the trapezoidal icositetrahedron (top figure). Thys polyhedron is called 'trapezoidal icositetrahedron' due to its general shape and the shape of its faces. Etymologically it would be (eicosi = twenty; tetra = four; edra = face) and (trapezoidal = quadrilateral polygon analogous to the trapezium) Figure having 24 trapezoidal faces.
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
PLATE 6: Solid Rhombic Dodecahedron
The easiest way to obtain the rhombic dodecahedron is by applying several consecutive geometrical operations to two regular polyhedra.^{1} These two polyhedra are reciprocal, the hexahedron and the octahedron (lower figure). The central figure is composed of the two lower polyhedra interlinked, which assume the appearance of the starred cuboctahedron. But here, too, as in the case of the preceding Plate, the real starred cuboctahedron has quite different features, for 8 of its 14 vertices are more expanded (see Plate 3, central figure). By properly joining the 14 vertices of the interpenetrated polyhedra, with 24 segments of straight lines of equal length, we obtain the desired rhombic dodecahedron, which literally means that which is formed by 12 rhomboids. The dual of the rhombic dodecahedron is the cuboctahedron, and only in an indirect way are they conjugates of each other. Their second geometric solution is remote and cyclical.(See page 9 [i.e. the "Prospectus", above]).
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 
________________
^{1}This way of obtaining the rhombic dodecahedron is not new, but it is used here with a new arrangement because the author wanted to apply the same principle he used in developing the method of the other 12 plates, thus obtaining a coherence of style.
This figure is also found in nature as "common garnet".

6
horizontal
Plate 6 
7
vertical
Plate 7
PLATE 7: Solid Rhombic Triacontahedron
The solid rhombic triacontahedron is obtained, like the preceding ones, by applying simple geometrical operations to two reciprocal Platonic polyhedra. The core of the two polyhedra is the icosidodecahedron, but here, too, the real starred icosidodecahedron has quite different features. (Compare with Plate 9, central figure). These polyhedra are the regular dodecahedron and the regular icosahedron (see lowest figure). The central figure is composed of the two lower polyhedra interlinked, which assume the appearance of a stellated icosidodecahedron. By properly joining its vertices with 60 segments of straight lines, we obtain the desired figure. It is called rhombic triacontahedron due to the 30 rhombs which form its surface (see top figure). Its dual is the icosidodecahedron but they are not conjugates to each other. Consequently the geometric solution of this plate is the only possible one, as seen in the prospectus on page 9.[See "Prospectus", above].
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
PLATE 8: Solid Hexakis Octahedron
The solution of the hexakis octahedron may be obtained by the application of three consecutive geometric operations constructed on the surface of the small rhombicuboctahedron, which is its archetype (see lowest figure). The starred small rhombicuboctahedron (central figure) has at its core the small rhombicuboctahedron, which is no longer visible. Its surface consists of 8 trigonal pyramids and 18 tetragonal ones constructed of the faces of the original polyhedron. The bases of the 18 tetragonal pyramids are squares, while the bases of the 8 trigonal are equilateral triangles, but the height of the former are qual to the length of the sides of the square or those of the equilateral triangles. This is an important factor to remember, for it contributes to produce the correct result. This principle is clearly visible at the center figures of the two preceding plates, where the length of the sides of trigonal pyramids correspond to the height of the tetragonal and pentagonal pyramids.
By properly joining the 26 vertices of the starred small rhombicuboctahedron with straight lines, of unequal length, we generate the final solution of the problem, represented in perspective by the top figure in this plate. It is known as hexakis octahedron, which literally means that each face of the octahedron is now composed of 6 supplementary faces. The hexakis octahedron is the dual of the great rhombicuboctahedron and they are each other conjugated with direct result for the second method mentioned on page 9 [i. e. in the "Prospectus"].
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
PLATE 9: Solid Pentakis Dodecahedron
The solution of this polyhedron is obtained by the application of three consecutive geometric operations constructed on the faces of the icosidodecahedron (lowest figure in the plate). This is the archetype of the pentakis dodecahedron. By constructing the stars on each triangular and pentagonal face, we obtain the starred icosidodecahedron (central figure). here again, as we did in the preceding two problems, it is necessary to determine the correct height of the 12 pentagonal pyramids in relation to the 20 triangular pyramids. While the vertices of the latter are determined by the length of their lateral edges which are equal to the length of their bases, the vertices of the former are determined by sing the length of their bases which are equal to their height. Thus, the 12 stars of the pentagonal pyramids protrude further than the 20 stars of the triangular pyramids. By properly connecting the vertices of this starred polyhedron with 60 straight lines, we form a new polyhedron composed of 60 triangular faces (see top figure).^{1} This is the pentakis dodecahedron, so called because each face of the dodecahedron is replaced by 5 complementary faces. The dual of this polyhedron is the truncated icosahedron and they are reciprocal conjugates of each other.
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. 
_________________
^{1}Notice that this polyhedron is very similar to the starred dodecahedron (Plate 4, central figure), but it is not the same. In fact the vertices of the pentakis dodecahedron produce a figure having only convex pentahedral anguloids; while the figure in Plate 4 is composed anguloids alternately convex and concave.

9
horizontal
Plate 9 
10
vertical
Plate 10
PLATE 10: Solid Trapezoidal Hexecontahedron
The solid trapezoidal hexecontahedron is obtained by applying several consecutive geometrical operations to two interlocked polyhedra. The two archetypal polyhedra are reciprocals—the icosidodecahedron and the rhombic triacontahedron respectively (lowest figure, left and right). The central figure is composed of the two lower polyhedra, interpenetrated, which assume the appearance of a starred polyhedron, whose core is an irregular solid. It is irregular because , while the edges of the icisidodecahedron are intersected at their midpoint, the edges of the rhombic triacontahedron are not intersected at midpoint by the former. The reason for the irregularity of the core polyhedron is that it includes rectangular faces. Needless to say, the Archimedean polyhedra do not include faces of rectangular shape. By properly joining its vertices, we obtain another figure which is known as the trapezoidal hexecontahedron (top figure). This is the dual of the small rhombicosidodecahedron and they are each other reciprocally conjugated as see in the prospectus on page 9 [View "Prospectus", above, in POLYHEDRA IMAGES section]. The Trapezoidal hexacontrahedron literally means that the polyhedron is composed of 60 faces of trapezoidal shape.
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
PLATE 11: Solid Hexakis Icosahedron
The solution of this problem is obtainable by the application of three consecutive geometric operations constructed on the faces of the small rhombicosidodecahedron (lowest figure in the plate). This is the archetype of the hexakis icosahedron. By costructing the stars on each of the triangular, quadrilateral and pentagonal faces of this figure, we obtain the starred small rhomboicosidecahedron pictured in perspective in the plate (see central fugure). This is a more complicated solution than the preceding ones, since it involves determining the correct height of the pentagonal pyramids in relation to the triangular and tetragonal ones. While the vertices of the latter are determined by the length of their lateral edges equal to the length of their bases, the vertices of the former are determined by using the length of their bases which are equal to their height. In this way the 12 pentagonal stars protrude further than the 50 stars of the trigonal and tetragonal pyramids. By properly connecting the vertices of this starred polyhedron, with straight lines of equal length, we form a new polyhedron composed of 120 triangular faces, which is known as the hexakis icosahedron (see top figure). Its proper name means that for each face of the icosahedron there are 6 supplementary faces. The dual of the hexakis icosahedron is the great rhombicosidodecahedron and they are reciprocally conjugated to each other, as outlined on page 9 [See "Prospectus", above, in POLYHEDRA IMAGES section].
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
PLATE 12: Solid Pentagonal Icositetrahedron^{1}
The solution of the polyhedron shown in this plate is obtainable by applying several consecutive geometrical operations constructed on the faces of the dextro snub hexahedron (lower figure in the plate) which is the archetype of the laevo pentagonal icosistetrahedron. The starred dextro snub hexahedron (in the center of the plate) consists of 36 trigonal and tetragonal pyramids constructed on the faces of the original polyhedron. This solution involves a more complicated problem than some of the preceding ones: that of building up the six tetragonal pyrmids in such a way that their height (from the vertex to the center of the base) is equal to the length of the base of the pyramid. In this way the vertices of the tetragonal pyramids protrude further than the 32 vertices of the trigonal pyramids, thus creating six new points which, when connected with the vertices of the trigonal pyramids, form 24 new planes. These planes are irregular pentagons which, altogether, form the surface of the polyhedron called laevo pentagonal icositetrahedron (see top figure). The dual of the laevo pentagonal icositetrahedron is the dextro snub hexahedron depicted in this very same plate in the lowest figure.
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. 3^{4}. 4. 
_____________________
^{1}It can be constructed either dextro (i.e. righthanded) or laevo (lefthanded). In this Plate, the unpredictable beautiful geometry shows that the lowest and central figures are righthanded and its dual ressult as being lefthanded. This principle is valid for the polyhedra represented in the following plate, too.

12
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Plate 12 
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Plate 13
PLATE 13: Solid Pentagonal Hexecontrahedron^{1}
The solid laevo pentagonal hexecontahedron is obtained by the application of several consecutive geometric operations constructed on the surface of the dextro snub dodecahedron (lower figure in the plate), which is the archetype of the pentagonal hexecontahedron. The starred snub dodecahedron (central figure), having as its core the dextro snub dodecahedron, consists of 80 trigonal and pentagonal pyramids, constructed on the faces of the original polyhedron. The complexity of this solution is identical with that of the preceding plate, however. The 12 pentagonal pyramids protrude further than the trigonal ones. By properly joining the vertices of the starred figure, we obtain a new polyhedron called the laevo pentagonal hexecontahedron(top figure). Its dual is the dextro snub dodecahedron depicted on the same plate.
This polyhedron is called pentagonal hexecontahedron due to the shape of its surface. Etymologically, it would be ('hexeconta' = sixty; 'hedron' = base) 60 face figure with polygons analogous to the pentagon.
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. 3^{4}. 5. 
______________________
^{1}We refer the reader to the footnote on Plate 12.

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Plate 13 