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 
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
horizontal
Plate 12

13
vertical
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.

13
horizontal
Plate 13
