[1] [1] [1] [1]
Visual Techniques For Computing Polyhedral
Volumes
2001/08/0221:39:06 UTC
Abstract
The volume of regular polyhedra have been a source of interest to
geometers since the time of Plato and Aristotle, and formulae for
computing the volumes of the dodecahedron and icosahedron can be
traced back to ancient Greece. In this paper, we revisit these
volumes from a slightly different perspective we illustrate
various constructions that permit the final formulae to be derived
by simple visual inspection.
In presenting these techniques, we gain a fresh perspective on
the relationship between the dodecahedron, icosahedron, cube, and
the golden ratio f. The visual nature of
these computational techniques in combination with zome models make
these proofs easily accessible.
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1 Introduction
The volume of regular polyhedra have been a source of interest
to geometers since the time of Plato and Aristotle, and formulae
for computing the volumes of the dodecahedron and icosahedron can
be traced back to ancient Greece [Coe89,Coe73,Wei98]. In this paper, we revisit these volumes
from a slightly different perspective. We illustrate various
constructions that permit the final formulae to be derived by
simple visual inspection.
In presenting these techniques, we gain a fresh perspective on
the relationship between the dodecahedron, icosahedron, cube, and
the golden ratio f. All of the
techniques described in this paper are made tangible using Zome
systems' polyhedra building kit. The visual nature of our
computational technique along with zome models make these proofs
accessible to students of all levels.
The techniques described in this paper demonstrate the value of
picking an appropriate frame of reference when solving problems in
mathematics. Using Zome Systems building kit, our basic units of
measure are 1, sin60, sin72, Ö2,
[1/( Ö2)] and these same lengths
scaled by powers of the Golden Ratio f.
These units make for easy computation of expressions that would
otherwise be computationally awkward. In this context, selecting
the right basic units of measure is equivalent to picking an
appropriate coordinate system or equivalently, the right set of
basis vectors.
1.1 Basic Lengths
Zome Systems http://www.zometool.com ([HP01]) leverages the following
mathematical facts:
 The symmetry of the dodecahedron and icosahedron.
 The golden ratio and its scaling property.
Nodes in the zome kit have all the directions needed to
build the various Platonic solids. Zome struts come in four
different colors (blue, red, yellow, green) with struts of a given
color corresponding to a given symmetry axis. Struts of each color
come in three different sizes, and successive struts of the same
color are in the golden ratio. In the rest of this paper,
we refer to zome struts by symbols made up of the first letter of
the color, suffixed by (1, 2, ...); see table 1 for a table of all the zome lengths. The
table also shows the mathematical significance of these lengths in
brief; the rest of the paper builds on these properties.
Color 
Significance 
1 
2 
3 
Blue 
Unity 
1 
f 
f^{2} 
Red 
Radius of I_{1} 
sin72 
fsin72 
f^{2}sin72 
Yellow 
Radius of C_{1} 
sin60 
fsin60 
f^{2} sin60 
Green 
Face diagonal of C_{1} 
Ö2 
fÖ2 
f^{2}Ö2 
green 
Radius of a C_{1} face 
[1/( Ö2)] 
f[1/( Ö2)] 
f^{2}[1/( Ö2)] 
Table 1: Basic zome lengths. Zome struts come in different
colors and sizes. The sizes are mathematically significant. Here,
I_{1} denotes the unit icosahedron and C_{1}
denotes the unit cube.
1.2 Identities Of The
Golden Ratio
This section derives some useful identities involving the golden
ratio f and the basic lengths in the
zome system.
Successive Powers
Successive powers of the golden ratio form a Fibonacci
sequence.
Golden ratio and the diagonal of the pentagon.
Observe one of the small isosceles triangles in figure 1 it has base B_{2} = f and sides B_{1} = 1. Dropping a
perpendicular from the apex of this triangle to its base bisects
its base B_{2} = f. From the
righttriangles that result, we get
ff
Figure 1: Diagonal of a unit pentagon has length f. Drawing 2 diagonals incident on a vertex of the
pentagon divides it into 3 triangles, and the pentagonal area can
be computed by summing these triangles to get sin72(1+[(f)/ 2]).
Identity of the golden rhombus.
Observing that R_{1} lines constitute the radii of a
golden rectangle gives R_{1} in terms of the golden ratio.
R_{1} can be computed by observing the righttriangle
marked ABC in figure 2.
(2,2)(2,2) 0,0) 0.5,0.809016994375)
(0.5,0.809016994375)A5,0.809016994375)(0.5,0.809016994375)
(0,0.809016994375)B_{1}.5,0.809016994375)(0.5,0.809016994375)
(0.5,0)B_{2} 0.5,0.809016994375) (0.5,0.809016994375)B
.5,0.809016994375)(0.5,0.809016994375)
(0,0.809016994375)10.5,0.809016994375) (0.5,0.809016994375)C
5,0.809016994375)(0.5,0.809016994375) (0.5,0.0)f .5,0.809016994375)(0.5,0.809016994375)
5,0.809016994375)(0.5,0.809016994375) ,0)(0.5,0.809016994375)
.5,0.809016994375)(0,1.61803398875)
1.61803398875)(0.5,0.809016994375) 5,0.809016994375)(1,0)
0)(0.5,0.809016994375) 5,0.809016994375)(0,1.61803398875)
1.61803398875)(0.5,0.809016994375) .5,0.809016994375)(1,0)
Figure 2: Relation between the golden rectangle and golden
rhombus. Observe righttriangle marked ABC to derive the identity
given by equation 1.6.
R_{1} in terms of trigonometric ratios
The length of R_{1} can be expressed in terms of
trigonometric ratios by putting together the identities derived so
far.



(6) 


(7) 


(8) 
\intertextCombining these
givessin^{2} 72 


(9) 


(10) 


(11) 


(12) 



Powers of the golden ratio and trigonometry.
Construct a righttriangle of sides B_{1} and
B_{3} its hypotenuse is the result of joining the
B_{1} and B_{3} lines using 2Y_{2} = fÖ3 struts
see figure 3. This
gives the identities
(2,2)(2,2) 0,0) 0.5,1.30901699438)
(0.5,1.30901699438)A5,1.30901699438)(0.5,1.30901699438)(0.0,1.30901699438)B_{
1}.5,1.30901699438)(0.5,1.30901699438) (0.5,0)B_{3}
0.5,1.30901699438) (0.5,1.30901699438)B
.5,1.30901699438)(0.5,1.30901699438)
(0,1.30901699438)10.5,1.30901699438) (0.5,1.30901699438)C
5,1.30901699438)(0.5,1.30901699438) (0.5,0)f^{2} .5,1.30901699438)(0.5,1.30901699438)
5,1.30901699438)(0.5,1.30901699438) (0,0)2Y_{2} = fÖ3
,0)(0.5,1.30901699438) .5,1.30901699438)(0,1.61803398875)
1.61803398875)(0.5,1.30901699438) 5,1.30901699438)(1,0)
0)(0.5,1.30901699438) 5,1.30901699438)(0,1.61803398875)
1.61803398875)(0.5,1.30901699438) .5,1.30901699438)(1,0)
Figure 3: This figure shows a Y_{2} yellow rhombus and
its relation to a rectangle of sides B_{1}
×B_{3} rectangle. It is used in deriving the identity
shown in equation 1.13 observe
that this figure is the same as figure 2 scaled by a factor of f in the Y direction.
Computing other trigonometric ratios
From the identities for sin72 and cos36, we can derive sin36 as
follows:
2 Locating Vertices Of
Various Polyhedra
This section locates the vertex coordinates of regular polyhedra
using zome models. Constructions used in this exercise prove useful
in computing distances needed for the volume computations.
2.1 Locating Vertices Of
The Cube
Construct a unit cube of side B_{1} = 1. Let (0,0,0) be
the center of the cube. The coordinates of the cube vertices
are
{(± 
1
2

, ± 
1
2

, ± 
1
2

)}. 

Notice that by the choice of the basic zome lengths, the center
of the cube can be connected to the 8 cube vertices by
Y_{1} struts. Thus, the radius of the unit cube is [(Ö3)/ 2] = sin60 = Y_{1}.
2.2 Locating Vertices Of
The Tetrahedron
Construct a unit B_{1} cube. Pick the cube vertex that
lies in the first octant having coordinates T =
(^{1}/_{2},^{1}/_{2},^{1}/_{
2}). Draw the 3 face diagonals of the cube incident on vertex
T. Each diagonal has length G_{1} = Ö2. Finally, draw the face diagonals of the
cube that connect the endpoints of the 3 diagonals just drawn.
This constructs a tetrahedron of side G_{1} = Ö2 inside the B_{1} cube.
Let (0,0,0) be the center of the cube notice that it is also
the center of the G_{1} tetrahedron. The vertex coordinates
of this G_{1} tetrahedron can be read from this model
as:

\hline( 
1
2

, 
1
2

, 
1
2

) 








As shown in section 2.1, the
center of the above model can be joined to the cube vertices using
Y_{1} = sin60 yellow lines. From this it follows that the
center of the G_{1} tetrahedron can be joined to the
tetrahedral vertices using Y_{1} = sin60 lines. Observe
that drawing these radii divides the interior of the tetrahedron
into 4 congruent pyramids with a G_{1} equilateral triangle
as base and vertical sides Y_{1}.
2.3 Locating Vertices Of
The Octahedron
Construct an octahedron of side G_{1} = Ö2. View this model placed on one of its
vertices so that the opposite vertex is directly above the
chosen vertex. Let (0,0,0) be the center of the octahedron. Let the
symmetry axis formed by joining the top and
bottom vertices denote the Z axis. Connect these vertices
to the center using B_{1} = 1 struts. Similarly, locate
opposite pairs of vertices and draw the X and Y axes. From this
model, the vertex coordinates of the octahedron are given by:
{(±1,
0, 0), (0,±1, 0), (0, 0, ±1)}. 

2.4 Locating Vertices Of
The Rhombic Dodecahedron
Construct a B_{1} unit cube. Construct pyramids of side
Y_{1} = sin60 on each face of this cube. Removing the cube
edges leaves a rhombic dodecahedron of side Y_{1}. The
rhombic dodecahedron has 12 faces and 14 vertices.
Vertices of the rhombic dodecahedron can be categorized as:

 Vertices of the unit cube.
 The apex of each of the 6 pyramids described above.
Let (0,0,0) be the center of this rhombic dodecahedron hence
the center of the cube. This gives the coordinates of the 8 of the
14 vertices to be
Next, observe that by construction, the Y_{1} pyramids
built on each face are congruent to the pyramids constructed by
connecting the center of the unit cube to its vertices see
section 2.1. Thus, the height
of these pyramids is half the side of the cube and therefore
^{1}/_{2}.
The coordinates of these 6 vertices of the rhombic dodecahedron
are then given by
{(±1,
0, 0), (0, ±1, 0), (0, 0, ±1)}. 

Notice that from the above, the rhombic dodecahedron has all the
vertices of a B_{1} cube and the dual G_{1}
octahedron a fact that will be used later in the techniques for
computing the volume of the rhombic dodecahedron.
2.5 Locating Vertices Of
The Cubeoctahedron
The cubeoctahedron is the dual to the rhombic dodecahedron
described in section 2.4. The
rhombic dodecahedron was shown to have the vertices of the cube and
the octahedron; by duality, the cubeoctahedron has the faces of
the cube and the octahedron.
Construct a 2B_{1} cube. Join the midpoints of adjacent
edges using G_{1} = Ö2
green lines. Removing the cube edges leaves a G_{1}
cubeoctahedron.
Let (0,0,0) be the center of the 2B_{1} cube, hence the
center of the cubeoctahedron. From this model, the vertex
coordinates can be read as:
{(0, ±1, ±1),(±1,
0, ±1),(±1,±1,0)}. 

2.6 Locating Vertices Of
The Dodecahedron
Consider the unit dodecahedron with sides B_{1}. Each
face is a unit pentagon. A diagonal of a pentagon is in the golden
ratio to the length of its side; consequently, any face diagonal
can be drawn by using a B_{2} strut  see figure 1. Place this dodecahedron on one of its
edges; this base edge can be connected to its opposite
(top) edge using a pair of B_{3} struts to form a
B_{3} ×B_{1} rectangle. View this model with
the B_{3} ×B_{1} rectangle just constructed
lying in the ZX plane. Locate opposing pairs of edges of the
dodecahedron to similarly construct B_{3}
×B_{1} rectangles in the XY and YZ planes. This
consumes 12 of the 20 vertices.
View this model with the XY, YZ and ZX planes in place; the
remaining 8 vertices form a cube whose sides are face diagonals of
the dodecahedron and therefore of length B_{2} = f.
Let (0,0,0) be the center of the dodecahedron. Then the 8
vertices making up the cube have coordinates (±[(f)/ 2],±[(f)/
2],±[(f)/ 2]).
The 4 vertices of the B_{3} ×B_{1}
rectangle in the XY plane have coordinates (±[(f^{2})/ 2], ±^{1}/_{2}, 0). The
remaining coordinates can be located in an analogous manner by
examining the rectangles in the ZX and YZ planes notice that these
are just rotated copies of the B_{3}×B_{1}
rectangle in the XY plane see figure 4 for a view of one face of the
dodecahedron from this model.
Figure 4: This view shows the threedimensional perspective
of a face of the dodecahedron with one of its face diagonals. The
face is tilted by 58.282525589deg = arctanf.
2.7 Locating Vertices Of
The Icosahedron
Construct a unit icosahedron of side B_{1}. Place it on
one of its edges, and notice that the bottom edge can be
connected to its opposite (top) edge using a pair of
B_{2} = f struts. View this
model so the B_{2}×B_{1} golden rectangle
just constructed lies in the ZX plane. Locate opposing pairs of
edges to construct B_{2} ×B_{1} rectangles in
the XY and YZ planes. This accounts for the 12 vertices of the
icosahedron.
Let (0, 0, 0) be the center of the icosahedron. The 4 vertices
of the B_{2} ×B_{1} rectangle in the XY plane
have coordinates (±[(f)/ 2], ±^{1}/_{2}, 0). The
coordinates of the remaining 8 vertices of the icosahedron can be
similarly read off the model  notice that they are the appropriate
rotation of the 4 vertices shown above. Thus, we get the
coordinates for the 12 vertices of the icosahedron to be:

\hline(± 
f
2

, ± 
1
2

, 0) 







(18) 
Notice that this construction also reveals that a unit
icosahedron can be packed inside a cube of side f a fact that will be used later in one of
the techniques for computing the volume of the icosahedron.
As shown in the table of basic lengths (see table 1), the R_{1} zome strut is the
radius of the B_{1} icosahedron. Observe that these 12
radii draw the diagonals of the B_{2}×B_{1}
golden rectangles in the XY, YZ and ZX planes in the above
model.
2.8 Locating Vertices Of
The Rhombic Triacontahedron
The rhombic triacontahedron is an Archimedian polyhedron with 30
faces and 32 vertices. In locating its vertices, we will see that
it has the vertices of the dodecahedron and its dual
icosahedron.
Construct a B_{1} dodecahedron. Construct red pyramids
of vertical side R_{1} = sin72 on each of the faces of the
dodecahedron. The edges of the dodecahedron form the short diagonal
of each rhombic face removing these edges of the dodecahedron
leaves a red rhombic triacontahedron of side R_{1} =
sin72.
Next, observe that the long diagonal of each rhombic face can be
drawn using B_{2} = f. Thus, the
diagonals of the rhombic faces are in the golden ratio and each
face of the rhombic triacontahedron is a golden rhombus. Finally,
observe that drawing the B_{2} diagonals and removing the
red edges would leave a B_{2} icosahedron. See
figure 2 for an illustration
showing the relation between the golden rhombus and the fact that
the radii of a B_{1} icosahedron draw the diagonals of a
golden rectangle as described in 2.7.
2.9 Locating Vertices Of
The Icosidodecahedron
The icosidodecahedron has 32 faces and 30 vertices, and is dual
to the rhombic triacontahedron described in 2.8. Since the rhombic triacontahedron has
all the vertices of the dodecahedron and its dual icosahedron, by
duality it follows that the icosidodecahedron has all the faces of
the dodecahedron and icosahedron.
Observe that a zome node has 30 holes that can take blue struts.
Inserting B_{2} struts into each of these 30 holes and
connecting their endpoints with B_{1} struts constructs an
B_{1} icosidodecahedron. By construction, the radius of the
icosidodecahedron of side B_{1} = 1 is B_{2} =
f.
Let (0,0,0) be the center of the icosidodecahedron. Place this
model on one of its vertices, and observe that the B_{2}
struts connecting the center to the bottom and
top vertices can be viewed as the Z axis. Locate the X and
Y axes in a similar manner to see that 6 of the 30 vertices of the
icosidodecahedron are also vertices of an octahedron of side
G_{2} = fÖ2. This gives 6 of the 30 vertices to be
{(±f,0,0), (0,±f, 0),
(0,0,±f)}. 

We obtained the above fact by placing the icosidodecahedron on
any one of its vertices it therefore follows that the
remaining 24 vertices can in turn be divided into disjoint sets of
6 vertices each, with each set corresponding to the vertices of a
rotated copy of the G_{2} = fÖ2 octahedron.
This leads to an important result the unit icosidodecahedron
can be wrapped around the compound of 5 concentric
octahedra of side fÖ2.
To compute the coordinates of the remaining vertices of the
icosidodecahedron, consider the model built in section 2.6 where we constructed a B_{1}
icosahedron. Scale this model by 2 to obtain a 2B_{1}
icosahedron.
Let (0, 0, 0) be the center of this model. By scaling all values
computed in equation 2.1, we first
locate the 12 vertices of the 2B_{1} icosahedron to be:
Next, observe the 2B_{1} ×2B_{2} golden
rectangles in the XY, YZ and ZX planes, and consider the midpoints
of the 2B_{1} sides. These have coordinates {(±f,0,0), (0,±f, 0),
(0,0,±f)}. Thus, the midpoints of 6 of the 30 edges of
the 2B_{1} icosahedron give the vertices of the
G_{2} octahedron. By symmetry, it follows that the 30 edges
of the 2B_{1} icosahedron can be partitioned into 5
disjoint sets of 6 edges each, where the midpoints of edges in any
given partition form a rotated copy of a G_{2}
octahedron.
By combining the above with the earlier result that the vertices
of the icosidodecahedron are the same as the vertices of the
compound of 5 concentric G_{2} octahedra, we can compute
the coordinates of all 30 vertices by reading off the midpoints of
the 30 edges of the 2B_{1} icosahedron.
Observe that by construction the 2B_{1} icosahedron as
oriented is symmetric about the coordinate axis. Therefore, we need
only compute the coordinates of the remaining 24 vertices in one of
the octants.
Consider the 3 vertices of the icosidodecahedron in the first
octant. By construction, these are the midpoints of the sides of
the 2B_{1} triangle shown in figure 5.
The coordinates of of the 30 vertices of the icosidodecahedron
are therefore:


\hline(± 
f
2

, ± 
1+f
2

, ± 
1
2

) 

(± 
1+f
2

, ± 
1
2

, ± 
f
2

) 

(± 
1
2

, ± 
f
2

, ± 
1+f
2

) 





(20) 
(2,2)(2,2)
(1,0)(0,0)(0,0)(1,0)(1,0)(0.5,0.866025403785)(0.5,0.866025403785)(0,1.73205080757)(0,1.73205080757)(0.5,0.866025403785)(0.5,0.866025403785)(1,0)1,0)
(1,0)(f, 1, 0) 1,0) (1,0)(0,f,1) 0,1.73205080757) (0,1.73205080757)(1,0,f) 0,0) (0,0)( [(f)/ 2], [(1+f)/ 2],
^{1}/_{2})
0.5,0.866025403785)(0.5,0.866025403785)( [(1+f)/ 2],^{1}/_{2}, [(f)/ 2]) 0.5,0.866025403785) (0.5,0.866025403785)
(^{1}/_{2}, [(f)/ 2],
[(1+f)/ 2])
Figure 5: Face of the 2B_{1} icosahedron in the first
octant. Its midpoints give 3 vertices of the icosidodecahedron of
side B_{1}, and by symmetry,, these help locating the
vertices of the icosidodecahedron that do not lie on the coordinate
axes.
Finally, observe that this construction has shown how the
icosidodecahedron can be wrapped around the compound of 5
concentric octahedra. Applying duality to this result, and using
the fact that:

 Vertices map to faces in the dual.
 The inside and outside reverse roles in the
dual.
the rhombic triacontahedron which is dual to the icosahedron can
be seen to have each of its 30 rhombic faces on each of the 30 cube
faces of the compound of 5 concentric cubes.
3 Using The Cube To Compute
Volumes
3.1 Volume Of The
Tetrahedron
Consider the G_{1} tetrahedron constructed in
section 2.2. From this
model, the interior of the B_{1} cube can be decomposed
into a G_{1} tetrahedron and 4 pyramids having a green base
and blue vertical sides. Place the model on one of the cube faces,
and observe one of these pyramids. It has a right triangle of sides
B_{1}, B_{1}, G_{1} as base, and a side of
the B_{1} cube as its height. This gives the volume of this
pyramid to be:
Subtracting 4 copies of this pyramid from the cube gives the volume
of the G_{1} tetrahedron V_{T} to be:
3.2 Volume Of The
Octahedron
Consider the G_{1} octahedron constructed in
section 2.3. Place it on
one of its triangular faces. Construct a G_{1} tetrahedron
and observe that the tetrahedron has the same face as the
octahedron. Take 4 copies of this G_{1} tetrahedron, and
place them on 4 faces of the octahedron to form a 2G_{1}
tetrahedron. This shows that the 2G_{1} tetrahedron can be
decomposed into an octahedron and 4 tetrahedra.
We computed the volume of the G_{1} tetrahedron to be
^{1}/_{3} in section 3.1. By applying the scaling rule, the volume
of the 2G_{1} tetrahedron is ^{8}/_{3}.
Subtracting 4 copies of the G_{1} tetrahedron from the
2G_{1} tetrahedron gives the volume of the G_{1}
octahedron V_{O}:
3.3 Volume Of The Rhombic
Dodecahedron
Consider the rhombic dodecahedron constructed in section 2.4. Its volume can be decomposed into
the unit cube and 6 pyramids having a B_{1} square base and
Y_{1} vertical sides. Section 2.4 also showed the height of this pyramid to
be ^{1}/_{2}. This gives the volume of this pyramid
V_{P} to be:
The volume of the rhombic dodecahedron V_{\textRD} is
therefore:



(34) 


(35) 

= \textTwice the volume of the unit
cube. 

(36) 



This can also be seen by realizing that the yellow pyramid
constructed on each face of the B_{1} cube is congruent to
the pyramid constructed by joining the center of the cube to the
vertices of a given face. Thus, the 6 pyramids constructed
outside the cube can be packed into the interior of the
cube, giving the volume of the Y_{1} rhombic dodecahedron
to be twice the volume of the unit cube.
Finally, consider once again the model of the rhombic
dodecahedron of side sin60 and draw the longer diagonal of each
face. By the choice of zome lengths, this is a G_{1} =
Ö2 green line. Drawing the long
diagonal of all 12 faces gives a G_{1} octahedron. This
decomposes the rhombic dodecahedron into an octahedron and 8
pyramids having a G_{1} equilateral triangle as base and
Y_{1} vertical sides. We showed in section 2.2 that this pyramid is 1/4 the
volume of the G_{1} tetrahedron. From this we can compute
the volume of the rhombic dodecahedron V_{\textRD} to
be
See figure 6 for a visual
representation of this relationship.
(1,1)(1,1) .5,0)(0,0.866025403785)
(0.25,0.433)Y_{1}0.866025403785)(0.5,0)
(0.25,0.433)sin605,0)(0,0.866025403785)
(0.25,0.433)sin600.866025403785)(0.5,0)
(0.25,0.433)Y_{1} .5,0)(0.5,0) (0.0,0)B_{1} = 1
0.866025403785)(0,0.866025403785) (0,0)G_{1} = Ö2linecolor=black](0,0)
Figure 6: The yellow rhombus of side Y_{1} = sin60 has a
short B_{1} = 1 diagonal and a long G_{1} = Ö2 diagonal. This relation leads to two
equivalent ways of constructing a rhombic dodecahedron of side
Y_{1}.
3.4 Volume Of The
Cubeoctahedron
Consider the cubeoctahedron constructed in section 2.5. This shows that the
2B_{1} cube can be decomposed into a cubeoctahedron and 8
pyramids. Notice that these pyramids are the same that occurred in
section 3.1 while computing the
volume of the tetrahedron we computed this to be
^{1}/_{6}. Thus, the volume of the cubeoctahedron
V_{\textCO} is:
4 Volume Of A
Dodecahedron
Consider the model built in section 2.6 in locating the vertices of the
dodecahedron. A dodecahedron can be viewed as the result of adding
6 roof structures to a cube this construction was known
to Euclid. This decomposition of the dodecahedron can be used in
computing its volume.
From the model built in 2.6, the
cube has sides B_{2} = f and
therefore has volume f^{3}. It
only remains to compute the volume of the roof
structures.
Each roof has a square base of side B_{2}. The
vertical faces are a pair of triangles and trapezium. Consider one
of these trapezoidal faces; the parallel sides have length
B_{2} = f and B_{1} =
1.
A trapezium can be viewed as the sum of a triangle and a
parallelogram. Applying this decomposition to the roof
structure, it can be decomposed into a pyramid and a triangular
crosssection.
Pyramid volume.
This pyramid has a rectangular base with sides B_{2} and
B_{2}  B_{1}. Applying identities of the golden
ratio, B_{2}B_{1} = f1
= [1/( f)], giving the area of the base
A_{\textbase} to be
From the model constructed in 2.6,
the height of this pyramid is half of B_{3}B_{2}.
Since successive zome lengths are in the golden ratio,
The volume of this pyramid using 4.1 and 4.2 is therefore:
Next, consider the triangular crosssection. Its face is a
triangle of side B_{2} whose height is the same as the
height of the pyramid computed above in equation 4.2. The area of the triangular face is
therefore [(f)/ 4]; the length of the
crosssection is B_{1} = 1, giving its volume to be
The dodecahedron as constructed is equal to the cube plus 6
roof structures one on each face of the cube. Thus, the
volume of the dodecahedron is
V_{D} = f^{3} + 6( 
f
4

+ 
1
6

). 

(55) 
Radius Of The Dodecahedron
Consider once again the cube of side B_{2} identified in
the model built in 2.6. As shown in 1, the radius of this cube is
Y_{2} = fsin60. We identified
the vertices of this cube of side B_{2} = f by first placing the dodecahedron on any
one of its edges. Therefore there is nothing special about
these 8 of the 20 vertices of the dodecahedron, and it follows by
symmetry that all vertices of the dodecahedron are a distance
Y_{2} = fsin60 from its center.
Notice that drawing these 20 radii decomposes the interior of the
dodecahedron into 12 congruent pyramids that have a unit pentagon
as the base and Y_{2} as the vertical sides. This
construction in turn leads to an alternative technique for
computing the volume of the dodecahedron.
5 Volume Of The
Icosahedron
The icosahedron is the dual to the dodecahedron. This
duality when applied to the technique described in the previous
section leads to a solution for computing the volume of the
icosahedron.
5.1 Volume Of The
Icosahedron Part I
We computed the volume of the dodecahedron by building a cube
inside the dodecahedron. The dual to this solution is to
build an octahedron (dual to the cube) around the
icosahedron. Observe that when we take the dual the inside
comes out.
View the model built in 2.7 placed
on one of the icosahedral edges and oriented so the B_{2}
×B_{1} rectangles lie in the XY, YZ and ZX planes.
Construct a righttriangle with the top B_{1} edge
of the icosahedron as its hypotenuse in the ZX plane  in the zome
model, this triangle has g_{1} = [1/( Ö2)] green legs. Thus, this
righttriangle has B_{1} = 1 as the hypotenuse and sides
g_{1} = [1/( Ö2)]. Repeat
this construction on the bottom B_{1} edge.
Finally, construct two more righttriangles each with one of the
B_{2} = f sides as the
hypotenuse. These righttriangles have sides g_{2} =
[(f)/( Ö2)]. The above constructs a square of side
[(1+f)/( Ö2)] around the golden rectangle
B_{2}×B_{1} in the ZX plane see
figure 5.1.
(2,2)(2,2)
.30901699438,0)(0.5,0.809016994375) (0.9,0.4050)g_{2}
.5,0.809016994375)(0,1.30901699438)
(0.25,1.15)g_{1}1.30901699438)(0.5,0.809016994375)
(0.25,1.15)[1/( Ö2)]5,0.809016994375)(1.30901699438,0)
(0.9,0.4045)[(f)/( Ö2)]30901699438,0)(0.5,0.809016994375)
(0.9,.4045)g_{2} 5,0.809016994375)(0,1.30901699438)
(0.25,1.15)g_{1}1.30901699438)(0.5,0.809016994375)
(0.25,1.15)g_{1}.5,0.809016994375)(1.30901699438,0)
(0.9,0.4045)g_{2} .5,0.809016994375)(0.5,0.809016994375)
(0.0,0.8090)B_{1}.5,0.809016994375)(0.5,0.809016994375)
(0,0.5)1.5,0.809016994375)(0.5,0.809016994375)
(0.5,0.0)B_{2} 5,0.809016994375)(0.5,0.809016994375)
(0.5,0)f
nestyle=dotted,linecolor=black](1.30901699438,0)(1.30901699438,0)nestyle=dotted,linecolor=black](0,1.30901699438)(0,1.30901699438)0,0)
Figure 7: This figure shows a green square constructed around a
blue golden rectangle.
Repeat this construction for the
B_{2}×B_{1} rectangles in the XY and YZ
planes. The result is to construct an octahedron of side
The volume of an octahedron of side Ö2 is ^{4}/_{3} as shown in 3.2. The volume of the octahedron of side
[(f^{2})/( Ö2)] constructed above has its side scaled by
[(f^{2})/ 2] and its volume by
the scaling rule is:
Next, we compute the volume of the pyramids we added to
the icosahedron in constructing the octahedron. View the model of
the octahedron around the icosahedron with the
B_{2}×B_{1} rectangles lying in the XY, YZ
and ZX planes. Observe one of the g_{1}, g_{1},
B_{1} righttriangle in the XY plane, and consider the
obtuse pyramid that has this triangle as its base. The
apex of this pyramid is a vertex of the top edge of the
icosahedron with Z coordinate [(f)/ 2]
which is also the height of this pyramid. This gives the
volume of the obtuse pyramid to be:
There are two copies of this pyramid at each of the 6 vertices of
the octahedron. Thus, the volume of the icosahedron is:



(68) 

= 
5f^{2}
6

\text(using 1.1) 

(69) 



(2,0.5)(2,2)
.92561479341,0)(0.218508012224,0)
(0.355,0)g_{2}218508012224,0)(0.92561479341,0)
(0.565,0)g_{1}92561479341,0)(0.353553390593,0.990839414728)
(0.635,0.495)g_{2}353553390593,0.990839414728)(0,1.60321185042)
(0.1765,1.325)g_{1}1.60321185042)(0.572061402818,0.612372435695)
(0.2875,1.11)g_{2}.572061402818,0.612372435695)(0.92561479341,0)
(0.745,0.306)g_{1}
218508012224,0)(0.353553390593,0.990839414728)
(0.33,0.495)B_{1}353553390593,0.990839414728)(0.572061402818,0.612372435695)
(0.11,0.8)B_{1}.572061402818,0.612372435695)(0.218508012224,0)
(0.18,0.305)B_{1}
Figure 8: This figure shows a face of the compound of the
octahedron and icosahedron constructed in section 5.1. The green triangle is the octahedral
face, and the embedded blue triangle is a face of the icosahedron.
The blue triangle divides the sides of the green triangle in the
golden ratio.
5.2 Volume Of The
Icosahedron Part II
The volume of the unit icosahedron can also be computed by
packing it in a B_{2}cube, and subtracting the
volume of the space between the cube and the icosahedron from the
volume of the cube.
Place the B_{1} icosahedron on one of its edges as
before. We will construct a B_{2} cube around this
icosahedron so that each face of the cube contains a corresponding
edge of the icosahedron so for instance, the top edge of
the icosahedron lies within the top face of this cube.
Figure 9 shows one such face of
the cube along with the contained icosahedral edge.
(2,2)(2,2)
.809016994375,0.809016994375)(0.809016994375,0.809016994375)
(0.0,0.8090)B_{2}
.809016994375,0.809016994375)(0.809016994375,0.809016994375)
(0.0,0.8090)f
.809016994375,0.809016994375)(0.809016994375,0.809016994375)
(.8090,0)B_{2}
809016994375,0.809016994375)(0.809016994375,0.809016994375)
(0.8090,0)f .5,0)(0.5,0)
(0,0)B_{1} = 1 .809016994375,0.809016994375)(0.5,0)
(0.65,0.4045)Y_{1} 809016994375,0.809016994375)(0.5,0)
(0.65,0.4045)sin60 .809016994375,0.809016994375)(0.5,0)
(0.65,0.4045)Y_{1} 809016994375,0.809016994375)(0.5,0)
(0.65,0.4045)sin60 5,0)(0.809016994375,0) (0.65,0)[1/( 2f)] 0)(0,0.809016994375) (0,0.4045)[(f)/ 2]
Figure 9: One face of the B_{2} cube with the contained
B_{1} icosahedral edge. The icosahedral edge is connected
to the vertices of the surrounding B_{2} square using
Y_{1} = sin60 lines. The vertical dotted lines show the
height of the trapezoidal base; the horizontal dotted lines show
the height of the trapezoidal pyramid.
For each face of the B_{2} cube, join the vertices of
the contained icosahedral edge to the vertices of that face using
Y_{1} = sin60 lines as shown in figure 9. This constructs yellow triangular
pyramids with vertical side Y_{1} on 8 faces of the
icosahedron, one per vertex of the cube. Consider one such pyramid;
its base is a face of the icosahedron and therefore a B_{1}
equilateral triangle, and its apex the corresponding vertex of the
cube. The vertical sides of this pyramid are the yellow lines shown
in figure 9 and have length
Y_{1} = sin60.
Observe further that for each face of the cube, there are two
trapezoidal pyramids inside the cube. These have the B_{1},
Y_{1}, B_{2}, Y_{1} trapezium appearing as
part of figure 9 as base.
Consider one of the trapezoidal pyramid with its base contained in
the front face of the B_{2} cube for now,
consider the trapezoidal pyramid whose base appears in the bottom
half of figure 9. The apex of
this pyramid then lies on the bottom face of the cube. The height
of this pyramid is the height of the Y_{1}, Y_{1},
B_{1} triangle with base B_{1} shown in
figure 9 by horizontal dotted
lines. The height of this trapezoidal pyramid is therefore
Now, observe that the volume of the B_{2} cube can be
decomposed into the following disjoint pieces:
 The volume of the unit icosahedron.
 the 8 triangular pyramids one for each of the 8 vertices of
the cube.
 12 trapezoidal pyramids 2 for each of the 6 faces of the
cube.
Volume of the triangular pyramid.
To calculate the volume of a triangular pyramid, we first
compute its height. We do this by applying the Pythogorian theorem
to a righttriangle constructed by dropping a perpendicular from
the apex of this pyramid to its base it passes through the
centroid of the B_{1} equilateral triangle. For a
unit equilateral triangle, the centroid is at a distance [1/( Ö3)] from its vertices. The
righttriangle therefore has base [1/( Ö3)] and and hypotenuse Y_{1} =
[(Ö3)/ 2] . The height of the
pyramid is therefore
The area of the triangular base is
The volume of the triangular pyramid using 5.4 is:
Volume of the trapezoidal pyramid.
We first compute the area of the B_{1}, Y_{1},
B_{2}, Y_{1} trapezium that is the base of the
trapezoidal pyramid  see figure 9. The height of this trapezium shown by
the vertical dotted line in figure 9is [(f)/ 2].
The area of the base trapezium is therefore
As shown in figure 9, the
height of the trapezoidal pyramid is [1/( 2f)] = [(f1)/ 2].
Using 5.5 the volume of the
trapezoidal pyramid is therefore
Putting the pieces together, we sum the volumes of the 8
triangular pyramids and 12 trapezoidal pyramids to get
The volume of the icosahedron is given by subtracting the residue
shown in equation 5.6
from f^{3}, the volume of the
B_{2} cube. Writing f^{3} = f^{2}+f, we get


= f^{2}+f 
2 f1
3

 
f^{2}
2



(89) 


(90) 



6 Volume Of The Rhombic
Triacontahedron
We showed in section 2.8 that
the vertices of the rhombic triacontahedron are the vertices of
dodecahedron and its dual icosahedron. From the model constructed
in section 2.8, the Rhombic
triacontahedron consists of a B_{2} icosahedron and 20
R_{1} pyramids constructed on each of the triangular
icosahedral faces. These pyramids have a B_{2} equilateral
triangle as their base the vertical sides of the pyramids are the
R_{1} = sin72 edges of the rhombic triacontahedron.
The volume of the rhombic triacontahedron can therefore be
written as the sum of the volume of the B_{2} icosahedron
and 20 pyramids.
Volume of a Triangular Pyramid
By following the method used in section 5.2, and applying identities of the golden
ratio, we compute the height of this pyramid to be



(92) 


(93) 


(94) 

= 
Ö3
6f

\text(using 1.13) 

(95) 

= 
f
24

\text(using the height from the above equation). 

(96) 



Summing The Parts
The volume of the B_{1} = 1 icosahedron is [(5f^{2})/ 6] see section 5.1. The volume of the B_{2} =
f icosahedron is obtained by the scaling
rule to be:
The volume of the 20 triangular pyramids is
Using the identity f^{4}+1 =
3f^{2}, we get the volume to be:



(98) 


(99) 

= 
5f^{3}
2

\text(Using 1.13). 

(100) 



7 Volume Of The
Icosidodecahedron
We showed in section 2.9 that
the vertices of an icosidodecahedron were the edge midpoints of a
2B_{1} icosahedron. From this, it follows that an
icosidodecahedron can be constructed by truncating a 2B_{1}
icosahedron to its edge midpoints. Truncating the 2B_{1}
icosahedron results in removing 12 pyramids, each having a
B_{1} pentagonal base and B_{1} vertical sides.
The volume of the B_{1} icosahedron is [(5f^{2})/ 6]; see section 5.1 by applying the scaling rule, the
volume of the 2B_{1} icosahedron is 8[(5f^{2})/ 6].
Subtracting 12 copies of the pentagonal pyramid from the
2B_{1} icosahedron gives the volume of the B_{1}
icosidodecahedron.
7.1 Area of the Unit
Pentagon
Consider a unit pentagon of side B_{1} = 1. Pick any
vertex, and construct the B_{2} = f diagonals incident on that vertex see
figure 1. This divides the area of
the pentagon into 3 triangles, two of which are congruent see
figure 1.
Consider one of these 2 congruent triangles, and observe the
B_{1} = 1 adjacent sides with an included angle of
108^{°}. The area of this
triangle is
This gives the area of the 2 congruent triangles to be sin72.
Next, consider the triangle with sides {B_{2},
B_{2}, B_{1}}, and observe the adjacent sides of
length B_{1}, B_{2} with included angle
72^{°}. Its area is given
by
Summing the parts, the area of the unit pentagon is



(102) 

= 
(1+ f)^{3/2}
4

\text(Using 1.6). 

(103) 



7.2 Volume of the
Pentagonal Pyramid
Consider the center of the pentagon, and observe the triangle
formed by connecting it to 2 adjacent vertices of the pentagon. Let
r be the radius of the unit pentagon. The central angle is [360/ 5]
= 72^{°}, and the base
angles of this isosceles triangle is sin54. By the sine rule, we
have
Dropping a perpendicular from the apex of the pyramid to its base,
and applying the Pythogorian theorem as in section 5.2, the height of this pyramid is



(109) 

= 
æ
Ö



4 sin ^{2} 36 1
4 sin ^{2} 36



, 

(110) 

= 
2 sin36

\text(Using 1.15). 

(111) 



Rewriting sin36 and cos36 in terms of f, using equations ( 1.4 and 1.15), and
using the identity Ö{(3f^{2})} = [1/( f)]), we get the height to be
The volume of the pentagonal pyramid is
Summing the parts.
The The volume of icosidodecahedron is given by


= 
5
6

f^{2}
8  (1+f^{2}) 

(120) 


(121) 



8 Conclusion
To conclude, here is a table listing the various formulae
derived in this paper.
Polyhedron 
Side 
Vol 
Alt 
V 
Dodecahedron 
1 
f^{3} + 6([(f)/ 4] + ^{1}/_{6}) 

2 + [(7f)/ 2] 
Icosahedron 
1 
f^{3}12[(f)/ 24] 8 [(Ö5)/
24] 
[(f^{6})/ 6]  12 [(f)/ 24] 
5[(f^{2})/ 6] 
Tetrahedron 
Ö2 
[(1^{3})/ 3] 
1/3 unit cube volume 
Octahedron 
Ö2 
1^{3} + [(1^{3})/ 3] 
4 times the tetrahedron volume 
Rhombic Dodecahedron 
sin60 
1^{3} + 6 [(1^{3})/ 6] 
Twice unit cube 
Cubeoctahedron 
Ö2 
[20/ 3] 
Chamfered cube 
RT^{1}

sin72 
5[(f^{2})/ 6](f^{3} +Ö{2
 f}) 
[(5f^{3})/ 2] 
Table 2: Volumes of regular polyhedra. The volume for each
polyhedron is expressed in forms that make the decomposition
obvious.
Finally, here are the 5 platonic solids drawn using package
Metapost.
Figure
Figure 10: The Tetrahedron has 4 vertices and 4 faces, and
is dual to itself.
Figure
Figure 11: The cube has 8 vertices and 6 faces.
Figure
Figure 12: The octahedron has 6 vertices, 8 faces, and is
dual to the cube.
Figure
Figure 13: The dodecahedron has 12 faces and 20 vertices.
Figure
Figure 14: The icosahedron has 12 vertices, 20 faces, and
is dual to the dodecahedron.
9 Acknowledgements
We would like to thank Zome Systems for designing a truly
wonderful educational kit. The potential for teaching polyhedral
Geometry using the zome kit was first brought to our attention by
the work of author George Hart through his WWW site, (see http://www.georgehart.com) and his
books on the topic see [HP01].
This paper was authored in L^{A}T_{E}X on the
Emacspeak audio desktop^{2} with Emacs package AucTeX providing
authoring support. At a time when most of the attention around Open
Source Software is focused on the operating system, we would like
to draw readers' attention to the wonderful array of highquality
open source authoring and document preparation tools created over
the last 25 years by the (La)TeX community. All figures in
this paper were drawn using declarative authoring packages
pstricks and metapost that enabled the first
author to reliably draw these diagrams without having to look at
the final output. The highlevel markup also makes this content
longlived and reusable an immediate advantage is that the content
can be easily made available using a variety of access modes
ranging from highquality print and online hypertext to
highquality audio renderings^{3} see [Ram98]. The document preparation tools
used to prepare this paper are well described in [GMS94,GRM97,GR99] and the first author would like to thank
author Sabastian Rahtz and their publisher Addison Wesley for
providing access to the (La)TeX sources to these books. We
would like to thank author John Hobby for his work on Metapost and
acknowledge author Denis Roegel for his metapost macros for drawing
threedimensional polyhedra. Finally, we would like to thank Donald
E Knuth for the T_{E}X typesetting system.
References
 [Coe73]
 H. S. M. Coexter. Regular Polytopes  Third
Edition. Dover, 1973.
 [Coe89]
 H. S. M. Coexter. Introduction to Geometry.
John Wiley and Sons, 1989.
 [GMS94]
 Michel Goossens, Frank Mittelbach, and Alexander Samarin.
The L^{A}T_{E}X Companion. AddisonWesley,
Reading, MA, 1994.
 [GR99]
 Michel Goossens and Sebastian Rahtz. The
L^{A}T_{E}X Web Companion Integrating
T_{E}X, HTML, and XML. AddisonWesley, Reading, MA,
1999.
 [GRM97]
 Michel Goossens, Sebastian Rahtz, and Frank Mittelbach. The
L^{A}T_{E}X Graphics Companion: Illustrating
Documents wth T_{E}X and PostScript. AddisonWesley,
Reading, MA, 1997.
 [HP01]
 George Hart and Henri Picciotto. Zome Geometry: Handson
Learning with Zome Models. Key Curriculum Press, 2001.
 [Ram98]
 T. V. Raman. AsTeR Audio System For Technical
Readings. Lecture Notes In Computer Science. Springer Verlag,
December 1998.
 [Wei98]
 Eric W. Weisstein. The CRC Concise Encyclopedia of
Mathematics. CRC Press, 1998.
Footnotes:
^{1}
Rhombic Triacontahedron
^{2}
http://emacspeak.sf.net
^{3}
AsTeR http://www.cs.cornell.edu/home/raman/aster/astertoplevel.html
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On 2 Aug 2001, 14:45.