Visual Techniques For Computing Polyhedral Volumes

T. V. Raman
Email: raman@cs.cornell.edu
http://www.cs.cornell.edu/home/raman
, M. S. Krishnamoorthy
Email: moorthy@cs.rpi.edu
http://www.cs.rpi.edu/~moorthy

2001/08/0221:39:06 UTC

Abstract

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.

This paper was authored and typeset using the rich suite of Open Source tools available on the Linux desktop. 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 high-quality Open Source authoring and document preparation tools created over the last 25 years by the (La)TeX community. Figures in this paper were drawn using declarative authoring packages that enabled the first author to draw reliably without having to look at the final output. The high-level markup also makes this content long-lived.

© This work is being made available under the same copyright as that used by the Linux Documentation Project - see http://www.linuxdoc.org/LDP-COPYRIGHT.html.

This HTML is for WWW search engines. The typeset version should be considered difinitive, and can be found at paper.pdf

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.

Table 1: Basic zome lengths. Zome struts come in different colors and sizes. The sizes are mathematically significant. Here, I₁ denotes the unit icosahedron and C₁ 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 B₂ = f and sides B₁ = 1. Dropping a perpendicular from the apex of this triangle to its base bisects its base B₂ = f. From the right-triangles 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 R₁ lines constitute the radii of a golden rectangle gives R₁ in terms of the golden ratio. R₁ can be computed by observing the right-triangle 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)B₁.5,0.809016994375)(-0.5,-0.809016994375) (-0.5,0)B₂ -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 right-triangle marked ABC to derive the identity given by equation 1.6.

R₁ in terms of trigonometric ratios

The length of R₁ can be expressed in terms of trigonometric ratios by putting together the identities derived so far.

Powers of the golden ratio and trigonometry.

Construct a right-triangle of sides B₁ and B₃ -its hypotenuse is the result of joining the B₁ and B₃ lines using 2Y₂ = 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)B₃ -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)f² .5,1.30901699438)(0.5,-1.30901699438) 5,1.30901699438)(-0.5,-1.30901699438) (0,0)2Y₂ = 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 Y₂ yellow rhombus and its relation to a rectangle of sides B₁ ×B₃ 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 B₁ = 1. Let (0,0,0) be the center of the cube. The coordinates of the cube vertices are

Notice that by the choice of the basic zome lengths, the center of the cube can be connected to the 8 cube vertices by Y₁ struts. Thus, the radius of the unit cube is [(Ö3)/ 2] = sin60 = Y₁.

2.2  Locating Vertices Of The Tetrahedron

Construct a unit B₁ 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 G₁ = Ö2. Finally, draw the face diagonals of the cube that connect the end-points of the 3 diagonals just drawn. This constructs a tetrahedron of side G₁ = Ö2 inside the B₁ cube.

Let (0,0,0) be the center of the cube -notice that it is also the center of the G₁ tetrahedron. The vertex coordinates of this G₁ tetrahedron can be read from this model as:

As shown in section 2.1, the center of the above model can be joined to the cube vertices using Y₁ = sin60 yellow lines. From this it follows that the center of the G₁ tetrahedron can be joined to the tetrahedral vertices using Y₁ = sin60 lines. Observe that drawing these radii divides the interior of the tetrahedron into 4 congruent pyramids with a G₁ equilateral triangle as base and vertical sides Y₁.

2.3  Locating Vertices Of The Octahedron

Construct an octahedron of side G₁ = 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 B₁ = 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:

2.4  Locating Vertices Of The Rhombic Dodecahedron

Construct a B₁ unit cube. Construct pyramids of side Y₁ = sin60 on each face of this cube. Removing the cube edges leaves a rhombic dodecahedron of side Y₁. 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

The coordinates of these 6 vertices of the rhombic dodecahedron are then given by

Notice that from the above, the rhombic dodecahedron has all the vertices of a B₁ cube and the dual G₁ 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 Cube-octahedron

The cube-octahedron 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 cube-octahedron has the faces of the cube and the octahedron.

Construct a 2B₁ cube. Join the mid-points of adjacent edges using G₁ = Ö2 green lines. Removing the cube edges leaves a G₁ cube-octahedron.

Let (0,0,0) be the center of the 2B₁ cube, hence the center of the cube-octahedron. From this model, the vertex coordinates can be read as:

2.6  Locating Vertices Of The Dodecahedron

Consider the unit dodecahedron with sides B₁. 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 B₂ 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 B₃ struts to form a B₃ ×B₁ rectangle. View this model with the B₃ ×B₁ rectangle just constructed lying in the ZX plane. Locate opposing pairs of edges of the dodecahedron to similarly construct B₃ ×B₁ 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 B₂ = 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 B₃ ×B₁ rectangle in the XY plane have coordinates (±[(f²)/ 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 B₃×B₁ rectangle in the XY plane -see figure 4 for a view of one face of the dodecahedron from this model.

2.7  Locating Vertices Of The Icosahedron

Construct a unit icosahedron of side B₁. 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 B₂ = f struts. View this model so the B₂×B₁ golden rectangle just constructed lies in the ZX plane. Locate opposing pairs of edges to construct B₂ ×B₁ 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 B₂ ×B₁ 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:

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 R₁ zome strut is the radius of the B₁ icosahedron. Observe that these 12 radii draw the diagonals of the B₂×B₁ 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 B₁ dodecahedron. Construct red pyramids of vertical side R₁ = 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 R₁ = sin72.

Next, observe that the long diagonal of each rhombic face can be drawn using B₂ = 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 B₂ diagonals and removing the red edges would leave a B₂ icosahedron. See figure 2 for an illustration showing the relation between the golden rhombus and the fact that the radii of a B₁ 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 B₂ struts into each of these 30 holes and connecting their end-points with B₁ struts constructs an B₁ icosidodecahedron. By construction, the radius of the icosidodecahedron of side B₁ = 1 is B₂ = f.

Let (0,0,0) be the center of the icosidodecahedron. Place this model on one of its vertices, and observe that the B₂ 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 G₂ = fÖ2. This gives 6 of the 30 vertices to be

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 G₂ = 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 B₁ icosahedron. Scale this model by 2 to obtain a 2B₁ 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 2B₁ icosahedron to be:

Next, observe the 2B₁ ×2B₂ golden rectangles in the XY, YZ and ZX planes, and consider the mid-points of the 2B₁ sides. These have coordinates {(±f,0,0), (0,±f, 0), (0,0,±f)}. Thus, the mid-points of 6 of the 30 edges of the 2B₁ icosahedron give the vertices of the G₂ octahedron. By symmetry, it follows that the 30 edges of the 2B₁ icosahedron can be partitioned into 5 disjoint sets of 6 edges each, where the mid-points of edges in any given partition form a rotated copy of a G₂ 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 G₂ octahedra, we can compute the coordinates of all 30 vertices by reading off the mid-points of the 30 edges of the 2B₁ icosahedron.

Observe that by construction the 2B₁ 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 mid-points of the sides of the 2B₁ triangle shown in figure 5.

The coordinates of of the 30 vertices of the icosidodecahedron are therefore:

(-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 2B₁ icosahedron in the first octant. Its mid-points give 3 vertices of the icosidodecahedron of side B₁, 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 G₁ tetrahedron constructed in section 2.2. From this model, the interior of the B₁ cube can be decomposed into a G₁ 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 B₁, B₁, G₁ as base, and a side of the B₁ cube as its height. This gives the volume of this pyramid to be:

3.2  Volume Of The Octahedron

Consider the G₁ octahedron constructed in section 2.3. Place it on one of its triangular faces. Construct a G₁ tetrahedron and observe that the tetrahedron has the same face as the octahedron. Take 4 copies of this G₁ tetrahedron, and place them on 4 faces of the octahedron to form a 2G₁ tetrahedron. This shows that the 2G₁ tetrahedron can be decomposed into an octahedron and 4 tetrahedra.

We computed the volume of the G₁ tetrahedron to be ¹/₃ in section 3.1. By applying the scaling rule, the volume of the 2G₁ tetrahedron is ⁸/₃. Subtracting 4 copies of the G₁ tetrahedron from the 2G₁ tetrahedron gives the volume of the G₁ 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 B₁ square base and Y₁ 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:

This can also be seen by realizing that the yellow pyramid constructed on each face of the B₁ 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 Y₁ 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 G₁ = Ö2 green line. Drawing the long diagonal of all 12 faces gives a G₁ octahedron. This decomposes the rhombic dodecahedron into an octahedron and 8 pyramids having a G₁ equilateral triangle as base and Y₁ vertical sides. We showed in section 2.2 that this pyramid is 1/4 the volume of the G₁ tetrahedron. From this we can compute the volume of the rhombic dodecahedron V_\textRD to be

(-1,-1)(1,1) .5,0)(0,0.866025403785) (-0.25,0.433)Y₁0.866025403785)(0.5,0) (0.25,0.433)sin605,0)(0,-0.866025403785) (0.25,-0.433)sin60-0.866025403785)(-0.5,0) (-0.25,-0.433)Y₁ .5,0)(0.5,0) (0.0,0)B₁ = 1 0.866025403785)(0,-0.866025403785) (0,0)G₁ = Ö2linecolor=black](0,0)

Figure 6: The yellow rhombus of side Y₁ = sin60 has a short B₁ = 1 diagonal and a long G₁ = Ö2 diagonal. This relation leads to two equivalent ways of constructing a rhombic dodecahedron of side Y₁.

3.4  Volume Of The Cube-octahedron

Consider the cube-octahedron constructed in section 2.5. This shows that the 2B₁ cube can be decomposed into a cube-octahedron 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 cube-octahedron 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 B₂ = f and therefore has volume f³. It only remains to compute the volume of the roof structures.

Each roof has a square base of side B₂. The vertical faces are a pair of triangles and trapezium. Consider one of these trapezoidal faces; the parallel sides have length B₂ = f and B₁ = 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 cross-section.

Pyramid volume.

This pyramid has a rectangular base with sides B₂ and B₂ - B₁. Applying identities of the golden ratio, B₂-B₁ = f-1 = [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 B₃-B₂. 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 cross-section. Its face is a triangle of side B₂ 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 cross-section is B₁ = 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

Radius Of The Dodecahedron

Consider once again the cube of side B₂ identified in the model built in 2.6. As shown in 1, the radius of this cube is Y₂ = fsin60. We identified the vertices of this cube of side B₂ = 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 Y₂ = 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 Y₂ 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 B₂ ×B₁ rectangles lie in the XY, YZ and ZX planes. Construct a right-triangle with the top B₁ edge of the icosahedron as its hypotenuse in the ZX plane - in the zome model, this triangle has g₁ = [1/( Ö2)] green legs. Thus, this right-triangle has B₁ = 1 as the hypotenuse and sides g₁ = [1/( Ö2)]. Repeat this construction on the bottom B₁ edge. Finally, construct two more right-triangles each with one of the B₂ = f sides as the hypotenuse. These right-triangles have sides g₂ = [(f)/( Ö2)]. The above constructs a square of side [(1+f)/( Ö2)] around the golden rectangle B₂×B₁ in the ZX plane -see figure 5.1.

(-2,-2)(2,2) .30901699438,0)(-0.5,0.809016994375) (-0.9,0.4050)g₂ .5,0.809016994375)(0,1.30901699438) (-0.25,1.15)g₁1.30901699438)(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)g₂ 5,-0.809016994375)(0,-1.30901699438) (0.25,-1.15)g₁-1.30901699438)(-0.5,-0.809016994375) (-0.25,-1.15)g₁.5,-0.809016994375)(-1.30901699438,0) (-0.9,-0.4045)g₂ .5,0.809016994375)(0.5,0.809016994375) (0.0,0.8090)B₁.5,-0.809016994375)(0.5,-0.809016994375) (0,-0.5)1.5,0.809016994375)(-0.5,-0.809016994375) (-0.5,0.0)B₂ 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 B₂×B₁ 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 [(f²)/( Ö2)] constructed above has its side scaled by [(f²)/ 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 B₂×B₁ rectangles lying in the XY, YZ and ZX planes. Observe one of the g₁, g₁, B₁ right-triangle 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:

(-2,-0.5)(2,2) .92561479341,0)(0.218508012224,0) (-0.355,0)g₂218508012224,0)(0.92561479341,0) (0.565,0)g₁92561479341,0)(0.353553390593,0.990839414728) (0.635,0.495)g₂353553390593,0.990839414728)(0,1.60321185042) (0.1765,1.325)g₁1.60321185042)(-0.572061402818,0.612372435695) (-0.2875,1.11)g₂.572061402818,0.612372435695)(-0.92561479341,0) (-0.745,0.306)g₁ 218508012224,0)(0.353553390593,0.990839414728) (0.33,0.495)B₁353553390593,0.990839414728)(-0.572061402818,0.612372435695) (-0.11,0.8)B₁.572061402818,0.612372435695)(0.218508012224,0) (-0.18,0.305)B₁

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 B₂cube, and subtracting the volume of the space between the cube and the icosahedron from the volume of the cube.

Place the B₁ icosahedron on one of its edges as before. We will construct a B₂ 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)B₂ .809016994375,-0.809016994375)(0.809016994375,-0.809016994375) (0.0,-0.8090)f .809016994375,0.809016994375)(-0.809016994375,-0.809016994375) (-.8090,0)B₂ 809016994375,0.809016994375)(0.809016994375,-0.809016994375) (0.8090,0)f .5,0)(0.5,0) (0,0)B₁ = 1 .809016994375,0.809016994375)(-0.5,0) (-0.65,0.4045)Y₁ 809016994375,0.809016994375)(0.5,0) (0.65,0.4045)sin60 .809016994375,-0.809016994375)(-0.5,0) (-0.65,-0.4045)Y₁ 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 B₂ cube with the contained B₁ icosahedral edge. The icosahedral edge is connected to the vertices of the surrounding B₂ square using Y₁ = 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 B₂ cube, join the vertices of the contained icosahedral edge to the vertices of that face using Y₁ = sin60 lines as shown in figure 9. This constructs yellow triangular pyramids with vertical side Y₁ 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 B₁ 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 Y₁ = sin60.

Observe further that for each face of the cube, there are two trapezoidal pyramids inside the cube. These have the B₁, Y₁, B₂, Y₁ 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 B₂ 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 Y₁, Y₁, B₁ triangle with base B₁ -shown in figure 9 by horizontal dotted lines. The height of this trapezoidal pyramid is therefore

Now, observe that the volume of the B₂ 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 right-triangle constructed by dropping a perpendicular from the apex of this pyramid to its base -it passes through the centroid of the B₁ equilateral triangle. For a unit equilateral triangle, the centroid is at a distance [1/( Ö3)] from its vertices. The right-triangle therefore has base [1/( Ö3)] and and hypotenuse Y₁ = [(Ö3)/ 2] . The height of the pyramid is therefore

The area of the triangular base is

Volume of the trapezoidal pyramid.

We first compute the area of the B₁, Y₁, B₂, Y₁ 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 9-is [(f)/ 2]. The area of the base trapezium is therefore

Putting the pieces together, we sum the volumes of the 8 triangular pyramids and 12 trapezoidal pyramids to get

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 B₂ icosahedron and 20 R₁ pyramids constructed on each of the triangular icosahedral faces. These pyramids have a B₂ equilateral triangle as their base -the vertical sides of the pyramids are the R₁ = sin72 edges of the rhombic triacontahedron.

The volume of the rhombic triacontahedron can therefore be written as the sum of the volume of the B₂ 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

Summing The Parts

The volume of the B₁ = 1 icosahedron is [(5f²)/ 6] see section 5.1. The volume of the B₂ = f icosahedron is obtained by the scaling rule to be:

The volume of the 20 triangular pyramids is

7  Volume Of The Icosidodecahedron

We showed in section 2.9 that the vertices of an icosidodecahedron were the edge mid-points of a 2B₁ icosahedron. From this, it follows that an icosidodecahedron can be constructed by truncating a 2B₁ icosahedron to its edge mid-points. Truncating the 2B₁ icosahedron results in removing 12 pyramids, each having a B₁ pentagonal base and B₁ vertical sides.

The volume of the B₁ icosahedron is [(5f²)/ 6]; -see section 5.1- by applying the scaling rule, the volume of the 2B₁ icosahedron is 8[(5f²)/ 6].

Subtracting 12 copies of the pentagonal pyramid from the 2B₁ icosahedron gives the volume of the B₁ icosidodecahedron.

7.1  Area of the Unit Pentagon

Consider a unit pentagon of side B₁ = 1. Pick any vertex, and construct the B₂ = 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 B₁ = 1 adjacent sides with an included angle of 108^°. The area of this triangle is

Next, consider the triangle with sides {B₂, B₂, B₁}, and observe the adjacent sides of length B₁, B₂ with included angle 72^°. Its area is given by

Summing the parts, the area of the unit pentagon is

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

Rewriting sin36 and cos36 in terms of f, using equations ( 1.4 and 1.15), and using the identity Ö{(3-f²)} = [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

8  Conclusion

To conclude, here is a table listing the various formulae derived in this paper.

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

Figure

Figure

Figure

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^AT_EX on the Emacspeak audio desktop² 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 high-quality 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 high-level markup also makes this content long-lived and reusable -an immediate advantage is that the content can be easily made available using a variety of access modes ranging from high-quality print and online hypertext to high-quality audio renderings³ 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 three-dimensional polyhedra. Finally, we would like to thank Donald E Knuth for the T_EX 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^AT_EX Companion. Addison-Wesley, Reading, MA, 1994.

[GR99]

Michel Goossens and Sebastian Rahtz. The L^AT_EX Web Companion -Integrating T_EX, HTML, and XML. Addison-Wesley, Reading, MA, 1999.

[GRM97]

Michel Goossens, Sebastian Rahtz, and Frank Mittelbach. The L^AT_EX Graphics Companion: Illustrating Documents wth T_EX and PostScript. Addison-Wesley, Reading, MA, 1997.

[HP01]

George Hart and Henri Picciotto. Zome Geometry: Hands-on 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/aster-toplevel.html