The Dodecahedron

The  Dodecahedron

Platonic Relationships

Exercise: Get to know the five Platonic solids and the relationships between them. Start by counting the number of faces, edges, and vertices found in each of these five models. Make a table with the fifteen answers and notice that only six different numbers appear in the fifteen slots.
Answer:


                faces edges vertices

tetrahedron       4     6     4

cube              6    12     8

octahedron        8    12     6

dodecahedron     12    30    20

icosahedron      20    30    12

Observe that every number which appears somewhere in this table appears at least twice. These are not mere numerical coincidences.



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Relationships Between the Platonic Solids

Every time that a number shows up in two different places in the above table there is a significant relationship to be understood. Stop and savor each of these connections, studying the accompanying figure until it is clear:


6 edges in a tetrahedron = 6 faces in a cube:

This is a consequence of the fact that a tetrahedron can be inscribed in a cube. This establishes a one-to-one relationship between faces of the cube and edges of the tetrahedron, so there must be the same number of each. Each of the 6 edges of the tetrahedron shows up as one of the diagonals of one of the 6 faces of the cube.
Note that there are two different ways in which 4 of the 8 cube vertices could be chosen as the tetrahedron vertices.



4 faces in a tetrahedron = 4 vertices in a tetrahedron:

This follows from the fact that in the tetrahedron, every face is directly opposite a vertex, so there is a one-to-one relation between faces and vertices. If there are 4 of one, there must be4 of the other. In the other four Platonic solids, faces are opposite faces and vertices are opposite vertices, so the number of faces does not need to equal the number of vertices. In other words, only the tetrahedron has the property that you can rest it face-down on a table and not have a face on top; instead, a vertex is on top.
Another way of characterizing the same property is that a tetrahedron can be superimposed with a copy of itself facing in the opposite direction. The two tetrahedra have a common center, so the 4 vertices of one tetrahedron are centered in the 4 faces of the other tetrahedron. No other Platonic solid has this property. When two tetrahedra are combined in this manner, the result is called the compound of two tetrahedra, or the stella octangula, Kepler's Latin term for eight-pointed star.


6 edges in a tetrahedron = 6 vertices in an octahedron:

This is a consequence of the fact that an octahedron can be inscribed in a tetrahedron. The 6 edge-midpoints of the tetrahedron are the 6 vertices of the octahedron. (The octahedron in this image is the intersection of the two components of the stella octangula.)


6 faces in a cube = 6 vertices in an octahedron,

8 vertices in a cube = 8 faces in an octahedron,

12 edges in a cube = 12 edges in an octahedron:

All three of these numerical identities can be seen if we examine a compound of a cube and an octahedron. In the center of each of the 6 faces in the cube is one of the 6 vertices of the octahedron. In the center of each of the 8 faces of the octahedron is one of the 8 vertices of the cube. Also, the 12edges of the cube and the 12 edges of the octahedron bisect each other at right angles. This special triple relationship between the cube and the octahedron is called duality, and has many important consequences.


12 edges in a cube = 12 faces in a dodecahedron:

This is a consequence of the beautiful fact that a cube can be inscribed in a dodecahedron. Note that each of the 12 faces of the dodecahedron contains one of the 12 edges of the cube. The cube's edges are diagonals of the pentagons. This figure also suggests how one can build a dodecahedron by adding six pyramid-like bumps to the six faces of the cube.



12 edges in an octahedron = 12 vertices in an icosahedron:

Another 12=12 equivalence, this one follows from a construction in which an icosahedron is inscribed in an octahedron. Each of the 12 edges of the octahedron contains one of the 12vertices of the icosahedron. Incidentally, the edges of the octahedron are divided according to the golden ratio.


12 faces of dodecahedron = 12 vertices of icosahedron,

20 vertices of dodecahedron = 20 faces of icosahedron,

30 edges of dodecahedron = 30 edges of icosahedron:

Again, a triple relationship of duality holds between two polyhedra. These three numerical identities can be clearly seen if we examine a compound of a dodecahedron and an icosahedron. In the center of each of the 12 faces of the dodecahedron is one of the 12 vertices of the icosahedron. And, in the center of each of the 20 faces of the icosahedron is one of the 20 vertices of the dodecahedron. Also, the 30 edges of the dodecahedron and the 30edges of the icosahedron cross each other at right angles at their midpoints.


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The last two may be the toughest. If you've followed the above, work on these as exercises:


12 edges in a octahedron = 12 faces in a dodecahedron:

Exercise: Figure out how to construct a model of an octahedron inscribed in a dodecahedron. Study it to directly see a one-to-one relationship between octahedron edges and dodecahedron faces.
Hint: Combine this idea and this idea, then erase the cube. The octahedron vertices will lie on the midpoints of six of the dodecahedron edges.
Answer: Look at the answer face-on and see how one octahedron edge lies directly behind each dodecahedron face.


12 edges in an cube = 12 vertices in an icosahedron:

Exercise: Figure out how to construct a model of an icosahedron inscribed in a cube.
Hint: Combine this idea and this idea, then erase the dodecahedron and enlarge the cube.
Answer: Answer.




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Now it is time to observe a deeper relationship hidden in all five rows of the table at the top of this page. For any given polyhedron, let V be the number of vertices, let E be the number of edges, and let F be the number of faces.
Exercise: From the entries in the table, compute F-E+V for each of the five Platonic solids.
Answer: Notice a simple consistent answer.

The dodecahedron has 30 edges, 20 vertices and 12 faces. Dodeca is a prefix meaning “twelve.”

The dodecahedron is the only polyhedron I know of which is composed entirely of pentagons.
Buckminster Fuller describes what he calls a '6 pentagonal tensegrity sphere' in Synergetics I, 726.01, and which is shown at Marvin Solit's website at www.fnd.org/geo.htm, but I don't believe that structure can be built without tensegrity struts and tension wires.

 
Figure 1
The dodecahedron is pentagonal both inside and out, as can be seen from Figure 1. Like the icosahedron, it has many golden section relationships, which we shall see.
The dodecahedron is even more versatile then the icosahedron. The icosahedron contains  and  geometry, but the dodecahedron contains ,  and  geometry!

Figure 2
Cube and tetrahedron  in dodecahedron. Cube in gray, tetrahedron in green
This view of the dodecahedron is significant in that it shows the 2 dimensional shadow of the decagon. The decagon itself is based upon the pentagon, the building block of the dodecahedron. See Pentagon and Decagon for more information.

Figure 2A -- Another view of tetrahedron (green) inside cube (blue) inside dodecahedron (orange)

Figures 2 and  2A show how a cube and a tetrahedron can be placed inside a dodecahedron. These placements are 'nice' meaning that the vertices of the placed-in solids are all vertices of the dodecahedron.

The cube, octahedron and tetrahedron are all based on root 2 and root 3 geometry: The relationship of the side of the cube to the radius of its enclosing sphere  is r = sqrt(3) / 2.
For the  tetrahedron,  For the octahedron, 
The dodecahedron is capable of elegantly sustaining these  and  relationships, along with its own many  relationships.

Figure 3 -- octahedron  inside tetrahedron
Notice that the octahedron fits precisely on the bisected sides of the tetrahedron.

The icosahedron cannot contain any of the other 5 solids 'nicely' on its vertices.
The icosehedron and the dodecahedron are 'duals' (as are the cube and the octahedron). By 'dual' is meant that if you put a vertex in the middle of all of the faces and connect the lines, you get the dual.
By placing a vertex at the middle of  all the faces of the dodecahedron you get an icosahedron, and vice-versa. Figure 4 shows the dual nature of the icosahedron and dodecahedron.

Figure 4  Duals ----  dodecahedron inside icosahedron
Notice that to create the dodecahedron, all we did was draw lines from each vertex of the icosahedron to every other vertex. The vertices of the dodecahedron are at the intersection points. We could just as easily have found the vertices of the dodecahedron by drawing lines on every triangular face of the icosahedron. Where those lines intersect is the center of the face, and a vertex of the dodecahedron. That occurs because the dodecahedron has 12 faces and the icosahedron has 12 vertices.

Now for the standard analysis:
What is the volume of the dodecahedron?
We will use the pyramid method.

There are 12 pentagonal pyramids, 1 for each face, each pyramid beginning at O, the centroid. See Figure 1 and Figure 5.
The volume of any n-sided pyramid is 1/3 * area of base * pyramid height.

First we need to get the area of the base, which is the area of each pentagonal face:

Figure 5   One pyramid on face BCHLG

The area of the pentagon is the area of the 5 triangles which compose it.
From  Area of Pentagon  we know
 This is approximately 1.720477401 .

Now we need to find the height of the pyramid, OU. To do that, we need to find the distance from O to a vertex, lets say, OH. This distance will be the hypotenuse of the right triangle OUH. Since we already know UH, we can then get OU from the good ol’ Pythagorean Theorem.
Imagine a sphere surrounding the dodecahedron and touching all of its vertices. OH is just the radius of the enclosing sphere. If you look at Figure 6, HOZ = GON = diameter. There is a line through HOZ to show the diameter.

                             Figure 6

Now look at the rectangle MIFK. The diagonal of it, MF, is also a diameter (MF = HZ). Notice that the long sides of the rectangle, MK, and IF, are diagonals of the two large pentagons ADMQK and FEIRP.
We know from  Composition of the Pentagon  that the diagonal of a pentagon is  * side of pentagon.
We also can see from Figure 6 and more clearly in Figure 1 that the sides of the large pentagons themselves are diagonals of the pentagonal faces of the dodecahedron! (For instance, DA is a diagonal of the face ABCDE).  That means each side of the large pentagons is  * s and that MK (or any of the diagonals of a large pentagon) is  *  * s.
So MK =  s.
In fact, like the icosahedron, the dodecahedron is composed of rectangles divided in Extreme and Mean Ratio. In the icosahedron, we found these rectangles to be  rectangles.
In the dodecahedron, they are   rectangles.

In rectangle MKIF, MK = IF = , MI = KF = , as shown in figure 7.

 Figure 7 -- showing vertices of the  rectangle MKIF.

There are 30 sides to the dodec, and therefore 15 different  rectangles.

All of this as explanation of finding the distance OH from Figure 5! Because we are not using trigonometry, we need OH in order to get the pyramid height, OU in Figure 5. Notice that MHFZ is also a  rectangle and that HZ is the diagonal of it. If we can find HZ, then OH is just 1/2 of that.
d² = HZ²  =  MZ²  +  HM²   =  
diameter = HZ =  

Now we can find OU, the height of the pyramid.
From  Area of Pentagon  we know the distance mid-face to any vertex of a pentagon = 
So UH =  

                         

The volume of 1 pyramid = 1/3 * area base * pyramid height =
                                      

                                               

 Or

Note that (from Figure 5) OU  / UX   = .

What is the surface area of the dodecahedron? It is
12 faces * area of face = , or

What is the central angle of the dodecahedron?
From Figure 5, this is (for example)  HOC:

Figure  8 --  dodecahedron central angle
HC = side of dodecahedron, so XH = (1/2)s.

OH = radius = one half HZ = .
sin(  XOH) = XH / OH  =   
 XOH =  = 20.90515744°
 HOC = central angle = 41.81031488°
Since each face of the dodecahedron is a pentagon, the surface angle = 108°

While we're at it, lets get OX, the distance from the centroid to any mid- edge.
  
OX  =  

What is the dihedral angle of the dodecahedron?

                     Figure 9

The dihedral angle is  AXH.  AH is one of the long sides of any of the 15   rectangles which compose the dodec. AX and HX are the height h of the pentagon.
We know from   Construction of the Pentagon Part 2  that the height h of the pentagon is: 
We know from Figure 7 that AH is , so IH = 
sin(  IXH)  = IH / XH  =  
We recognize this ratio as our old friend the Phi Triangle with sides in ratio of 
 IHX =  = 58.28252558°.
Dihedral angle AXH = 2 *  IXH,
So dihedral angle AXH = 116.5650512°.

Let's compare distances:
Distance from centroid to mid-face (h) =  = 1.113516365s.
Distance from centroid to mid-edge =  = 1.309016995.
Distance from centroid to vertex =   = 1.401258539.

Go back to Figure 6.  We have colored the 4 internal pentagonal planes of the dodecahedron. U,X, W and V are the centers of these 4 planes which line up with the centroid O.
What is the distance  UX = WV?  What is XW?
If we can find these out we can figure out more deeply how the dodecahedron is constructed.

Figure 6, repeated

In Figure 6 we can see that UH on the top plane is the distance from the pentagon center to a vertex.
On plane ADMQK,  XM is parallel to UH and is also  the distance from that pentagon center to a vertex. UH is connected to XM by  HM, a side of the pentagon face CDIMH.
So we have a quadrilateral UHMX, with UH parallel to XM. From here we can derive UX, the distance between the two planes.

Figure 10  -- dodecahedron planar distance.

We know from Construction of the Pentagon Part 2  that the distance from the center of pentagon to a vertex = 
Therefore UH = 
The side of the large pentagon ADMQK in Figure 6 is, as we have seen, a diagonal of a dodecahedron face and so the side of the large pentagon =   Therefore, XM =  
Therefore, XM =  *UH, and XM is divided in Mean and Extreme Ratio at N (See Figure 10).
 =  
HM = s. Triangle MNH is right by construction.
So  = 
HN =  

Notice: UH = UX. So the dodecahedron is designed such that the distance to the 2 large pentagonal planes from the top or bottom faces is exactly equal to the distance between the center and a vertex of any of the faces of the dodecahedron.
This relationship is precisely what we saw in the icosahedron! That makes sense because the two are duals of each other.

The difference is that the dodecahedron is entirely pentagonal, both internally, and externally, on its faces.

What is the distance XW between the 2 large pentagonal planes ADMQK and FEIRP?
Figure 6 is misleading, it looks like the distance must be MI or KF, the dodecahedron side, but it isn't.
We already have enough information to establish this distance.
UX = VW.  UV = 2*height of any pyramid = 
So XW = UV - 2*UX =    
     XW                    =  

Notice UX / XW = .
UW is divided in Mean and Extreme Ratio at X.
XV / WV =  .   XV is divided in Mean and Extreme Ratio at W.

Let's make a chart of these planar distances along the diameter of the dodecahedron as we did with the icosahedron:  (see Figure 6):

Relative Chart of Distances –– Pentagonal Planes of Dodecahedron
Relative to the side of the dodecahedron

(Available in the book)     

Here is a table of these relationships, letting UX = 1:

(Available in the book)

Note that the diameter of the enclosing sphere is HZ, not UV.
We already know that the diameter is, from page 92, 
So what is the distance from U to the top of the sphere, and from V to the bottom of the sphere?  Let T’ be the top of the sphere and B’ be the bottom of the sphere. Refer to Figure 6.
If the radius is   and the distance OU is , then
UT’ =VB’ =   

Finally, let us demonstrate how the dodecahedron may be constructed from the interlocking vertices of 5 tetrahedron. We have already seen how the cube fits inside the dodecahedron, and how 2 interlocking tetrahedron may be formed from the diagonals of the cube. As Buckminster Fuller has pointed out, however, the cube and the dodecahedron are structurally unsound unless bolstered by the additional struts supplied by the tetrahedron. Fuller concludes logically that the tetrahedron is the basic building block of Universe; yet it is the dodecahedron that provides the blueprint and forms the structure for the interlocking tetrahedrons. The dodecahedron unites the geometry of crystals and lattices (root 2 and root 3) with the geometry of Phi (root 5), found in the biology of organic life.

 

Conclusions:

The dodecahedron is entirely pentagonal, consisting of the  geometry of Phi. Yet it contains the  and  geometry of the cube, tetrahedron, and octahedron.
Remarkably, the sides of the cube are  * side of the dodecahedron, because the cube side is the diagonal of a pentagonal face. Here is the key to the relationship of the first three Regular Solids and the much more complex icosahedron and dodecahedron.

Later on in this book we will discover a remarkable polyhedron that defines the relationship and provides the proper nesting for all 5 Platonic Solids, including the icosahedron, directly on its vertices. If a polyhedron could be called exciting, this one is IT! If you can’t wait, go to the last chapter of the book.

Dodecahedron Reference Tables
(included in the book)