Download Polyhedra and Geodesic Structures

Polyhedra and Geodesic
Structures
Vincent J. Matsko
Revised 2011
Chapter 0
Trigonometry
A brief summary of trigonometric relationships used in the text is included
below.
The trigonometric functions cosine and sine may be defined as follows:
the point on the unit circle x2 + y 2 = 1 (in the Cartesian plane) making
an angle θ with the positive x-axis has coordinates (cos θ, sin θ) (see Figure
0.1).
(cos θ, sin θ)
θ
Figure 0.1
The remaining trigonometric functions are defined as follows:
tan θ :=
sin θ
,
cos θ
cot θ :=
cos θ
,
sin θ
sec θ :=
1
1
,
cos θ
csc θ :=
1
.
sin θ
(0.1)
2
CHAPTER 0. TRIGONOMETRY
Note that tan θ and sec θ are not defined when cos θ = 0 (i.e., at odd multiples of π2 ), while cot θ and csc θ are not defined when sin θ = 0 (i.e., at even
multiples of π2 ).
Since the circle in Figure 0.1 has radius 1, it is evident that the Pythagorean
Theorem implies that
cos2 θ + sin2 θ = 1.
(0.2)
If cos θ 6= 0 (or sin θ 6= 0), we may divide (0.2) by cos2 θ (or by sin2 θ) to
obtain the additional Pythagorean identities:
1 + tan2 θ = sec2 θ,
cot2 θ + 1 = csc2 θ.
(0.3)
(cos θ, sin θ)
θ
−θ
(cos(−θ), sin(−θ))
Figure 0.2
Examination of Figure 0.2 reveals that
cos(−θ) = cos θ,
sin(−θ) = − sin θ.
(0.4)
These relationships indicate that cosine is an even function, while sine is an
odd function. They further imply (see (0.1)) that
tan(−θ) = − tan θ, cot(−θ) = − cot θ, sec(−θ) = sec θ, csc(−θ) = − csc θ,
(0.5)
so that the secant function is even, while the tangent, cotangent, and cosecant functions are odd.
Recall that the six trigonometric functions may also be defined in terms
of an arbitrary right triangle, as in Figure 0.3.
3
r
c
b
θ
p
q
a
Figure 0.3
Here, we have
cos θ =
a
b
b
c
c
a
, sin θ = , tan θ = , sec θ = , csc θ = , cot θ = . (0.6)
c
c
a
a
b
b
It is evident that the measure of ∠prq is
π
− θ = sin θ,
2
π
sec
− θ = csc θ,
2
cos
π
π
2
− θ, so that (0.6) implies
− θ = cos θ,
2π
csc
− θ = sec θ,
2
sin
π
− θ = cot θ,
π2
cot
− θ = tan θ. (0.7)
2
tan
Although (0.7) was derived in the case 0 < θ < π2 , these relationships are
valid for all values of θ.
Virtually all other useful relationships may be derived from the identity
cos(α + β) = cos α cos β − sin α sin β.
(0.8)
This result is so important that we include a brief proof here.
Let angles α, β, and α + β be drawn in a unit circle as in Figure 0.4.
Here, α is the measure of ∠aOc, β is the measure of ∠cOd, and hence
α + β is the measure of ∠aOd. Points b and e are such that both bc and
ed are perpendicular to the x-axis. The point f is on ed such that f c is
perpendicular to ed. Clearly,
cos(α + β) = [Ob] − [eb].
(0.9)
First, note that [Oc] = cos β, so that [Ob] = cos α cos β. Moreover,
[dc] = sin β.
4
CHAPTER 0. TRIGONOMETRY
(cos(α + β), sin(α + β)) d
(cos α, sin α)
f
c
β
α
e
O
b
a
Figure 0.4
Since f c is parallel to eb, ∠Ocf must have measure α. But ∠f dc and
∠Ocf are both complementary to ∠f cd, so ∠f dc must also have measure
α. Thus
[f c]
sin α =
,
[dc]
so that [f c] = [dc] sin α = sin α sin β. But clearly we have [f c] = [eb], so that
[eb] = sin α sin β. Substituting into (0.9), we see that (0.8) is valid.
Although proved only in the case that α and β are acute and α + β is
obtuse, it turns out that (0.8) is valid for all α and β.
Writing α − β as α + (−β) and using (0.8) and (0.4) gives
cos(α − β) = cos α cos β + sin α sin β.
(0.10)
The relationships in (0.8) and (0.10) are sometimes written together:
cos(α ± β) = cos α cos β ∓ sin α sin β,
(0.11)
where either the upper signs or lower signs may be taken.
To find analogous formulas for sin(α + β) and sin(α − β), note that (0.7)
implies
π
π
sin(α + β) = cos
− (α + β) = cos
−α −β .
2
2
5
Hence using (0.11) and (0.7) results in
sin(α + β) = sin α cos β + cos α sin β.
As above, we may write α − β as α + (−β), thus obtaining the relationships
sin(α ± β) = sin α cos β ± cos α sin β.
(0.12)
To find an expression for tan(α ± β), we may write
tan(α ± β) =
sin(α ± β)
,
cos(α ± β)
substitute from (0.11) and (0.12), and then divide both numerator and denominator by cos α cos β to obtain
tan(α ± β) =
tan α ± tan β
.
1 ∓ tan α tan β
(0.13)
This formula is valid whenever tan α, tan β, and tan(α ± β) are all defined.
Of frequent use are double-angle and half-angle formulas. Writing (0.8)
with β := α gives
cos 2α = cos2 α − sin2 α,
which in combination with (0.2) may be written in three forms:
cos 2α = 2 cos2 α − 1 = cos2 α − sin2 α = 1 − 2 sin2 α.
(0.14)
Now the expressions cos 2α = 2 cos2 α − 1 and cos 2α = 1 − 2 sin2 α may be
solved for cos α and sin α, respectively, resulting in
r
r
1 + cos 2α
1 − cos 2α
cos α = ±
, sin α = ±
.
(0.15)
2
2
Here, the “±” indicates that the appropriate sign must be chosen depending
upon the quadrant in which α lies. These formulas are often written with
α = θ/2, hence the name “half-angle” formulas:
r
r
θ
1 + cos θ
θ
1 − cos θ
cos = ±
, sin = ±
.
(0.16)
2
2
2
2
Writing (0.12) with β := α and taking the upper signs yields
sin 2α = 2 sin α cos α,
(0.17)
6
CHAPTER 0. TRIGONOMETRY
while writing (0.13) with β := α and taking the upper signs results in
tan 2α =
2 tan α
1 − tan2 α
(0.18)
wherever tan α and tan 2α are defined.
To find an expression for tan θ/2, use (0.16) with (0.2) to obtain
r
r
r
θ
1 − cos θ
1 − cos θ
(1 − cos θ)2
1 − cos θ
tan = ±
=±
·
=±
,
2
2
1 + cos θ
1 − cos θ
sin θ
sin θ
or alternatively
s
r
r
1 − cos θ
1 + cos θ
sin2 θ
sin θ
θ
·
=±
=±
.
tan = ±
2
2
1 + cos θ
1 + cos θ
(1 + cos θ)
1 + cos θ
It happens that the choice of “+” is always correct in these cases, so that
tan
θ
1 − cos θ
sin θ
=
=
.
2
sin θ
1 + cos θ
(0.19)
These trigonometric relationships are those most commonly used. Trigonometry is also used in conjunction with triangles, so a few important results
are derived here.
Consider the triangle shown in Figure 0.5:
r
a
p
c
h
θ
s
q
b
Figure 0.5
Recall that the area A of the triangle is given by
1
A = bh.
2
(0.20)
If θ is the angle between the sides with lengths a and b, then h = a sin θ, so
that
1
A = ab sin θ.
(0.21)
2
7
Also note that
c2 = [rs]2 + [sq]2 .
But [rs] = h = a sin θ and [sq] = b − [ps] = b − a cos θ. Hence
c2 = (a sin θ)2 + (b − a cos θ)2 .
Expanding and using (0.2) gives
c2 = a2 + b2 − 2ab cos θ,
(0.22)
often referred to as the cosine law for triangles.
Note that when θ is obtuse, as shown in Figure 0.6, we have
h = a sin(π − θ) = sin θ.
Thus (0.21) remains valid in this case.
r
c
a
h
π−θ
s
θ
p
q
b
Figure 0.6
To see that (0.22) is also valid when θ is obtuse, observe that
[sq] = b + [sp] = b + a cos(π − θ) = b − a cos θ.
The rest of the proof remains unchanged.
8
CHAPTER 0. TRIGONOMETRY
Chapter 1
Angles and Constructions
Throughout this chapter, the reader is assumed to be familiar with the usual
angles in the plane, basic geometric constructions, and basic trigonometry.
A brief review of these ideas is included in Chapter 0 and the appendices.
1.1
Angles
In this Chapter, all angles will be measured in degrees. Other units of
measurement may be employed later on; the reader will be given sufficient
notice when this occurs.
Table 1.1 lists some commonly used values of trigonometric functions.
With formulae such as sin(180◦ − α) = sin α, cos(180◦ − α) = − cos α, etc.,
we may easily obtain other values as necessary. For simplicity, we use the
abbreviation
√
τ = (1 + 5)/2.
(1.1)
τ is called the “golden ratio.” We undertake a brief geometric exploration
to see why this ratio is of such significance.
Consider the pentagon in Figure 1.1. We leave it to the reader to see
that ∆pqr and ∆qst are similar isosceles triangles with apex angle 36◦ and
base angles of 72◦ . Call the ratio of the length of the longer sides to that
of the shorter side of either triangle ρ. The notation “[pq]” denotes the
length of the segment pq; with x = [pq] = [qr] = [qt] and y = [pr] = [rs],
consideration of these similar triangles reveals that
ρ=
x
x+y
y
1
=
=1+ =1+ .
y
x
x
ρ
9
10
CHAPTER 1. ANGLES AND CONSTRUCTIONS
Multiplying through by ρ yields ρ2 = ρ+1 and hence the quadratic equation
ρ2 − ρ − 1 = 0. By applying the usual quadratic formula, we see that this
equation has one positive and one √
negative root. Since our ratio is positive,
we choose the positive root, (1 + 5)/2. This number is often referred to
by the Greek letter, τ .
It seems that τ is one of those numbers which pops up everywhere in
mathematics. In geometry, whenever regular pentagons or decagons are
mentioned, you can be sure that τ is lurking nearby.
cos 0◦ = 1,
cos 18◦ =
cos 30◦ =
cos 36◦
sin 0◦ = 0,
1
2
q
√
5+ 5
2
=
√
=
5+1
4
cos 45◦ =
√1 ,
2
cos 54◦ =
1
2
√
√
τ + 2,
sin 18◦ =
√
5−1
4
=
1
2τ
= 12 (τ − 1),
sin 30◦ = 12 ,
3
2 ,
√
1
2
=
τ
2,
3 − τ,
sin 36◦
=
sin 45◦ =
1
2
q
√
5− 5
2
√1 ,
2
sin 54◦ = τ2 ,
√
cos 60◦ = 21 ,
sin 60◦ =
cos 72◦ = 21 (τ − 1),
sin 72◦ =
cos 90◦ = 0,
sin 90◦ = 1.
Table 1.1
3
2 ,
1
2
√
τ + 2,
=
1
2
√
3 − τ,
1.2. REGULAR POLYGONS
11
p
q
r
s
t
Figure 1.1
1.2
Regular Polygons
As an application of these ideas, we examine a few constructions in the
plane. The first is the construction of the regular pentagon (see Figure 1.2).
We proceed as follows:
1. Construct a circle with center O.
2. Construct perpendicular lines through O; label two of the intersections
of these lines with the circle a and c (as in Figure 1.2).
3. Construct b as the midpoint of Oa. (A midpoint is usually constructed
using a perpendicular bisector, hence the vertical segment through b
in Figure 1.2.)
←→
4. Construct d on Oa such that [bd] = [bc].
5. [cd] is the length of the side of a regular pentagon inscribed in the
circle.
Why does this construction yield a regular pentagon? Consider for a moment
a regular pentagon inscribed in a circle of radius 1. Let s denote the length
of the sides of the pentagon. Since each edge of the pentagon subtends an
angle of 72◦ with the center of the circle, we may apply the cosine law for
triangles, yielding
s2 = 12 + 12 − 2 · 1 · 1 · cos 72◦ .
12
CHAPTER 1. ANGLES AND CONSTRUCTIONS
c
b
a
d
O
e
Figure 1.2
Using√the value of cos 72◦ from Table 1.1, we find that s2 = 3 − τ , so
that s = 3 − τ .
√
Now we must verify that our construction results in [cd] = 3 − τ ,
assuming that [Oa] = [Oc] = 1. Since [Ob] = 21 , an application of the
√
√
Pythagorean theorem to ∆bOc gives
√ [bc] = 5/2. Hence [bd] = 5/2, and
therefore [Od] = [bd] − [bO] = ( 5 − 1)/2. Another application of the
Pythagorean theorem, this time to ∆cOd, yields
√
2
2
2
2
[cd] = [Oc] + [Od] = 1 +
5−1
2
!2
√
√
6−2 5
5− 5
= 1+
=
= 3 − τ.
4
2
√
Thus, [cd] = 3 − τ , and our construction is therefore valid.
The construction of the regular decagon is closely related. In Figure 1.3,
we assume that a regular decagon has been inscribed in a circle of radius 1.
To find s, the length of the sides of the decagon, we again apply the cosine
law for triangles. This yields
s2 = 12 + 12 − 2 · 1 · 1 · cos 36◦ = 2 − 2 ·
so that s =
√
2 − τ.
τ
= 2 − τ,
2
1.3. POLYGON VARIATIONS
13
s
36◦ 1
Figure 1.3
√
What is 2 − τ ? It happens (see §1.5 for a thorough discussion) that
2 − τ = τ −2 , so that
√
1
5−1
−1
s=τ = =
.
τ
2
We remark here that any expression involving
into
√ “τ ” may be converted
√
an expression involving only rationals and “ 5” by using τ = ( 5 + 1)/2
and rationalizing
the denominator as necessary. Conversely, expressions
√
involving “ √
5” may be written as expressions involving “τ ” by using the
relationship 5 = 2τ − 1.
√
Now recall that in Figure 1.2, we have [Od] = ( 5 − 1)/2. Thus, in
creating our regular pentagon, we have created our decagon as well! This
might not have been √
so obvious if we failed to convert s = τ −1 into an
expression involving “ 5.”
1.3
Polygon Variations
In addition to the regular polygons, there are a number of other polygons
which will be discussed later on. Those we will consider happen to be
equiangular, although the lengths of the sides may vary.
—A hexagon variation.
One such polygon is a hexagon, each of whose angles has measure 120◦ ,
but whose sides alternate in the ratios τ and 1. Such a hexagon is called a
τ : 1-hexagon. As can be seen in Figure 1.4, there are two approaches to
14
CHAPTER 1. ANGLES AND CONSTRUCTIONS
constructing such a hexagon from an equilateral triangle – either inscribe
the shorter sides of the hexagon in the triangle or inscribe the longer sides.
We examine a construction using the latter approach, relegating the former
approach to the Exercises.
Figure 1.4
This hexagon variation may be constructed as follows (see Figure 1.5).
To avoid clutter on the diagram, many of the auxiliary arcs necessary for
the construction are omitted.
1. Create an equilateral triangle ∆abc.
2. Perpendicularly bisect bc at d.
3. With this same radius [bd], draw an arc centered at d from b counterclockwise to c.
−→
4. With center b and radius [bd], draw an arc counterclockwise from ab .
5. Extend cb past b and draw a perpendicular to this line at b. Denote
by g the intersection of this perpendicular with the arc drawn in Step
4.
6. Draw the segment cg to intersect the arc drawn in Step 3 at h.
7. Construct an arc centered at c with radius [ch] to intersect ab at j and
k. j and k are vertices of the τ : 1-hexagon.
1.3. POLYGON VARIATIONS
15
8. Draw similar arcs from a and b to complete the construction of the
vertices of the hexagon, and then join the vertices to form the sides.
c
d
h
g
a
j
k
b
Figure 1.5
Why does this construction yield a τ : 1-hexagon? To see why, begin
with the assumption that [ab] = [bc] = [ca] = s. From Step 5, it follows
that ∠cbg is a right angle. From Steps 4 and 5, it follows that bd and bg are
radii of the same circle, and hence [bg] = [bd] = 12 s. From the Pythagorean
√
theorem, we see that [cg] = 25 s.
Now ∠chb is a right angle as it is inscribed in a semicircle. Thus ∆cbg
and ∆chb are similar triangles, giving
[ch]
[cb]
=
.
[cb]
[cg]
With data from above, we solve for [ch] to see that [ch] =
√2 s.
5
√2 s,
5
from Step 7 that [cj] =
We now apply the cosine law for triangles to ∆cbj, yielding
[cj]2 = [cb]2 + [jb]2 − 2[cb][jb] cos 60◦ .
Using [cj] =
√2 s
5
and [cb] = s, this equation becomes
1
[jb]2 − [jb]s + s2 = 0.
5
and hence
16
CHAPTER 1. ANGLES AND CONSTRUCTIONS
We wish to establish that [kj]/[jb] = τ , so we use the previous equation
with the observation that s = [kj] + 2[jb]. This substitution gives
1
[jb]2 − [jb]([kj] + 2[jb]) + ([kj] + 2[jb])2 = 0,
5
which after some rearrangement becomes
[kj]2 − [kj][jb] − [jb]2 = 0.
Dividing through by [jb]2 yields
[kj] 2 [kj]
−
− 1 = 0.
[jb]
[jb]
This implies that
√
[kj]
1+ 5
=
=τ
[jb]
2
or
√
[kj]
1− 5
=
= 1 − τ.
[jb]
2
Since the ratio must be positive, we see that [kj]/[jb] = τ , as desired. Hence,
our construction does indeed yield a τ : 1-hexagon.
—Decagon and decagram variations.
Happily, it is much easier to construct a τ : 1-decagon – that is, a decagon
all of whose angles have measure 144◦ but whose sides alternate in the ratios
τ and 1. One simply constructs a regular pentagon, trisects the sides, and
“connects the dots” (see Figure 1.6(b)). That this yields a τ : 1-decagon
is a consequence of the fact that the dashed lines in this figure are part of
36◦ –108◦ –36◦ triangles, and the ratio of the longer to the shorter sides of
such triangles is τ : 1.
a
c
e
d
b
(a)
(b)
Figure 1.6
f
1.3. POLYGON VARIATIONS
17
By joining every third vertex in a regular decagon, a regular decagram
is formed (see Figure 1.6(a)). The suffix “-gram” is used when the regular
polygon is star-shaped, and hence nonconvex. Similarly, one may construct
a decagram by connecting every third vertex of a τ : 1-decagon (see Figure
1.6(b)). One
√ may show (see Exercise 4) that the sides of this decagram are in
the ratio 5 : 2. By examining the bold pentagon
in Figure 1.6(b), one sees
√
an alternative method for constructing a 5 : 2-decagram – simply extend
the sides of a pentagon and extend the segments joining the midpoints of
adjacent sides of the pentagon.
—Octagon and octagram variations.
A regular octagon
may be formed by dividing the sides of a square in
√
the ratios 1 : 2 : 1 (see Figure 1.7(a)). To obtain an octagon variation,
the sides of a square may be divided into the ratios 1 : 1 : 1 (see Figure
1.7(b)). It is easy to see that the ratio of √
the length of the longer sides of
this octagon to that of the shorter sides is 2 : 1.
As with the decagon, a regular octagram may be inscribed in a regular
octagon by connecting every third
√ vertex (see Figure 1.7(a)). Likewise, an
octagram may be inscribed in a 2 : 1-octagon by connecting every third
vertex (see Figure 1.7(b)). We see that the sides of the octagram variation
come in two different lengths – in Figure 1.7(b), the ratio of √
the length of
the “vertical
”
sides
to
the
length
of
the
“oblique”
sides
is
3
:
2
√
√ 2, resulting
in a 3 : 2 2-octagram. A simple construction for creating a 2 : 1-octagon
is included in the Exercises.
(a)
(b)
Figure 1.7
Polygons described in this section; that is, equiangular polygons whose
sides alternate in length, are called quasi-regular polygons.
18
CHAPTER 1. ANGLES AND CONSTRUCTIONS
1.4
Rating a Construction
The following scheme may be used to score a particular construction according to its complexity – every time a procedure with straightedge and
compass is done, the corresponding number of points is added to the total.
Suggested values are as follows:
1. Drawing an arbitrary line (i.e., not specifying any points on the line)
– 0 points.
2. Drawing a line through one specified point (such as drawing a diameter
of a circle) – 1 point.
3. Drawing a line through two specified points – 2 points.
4. Extending a given line – 1 point.
5. Opening a compass to a random setting (such as may be needed in a
bisection operation) – 0 points.
6. Opening a compass to a specified setting (such as the length of a
particular segment) – 2 points.
7. Drawing any arc of a circle with an unspecified center, such as drawing
an initial circle in a construction (assuming that the compass is already
set to the appropriate radius) – 0 points.
8. Drawing any arc of a circle with a specified center; that is, specifying
a point or requiring that the center lie on a given line (assuming the
compass is already set to the appropriate radius) – 1 point.
This point system was designed in an attempt to give the simplest, most
accurate constructions the lowest values. For example, in setting the compass to the length of a specified segment in the plane, there is some room
for human error – and hence this is an “expensive” operation. Drawing
an initial line for a construction, however, introduces practically no error –
unless your “straightedge” is crooked! Of course, one is free to modify the
point system as one sees fit. One might, for example, in order to encourage
creativity in using the compass, increase the point value for drawing a line
through two points to 10 points.
We illustrate with a sample√problem: given a segment of unit length,
construct a segment of length 15. Let us use the Pythagorean theorem
and the fact that any triangle inscribed in a semicircle is a right triangle to
consider the following construction (see Figure 1.8):
1.5. POWERS OF τ
19
f
a
d0
c
d
e
b
g
1
Figure 1.8
1. Set your compass to the unit length – 2 points.
←→
2. Draw an arbitrary line ab – 0 points.
←→
3. Choose an arbitrary point c on ab , and create a semicircle through d
and d0 with preset compass – 1 point.
4. With center at d, create e on ab – 1 point.
5. Reset the compass to length [ce] – 2 points.
6. Draw a circle with center e with preset compass – 1 point.
7. With f as the intersection of the semicircle in Step 3 and the circle in
Step 6, draw f g – 2 points.
Total for this construction – 9 points.
Since [cg] = 4, [cf ] = 1, and ∠cf g is a right angle (being inscribed
in a
√
semicircle), it follows from the Pythagorean theorem that [f g] = 15.
1.5
Powers of τ
Recall that when discussing a construction of the regular decagon earlier,
it was helpful to know that 2 − τ = τ −2 . It is now time to look at such
calculations in more detail.
We first examine powers of τ . Since τ 2 = τ + 1, we know that
τ 3 = τ 2 + τ = (τ + 1) + τ = 2τ + 1.
Likewise,
τ 4 = τ 3 + τ 2 = (2τ + 1) + (τ + 1) = 3τ + 2.
20
CHAPTER 1. ANGLES AND CONSTRUCTIONS
Of course, we may keep going and calculate τ 5 , τ 6 , etc. On the other hand,
dividing the relationship τ 2 = τ + 1 by τ results in τ = 1 + 1/τ , so that
1
= τ −1 = τ − 1.
τ
Proceeding as above, we have 1 = τ −1 + τ −2 , so that
τ −2 = 1 − τ −1 = 1 − (τ − 1) = 2 − τ.
Again, we may calculate τ −3 , τ −4 , etc. A short table of values for powers
of τ is listed below.
τ1 = τ,
τ −1 = τ − 1,
τ 2 = τ + 1,
τ −2 = 2 − τ ,
τ 3 = 2τ + 1,
τ −3 = 2τ − 3,
τ 4 = 3τ + 2,
τ −4 = 5 − 3τ ,
τ 5 = 5τ + 3,
τ −5 = 5τ − 8,
τ 6 = 8τ + 5,
τ −6 = 13 − 8τ .
Table 1.2
The reader familiar with the famous Fibonacci sequence will no doubt
find this table intriguing. For those unfamiliar, the Fibonacci sequence is
defined as follows: put F0 = 0, F1 = 1, and calculate the remaining members
of the sequence by the recurrence relation
Fn+2 = Fn+1 + Fn
when n ≥ 0. Thus, when n = 0, we have F2 = F1 + F0 = 1 + 0 = 1; when
n = 1, we have F3 = F2 + F1 = 1 + 1 = 2, etc. This results in the sequence
F0 , F1 , F2 , F3 , F4 , F5 , F6 , F7 , ... = 0, 1, 1, 2, 3, 5, 8, 13, ...
1.6. EXERCISES
21
Of course these are precisely the numbers occurring in Table 1.2. In fact,
the whole of this table may be summarized by the following rules, valid for
n ≥ 1:
τ n = Fn τ + Fn−1 ,
τ −n = (−1)n+1 (Fn τ − Fn+1 ).
Thus, any power of τ , whether positive or negative, may be written in the
form aτ + b, where a and b are integers.
1.6
Exercises
1. A regular 15-sided polygon is called a pentadecagon, or 15-gon.
(a) Show that the angle subtended by an edge of a pentadecagon
from its center is 24◦ .
(b) Using constructions of angles learned in this chapter, construct a
24◦ angle, and then construct a pentadecagon.
2. Consider the following construction (see Figure 1.9) of a regular pentadecagon:
(a) Draw a circle with center O.
(b) Draw a horizontal diameter of this circle.
(c) Bisect Oa perpendicularly in b (use the compass set to Oa for
simplicity).
(d) Create a diameter of the circle perpendicular to the one constructed in (b).
(e) Create d so that [bc] = [bd] (the pentagon construction again).
(f) Bisect Od at e.
(g) Create f so that [ef ] = [Od].
(h) Create g so that [eg] = [Oa].
(i) f g is an edge of the 15-gon.
22
CHAPTER 1. ANGLES AND CONSTRUCTIONS
c
g
f
a
b
e
d
O
Figure 1.9
Use the following outline to prove that this construction is correct.
Assume throughout that [Oa] = 1.
(a) Show that [Oe] = cos 72◦ and [Og] = sin 72◦ .
(b) √
Show that ∠Oef has measure 60◦ , and conclude that [Of ] =
3 cos 72◦ .
(c) Show that the sides of a regular 15-gon inscribed in a circle of
radius 1 have length 2 sin 12◦ .
(d) Since 12◦ = 72◦ − 60◦ , we may write
2 sin 12◦ = 2 sin (72◦ − 60◦ ) = 2 (sin 72◦ cos 60◦ − cos 72◦ sin 60◦ ) .
Using the results of (a) and (b), show that [Og] = 2 sin 72◦ cos 60◦
and [Of ] = 2 cos 72◦ sin 60◦ . Finally, show that [f g] = 2 sin 12◦ .
(e) (Due to Sara Fessler) Alternatively, consider ∆ef g. Knowing
∠ef g, ∠f eg, and [eg], conclude that [f g] = 2 sin 12◦ .
3. Consider the following construction (see Figure 1.10), where auxiliary
construction lines are omitted for clarity. For simplicity, use [ab] = 2.
1.6. EXERCISES
23
(a) Create equilateral triangle ∆abc.
(b) Create the perpendicular bisector of ac at d.
(c) Extend ac past a.
(d) With compass set to [ad], create e on the perpendicular bisector
of ac so that [ad] = [de], and f past a so that [ad] = [af ].
(e) With compass set to [f e] and centered at f , create g on ac so
that [f e] = [f g].
(f) With compass set to [cg] and centered at c, create h on cb so that
[cg] = [ch].
(g) With compass still set to [cg], draw an arc centered at b to create
j and k, and an arc centered at a to create m and n.
f
a
e
m
n
d
k
g
c
b
j
h
Figure 1.10
Show that ghjkmn is a τ : 1-hexagon.
4. Recall from §1.1 that the ratio of the length of the longer sides to that
of the shorter side of an isosceles triangle with apex angle 36◦ (such
as ∆abc in Figure 1.6(b)) is τ . Argue geometrically that the ratio of
the length of the longer side to that of the shorter sides of an isosceles
triangle with apex angle 108◦ (such as ∆bcd in Figure 1.6(b)) is also
24
CHAPTER 1. ANGLES AND CONSTRUCTIONS
τ . Use these facts to show that the ratio of the lengths of the sides of
the decagram in Figure 1.6(b) is
√
[ef ]
5
=
.
[ae]
2
√
5. Consider the following construction of a 2 : 1-octagon. Begin by
creating a, b, and c as in the construction of the regular pentagon (see
Figure 1.2). Draw a line through c and b, intersecting
the circle again
√
at f . Show that af and f e are two sides of a 2 : 1-octagon; that is,
[f e] √
= 2.
[af ]
6. Consider the following construction “system”, given a segment of unit
length (see Figure 1.11):
g
a
b
c
k
j
h
d
m
e
f
n
Figure 1.11
(a) Set your compass to a unit length.
(b) Draw a circle with the compass at this setting (this circle is the
one centered at b in Figure 1.11).
(c) Draw a (long) horizontal line through the center of this circle (as
in Figure 1.11).
(d) Draw a circle with center c and unit radius.
(e) Draw a circle with center d and unit radius.
(f) Draw a circle with center e and unit radius.
Note that this “system” of circles may be extended indefinitely.
1.6. EXERCISES
(a) Show that [gk] =
25
√
3.
(b) There are ten segments (that is, segments with
√ endpoints among
the points a–n) in Figure 1.11 with length 7. Find them.
(c) With the result of (b) in mind, describe and
√ rate a construction
which would yield a segment of length 3 + n2 , where n is a
positive integer.
√
(d) Find a segment in Figure 1.11 of length 13.
√
(e) Keeping in mind the result of (d), show that n2 + n + 1 (where
n is a positive integer) may be produced by a construction whose
rating is n + 5 points. (Hint: Keep constructing more circles until
you have enough....)
√
√
7. Choose a length at random, say 19 or 23. Divide into teams, and
see who comes up with the most efficient construction using the rating
system described in §1.4, or devise a rating system of your own.
26
CHAPTER 1. ANGLES AND CONSTRUCTIONS
Chapter 2
The Platonic Solids
2.1
Some Definitions
We now embark on a discussion of perhaps the best-known and most celebrated of all polyhedra – the Platonic solids. Before doing so, however, a
word about convexity is in order.
A polygon or polyhedron is said to be convex if, informally, it has
no “dents” (see Figure 2.1). More formally, convexity is described by the
property that for any two points in the polygon (or polyhedron), the segment
joining these two points lies wholly within the polygon (or polyhedron). The
dashed lines in Figure 2.1 illustrate the nonconvexity of the corresponding
polygons – parts of the dashed lines lie outside the figures. Further examples
of nonconvex polygons and polyhedra are discussed in Chapter 12.
convex polygons
nonconvex polygons
Figure 2.1
27
28
CHAPTER 2. THE PLATONIC SOLIDS
An interesting relationship commonly known as “Euler’s formula” is
valid for all convex polyhedra – if V denotes the number of vertices of a
polyhedron, E the number of edges, and F the number of faces, then
V − E + F = 2.
(2.1)
Take, for example, a cube, which has eight vertices (or corners), twelve
edges, and six square faces. Then we see that
V − E + F = 8 − 12 + 6 = 2.
The stage is now set for a discussion of the Platonic solids. A Platonic
solid is a polyhedron with the following properties:
(P1 ) It is convex.
(P2 ) Its faces are all the same regular polygon.
(P3 ) The same number of polygons meet at each of its vertices.
Note that since a Platonic solid is convex, the polygons referred to in
(P2 ) must also be convex.
Which polyhedra satisfy properties (P1 )–(P3 )? We provide two different
approaches to answering this question.
2.2
A Geometric Enumeration
We take an incremental approach here, and begin by asking, “Which Platonic solids have equilateral triangular faces?” Now the fewest number of
triangles which may meet at a vertex is three. It is a simple matter to construct a polyhedron with three triangles meeting at each vertex – just take
a triangular pyramid. Four triangles are required (see Figure 2.2(a)), and
hence this polyhedron is called a tetrahedron.
What about four triangles meeting at a vertex? We know from our
Egyptian studies that four triangles meet at the apex of a square pyramid,
while two triangles meet at each vertex of the square base. This implies
that if we take two square pyramids and join them base-to-base, the squares
“disappear,” leaving 2+2 = 4 triangles at each vertex of the interior squares.
The result is a convex polyhedron with four equilateral triangles meeting at
each vertex. Since two pyramids were used, 2 × 4 = 8 triangles are required
(see Figure 2.2(b)), and hence this polyhedron is called an octahedron.
2.2. A GEOMETRIC ENUMERATION
(a)
29
(b)
(c)
Figure 2.2
Five triangles at a vertex is a bit trickier. All in all, twenty triangles
are required (see Figure 2.2(c)), and hence this polyhedron is called an
icosahedron (“icosi” is the Greek prefix for “20”). The best way to see
how these triangles fit together is to build an icosahedron yourself. (In the
final analysis, there is no substitute for hands-on construction.) Should this
be momentarily inconvenient, an alternative description follows.
Suppose we try at first to extend our base-to-base square pyramid construction of an octahedron to a base-to-base pentagonal pyramid construction of a polyhedron with five triangles meeting at each vertex. Of course
five triangles meet at the apex (and the vertex opposite) but, as with the
square pyramid, only four triangles meet at each vertex of the “disappearing” pentagonal bases (see Figure 2.3(a)). Because of its construction, the
polyhedron in Figure 2.3(a) is called a pentagonal bipyramid.
Happily, this situation may be remedied. Consider for a moment the
pentagonal “toy drum” of Figure 2.3(b). The top and bottom pentagons
are out of phase by 36◦ (see Figure 2.3(c) for a top view of Figure 2.3(b)),
and the intervening space is filled by a zig-zag of ten equilateral triangles.
The salient anatomical feature of this “drum” (also called a pentagonal
30
CHAPTER 2. THE PLATONIC SOLIDS
antiprism) is that exactly three triangles (and one pentagon) meet at each
vertex. As a result, appending a pentagonal pyramid to both the top and
bottom of this antiprism yields a polyhedron with precisely five triangles
meeting at each of its twelve vertices (see Figure 2.4). The pentagonal
antiprism contributes ten equilateral triangles, and each of the pentagonal
pyramids contributes five, for a total of twenty triangles – hence the name
“icosahedron.”
(a)
(b)
(c)
Figure 2.3
Our search for Platonic solids with triangular faces ends here, for one is
easily convinced that the angles of six equilateral triangles comprise 360◦ ,
and hence any vertex with six equilateral triangles would be “flat.” This,
of course, does not result in a convex polyhedron, but rather a tiling of the
plane by equilateral triangles.
So now we have enumerated all possible Platonic solids with equilateral
triangles as faces. What about the next regular polygon, the square? Three
squares at a vertex results in our old friend the cube (sometimes called a
hexahedron), and at four squares we are already flat.
2.2. A GEOMETRIC ENUMERATION
(a)
31
(b)
Figure 2.4
What about regular pentagons? As it happens, there is a Platonic solid
with three pentagons at each vertex. As with the icosahedron, the best way
to understand this polyhedron is to build it; barring that, we press on...
(a)
(b)
(c)
Figure 2.5
Perhaps the best way to imagine such a solid is to begin with an arrangement of six pentagons as shown in Figure 2.5(a). Five of these pentagons
may be folded up to yield a bowl-like shape; as it happens, two such bowls
fit exactly together. The result, as it requires precisely twelve pentagons, is
called a dodecahedron (or sometimes a pentagonal dodecahedron).
32
CHAPTER 2. THE PLATONIC SOLIDS
Now each angle of a regular pentagon had measure 108◦ . Hence four
such angles have measure 432◦ > 360◦ , and hence it is impossible to fashion
a vertex of a convex polyhedron with four (or more) pentagons at a vertex.
So on to regular hexagons. With three hexagons at each vertex we are
already flat, yielding a hexagonal tiling of the plane. As a result, there are
no Platonic solids with regular hexagonal faces.
Our search stops here. Since three hexagons result in a flat vertex, three
regular polygons with more than six sides, if they met at a vertex, would
comprise more than 360◦ . So as with the case of four pentagons, no convex
polyhedron may be formed.
2.3
An Algebraic Enumeration
The foregoing is not the only approach to an enumeration of the Platonic
solids. We now embark on an algebraic description, making use of Euler’s
formula (see (2.1)). Our attack is to translate (P1 )–(P3 ) into algebraic analogues. To begin, suppose that we have a Platonic solid before us, with the
number of vertices, edges, and faces being denoted by V , E, and F , respectively. We see from (P2 ) that all faces of this solid are the same regular
polygon. Now suppose that this polygon has p sides.
Then pF is simply the total number of sides on all faces of the Platonic
solid. Because each edge of the solid is the meeting place of exactly two
faces (and hence two of the sides among the pF ), we find that our count
includes every edge of the solid exactly twice. Thus, we have
(P20 ) pF = 2E.
(If you have trouble visualizing this, take hold of the nearest Platonic solid
and work through the previous paragraph.)
From (P3 ), we see that the solid has the same number of polygons meeting at each vertex; this number shall be denoted by q. Then there must also
be exactly q edges meeting at each vertex as well, so that qV counts the
total number of edges incident at all vertices of the polyhedron. But each
edge is incident at exactly two vertices, so that qV counts each edge of our
solid exactly twice. Thus, we have
(P30 ) qV = 2E.
Finally, because our Platonic solid is convex, we replace (P1 ) with Euler’s
formula:
(P10 ) V − E + F = 2.
We can now solve for V and F from (P30 ) and (P20 ) and substitute into
2.3. AN ALGEBRAIC ENUMERATION
(P10 ), yielding
33
2E
2E
−E+
= 2.
q
p
A little algebra yields
1
1
1 1
+ = + ,
p q
2 E
(2.2)
which must be valid for any Platonic solid.
Now p and q are integers 3 or greater, and E is a positive integer. So if
p ≥ 4 and q ≥ 4, we would have 1/p + 1/q ≤ 1/2, making (2.2) impossible.
As a consequence, we must have p = 3 or q = 3 (or possibly both).
Assume for the moment that p = 3. Then (2.2) becomes
1
1
1
= + .
q
6 E
(2.3)
Since E is a positive integer, this means that 1/q > 1/6, or equivalently,
q < 6. Since at least three polygons must meet at the vertex of a polyhedron,
the only possibilities are q = 3, q = 4, or q = 5. With each choice of q, E
may be determined from (2.3). V may then be determined from (P30 ), and
F from (P20 ).
Since (2.2) is symmetric in the occurrence of “p” and “q” (each playing
precisely the same role), the reader is invited to make an analogous argument
with the assumption that q = 3. When this is done, Table 2.1 is obtained,
which includes all possibilities for p and q as described in the previous few
paragraphs.
As a final note, the polyhedron with regular faces of p sides, where q are
assembled at each vertex, is sometimes denoted by {p, q}. This notation for
referring to a polyhedron is called a Schläfli symbol.
p
q
E
V
F
Platonic Solid
{p, q}
3
3
6
4
4
Tetrahedron
{3, 3}
3
4
12
6
8
Octahedron
{3, 4}
4
3
12
8
6
Cube
{4, 3}
3
5
30
12
20
Icosahedron
{3, 5}
5
3
30
20
12
Dodecahedron
{5, 3}
Table 2.1
34
CHAPTER 2. THE PLATONIC SOLIDS
Notice that there are exactly five possibilities, each corresponding to one
of the Platonic solids described in the previous section. As expected, an
algebraic approach yields the same set of Platonic solids as an incremental
geometric approach.
2.4. EXERCISES
2.4
35
Exercises
1. Build the five Platonic solids using the nets provided at the end of
the chapter. (An arrangement of polygons which may be folded to
produce a polyhedron is called a net for that polyhedron.)
2. Find all possible nets for the cube. In other words, find all arrangements of six contiguous squares in the plane which may be folded to
obtain a cube. (Two nets are considered the same if one may be obtained from the other by a rotation and/or a reflection.)
3. Find all possible nets for the octahedron.
4. Color the faces of an octahedron with four colors so that all four colors
are incident at each vertex. Then color the faces of an icosahedron with
five colors so that all five colors are incident at each vertex.
5. Number the vertices of a dodecahedron with the numbers 1 through 5
so that each pentagonal face has five differently numbered vertices.
6. Color the edges of an icosahedron with three different colors so that
each face of the icosahedron has three differently colored edges.
7. Build two square pyramids and then arrange them base-to-base to
form an octahedron.
8. Construct an icosahedron by constructing two pentagonal pyramids
and a pentagonal antiprism, and then arranging them appropriately.
9. (a) Build four tetrahedra and one octahedron so that all of the polyhedra have the same edge length. Arrange them to form a larger
tetrahedron. Use this construction to find the ratio of the volume
of an octahedron to the volume of a tetrahedron with the same
edge length as the octahedron.
(b) Build eight tetrahedra and one octahedron so that all polyhedra
have the same edge length. On each face of the octahedron,
affix a tetrahedron. What Platonic solid do the exposed vertices
form? The resulting figure is called a stella octangula, and may
be thought of as two large tetrahedra intersecting in a common,
smaller octahedron.
36
CHAPTER 2. THE PLATONIC SOLIDS
10. A tetrahedron may be cut into two congruent parts by a plane parallel
to and midway between a pair of opposite edges of the tetrahedron
(see Figure 2.6). Build two of these pieces and arrange them to form
a tetrahedron.
Figure 2.6
11. A cube may be cut into two congruent parts by a plane perpendicularly
bisecting a long diagonal of the cube (see Figure 2.7). Build two of
these pieces and arrange them to form a cube.
Figure 2.7
Figure 2.8
12. A cube may be inscribed in a dodecahedron as in Figure 2.8. Thus,
we see that a dodecahedron may be formed by affixing six “roofs” on
the faces of a cube (one such roof is shown in a darker orange). Build
a cube and six roofs, and arrange them to form a dodecahedron.
Chapter 3
Spherical Trigonometry
3.1
Spherical Triangles
What is a spherical triangle? Succinctly put, a spherical triangle is a
triangle on the surface of a sphere, the sides of which are arcs of great circles
of the sphere. We find examples of great circles on a sphere by considering
lines of longitude and the equator on a spherical globe. We may trace out a
spherical triangle on such a globe by beginning at the North Pole, following
0◦ of longitude to the equator, travelling east until we hit 30◦ of longitude,
and following 30◦ of east longitude north back to the North Pole, as is shown
in Figure 3.1.
30◦
Figure 3.1
Now in the plane, we measure the six “parts” of a triangle by measuring
the lengths of the sides (relative to some standard unit), and the measures
of the angles between adjacent sides. The situation is somewhat different
for spherical triangles, where all “parts” are, in fact, angles.
37
38
CHAPTER 3. SPHERICAL TRIGONOMETRY
Recall that each of the sides of a spherical triangle is an arc of a great
circle – and hence can readily be measured in degrees relative to that great
circle (which has the same radius as the sphere). As for the angles between
adjacent sides of a spherical triangle – called vertex angles – we note that
a great circle of a sphere may be imagined as the intersection of a plane
passing through the center of the sphere and the surface of that sphere.
Thus two adjacent sides, being arcs of great circles, determine two great
circles, which in turn determine two planes passing through the center of
the sphere. The angle between these two planes, then, is the vertex angle of
the spherical triangle between these two adjacent sides. The reader should
take a moment to verify that the sides of the spherical triangle in our initial
example have measures 30◦ , 90◦ , and 90◦ , and that the three vertex angles
have measures 30◦ , 90◦ , and 90◦ as well (see Figure 3.1). This coincidence
is an accident of this particular example, and will not occur in most of the
subsequent examples.
The reader may have noticed that the measures of the vertex angles of
the foregoing spherical triangle sum to 90◦ + 90◦ + 30◦ > 180◦ – unusual in
that we are used to the angles of a plane triangle summing to precisely 180◦
regardless of the shape of the triangle. In fact, the sum of the vertex angles
of a spherical triangle is always greater than 180◦ . If we denote this sum
1
◦
by Σ, we find, moreover, that 720
◦ (Σ − 180 ) is exactly the fraction of the
surface of the sphere occupied by the spherical triangle. Thus, our triangle
1
1
◦
◦
occupies 720
◦ (210 − 180 ) = 24 of the surface of the sphere, a fact which
may be verified by studying the geometry in Figure 3.1.
Although this phenomenon will be addressed more thoroughly in Chap1
◦
ter 13, we note one consequence here. Since 720
◦ (Σ − 180 ) is the fraction
of the surface of a sphere occupied by a spherical triangle, we must have
0<
1
(Σ − 180◦ ) < 1,
720◦
from which it readily follows that 180◦ < Σ < 900◦ . In other words, not
only must the vertex angles of a spherical triangle sum to more than 180◦ ,
they must also sum to less than 900◦ .
For our purposes, we shall always assume that the vertex angles of a
spherical triangle are less than 180◦ in measure. It is also possible to specify
that the three arcs described above determine a large triangle enclosing the
◦
◦
◦
other 23
24 of the sphere whose vertex angles are 330 , 270 , and 270 . The
necessity of always making such distinctions is avoided with this simplifying
assumption.
3.2. FUNDAMENTAL RELATIONSHIPS
3.2
39
Fundamental Relationships
We now wish to discover a few of the relationships between the various
angles of a spherical triangle. To this end, consider a spherical triangle as
in Figure 3.2, where O is the center of the sphere on which lie p, q, and r.
Let p0 be any point on Op, and select q 0 on Oq and r0 on Or so that both
p0 q 0 and p0 r0 are orthogonal to Op0 .
c
q
r
B
a
A
C
b
p
p0
r0
q0
O
Figure 3.2
Figure 3.3 shows an “unfolded” version of this spherical wedge formed
by O, p, q, and r. Since unfolding Figure 3.2 makes “∆Oq 0 r0 ” ambiguous,
we will mean by ∆Oq 0 r0 the triangle containing the angle c (rather than the
triangle containing a + b).
To make our calculations easier, let us assume that [Op0 ] = 1. Then
by considering right triangles ∆Op0 q 0 and ∆Op0 r0 , we have the following
relationships:
[Op0 ] = 1,
[p0 q 0 ] = tan b,
[Oq 0 ] = sec b,
[p0 r0 ] = tan a,
We see by examining ∆Oq 0 r0 that
[q 0 r0 ]2 = [Or0 ]2 + [Oq 0 ]2 − 2[Or0 ][Oq 0 ] cos c,
[Or0 ] = sec a.
(3.1)
40
CHAPTER 3. SPHERICAL TRIGONOMETRY
where substitution of the various values from (3.1) yields
[q 0 r0 ]2 = sec2 a + sec2 b − 2 sec a sec b cos c.
(3.2)
p
r
q
q0
p0
r0
r
r0
b a
c
O
Figure 3.3
Now if we imagine slicing the spherical wedge by a plane which passes
through points p0 , q 0 , and r0 , we would expose ∆p0 q 0 r0 . Since p0 q 0 and p0 r0 are
both orthogonal to Op, this slicing plane is orthogonal to Op, and therefore
∠q 0 p0 r0 = C (which is the vertex angle between the sides pq and pr in Figure
3.2). Considering ∆p0 q 0 r0 results in
[q 0 r0 ]2 = [p0 r0 ]2 + [p0 q 0 ]2 − 2[p0 r0 ][p0 q 0 ] cos C,
into which we may substitute values from (3.1) to yield
[q 0 r0 ]2 = tan2 a + tan2 b − 2 tan a tan b cos C.
(3.3)
Substituting the value for [q 0 r0 ]2 from (3.2) into (3.3) results in
sec2 a + sec2 b − 2 sec a sec b cos c = tan2 a + tan2 b − 2 tan a tan b cos C.
Rearranging terms yields
2 sec a sec b cos c = (sec2 a − tan2 a) + (sec2 b − tan2 b) + 2 tan a tan b cos C.
3.3. EDGE ANGLES OF PLATONIC SOLIDS
41
Using the fact that sec2 a − tan2 a = sec2 b − tan2 b = 1 and then dividing by
2 gives
sec a sec b cos c = 1 + tan a tan b cos C.
Finally, multiplying through by cos a cos b results in
cos c = cos a cos b + sin a sin b cos C.
(3.4)
Of course, the same derivation could be used to find formulas for cos a or
cos b; we would find that
cos a = cos b cos c + sin b sin c cos A,
cos b = cos a cos c + sin a sin c cos B.
(3.5)
There are three other relationships among the various angles of the spherical triangle ∆abc which will also be important to us, but whose derivation
is not so straightforward. Thus, we include the results but omit their derivation here.
cos A = − cos B cos C + sin B sin C cos a,
cos B = − cos A cos C + sin A sin C cos b,
(3.6)
cos C = − cos A cos B + sin A sin B cos c.
3.3
Edge Angles of Platonic Solids
We now wish to apply our results from spherical trigonometry to the derivation of angular properties of the Platonic solids. Our first task is to calculate
the “edge angle” of the regular icosahedron; that is, the angle subtended by
an edge when its endpoints are connected to the center of the polyhedron
(see Figure 3.4). To do this, we first imagine the icosahedron to be inscribed
in a sphere. Three vertices of a triangular face will lie on the sphere, which
we may imagine as the vertices of a spherical triangle. We think of centrally
projecting a face of the icosahedron onto the sphere to obtain this triangle. Note that all edge angles are equal as are all vertex angles due to the
symmetry of the icosahedron.
42
CHAPTER 3. SPHERICAL TRIGONOMETRY
vertex of
icosahedron
a
vertex of
icosahedron
A
a
A
a
A
vertex of
icosahedron
center of icosahedron
Figure 3.4
We first observe from (3.6) that we have the relationship
cos A = − cos A cos A + sin A sin A cos a,
which we may write as
cos A = − cos2 A + sin2 A cos a.
Using the fact that sin2 A = 1 − cos2 A, we may solve this equation for cos a,
yielding
cos a =
cos A(1 + cos A)
cos A(1 + cos A)
cos A
=
=
,
2
1 − cos A
(1 − cos A)(1 + cos A)
1 − cos A
remarking that the factors of “1 + cos A” could be cancelled as cos A 6= −1
– that is, A is not 180◦ (a “straight” angle).
What is A? By looking at five of these spherical triangles “from the top”,
we see (as in Figure 3.5) that five of the angles A comprise one complete
revolution, so that A = 72◦ . Using Table 1.1, we find that
cos a =
cos 72◦
τ −1
1
1
=
= (2τ − 1) = √ .
◦
1 − cos 72
3−τ
5
5
3.3. EDGE ANGLES OF PLATONIC SOLIDS
43
A A
A A
A
Figure 3.5
−1
Now how was τ3−τ
reduced to 51 (2τ − 1) above? Rather than digress too
far off task, this discussion is postponed until §3.7, where all the details are
presented.
√
Continuing our discussion of the icosahedron, we saw that cos a = 1/ 5,
so that a ≈ 63.4◦ . Note that a is slightly larger than 60◦ , and thus the tetrahedron formed by a face of the icosahedron and the center of the icosahedron
is not quite regular.
We call the angle a just obtained E3,5 to symbolize that it is an edge
angle of the icosahedron. Similarly, we denote by Ep,q the edge angle of the
Platonic solid {p, q} as enumerated in Table 2.1. It is left to the reader to
verify that the procedure used to find E3,5 may be used to find E3,3 and E3,4 ,
resulting in
cos 120◦
1
cos E3,3 =
=−
1 − cos 120◦
3
and
cos 90◦
cos E3,4 =
= 0.
1 − cos 90◦
We now wish to tackle the task of finding E5,3 . Since the faces of the
dodecahedron are not triangles themselves, we “create” a spherical triangle
by projecting the center of a pentagonal face onto the circumscribing sphere.
Since the ends of the edges (i.e., the vertices of the dodecahedron) lie on the
sphere, the spherical triangle ∆rst in Figure 3.6 is obtained. The angle A
has measure 72◦ (see Figure 3.5), and the angles across from A have measure
60◦ as six such equal angles comprise a complete revolution about a vertex
of the dodecahedron when projected onto the sphere.
We may find E5,3 using (3.6), yielding
cos A = − cos 60◦ cos 60◦ + sin 60◦ sin 60◦ cos E5,3 .
44
CHAPTER 3. SPHERICAL TRIGONOMETRY
t
E5,3
60◦
60◦
s
A
r
Figure 3.6
Since cos 60◦ =
1
2
and sin 60◦ =
√
3
2 ,
cos E5,3 =
we may rewrite this expression as
4 cos A + 1
.
3
Since A = 72◦ , we find using Table 1.1 that
cos E5,3
√
2τ − 1
5
=
=
.
3
3
We may likewise find cos E4,3 by similar methods, the major difference
being that A = 90◦ rather than A = 72◦ . This results in
1
cos E4,3 = .
3
Details are left to the Exercises.
It may be remarked that the methods used to find E5,3 and E4,3 may
be used to find E3,3 , E3,4 , and E5,3 as well. This results in the following
generalization: if the Platonic solid has faces with p edges and there are q
such faces about each vertex, then the edge angle Ep,q is given by
cos Ep,q =
2π
cos 2π
p + cos q
sin2 πq
=
2π
1 + cos 2π
q + 2 cos p
1 − cos 2π
q
=2
1 + cos 2π
p
1 − cos 2π
q
− 1, (3.7)
where angles are given in radians. It is customary when presenting such
formulas to use radian measure for angles; this custom will be adopted in
the text.
Details are left to the Exercises.
3.4. CIRCUMRADII OF PLATONIC SOLIDS
3.4
45
Circumradii of Platonic Solids
Due to its symmetry, it happens that each Platonic solid may be inscribed
in a sphere. Such a sphere is called the circumsphere of a Platonic solid,
and its radius, the circumradius. We now seek to find the circumradii of
the Platonic solids. Of course, one may imagine that larger Platonic solids
have larger circumspheres, and consequently greater circumradii. It has
become common, therefore, to “fix” the size of the Platonic solids in order
to calculate their circumradii. The usual convention is to assume that the
edges of the Platonic solids have length 2. We adopt this convention here.
m
r
ρ
s
ρ
E E
2 2
O
Figure 3.7
Happily, the calculation of the edge angles in the previous section will
come in quite handy. Take, for example, Figure 3.7. Here, O is the center
of the Platonic solid, and r and s are adjacent vertices, so that ∠rOs is
the edge angle E of the polyhedron and ρ its circumradius. (We write “E”
rather than “Ep,q ” here as some of the formulas will be valid even if E does
not correspond to the edge angle of a Platonic solid, but rather some other
polyhedron.) Also, m is the midpoint of rs, so that [rm] = [ms] = 1 and
Om bisects E. It is easy to see in the right triangle ∆Oms that
1
1
sin E = .
2
ρ
Since we have data concerning cos E already, we employ the appropriate
trigonometric identity to obtain
r
1
2
ρ = csc E =
.
(3.8)
2
1 − cos E
46
CHAPTER 3. SPHERICAL TRIGONOMETRY
Hence knowing one of ρ and E, we may easily find the other using (3.8).
Data are summarized for reference in Table 3.1 at the end of §3.5. Note
that for typographical clarity, ρ2 is included in the table rather than ρ.
3.5
Dihedral Angles of Platonic Solids
A dihedral angle of a polyhedron is an angle between two faces of a polyhedron. We now endeavor to calculate these angles for all of the Platonic
solids. Again, spherical trigonometry will play an important role.
108◦
D5,3
b
D5,3
c
108◦
D5,3
108◦
d
a
Figure 3.8
We begin with the dodecahedron. We alter our attack slightly by considering a vertex of the dodecahedron as the center of a sphere on which lies
our spherical triangle, so that in Figure 3.8, a is a vertex of the dodecahedron (and center of the sphere), while b, c, and d are the three vertices of
the dodecahedron adjacent to a. Note that all edge angles are equal, as are
all vertex angles of this spherical triangle. The edge angles have measure
108◦ , the angle in a regular pentagon, and the dihedral angle is represented
by D5,3 , the dihedral angle of a dodecahedron. Using (3.4), we find that
cos 108◦ = cos 108◦ cos 108◦ + sin 108◦ sin 108◦ cos D5,3 .
Solving for cos D5,3 yields
cos D5,3 =
1 − 2τ
1
= −√ .
5
5
As before, spherical trigonometry may be used to calculate D3,3 , D3,4 ,
D3,5 , and D4,3 as well. As before, we wish to offer a generalization which includes all possibilities. It may be shown that given a spherical polygon with
3.5. DIHEDRAL ANGLES OF PLATONIC SOLIDS
47
q edges, all of which are equal to some angle ϕ, then with the assumption
that all vertex angles are equal to D, we obtain the following formula:
cos D = 1 − 2
1 + cos 2π
q
1 + cos ϕ
.
(3.9)
A proof is given in the Exercises.
Now let us consider a Platonic solid whose faces have p sides, q of which
meet at a vertex, and whose dihedral angle is Dp,q . If we proceed as with
the dodecahedron, we obtain a spherical polygon with q edges of measure
ϕ, where ϕ represents the angle of a face of the Platonic solid. Of course,
the vertex angles of this polygon have measure Dp,q .
We recall that in a polygon of p sides, the sum (in radian measure) of
the angles of the polygon is (p − 2)π. When the polygon is regular, each of
the p angles has the same measure, so that
ϕ=
1
2π
(p − 2)π = π −
.
p
p
Now
2π
2π
2π
2π
cos ϕ = cos π −
= cos π cos
+ sin π sin
= − cos ,
p
p
p
p
(3.10)
so that substituting this value in (3.9) yields
cos Dp,q = 1 − 2
1 + cos 2π
q
1 − cos 2π
p
.
(3.11)
A summary of edge and dihedral angles and circumradii for the Platonic solids is given below (where, for example, we write “D” for “Dp,q ” for
formatting purposes):
Polyhedron
p
q
cos E
E
cos D
tan 21 D
D
ρ2
Tetrahedron
3
3
− 13
109.5◦
1
3
√1
2
70.5◦
3
2
Octahedron
3
4
0
90◦
− 31
109.5◦
2
Cube
4
3
1
3
70.5◦
0
1
90◦
3
√1
5
√
5
3
63.4◦
−
τ2
138.2◦
τ +2
41.8◦
− √15
τ
116.6◦
3τ 2
Icosahedron
3
5
Dodecahedron
5
3
Table 3.1
√
5
3
√
2
48
CHAPTER 3. SPHERICAL TRIGONOMETRY
Several interesting relationships arise out of Table 3.1. In particular, we see
that
E3,3 + D3,3 = π,
E5,3 + D3,5 = π,
E4,3 + D3,4 = π,
E3,5 + D5,3 = π,
E3,4 + D4,3 = π.
These relationships may be summarized as follows: if {p, q} represents a
Platonic solid, then
Ep,q + Dq,p = π.
(3.12)
These are easily seen by observing that two angles, each less than π, are
supplementary only when their cosines have opposite signs. Implicit in these
relationships is the duality between pairs of Platonic solids, which will be
discussed in Chapter 9.
We remark that combining the previous formulas for cos Dp,q and cos Ep,q
results in the following formula relating p, q, Dp,q , and Ep,q :
sin
2π
1
2π
1
π
π
sin Dp,q = sin
cos Ep,q = 2 cos cos .
p
2
q
2
p
q
(3.13)
Details are left to the Exercises.
3.6
Non-Euclidean Geometry
This chapter has so far served as an introduction to the geometry and
trigonometry of triangles on the surface of a sphere. As seen in §3.1, we
encounter some unusual phenomena, such as the fact that the three angles
in a spherical triangle always sum to more than 180◦ . Since our knowledge
of Euclidean plane geometry is not directly applicable, we refer to spherical
geometry as an example of a non-Euclidean geometry.
Most systems of geometry studied by mathematicians are non-Euclidean.
When studying non-Euclidean geometry for the first time, the difficulty lies
with “forgetting” everything you know about Euclidean geometry, for often
very little of it is applicable.
So let’s continue exploring our spherical world. Typically, a mathematical system if called a geometry when it refers to objects called points and
lines, and a relation between points and lines called incidence. What are
points and lines on a sphere?
Points apparently cause no difficulty, but lines? We take some guidance
from physics. On a curved surface, although there are not usually “straight”
3.6. NON-EUCLIDEAN GEOMETRY
49
lines, there is typically a shortest path between two given points. Such a
path is called a geodesic path.
On a sphere, the shortest path between two points is an arc of a great
circle, which would be a geodesic path. Note that there are usually two such
arcs, one measuring less than 180◦ and one measuring more – the geodesic
path is the one measuring less.
y
q
p
G
x
Figure 3.9
Now a geodesic path G has in interesting property: given two points p
and q on G, the shortest path between p and q is contained in G. This should
be clear by thinking about Figure 3.9, where G is the geodesic path between
x and y, and p and q are on G. If there were a shorter path between p and
q, say H, we could then easily find a shorter path between x and y: simply
go from x to p along G, from p to q along H, and then from q to y along
G. But since G is already the shortest path from x to y, the existence of a
shorter path from p to q is not possible.
Let’s return to the plane for a moment to see these ideas in a familiar
context. Since the shortest distance between two points is obtained by the
line segment between those points, then the geodesic paths in the plane are
just the line segments. Also, if p and q lie on the line segment xy (as in
Figure 3.10), then the segment xy contains the segment pq.
x
p
q
y
Figure 3.10
We know that any line segment actually lies along a straight line. In
other words, we may extend a segment in either direction to produce a line.
Of course a line may not be extended any further – it is already as “long”
as possible. A line also has the shortest path property discussed earlier:
given any two points x and y on a line, the segment xy (that is, the geodesic
50
CHAPTER 3. SPHERICAL TRIGONOMETRY
path between x and y) lies along the line. Thus a line is called a maximal
geodesic path: maximal because it cannot be extended any further (that is,
no points may be added) while still maintaining the shortest path property.
This may all seem obvious in the plane, but remember that most geometries are non-Euclidean. So it is vitally important to be able to talk
about lines without using concepts that apply only in a Euclidean realm.
For example, although we may imagine a line as a straight line segment extended infinitely in both directions, we may also think of a line as a maximal
geodesic path. The advantage of the latter description is that it is applicable
in other domains, such as on a curved surface (like a sphere).
So what are maximal geodesic paths on the sphere? Since shortest paths
on a sphere are always arcs of great circles, it follows that maximal geodesic
paths are the great circles themselves. In our geometry, we shall refer to
maximal geodesic paths as lines.
This may seem like a long-winded way of defining a line, but it has the
advantage of being applicable on most curved surfaces usually encountered.
It is now a simple matter to define a line segment as part of a line bounded
by two distinct points. And, as we have seen, a triangle consists of three
line segments, a quadrilateral four, etc. In other words, many of the familiar
definitions may be recast in a non-Euclidean context.
But much is different. Below we highlight several differences between
geometry in the Euclidean plane and geometry on the sphere.
1. Two distinct points do not necessarily determine a line. Just consider
the North and South Poles on a globe. Every line of longitude passes
through these two points. Two such points – that is, points at opposite ends of the diameter of a sphere – are called antipodes. We
must reformulate our result as follows: If two distinct points are not
antipodes, there is exactly one line passing through both points.
2. There are no parallel lines. In other words, every pair of distinct
lines intersects in two antipodal points. This can readily be seen by
observing that two distinct lines (great circles) lie in two distinct planes
passing through the center of the sphere. These planes intersect in a
(straight) line passing through the center of the sphere, and this line
intersects the sphere in a pair of antipodes. These antipodes are the
intersection of the great circles.
3. Lengths of line segments are measured in radians (or degrees). The
lengths of the sides of a spherical triangle (that is, line segments) are
measures of angles. Note that in deriving (3.4) and (3.6), the radius of
3.7. SIMPLIFYING EXPRESSIONS INVOLVING τ
51
the sphere was not needed. We were able to calculate ρ in §3.4 only in
reference to a Platonic solid of edge length 2. Defining a Euclidean arc
length (such as for calculating distances on the surface of the Earth)
requires knowing the radius of the sphere. Most of the important
results we encounter in spherical geometry and trigonometry will be
independent of the radius of the sphere.
4. The sum of the angles in a spherical triangle is always greater than π
(or 180◦ ) and this sum may be different for different triangles. Thus,
knowing two angles of a spherical triangle does not allow for an easy
determination of the third angle. Relationships such as (3.6) are necessary.
5. Similarity and congruence are the same concept. In the Euclidean
plane, two triangles may have the same angles – and hence be similar
– but have different side lengths. But on the sphere, we see from (3.6)
that the vertex angles of a spherical triangle determine the sides of the
triangle. Thus, if two spherical angles have the same vertex angles,
they have the same sides, and hence are congruent.
We will encounter further differences between Euclidean and spherical
geometry in Chapter 4. Remember that most important geometries in
mathematics and physics are non-Euclidean. So it is worth our time to
briefly highlight the major differences between Euclidean and spherical geometry and trigonometry in order to accustom our minds to thinking in
non-Euclidean contexts.
Another short digression on the subject of τ is in order. Indeed, how was
τ −1
1
3−τ reduced to 5 (2τ − 1)? Reciprocation of expressions involving τ must be
discussed.
3.7
Simplifying Expressions Involving τ
So let us assume that we have an expression of the form 1/(rτ + s), where r
and s are rational, and we wish to write this expression in the form uτ +√v,
where u and v are also rational. We may proceed by (1) substituting 1+2 5
√
for τ ; (2) rationalizing the denominator; and (3) substituting 2τ − 1 for 5
and rearranging terms. Beginning with steps (1) and (2), we see that
√
√
1
r + 2s − 5r
r 5 − r − 2s
2
√ √ ·
√ =
=
.
2(r2 − rs − s2 )
r + 2s + 5r r + 2s − 5r
r 1+2 5 + s
52
CHAPTER 3. SPHERICAL TRIGONOMETRY
Substituting
√
5 = 2τ − 1 as in step (3) yields
1
r+s
r
τ− 2
,
= uτ + v = 2
2
rτ + s
r − rs − s
r − rs − s2
so that
r
r+s
,
v=− 2
.
(3.14)
r2 − rs − s2
r − rs − s2
In particular, to find 1/(3 − τ ), we have r = −1 and s = 3. Using the
formulas just derived, we have u = 15 and v = 25 , and hence
u=
1
1
= (τ + 2).
3−τ
5
Thus, to simplify
τ −1
3−τ
τ −1
3−τ ,
=
=
we have
1
1
(τ + 2)(τ − 1) = (τ 2 + τ − 2)
5
5
1
1
1
(τ + 1 + τ − 2) = (2τ − 1) = √ .
5
5
5
Such simplifications will occur frequently, and the reader may duplicate the
above procedure in order to verify them.
3.8
Exercises
1. In the same manner in which it was shown that cos E5,3 =
§3.3, show that cos E4,3 = 1/3.
√
5/3 in
2. Proceed as follows to derive equation (3.9).
We saw in Figure 3.5 how a spherical pentagon may be decomposed
into five spherical triangles. We do the same with a spherical polygon
of q sides, each of measure ϕ, as in Figure 3.11. Note that our subdivision cuts the vertex angle D in half, so that the corresponding vertex
angles have measure D/2 in Figure 3.11.
(a) Apply (3.6) to a triangle in Figure 3.11 to obtain
cos
2π
D
D
+ cos2 = sin2 cos ϕ.
q
2
2
(b) Using the appropriate trigonometric identities, transform the previous equation into (3.9).
3.8. EXERCISES
53
ϕ
ϕ
2π 2π 2π
q
q
q
D
D
ϕ
ϕ
D
D
ϕ
Figure 3.11
3. Proceed as follows to show (3.7); that is, show that if a Platonic solid
has faces with p edges and there are q such faces about each vertex,
then the edge angle Ep,q is given by
cos Ep,q =
2π
cos 2π
p + cos q
sin2 πq
=
2π
1 + cos 2π
q + 2 cos p
1 − cos 2π
q
=2
1 + cos 2π
p
1 − cos 2π
q
− 1.
Our analysis is similar to that of the previous Exercise. The reader
should verify that projecting the Platonic solid onto the sphere and
forming a spherical triangle whose vertices are the center of a spherical
face and the endpoints of an edge of that face results in Figure 3.12.
center of face
(projected
onto sphere)
2π
p
π
q
center of polyhedron
Figure 3.12
Ep,q
π
q
54
CHAPTER 3. SPHERICAL TRIGONOMETRY
(a) Show that applying (3.6) results in
cos
2π
π
π
= − cos2 + sin2 cos Ep,q .
p
q
q
(b) Derive the above formula by using the appropriate trigonometric
simplifications.
4. This Exercise develops another method for calculating ρ.
(a) Using (3.7) and (3.8), show that
sin πq
.
ρ= q
sin2 πq − cos2 πp
(b) Let p, q, and r be related as in Table 3.2.
p
3
3
4
3
5
q
3
4
3
5
3
r
4
3
3
5
2
5
2
Table 3.2
By direct calculation (using data from Table 1.1), show that
cos2
π
π
π
+ cos2 + cos2 = 1.
p
q
r
(3.15)
(c) Use the result from (b) with (a) to show that
ρ = sin
π
π
sec .
q
r
(3.16)
5. We now wish to derive formula (3.13) at the end of §3.5; that is,
sin
2π
1
2π
1
π
π
sin Dp,q = sin
cos Ep,q = 2 cos cos .
p
2
q
2
p
q
(a) Using the identity sin2 D2 =
21
sin
2
1−cos D
,
2
Dp,q =
show from (3.9) that
1 + cos 2π
q
1 − cos 2π
p
.
3.8. EXERCISES
55
(b) In a similar fashion, show from (3.7) that
1 + cos 2π
1
p
cos2 Ep,q =
.
2
1 − cos 2π
q
(c) Show that
2 2π
sin
sin2
and
p
21
sin
2
Dp,q
2π
1
cos2 Ep,q
q
2
2π
2π
1 + cos
,
= 1 + cos
p
q
2π
2π
1 + cos
,
= 1 + cos
p
q
2π
2π
= 1 + cos
1 + cos
,
4 cos cos
p
q
p
q
2π
2π
concluding that
sin2
2π
1
2π
1
π
π
sin2 Dp,q = sin2
cos2 Ep,q = 4 cos2 cos2 .
p
2
q
2
p
q
1
2π
1
π
π
(d) Argue that sin 2π
p , sin 2 Dp,q , sin q , cos 2 Ep,q , cos p , and cos q are
all positive, and hence square roots may be extracted to yield the
desired result.
6. Using the identity sin 2θ = 2 sin θ cos θ, conclude from the previous
exercise that
π
π
1
sin Dp,q sin = cos
2
p
q
and
1
π
π
cos Ep,q sin = cos .
2
q
p
7. By building a geometrical model, show that D3,3 + D3,4 = π.
8. Carry out the following procedure:
(a) Accurately construct a “fan” as in Figure 3.3.
(b) Measure the angles a, b, and c in your particular construction.
(c) Use the data from (b) to calculate C using (3.4).
(d) Cut out your wedge along segments Or0 , r0 q 0 , q 0 r0 , and r0 O; that
is, cut out the quadrilateral bounded by “both” q 0 r0 ’s and “both”
Or0 ’s.
56
CHAPTER 3. SPHERICAL TRIGONOMETRY
(e) Fold along Oq 0 and Op0 and make a wedge.
(f) Place ∆p0 q 0 r0 flat on a piece of paper (so that the apex O is
pointing towards the ceiling) and trace ∠q 0 p0 r0 on your paper.
(g) Measure ∠q 0 p0 r0 . How close did you come to your calculation in
(c)?
9. Imagine a regular tetrahedron being decomposed into four congruent,
smaller tetrahedra, each of which is constructed as follows: choose a
face of the tetrahedron as a base, and join the three vertices of this
triangular face to the center of the tetrahedron. Each face yields a
“squat” tetrahedron. Using the data in Table 3.1, build four such
tetrahedra and assemble them to form a regular tetrahedron. For an
extra challenge, do the construction without the aid of a protractor,
using straightedge and compass alone.
10. Analogous to the construction in the previous exercise, build six square
pyramids which, when assembled appropriately, form a cube. Each
pyramid has as its base a face of the cube and as its apex the center
of the cube. Use data from Table 3.1 as necessary.
11. Consider the spherical pentagon shown in Figure 3.13. Show that as
a shrinks to 0, the pentagon becomes planar in the sense that 2θ gets
closer and closer to 108◦ . In other words, show that
lim cos 2θ = −
a→0
a
θ
θ
72◦
Figure 3.13
1
.
2τ
Chapter 4
Geodesic Structures
The concept of a geodesic dome was first described by Walter Bauersfeld
in Jena, Germany in 1922 and later popularized by R. Buckminster Fuller.
This chapter begins an exploration of the basic mathematics underlying
these fascinating structures.
4.1
A Geodesic Icosahedron
We begin with an example. Let us inscribe an icosahedron in a sphere of
radius ρ. This may be done so that the icosahedron touches the sphere only
at its vertices. If we focus attention on a particular face of the icosahedron,
we see that its three vertices determine a spherical triangle. All we need to
do is join pairs of the vertices by arcs of great circles. (Recall that a great
circle on a sphere is a circle whose center is the center of the sphere, such
as the equator on a spherical globe.) Call the vertices of this triangle p, q,
and r, and call the center of the sphere O.
Now each side of this spherical triangle is an arc of a circle with radius
ρ (the same as the radius of the sphere), and hence we may unfold this triangle as in Figure 4.1, where the outermost arc corresponds to the spherical
triangle and the vertices of the icosahedral face are labelled. Since pq is an
edge of the icosahedron, then ∠pOq has the same measure as E3,5 , which is
approximately 63.435 degrees.
57
58
CHAPTER 4. GEODESIC STRUCTURES
r
q
c
b
E3,5
E3,5
E3,5
p
O
a
p
a
Figure 4.1
We may widen the arc pqrp by including an auxiliary arc abca and a tab
at segment pd so that this thick arc may be cut out, folded along segments
bq and cr, and then glued to form a thick spherical triangle. Since an icosahedron has twenty faces, we need just twenty of these spherical triangles,
glued back-to-back, to form a geodesic icosahedron (as in Figure 4.2).
Figure 4.2
Recall (see §3.6) that given any two points on a sphere, the shortest path
between them (on the sphere) is the arc of a great circle passing through
the points. (The analogue in the plane is that the shortest path between
two points is the line segment joining them.) Since such shortest paths
are geodesics, and since such paths were used to construct our spherical
icosahedron, we describe the resulting structure as “geodesic.”
4.2. A GEODESIC DODECAHEDRON
4.2
59
A Geodesic Dodecahedron
We may apply similar ideas to create a geodesic model of a dodecahedron.
Of course we immediately encounter the fact that the faces of the dodecahedron are pentagons rather than triangles, and although we could work
with spherical pentagons, we shall opt for working with spherical triangles
as they produce a sturdier structure.
To realize the dodecahedron as a structure composed of spherical triangles, we look again to Figure 3.6, where A = 72◦ in our case. Thus, we see
that five triangles such as that in Figure 4.3(a) may be arranged about p to
yield a spherical pentagon corresponding to one face of the dodecahedron
(compare with Figure 3.6). Since five triangles are needed for each of twelve
spherical pentagons, we see that sixty triangles (an unfolded version of one is
shown in Figure 4.3(b)) are required for a finished model (see Figure 4.3(c)).
q
r
E5,3
q
p
60◦
60◦
b
72◦
r
p
b
E5,3
O
O
(a)
(b)
(c)
Figure 4.3
b
p
60
CHAPTER 4. GEODESIC STRUCTURES
We see from the considerations in §3.3 that ∠qOr has measure E5,3 , and
the symmetry of the configuration requires that ∠pOq and ∠rOp have the
same measure, which we shall call b. Of course, to find b is simply a matter
of applying (3.6) once again, yielding
cos 60◦ = − cos 60◦ cos 72◦ + sin 60◦ sin 72◦ cos b,
from which it follows that
√
√
3
τ +2
τ +1 τ +2
1
(τ + 2) 2
√
cos b = p
= √
= √ (τ + 1) (3 − τ ) =
.
5
3 τ +2
3
5 3
3(τ + 2)
τ +1
Thus, to create a triangle as in Figure 4.3(b), we must create angles
E5,3 ≈ 41.8◦ ,
b ≈ 37.4◦ .
Building this model is a good first project for those interested in creating
more complex geodesic structures.
4.3
The 2-frequency Icosahedron
Our next example is based on our first. Consider first a spherical triangle
corresponding to one face of an icosahedron. Then divide each side into
two equal arcs, and connect the midpoints of these arcs by arcs of great
circles, as in Figure 4.4(a). Because of symmetry, we see that the three
“vertex triangles” are congruent, while the central triangle is of a somewhat
different character. Having migrated from plane to spherical geometry, gone
are the days when joining the midpoints of the sides of a triangle yields four
smaller, congruent triangles.
r
72◦
u
t
q
s
p
(a)
(b)
Figure 4.4
4.4. 4-FREQUENCY ICOSAHEDRA
61
Because of the initial arc-bisection construction, it is evident that arcs
tr and ru in Figure 4.4(a) have measure 12 E3,5 , while the dihedral angle at
r still has measure 72◦ . Hence we see from (3.4) that
1
1
cos a = cos2 E3,5 + sin2 E3,5 cos 72◦ .
2
2
Recalling (see (0.16)) that
cos2
θ
1 + cos θ
=
,
2
2
sin2
θ
1 − cos θ
=
,
2
2
we find after substitution and some algebra that
1
1
1
1
τ
1
1+ √
+
1− √
· (τ − 1) = .
cos a =
2
2
2
2
5
5
Thus
1
E3,5 ≈ 31.7◦ , a = 36◦ .
2
This information is sufficient to create the four smaller triangles in Figure
4.4(a), and hence the entire model (shown in Figure 4.4(b)). Note that the
purple spherical triangles are the same as those shown in Figure 4.2, while
the yellow triangles are the ones inscribed inside them.
This structure is called a 2-frequency icosahedron because it is obtained by dividing the sides of an icosahedral spherical triangle into two
congruent arcs. Higher frequency icosahedra are possible, and the reader is
invited to peruse Magnus Wenninger’s excellent Spherical Models for examples of these structures and many others.
4.4
4-Frequency Icosahedra
In this and the next few sections, we begin a discussion of higher frequency
geodesic structures. Many important ideas may be illustrated by considering the 4-frequency icosahedron, and so this structure will be the focus of
attention. Various subtleties present themselves, however, arising from the
differences between plane and spherical geometries.
As an example, consider a division of a plane equilateral triangle as in
Figure 4.5(a). By dividing the sides into equal fourths and drawing segments
pq, rs and tu as in the Figure, we find that they intersect at a common point,
a. This phenomenon allows us to completely subdivide the original triangle
into 16 smaller, congruent equilateral triangles.
62
CHAPTER 4. GEODESIC STRUCTURES
r
u
p
q
a
t
s
(a)
(b)
Figure 4.5
Let us turn now to the spherical analogue in Figure 4.5(b). We divide
the arcs of an equilateral spherical triangle into equal fourths, and join the
analogous points by arcs of great circles. To our dismay, these arcs do
not intersect in common points, but rather form additional, small spherical
triangles (as shown in the Figure). The sides of these small triangles are
approximately 3◦ each, verification of which is left to the Exercises. As a
result, there is no simple way to subdivide the triangles of a spherical icosahedron. (This difficulty was foreshadowed in Figure 4.4, where we observed
that joining the midpoints of the sides of a spherical icosahedral triangle
yielded two different types of triangles.) The issue of the “small triangles”
of Figure 4.5(b) may be addressed in various ways, each yielding a different
subdivision of the spherical icosahedral triangle. Our task in this and the
next few sections will be to consider various such strategies for subdivision.
The first strategy is based on the 2-frequency icosahedron of the previous section. We begin with the 2-frequency subdivision of Figure 4.4, and
further divide each of these four triangles by bisecting the arcs of the sides of
these triangles, as in Figure 4.6. This results in arcs of five different lengths,
labelled in Figure 4.6 with a, b, c, d and e. (In the interest of making the
diagrams simpler, a planar subdivision of an equilateral triangle is used to
represent the subdivided spherical triangle.) For reference in the forthcoming calculations, angles A, B, D, E and F are also labelled. The algebraic
details of the calculations are rather tedious and are not particularly revealing, and so results will be recorded numerically in degrees (for those
desiring to build such a 4-frequency model). Should the reader attempt
to verify these data, it is recommended to record intermediate calculations
4.4. 4-FREQUENCY ICOSAHEDRA
63
with as many decimal places of accuracy as feasible, using your calculator’s
memory capabilities if possible, so as not to introduce significant errors into
the results.
We now proceed to calculate the arcs labelled in Figure 4.6. We note first
that a = 14 E3,5 ≈ 15.859◦ . Then b may be calculated by applying (3.4) to
spherical ∆wqv, noting that the dihedral angle wqv has measure 72◦ (recall
that ∆pqr is an icosahedral spherical triangle). This results in b ≈ 18.486◦ .
Of course c = 18◦ , being half the arc a = 36◦ of Figure 4.4.
r
a
w
a
q
b
a
t
AB
y
u
c
d
e x
d
D F
E
e
c
A
v
a
s
z
p
Figure 4.6
Now B may be found by applying (3.4) to spherical ∆stu, resulting in
B ≈ 63.435◦ (it happens that B has the same measure as E3,5 ). Thus, we
have A = 12 (180◦ − B) ≈ 58.283◦ . Then d may be found by applying (3.4)
to spherical ∆txy, and e by applying (3.4) to ∆txw. We obtain d ≈ 18.699◦
and e ≈ 16.412◦ .
As a result, D + E + F ≈ 184.197◦ (see Figure 4.6), so that the arcs e,
d and e joining w and z do not all lie on the same great circle. This may
easily be verified using (3.4), as all sides of all the small triangles are now
known. One method of remedying this defect is found in Exercise 6, but at
the price of arcs tx and xs no longer lying on the same great circle.
Another method of remedying this defect is, in some sense, to force
D + E + F = 180◦ by taking the arc d and extending it, so that e lies along
the same great circle as d. But here is the problem: the arc e, upon its
extension from d, intersects arc qt at some point different from w, so that qt
64
CHAPTER 4. GEODESIC STRUCTURES
is no longer bisected by an end of e. This situation is illustrated in Figure
4.7(a). A, B, c, d, and F are as before. But since e0 is an extension of
d (we will indicate all objects which change with this approach by using a
superscript 0 or 00), the angle previously labeled D is now actually 180◦ − 2F
(notice the “vertical” angles meeting at x). This means that the angle
previously labeled E is in fact F .
r
a00
w0
a0
q
b0
a0
t
AB
y
c
e0 x
F
d
d
F
u
z
c
e0
G
A
0
00
s
v
a
(a)
p
(b)
Figure 4.7
Using (3.4) and (3.6) as necessary, it may be verified that in this instance,
we have G ≈ 64.466◦ , a0 ≈ 14.545◦ , and a00 = 21 E3,5 − a0 ≈ 17.172. The
reader should take a moment to imagine exactly what spherical subdivision
was just found: nothing more than Figure 4.5(a) – inscribed on a face of the
icosahedron – projected onto the icosahedron’s circumsphere. Note that the
symmetry of the spherical triangle along qu implies that after projection, x
is in fact the midpoint of ts. But there is not sufficient symmetry to force
w0 to be the midpoint of qt. Figure 4.7(b) shows the completed 4-frequency
icosahedron.
These subdivisions of spherical triangles are called Class I subdivisions
in the literature (see the works of Joe Clinton and Hugh Kenner) because
the planar subdivisions upon which the spherical subdivisions are based are
formed by lines parallel to the sides of the original triangle. A discussion of
the case where additional lines are perpendicular to the sides of the original
triangle takes place in Chapter 10.
4.5. ANOTHER 4-FREQUENCY ICOSAHEDRON
4.5
65
Another 4-Frequency Icosahedron
In §4.4, we discussed a 4-frequency icosahedral triangle (see Figure 4.6)
which, were one to construct nets for such a triangle, would require five
different angles (a, b, c, d and e in Figure 4.6). In this section, a method is
described which requires only four different angles – but at the cost of the
arcs analogous to tx and xs of Figure 4.6 no longer lying on the same great
circle. Thus, we move x so as to force d = e, but this requires that x be
moved off arc ts. The calculations involved are heavily trigonometric; this
section may be skipped over without loss of continuity with regard to that
which follows.
We begin with Figure 4.8, which is similar in many respects to Figure
4.6 – similarly labelled components are analogous. For example, arc qw has
measure a in both figures. In addition, n is the center of the spherical ∆pqr,
h is the angle between n and x (recall that x will be moved, but due to the
symmetry of the subdivided triangle, x must lie on arc qn), and f is the
measure of arc pn (and hence also of qn).
r
e
w
n
h
x d
f −h
a
e
l
z
f
q
p
Figure 4.8
Now applying (3.4) to spherical ∆xnl yields
cos d = cos2 h + sin2 h cos 120◦ ,
which may be rewritten as
cos d =
3
1
cos2 h − .
2
2
(4.1)
66
CHAPTER 4. GEODESIC STRUCTURES
Again applying (3.4), this time to spherical ∆wqx, yields
cos e = cos a cos(f − h) + sin a sin(f − h) cos 36◦ .
The cos(f − h) and sin(f − h) terms in this expression may be expanded,
giving
τ
τ
cos e = sin h cos a sin f − sin a cos f + cos h cos a cos f + sin a sin f .
2
2
1
Now a = 4 E3,5 , and f may be found by applying (3.4) to spherical ∆qnp,
yielding
1
2
cos2 f = cos E3,5 + .
3
3
So by defining
τ
τ
A := cos a sin f − sin a cos f, B := cos a cos f + sin a sin f,
(4.2)
2
2
we may write
cos e = A sin h + B cos h,
where A and B are known.
Recall that we were attempting to force d = e, so setting this expression
for cos e equal to that for cos d in (4.1) results in
3
1
cos2 h − = A sin h + B cos h,
2
2
or equivalently,
2A sin h = 3 cos2 h − 1 − 2B cos h.
To obtain an equation which may be solved for cos h, we may square both
sides of the previous equation, substitute sin2 h = 1 − cos2 h, and rearrange
terms, resulting in
9 cos4 h − 12B cos3 h + (4A2 + 4B 2 − 6) cos2 h + 4B cos h + 1 − 4A2 = 0. (4.3)
Now A and B may be determined by (4.2). These values may be substituted into (4.3) so that a numerical solution for cos h satisfying 0 < cos h < 1
may be found by solving (4.3) using a calculator or other appropriate mathematical software. This results in
cos h ≈ 0.985057,
h ≈ 9.917◦ .
(The analogous distance from the center of the icosahedral spherical triangle
to x in Figure 4.6 is approximately 10.812◦ , as can be verified by the reader.)
Now a and b are the same as in §4.4, but values of c, d and e are given by
(compare with §4.4)
c ≈ 18.021◦ ,
d = e ≈ 17.156◦ .
4.6. STILL ANOTHER 4-FREQUENCY ICOSAHEDRON
4.6
67
Still Another 4-Frequency Icosahedron
In this section, we discuss yet another method for building a 4-frequency
icosahedron. This method is one used by Wenninger in Spherical Models
(see, for example, pp. 92–97). Although only an approximation (made by
ignoring the small triangles in Figure 4.5(b)), it has two points in its favor.
First, it may easily be extended to designing geodesic models at higher frequencies. Secondly, models built using such approximations yield excellent
results. (Wenninger has used this method to build 10-frequency geodesic
icosahedra with 2000 spherical triangles!) It seems that the errors seem to
cancel each other out (although more work should be done to see exactly
why!).
We pause for a moment to introduce a few relationships between parts
of spherical triangles. These are not relationships we will need to use very
often, but they are useful here. A proof is outlined in the Exercises. With
the spherical parts as labelled in Figure 3.2, we have
sin A
sin B
sin C
=
=
,
sin a
sin b
sin c
(4.4)
sin a
sin b
sin c
=
=
.
sin A
sin B
sin C
(4.5)
or, equivalently,
We implicitly assume that all angles are between 0◦ and 180◦ to avoid any
possibility of division by zero.
Now the idea is as follows. Consider an icosahedral spherical triangle
whose sides are divided into equal fourths (as in Figure 4.9(a)). Subdivide
this triangle by arcs of great circles as suggested in Figure 4.5; in our case,
join p and q, r and s, t and u, and the analogous remaining arcs. Our
idealization is as follows: assume that three arcs intersect at points such
as v in Figure 4.9 (rather than form a small triangle as in Figure 4.5(b)),
and assume that such intersections divide these arcs into equal pieces. For
example, arc rs would be cut in half, while arc tu would be divided by equal
thirds by these intersections.
We use these idealizations as illustrated in Figure 4.9(b) in order to find
d. Here, x is the midpoint of arc tu, so that the arc of the great circle joining
w and x forms a right angle at x. Note that since ∠twu = 72◦ , we see that
∠xwu = 36◦ . Hence, we may apply (4.5) to spherical ∆wxu to obtain
sin 3a =
sin 3a
sin f
=
,
◦
sin 90
sin 36◦
68
CHAPTER 4. GEODESIC STRUCTURES
so that sin f = sin 36◦ sin 3a. Since a = 14 E3,5 ≈ 15.859◦ , we may solve this
equation for f . Now it is clear from Figure 4.8(b) that 2f = 3d, so that
d = 23 f . In this case, we find that d ≈ 17.143◦ . Using this method to find
b and c, we get b ≈ 18.486◦ and c = 18◦ . This technique may be applied
to a subdivision of a spherical icosahedral triangle of arbitrary frequency, as
seen in the Exercises.
w
a
b
p
q
a
c
r
c
s
a
t
d
d
d
u
w
a
36◦
(a)
3a
t
d
x
d
f = 32 d
d
u
a
(b)
Figure 4.9
We again emphasize that the data obtained by this technique are not
quite exact, but rather are approximations which are sufficiently accurate
for the purpose of building paper geodesic models.
4.6. STILL ANOTHER 4-FREQUENCY ICOSAHEDRON
§4.4
§4.5
§4.6
a
15.859◦
15.859◦
15.859◦
b
18.486◦
18.486◦
18.486◦
c
18◦
18.021◦
18◦
d
18.699◦
17.156◦
17.143◦
e
16.412◦
17.156◦
17.143◦
69
Table 4.1
For easy reference, with angles as labelled in Figure 4.10, a chart of
values for those angles comparing the different methods of solution is given
in Table 4.1. Of course, there are even more ways to subdivide a spherical
icosahedral triangle; one is given in Exercise 6, and the interested reader
is referred to Wenninger’s Spherical Models (or Chapter 10 of the current
treatise) for still others. Or perhaps the creative reader will devise a yet
undiscovered method....
a
a
b
a
e
e
a
c
a
c
e
a
b
a
e
a
c
d
d
d
c
c
c
a
e
e
a
Figure 4.10
a
b
a
70
4.7
CHAPTER 4. GEODESIC STRUCTURES
Non-Euclidean Considerations
Again, we take a brief moment to review some differences between Euclidean
and spherical geometries.
In the Euclidean plane, all equilateral triangles are similar and vary only
in the length of their sides. On the sphere, as the sides of an equilateral
triangle get larger, so do the angles between the sides. As the sides get
smaller, tending to 0◦ , the angles get closer and closer to 60◦ .
This difference is clearly seen in Figure 4.4. Both ∆pqr and ∆stu are
equilateral spherical triangles. Since ∆stu is the smaller of the two triangles,
its angles must be less that 72◦ . Moreover, regardless of the size of the larger
equilateral triangle, the three triangles at the vertices (such as ∆rtu) can
never be equilateral.
This is easily seen by referring to (3.6). Note that once the angles of
an equilateral triangle are determined, so are the sides: simply solve the
first equation of (3.6) for cos a. Referring back to Figure 4.4, we see that
if an equilateral triangle has angles of 72◦ , the sides have measure E3,5 (the
measure of arc qr). Hence a side length of tr (which is 12 E3,5 ) cannot produce
an equilateral triangle ∆rtu.
The consequences of this observation have been seen throughout the
entire chapter: in subdividing an equilateral spherical triangle, nothing can
be taken for granted. Symmetry must be reconsidered in each case. All new
angles and sides must be calculated.
We saw in Figure 4.5 another important difference between Euclidean
and spherical geometry. The subdivision of an equilateral triangle in Figure
4.5(a) is familiar. But the analogous subdivision on the sphere – that is,
subdividing the sides of an equilateral triangle into equal parts and joining
parts by arcs of great circles – does not result in a neat subdivision, but
produces small triangular “windows” as shown in Figure 4.5(b).
As we saw throughout this chapter, this phenomenon wreaks havoc with
trying to find a high-frequency geodesic subdivision of a spherical triangle.
Recall that the rather simple method described in §4.6 is only an approximation, but well-suited to model building as human beings can only cut,
score, fold, and glue so accurately. In constructing a geodesic house or
other structure, however, such approximations cannot be used. Computer
programs may be written to calculate angles to great accuracy – accuracy
precise enough for large-scale work. For in engineering applications, it is
frequently desired to obtain as few differently sized arcs as possible – and
we saw in §4.5 the need for numerically finding roots of a cubic equation
to find a subdivision for a 4-frequency icosahedron using only four different
4.8. EXERCISES
71
arcs (rather than the five of §4.4). Of course, only four different arcs were
required in §4.6, but this section produced only an approximate subdivision. The subdivisions in §4.4 and§4.5 were based on exact computations
using formulas of spherical trigonometry (although for calculation purposes,
approximations were found).
Thus finding precise values for high-frequency subdivisions of spherical
triangles is not for the mathematically timid. It is here that one comes
face-to-face with the consequences of working with non-Euclidean geometry.
4.8
Exercises
1. Build a geodesic icosahedron with the template provided.
2. Build a geodesic dodecahedron with the template provided.
3. Design and construct a 2-frequency icosahedron.
4. This exercise lays the groundwork for the following two. Here, ∆pqr
is a spherical icosahedral triangle whose sides are divided into equal
fourths. st, tu and uv are arcs of great circles, with st and uv intersecting at w. The goal of this exercise is to determine values of all the
labelled angles in Figure 4.11. The reader may attempt to calculate
these on his or her own, or proceed according to the following outline.
r
72◦
s
x
a
f = ts
c
w
D
E
a
c
b
D
◦
q 72 B A
a
a
t
u
v
y
Figure 4.11
72◦ p
72
CHAPTER 4. GEODESIC STRUCTURES
(a) Find f , and then A, by applying (3.4) to spherical ∆pst.
(b) Find b, and then B, by applying (3.4) to spherical ∆utq.
(c) Knowing A and B at t, calculate D.
(d) Use (3.6) applied to spherical ∆uwt in order to find E, and use
either (3.4) or (3.6) with this same triangle in order to find c.
5. We begin with Figure 4.11 and join x rand y by an arc of a great circle,
as shown in Figure 4.12(a). As foreshadowed in §4.4, this arc does
72◦rather creates a small triangle
not intersect arcs st and uv at w, but
shown in Figure 4.12(a) and enlarged in Figure 4.12(b). The symmetry
of the construction in Figure 4.12(a) should convince the reader that
s
this small triangle is indeed isosceles. Here, we calculate the remaining
angles and sides of this triangle. The following outline may be used.
x
a
c
u
a
q
b
72◦
B
a
D
E
w
z
g G
A
G
k
l
F
a
t
h
y
(a)
v
w
E
h=c−g
G
z
p
(b)
Figure 4.12
(a) Recall that arc xy has measure 36◦ . Use (3.4) with spherical
∆xyq to calculate F . (Alternatively, note that F is the same
angle as A in Figure 4.6.)
(b) Use (3.6) with spherical ∆tzy in order to find G. Then apply
(3.6) again to determine g. (Note that g is less that c of Figure
4.10; it is for this reason that we know that arc xy intersects arc
tw rather than arc ws.)
(c) Note that h = c − g. Check this calculation by using (3.6) with
the triangle in Figure 4.11(b) to find h. How close is your result?
4.8. EXERCISES
73
(d) Finally, use (3.6) with this same triangle in order to calculate k.
(e) To confirm the result in (d), find l by applying (3.4) to ∆zty, and
note that k + 2l = 36◦ .
6. In this Exercise, a 4-frequency subdivision of an icosahedral spherical
triangle is formed by appropriately completing Figure 4.11 (see Figure
4.13).
(a) Find d, the measure of arc wz, as shown in Figure 4.13.
(b) Similarly, find g, the measure of arc wy.
(c) In §4.4, we saw that the arcs analogous to xw and wy laid along
the same great circle (see arcs tx and xs of Figure 4.6). In Figure
4.13, this would be equivalent to the statement 2H + E = 180◦ .
Recall, however (see the end of §4.4), that since we joined u
and v by a single arc of a great circle, we “sacrificed” this ideal.
Calculate H, and determine the extent of the sacrifice.
s
j
z
x
d
H w
E
H
u
t
v
g
y
k
Figure 4.13
7. Prove the relationships in (4.4), using the following outline if desired.
Begin with the following two versions of (3.4):
cos a − cos b cos c = sin b sin c cos A,
cos b − cos a cos c = sin a sin c cos B.
74
CHAPTER 4. GEODESIC STRUCTURES
(a) Square both sides of each equation, and solve both equations for
2 cos a cos b cos c.
(b) By setting the two expressions obtained in (a) for 2 cos a cos b cos c
equal to each other and replacing all cosines by sines using the
relationship cos2 θ = 1 − sin2 θ, deduce that
sin2 A
sin2 B
=
.
2
sin a
sin2 b
(c) With the assumption that all angles are between 0◦ and 180◦ ,
extract square roots to obtain part of (4.4).
(d) Without repeating these calculations, complete a proof of (4.4).
8. Use the techniques described in §4.6 to design a 6-frequency icosahedron. (Hint: You should obtain angles such as 10.572◦ , 12.382◦ ,
12.241◦ , 12◦ , 11.649◦ and 11.177◦ .)
Chapter 5
The Archimedean Solids
5.1
Truncation
The process of truncation – that is, “chopping off” vertices of a polyhedron
to produce a different polyhedron – results in many interesting polyhedra
when applied to the Platonic solids. Such polyhedra are the focus of this
section.
—Partial truncation.
The first method of truncation results in the edges of the original Platonic solid remaining partially intact. Such “partial” truncation, applied to
a tetrahedron, is illustrated in Figure 5.1. As long as we truncate “evenly”
– that is, the plane of truncation cuts off equal segments from all the edges
which it intersects – we find that four small regular tetrahedra have been
removed from the original tetrahedron, leaving equilateral triangles where
the vertices once were. Note also that the truncation process transforms the
triangular faces of the original tetrahedron into irregular (although equiangular) hexagons.
Figure 5.1
75
76
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
A moment’s thought suggests that if we truncate our tetrahedron a little
further down – that is, cut off slightly larger tetrahedra at the vertices – then
the resulting hexagons are indeed regular (see Figure 5.2). In other words, we
have created a polygon with regular polygons as faces (in this case, triangles
and hexagons) such that at each vertex are incident exactly two hexagons
and one triangle. Such a polyhedron – that is, a convex polyhedron whose
faces are regular polygons with the same set of polygons meeting at each
vertex – is called an Archimedean solid.1
Figure 5.2
The polyhedron in Figure 5.2 is referred to as the truncated tetrahedron. We note that the truncation process divided the edges of the original
tetrahedron into equal thirds. The truncation process which transforms the
octahedron and icosahedron into Archimedean solids also divides the edges
of the original polyhedra into equal thirds, yielding the truncated octahedron and the truncated icosahedron (see Figure 5.3). Since the octahedron and icosahedron have four and five triangles, respectively, meeting
at a vertex, truncation yields squares and pentagons rather than triangles
where the vertices of these polyhedra once were. Note that the truncated
icosahedron has the same arrangement of pentagons and hexagons found on
a soccer ball.
1
This definition is not quite precise; a more accurate definition of an Archimedean solid
would involve the notion of a symmetry group, upon which we shall not elaborate at this
point.
5.1. TRUNCATION
77
(a)
(b)
Figure 5.3
As far as a partial truncation of the cube goes, we see that each vertex
of the cube becomes an equilateral triangle as three squares meet at each
vertex of the cube. However, regular octagons are not created in the square
faces by dividing
the edges of the cube into equal thirds, but rather the
√
ratios 1 : 2 : 1 (see Figure 1.7(a)). The truncated cube is shown in
Figure 5.4.
Figure 5.4
What of the truncated dodecahedron? Again, equilateral triangles
replace the vertices of the dodecahedron (since three pentagons meet at a
vertex), and regular decagons replace the pentagonal faces. As with the
cube, a division of the edges of a regular pentagon into equal thirds does
not result in a regular decagon (although it does result in the decagon in
Figure 1.6(b)). To see how the edges must be divided, consider the triangle
in Figure 5.5(b), which is one of the corners “cut off” of a pentagonal face
of the dodecahedron in the truncation process. The cosine law for triangles
reveals that
√
3+ 5
2
2
2
◦
= 1 + τ = τ 2,
s = 1 + 1 − 2 · 1 · 1 · cos 108 =
2
78
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
so that s = τ . Thus, dividing the edges of a pentagon in the ratios 1 : τ : 1
results in a regular decagon.
1
s
s
1
1
1
s
(a)
(b)
(c)
Figure 5.5
—Complete truncation.
A truncation process which bisects the edges of a Platonic solid yields
a complete truncation of that polyhedron – complete in the sense that
nothing remains of the original edges of the solid except their midpoints.
Because of the great symmetry of the Platonic solids, complete truncation of one of them invariably yields an Archimedean solid.
Our first example is the tetrahedron. Complete truncation results in slicing off tetrahedra whose edge length is half that of the original tetrahedron
(see Figure 5.6(a)). The result is nothing more than an octahedron! This
5.1. TRUNCATION
79
comes not so much as a surprise when we realize that complete truncation
replaces the triangular faces of the original tetrahedron with not hexagons,
but rather smaller triangles (see Figure 5.6(b)). Of course, the vertices of the
original tetrahedron are still replaced by triangles, making eight triangles in
all, and hence the familiar octahedron.
(a)
(b)
Figure 5.6
Let us now examine a complete truncation of the cube. Again, the vertices are replaced by equilateral triangles, but the square faces are replaced
with smaller squares (see Figure 5.7(b)) rather than octagons. The resulting
polyhedron (see Figure 5.7(a)) is called the cuboctahedron.
(a)
(b)
Figure 5.7
80
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
What of the complete truncation of the octahedron? As with the partially truncated octahedron (see Figure 5.3), vertices of the original octahedron are replaced by squares, but the faces are replaced by smaller triangular
faces (as in Figure 5.6(b)). The result is, remarkably, the cuboctahedron!
(See Figure 5.8.) This “coincidence” is closely related to the subject of
duality, which shall be explored in Chapter 9. This also gives the cuboctahedron its name, its faces being the six smaller squares inscribed in the
faces of a cube, together with the eight smaller triangles inscribed in the
faces of an octahedron.
Figure 5.8
Given the foregoing discussion, perhaps it is not so surprising that complete truncation, when applied to the icosahedron and dodecahedron, yields
the same polyhedron – aptly called the icosidodecahedron2 (see Figure
5.9). Again, we shall return to a discussion of this polyhedron in our discussion of duality.
Figure 5.9
2
If “icosi” is the Greek prefix for 20, why is “icosahedron” spelled with an “a”? It
happens that “icosihedron” would, in Greek, result in two consecutive “e” sounds, and
hence the prefix is modified to “icosa” in this case.
5.2. TRUNCATION DEFORMATION
5.2
81
Truncation Deformation
We were introduced to two new polyhedra in our discussion of complete
truncation of the Platonic solids – the cuboctahedron and the icosidodecahedron. Let us take a moment to indulge a geometrical whim to apply the
process of partial and complete truncation to these two polyhedra.
We begin with the cuboctahedron. Since the cuboctahedron has eight
triangular faces, we first attempt to partially truncate so as to divide the
edges of the cuboctahedron into equal thirds – thus leaving regular hexagons
in place of the triangular faces (see Figure 5.10(a)). However, in so doing,
we find that the octagons thereby produced are not regular, and the vertices
of the cuboctahedron have been
√ replaced by rectangles, the lengths of the
sides of which are in the ratio 2 : 1 as can be seen by unfolding a vertex
of the cuboctahedron as in Figure 5.10(b). Even though the vertices of the
resulting polyhedron are regular in that precisely one rectangle, one hexagon,
and one octagon meet at each vertex, the faces themselves are not regular,
and thus the solid is not Archimedean. As the truncation of the vertices of
the cuboctahedron will always
√ result in a rectangle, the lengths of the sides
of which are in the ratio 2 : 1, no partial or complete truncation of the
cuboctahedron can produce an Archimedean solid.
(a)
(b)
Figure 5.10
Happily, all is not lost. In this case, it is possible to deform our partially
truncated cuboctahedron into an Archimedean solid by moving the regular
hexagons in slightly toward the center of the polyhedron, thereby scrunching
the rectangles into squares and creating regular octagons in the process. The
result is the polyhedron in Figure 5.11, often called the rhombitruncated
82
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
cuboctahedron. “Rhombitruncated” alludes to the square faces obtained
by altering the rectangles formed by truncation, the “cub” syllable to the
six octagons lying in the facial planes of a cube, and the “octa” refers to the
eight hexagons lying in the facial planes of an octahedron.
Figure 5.11
We expect a similar situation to arise upon considering a complete truncation of the cuboctahedron, and indeed it does. A complete truncation
of the cuboctahedron which results in its edges being bisected is shown in
Figure √
5.12(a). Here, we see eight equilateral triangles, six squares, and
twelve 2 : 1 rectangles. Like the polyhedron in Figure 5.10(a), this solid
is vertex-regular in that the same assortment of polygons is present at each
vertex, but as not all the rectangles are squares, it is not Archimedean.
Again, happily, we may deform this solid into one which is Archimedean by
moving the triangles slightly toward the center of the polyhedron, thereby
scrunching the rectangles into squares, and the six squares into six slightly
smaller squares. The result is shown in Figure 5.12(b), and is called the
rhombicuboctahedron.
(a)
(b)
Figure 5.12
5.3. SNUB POLYHEDRA
83
Of course, the same procedure may be applied to the icosidodecahedron,
yielding the rhombitruncated icosidodecahedron and the rhombicosidodecahedron, shown in Figure 5.13. The ambitious reader may wish to
investigate the details of creating these solids as was done for the cuboctahedron and its derivatives.
(a)
(b)
Figure 5.13
The process by which the preceding four Archimedean solids were created
inspire one to call them truncation deformations. The curious reader
may inquire as to why this process of truncation deformation was not also
applied to the partially truncated Platonic solids of §5.1. As it happens,
regardless of the effort invested, no Archimedean solids can be produced by
deforming truncations of the partially truncated Platonic solids. Such are
the perils of the ambitious geometer.
5.3
Snub Polyhedra
There are two Archimedean solids with a property not shared by any of the
others – that of existing in enantiomorphic pairs. This essentially means
that these polyhedra come in left-handed and right-handed pairs, or pairs
of mirror images. We clarify this property with an example.
Let us first consider our old friend the cuboctahedron. We may imagine
this solid to be created as follows: begin with a cube, expand the six faces
radially outward (as in Figure 5.14(b)), rotate each square 45◦ and then
move the squares radially inward. The result is the six square faces of the
cuboctahedron (see Figure 5.14(c)). By filling in the triangular gaps, the
cuboctahedron is formed.
84
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
(a)
(b)
(c)
Figure 5.14
Now let us begin with a cube, expand the faces somewhat and rotate
each of the faces a bit less than 45◦ as in Figure 5.15(a) (the calculation of
the precise distance and angle is somewhat involved and is not necessary for
our purposes). The squares do not meet in this instance, but done properly,
there is just enough room to weave interlocking rings of triangles around
these squares, as in Figure 5.15(b). The result is called the snub cube,
“snub” referring to the slight twist of the square faces.
It was important that our rotations of the squares were consistent – were
six individuals small enough to stand on the outside of the square faces, they
would all have experienced a slight rotation to their right. However, had we
chosen to rotate slightly to the left, a slightly different polyhedron would
have been formed – the mirror image of the first (shown in Figure 5.15(c)).
In other words, one would see Figure 5.15(c) if one were to look at a reflection
in a mirror of Figure 5.15(b). These are different polyhedra in the sense that
if we tried to superimpose Figure 5.15(b) onto Figure 5.15(c) such that one
of the square faces of the former coincided with one of the latter, not all of
the rest of the faces would coincide in pairs. It is for this reason that we
say that the snub cube exists in enantiomorphic pairs. None of the other
Archimedean solids discussed up to this point have this property.
(a)
(b)
Figure 5.15
(c)
5.3. SNUB POLYHEDRA
85
The other Archimedean solid with this property is the snub dodecahedron, formed by a process similar to that illustrated in Figure 5.15, except
that we begin with a dodecahedron, expand outward and rotate slightly its
pentagonal faces, and surround these pentagons with rings of interlocking
triangles. The result, an enantiomorphous pair (depending upon whether
the rotation of the pentagonal faces is clockwise or counterclockwise), is
shown in Figure 5.16(a),(b).
(a)
(b)
Figure 5.16
As a final remark, we note that the same process of expansion and rotation may be carried out with the tetrahedron as well. The result, interestingly enough, is our friend the icosahedron! In this case, the enantiomorphism is not apparent unless the faces of the icosahedron are colored.
For example, the faces of the tetrahedron may be colored, say red, before
their expansion. If the surrounding rings of triangles are, say yellow, then
the clockwise and counterclockwise rotations of the red faces will yield two
enantiomorphously colored icosahedra (see Figure 5.17).
(a)
(b)
Figure 5.17
86
5.4
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
Prisms and Antiprisms
We are almost at the conclusion of our geometrical study of the Archimedean
solids. However, the last word is not had by merely a few remaining polyhedra, but rather two infinite families of polyhedra.
An example should make this clear. Begin with your favorite regular
polygon – say a hexagon. Proceed to make a “sandwich” with two hexagons
as the slices of bread, the filling consisting of a ring of squares, as in Figure
5.18. The result is called a hexagonal prism, and has one hexagon and
two squares incident at each of its vertices.
Figure 5.18
But, of course, there is no particular reason why we needed to begin
with a hexagon – any regular polygon would have sufficed. As a result,
each regular polygon generates a prism as described above (note that a
square prism is just a cube). Since there are an infinite number of regular
polygons (one for each integer three or greater), an infinite family of prisms
is generated.
The other infinite family of Archimedean solids is the antiprisms, one
of which – the pentagonal antiprism – we have met in Figure 2.4. And, as
evidenced by the preceding discussion, there is no particular reason why we
must begin with a pentagon – any regular polygon will do. In fact, we are
familiar with a second member of this family – the reader will note that
an octahedron is in fact a triangular antiprism. By sitting the octahedron
on one of its faces, the ring of triangles (pointing, alternately, “up” and
“down”) can readily be observed. Three triangles and one of the “base”
polygons are incident at each vertex of an antiprism.
So this concludes our purely geometrical observations. One final remark:
one often hears of the “thirteen Archimedean solids” or the “thirteen semiregular polyhedra;” the reader should be cautioned that references to the
5.5. AN ALGEBRAIC ENUMERATION
87
Archimedean solids often neglect the two infinite families discussed in this
section (or perhaps mention them in a parenthetical remark).
5.5
An Algebraic Enumeration
This section is analogous to §2.3 in that we seek an algebraic enumeration of
the Archimedean solids. The task is a bit more involved in the Archimedean
case as the requirements for a polyhedron to be an Archimedean solid are
less strict than those for a Platonic solid. As such, we approach a bit more
cautiously – we first investigate a particular case, and then proceed to the
general case.
—A particular case.
We begin with the following question – which Archimedean solids have
faces consisting of both equilateral triangles and squares? Our analysis is
modelled after the discussion in §2.3.
Now we know that at each vertex there is an arrangement of triangles and
squares, but we do not know how many of each there are – these unknowns
we denote by t and s, respectively. We denote, as usual, the number of
vertices, edges, and faces of the Archimedean solid by V , E, and F .
How many triangles are there on this polyhedron? Let us denote this
number by T . Then the total number of sides contributed by all of the
triangles is 3T . Now at each vertex meet 2t sides of triangles (two for each),
and hence we have 2tV for the entire polyhedron. But this counts each side
twice, for each side of each triangle is incident at two vertices. Hence, tV is
also the total number of sides of all triangles of the polyhedron, and hence
3T = tV . (Note how this differs from the Platonic case – since we do not
know whether two triangles, one triangle and one square, or two squares
meet at a particular edge of the polyhedron, we cannot count the number
of edges of the polyhedron contributed by the triangular faces, but we must
rather count the number of sides of the contributing triangles.)
Denoting the number of squares on the polyhedron by S, a similar analysis yields that 4S = sV . Since our polyhedron only has triangles and
squares as faces, we see that F = T + S; using the relationships just derived,
it follows that
tV
sV
F =
+
.
(5.1)
3
4
(Note that (P 20 ) and (P 30 ) of §2.3 imply that pF = qV and hence F =
qV
p ; since we have two different polygons as faces, (5.1) is an Archimedean
analogue of this relationship.)
88
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
Our next argument parallels that concerning the derivation of (P 30 ).
For at each vertex meet t + s edges of the polyhedron (not merely sides of
polygons, for we are counting the contributions from all of the faces of the
polyhedron), and hence (t + s)V counts all edges meeting at all vertices of
the polyhedron. Yet as each edge is incident at exactly two vertices, this
counts each edge exactly twice, and hence we see that
(t + s)V = 2E.
(5.2)
This, then, is an Archimedean analogue of (P 30 ).
For completeness, we include Euler’s formula (since all Archimedean
solids are convex):
V − E + F = 2.
(5.3)
In this case, as both E and F may easily be expressed in terms of s, t, and
V as in (5.1) and (5.2), we substitute these expressions into (5.3), yielding
V −
(t + s)V
tV
sV
+
+
= 2,
2
3
4
or after a little algebra,
t
s
2
+ +
= 1.
6 4 V
(5.4)
We now wish to find all possible solutions to (5.4). Since s represents the
number of squares at each vertex, then certainly s < 4 (or else the vertex
would be flat); moreover, since we assumed that our polyhedron has square
faces, it follows that s ≥ 1. We examine the three possibilities.
1. When s = 1, (5.4) becomes
t
2
3
+
= .
6 V
4
Since we must have at least three polygons at each vertex, then t ≥ 2;
it also follows that t < 5 (else the sum of the angles around a vertex
would exceed 360◦ and our polyhedron would not be convex). When
t = 2, we have V = 24
5 , an impossibility. (Note that a square pyramid
has one square and two triangles at each vertex of its base, but this
results in four triangles at its apex.) When t = 3, we have V = 8,
resulting in a square antiprism. And when t = 4, we have V = 24,
giving a snub cube.
5.5. AN ALGEBRAIC ENUMERATION
89
2. When s = 2, (5.4) becomes
t
2
1
+
= .
6 V
2
Note that t ≥ 1 yields at least three polygons at each vertex; but we
must have t < 3 to avoid a flat vertex. When t = 1, we obtain V = 6
– the triangular prism, and when t = 2, we find that V = 12 – the
cuboctahedron.
3. When s = 3, (5.4) becomes
t
2
1
+
= .
6 V
4
Here, our only option is t = 1, yielding V = 24 and the small rhombicuboctahedron.
—The general case.
Now that we have examined a particular case in some detail, we look at
the general case. Our first observation is that there can be at most three
different types of polygon about each vertex. If there were at least four, then
the smallest possible sum for the angles of these polygons at a given vertex
would be 60◦ + 90◦ + 108◦ + 120◦ , as the triangle, square, pentagon, and
hexagon have the smallest angles among the regular polygons. But this sum
is 378◦ , and hence cannot correspond to a convex polyhedron. Therefore, at
most three different polygons may possibly meet at each vertex.
So let s1 , s2 , and s3 denote the number of sides these polygons have, let
F1 , F2 , and F3 denote the number of these polygons occurring as faces on
the Archimedean solid, and let n1 , n2 , and n3 indicate the number of each of
these polygons incident at each vertex of the polyhedron. (If there happen
to be only two different types of polygon at each vertex, the analysis which
follows may be easily modified by putting n3 := 0.)
The foregoing analysis of the particular case has a direct analogue here.
It is readily seen, using the same argument as above, that s1 F1 = n1 V ,
s2 F2 = n2 V , and s3 F3 = n3 V . It is also clear that F = F1 + F2 + F3 , so
that the analogue of (5.1) becomes
F =
n1 V
n2 V
n3 V
+
+
.
s1
s2
s3
(5.5)
Again, similar argumentation gives the analogue of (5.2); namely
(n1 + n2 + n3 )V = 2E.
(5.6)
90
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
And, of course, Euler’s formula (5.3) is still valid.
As before, we substitute into Euler’s formula for F using (5.5), and for
E from (5.6), giving, after some algebra, the analogue of (5.4):
1=
2
+ n1
V
1
1
−
2 s1
+ n2
1
1
−
2 s2
+ n3
1
1
−
2 s3
.
(5.7)
Thus, we seek all solutions of (5.7) in integers (where s1 , s2 , s3 ≥ 3
and V ≥ 4) which also yield constructible convex polyhedra. For example,
although
2
1=
+1
24
1 1
−
2 3
+1
1 1
−
2 5
+1
1
1
−
2 20
,
there is no Archimedean solid possible with one triangle, one pentagon, and
one 20-gon at each of its 24 vertices. Therefore, each solution of (5.7) yields
a candidate for an Archimedean solid, not a guarantee that we have found
one. The situation is decidedly more complex than our algebraic discussion
of the Platonic solids – the ambitious reader is encouraged to consult the
Exercises for a thorough investigation.
Two particular families of solutions are noteworthy; both involve only
two types of polygons at each vertex. For the first, we assume that two
squares and one other polygon meet at each vertex. Then s1 = 4, n1 = 2,
and n2 = 1. Substituting these values into (5.7) yields V = 2s2 , which corresponds to the infinite family of prisms. Each polygon of s2 sides generates
a prism with 2s2 vertices. The second is the infinite family of antiprisms,
where substituting s1 = 3, n1 = 3, and n2 = 1 into (5.7) also yields V = 2s2 ,
thus giving one antiprism for each regular polygon.
5.6
Exercises
This set of Exercises enumerates the Archimedean solids by finding all algebraic solutions to (5.7) which correspond to geometrically realizable polyhedra.
1. We first consider the case where n1 = n2 = n3 = 1, so that three
different polygons meet at each vertex.
5.6. EXERCISES
91
s3
s2
s2
s3
s1
s1
s2
Figure 5.19
Our first observation is that s1 , s2 , and s3 must be even. Consider, if
you will, Figure 5.19, and let p be a vertex of the polyhedron which is
on a particular face with s1 sides (as in the Figure). Because one of
each type of face is incident at each vertex, we may begin at p with,
say, an s2 -gon, and proceed clockwise, encountering faces of s2 , s3 , s2 ,
s3 ,... sides. We must end up at p with an s3 -gon, else we would have
an s1 -gon and two s2 -gons incident at p. This, however, implies that
s1 is even, as we encountered successive pairs of s2 , s3 -gons along the
way. In a similar way, one may show that s2 and s3 are even as well.
With this in mind, find two solutions to (5.7) when n1 = n2 = n3 = 1
and s1 , s2 , and s3 are all different even integers greater than or equal
to 4.
2. We now consider the case when there are three polygons meeting at
a vertex, but of only two different types, so that we may assume that
n1 = 2, n2 = 1, and n3 = 0.
(a) Using an argument similar to that used in (1) above, show that
s1 must be even.
(b) Consider the case s1 = 4 separately, resulting in the family of
prisms discussed at the end of §5.5.
(c) Find all possible solutions when s1 ≥ 6. What type of Archimedean
solid is generated? (Hint: Since s1 is even, write s01 = 12 s1 . Show
that (5.7) becomes
1
1
1
2
+
= + .
0
s1 s2
2 V
Compare this with equation (2.2).)
92
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
3. We now look at possibilities which involve four polygons meeting at
each vertex. The first we shall consider is when there are three different
types of polygons at each vertex (there cannot be four, as was seen in
§5.5). Hence there must be two of one type, and one of each of the
others, so that we may assume that n1 = 2, n2 = 1, and n3 = 1.
(a) Using an argument similar to that used in (1), show that s1 must
be even.
(b) Begin with the case s1 = 4, so that s2 and s3 are different and
not equal to 4. (You should find only one solution.)
(c) Argue algebraically that when s1 ≥ 6, no other solutions to (5.7)
are possible.
4. Now consider the case where four polygons meet at a vertex, but there
are only two different types of polygons.
(a) Assume first that n1 = n2 = 2 and n3 = 0. Show that (5.7)
reduces to
1
1
1
1
+
= +
s1 s2
2 V
(compare this to (2.2)). Find the two Archimedean solids corresponding to solutions where s1 6= s2 .
(b) Secondly, assume that there are three of one type of polygon and
one of another incident at each vertex, so that we may assume
that n1 = 3, n2 = 1, and n3 = 0.
i. When s1 = 3, show that the family of antiprisms is generated
(see the final discussion of §5.5).
ii. Find the only possible solution corresponding to s1 = 4.
iii. Argue that no solution is possible when s1 ≥ 5.
5. Finally, consider the case where five polygons meet at a vertex (it is
easy to show that six are impossible). Argue that at least four of these
polygons must be triangles, and find the remaining two Archimedean
solids.
6. Verify Descartes’ rule of deficiency for Archimedean solids. Begin by noting that since Archimedean solids are convex, the sum of
the angles incident at a vertex of an Archimedean solid must be less
than 360◦ . Subtracting this sum from 360◦ , then, results in a positive number of degrees, called the angular deficiency at that vertex.
5.6. EXERCISES
93
Show that if the angular deficiencies are summed over all vertices of
an Archimedean solid, the result is always 720◦ . (Hint: Use (5.7).)
7. This problem requires a knowledge of linear algebra, and may safely
be omitted by the uninitiated.
The vector (V, E, F ) is often called an f -vector in a combinatorial
theory of polyhedra. The next few problems are concerned with f vectors. Assume that all polyhedra under consideration are convex, so
that Euler’s formula is always applicable.
(a) Consider a polyhedron P whose f -vector is f = (V, E, F ). Now
consider a partial truncation (see §5.1) Pt of P such that each
edge of P generates two vertices of Pt . Suppose that ft = (Vt , Et , Ft )
is the f -vector for Pt . Find a 3 × 3 matrix Mt such that
ft = Mt f .
(b) Find the analogous matrix MT for the process of complete truncation, i.e., each edge of the original polyhedron generates just
one vertex of its complete truncation (see §5.1).
(c) Consider the following expansion procedure. As an example,
consider a cube. Now move the faces of the cube radially outward (as in Figures 5.14 and 5.15) but do not rotate them. Thus,
the cube is transformed into a rhombicuboctahedron, a tetrahedron would become a cuboctahedron, while a dodecahedron
would generate a rhombicosidodecahedron. Analogous to (a) and
(b), find the matrix Me corresponding to the expansion process.
(d) Find the analogous matrix Md for the process of taking the dual
of a polyhedron (as described in Chapter 9).
(e) Using matrix multiplication, show that Mc Md = Mc . As a result, for any polyhedron P with f -vector f , we have
Mc Md f = Mc f .
This suggests that the polyhedron obtained by completely truncating P (i.e., finding Mc f ) is the same “shape” as that obtained
by first taking the dual of P (i.e., finding Md f ), and then completely truncating this polyhedron (i.e., finding Mc (Md f )).
Is this suggestion valid? Provide an argument or exhibit a counterexample to show that it is invalid.
94
CHAPTER 5. THE ARCHIMEDEAN SOLIDS
(Roughly speaking, two polyhedra have the same “shape” if their
faces may be matched one for one such that corresponding faces
have the same number of sides and the adjacency of faces corresponds as well. Thus, a cube and a rectangular prism have the
same “shape,” as do a cube and a parallelepiped.)
(f) Show that Mt Md = Mt , and proceed as in (e).
(g) Show that Me Md = Me , and proceed as in (e).
(h) Show that Mc Mc = Me , and proceed as in (e).
(i) Find a formula for Mnt , where n is a positive integer, and find
the eigenvalues and eigenvectors for Mnt .
(j) Proceed as in (i) for Mnc .
(k) Proceed as in (i) for Mne . (Hint: Use the results of (h)).
Chapter 6
Angles and Archimedeans
In Chapter 3, we saw how spherical trigonometry can be applied to the geometry of the Platonic solids in order to calculate edge angles and dihedral
angles of these polyhedra. We undertake a similar endeavor in this chapter, applying spherical trigonometry to the Archimedean solids. We will,
however, postpone a discussion of the snub polyhedra until Chapter 8.
6.1
Partially Truncated Platonic Solids
—Dihedral angles.
Consider for a moment the truncated tetrahedron. With model in hand,
one readily sees that the dihedral angles between the hexagonal faces are
acute (being the same as the dihedral angles of the tetrahedron itself), while
the dihedral angles between the triangular and hexagonal faces are obtuse.
A similar phenomenon occurs with all of the partially truncated Platonic
solids – the dihedral angle is maintained at the corresponding edges of the
truncated polyhedron, but new dihedral angles are introduced abutting the
edges of the new faces which are obtained as a result of truncating the
Platonic solid.
The measures of these new angles are fairly straightforward to calculate.
The approach is identical to that used at the beginning of §3.4 to compute
D5,3 – we center a sphere at a vertex of the truncated tetrahedron, thereby
forming the spherical triangle shown in Figure 6.1. An application of (3.4)
yields
cos 120◦ = cos 120◦ cos 60◦ + sin 120◦ sin 60◦ cos Dt3,3 ,
95
96
CHAPTER 6. ANGLES AND ARCHIMEDEANS
120◦
60◦
120◦
Dt3,3
vertex of truncated tetrahedron
Figure 6.1
from which we calculate
1
cos Dt3,3 = − .
3
Here, the superscript “t” stands for “truncation,” so that Dtp,q is the dihedral
angle of the partially truncated Platonic solid {p, q} which abuts the new
faces of the truncated polyhedron. Of course, the dihedral angles of the
other partially truncated solids may be similarly calculated; the results are
summarized in Table 6.1 (located after the Exercises).
Given the Platonic solid with faces of p sides each, q of these meeting at
each vertex, we have
tan
π
π
tan cos Dtp,q = −1.
p
q
(6.1)
Details of the derivation of this formula are left to the Exercises.
—Edge angles.
Finding the edge angles of the partially truncated Platonic solids is somewhat more involved than finding the dihedral angles. However, because we
know that all edges of such a polyhedron have the same length, and, due
to the symmetry of the truncation process, that all vertices of such a polyhedron lie on the same sphere, it follows that all edges of the partially
truncated Platonic solid subtend the same angle at the center of the sphere.
Thus, there is just a single edge angle to find.
One procedure is illustrated here by finding the edge angle of the truncated tetrahedron. Consider Figure 6.2, where O is the center of a tetrahedron, ps is an edge of that tetrahedron, and hence qr is an edge of the
truncated tetrahedron (recall that [qr] = 13 [ps]). E3,3 denotes, as before,
the edge angle of the tetrahedron, and Et3,3 denotes the edge angle of the
truncated tetrahedron. The midpoint of qr is denoted by m.
6.1. PARTIALLY TRUNCATED PLATONIC SOLIDS
p
q
m
97
r
s
Et3,3
E3,3
O
center of truncated tetrahedron
Figure 6.2
Now since
[mr]
[qr]
1
=
= ,
[ms]
[ps]
3
it follows upon considering right triangles ∆Omr and ∆Oms that tan 12 Et3,3 =
1
1
1
21 t
21
3 tan 2 E3,3 , and hence tan 2 E3,3 = 9 tan 2 E3,3 . Now
tan2 12 θ = sec2 12 θ − 1 =
so that
1
cos2 12 θ
−1=
2
1
−1=
t
1 + cos E3,3
9
1
1+cos θ
2
−1=
2
1+cos θ
− 1,
2
−1 .
1 + cos E3,3
Solving algebraically for cos Et3,3 yields
cos Et3,3 =
5 cos E3,3 + 4
.
4 cos E3,3 + 5
(6.2)
7
Since cos E3,3 = − 31 , we calculate that cos Et3,3 = 11
. Of course, since partial
truncation of the octahedron and icosahedron also results in the trisection
of the edges of the original polyhedron, the same formula is valid if the
subscript “3,3” is replaced by “3,4” or “3,5”; i.e.,
cos Et3,4 =
5 cos E3,4 + 4
.
4 cos E3,4 + 5
Now the edge angles of the truncated cube and the truncated dodecahedron must be calculated differently, as we recall that the truncation process
does not divide the edges of the original polyhedron into three equal parts
(see Figure 1.7(a) and Figure 5.5(a)). For example, if Figure 6.3 represented
a diagram for calculating the edge angle of a truncated cube, we would have
√
√
[qr]
2
=√
= 2 − 1.
[ps]
2+2
98
CHAPTER 6. ANGLES AND ARCHIMEDEANS
p
q
r
s
Etp,q
Ep,q
O
center of truncated polyhedron
Figure 6.3
In general, if the Platonic solid {p, q} is partially truncated to yield an
Archimedean solid, and if λ = [qr]/[ps] (as illustrated in Figure 6.3) and
µ = (1 − λ2 )/(1 + λ2 ), then it may be shown that
cos Etp,q =
(1 + λ2 ) cos Ep,q + 1 − λ2
cos Ep,q + µ
=
.
2
2
(1 − λ ) cos Ep,q + 1 + λ
µ cos Ep,q + 1
(6.3)
The derivation of this formula is identical to that of (6.2) except that “λ”
is used in place of “ 13 .” Details are left to the Exercises. Values for the
cosines of edge angles of the partially truncated Platonic solids are included
in Table 6.2 (located after the Exercises).
6.2
Completely Truncated Platonic Solids
In this section, we consider the cuboctahedron and icosidodecahedron. A superscript “T ” denotes complete truncation, so that ET3,5 (or ET5,3 ) denotes the
edge angle of the icosidodecahedron. Note that for each of these polyhedra,
all dihedral angles have the same measure, as do all edge angles.
—Dihedral angles.
Let us consider the truncated cube for a moment. We see that moving
the triangular faces radially inward (i.e., furthering the truncation process)
until the vertices meet yields a cuboctahedron. However, throughout this
process, the dihedral angle between the triangular and octagonal faces does
6.2. COMPLETELY TRUNCATED PLATONIC SOLIDS
99
not change. As a result, the dihedral angle of the cuboctahedron is the same
as Dt4,3 . Viewing the cuboctahedron from the perspective of completing the
truncation process on the truncated octahedron as well, we see that
Dt4,3 = DT3,4 = Dt3,4 .
(6.4)
By considering an analogous situation involving the icosidodecahedron, a
truncated dodecahedron, and a truncated icosahedron, we see that
Dt5,3 = DT3,5 = Dt3,5 .
There is an alternative way to calculate DT3,4 and DT3,5 ; while more computationally intense, it is of independent geometrical interest. We illustrate
by using this method to find DT3,5 .
To begin, we cut an icosidodecahedron in half as in Figure 6.4. This
exposes an equatorial decagon, with exactly one triangle, one pentagon, and
one decagon meeting at each vertex of the decagon. By centering a sphere
at one such vertex, we create a spherical triangle with edge angles 60◦ , 108◦ ,
and 144◦ – with DT3,5 being opposite the edge with measure 144◦ . Applying
(3.4) gives
cos 144◦ = cos 60◦ cos 108◦ + sin 60◦ sin 108◦ cos DT3,5 ,
from which we may calculate cos DT3,5 .
Figure 6.4
In a similar fashion, we may slice a cuboctahedron to expose two equatorial hexagons, using a vertex of one of them to calculate DT3,4 . Details are
left to the Exercises.
100
CHAPTER 6. ANGLES AND ARCHIMEDEANS
—Edge angles.
Of course, consideration of these equatorial decagons and hexagons of the
icosidodecahedron and cuboctahedron immediately yields that ET3,5 = 36◦
and ET3,4 = 60◦ . We shall use spherical trigonometry to verify these results.
We illustrate by finding ET3,4 . In Figure 6.5(a), pr is an edge of a cuboctahedron inscribed in a cube, q is a vertex of this cube, and O is the center
of the cube. The spherical version of ∆pqr is shown in Figure 6.5(b). Since
three squares meet at q, the dihedral angle at q must be 120◦ ; since p and
r are midpoints of edges of the cube, sides pq and qr must have measure
1
2 E4,3 . Therefore, we may apply (3.4) to yield
cos ET3,4 = cos2 21 E4,3 + sin2 12 E4,3 cos 120◦ = 12 ,
and hence ET3,4 = 60◦ . The same result would have been obtained were we
to have considered the cuboctahedron as being inscribed in an octahedron
rather than a cube.
q
1
2 E4,3
p
q
120◦
p
ET3,4
O
1
2 E4,3
r
r
O
(a)
(b)
Figure 6.5
This same procedure may be applied to an icosahedron or dodecahedron
in order to find ET3,5 . Details are left to the Exercises.
6.3. TRUNCATION DEFORMATION
6.3
101
Truncation Deformation
—Partially truncated forms.
We first consider the rhombitruncated cuboctahedron (Figure 5.11(b)).
To find the dihedral angles is not difficult since exactly three faces meet at
each vertex, yielding the spherical triangle shown in Figure 6.6. Note that
the octagons and hexagons are in a cuboctahedral arrangement, so that the
dihedral angle between these faces is simply DT3,4 .
120◦
DE
3,4
90◦
135◦
DT3,4
DE
4,3
vertex of rhombitruncated cuboctahedron
Figure 6.6
On the other hand, the other two dihedral angles have not been encountered before. The label “DE
3,4 ” in Figure 6.6 indicates that this is the dihedral
angle between the squares and the octahedral (“{3, 4}”) faces, which happen
to be regular hexagons. The superscript “E” represents the fact that the
rhombitruncated cuboctahedron may be formed by a process of complete
expansion. In other words, the triangles of the octahedron are expanded
radially outward, then truncated to form hexagons. Octagons and squares
are then situated so that they correspond to the vertices and edges, respectively, of the original octahedron. This process results in a complete
expansion of the edges of the original octahedron in that although all twelve
edges are evident in the rhombitruncated cuboctahedron, none is adjacent
to any other.
Likewise, DE
4,3 is the dihedral angle between the square faces and the
cubical (octagonal) faces. A moment’s thought reveals that expanding the
cube in a manner analogous to that just described for the octahedron also
produces the rhombitruncated cuboctahedron.
E
Calculation of DE
3,4 and D4,3 is a straightforward application of (3.4) to
the spherical triangle in Figure 6.6; details are left to the reader. We will
occasionally adopt a notation such as “DE
4,3 [4-8]” to reinforce the idea that
102
CHAPTER 6. ANGLES AND ARCHIMEDEANS
DE
4,3 is the dihedral angle between the square and octagonal (4-sided and
8-sided) faces of the rhombitruncated cuboctahedron. The edge angle EE
3,4
is included in Table 6.2 for completeness; its derivation is left as an Exercise.
A similar methodology may be applied to the icosidodecahedron to obtain the edge angle EE
3,5 of the rhombitruncated icosidodecahedron, the diE
hedral angle D3,5 [4-6] between the square and hexagonal faces of this polyhedron, and the dihedral angle DE
5,3 [4-10] between the square and decagonal
faces. Analogous to the cuboctahedral case, the dihedral angle between the
hexagonal and decagonal faces is just DT3,5 .
—Completely truncated forms.
Calculation of the dihedral angles of the rhombicuboctahedron and the
rhombicosidodecahedron is not a direct application of (3.4) as four faces meet
at each vertex. However, as with the cuboctahedron and icosidodecahedron
in §6.2, we may appropriately look at “caps” for a solution. In this section,
we use the superscript “e” to indicate a partial expansion process. In the case
of the rhombicuboctahedron, the triangles of the octahedron are expanded
radially outward, but are not truncated. Squares are then situated so that
they correspond to the vertices and edges of the original octahedron. All of
the triangular faces of the octahedron are still evident, so that the expansion
is not yet “complete,” but only “partial.”
Let us first consider the rhombicuboctahedron. Upon considering the
equatorial octagons of this polyhedron, it is evident that the dihedral angle
De4,3 [4-4] between adjacent squares has measure 135◦ , the same as the measure of the angles in a regular octagon. To find the dihedral angle De3,4 [4-3]
between the square and triangular faces, we look at an octahedral cap of the
rhombicuboctahedron, as in Figure 6.7. The base of this cap is a regular
octagon, at the vertices of which meet exactly one triangle, one square, and
the base octagon. We may center a sphere at one such vertex to obtain a
spherical triangle with
sides 60◦ , 90◦ , and 135◦ ; an application of (3.4) now
√
√
yields cos De3,4 = − 2/ 3.
Figure 6.7
6.3. TRUNCATION DEFORMATION
103
The case of the rhombicosidodecahedron is slightly more complicated.
By analogously considering
a decagonal cap of this polyhedron, we see that
√
cos De3,5 [4-3]= −τ / 3. However, there is no direct analogue of the equatorial
octagons of the rhombicuboctahedron. So we take the following approach,
illustrated in Figure 6.8. Here, q and r are midpoints of opposite sides of
a square face, and p and s are the centers of the pentagonal faces adjacent
to that square face (see Figure 6.8(a)). A cross-sectional view is taken in
Figure 6.8(b), where the segments pq and sr are extended to meet at t.
Since extending the pentagonal faces of the rhombicosidodecahedron results
in a dodecahedron, it follows that the angle at t must have measure D5,3 .
Similarly, by examining Figure 6.8(a), it is evident that ∠pqr has the same
measure as De5,3 [4-5]. With α denoting the measure of ∠tqr, it is clear that
2α + D5,3 = 180◦ , so that α = 90◦ − 12 D5,3 . But then De5,3 = 180◦ − α =
90◦ + 12 D5,3 . Thus
cos De5,3
s
1 + √15
1
1
◦
= cos 90 + D5,3 = − sin D5,3 = −
2
2
2
s
√
√
1+ 5 1
τ
· √ = −√
.
=−
4
2
5
5
Calculation of the edge angle of the rhombicuboctahedron is left to the
Exercises.
t
q α
De5,3
D5,3
r
p
(a)
s
(b)
Figure 6.8
104
6.4
CHAPTER 6. ANGLES AND ARCHIMEDEANS
Exercises
1. Derive (6.1) according to the following outline.
Consider the Platonic solid {p, q}. A vertex of this polyhedron is
truncated as in Figure 6.9. Sinceα is an angle of a polygon with q
edges, we know that α = π 1 − 2q .
ϕ
Dtp,q
α θ
θ
Figure 6.9
(a) Show that θ = π 1 − p1 .
(b) Apply (3.4) to the spherical triangle formed by centering a sphere
at a vertex of the truncated polyhedron (the edge angles of this
spherical triangle are θ, θ, and α), yielding
2π
π
π
2π
− 1 + cos
cos = sin sin
cos Dtp,q .
q
p
p
q
(c) Using the appropriate trigonometric identity, conclude that
tan
π
π
tan cos Dtp,q = −1.
p
q
2. Derive formula (6.3) as indicated in the text.
3. As suggested in §6.2, calculate DT3,4 by slicing a cuboctahedron in half
and considering a vertex of one of the resulting equatorial hexagons.
4. Find ET3,5 as suggested at the end of §6.2.
6.4. EXERCISES
105
5. Use the following outline to find the edge angle of the rhombitruncated
cuboctahedron. Assume throughout that an edge of this polyhedron
has length 1.
r
s
u
1
s
1
r
O
r
O
t
(a)
(b)
(d)
v
1
s
EE
3,4
O
(c)
(e)
Figure 6.10
(a) Show that
√ the diagonal tu of the octagon in Figure 6.10(a) has
length 2 + 1.
(b) A typical octagonal face is shown in Figure 6.10(a). Show that
the distance from the center of such a face, r, to one of its vertices,
s, is given by
s
1
[rs] = 1 + √ .
2
(c) Consider an octagonal “equator” of the rhombitruncated cuboctahedron as shown by the white line in Figure 6.10(c) and flattened out in Figure 6.10(b). Show that the distance from the
106
CHAPTER 6. ANGLES AND ARCHIMEDEANS
center of the polyhedron,
O, to the center of an octagonal face,
√
1
r, is [Or] = 2 + 2.
(d) By considering the right triangle in Figure 6.10(d), show that the
distance between the center of the rhombitruncated cuboctahedron, O, and one of its vertices, s, is
q
√
1
13 + 6 2.
[Os] =
2
(e) By considering the triangle in Figure 6.10(e), where s and v are
adjacent vertices of the rhombitruncated cuboctahedron, show
that
√
71 + 12 2
E
cos E3,4 =
.
97
6. Using a methodology similar to that in the previous exercise, find
cos Ee3,4 .
7. Explain all the duplications in the “cos D” and “D” columns in Table
6.1 which are not discussed in the text.
6.4. EXERCISES
Polyhedron
107
Symbol
cos D
Trunc. Tetra’n
Dt3,3
− 13
Trunc. Octa’n
Dt3,4
Trunc. Icosa’n
Dt3,5
− √1
q 3
− 4τ15+3
Trunc. Cube
Dt4,3
Trunc. Dodeca’n
Dt5,3
Cuboctahedron
DT3,4
Icosidodeca’n
DT3,5
Rh’truncated
DE
3,4 [4-6]
Cuboctahedron
DE
4,3 [4-8]
Rh’truncated
DE
3,5 [4-6]
Icosidodeca’n
DE
5,3 [4-10]
Rh’cubocta’n
De3,4 [4-3]
De4,3 [4-4]
Rh’icosidodeca’n
De3,5 [4-3]
De5,3 [4-5]
− √1
q 3
− 4τ15+3
− √1
q 3
− 4τ15+3
√
− √23
− √12
− √τ3
√
τ
−√
4
5
√
2
− √3
− √12
− √τ3
√
τ
−√
4
5
Table 6.1
tan D
√
−2 2
√
− 2
D (deg)
−2τ −2
√
− 2
142.623
−2τ −2
√
− 2
142.623
−2τ −2
√
−1/ 2
142.623
−1
135.000
−τ −2
159.095
−τ −1
√
−1/ 2
148.283
−1
135.000
−τ −2
159.095
−τ −1
148.283
109.471
125.264
125.264
125.264
144.736
144.736
108
CHAPTER 6. ANGLES AND ARCHIMEDEANS
Polyhedron
Symbol
cos E
E (deg)
Trunc. Tetra’n
Et3,3
7
11
50.479
Trunc. Octa’n
Et3,4
4
5
36.870
Trunc. Icosa’n
Et3,5
Trunc. Cube
Et4,3
Trunc. Dodeca’n
Et5,3
√
80+9 5
109
√
3+8 2
17
√
24+15 5
61
Cuboctahedron
ET3,4
1
2
60.000
Icosidodeca’n
ET3,5
τ
2
36.000
Rh’trunc. Cubocta’n
EE
3,4
Rh’trunc. Icosidodeca’n
EE
3,5
Rh’cubocta’n
Ee3,4
Rh’icosidodeca’n
Ee3,5
Table 6.2
√
71+12 2
97
√
179+24 5
241
√
7+4 2
17
√
19+8 5
41
23.281
32.650
19.387
24.918
15.112
41.882
25.879
Chapter 7
Geodesic Structures, II
7.1
Right Triangles
We recall that in the plane, several formulas involving the various parts of a
triangle become simpler when the triangle contains a 90◦ angle. The same
is true of spherical triangles; several simple formulas are apparent when one
of the dihedral angles measures 90◦ . Since right spherical triangles occur
frequently in the design and construction of geodesic structures, it is worth
taking some time to derive such formulas.
So consider the right spherical triangle in Figure 7.1. We now undertake
the derivation of several results alluded to above.
B
a
c
A
b
Figure 7.1
First, applying (3.4) yields
cos c = cos a cos b + sin a sin b cos 90◦ ,
109
110
CHAPTER 7. GEODESIC STRUCTURES, II
so that
cos c = cos a cos b.
(7.1)
We may apply (3.6) in two different ways. The first results in
cos 90◦ = − cos A cos B + sin A sin B cos c,
which after a little algebra becomes
tan A tan B cos c = 1.
(7.2)
The second yields
cos A = − cos B cos 90◦ + sin B sin 90◦ cos a,
which together with the companion relationship obtained by reversing the
roles of A and B, results in
cos A = sin B cos a,
cos B = sin A cos b.
(7.3)
Now we turn our attention to (4.5), so that
sin a
sin b
sin c
=
=
,
sin A
sin B
sin 90◦
which gives the companion relationships
sin a = sin A sin c,
sin b = sin B sin c.
(7.4)
Multiplying each equation in (7.3) by sin c and substituting from (7.4)
yields
cos a sin b = cos A sin c, cos b sin a = cos B sin c.
(7.5)
Dividing the equations in (7.5) by (7.1), as in
cos a sin b
cos A sin c
=
,
cos a cos b
cos c
results in
tan b = cos A tan c,
tan a = cos B tan c.
(7.6)
Finally, we divide the equations in (7.4) by those in (7.5), as in
sin a
sin A sin c
=
,
cos a sin b
cos A sin c
to obtain the companion relationships
sin b tan A = tan a,
sin a tan B = tan b.
(7.7)
None of these is especially difficult to derive, but it will prove useful to
have them handy for future reference.
7.2. VERTEX ANGLES
7.2
111
Vertex Angles
We wish to extend the analysis begun in §4.2 to all of the Archimedean
solids. Thus, given a face of an Archimedean solid as in Figure 7.2, we wish
to project it onto a sphere and join arcs from the center of the resulting
spherical polygon to each of its vertices. We seek, then, arcs such as χ in
Figure 7.2 (or such as b in Figure 4.3(a)). However, we encounter a difficulty
here which was not present in the case of the Platonic solids. Recall that in
§4.2, in the case of the dodecahedron, the dihedral angle A (as labelled in
Figure 7.2) was conveniently 60◦ . This resulted directly from the fact that
precisely three pentagons met at each vertex, and thus the symmetry of the
dodecahedron could be exploited. With the Archimedeans, however, we find
that at least two different types of polygons meet at each vertex, rendering
analogous symmetry considerations inapplicable.
A
E
2π
p
χ
Figure 7.2
As a result, we must consider a different approach. Although in general
we may not know A in Figure 7.2, we do know the edge angle E of the
Archimedean solids, giving the spherical triangle shown in Figure 7.3. We
may then apply (3.4), yielding
cos E = cos2 χ + sin2 χ cos
2π
.
p
Using the relationship sin2 χ = 1 − cos2 χ, this may be rewritten as
2
cos χ =
cos E − cos 2π
p
1 − cos 2π
p
We also have
sin χ =
sin E2
,
sin πp
.
(7.8)
112
CHAPTER 7. GEODESIC STRUCTURES, II
which follows immediately from bisecting the angle
applying (7.4.).
2π
p
in Figure 7.3(a) and
2π
p
χ
χ
E
(a)
(b)
Figure 7.3
√
√
Thus, for our dodecahedron, where cos E5,3 =
5−1
4 ,
5
3
◦
and cos 2π
p = cos 72 =
we find that
√
2
cos χ =
5
3
√
5−1
√ 4
5−1
4
−
1−
√
√
5+3
5+2 5
√ =
.
=
15
3(5 − 5)
Although in a different algebraic form, this result does agree with that of
§4.2, so that we find in this case that χ ≈ 37.4◦ .
It is evident that this procedure may be applied to each face of each
Archimedean solid, for all that is needed are the values for p and E; p is
simply the number of sides on the face in question and E is the edge angle
of the Archimedean solid, whose cosine is available in Table 6.1. A complete
set of data is given in Table 7.1 (located after the Exercises). These data
may be used to construct geodesic versions of the Archimedean solids as
was done for the dodecahedron in §4.2; one example is the rhombitruncated
cuboctahedron shown in Figure 7.3(b). Since the area of the squares is so
much smaller than the areas of the octagons and hexagons, the squares on
the geodesic model are sometimes left intact to make a more aesthetically
pleasing sphere.
We remark that for building models, only numerical data are needed.
As our approach is to also find algebraic expressions for the various angles,
somewhat more effort is involved.
7.3. MIDEDGE ANGLES
7.3
113
Midedge Angles
By subdividing faces of the Archimedean solids still further, we create another type of geodesic structure. In addition to joining the center of a
spherical polygon to each of its vertices, we may also join the center to the
midpoints of the edges as in Figure 7.4. This is the approach taken in Section
II of Spherical Models.
m
Figure 7.4
Figure 7.5(a) shows an enlargement of one of the component triangles of
Figure 7.4. χ is as in Figure 7.3, but the angles E and 2π
p of Figure 7.3 are
bisected in this case. Moreover, as a result of our construction, we see that
the dihedral angle created at m has measure π2 . We seek ψ, the measure
of the edge from the center of the spherical polygon to m. Figure 7.5(b)
shows a rhombitruncated cuboctahedron based on such triangles, although
squares, hexagons, and octagons are subdivided rather than pentagons.
π
p
χ
ψ
m
E
2
(a)
(b)
Figure 7.5
114
CHAPTER 7. GEODESIC STRUCTURES, II
To find ψ, we apply (7.1) to the spherical triangle in Figure 7.5(a),
yielding
E
cos χ = cos ψ cos .
2
To transform this relationship into a more useable form, we square both
sides, resulting in
cos2 χ = cos2 ψ cos2
so that
cos2 ψ =
1 + cos E
E
= cos2 ψ
,
2
2
2 cos2 χ
.
1 + cos E
(7.9)
Thus, for our dodecahedral example, we would have
2
cos ψ =
2·
1
√
5+2 5
15
√
+ 35
√
√
5+ 5
10 + 4 5
√ =
=
,
10
15 + 5 5
from which we calculate ψ ≈ 31.7◦ . We may use the data in Table 7.1
for χ to calculate cos2 ψ for all of the Archimedean solids; the results are
also collected in Table 7.1. The variables “χ” and “ψ” were chosen in this
section and the last so as to be consistent with Table 3 on p. 53 of Spherical
Models. Wenninger’s “φ” in Table 3 is simply our E/2. This notation is
also consistent with Coxeter’s use of “χ,” “ψ,” and “φ” in Regular Polytopes
(see §7.9, where equation 2.52 is referenced) in describing a characteristic
simplex on a sphere.
7.4
Schwarz Triangles
Certain spherical triangles occur frequently in the literature on polyhedra
and geodesic structures. We begin a discussion of one such group of triangles,
Schwarz triangles, with a fresh look at some old data. A few notations are
helpful here: we denote by χp,q and ψp,q the angles in Table 7.1 (as well
as Figure 7.5) which correspond to the Platonic solid {p, q}. Finally, we
assemble a small table using data from Tables 3.1 and 7.1.
A cursory glance at Table 7.2 reveals the rather interesting relationships
1
Ep,q = ψq,p ,
2
χp,q = χq,p .
(7.10)
We seek to explore these relationships both algebraically and geometrically.
7.4. SCHWARZ TRIANGLES
115
Polyhedron
{p, q}
cos2 12 Ep,q
cos2 χp,q
cos2 ψp,q
Tetrahedron
{3, 3}
1
3
1
9
1
3
Octahedron
{3, 4}
1
2
1
3
2
3
Cube
{4, 3}
2
3
1
2
Icosahedron
{3, 5}
Dodecahedron
{5, 3}
1
3
√
5+2 5
15
√
5+2 5
15
1
5 (τ
+ 2)
τ2
3
τ2
3
1
5 (τ
+ 2)
Table 7.2
Recall that we may use (7.8) to find χ for any Archimedean solid. In
restricting our attention for the moment to Platonic solids, we may substitute the last expression for cos Ep,q in (3.7) into (7.8); with a little algebra,
we obtain
2π
1 + cos 2π
1
+
cos
p
q
.
cos2 χp,q = 2π
1 − cos 2π
1
−
cos
p
q
Using the appropriate trigonometric identities and extracting square roots
(since all angles are in suitable ranges) yields
π
π
(7.11)
cos χp,q = cot cot .
p
q
It is evident that this expression for cos χp,q is symmetric in p and q, so that
cos χp,q = cos χq,p , from which we deduce that χp,q = χq,p . We leave the
verification of the other half of (7.10) as an Exercise.
a
π
4
χ4,3
ψ4,3
m
π
2
π
3
1
2 E4,3
Figure 7.6
b
116
CHAPTER 7. GEODESIC STRUCTURES, II
What is the geometrical significance of (7.10)? Let us consider a specific
example, say, the cube. In this case, Figure 7.5 may be redrawn as Figure
7.6. Recall that a is the projection of the center of a face of the cube onto
its circumsphere, m is the projection of the midpoint of an edge, and b is
a vertex of the cube (which lies on the circumsphere). In how many ways
might we project such a triangle onto the circumsphere?
r
s
t
u
(b)
(a)
(c)
(d)
Figure 7.7
Consider two triangles as shown in Figure 7.6, one the reflection of the
other, as they would appear drawn on a face of the cube (see Figure 7.7(a)).
We see that 8×6 = 48 such triangles would tile the cube, as shown in Figure
7.7(b). If these triangles are projected simultaneously onto the circumsphere
of the cube, we obtain a tiling of a sphere by 48 congruent right spherical
triangles, as shown in Figure 7.7(d). (The term “congruent” as used here
7.4. SCHWARZ TRIANGLES
117
includes the possibility that two triangles may have all angles the same, but
be mirror images of each other, as would triangles ∆rst and ∆rut in Figure
7.7(a) when projected onto the circumsphere.)
It is apparent that this procedure may be carried out with any of the
Platonic solids. But one case is of special interest to us at the moment: the
octahedron. In view of (7.10), we see that the result is the same tiling of
the sphere obtained for the cube! This result is not altogether surprising,
as we hope to demonstrate.
Consider the rhombic dodecahedron (see §9.1 and Figure 9.1), and divide each rhombic face into four congruent right triangles as in Figure 7.8.
Now project these 48 triangles onto the circumsphere of the rhombic dodecahedron (although note that only the six outermost vertices would be on
the circumsphere).
4
3
3
4
Figure 7.8
If we look as the six tetravalent vertices of the rhombic dodecahedron
(that is, vertices where four faces are incident and marked with a “4” in
Figure 7.8), we see that eight spherical right triangles meet at such vertices,
and that these triangles form spherical squares identical to those obtained
from projecting the square faces of a cube onto the circumsphere.
On the other hand, if we examine the projections of the eight trivalent
vertices (marked with a “3” in Figure 7.8) of the rhombic dodecahedron onto
the circumsphere, we see that six spherical triangles meet at such projected
vertices, and that these triangles form spherical triangles identical to those
obtained by projecting the triangular faces of the octahedron (as shown in
Figure 7.7(c)) onto the circumsphere.
118
CHAPTER 7. GEODESIC STRUCTURES, II
This analysis may be made for the other dual pairs of Platonic solids
(where the tetrahedron is considered to be dual to itself). In general, the
number of triangles in the spherical tiling is four times the number of edges
on the Platonic solid. The component triangles of such tilings are often
called Schwarz triangles, and are discussed much more fully in Uniform
Polyhedra by Coxeter, Longuet-Higgins, and Miller.
7.5
Wythoff Symbols
There is a convenient way to exploit the apparent “coincidence” which arose
in the previous section; namely, that the cube and octahedron generated the
same tiling of the sphere by spherical triangles. To do so, we introduce the
Wythoff symbol. We content ourselves with a brief introduction; the
devoted reader will certainly not fail to seek out §3 of Uniform Polyhedra
by Coxeter, Longuet-Higgins, and Miller.
Recall that the spherical triangle in Figure 7.6 had dihedral angles π/2,
π/3, and π/4. Imagine 48 of these triangles tiling the sphere, and pose the
question: how many points corresponding to the point labelled b would there
be on the sphere, and what polyhedron has its vertices consisting of these
points?
It should be apparent from our previous discussion that there would be
eight such points on the sphere, and that these points comprise the vertices
of a cube. This may be expressed as
π π π
π π π
= ,
3 2 4
3 4 2
where the three dihedral angles are written, and the vertical slash following
π/3 indicates that it is the vertex of the dihedral angle π/3 which currently
holds our interest.
But Wythoff did not use the radian measure of the angles in notating
this phenomenon; he used the notation p | q r for an analogous polyhedron
whose generating triangle had dihedral angles π/p, π/q, and π/r, and whose
vertices were those at the dihedral angles π/p. Thus, Wythoff would have
described the cube with the expression
3 | 2 4 = 3 | 4 2.
It is perhaps clear, given the discussion in the previous section, that the
octahedron is given by
4 | 2 3 = 4 | 3 2.
7.5. WYTHOFF SYMBOLS
119
The curious reader is no doubt wondering about
2 | 3 4 = 2 | 4 3.
Examination of Figure 7.8 suggests that points such as m (see Figure 7.6) are
projections of the centers of the rhombic faces of a rhombic dodecahedron;
thus, we suspect (see Figure 9.1), and are indeed correct in our suspicion,
that 2 | 3 4 is the cuboctahedron.
There is more to our story, however. Let us now bisect the dihedral angle
at m (see Figure 7.9) so that the point c on side ab is such that ∠amc and
∠cmb both have measure π/4. When c on side ab is determined in this way,
it may be shown that points such as c on our tiling of the sphere comprise
the vertices of yet another Archimedean solid, the rhombicuboctahedron. To
indicate that c is opposite the dihedral angle π/2 (or “2” in our symbology),
we describe the rhombicuboctahedron with the symbol
3 4 | 2 = 4 3 | 2.
In a similar fashion, we may describe the truncated cube by
2 3 | 4 = 3 2 | 4,
and the truncated octahedron by
2 4 | 3 = 4 2 | 3.
a
π
4
c
π
4
m
π/4
π
3
Figure 7.9
b
120
CHAPTER 7. GEODESIC STRUCTURES, II
Yes, the story continues. There is exactly one point d in the interior of
spherical ∆amb, called its incenter, such that the three arcs drawn from d
perpendicular to the three sides of the spherical triangle all have the same
measure. There is one such point for each of the 48 triangles tiling our
sphere; these 48 points are vertices of a rhombitruncated cuboctahedron,
notated by 2 3 4 | (and the five other permutations of this symbol).
And finally, we denote by | 2 3 4 (and its other five permutations) the
snub figure corresponding to spherical ∆amb, the snub cube.
Regrettably, this concludes our brief introduction. Again, the reader is
invited to peruse the reference cited above for a decidedly more thorough
discussion of the description of polyhedra using Wythoff symbols.
7.6
Exercises
1. Using the notations of §7.1, show that for any right spherical triangle
(see Figure 7.1), we have
sin a sin b
tan a tan b
=
= sin c tan c.
cos A cos B
sin A sin B
2. In a right isosceles spherical triangle, a = b and A = B in Figure 7.1.
In this case, show that
sin 2a
sin 2A
=
.
sin a
sin2A
3. Bisect the isosceles spherical triangles in Figures 3.11 and 3.12 to obtain right triangles. Use these triangles together with (3.10) and relationships from §7.1 as appropriate to derive first Exercise 6, then
Exercise 5, of Chapter 3.
4. Consider the quadrilateral as shown in Figure 7.10. Note that B > 90◦ ,
since the sum of the angles in a spherical quadrilateral is greater than
360◦ .
(a) Show that
tan b = sin a and
cos B = − sin2 a.
(b) Let d1 and d2 be the measures of the two diagonals of this quadrilateral in the case a = 30◦ . Show that
√
√
27 + 14
√
cos(d2 − d1 ) =
.
80
7.6. EXERCISES
121
b
B
a
b
a
Figure 7.10
5. Consider the quadrilateral in Figure 7.11(a). By subdividing it into
three right triangles as in Figure 7.11(b), use the outline below to
determine x in terms of a and b. Assume that the measures of a and
b are less than 90◦ .
b
b
a
a
a
x
a
d
(a)
f
(b)
Figure 7.11
(a) Beginning with (7.6), ultimately show that tan d = sec a tan b.
Use this to conclude that
cos2 d =
1
cos2 a cos2 b
=
.
1 + sec2 a tan2 b
1 − sin2 a cos2 b
(b) Employing (7.1) three times, show that
cos f =
cos d
.
cos b
122
CHAPTER 7. GEODESIC STRUCTURES, II
(c) Using (a), derive
sin b
sin d = p
.
1 − sin2 a cos2 b
(d) From (a) and (b), show that
sin a sin b
sin f = p
.
1 − sin2 a cos2 b
(e) Using (a)–(d), conclude that
cos x = cos(d + f ) =
cos b − sin a
.
1 − sin a cos b
(f) Finally, show that
sin x =
cos a sin b
.
1 − sin a cos b
6. Using (7.8) and trigonometric identites as appropriate, derive
sin χ =
sin E2
.
sin πp
7. Using results from §7.1 and §7.2, show that
sin
2π
E
cos ψ tan χ = tan .
p
2
8. Verify the data in Table 7.1.
9. Verify the first relationship in (7.10) as follows. Recall from Exercise
6 of Chapter 3 that
1
π
π
cos Ep,q = cos csc .
2
p
q
(a) By substituting into (7.9) as appropriate, show that
cos ψp,q = csc
π
π
cos ,
p
q
and hence conclude that
1
Ep,q = ψq,p .
2
7.6. EXERCISES
123
(b) By using the results of this Exercise with (7.11), show that
1
cos Ep,q cos ψp,q = cos χp,q .
2
(c) Deduce the result in (b) by applying results from §7.1 to the
spherical right triangle in Figure 7.5. Note, as a result, that the
result in (b) (omitting the subscripts) is valid for any Archimedean
solid, not merely Platonic solids.
10. The tetrahedron generates a tiling of the sphere by 24 congruent spherical triangles. Show that each of these triangles may be decomposed
into two congruent spherical triangles, each congruent to the spherical
triangle in Figure 7.6.
The icosahedron generates a tiling of the sphere into 120 congruent
spherical triangles. Can five of these be arranged so as to produce one
of the 24 triangles obtained from the tetrahedral tiling of the sphere?
Why or why not?
11. Build a geodesic truncated icosahedron using the templates provided
at the end of the chapter. Use Tables 6.1 and 7.1 to help you see how
the triangles need to fit together.
Polyhedron
p
cos2 χ
Tetrahedron
3
1
9
70.529
1
3
54.736
Octahedron
3
1
3
√
5+2 5
15
54.736
2
3
35.264
37.377
τ2
3
20.905
54.736
1
2
45.000
+ 2)
31.717
χ (deg)
cos2 ψ
ψ (deg)
Icosahedron
3
Cube
4
Dodecahedron
5
1
3
√
5+2 5
15
Truncated
3
25
33
29.496
25
27
15.793
Tetrahedron
6
3
11
58.518
1
3
54.736
Truncated
4
4
5
26.565
8
9
19.471
Octahedron
6
3
5
39.232
2
3
35.264
37.377
Table 7.1
1
5 (τ
124
CHAPTER 7. GEODESIC STRUCTURES, II
Polyhedron
Truncated
p
5
Icosahedron
6
Truncated
3
Cube
8
Truncated
3
cos2 χ
√
445+16 5
545
√
51+18 5
109
√
23+16 2
51
√
5+2 2
17
√
109+30 5
183
√
25+8 5
61
Dodecahedron
10
Cuboctahedron
3
2
3
4
3
Icosidodecahedron
5
Rh’truncated
4
Cuboctahedron
6
8
Rh’truncated
4
Icosidodeca’n
6
10
Rh’cubocta’n
3
4
Rh’icosidodeca’n
3
4
5
χ (deg)
20.077
23.800
cos2 ψ
√
765−9 5
810
τ2
18.939
3
√
3+2 2
6
47.266
1
2
11.211
1
5 (τ
16.472
20.905
9.736
45.000
5.660
+ 2)
31.717
35.264
8
9
19.471
1
2
45.000
2
3
35.264
τ2
3
20.905
1
5 (τ
+ 2)
√
71+12 2
97
√
45+24 2
97
√
69−2 2
97
√
179+24 5
241
√
117+48 5
241
√
175+10 5
241
√
31+8 2
51
√
7+4 2
17
√
79+16 5
123
√
19+8 5
41
√
135+18 5
205
33.017
√
23+3 5
30
ψ (deg)
4
15 (τ
+ 2)
10.812
4
5
√
10+ 2
12
√
2+ 2
4
√
10− 2
12
√
25+2 5
30
√
5+2 5
10
25.565
24.095
18.462
5
6
√
10+ 2
12
√
2+ 2
4
√
25+2 5
30
√
5+2 5
10
22.393
9
10
18.435
31.717
17.764
25.561
34.316
10.717
15.248
25.184
24.374
30.361
14.985
Table 7.1 (continued)
12.764
22.500
32.236
7.623
13.283
12.764
22.500
7.623
13.283
Chapter 8
Antiprisms and Snub
Polyhedra
Both antiprisms and the snub polyhedra have more than three faces meeting
at each vertex. None of the techniques used so far allows us to easily calculate
their edge and dihedral angles, so we need a fresh approach. If one imagines
a sphere centered at the vertex of, say, a square antiprism (peek ahead to
Figure 8.3), it is evident that a spherical isosceles trapezoid is formed. So
we begin by deriving some useful formulas for these spherical polygons.
8.1
Spherical Isosceles Trapezoids
Let the sides and dihedral angles of a spherical trapezoid be labelled as in
Figure 8.1(a). We consider the decomposition of this trapezoid shown in
Figure 8.1(b). Here, r and s divide the arcs on which they lie into two equal
pieces, so that the dihedral angle qrs is a right angle.
Considering spherical ∆qps, we may use (3.4) to obtain both
cos δ = cos β cos
α
α
+ sin β sin cos D1
2
2
(8.1)
cos β = cos δ cos
α
α
+ sin δ sin cos E.
2
2
(8.2)
and
125
126
CHAPTER 8. ANTIPRISMS AND SNUB POLYHEDRA
γ
D2
γ
q
r
D2
β
β
β
δ
β
δ
F
D1
D1
α
p
D1
E
α
2
(a)
D1
E
s
α
2
(b)
Figure 8.1
We may apply (7.4) to spherical ∆qrs, which yields, since E and F are
complementary,
sin γ2
sin γ2
=
.
sin δ =
sin F
cos E
This relationship may be substituted into (8.2) which, after slightly rearranging terms, yields
α
α
γ
cos δ cos = cos β − sin sin .
2
2
2
Now we may multiply both sides of (8.1) by cos α2 ; equating the expression
for cos δ cos α2 thereby obtained with that of the previous equation yields
cos β cos2
α
α
α
α
γ
+ sin β sin cos cos D1 = cos β − sin sin .
2
2
2
2
2
Replacing cos2 α2 by 1 − sin2 α2 in the previous equation, simplifying, and
solving for D1 yields
cos D1 =
cos β sin α2 − sin γ2
.
sin β cos α2
(8.3)
Of course a parallel argument may be used to obtain cos D2 , resulting in
cos D2 =
cos β sin γ2 − sin α2
.
sin β cos γ2
(8.4)
It will also be useful later on to find similar formulae in the case that
the diagonals of the trapezoid, rather than the sides, are given (as in Figure
8.2). Note that the symmetry of the figure implies that w is the midpoint
of arc qs and u is the midpoint of arc pt so that both ∠swv and ∠puv are
right angles. Thus, u, v and w all lie on the same great circle which bisects
Dx .
8.1. SPHERICAL ISOSCELES TRAPEZOIDS
γ
w
s
D2
β−δ
127
q
β−δ
v
δ
Dx
u
α
p
δ
D1
t
Figure 8.2
We first consider spherical triangle ∆tuv, where we may apply (7.4) to
obtain
sin α2
sin δ =
.
(8.5)
sin 12 Dx
Similarly, using (7.4) with spherical triangle ∆swv results in
sin γ2
= sin(β − δ) = sin β cos δ − cos β sin δ,
sin 12 Dx
(8.6)
where β is the measure of the diagonals pq and st. We may solve (8.5) for
sin 12 Dx , substitute into (8.6), and rearrange terms, obtaining
sin α2 cos δ
cos β sin α2 + sin γ2
=
.
sin δ
sin β
(8.7)
Now considering spherical triangle ∆ptv, we may apply (3.4), giving
cos δ = cos α cos δ + sin α sin δ cos D1 .
Solving this equation for cos D1 and using the fact that
sin α2
1 − cos α
α
= tan =
,
sin α
2
cos α2
we have
cos D1 =
sin α2 cos δ
(1 − cos α) cos δ
=
.
sin α sin δ
cos α2 sin δ
Finally, we substitute from (8.7) into this equation, yielding
cos D1 =
cos β sin α2 + sin γ2
.
sin β cos α2
(8.8)
128
CHAPTER 8. ANTIPRISMS AND SNUB POLYHEDRA
Similar reasoning may be used to find an expression for cos D2 , resulting in
cos D2 =
cos β sin γ2 + sin α2
.
sin β cos γ2
(8.9)
We now wish to determine cos Dx . Applying (3.6) to spherical triangle ∆pvt
yields, since sin2 D1 = 1 − cos2 D1 ,
cos Dx = − cos2 D1 + sin2 D1 cos α = cos α − (1 + cos α) cos2 D1 .
Using the relationship 1 + cos α = 2 cos2 α2 , we may substitute (8.8) into this
equation, which results in
2(cos β sin α2 + sin γ2 )2
.
cos Dx = cos α −
sin2 β
Expanding this expression for cos Dx and simplifying using the relationship
2 sin2 2θ = 1 − cos θ gives
cos Dx =
cos α + cos γ − 1 − cos2 β − 4 sin α2 sin γ2 cos β
.
sin2 β
(8.10)
Thus, we have found means of calculating cos D1 , cos D2 and cos Dx .
8.2
Dihedral Angles of the Antiprisms
We now embark upon the task of calculating the edge and dihedral angles of
the antiprisms. In doing so, we prepare for a discussion of the snub figures.
Let us consider an antiprism whose faces, aside from the triangular ones,
have p sides. Thus, one p-sided polygon and three triangles meet at a vertex
of this Archimedean solid. (See, for example, the pentagonal antiprism of
Figure 2.3(b)). To find the dihedral angles, we proceed as usual to center
a sphere at a vertex of the antiprism,
resulting
in a spherical quadrilateral
2
as in Figure 8.3. Here, ϕ = π 1 − p is an angle of a p-sided polygon,
Dap
3 is the dihedral angle between two triangular faces (the superscript “ap”
indicates an antiprism (“a”) with p-sided (“p”) bases), while Dap
p represents
the dihedral angle between a triangular and a p-sided face.
8.3. EDGE ANGLES OF THE ANTIPRISMS
129
π
3
Dap
3
Dap
3
π
3
π
3
Dap
p
Dap
p
ϕ = π 1 − p2
Figure 8.3
Because of the symmetry of Figure 8.3, the spherical quadrilateral is in
fact an isosceles trapezoid, so that our recent work is applicable. Applying
(8.3) and (8.4) immediately yields
cos Dap
p
cos π3 sin ϕ2 − sin π6
1 ϕ
ϕ
= √ tan − sec
=
ϕ
π
sin 3 cos 2
2
2
3
and
cos Dap
3 =
cos π3 sin π6 − sin ϕ2
1
ϕ
=
1
−
4
sin
.
sin π3 cos π6
3
2
(8.11)
(8.12)
We may rewrite (8.11) and (8.12) in terms of p rather than ϕ, the result
being
1
π
1
π
π
ap
ap
cos Dp = √
, cos D3 =
1 − 4 cos
.
(8.13)
cot − csc
p
p
3
p
3
We remark that an apparent ambiguity arises in the case p = 3; in this
case both expressions in (8.13) represent cos Da3
3 . Recall, however, that
a triangular antiprism is in fact an octahedron (see §5.4), so that both
1
expressions in (8.13) yield cos Dap
3 = cos D3,4 = − 3 . The ambiguity is only
apparent.
8.3
Edge Angles of the Antiprisms
We now proceed to find the edge angle of the antiprism described at the
beginning of the last section. We begin by considering two adjacent triangular faces, as shown in Figure 8.4(a). Here, p and r are centers of two
adjacent triangular faces, with q the midpoint of the common edge, and O
is the center of the Archimedean solid. A cross-sectional view is taken in
130
CHAPTER 8. ANTIPRISMS AND SNUB POLYHEDRA
Figure 8.4(b). Clearly, [pq] = [qr], ∠qpO and ∠qrO are both right angles,
and ∠pqr has the same measure as Dap
3 . If γ is the measure of angles ∠pOq
π
1 ap
and ∠rOq, we see that γ = 21 (π − Dap
3 ) = 2 − 2 D3 , so that
π 1 ap
1
− D3 = sin Dap
.
(8.14)
cos γ = cos
2 2
2 3
t
q
p q
r
Dap
3
p
t̄
C
r
1 ap
2E
γ γ
O
O
(a)
p̄
(b)
π
3
γ
q̄
(c)
Figure 8.4
With this information, we now consider the spherical triangle ∆p̄q̄ t̄ in
Figure 8.4(c), which is the spherical version of ∆pqt of Figure 8.4(a). Here,
Eap is the edge angle of the antiprism, with 21 Eap corresponding to the halfedge qt in Figure 8.4(a).
We now apply (7.7) to the right spherical triangle ∆p̄q̄ t̄ to obtain
sin γ tan
π √
1
= 3 sin γ = tan Eap .
3
2
Using this relationship with the appropriate trigonometric identities, we find
that
1
1
2(1 + 2 cos Eap )
cos2 γ = 1 − sin2 γ = 1 − tan2 Eap =
.
3
2
3(1 + cos Eap )
From (8.14), we also find that
1
1
ap
cos2 γ = sin2 Dap
3 = (1 − cos D3 ).
2
2
Equating these expressions for cos2 γ and solving for cos Eap , we see that
cos Eap =
4
− 1.
5 + 3 cos Dap
3
(8.15)
8.4. DIHEDRAL ANGLES OF THE SNUB POLYHEDRA
131
Finally, substituting for cos Dap
3 from (8.13) yields
cos Eap =
8.4
2
− 1.
3 − 2 cos πp
(8.16)
Dihedral Angles of the Snub Polyhedra
Calculation of the dihedral angles of the snub cube and the snub dodecahedron poses another challenge. The general situation is illustrated in Figure
8.5(a), where the dihedral angles between triangular faces are all equal –
note the similarity to Figure 8.3 – but for the snub polyhedra, we have an
extra triangular face at each vertex.
π
3
π
3
D2
π
3
D2
u
D2
π
3
π
3
D1
D1
π
3
p
ϕ
(a)
t
D2
π
3
α
β
ϕ
r
π
3
D1 q
(b)
Figure 8.5
A fruitful approach is to dissect the spherical pentagon as shown in
Figure 8.5(b). What is useful here is that the spherical quadrilaterals putr
and purq are in fact isosceles trapezoids, so again all of our work from §8.1
is available to us.
We begin by considering the triangle and trapezoid which share the
common side α. Applying (3.4) to spherical ∆utr yields
cos α = cos2
π
π
1
+ sin2 cos D2 = (1 + 3 cos D2 ).
3
3
4
(8.17)
Applying (8.3) to spherical trapezoid purq results in
cos D1 =
cos π3 sin ϕ2 − sin α2
sin ϕ2 − 2 sin α2
√
=
.
sin π3 cos ϕ2
3 cos ϕ2
With the abbreviation
λ :=
√
3 cos
ϕ
ϕ
cos D1 − sin ,
2
2
(8.18)
132
CHAPTER 8. ANTIPRISMS AND SNUB POLYHEDRA
this relationship may be rewritten as
λ = −2 sin
α
< 0.
2
(8.19)
We may substitute into sin2 α2 = 21 (1 − cos α) from (8.17) and (8.19), obtaining
2
cos D2 = 1 − λ2 .
(8.20)
3
We now look at the triangle and trapezoid which share the common side
β. Applying (3.4) to spherical ∆pqr results in
cos β = cos
√
π
1
π
cos ϕ + sin sin ϕ cos D1 = (cos ϕ + 3 sin ϕ cos D1 ). (8.21)
3
3
2
Considering the spherical trapezoid putr, we see that D2 and β have the
same relationship as Dap
3 and ϕ of Figure 8.3, so that (8.12) becomes
1
β
cos D2 =
1 − 4 sin
.
(8.22)
3
2
We may substitute into sin2 β2 = 12 (1−cos β) from (8.21) and (8.22), resulting
in
√
9 cos2 D2 − 6 cos D2 − 7 + 4 cos ϕ + 4 3 sin ϕ cos D1 = 0.
(8.23)
Solving for cos D1 from (8.18), we have
1
ϕ
ϕ
√
cos D1 =
sec
λ + sin
.
2
2
3
(8.24)
Substituting for cos D1 from (8.24) and cos D2 from (8.20) into (8.23) yields,
since λ 6= 0,
ϕ
λ3 − 2λ + 2 sin = 0.
(8.25)
2
Once λ is found by solving this cubic equation, cos D1 may be found using
(8.24).
This cubic also yields a cubic equation which may be solved to obtain
cos D2 . From (8.25), we see that
ϕ 2
(λ3 − 2λ)2 = −2 sin
,
2
so that
λ6 − 4λ4 + 4λ2 − 4 sin2
ϕ
= 0.
2
8.5. EDGE ANGLES OF THE SNUB POLYHEDRA
133
Since all powers of λ are even, we may substitute for λ2 from (8.20), yielding
27 cos3 D2 − 9 cos2 D2 − 15 cos D2 + 13 − 16 cos ϕ = 0.
(8.26)
Because of the relationship (8.20), it is evident that only one of (8.25) and
(8.26) needs to be solved in order to find cos D1 and cos D2 .
We use a notation for the dihedral angles of the snub polyhedra similar
to that developed in §8.2 for the antiprisms. A superscript “s” is used to
denote a snub figure, so that Ds4,3 [4-3] is the dihedral angle between the
square and triangular faces of the snub cube (the subscript denoting the six
square faces in the facial planes of the cube), and Ds3,4 [3-3] is the dihedral
angle between the triangular faces (the subscript relating to the triangular
faces, eight of which lie in the facial planes of an octahedron). Ds5,3 [5-3]
and Ds3,5 [3-3] have similar connotations as far as the snub dodecahedron is
concerned.
We may use Cardan’s formula (which is summarized immediately preceding the Exercises) to solve (8.26); for the snub cube, we have ϕ = 90◦ ,
from which we calculate
q
q
√
√
2 3
1
3
s
cos D3,4 =
−19 + 297 + −19 − 297 +
≈ −0.892858.
9
2
3
For the snub dodecahedron, we have ϕ = 10
, and hence
√
1 √
1
3
3
s
cos D3,5 =
u+v+ u−v+
≈ −0.962101,
3
3
where
r
√
√
2
2 2
u = − (49 + 27 5), v =
(93 + 49 5).
27
3 3
The other dihedral angles may then be found as described above. Data are
summarized in Table 8.1.
8.5
Edge Angles of the Snub Polyhedra
Briefly reviewing the procedure in §8.3 for calculating the edge angles of
the antiprisms, it is apparent that the same procedure may be followed for
calculating the edge angles of the snub polyhedra. Thus, (8.15) is at our
disposal; denoting the edge angles of the snub cube and snub dodecahedron
by Es3,4 and Es3,5 , respectively, we see from (8.15) that
cos Es3,4 =
4
− 1,
5 + 3 cos Ds3,4
cos Es3,5 =
4
− 1.
5 + 3 cos Ds3,5
(8.27)
134
CHAPTER 8. ANTIPRISMS AND SNUB POLYHEDRA
Thus, these edge angles may easily be found numerically once the dihedral
angles are obtained. We proceed with a more algebraic approach.
For brevity, we write (8.27) as
cos E =
4
− 1,
5 + 3 cos D
and hence we will omit subscripts and superscripts for now. This relationship
may be solved for cos D, yielding
1 5 cos E + 1
cos D = −
.
3 cos E + 1
Recall, however, that cos D satisfies the cubic equation (8.26) since D2 in
that equation represents the dihedral angle between triangular faces. Thus,
we may substitute the previous expression for cos D into (8.26). After multiplying this equation by −(cos E + 1)3 , expanding, grouping like terms, and
dividing by 16, we obtain
(cos ϕ + 7) cos3 E + (3 cos ϕ + 1) cos2 E + (3 cos ϕ − 3) cos E + cos ϕ − 1 = 0.
Since ϕ 6= 0 and hence 1 − cos ϕ 6= 0, we may divide through by 1 − cos ϕ
and use the identity 1 − cos ϕ = 2 sin2 ϕ2 , which results in
ϕ
ϕ
(8.28)
4 csc2 − 1 cos3 E + 2 csc2 − 3 cos2 E − 3 cos E − 1 = 0.
2
2
When ϕ = 90◦ , we obtain the following equation for cos Es3,4 :
7 cos3 Es3,4 + cos2 Es3,4 − 3 cos Es3,4 − 1 = 0.
(8.29)
When ϕ = 108◦ , the equation for cos Es3,5 becomes
√
√
(23 − 8 5) cos3 Es3,5 + (9 − 4 5) cos2 Es3,5 − 3 cos Es3,5 − 1 = 0.
(8.30)
These may be solved using Cardan’s formula, if desired. The details are left
as an Exercise.
8.6. SOLVING A CUBIC
135
Polyhedron
Triangular Antiprism
θ
cos θ
θ
θ (deg)
Da3
3
− 13
D3,4
109.471
Ea3
0
E3,4
90.000
Da4
4 [3-4]
Square Antiprism
Da4
3 [3-3]
Ea4
Pentagonal Antiprism
√
√1 (1 − 2)
3
√
1
3 (1 − 2 2)
√
2 2−1
7
103.836
√τ −2
9−3τ
√
− 35
100.812
D3,5
138.190
√1
5
E3,5
63.435
Da5
5 [3-5]
Da5
3 [3-3]
Ea5
Hexagonal Antiprism
Snub Dodecahedron
74.858
√2
3
√
1−2 3
3
Da6
6 [3-6]
1−
Da6
3 [3-3]
Snub Cube
127.552
98.899
145.222
Ea6
√1
3
Ds3,4 [3-3]
−0.892858
153.235
Ds4,3 [4-3]
−0.798461
142.983
Es3,4
0.723078
43.691
Ds3,5 [3-3]
−0.962101
164.175
Ds5,3 [5-3]
−0.890451
152.930
Es3,5
0.892418
26.821
54.736
Table 8.1
8.6
Solving a Cubic
Here, we summarize a standard method for finding the roots of a cubic
equation, often referred to as Cardan’s formula.
So let the cubic equation we wish to solve have the form
x3 + ax2 + bx + c = 0.
Define α and β by
1
α = (3b − a2 ),
3
β=
1
(2a3 − 9ab + 27c).
27
136
CHAPTER 8. ANTIPRISMS AND SNUB POLYHEDRA
Also, put
∆=
Now define
β 2 α3
+
.
4
27
r
β √
β √
3
+ ∆, δ = − − ∆.
2
2
Then the three roots of the cubic equation are
r
γ=
a
γ+δ− ,
3
−
3
−
a
γ + δ γ − δ√
−3 − ,
+
2
2
3
−
a
γ + δ γ − δ√
−3 − .
−
2
2
3
Moreover, if a, b, and c are all real numbers, then if ∆ > 0, there is one real
root and two conjugate complex roots. If ∆ = 0, there are three real roots
with at least two equal. Finally, if ∆ < 0, there are three distinct real roots.
We illustrate with the example of finding cos Ds3,4 as given in §8.4. So
with x = cos Ds3,4 and ϕ = 90◦ , we see that (8.26) becomes
27x3 − 9x2 − 15x + 13 = 0.
Dividing through by 27 yields
1
5
13
x3 − x2 − x +
= 0,
3
9
27
so that
1
a=− ,
3
5
13
b=− , c= .
9
27
5 1
16
1
− −
=−
α=
3
3 9
27
Then
and
1
β=
27
1
5
304
2 −
−9·
+ 13 =
.
27
27
729
This gives
704
26 · 11
∆=
=
= 297 ·
19683
39
6
2
.
9
Also, we have
β
152
− =−
= −19 ·
2
729
3
2
.
9
Writing ∆ and −β/2 in this way immediately gives
q
q
√
√
23
23
γ=
−19 + 297, δ =
−19 − 297.
9
9
8.7. EXERCISES
137
The only real root of this cubic is then
q
q
√
√
1
a
2 3
3
−19 + 297 + −19 − 297 +
,
γ+δ− =
3
9
2
agreeing with the result at the end of §8.4.
8.7
Exercises
1. Derive (8.25) as suggested in the text.
2. Derive (8.26) as suggested in the text.
3. Show that with the substitution
z=
1 − 3 cos D2
,
2
(8.26) becomes
z 3 − z 2 − z − 1 + 2 cos ϕ = 0.
Also show that with λ as given by (8.20), we have
λ2 = z + 1.
4. Derive (8.28) as suggested in the text.
5. By using Cardan’s formula to solve the cubic equation (8.25), and then
using (8.24), show that
q
q
√
√
1
3
3
s
cos D4,3 = √
−54 + 6 33 + −54 − 6 33 + 3 ≈ −0.798461.
3 3
and
√
√
2
τ
3
cos Ds5,3 = p
u+v+ 3u−v+
≈ −0.890451,
2
3(3 − τ )
where
τ
u=− ,
2
1
v=
6
r
27τ − 5
.
3
6. With E and D1 representing either Es3,4 and Ds4,3 , or Es3,5 and Ds5,3 ,
respectively, use the following outline to show that
√
ϕ
3 cos E cos D1 cot = −1.
(8.31)
2
138
CHAPTER 8. ANTIPRISMS AND SNUB POLYHEDRA
(a) Show that (8.26) may be rewritten as
(9 cos2 D2 − 6 cos D2 − 3)(3 cos D2 + 1) + 16(1 − cos ϕ) = 0.
(b) Solve this relationship for 9 cos2 D2 − 6 cos D2 and substitute the
result into (8.23).
(c) Finally, solve (8.27) for cos D2 (which may be either cos Ds3,4 or
cos Ds3,5 ) and substitute into the result from (b). Using trigonometric identities as appropriate, simplify to obtain (8.31).
7. Use Cardan’s formula to solve (8.29), yielding
q
q
√
√
1
3
3
s
cos E3,4 =
566 + 42 33 + 566 − 42 33 − 1 .
21
8. The results of this chapter may all be used to find the dihedral and
edge angles of the icosahedron by choosing ϕ = 60◦ .
(a) In this case, show that (8.25) becomes
1
= 0,
(λ + τ )(λ − 1) λ −
τ
and use the
negative root λ = −τ with (8.24) to show that
√
5
cos D1 = − 3 .
(b) Show that (8.26) may be written
(3 cos D2 − 1)(9 cos2 D2 − 5) = 0,
√
which also yields cos D2 = −
root.
5
3 ,
as expected, as its only negative
(c) Show that (8.28) may be written
(5 cos2 E − 1)(3 cos E + 1) = 0,
which yields cos E =
√1
5
as its only postive root.
9. Note that for the derivation of (8.26), the assumption that ϕ > 0
was not needed. Show that when ϕ = 0, cos D2 = − 13 is a root of
(8.26), and when ϕ = −60◦ , cos D2 = 31 is a root of (8.26). Note
that these correspond to the dihedral angles of the octahedron and
the tetrahedron. Why do you think that this is the case? Note also
that (8.27) remains valid for the octahedron and tetrahedron, with
superscripts and subscripts appropriately modified.
8.7. EXERCISES
10. Note that when ϕ = 2 ·
Explain.
139
π
3
=
2π
3 ,
then cos D2 = −1 is a root of (8.26).
140
CHAPTER 8. ANTIPRISMS AND SNUB POLYHEDRA
Chapter 9
Duality
We are now prepared to address duality, a topic central to the study of
polyhedra. Our approach will not be the most general possible, but will
suffice for a discussion of the Archimedean solids and their duals. Many
interesting polyhedra shall be encountered during our investigation.
9.1
Basic Concepts
We begin by examining the process by which a dual polyhedron is created
from its parent polyhedron. Consider first the cuboctahedron as in Figure
9.1(a). On each edge, find that point which, when joined to the center
of the polyhedron, results in a segment perpendicular to that edge. (Note
that in the case of the Archimedean solids, this is always the midpoint of
the edge.) Then the line in space perpendicular to both this segment and
the original edge contains an edge of the dual polyhedron. By executing
this procedure at each edge of the cuboctahedron, one finds that the lines
formed as perpendiculars to the edges soon begin intersecting each other
at vertices of the dual polyhedron. The completion of the process for the
cuboctahedron yields the rhombic dodecahedron of Figure 9.1(b), whose
name is derived from the fact that each of its twelve faces is a rhombus.
The rhombic dodecahedron by itself is shown in Figure 9.2.
In examining Figures 9.1 and 9.2, one may make several observations
which happen to be valid for any Archimedean solid and its dual. For
example:
1. The faces of the Archimedean dual are congruent to one another. This
results from the fact that there is the same arrangement of faces at
each vertex of an Archimedean solid (see §5.1).
141
142
CHAPTER 9. DUALITY
(a)
(b)
Figure 9.1
Figure 9.2
2. Upon superposing the dual on its parent (seen in Figure 9.1(b)), one
sees that each edge of the dual is perpendicular to an edge of the
parent. Moreover, each edge of the Archimedean dual bisects the corresponding edge of its parent. The converse, however, is not true –
the edges of the parent do not necessarily bisect those of the dual. For
example, one may show that the points where the edges of the cuboctahedron meet those of the rhombic dodecahedron divide the edges
of the rhombic dodecahedron in the ratio 1 : 2 (see Figure 9.3). In
addition, it is these points (and not the midpoints, as was the case
for the cuboctahedron) on the edges of the dual which, when joined
to the center of the dual, form segments perpendicular to the edges of
the dual.
3. As a result of the process for constructing the dual, we see that the
number of edges on an Archimedean solid is the same as the number
9.2. FURTHER EXAMPLES
143
of edges on its dual. Moreover, it is also evident from the construction
process that the number of faces of the parent is the same as the
number of vertices of the dual, while dually, the number of vertices
on the parent is the same as the number of faces on the dual. These
“reciprocal” relationships are of the type encountered in a discussion
of duality in almost any branch of mathematics.
1
2
Figure 9.3
4. It is evident from the construction process that taking the dual of
an Archimedean dual results in the parent Archimedean solid. For
example, the dual of the rhombic dodecahedron (itself an Archimedean
dual) is in fact a cuboctahedron.
9.2
Further Examples
Before going on to discuss the calculation of edge and dihedral angles of
duals, we look at several examples of duals of Archimedean solids. We first
consider the tetrahedron. As is seen in Figures 9.4(a) and (b), a tetrahedron may be inscribed in a cube in two different ways; in each case the six
edges of the tetrahedron consist of one diagonal from each face of the cube.
Superposing these two tetrahedra yields Figure 9.4(c), where it is evident
that these tetrahedra are, in fact, duals of each other. Indeed, the centers
of the tetrahedron and cube coincide, and the segment joining this common
center and a center of a face of the cube (such as m in Figure 9.4(c)) is
orthogonal to both of the face diagonals (edges of the tetrahedra) which are
perpendicular at m. Note that the edges of the dual tetrahedron are in fact
bisected by those of the parent (in contrast to Observation 3 in the previous
section). Such will be the case only when the parent solid is a Platonic solid.
144
CHAPTER 9. DUALITY
m
(a)
(b)
(c)
Figure 9.4
The polyhedron in Figure 9.4(b) is often called a stella octangula (Exercise 6(b) of Chapter 2 and model 19 in Polyhedron Models), indicating
that these two mutually intersecting tetrahedra may be formed by affixing
a small tetrahedron to each face of an octahedron. Because the polyhedron
dual to a tetrahedron is another tetrahedron, the tetrahedron is sometimes
said to be self-dual.
What of the other Platonic solids? It is clear from Figure 9.5 that the
octahedron is dual to the cube (and vice versa), and from Figure 9.6 that the
icosahedron is dual to the dodecahedron (and vice versa). (See also models
43 and 47, respectively, in Polyhedron Models.)
Figure 9.5
Figure 9.6
In fact, due to the highly symmetrical nature of the Platonic solids, it is
the case that among the group of Platonic and Archimedean solids:
1. The Platonic solids are the only solids whose duals are among the
group as well, and
9.2. FURTHER EXAMPLES
145
2. The Platonic solids are the only solids among the group all of whose
dual edges are bisected by the parent solid.
For completeness, we include the duals of all the Platonic and Archimedean
solids below. Names for these polyhedra are given in Table 9.1. A few explanatory remarks are in order, however.
1. A name such as “pentakis dodecahedron” indicates that this polyhedron may be formed by placing a pentagonal (hence “pentakis”) pyramid on each face of a dodecahedron (here, the triangular faces of the
pyramids are not necessarily equilateral).
2. Adjectives such as “rhombic,” “pentagonal,” and “trapezial” indicate
the shape of the faces of the polyhedron. Here, “pentagonal” does
not necessarily imply regular pentagons. A trapezium is a quadrilateral with no two sides parallel, hence the adjective “trapezial.” Many
authors use “trapezoidal” rather than “trapezial;” it happens, according to my dictionary, that the British interchange the meanings of
“trapezium” and “trapezoid.”
3. “Icositetra” is a prefix indicating “24,” “triaconta” denotes “30,” and
“hexeconta” means “60.” These prefixes refer to the number of faces
on the polyhedron.
Polyhedron
Dual Polyhedron
Tetrahedron
Tetrahedron
Cube
Octahedron
Octahedron
Cube
Icosahedron
Dodecahedron
Dodecahedron
Icosahedron
Truncated tetrahedron
Triakis tetrahedron
Truncated octahedron
Tetrakis hexahedron
Truncated icosahedron
Pentakis dodecahedron
Truncated cube
Triakis octahedron
Table 9.1
146
CHAPTER 9. DUALITY
Polyhedron
Dual Polyhedron
Truncated dodecahedron
Triakis icosahedron
Cuboctahedron
Rhombic dodecahedron
Icosidodecahedron
Rhombic triacontahedron
Rhombitruncated cuboctahedron
Hexakis octahedron
Rhombicuboctahedron
Trapezial icositetrahedron
Rhombitruncated icosidodecahedron
Hexakis icosahedron
Rhombicosidodecahedron
Trapezial hexecontahedron
Snub cube
Pentagonal icositetrahedron
Snub dodecahedron
Pentagonal hexecontahedron
Table 9.1 (continued)
9.3
Edge Angles
How can the edge angles of an Archimedean dual be calculated from information about the parent Archimedean solid? This section will focus on
answering that question.
Let us consider for a moment Figure 9.7(a), where we see a truncated
tetrahedron and a face rtu of its dual, a triakis tetrahedron. Important
points are labelled in the cross-section shown in Figure 9.7(b): O is the
center of the truncated tetrahedron, p is the center of a hexagonal face of
the truncated tetrahedron, and q is the center of a triangular face. Because
of the symmetry of the truncated tetrahedron about this triangular face,
it follows that the vertex of the triakis tetrahedron above this triangular
face – labelled r in Figure 9.7(a) and formed as the intersection of the
perpendicular bisectors of the edges of this face created in the process of
−→
dualizing the truncated tetrahedron – lies on the ray Oq. For similar reasons,
the vertex t of the triakis tetrahedron above the hexagonal face of which p
−→
is the center must lie on the ray Op. Finally, s is the point on the edge rt of
the triakis tetrahedron which is the midpoint of the edge of the truncated
tetrahedron in which it lies, and hence the segment Os is perpendicular to
the edge rt.
Since s is the midpoint of the edge incident to the faces of which p and
9.3. EDGE ANGLES
147
q are centers, it follows that ∠qsp has the same measure as the dihedral
angle Dt3,3 of the truncated tetrahedron. Again, due to the symmetry of the
truncated tetrahedron, the segment Op is perpendicular to the hexagonal
face of which p is the center. As a result, we see that segment Op and ps in
Figure 9.7(b) are perpendicular. Similarly, Oq and qs are also perpendicular.
Finally, since rt is an edge of the triakis tetrahedron (whose center is O),
we see that ∠rOt is an edge angle of the triakis tetrahedron, which shall be
denoted by Ēt3,3 .
r
t
u
r
q
s
t
Dt3,3
(a)
Ēt3,3
p
O
(b)
Figure 9.7
It is evident from examining the quadrilateral Oqsp that Dt3,3 and Ēt3,3
are supplementary, so that Dt3,3 + Ēt3,3 = π, and hence cos Ēt3,3 = − cos Dt3,3 .
This phenomenon is not peculiar to this pair of duals; in general, we see
that
(D1 )
The edge angle subtended by an edge of an Archimedean dual
is supplementary to that dihedral angle of the Archimedean
solid whose edge is bisected by the dual edge.
This relationship suggests the notation Ēt3,3 for the edge angle of the
triakis tetrahedron subtended by the edge rt. Thus, if D∗p,q represents a dihedral angle of an Archimedean solid, then Ē∗p,q represents the corresponding
148
CHAPTER 9. DUALITY
edge angle of its dual, and we have cos Ē∗p,q = − cos D∗p,q . We see, then, that
Ē3,3 is the edge angle of the edge tu of the triakis tetrahedron (see Figure
9.7(a)) which intersects an edge of the truncated tetrahedron at which the
dihedral angle is D3,3 , the same as that of the regular tetrahedron. It follows, therefore, that tu would also be an edge of the tetrahedron dual to
the tetrahedron which has been truncated in Figure 9.7(a). This gives the
triakis tetrahedron its name, for it may be imagined as a tetrahedron with
squat triangular (“triakis”) pyramids affixed to its faces.
We now proceed to investigate Figure 9.7(b) in more detail. Upon
consideration of the similar right triangles ∆Oqs and ∆Osr, we see that
[Oq]/[Os] = [Os]/[Or]; likewise, we note that the similarity of right triangles ∆Ops and ∆Ost implies that [Op]/[Os] = [Os]/[Ot]. These two
relationships together imply that
[Oq][Or] = [Os]2 = [Op][Ot].
Such relationships can be written for any Archimedean solid and its dual;
they can be imagined as a result of creating Archimedean duals by the
process of polar reciprocation, a topic which will not be addressed here
(but see Wenninger’s Dual Models, pp. 1-5).
9.4
Faces of Archimedean Duals
Upon looking at the triakis tetrahedron, for example, one observes that the
faces are not equilateral triangles – in fact, far from it! How can we learn
more about the shape of these triangular faces? We will investigate a useful
procedure called the “Dorman Luke construction” by Wenninger in Dual
Models (p. 30). (Actually, this procedure derives from the process of polar
reciprocation.)
We begin with an expanded view of the face rtu of the triakis tetrahedron, as in Figure 9.8(a). Here, in addition to the points labelled in Figure
9.7(a), are the midpoints w and x of two other edges of the truncated tetrahedron, and a vertex v of the truncated tetrahedron. As a result of the
dualizing process, we see that triangles ∆Osv, ∆Owv, and ∆Oxv are all
right triangles (recall that O is the center of the truncated tetrahedron).
Since these triangles all share the same hypotenuse Ov, and since any
triangle inscribed in a semicircle is a right triangle, it follows that these
triangles may be inscribed in a sphere with diameter Ov (shaded in gray
in Figure 9.8(a)). Moreover, since [vs] = [vw] = [vx], this common value
being one-half the edge length of the truncated tetrahedron, it follows that
9.4. FACES OF ARCHIMEDEAN DUALS
149
s, w, and x all lie on a circle which is the intersection of the sphere and a
plane which is perpendicular to Ov (intersecting Ov at a point we call z)
and which passes through s, w, and x. This circle, whose center is z, is
shown in Figure 9.8(b). Because of the perpendicularity of the dual edges
and the parent edges, we see that the sides of the face rtu of the triakis
tetrahedron are in fact tangent to this circle at s, x, and w. Note that the
lengths of ws, sx, and xw may be easily determined as s, w, and x are
the midpoints of edges of the truncated tetrahedron (a flattened vertex of
which is shown in Figure 9.8(c)), where we assume for ease of calculation
that the edge length is 2. The triangle ∆swx is called a vertex figure of
the truncated tetrahedron.
r
v s
w
t
x
u
w
1
s
√
w
3
x
(a)
r
w
u
s
1
z √
h
3
(c)
t
x
(b)
Figure 9.8
To determine the shape of ∆rtu, we see that [ws] = 1 (ws being as long
as half an edge of the truncated tetrahedron), and hence from Figure 9.8(c)
150
CHAPTER 9. DUALITY
√
that [wx] = [xs] = 3. It may be shown that given a triangle whose sides
have length a, b, and c, then the radius R of the circumcircle (i.e., the circle
circumscribing the triangle) satisfies
R2 =
a2 b2 c2
.
(a + b + c)(−a + b + c)(a − b + c)(a + b − c)
√
2
With
√ a = b = 3 and c = 1, we find that R = 9/11 and hence R = [zx] =
3/ 11. It follows (see Figure 9.8(b)) that
√
√
3/2
11
[hx]
cos(∠zux) = sin(∠hzx) =
= √ = √ ,
R
3/ 11
2 3
and hence
cos(∠rut) = 2 cos2 (∠zux) − 1 = 2 ·
11
5
−1= .
12
6
One may similarly conclude that cos(∠urt) = −7/18.
We may summarize our observations succinctly as follows:
(D2 ) The sides of a face of an Archimedean dual are tangent to the
circumcircle of a vertex figure of the Archimedean parent at
the vertices of this figure.
We remark in general that a vertex figure of an Archimedean solid is
obtained by selecting a vertex of the solid and then moving along each edge
incident at that vertex the same distance in order to obtain the vertices
of the vertex figure.√ Thus, as a further example, the vertex figure for a
cuboctahedron is a 2 : 1 rectangle, and thus a face of its dual rhombic
dodecahedron may be found as in Figure 9.9.
1
√
2
Figure 9.9
9.5. DIHEDRAL ANGLES
151
Although the construction (D2 ) of the faces of Archimedean duals is
geometrically intriguing and involves many of the important relationships
between the Archimedean solids and their duals, it is a somewhat arduous
task in an algebraic sense. A simpler algebraic method will be explored in
§9.6.
9.5
Dihedral Angles
We wish to extend our analysis of the previous section in order to calculate dihedral angles of Archimedean duals. We will soon discover, happily,
that these calculations are rather simple, thereby revealing a straightforward
method for finding the angles of the faces of the duals.
Let us first attempt to find the dihedral angle at the edge tu of the triakis
tetrahedron in the previous section (see Figures 9.8(a), (b)). To do so, we
consider the other face of the triakis tetrahedron of which tu is an edge (see
Figure 9.10(a)), where r0 is the other vertex of that face and z 0 is the center
of the circle to which the sides of ∆r0 tu are tangent. (Note: z and z 0 are
not shown in Figure 9.10(a).)
r
w
x
v
v s
z
D̄t3,3
v0
z0
t
x
Et3,3
u
v0
r0
O
(a)
(b)
Figure 9.10
A cross-section of Figure 9.10(a) is shown in Figure 9.10(b). Here, v is
−→
the vertex of the truncated tetrahedron on the ray Oz (see Figure 9.8(a)),
−→
and similarly, v 0 is the vertex of the truncated tetrahedron on the ray Oz 0 .
Recall (Figures 9.8(a),(b)) that Ov is perpendicular to the plane containing
the circle of which z is the center, so that ∠Ozx must be a right angle; one
152
CHAPTER 9. DUALITY
similarly argues that ∠Oz 0 x must likewise be a right angle. Since a tangent
to a circle is perpendicular to a radius drawn to the point of tangency, it
follows that both zx and z 0 x are perpendicular to tu (as in Figure 9.10(a)),
and therefore ∠zxz 0 (see Figure 9.10(b)) must have the same measure as the
dihedral angle of the triakis tetrahedron, which we denote by D̄t3,3 . Since O
is the center of the truncated tetrahedron as well, we see that ∠vOv 0 has
measure Et3,3 , the edge angle of the truncated tetrahedron.
These considerations lead us, upon examining the quadrilateral z 0 Ozx in
Figure 9.10(b), to the conclusion that Et3,3 and D̄t3,3 are supplementary; i.e.,
that Et3,3 + D̄t3,3 = π. But had we considered the dihedral angle at the edge
rt instead, we would nonetheless have arrived at a figure similar to Figure
9.10(b), and would have inevitably concluded that this dihedral angle was
supplementary to Et3,3 . We make the following generalization:
(D3 ) The dihedral angles of an Archimedean dual all have the same
measure, all being supplementary to the edge angle of the parent Archimedean solid.
We use similar notation for the other duals; therefore, if E∗p,q represents
the edge angle of some Archimedean solid, then D̄∗p,q represents the dihedral
angle of its dual, so that E∗p,q + D̄∗p,q = π and hence cos D̄∗p,q = − cos E∗p,q .
9.6
Faces Revisited
Since all of the dihedral angles of the triakis tetrahedron have the same
measure, and since all the angles of faces incident at the same vertex have
the same measure, calculations of the facial angles is a straightforward application of (3.9).
Consider, as an example, the acute angle ϕ (such as ∠rut in Figure
9.8(c)) of the faces of the triakis tetrahedron. Solving (3.9) for ϕ yields
cos ϕ = 2
1 + cos 2π
q
1 − cos D
− 1.
(9.1)
In our case, q = 6 since six acute angles meet at each of four vertices
of the triakis tetrahedron, and of course D = D̄t3,3 . These values may be
substituted into (9.1) (where Table 6.1 is used to find cos D̄t3,3 = − cos Et3,3 ),
which results in cos ϕ = 5/6.
If ϕ represents the obtuse angle of the face of the triakis tetrahedron
(such as ∠urt in Figure 9.8(c)), then as three of these angles meet at each of
9.6. FACES REVISITED
153
four vertices of the triakis tetrahedron, we find upon using (9.1) with q = 3
that cos ϕ = −7/18.
Values of ϕ for the remainder of the Archimedean duals may be found
in Table 9.2.
Polyhedron
q
cos ϕ
ϕ (deg)
Triakis Tetrahedron
3
7
− 18
112.885
6
5
6
33.557
4
1
9
83.621
6
48.190
8
2
3
√
9 5−7
36
√
9− 5
12
√
1−2 2
4
√
2+ 2
4
3
3
− 10
τ
Tetrakis Hexahedron
Pentakis Dodecahedron
5
6
Triakis Octahedron
3
Triakis Icosahedron
10
Rhombic Dodecahedron
Rhombic Triacontahedron
Hexakis Octahedron
+ τ)
3
− 13
4
1
3
3
− √15
5
√1
5
√
2− 2
12
√
6− 2
8
√
1+6 2
12
√
5−2 5
30
√
15−2 5
20
√
9+5 5
24
4
6
8
Hexakis Icosahedron
1
10 (7
4
6
10
Table 9.2
68.619
55.691
117.201
31.400
119.039
30.480
109.471
70.529
116.565
63.435
87.202
55.025
37.773
88.992
58.238
32.770
154
CHAPTER 9. DUALITY
Polyhedron
q
Trapezial Icositetrahedron
5
3
−0.419643
114.812
4
0.160713
80.752
3
−0.471576
118.137
5
0.383433
67.453
4
3
4
Pentagonal Icositetrahedron
Pentagonal Hexecontahedron
ϕ (deg)
√
− 2+8 2
√
2− 2
4
√
5+2 5
− 20
√
5−2 5
10
√
9 5−5
40
3
Trapezial Hexacontahedron
cos ϕ
115.263
81.579
118.269
86.974
67.783
Table 9.2 (continued)
9.7
Exercises
1. With the nets provided, build the rhombic dodecahedron, the triakis
octahedron, and the trapezial icositetrahedron.
2. As in Figure 9.11, we see that a triakis tetrahedron may be imagined
as a regular tetrahedron with four squat tetrahedra affixed to its faces.
Here, e is on df and is the center of a triangular face of the regular
tetrahedron. In this exercise, we wish to find the ratio [ed]/[ef ]; that
is, the ratio of the height of the regular tetrahedron to the height of
the squat tetrahedra. Assume that the edges of the tetrahedron have
length 2, so that [cd] = 2.
(a) Using data from Table 9.2, show that
6
[cf ] = .
5
(b) Find [ce], and then by considering right triangles ∆ced and ∆cef ,
show that
[ed]
= 5.
[ef ]
9.7. EXERCISES
155
(c) Using (b), show that the ratio of the volume of the triakis tetrahedron relative to its base tetrahedron is 9/5.
h
f
e
c
g
d
Figure 9.11
(d) Show also that ∠dcf has the same measure as D3,3 .
3. State and demonstrate results analogous to those of the previous exercise for the triakis octahedron and the triakis icosahedron.
4. A tetrakis hexahedron may be imagined as a cube with six square
pyramids affixed to its faces. Using a procedure analogous to (a)–
(c) of Exercise 2, show that the ratio of the volume of the tetrakis
hexahedron to that of its base cube is 32 .
5. A rhombic dodecahedron may also be imagined as a cube with six
square pyramids affixed to its faces. Describe this relationship, then
calculate that the ratio of the volume of the rhombic dodecahedron to
that of its base cube is 2.
6. Consider the faces of a rhombic dodecahedron, and imagine inscribing
a square in each rhombus, as in Figure 9.12. Using data from Table
9.2, show that this may be accomplished by dividing the edges of each
rhombus in the ratio
√
y
= 2.
x
156
CHAPTER 9. DUALITY
y
x
Figure 9.12
(Note: The squares thus formed are twelve square faces of the rhombicuboctahedron, so that the procedure just described shows that a
rhombicuboctahedron may be inscribed in a rhombic dodecahedron.)
7. Show the following result, which is a generalization of the result in the
previous exercise: a square may be inscribed in a rhombus (see Figure
9.13) by dividing the edges of the rhombus in the ratio
y
A
= cot ,
x
2
where A is the smaller of the interior angles of the rhombus.
x
y
A
A/2
Figure 9.13
(a) What is this ratio for a rhombic triacontahedron?
(b) As a result, what Archimedean solid may be inscribed in a rhombic triacontahedron using this procedure?
8. In this exercise, we look at at alternative way of deriving (8.24). To
do so, we consider the dual of our snub polyhedron. Since our snub
polyhedron has four triangles and one p-gon at each vertex, its dual
9.7. EXERCISES
157
has faces which are pentagons – at four of its vertices, three faces
meet, while p meet at the remaining vertex. Let ϕ3 denote the interior
angles at the vertices of the pentagon where three faces meet, and let
ϕp denote the interior angle at the remaining vertex.
Since the interior angles of a convex pentagon sum to 3π, it is evident
that 4ϕ3 + ϕp = 3π, so that
cos 4ϕ3 = cos(3π − ϕp ) = − cos ϕp .
Substitute for cos ϕ3 and cos ϕp from (9.1) into this expression, and
thus produce an alternate derivation of (8.24).
158
CHAPTER 9. DUALITY
Chapter 10
Geodesic Structures, III
Now that we have explored geodesic structures based on the Archimedean
solids, we return to the subject of Chapter 4: the icosahedron. The structures discussed here may be found in Wenninger’s Spherical Models. In this
chapter, we fill in many of the details that were left to the reader there.
10.1
A 4-frequency Icosahedron
There are many ways to subdivide a planar equilateral triangle upon which
a geodesic structure may be based. In Chapter 4, an equilateral triangle was
subdivided by interior segments parallel to the sides of the triangle. Here,
we subdivide using segments perpendicular to the sides of the triangle, as
in Figure 10.1. Geodesic icosahedra based on such a subdivision are called
Class II icosahedra. Thus, Figure 10.1 describes a 4-frequency Class II
icosahedron.
As we saw in Chapter 4, one planar subdivision may result in several
different geodesic structures. Recall the reason for this: although three
segments may intersect in a point in the planar figure, the corresponding
great circles do not, in general, intersect in a single point, but rather form
a small spherical triangular “window.” These windows are too small to
construct accurately on a small scale; different methods of compensating for
this fact determine various possible structures.
In this section, we discuss the particular geodesic structure described on
p. 110 of Spherical Models; Figure 55 there is analogous to our Figure 10.1.
Denote by A the measure of the dihedral angles of the spherical equilateral
triangle in Figure 10.1; ∠pqr is one example. Let E be the measure of the
sides of the spherical triangle, and put s = 12 E, so that arc rq has measure
159
160
CHAPTER 10. GEODESIC STRUCTURES, III
s. In the icosahedral case, of course A = 72◦ and s = 12 E3,5 , but we will
develop more general results which will enable us to subdivide any equilateral
spherical triangle in the manner illustrated in Figure 10.1.
p
a
1
2b
1
2b
c
c
d
d
d
d
d
c
c
1
2b
1
2b
d
a
1
2b
c
c
a
t
q
u
Figure 10.1
Before we begin, we recall a few relationships between s and A which
will be useful later on. Employing (3.9) and substituting A, 3, and 2s,
respectively, for D, q, and ϕ yields
cos A =
cos 2s
,
1 + cos 2s
cos 2s =
cos A
.
1 − cos A
(10.1)
First, we find the measure of the arc labelled d in Figure 10.1. It is apparent
by symmetry considerations that ∠qOr has measure 60◦ . Applying (7.7),
then, to right spherical triangle ∆Orq results in sin d tan 60◦ = tan s. In
addition, cos d may be found by employing the identity sin2 d + cos2 d = 1,
so that
√
1
4 cos2 s − 1
sin d = √ tan s, cos d = √
.
(10.2)
3
3 cos s
10.1. A 4-FREQUENCY ICOSAHEDRON
161
These formulas are valid when s ≤ 60◦ , or equivalently, E ≤ 120◦ . This
poses no difficulty, as none of the polyhedra we have discussed thus far has
an edge angle exceeding 120◦ .
Next, we seek the measure of the arc labelled a. Our strategy will be
first to find the measure of the arc pr, which we call e, and then observe
that a + 2d = e.
Considering right spherical triangle ∆prq, we may apply (7.1), thus obtaining cos 2s = cos e cos s. Thus
cos e =
cos 2s
2 cos2 s − 1
=
.
cos s
cos s
We may also find sin e by employing the identity sin2 e + cos2 e = 1 and some
algebra, resulting in
cos e =
2 cos2 s − 1
,
cos s
p
sin e = tan s 4 cos2 s − 1.
(10.3)
Again, these are valid for s ≤ 60◦ .
Since a + 2d = e, we have a = e − 2d. But d and e are fully described by
(10.2) and (10.3). Thus, we may employ the usual trigonometric relationships, in addition to a good deal of algebra, to obtain
cos a =
2 cos2 s + 1
,
3 cos s
sin a =
p
1
1
sin e = tan s 4 cos2 s − 1.
3
3
(10.4)
Now it is an easy matter to find c. We may apply (3.4) to spherical
∆Ors and use (10.2) to obtain, after a little algebra,
cos c = 1 −
1
tan2 s.
6
(10.5)
The final piece of the puzzle is arc st. In the actual construction of the
geodesic model, the exterior edges of the equilateral triangle (such as pq)
are not present. Instead, triangles such as ∆squ are used which straddle
the exterior edges. Were the exterior edges present, they would cut such
triangles exactly in half.
So let b represent the measure of arc su. We might be tempted to find
b by extending the arc labelled c to rq to create st, and then doubling our
result. However, because of the way plane triangles project onto spherical
triangles, the angle ∠stq would in fact not be 90◦ , and we would need to
separately construct triangles ∆stq and ∆utq.
162
CHAPTER 10. GEODESIC STRUCTURES, III
Instead, we find b by considering ∆squ and applying (3.4), noting that
∠squ has measure A, being twice ∠sqt, which has measure 21 A. An application of (3.4) yields
cos b = cos2 a + sin2 a cos A.
Substituting for a and A from (10.4) and (10.1), a bit of algebra reveals that
cos b =
22 cos4 s − 5 cos2 s + 1
1
=1+
tan2 s sec2 s(1 − 4 cos2 s).
4
18 cos s
18
(10.6)
Finally, then, the subdivision of the spherical equilateral triangle in Figure 10.1 is completely described. When this figure describes a 4-frequency
Class II icosahedron, we see that
a ≈ 16.472◦ ,
b ≈ 19.188◦ ,
c ≈ 20.554◦ ,
d ≈ 20.905◦ .
A completed model is shown below. The red icosahedral arcs are not part
of the subdivision, but are included for reference.
Figure 10.2
10.2
A 6-frequency Icosahedron
We may continue our discussion of geodesic structures based on subdivisions
of a planar equilateral triangle by considering Figure 10.2. This subdivision
yields a 6-frequency Class II icosahedron.
To work out explicit expressions for all the arcs in Figure 10.2 as was
done in the previous section for the 4-frequency Class II icosahedron would
10.2. A 6-FREQUENCY ICOSAHEDRON
163
be a daunting task. Instead, a technique called gnomonic projection
is employed. This technique is used extensively by Magnus Wenninger in
Spherical Models, and although only an approximation, it provides data
which allow one to build beautiful models. Here, we will describe how the
technique of gnomonic projection may be applied in our scenario. (Note:
the data given are exact (up to the accuracy of your calculating device),
but for the model which is a projection of Figure 10.3 onto a sphere. Keep
in mind (see Figure 4.5) that the edges of the icosahedron, when projected
on the sphere, will not be divided into equal sixths (as suggested by Figure
10.3). It is in this sense that we use the term approximation.)
a
1
2b
l3
1
2b
c
t
c
l2
d
e
e
f
f
g
e
g
r
e
f
g
e
f
d
g
c
f
d
c
e
1
2b
l1
1
2h
g
g
e
a
e
s
1
2h
1
2b
e
c
1
2b
e
e
1
2h
e
c
1
2b
a
Figure 10.3
So begin with Figure 10.4, analogous to Fig. 56a in Spherical Models.
This Figure is obtained by slicing the icosahedron and its circumsphere
with a plane containing O (the center of the icosahedron) and the line l1 of
Figure 10.3. Thus, segments labelled g, d, and a in Figure 10.3 are projected
164
CHAPTER 10. GEODESIC STRUCTURES, III
onto arcs of a circle with radius ρ in Figure 10.4, where ρ is the radius of the
sphere circumscribing the icosahedron (see (3.8)). Recall that in deriving
the formula (3.8), it was assumed that the edge of the Platonic solid in
question had length 2. Thus, we will assume that the equilateral triangle of
Figure 10.3 has edge length 2.
g
g
1
2h
d
a
λ
λ
q
λ
1
2λ
λ
p
ρ
w
l1
ρ
O
Figure 10.4
In order to produce results which may be easily generalized, we perform
the relevant calculations in terms of “n,” where 2n is the frequency of our
Class II icosahedron. In the case illustrated in Figure 10.3, we have n = 3.
Returning to Figure 10.4, note that tan g = λ/w. Thus, it will be useful
to find λ and w. Now λ is simply the length of the segments labelled g (or
d or a) in Figure 10.3. By observing that the edge of length 2 of the large
equilateral triangle is subdivided into 2n pieces, each of which is the height
of a smaller equilateral triangle of edge length λ, we see that
2
λ= √ .
3n
(10.7)
Note that w is the length of a leg of the right triangle ∆Opq, so that
w2 + (nλ)2 = ρ2 ,
10.2. A 6-FREQUENCY ICOSAHEDRON
165
which yields
w=
p
ρ2 − n2 λ2 .
(10.8)
So now we may find g;
g = arctan
λ
.
w
Having found g, we note that in Figure 10.4,
tan(g + d) =
so that
d = arctan
2λ
,
w
λ
2λ
− arctan .
w
w
Using the relationship tan(α − β) =
tan d =
tan α−tan β
1+tan α tan β
(see (0.13)), we find that
wλ
.
w2 + 2λ2
Finally, the same methodology can be used to find a, yielding
tan a =
wλ
.
w2 + 6λ2
These results may readily be generalized. If instead of writing the angles
sought as g, d, a, we label them α1 , α2 , α3 , ..., αn (since there must be n such
angles), we find that if 1 ≤ k ≤ n,
tan αk =
wλ
.
w2 + k(k − 1)λ2
(10.9)
The proof of this result is left as an Exercise.
Finally, we note that h may be found by observing that
3
λ
1
tan g + h = 2 ,
2
w
and proceeding as above.
Up to this point, we have found a, d, g, and h. To find e and f , we carry
out the same procedure for line l2 that we did for l1 . This is illustrated
in Figure 10.5. The only obstacle preventing our solving for e and f using
the above methodology is that the length of st, and hence w0 , is yet to be
determined.
166
CHAPTER 10. GEODESIC STRUCTURES, III
f
e
λ
e
1
2λ
1
2λ
λ
s
l2
t
w0
ρ
O
Figure 10.5
It is not difficult, though, to determine [st]. Observing where s and t are
located in Figure 10.3 and considering right triangle ∆rst results in
[rs]2 + [st]2 = [rt]2 .
√
But [rs] is related to λ by [rs]/λ = 3/2, and [rt] is simply nλ (the radius
of the circle in Figure 10.3). Substituting these in the previous relationship
and using (10.7) yields
r
4
1
[st] =
− 2.
(10.10)
3 n
Considering right triangle ∆Ost in Figure 10.5, then, results in
r
p
4
1
w0 = ρ2 − [st]2 = ρ2 − + 2 .
(10.11)
3 n
We now have sufficient information to find e and f in Figure 10.5 using the
above methodology.
An analogous approach may be used to find b and c on l3 . Moreover,
it is easy to see how this methodology extends to higher-frequency Class II
icosahedra.
For the sake of completeness, the measures of the arcs obtained for the
6-frequency Class II icosahedron are summarized in the following table:
10.3. EXERCISES
167
a
10.388◦
e
13.368◦
b
12.168◦
f
14.175◦
c
12.297◦
g
14.286◦
d
12.703◦
h
13.238◦
Table 10.1
Moreover, a completed icosahedron is shown below (where, as before,
the red icosahedral arcs are only for reference).
Figure 10.6
10.3
Exercises
1. Derive the expression for cos d in (10.2) as suggested in the text.
2. Derive the expression for sin e in (10.3) as suggested in the text.
3. As indicated in the text, derive the relationships in (10.4).
4. Derive (10.5) as suggested in the text.
5. Obtain (10.6) as indicated in the text.
6. Let B represent the measure of angle ∠rsO in Figure 10.1. By considering spherical triangle ∆rOs, show that
3 cos 2s
sin B = √
.
13 cos2 2s − 1
(Hint: Use (4.4) and find sin c using (10.5).)
168
CHAPTER 10. GEODESIC STRUCTURES, III
7. Using the methods in this chapter, calculate the various angles required
to build a 4-frequency octahedron and a 6-frequency octahedron.
8. Derive (10.9) as follows (see Figure 10.7). Define
Ak = α1 + α2 + · · · + αk =
k
X
αi
i=1
when 1 ≤ k ≤ n. Using the observation that when k > 1,
αk = Ak − Ak−1 ,
find tan αk by employing the appropriate trigonometric relationship.
λ
λ
λ
λ
···
αn
α1
α2
α3
w
Figure 10.7
9. By extending the ideas in §10.2, design an 8-frequency icosahedron.
Chapter 11
Deltahedra
In this chapter, deltahedra are examined from algebraic, geometric, and
trigonometric perspectives.
11.1
Algebraic Preliminaries
Convex polyhedra whose faces consist entirely of equilateral triangles are
called deltahedra. (Some authors allow the term deltahedron to include
nonconvex polyhedra, but we will be considering only the convex cases here.)
For example, the tetrahedron, octahedron, and icosahedron are all deltahedra. It is a relatively pleasant task to describe several deltahedra, and
further, to show that our enumeration is complete. In addition, finding the
dihedral angles of the deltahedra will again allow us to employ our expanding knowledge of spherical trigonometry. We first examine some algebraic
considerations.
As in previous chapters, V , E, and F will denote the number of vertices,
edges, and faces, respectively, on a given deltahedron. Since the deltahedra we are considering are convex, Euler’s formula (2.1) is still valid; for
convenience, we reproduce it here:
V − E + F = 2.
(11.1)
We also have a version of (P20 ) in §2.3 for the same reasoning included
there. Specifically, since all faces of a deltahedron are triangles, 3F is the
total number of sides on all faces of a deltahedron. Moreover, each edge of
a deltahedron corresponds to the meeting of two faces, so that 3F counts
the number of edges exactly twice. This results in
3F = 2E.
169
(11.2)
170
CHAPTER 11. DELTAHEDRA
Our task is greatly simplified by a few new notations. Let ν3 , ν4 , and
ν5 denote the number of trivalent, tetravalent, and pentavalent vertices,
respectively; that is, the number of vertices where three triangles meet, four
triangles meet, and five triangles meet. Recall, as in §2.2, that six equilateral
triangles may not meet at a vertex, for this would result in a flat vertex.
More than six equilateral triangles at a vertex would violate the convexity
condition.
With these notations, it is immediately apparent that
ν3 + ν4 + ν5 = V,
(11.3)
since each vertex on a deltahedron must be trivalent, tetravalent, or pentavalent.
To obtain our final relationship, we count the total number of ends of
all the edges on a deltahedron. On one hand, this is clearly 2E, since each
edge has two ends.
On the other hand, the ends of each edge are incident to exactly two
vertices. The number of ends accounted for by the trivalent vertices is 3ν3 ,
since three faces, and hence three edges, meet at a trivalent vertex. Similarly,
the number of edges accounted for by the tetravalent vertices is 4ν4 and the
number accounted for by the pentavalent vertices is 5ν5 . Since these three
cases account for all ends of all edges of a deltahedron, we have
3ν3 + 4ν4 + 5ν5 = 2E.
(11.4)
Thus, (11.1)–(11.4) give us four relationships in six variables. It is our
task to find all possible solutions to these simultaneous equations which are
realizable by actual deltahedra.
There are many possible approaches to such a problem. Ours will be to
eliminate one variable from our system by considering all possible values that
ν3 may take on. Once this is done, we may easily enumerate the remaining
solutions to our algebraic problem.
To this end, let us begin by assuming that our deltahedron has at least
one trivalent vertex. We will conduct our examination by considering possible valencies of vertices adjacent to that trivalent vertex.
So let p be a trivalent vertex of a deltahedron. Note that since three
equilateral triangles meet at p, this vertex looks like a vertex of a tetrahedron. In other words, such a vertex – that is, a vertex of a polyhedron where
three faces meet – is rigid. Dihedral angles may be determined by (3.4) and
(3.6), so that a trivalent vertex has no “flexibility.”
11.2. TETRAVALENT VERTICES
171
As a result, if there is a second trivalent vertex adjacent to p, the deltahedron must be a tetrahedron. For if we begin with p, and then create a
trivalent vertex adjacent to p, we are simply adding the fourth face to a
tetrahedron. Thus, a deltahedron with adjacent trivalent vertices must be
a tetrahedron.
The tetrahedron, then, results in the following solution to (11.1)–(11.4):
ν3
ν4
ν5
V
E
F
4
0
0
4
6
4
Is it possible to find a deltahedron where a trivalent vertex is adjacent
to a tetravalent vertex? If we begin with p and try to create an adjacent
tetravalent vertex, we are again left with just one option, the triangular
dipyramid; that is, two tetrahedra joined together by a common face. This
results in the following solution to (11.1)–(11.4):
ν3
ν4
ν5
V
E
F
2
3
0
5
9
6
Can a trivalent vertex ever be adjacent to a pentavalent vertex on a
deltahedron? As it happens, this situation is not possible. We will look at
one justification for this claim in §11.3.
And thus, we have at this point found the two deltahedra with trivalent
vertices. This will allow an easy solution to (11.1)–(11.4). But that must
wait until after the next section.
11.2
Tetravalent Vertices
Before we embark on an algebraic enumeration of the remaining deltahedra,
we take a moment to explore the geometry of tetravalent vertices of deltahedra. Since all faces on a deltahedron are equilateral triangles, the vertex
figure of a tetravalent vertex is a spherical rhombus, with each side measuring 60◦ . It will prove useful to take a moment and derive some relationships
among the various angles of spherical rhombi.
So consider the spherical rhombus in Figure 11.1(a). The reader might
take a moment to see that opposite dihedral angles of a spherical rhombus
must have the same measure, just as in the planar case. To derive the most
general results, we consider the case where the sides of the rhombus all have
the same measure γ. Later, we will look at the particular case γ = 60◦ .
172
CHAPTER 11. DELTAHEDRA
B
γ
A
B
2
α
A
B
γ
A/2
A
β
B
2
B
(a)
γ
A/2
(b)
(c)
Figure 11.1
Applying (3.6) to the spherical triangle in Figure 11.1(b) (which is just
half our spherical rhombus), we see that
cos
B
B
B
= − cos cos A + sin sin A cos γ.
2
2
2
By employing the trigonometric identity tan 2θ =
relationship readily becomes
tan
sin θ
1+cos θ
(see (0.19)), this
A
B
tan = sec γ.
2
2
(11.5)
It is left as an exercise to derive from this the companion relationships
tan2 γ + (1 + sec2 γ) cos B
tan2 γ + (1 + sec2 γ) cos A
,
cos
B
=
−
.
1 + sec2 γ + tan2 γ cos B
1 + sec2 γ + tan2 γ cos A
(11.6)
By applying (3.4) to the same spherical triangle in Figure 11.1(b) and
θ
employing the trigonometric relationship tan 2θ = 1−cos
(see (0.19)), we
sin θ
obtain
α
B
tan = tan γ cos .
(11.7)
2
2
This and its companion relationship are summarized below, where β is the
measure of the other diagonal of the spherical rhombus (i.e., the diagonal
other than α; see Figure 11.1(c)):
cos A = −
tan
α
B
= tan γ cos ,
2
2
tan
β
A
= tan γ cos .
2
2
(11.8)
The companion relationship may be derived by observations of symmetry;
that is, replace α with β and B with A in (11.7). Or apply (3.4) to the spherical triangle obtained by cutting the spherical rhombus in Figure 11.1(a) in
half the other way.
11.3. THE REMAINING DELTAHEDRA
173
Applying (3.4) to the triangle in Figure 11.1(b) in a slightly different
way results in the companion relationships
cos α = cos2 γ + sin2 γ cos A,
cos β = cos2 γ + sin2 γ cos B.
(11.9)
Finally, we have
sec
α
β
sec = sec γ,
2
2
(11.10)
the derivation of which is left as an Exercise.
For easy reference, we include the foregoing results in the case γ = 60◦ .
tan
cos A = −
tan
3 + 5 cos B
,
5 + 3 cos B
α √
B
= 3 cos ,
2
2
1
cos α = (3 cos A + 1),
4
sec
11.3
A
B
tan = 2,
2
2
cos B = −
tan
(11.11)
3 + 5 cos A
,
5 + 3 cos A
β √
A
= 3 cos ,
2
2
1
cos β = (3 cos B + 1),
4
α
β
sec = 2.
2
2
(11.12)
(11.13)
(11.14)
(11.15)
The Remaining Deltahedra
We now return to the impossibility of our trivalent vertex p being adjacent
to a pentavalent vertex on a deltahedron. Let us consider the vertex figure
of such a pentavalent vertex q, as shown in Figure 11.2.
Since q is adjacent to the trivalent vertex p, one vertex of this vertex
figure lies on the edge of the deltahedron joining p and q. Let’s call this
vertex r. Recall that trivalent vertices are rigid; hence the dihedral angle
∠srt at r must have measure D3,3 .
174
CHAPTER 11. DELTAHEDRA
r
t
A
B
u
s
v
Figure 11.2
Now we know three parts of spherical ∆srt: two sides and an included
dihedral angle. Thus, with the aid of (3.4) and (3.6), the other parts of this
triangle may be determined. Carrying out this procedure, or simply noting
that the three parts already known correspond exactly to the vertex figure
of a tetrahedron, we see that st also has measure 60◦ , and dihedral angles
∠rst and ∠str also have measure D3,3 .
Since st has measure 60◦ , we see that stuv is a spherical rhombus with
sides measuring 60◦ . Hence the pentavalent vertex figure in Figure 11.2 may
be thought of as a spherical equilateral triangle and a spherical rhombus,
joined by a common side.
Thus, the dihedral angles at t and s have measures D3,3 +A and D3,3 +B,
respectively, where A and B are the dihedral angles of the spherical rhombus
(as in the previous section).
Since we are considering only convex deltahedra, the vertex figure in
Figure 11.2 must also be convex, so that both D3,3 + A and D3,3 + B must
measure less than 180◦ . Our work from the previous section will show that
this is not possible, so that we may infer the impossibility of a trivalent
vertex being adjacent to a pentavalent vertex on a deltahedron.
Now let’s see why D3,3 + A and D3,3 + B cannot both be less than 180◦ .
Since the sides of our rhombus measure 60◦ , (11.11)–(11.15) are valid, so
that
A
B
tan tan = 2.
2
2
√
When are A and B equal? In this case, tan A2 = tan B2 = 2, so that
employing a trigonometric relationship yields
cos A =
A
2
A
2
tan 2
1 − tan2
1+
1
=− .
3
11.3. THE REMAINING DELTAHEDRA
175
So, if A and B are equal, then A and B are both D3,4 (see Table 3.1), and our
spherical rhombus is the vertex figure of an octahedron. Since cos D3,3 = 13 ,
we see that D3,3 and D3,4 are supplementary, so that D3,3 + A and D3,3 + B
are both exactly 180◦ .
Now what happens when A and B are not equal? Then one of tan A2
√
√
and tan B2 must be greater than 2, and one must be less than 2 in order
for their product to be precisely 2. But since the function y = tan x2 is an
increasing function when x is between 0◦ and 180◦ , this means that one of
A and B is greater than D3,4 , and one is less. Finally, we conclude that one
of D3,3 + A and D3,3 + B is greater than 180◦ , and one is less than 180◦ .
Thus, both D3,3 + A and D3,3 + B cannot both be less than 180◦ , and so
the trivalent vertex p cannot be adjacent to the pentavalent vertex q.
Where does this leave us? In §11.1, we found all deltahedra where a
trivalent vertex is adjacent either to another trivalent vertex or to a tetravalent vertex. We just demonstrated that a trivalent vertex may not be adjacent to a pentavalent vertex. Thus, we have in fact enumerated all deltahedra which contain a trivalent vertex.
As a result, the remaining deltahedra contain only tetravalent and pentavalent vertices. Thus, we may put ν3 = 0 in (11.1)–(11.4), giving a system
of four equation in the five unknowns V , E, F , ν4 , and ν5 .
Of course, there are many ways to enumerate the solutions to such a
system. Our approach will be to solve the system in terms of ν4 , yielding
ν5 = 12 − 2ν4 ,
V = 12 − ν4 ,
E = 30 − 3ν4 ,
F = 20 − 2ν4 .
(11.16)
The derivation of this solution is left as an Exercise.
A simple way to analyze these solutions is to consider the various possible
values for ν4 . Since ν5 = 12 − 2ν4 , it must be the case that 0 ≤ ν4 ≤ 6,
since the number of pentavalent vertices cannot be negative. Thus, the only
values for ν4 which must be considered are the integers from 0 to 6, inclusive.
When ν4 = 0, we see from (11.16) that ν5 = 12. This results in the regular icosahedron, which we expected to see somewhere among our solutions.
In the case ν4 = 1, we have ν5 = 10. As is happens, this case cannot
be realized. Rather than give a rigorous analytical proof of this fact, we
consider a more informal justification. One may see the impossibility of this
case simply by trying to build one. First, begin with the tetravalent vertex.
The other four vertices thereby created must be pentavalent, since ν4 = 1.
In forcing these four vertices to be pentavalent, four new vertices and eight
new faces are created, bringing the total to nine vertices and twelve faces.
Since V = 12 − ν4 = 11, two more pentavalent vertices are needed, which
176
CHAPTER 11. DELTAHEDRA
requires at least eight additional triangular faces (since at most two faces
are incident at both of these vertices). This brings the number of triangular
faces to at least 20, but in fact F = 20 − 2ν4 = 18. This contradiction
demonstrates the impossibility of the case ν4 = 1.
When ν4 = 2, it follows from (11.16) that ν5 = 8. There is just one way
to realize this case; indeed, there is just ony way to realize each of the other
possible cases. The deltahedron obtained here may be thought of as a square
antiprism with square pyraminds affixed to each of its two square faces. This
polyhedron is called the gyroelongated square dipyramid (see Figure
11.3(a)). In 1966, Norman Johnson named all convex polyhedra with regular
polygons as faces. As it turns out, there are 92 such polyhedra (not counting
the Platonic and Archimedean solids) – a few too many to consider here.
The gyroelongated square dipyramid is number 17 in Johnson’s enumeration,
and hence is usually called J17 for brevity.
(a)
(b)
(c)
Figure 11.3
The case ν4 = 3 yields the triaugmented triangular prism (J51;
see Figure 11.3(b)). One way to visualize this deltahedron is to imagine
a triangular prism with square pyramids affixed to (and thus augmenting)
each of its three square faces.
Undoubtedly the most interesting case is ν4 = 4, the snub disphenoid
(J84; see Figure 11.3(c)). It cannot be decomposed as in the preceding two
examples; the reader is encouraged to build one in order to appreciate its
symmetry. A thorough investigation of the snub disphenoid will be undertaken in the next section.
11.4. THE SNUB DISPHENOID
177
Deltahedron
ν3
ν4
ν5
V
E
F
Tetrahedron
4
0
0
4
6
4
Triangular dipyramid (J12)
2
3
0
5
9
6
Icosahedron
0
0
12
12
30
20
Gyroelong. sq. dipyr. (J17)
0
2
8
10
24
16
Triaug. triang. prism (J51)
0
3
6
9
21
14
Snub disphenoid (J84)
0
4
4
8
18
12
Pentagonal dipyramid (J13)
0
5
2
7
15
10
Octahedron
0
6
0
6
12
8
Table 11.1
When ν4 = 5, a pentagonal dipyramid (J13) is obtained; that is, two
pentagonal pyramids joined at their bases. Finally, ν4 = 6 results in our
longtime companion, the octahedron.
For completeness, a table of all deltahedra is included above, along with
the corresponding solution to (11.1)–(11.4). The dihedral angles of the deltahedra will be discussed in the remainder of the chapter, as well as in the
Exercises.
11.4
The Snub Disphenoid
Recall that the snub disphenoid is the only deltahedron whose dihedral
angles are not simple combinations of known dihedral angles. As it turns
out, the situation is somewhat involved, although the presence of tetravalent
vertices will allow us to use the results derived in §11.2.
Our approach will be to consider the snub disphenoid as a union of two
identical parts, one of which is illustrated in Figure 11.4. Here, v and w
are two adjacent tetravalent vertices. (At this point, I would suggest that
the reader pause to construct two such units if he or she has not already
done so. This will greatly facilitate understanding the geometry of the snub
disphenoid.)
178
CHAPTER 11. DELTAHEDRA
r
t
u
s
v
w
Figure 11.4
One immediately sees that the spatial configuration of each part is completely determined by the dihedral angle at edge vw (see Figure 11.4), since
this dihedral angle determines all others at the tetravalent vertices v and w
by virtue of (11.11).
How large should this dihedral angle be? This question may be answered
by considering how the two identical parts are assembled to form the snub
disphenoid. By holding one part in each hand, the methodology is simple:
the protruding vertices r and t of one part must exactly fit into the sunken
vertices s and u of the other part.
There are at least two ways to find this dihedral angle. One method
employs spherical trigonometry, and is left as an Exercise. The second
reduces to a problem in plane geometry; it will be presented below.
So denote by A the measure of the dihedral angle at the edge vw. Because
v and w are tetravalent vertices, the dihedral angles at edges rv and tw also
have measure A. Employing the notation of §11.2, we see that the dihedral
angles at edges sv, sw, uv, and uw therefore have measure B. In addition,
the angles rvw and vwt must have measure β (as in Figure 11.1(c)).
Assume that the edges of the snub disphenoid have length 1, and denote
by x the midpoint of the edge vw. The above considerations result in the
following two diagrams:
11.4. THE SNUB DISPHENOID
179
r
t
s
u
√
1
1
v
β
β
1
√
3
2
3
2
A
w
(a)
x
(b)
Figure 11.5
Dropping perpendiculars from v and w onto rt in Figure 11.5(a) yields
[rt] = 1 + 2 sin(β − 90◦ ) = 1 − 2 cos β.
(11.17)
Using the cosine law for triangles with Figure 11.5(b) results in
3
[su]2 = (1 − cos A).
2
(11.18)
Finally, setting [rt]2 = [su]2 from (11.17) and (11.18) and substituting from
(11.14) for cos β yields
6 cos A = 5 + 6 cos B − 9 cos2 B.
(11.19)
All that remains is to substitute from (11.12) for cos A into (11.19),
multiply through by 5 + 3 cos B, and collect like terms. This results in the
following cubic equation for cos B:
27 cos3 B + 27 cos2 B − 75 cos B − 43 = 0.
(11.20)
Of course, we may also substitute from (11.12) for cos B into (11.19) to obtain a cubic equation for cos A. The following equation is thereby produced:
27 cos3A + 225 cos2A + 237 cos A + 23 = 0.
(11.21)
These cubic equations may be solved exactly (as in Chapter 8); approximate solutions are given by A ≈ 96.198◦ and B ≈ 121.743◦ . For further
exploration of these cubic equations, see the Exercises.
Finally, the remaining dihedral angle of the snub disphenoid, C, satisfies
the cubic equation
27 cos3 C − 45 cos2 C − 24 cos C + 44 = 0.
(11.22)
The derivation of this equation is rather involved, and is presented in
§11.6.
180
11.5
CHAPTER 11. DELTAHEDRA
Dihedral angles
Because of our previous work, it is not too difficult to find the dihedral angles
of the deltahedra. We briefly describe each case, postponing the discussion
of the third dihedral angle of the snub disphenoid until the next section.
The dihedral angles of the tetrahedron, octahedron, and icosahedron
were calculated in §3.5. The additional dihedral angle of the triangular
pyramid is simply 2D3,3 as two tetrahedra are joined together face-to-face,
so that
7
cos 2D3,3 = 2 cos2 D3,3 − 1 = − .
9
One dihedral angle of the gyroelongated square dipyramid is simply Da4
3 .
Another is D3,4 , since two square pyramids (half-octahedra) are joined to
a square antiprism. The third, which is formed where the squares of the
1
antiprism and half-octahedra meet, is then Da4
4 + 2 D3,4 . Using Table 3.1
and (8.13), we find that
√
√
1
4
1
cos Da4
1− 2−2 2 .
4 + 2 D3,4 =
3
(11.23)
It is easy to see that the dihedral angles of the triaugmented triangular
prism, in addition to D3,4 (from the half-octahedra), are 60◦ + D3,4 and
90◦ + 21 D3,4 . We have
√ 1
cos (60 + D3,4 ) = − 1 + 2 6 ,
6
◦
r
◦
cos 90 +
1
2 D3,4
=−
2
.
3
Finally, we consider the pentagonal bipyramid. Since two pentagonal
pyramids (along with a pentagonal prism) may also be used to build an
icosahedron, we see that one of the dihedral angles must be D3,5 . To find
the other dihedral angle, note that five vertices of the pentagonal bipyramid
are tetravalent, so §11.2 is applicable. Using (11.12), we see that the other
dihedral angle D satisfies
√
3 + 5 cos D3,5
4 5−5
cos D = −
=
.
5 + 3 cos D3,5
15
(11.24)
A tabular summary of the dihedral angles of all the deltahedra is given
below.
11.6. THE REMAINING DIHEDRAL ANGLE
181
Deltahedron
D1 (deg)
D2 (deg)
D3 (deg)
Tetrahedron
70.529
—
—
Triangular dipyramid
70.529
141.058
—
Icosahedron
138.190
—
—
Gyroelongated square dipyramid
109.471
127.552
158.572
Triaugmented triangular prism
109.471
169.471
144.736
Snub disphenoid
96.198
121.743
166.441
Pentagonal dipyramid
138.190
74.755
—
Octahedron
109.471
—
—
Table 11.2
11.6
The Remaining Dihedral Angle
In examining Figure 11.3 again, it is evident that we have yet to find that
dihedral angle C on the snub disphenoid at edges rs, st, tu, and ur. To
do so, recall the construction of the snub disphenoid. Two copies of Figure
11.3 are assembled so that the vertices r, s, t, u of one copy are made to
coincide with the vertices s, r, u, t, respectively, of the second copy. (It is
very useful to see this by actually constructing a model!) Thus, edge rs is
matched with edge sr.
Geometrically, this means that C is the sum of two dihedral angles. First,
we have the dihedral angle at rs when rs is considered as an edge of the
(irregular) tetrahedron with base triangle ∆suv and apex r; call this angle
D. Secondly, we have the dihedral angle at sr when sr is considered an edge
of the trapezoidal pyramid with base rtwv and apex s; call this dihedral
angle E. Thus, C = D + E. (Look at a model to convince yourself!)
Finding D is relatively straightforward. Consider, for a moment, the
spherical triangle ∆suv obtained by centering a sphere at r (see Figure
11.6). Note that sides uv and sv each have measure 60◦ , and the dihedral
angle at v has measure A. Thus, it is evident that spherical ∆suv is simply
half a spherical rhombus (see Figure 11.1(b)), so that D must have measure
1
2 B:
D = 21 B.
(11.25)
182
CHAPTER 11. DELTAHEDRA
u
60◦
A
v
60◦
s
Figure 11.6
Determining E is somewhat more involved. Since we intend to find cos C
using
cos C = cos 12 B + E = cos 12 B cos E − sin 12 B sin E,
(11.26)
we seek a way to substitute into this relationship.
We begin by drawing a spherical pentagon, as in Figure 11.7, by considering s as the center of a sphere. Here, y is the fifth vertex adjacent to
s on the snub disphenoid. By joining r and t with an arc of a great circle,
we divide ∠yrv into angles 12 B and E. We now turn our attention to the
rather complicated task of finding a cubic equation for cos C. (The reader
should feel free to devise a simpler way!)
y
A
r
B/2
B/2
E
E
60◦
t
60◦
v B
60◦
B w
Figure 11.7
Begin with Figure 11.8. Note that ∆ryt of Figure 11.7 and ∆vwt of
Figure 11.8 are halves of spherical rhombi (as discussed in §11.2), so that
11.6. THE REMAINING DIHEDRAL ANGLE
183
rt has measure α and vt has measure β. Applying (4.5) to spherical triangle
∆rvt yields
sin(E − 21 A)
sin E
.
=
sin β
sin 60◦
Expanding sin(E − 21 A) in this relationship, then substituting for sin 12 A
from (11.11) and for cos 12 A from (11.13), and finally multiplying through
by sin 12 B results in
√
3
sin E sin 21 B = sin β
2
r
1
1
1
1
1
√ sin E tan 2 β sin 2 B − 2 cos E cos 2 A cos 2 B .
3
α
E
E−
60◦
A
2 A
2
60◦
β
B−
v
t
A
2
A/2
B w
60◦
Figure 11.8
Now again substitute for cos 12 A from (11.13), use the relationship tan 12 β =
1−cos β
1
sin β , and solve for sin E sin 2 B:
sin E sin 12 B = 4 cos E cos 12 B
cos β − 1
.
2 cos β + 1
Substitute this expression for sin E sin 12 B into (11.26) and simplify, so that
cos C = cos E cos 21 B
5 − 2 cos β
.
2 cos β + 1
Now substitute for cos E by applying (3.4) to ∆vrt and for cos 21 B using
sin α
(11.13) and the identity tan 12 α = 1+cos
α , yielding
cos C =
2 cos β − cos α 5 − 2 cos β
·
.
3(1 + cos α) 2 cos β + 1
184
CHAPTER 11. DELTAHEDRA
Substitute for cos α and cos β from (11.14), and then substitute for cos A
from (11.12) and simplify:
cos C =
1 (9 cos2 B + 24 cos B + 7)(3 − cos B)
.
24
1 + cos B
(11.27)
The idea here is to get the 1 + cos B to cancel. One trick is to use (11.20).
We add the appropriate multiple of the left-hand of (11.20) (which is 0) to
the numerator of (11.27). As it happens, once does the job, so that
(9 cos2 B + 24 cos B + 7)(3 − cos B) + 27 cos3 B + 27 cos2 B − 75 cos B − 43 =
2(cos B + 1)(9 cos2 B + 6 cos B − 11),
Hence
1
(9 cos2 B + 6 cos B − 11).
(11.28)
12
Thus, we have cos C as a quadratic function of cos B. Using (11.20),
then, it is possible to find a cubic equation of which cos C is a root. The
task is not too complicated; we include it for completeness. The uninterested
reader may skip directly to (11.31).
By defining
λ = 3 cos B + 1, µ = 12 cos C,
(11.29)
cos C =
our algebraic work will be made somewhat less cumbersome. With these
definitions, (11.28) becomes
µ = λ2 − 12,
(11.30)
and with cos B = 13 (λ − 1), (11.20) becomes
λ3 − 28λ − 16 = 0.
Be rewriting this equation as λ3 − 28λ = 16 and squaring both sides of this
relation, we obtain
λ6 − 56λ4 + 784λ2 − 256 = 0.
From (11.30), we see that λ2 = µ + 12, so that
(µ + 12)3 − 56(µ + 12)2 + 784(µ + 12) − 256 = 0,
resulting in
µ3 − 20µ2 − 128µ + 2816 = 0.
Substituting from (11.29) and dividing through by 64 yields
27 cos3 C − 45 cos2 C − 24 cos C + 44 = 0.
(11.31)
11.7. EXERCISES
11.7
185
Exercises
1. As suggested in the text, derive equation (11.6) from (11.5).
2. Derive equation (11.10).
3. Obtain the equations in (11.16) as suggested in the text.
4. Using the notations of §11.2, assume that γ = 60◦ so that (11.11)–
(11.15) are applicable. Show that
cos 12 α sin 12 B = cos 12 A.
5. In this exercise, we derive (11.19) using spherical trigonometry. Begin
with half the snub disphenoid as in Figure 11.4. Center a sphere with
radius [sr] at s, so that the following spherical trapezoid is formed:
α
r
t
A
2
60◦
60◦
β
v
B 0 A/2
60◦
B w
Figure 11.9
Since B is the dihedral angle at w where two 60◦ arcs meet, spherical
∆vtw is half a spherical rhombus, so all notations and results from
Figure 11.1 and §11.2 are applicable. That arc rt must have measure α is evident from how the two halves of the snub disphenoid are
assembled. Let β divide the angle at v into two angles, one having
measure 12 A and the other B 0 .
Now apply (3.4) to spherical triangle ∆rvt, yielding
cos α = cos 60◦ cos β + sin 60◦ sin β cos B 0 .
Using the fact that B 0 = B − 21 A, expand cos(B − 12 A) and employ
(11.11)–(11.15) and trigonometric identities as appropriate to derive
(11.19).
186
CHAPTER 11. DELTAHEDRA
6. In this exercise, we find cos B as a quadratic expression in cos A, where
A and B are two of the dihedral angles of the snub disphenoid. Begin
by dividing both sides of (11.21) by 5 + 3 cos A. Then use this result
with (11.12) to derive
cos B = −
1
(9 cos2A + 60 cos A + 19).
24
(11.32)
7. Obtain the result given in (11.23).
8. Show that the dihedral angle D of the gyroelongated square pyramid
described by (11.23) is also described by
tan 21 D = √
4
1
.
2−1
Similarly, show that the dihedral angle D of the pentagonal dipyramid
described by (11.24) is also given by
√
tan 12 D = 3 − 5.
9. In this exercise, we find cos A and cos B as quadratic functions of cos C,
where A, B, and C are the dihedral angles of the snub disphenoid.
(a) Using (11.19) and (11.28), deduce that
2 cos B = cos A + 2 cos C + 1.
(11.33)
(b) With this result, use (11.32) to obtain
cos C = −
1
(9 cos2A + 72 cos A + 31).
24
(c) Square both sides of the previous equation and use (11.21) as
necessary to conclude that
cos2 C = −
1
(6 cos2A + 44 cos A − 18).
24
(d) Use the results of (b) and (c) to find that
1
cos A = (18 cos2 C − 12 cos C − 29).
3
11.7. EXERCISES
187
(e) Using (11.33), conclude finally that
1
cos B = (9 cos2 C − 3 cos C − 13).
3
10. Let A and B be two dihedral angles of the snub disphenoid, as in
§11.4. With a and b given by
a = 2 tan2 12 A,
b = 2 tan2 12 B,
show that
a3 + 39a2 − 256 = 0,
b3 − 39b − 16 = 0.
Further, show that a and b are related as follows:
a = b2 − 39,
16b = a2 + 39a.
188
CHAPTER 11. DELTAHEDRA
Chapter 12
Kepler-Poinsot Polyhedra
12.1
Star Polygons
Having spent some leisurely moments among the Archimedean solids, their
duals, and a few variations on these themes, it is time to adventure into a
starry geometrical realm where polyhedra are no longer convex.
How can this be? As it happens, the large majority of polyhedra considered by polyhedronists are in fact nonconvex. Before looking at a few
examples, let’s consider some two-dimensional analogues.
Of course, various stars come to mind, many of which we have encountered before. Such stars are still considered regular polygons as all sides are
the same length and all angles are the same, but of course they are nonconvex regular polygons. Yet the pentagram of Figure 1.1 is not referred to as
a pentagon, but rather as a 52 -gon. This is not entirely nonsensical if the
following definition is adopted:
(S)
A p-gon is a regular polygon (convex or nonconvex) whose
edges subtend an angle of 360◦ /p with the center of the polygon.
This is consistent with our intuition concerning convex polygons, and
allows for star polygons as well.
(S) gives the following construction, illustrated with p = 8/3 in Figure
1.7(a) – simply begin with a regular octagon, select any vertex, and proceed
to “connect the dots” by joining every third vertex to the one previous.
This procedure may easily be generalized. Suppose that r > s > 1 (when
s = 1, we obtain an ordinary convex polygon), and begin with an r-gon
189
190
CHAPTER 12. KEPLER-POINSOT POLYHEDRA
and connect every sth vertex. The result is referred to as an rs -gon. As an
additional example, recall the regular 10
3 -gon of Figure 1.6(a).
If r and s do not share a common factor, an rs -gon is unicursal; that
is, beginning with an edge of the rs -gon, the remaining r − 1 edges may be
drawn without lifting one’s pencil from the paper.
When r and s share a common factor, as with an 82 -gon (see Figure
12.1(a)), an rs -gon consists of overlapping polygons; in Figure 12.1(a), two
overlapping squares are evident. Although many geometers are currently
investigating polyhedra with such faces, we will not encounter them in our
discussion of uniform polyhedra.
r
Furthermore, we find that an rs -gon looks the same as an r−s
-gon, except
that the edges are perhaps drawn in the reverse order. For example, an 38 gon looks the same as an 85 -gon. So when considering p-gons, where p = r/s,
we usually assume that r > s > 1, r and s have no common factors, and
s < r/2. Thus, an 83 -gon is the only other regular polygon with eight sides
besides the octagon, and so we call it the regular octagram. Similarly, we see
that Figure 1.1 shows the regular pentagram, and Figure 1.6(a) illustrates
the regular decagram. Of course, there may be more than one type of
polygram for a given r; for example, when r = 7, we have both the 72 -gon
and the 73 -gon (see Figures 12.1(b), (c)).
(a)
(b)
(c)
Figure 12.1
12.2
Nonconvex Polyhedra
With these ideas in mind, we revisit the discussion of Platonic solids begun
in §2.1. What happens when the property (P1 ) is relaxed, and we only
search for polyhedra satisfying (P2 ) and (P3 )? Is our investigation fruitful?
As it happens, yes, but the situation is more delicate – for (P10 ) (Euler’s
formula) is not satisfied, in general, for nonconvex polyhedra. There are
methods to enumerate nonconvex polyhedra, but their thorough application
12.2. NONCONVEX POLYHEDRA
191
is not within the scope of this text. So for this chapter, we must be content
to have the enumeration completed for us. There is still plenty to do...
As it happens, four new polyedra await us – that is, polyhedra whose
faces are regular polygons, the same number of which are incident at each
vertex, but which are not convex. Three of them are relatives of the dodecahedron in the following sense. A glance at Figure 1.1 reveals that in the
center of the pentagram, we find a smaller pentagon. Thus, beginning with
a pentagon, we may extend its sides so that a pentagram is formed. If this
procedure is done to the twelve pentagons on a regular dodecahedron, we
find that the nonconvex polyhedron in Figure 12.2 is formed (20 in Polyhedron Models). This polyhedron is called a small stellated dodecahedron
(“stellated” referring to the process of creating the “stars” just described).
As its faces are pentagrams and five such meet at each vertex, we refer to
this polyhedron by the Schläfli symbol { 25 , 5}.
Figure 12.2
Figure 12.3
Let us continue. The knowledgeable geometer is well aware that joining adjacent vertices of a pentagram yields a pentagon which circumscribes
the pentagram. What happens when this procedure is undertaken with
the twelve pentagrams of the small stellated dodecahedron? Another nonconvex polyhedron, the great dodecahedron (see Figure 12.3 and 21 in
Polyhedron Models), is formed. It is so named as its faces consist of twelve
pentagons. Since five pentagons form a pentagram at each vertex, we refer
to this polyhedron by the Schläfli symbol {5, 25 }.
Why stop here? Let us extend the edges of the pentagons of the great
dodecahedron so that twelve larger pentagrams are formed. We are in luck,
and have just created the great stellated dodecahedron (see Figure 12.4
and 22 in Polyhedron Models). Its Schläfli symbol is { 25 , 3}.
192
CHAPTER 12. KEPLER-POINSOT POLYHEDRA
Tragically, this process of alternately creating pentagons and pentagrams
must come to an end. Circumscribing pentagons around the pentagrams of
the great stellated dodecahedron produces twelve pentagons which do not
close up. Reasons for this phenomenon lie within the realm of a theory
of stellations of polyhedra which, although fascinating (and perhaps the
subject of another treatise), will not be discussed here.
Figure 12.4
Figure 12.5
The fourth polyhedron awaiting us is obtained as a stellation of the
icosahedron (again, details shall be omitted), and is shown in Figure 12.5
(see 41 in Polyhedron Models). It is called the great icosahedron, and its
twenty triangular faces are situated pentagrammatically at each vertex, so
that its Schläfli symbol is {3, 52 }.
For historical reasons, these four nonconvex polyhedra are often called
the Kepler-Poinsot polyhedra.
12.3
Measurements
Now that we have created these polyhedra, the desire possesses us to calculate their edge angles, dihedral angles, and circumradii. We are extremely
fortunate in that (3.7), (3.8), (3.11), (3.12), and (3.13) remain valid for the
four new pairs of {p, q} values discussed above. This is not merely lucky; the
interested reader may verify that in the derivations of the aforementioned
formulas, much use is made of the fact that the angle subtended by an edge
of a p-gon at its center is 2π/p, a property that is valid when p = 5/2 as well
(as seen in (S) in §12.1). The appropriate extension of Table 3.1 is given in
Table 12.1. Recall that we fashioned three of the Kepler-Poinsot polyhedra
12.4. DIHEDRAL ANGLES
193
by successively building upon a dodecahedron; remember, however, that the
values for ρ given in Table 12.1 are valid when the edges of the polyhedra
are all of length 2. For formatting purposes, “E” and “D” are written for
“Ep,q ” and “Dp,q ,” respectively.
Polyhedron
p
q
cos E
E
cos D
tan 21 D
D
ρ2
Great Dodeca.
5
5
2
63.4◦
63.4◦
τ +2
5
2
5
τ
116.6◦
3−τ
Great Icosa.
3
5
2
√1
5
√
− 15
√
5
3
τ −1
Sm. St. Dodeca.
τ −2
41.8◦
3−τ
Gr. St. Dodeca.
5
2
3
√1
5
√
− 15
− √15
√
− 35
√1
5
τ −1
63.4◦
3(2 − τ )
116.6◦
116.6◦
138.2◦
Table 12.1
12.4
Dihedral Angles
Our previous discussion, then, allows for the completion of Table 3.2, which
is given in its complete form in Table 12.2. Due to the symmetry involved,
we see that (3.15) remains valid, and hence (3.16) is valid as well.
p
3
3
4
3
5
3
5
2
5
5
2
q
3
4
3
5
3
5
2
3
5
2
5
r
4
3
3
5
2
5
2
5
5
3
3
Table 12.2
These nine triplets of (p, q, r)-values form a set of sorts, representing the
nine polyhedra whose faces are the same regular polygon, where the same
number of such polygons meet at each vertex. Now that the set is complete,
some interesting results may be demonstrated. For example, for (p, q, r) as
in Table 12.2, we have
Dp,q + Dp,r = π.
(12.1)
To see this, we begin with a result from Exercise 6 of Chapter 3; namely,
that
1
π
π
sin Dp,q sin = cos .
2
p
q
194
CHAPTER 12. KEPLER-POINSOT POLYHEDRA
Solving this equation for sin 12 Dp,q , substituting it into the relationship cos Dp,q =
1 − 2 sin2 21 Dp,q , and finding a common denominator yields
cos Dp,q = 1 − 2 csc2
1 − cos2 πp − 2 cos2 πq
π
π
.
cos2 =
p
q
sin2 πp
(12.2)
Substituting from (3.15) for cos2 πq and rearranging terms results in
cos Dp,q
2π
2π
= − cos Dp,r ,
= − 1 − 2 csc cos
p
r
where the latter equality follows from inspection of (12.2). Now Dp,q and
Dp,r , being dihedral angles between faces of polyhedra, are both between
0 and π. But two such angles have opposite cosines only when they are
supplementary, which gives (12.1).
12.5
Summary of Important Relations
We summarize here the important relations valid for the regular polyhedra;
that is, the values of (p, q, r) listed in Table 12.2.
cos2
π
π
π
+ cos2 + cos2 = 1.
p
q
r
(12.3)
Dp,q + Dp,r = Ep,q + Dq,p = Ep,r + Eq,r = π.
cos Dp,q = 1 − 2
1 + cos 2π
q
1 + cos 2π
p
(12.4)
− 1.
(12.5)
1
π
π
sin Dp,q = csc cos ,
2
p
q
1
π
π
cos Dp,q = csc cos .
2
p
r
(12.6)
1
π
π
sin Ep,q = csc cos ,
2
q
p
1
π
π
cos Ep,q = csc cos .
2
q
r
(12.7)
1 − cos
2π ,
cos Ep,q = 2
p
ρ = sin
π
π
sec .
q
r
1 − cos 2π
q
(12.8)
Notice that (12.4) follows directly from (3.12) and (12.1). (12.5) is a
combination of (3.7) and (3.11). (12.6) and (12.7) follow from Exercise 6 of
Chapter 3 (see Exercise 2 below). (12.8) is simply (3.16).
12.6. EXERCISES
12.6
195
Exercises
1. Derive (12.1) as suggested in the text.
2. Using Tables 3.1 and 12.1, show that
E3, 5 = π − E3,5 ,
E5, 5 = E3,5 ,
2
2
E 5 ,5 = π − E3,5 ,
E 5 ,3 = π − E5,3 .
2
2
3. Show that the results of Exercises 2–6 of Chapter 3 remain valid for
the values of (p, q, r) found in Table 12.2.
4. Using (3.12) and (12.1), show that for all values of (p, q, r) from Table
12.2, we have
Ep,r + Eq,r = π.
5. Let us consider a moment the icosidodecahedron. Recall from Table
6.1 that its edge angle ET3,5 = 36◦ .
(a) Using (3.9), show that the angles between the sides of a spherical
pentagon derived from a pentagonal face of the icosidodecahedron
have measure D 5 ,5 .
2
(b) Again using (3.9), show that the angles between the sides of a
spherical triangle derived from a triangular face of the icosidodecahedron have measure D 5 ,3 .
2
(c) Use (a) and (b) to argue that the twelve spikes from a small stellated dodecahedron and the twenty spikes from a great stellated
dodecahedron may be assembled to form an icosidodecahedron.
(d) Build two spikes of each type – i.e., two triangular pyramids and
two pentagonal pyramids – to illustrate the result in (c).
196
CHAPTER 12. KEPLER-POINSOT POLYHEDRA
Chapter 13
Euler’s Formula
In this chapter, we give a proof of Euler’s formula and generalize Descartes’
rule of deficiency using spherical geometry. We then discuss solid angles at
vertices of the Platonic and Archimedean solids.
13.1
Area of a Spherical Polygon
In this section, we derive the formula for finding the area of a spherical
triangle (and other spherical polygons) alluded to in §3.1. We denote by ρ
the radius of the sphere; recall that the surface area of the sphere is then
4πρ2 .
Now consider the spherical triangle pqr as shown in Figure 13.1, whose
dihedral angles (in radians) are A, B, and C, with opposite sides a, b, and
c, respectively. Here, p, q, t, and s all lie on the great circle containing the
arc pq (with measure b), so that Figure 13.1 is an aerial view of a sphere.
In addition, arcs prt and qrs are semicircles containing the edges pr and
qr, respectively, of the spherical triangle. The areas of the four regions
into which the hemisphere is divided are denoted by K, L, M , and N for
convenience.
Since precisely half of the sphere is in view in Figure 13.1, it is evident
that
1
K + L + M + N = (4πρ2 ) = 2πρ2 .
2
197
(13.1)
198
CHAPTER 13. EULER’S FORMULA
b
K
p A
c
N
B
r
C
q
a
L
M
s
t
Figure 13.1
Now A is the dihedral angle of the lune pqtrp of the sphere, whose
shape may be likened to a section of orange or grapefruit. Since A/2π is
that fraction of a revolution occupied by the dihedral angle of the spherical
triangle at p, it follows that the lune pqtrp occupies the same fraction of the
surface area of the sphere, which relationship may be algebraically expressed
as
A
K +L=
(4πρ2 ) = 2Aρ2 .
(13.2)
2π
Similar considerations yield that
K + M = 2Bρ2 ,
K + N = 2Cρ2 .
(13.3)
Adding the relationships in (13.2) and (13.3) yields
3K + L + M + N = 2ρ2 (A + B + C).
Subtracting (13.1) from this equation yields
2K = 2ρ2 (A + B + C − π),
so that
K = ρ2 (A + B + C − π).
(13.4)
Thus, if A, B, and C are the dihedral angles of a spherical triangle, and we
write Σ = A + B + C, then the area of this spherical triangle is ρ2 (Σ − π).
Therefore the fraction of the sphere occupied by the triangle is
ρ2 (Σ − π)
1
1
=
(Σ − π) =
(Σ − 180◦ ),
2
4πρ
4π
720◦
13.1. AREA OF A SPHERICAL POLYGON
199
as mentioned in §3.1. (Note that Σ is used to represent the sum of the
angles in either degrees or radians as is appropriate.)
To find the area of a general spherical polygon, we recall a result from
plane geometry: the sum of the angles in a plane polygon of n sides is
(n − 2)π. This is seen by the simple device of drawing in diagonals of the
polygon so that n − 2 triangles are formed.
The same device can also be used in spherical geometry to find the area
of a spherical polygon. We include the details for the quadrilateral case, and
then state the general result.
So consider the spherical quadrilateral in Figure 13.2, which is divided
into two spherical triangles as shown. As suggested by the Figure, angles B
and D are divided by a diagonal of the quadrilateral, so that B = B1 + B2
and D = D1 + D2 . To find the area of the quadrilateral, it suffices to add
the areas of the two spherical triangles, which are
K1 = ρ2 (A + B1 + D1 − π)
and
K2 = ρ2 (C + B2 + D2 − π).
B1 B2
C
A
K1
K2
D1 D2
Figure 13.2
Thus,
K1 + K2 = ρ2 (A + (B1 + B2 ) + C + (D1 + D2 ) − 2π)
= ρ2 (A + B + C + D − 2π)
is the area of the spherical quadrilateral.
200
CHAPTER 13. EULER’S FORMULA
Now suppose that a spherical polygon of n sides is given, with dihedral
angles D1 , D2 , . . . Dn . Then the area of the spherical polygon is
K = ρ2 (D1 + D2 + · · · + Dn − (n − 2)π).
(13.5)
The ambitious reader may provide a straightforward argument for this result
using (13.4) and the principle of mathematical induction.
13.2
Euler’s Formula
We use the result of the previous section in presenting a proof of Euler’s
formula (2.1). This proof was first made known to me by J. J. Schäffer of
Carnegie Mellon University in the departmental lounge.
Let us begin with a convex polyhedron. As usual, let V , E, and F
denote the number of vertices, edges, and faces, respectively, on our polyhedron. Given that we will be employing (13.5), all angles will be measured
in radians.
As in our work with geodesic structures, project this polyhedron onto
a sphere. For simplicity, assume that this sphere has radius 1 so that we
may put ρ := 1 in (13.5). The result is a tiling of the sphere by spherical
polygons. Euler’s formula may be obtained by adding all of the dihedral
angles of all of these spherical polygons in two different ways.
To this end, consider a vertex projected on the sphere. The dihedral
angles around that vertex must sum to 2π. Thus, the sum of all the dihedral angles at all of the projected vertices is 2πV . Since each dihedral
angle of each of the spherical polygons is incident to exactly one projected
vertex, 2πV must be the sum of all the dihedral angles in all of the spherical
polygons.
Now we look at the sum in another way. Begin by considering a spherical
polygon P corresponding to a face of the polyhedron. Suppose that P has n
sides, and therefore n dihedral angles D1 , D2 , . . . Dn . We may rewrite (13.5)
(using ρ = 1) as
K + nπ − 2π = D1 + D2 + · · · + Dn .
(13.6)
We may write such an expression for each spherical polygon, and then sum
them all. We take a moment to consider what that sum would be.
Since K is the area of the spherical polygon P, summing the “K” terms
yields the sum of the areas of the spherical polygons which tile the sphere
onto which our polyhedron is projected. Since the polygons cover the entire
surface of the sphere, this sum is therefore 4π.
13.3. DESCARTES’ RULE OF DEFICIENCY
201
Since n is the number of sides on the spherical polygon P, summing the
“nπ” terms results in the total number of sides on all the spherical polygons,
multiplied by π. Since the spherical polygons are projections of the faces of
our polyhedron, this sum is the total number of sides on all the faces of the
polyhedron, multiplied by π. But this is just 2Eπ, since each edge of the
polyhedron corresponds to exactly two edges on two different faces of the
polyhedron. (Recall the analogous argument for Platonic solids in §2.3.)
Since there is a “−2π” term for each spherical polygon, and hence each
face of the polyhedron, this sum is just F (−2π) = −2F π.
Thus, the sum over all spherical polygons of the terms on the left-hand
side of (13.6) is
4π + 2Eπ − 2F π.
The corresponding sum of the terms on the right-hand side is the sum of all
dihedral angles of all the spherical polygons projected onto the sphere. But
just a moment ago, we saw that this sum is 2V π. Hence, we must have
4π + 2Eπ − 2F π = 2V π.
Dividing through by 2π and rearranging terms produces Euler’s formula
(2.1) as promised.
We remark on the necessity of assuming the polyhedron is convex. This
allowed us to project the convex faces of the polyhedron onto a sphere and
then add the areas of the spherical projections to obtain the surface area of
the sphere. A variation of (2.1) may be obtained for nonconvex polyhedra,
but we will not address that case here. The interested reader should consult
Coxeter’s Regular Polytopes.
13.3
Descartes’ Rule of Deficiency
We proved Descartes’ rule of deficiency for Archimedean solids in Exercise
6 of Chapter 5. Using a counting technique similar to that used in proving
Euler’s formula, we may now prove Descartes’ rule in the general case.
So consider a vertex v of the polyhedron. Suppose that there are n
angles at that vertex, α1 , α2 , . . . , αn , so that the deficiency δv at that vertex
is given by
δv = 2π − (α1 + α2 + · · · + αn ).
Now let’s sum this relationship over all the vertices of the polyhedron.
Summing the left-hand side is adding all the deficiencies at all the vertices of
the polyhedron; in other words, the total deficiency ∆ is obtained. Summing
202
CHAPTER 13. EULER’S FORMULA
the “2π” terms gives 2π for each vertex of the polyhedron, so that this sum
is 2V π. And finally, summing α1 + α2 + · · · + αn over all vertices of the
polyhedron is just the sum of all angles in all the faces of the polyhedron,
since each angle of a face is incident at exactly one vertex of the polyhedron.
Denote by S this sum of all angles in all faces of the polyhedron. Thus, we
have so far determined that
∆ = 2V π − S.
(13.7)
We look for another relationship involving S. To this end, consider a
face P of the polyhedron with n sides and n angles β1 , β1 , . . . βn . We know
that these n angles sum to (n − 2)π, so that we have
nπ − 2π = β1 + β2 + · · · + βn .
(Note the similarity to (13.6).)
Now let’s sum this relationship over all faces of the polyhedron. As we
just saw in proving Euler’s formula, the left-hand sides sum to 2Eπ − 2F π.
And the right-hand side is the sum of all angles in all the faces of the
polyhedron, which we called S. So we have obtained the relationship
2Eπ − 2F π = S.
(13.8)
Now we may substitute from (13.8) into (13.7) to obtain
∆ = 2V π − (2Eπ − 2F π) = 2π(V − E + F ).
But we know from Euler’s formula that V − E + F = 2, so that the total
deficiency is given by
∆ = 4π = 720◦ .
(13.9)
This proves Descartes’ rule of deficiency in the general case. Again, the
assumption that the polyhedron is convex is critical in making the proof
work.
13.4
Dihedral Angles Revisited
In this section, we will look at the duplications in Table 6.1 (see Exercise 7
of Chapter 6) with a more critical eye in anticipation of a discussion of solid
angles in the next section.
Referring to Table 6.1 (as we will continue to do without explicit reference each time), we see that the dihedral angles of the truncated octahedron,
13.4. DIHEDRAL ANGLES REVISITED
203
the truncated cube, and the cuboctahedron are all equal. This results not
from the particular shape of the faces meeting at a given edge, but rather
from the fact a facial plane of a cube is meeting the facial plane of an octahedron along that edge of the polyhedron. When this occurs, the angle
made by two such planes is about 125.264◦ .
What is the significance of 125.264◦ ? Think for a moment about truncating an octahedron. It should be clear that a half-octahedron (or square
pyramid) is being sliced off each vertex. It is also evident that the dihedral
angles at the square base of this pyramid must be 12 D3,4 . So the new dihedral
angle created by slicing off a half-octahedron must be
1
Dt3,4 = π − D3,4 ≈ 125.264◦ .
2
(13.10)
Now it is also clear from Exercise 6(a) of Chapter 2 (and Table 3.1) that the
dihedral angles of the tetrahedron and octahedron are supplementary, or
D3,4 = π − D3,3 ,
(13.11)
so that (13.10) may be rewritten
Dt3,4 = Dt4,3 = DT3,4 =
π 1
+ D3,3 .
2 2
Now recall from Exercise 6 of Chapter 9 that a rhombicuboctahedron
may be inscribed in a rhombic dodecahedron. This implies that twelve
squares of the rhombicuboctahedron and the twelve squares on the rhombitruncated cuboctahedron lie in the facial planes of a rhombic dodecahedron.
It is evident from Figure 6.10(c) that such planes meet the facial planes of
a cube at an angle of 43 π = 135◦ , so that this accounts for the fact that
3
De3,4 [4–4] = DE
3,4 [4–8] = π.
4
To account for De3,4 [4–3] = DE
3,4 [4–6], we must find the angle at which
the facial planes of a rhombic dodecahedron meet those of an octahedron.
Examine Figure 13.3, where the fact that a rhombicuboctahedron may be
inscribed in a rhombic dodecahedron is used. It is clear that if the vertices of
a rhombic dodecahedron are appropriately truncated (leaving squares as in
Figure 9.12), a rhombicuboctahedron remains. Because the dihedral angles
of the rhombic dodecahedron are 120◦ , it must be that the squat tetrahedra
sliced off the trivalent vertices of the rhombic dodecahedron are in fact onefourth tetrahedra. As a result, the dihedral angles at the equilateral bases of
204
CHAPTER 13. EULER’S FORMULA
these tetrahedra (which are the triangular faces of the rhombicuboctahedron
in octahedral facial planes) must be 12 D3,3 . And so the dihedral angles
remaining after these tetrahedra are sliced off have measure
1
◦
De3,4 [4–3] = DE
3,4 [4–6] = π − D3,3 ≈ 144.736 .
2
These relationships are summarized in the following table, where “C,”
“O,” and “R” represent cubical, octahedral, and rhombic dodecahedral facial
planes, respectively. The symmetry in the table is self-evident.
C
π
2
C
O
R
O
π
2
π
2
+ 21 D3,3
3
4π
+ 12 D3,3
R
3
4π
π − D3,3
π − 12 D3,3
π − 12 D3,3
2
3π
Table 13.1
We may verify our geometrical observations with explicit calculation in
the event we have made an erroneous assumption along the way. (Of course,
we may always check that the angles agree to several decimal places on our
calculators, but we include a trigonometric verification for the purist.)
For example, since π2 + 12 D3,3 is an obtuse angle,
r
1 − cos D3,3
1
1
π 1
t
+ D3,3 = − sin D3,3 = −
= −√ ,
cos D3,4 = cos
2 2
2
2
3
in agreement with Table 6.1.
Figure 13.3
13.5. SOLID ANGLES
205
Creating an analogous table for the Archimedean solids with icosahedral symmetry is a decidedly more complex issue. Details will be left to
the very ambitious reader. Here, “I,” “D,” and “R” represent icosahedral,
dodecahedral, and rhombic triacontahedral facial planes, respectively.
I
I
D
3
2π
D3,5
D
R
3
2π
− 12 (D3,5 + D5,3 )
− 21 (D3,5 + D5,3 )
π
2
R
D5,3
+ 12 D3,5
π
2
+ 12 D5,3
π
2
+ 12 D3,5
π
2
+ 12 D5,3
4
5π
Table 13.2
For convenient reference, the same table is given below with all values
in degrees.
I
D
R
I
138.190◦
142.623◦
159.095◦
D
142.623◦
116.454◦
148.283◦
R
159.095◦
148.283◦
144◦
Table 13.3
13.5
Solid Angles
As anyone who ever built with blocks or opened a box of sugar cubes knows,
cubes tile space; that is, arbitrarily large regions of space may be filled by
packing cubes so that they meet face-to-face with no gaps. In so doing, the
cubes meet eight at a vertex, so that it makes sense to say that a vertex of
a cube occupies 18 of the space around that vertex. We might notate this by
saying that the solid angle at a vertex of a cube is
1
S4,3 = .
8
But we can also see this in another way using (13.4). If we center a
sphere of radius ρ at a vertex of a cube, a spherical triangle is created. We
might determine the solid angle at that vertex by asking, “What fraction
206
CHAPTER 13. EULER’S FORMULA
of the surface area of the sphere does this spherical triangle occupy?” By
using (13.4), we see that the area of this triangle is
ρ
2
πρ2
π π
+ + −π =
.
2
2
2
2
π
Since the surface area of the sphere has area 4πρ2 , this fraction is simply
S4,3 =
πρ2
2
4πρ2
1
= .
8
We may generalize (13.4) by writing “S” for a generic solid angle:
S=
K
1
=
(A + B + C − π).
4πρ2
4π
(13.12)
Of course, if more than three faces meet at the vertex of our polyhedron, we
generalize (13.5):
S=
1
(D1 + D2 + · · · + Dn − (n − 2)π).
4π
(13.13)
Using (13.12) in the case of the tetrahedron, we find that
S3,3 =
3
1
D3,3 − .
4π
4
(13.14)
For the octahedron, we must use (13.13), so that together with (13.11), we
have
1 1
S3,4 = − D3,3 .
(13.15)
2 π
Now we may solve (13.14) for D3,3 , resulting in
1
4π
S3,3 +
.
D3,3 =
3
4
Substituting this in (13.15) gives
S3,4 =
1 4
− S3,3 .
6 3
Multiplying this equation through by 6 and rearranging terms yields
8 S3,3 + 6 S3,4 = 1.
(13.16)
13.6. EXERCISES
207
Surely this must be magic! But no – this simple relationship between the
solid angles at vertices of the tetrahedron and octahedron admits an elegant
geometrical explanation.
Consider the cuboctahedron. In view of the fact that the edge angle is
◦
60 , it is evident that if a triangular face is connected to the center of the
cuboctahedron, a regular tetrahedron is formed. If a square face is joined
to the center, a square pyramid with equilateral triangular faces is formed
– in other words, a half-octahedron is created.
Since a cuboctahedron has eight triangular faces and six square faces,
then these eight tetrahedra and six half-octahedra meet at the center of the
cuboctahedron. Moreover, due to our construction, there are no gaps, so
that these 14 polyhedra completely fill space at the center of the cuboctahedron.
But this is precisely what (13.16) validates: the fraction of space occupied by 8 tetrahedral and 6 octahedral vertices is 1, or 100%. This also
indicates that space may be filled with alternating tetrahedra and octahedra:
arrange tetrahedra and octahedra face-to-face so that no two tetrahedra are
ever adjacent, nor are two octahedra ever adjacent. (Of course those readers
already familiar with this tiling of space would understand the geometrical
content of (13.16) immediately.)
As a result, any tiling of space should be verifiable using (13.16). But
as a caution, finding a combination of solid angles which sum to 1 does not
necessarily guarantee that a face-to-face tiling of space exists. For example,
it is clear from (13.16) that
4 S3,3 + 3 S3,4 + 4 S4,3 = 1,
although tetrahedra, octahedra, and cubes can never tile space face-to-face
since the cubes can never be adjacent to the tetrahedra or octahedra.
For reference, solid angles of the vertices of the Platonic solids and the
nonsnub Archimedean solids are given in Table 13.4 (located after the Exercises). They may be used to verify that various other combinations of
polyhedra tile space; details will be left to the Exercises.
13.6
Exercises
1. Verify the angles given in Tables 13.1 and 13.2 by taking their cosines
and comparing with Table 6.1.
2. Each of the following combinations of polyhedra can arranged so that
they tile space. For each combination, determine the arrangement
208
CHAPTER 13. EULER’S FORMULA
necessary and write an analogue of (13.16) for each case. Verify your
results using Table 13.4.
(a) truncated octahedra alone;
(b) tetrahedra and truncated tetrahedra;
(c) octahedra and cuboctahedra;
(d) octahedra and truncated cubes;
(e) truncated tetrahedra, truncated octahedra, and cuboctahedra;
(f) tetrahedra, cubes, and rhombicuboctahedra;
(g) cubes, cuboctahedra, and rhombicuboctahedra;
(h) cubes, truncated octahedra, and rhombitruncated cuboctahedra;
(i) truncated tetrahedra, truncated cubes, and rhombitruncated cuboctahedra;
13.6. EXERCISES
209
Polyhedron
Symbol
S
S
Tetrahedron
S3,3
S3,3
0.043870
Octahedron
S3,4
Cube
S4,3
Trunc. Tetrahedron
St3,3
Trunc. Octahedron
St3,4
Trunc. Cube
St4,3
5
24
+ 13 S3,3
0.222957
Cuboctahedron
ST3,4
1
6
+ 23 S3,3
0.195913
Rh’trunc. Cubocta’n
SE
3,4
Rh’cubocta’n
Se3,4
Icosahedron
1
6
− 43 S3,3
1
8
1
6
− 13 S3,3
1
4
5
16
7
24
0.108173
0.125
0.152043
0.25
0.3125
− 13 S3,3
0.277043
S3,5
S3,5
0.209651
Dodecahedron
S5,3
S5,3
0.235688
Trunc. Icosahedron
St3,5
5
12
− 13 S5,3
0.338104
Trunc. Dodecahedron
St5,3
7
20
− 15 S3,5
0.308070
Icosidodecahedron
ST3,5
− 52 S3,5 − 23 S5,3
0.292348
Rh’trunc. Icosidodeca’n
SE
3,5
Rh’icosidodeca’n
Se3,5
8
15
3
8
7
30
Table 13.4
+ 51 S3,5 + 13 S5,3
0.375
0.353826
210
CHAPTER 13. EULER’S FORMULA
Chapter 14
Coordinates of Polyhedra
Now that we have thoroughly analyzed the convex and nonconvex regular
and semi-regular polyhedra using spherical trigonometry, we take some time
to look at polyhedra from a few different perspectives. An entire volume
could be written on these topics, so the reader will have to be content with
the brief introductions given here.
14.1
Introduction and Motivation
How do you solve a geometry problem?
That depends, of course, on the problem. Sometimes the use of various
theorems about congruent triangles, parallel lines, alternate interior angles,
intersecting chords of circles, etc., are all that is necessary. This is sometimes
referred to as a “synthetic” approach, and was once popular in high school
geometry courses.
If the problem is one of plane geometry, there is another method. By locating the geometrical objects in the plane and describing them by Cartesian
coordinates, a problem may be solved more algebraically then geometrically.
This is sometimes referred to as an “analytic” approach.
A simple example may help here. One recalls that any triangle inscribed
in a semicircle is a right triangle, as shown in Figure 14.1. This is often
proved as follows. Since ∆pqr is inscribed in a semicircle, the arc psr has
measure 180◦ . But since the measure of any angle inscribed in a circle is
half the measure of the intercepted arc (previously proved, of course!), ∠pqr
has measure 12 (180◦ ) = 90◦ , so that ∆pqr is a right triangle.
211
212
CHAPTER 14. COORDINATES OF POLYHEDRA
q
p
r
s
Figure 14.1
Let’s see how this may be shown analytically; that is, by imposing a
convenient coordinate system. It is simplest in this case to choose the center
of the circle as the origin (0, 0), and have p and r be opposite ends of a
diameter of a unit circle with coordinates (−1, 0) and (1, 0), respectively.
Thus, the circle is described by the equation x2 + y 2 = 1 in this coordinate
system. (When the choice of origin is not clear, it sometimes takes more
than one attempt to find the simplest approach.)
Now assign q the coordinates (x0 , y0 ); thus, x20 + y02 = 1. The slope
between points p and q is given by
mpq =
y0 − 0
y0
=
,
x0 − (−1)
x0 + 1
while the slope between r and q is given by
mrq =
y0
y0 − 0
=
.
x0 − 1
x0 − 1
(Here we must assume that y0 6= 0 so that both slopes are defined.)
Since x20 + y02 = 1, we have x20 − 1 = −y02 , resulting in
mpq mrq =
y0
y2
y0
·
= 0 2 = −1.
x0 + 1 x0 − 1
−y0
Thus, the product of the slopes of the lines containing the two sides of the
triangle is −1, so that these two sides must be perpendicular at q. Hence
∆pqr is a right triangle.
This analytic solution is rarely (if ever!) encountered since the synthetic
approach is so straightforward. The point here is that both approaches yield
correct results, but for one reason or another, one approach may be preferred
to the other.
14.2. THE CUBE
213
What is a good reason? In the previous example, seeking the briefest
solution might suggest the synthetic approach. But this presumes that all
necessary supporting theorems are already at hand.
In Chapter 3, we used spherical trigonometry to find edge and dihedral
angles of the Platonic solids. Why use spherical trigonometry? Recall that
a major focus has been the study and design of geodesic structures, and that
spherical trigonometry is really necessary for this.
But suppose we have other goals in mind. We might not care so much
about geodesic structures, but instead want to make some snazzy computer
graphics involving the Platonic solids. In this case, spherical trigonometry
is of little use since most graphics packages require that we describe various
geometrical objects using a three-dimensional coordinate system.
Thus, different approaches are suited to different applications. In this
chapter, we will tackle the problem of describing the Platonic solids by means
of three-dimensional coordinates for the budding computer graphicist. We
will then be able to use these coordinates to find the edge and dihedral
angles of the Platonic solids as well.
14.2
The Cube
The usual conventions for drawing a two-dimensional Cartesian coordinate
system are to have the positive x-axis pointing due east and the positive
y-axis pointing due north so that the axes are perpendicular. Thinking of
a coordinate system as a mathematical tool, nothing would be lost if the
positive x-axis pointed due west rather than due east. It just might look a
little unusual.
In creating a three-dimensional coordinate system, the third axis, called
the z-axis, is imagined to point “straight up” so that it is perpendicular to
the other two axes, just as walls in the corner of a room meet in mutually
perpendicular line segments. The difficulty lies in trying to draw a picture
of a corner – there is not enough room on a two-dimensional piece of paper
for three mutually perpendicular axes.
There are a few different conventions for drawing a three-dimensional
coordinate system in the plane. We will use the convention illustrated in
Figure 14.2, where the y-axis points due east, the z-axis points due north,
and the x-axis “comes out” in a southwesterly direction. When plotting a
point in this coordinate system, it is easiest to first move along the x-axis
according to the x-coordinate of the point, and then from this location, move
according to the y- and z-coordinates as in two dimensions.
214
CHAPTER 14. COORDINATES OF POLYHEDRA
Before giving coordinates for the cube, we review coordinates for a square
in two dimensions. One common coordinatization for a square lists vertices
as (0, 0), (1, 0), (1, 1), and (0, 1), describing a unit square with lower left
corner at the origin.
While there is nothing “wrong” with this choice of coordinates, it happens that for a discussion of symmetry and matrices in the next chapter, it
is important that the origin is the center of the square (or cube) rather than
one of its vertices.
So instead, we consider a square whose vertices are (1, 1), (−1, 1), (−1, −1),
and (1, −1). This square has sides of length 2, and the center of this square
is in fact (0, 0).
+z
(1, 2, 3)
+y
+x
Figure 14.2
It is now an easy task to make a cube. Thinking of our square as lying
in the xy-plane, moving it “up” one unit produces the square with vertices
(1, 1, 1), (−1, 1, 1), (−1, −1, 1), and (1, −1, 1) in the plane z = 1. Moving
the square “down” one unit results in the square with vertices (1, 1, −1),
(−1, 1, −1), (−1, −1, −1), and (1, −1, −1) in the plane z = −1. These translated squares are the top and bottom faces of the cube, as shown in Figure
14.3.
Note that the vertices of the cube consist of all lists of three coordinates,
each of which is either +1 or −1. Since there are two choices for each of
three coordinates, there are 23 = 8 vertices on a cube (as there should be!).
14.2. THE CUBE
215
Also note that two vertices are adjacent if and only if they differ in exactly
one coordinate; this coordinate indicates the axis to which the edge joining
these two vertices is parallel.
For the student more familiar with points, lines, and planes in a threedimensional coordinate system, it is instructive to find equations for the
lines containing the edges of the cube and the planes containing the faces of
the cube. Such discussions, however, will be relegated to the Exercises.
(−1, −1, 1)
(−1, 1, 1)
(1, −1, 1)
(1, 1, 1)
(−1, −1, −1)
(−1, 1, −1)
(1, −1, −1)
(1, 1, −1)
Figure 14.3
Now that we have coordinates for the vertices, how might we use them
to find the dihedral and edge angles of the cube? Some simple Euclidean
geometry is all that is needed. Consider Figure 14.4, where the ends of an
edge of the cube are connected to the origin to form the edge angle E. To
find E, all we need is the length of the sides of the given triangle; we know
that the top edge, being an edge of the cube, has length 2.
(1, −1, 1)
(1, 1, 1)
E
(0, 0, 0)
Figure 14.4
To find the other side lengths, we need the distance formula in three
dimensions. It turns out that this is a natural extension of the familiar
216
CHAPTER 14. COORDINATES OF POLYHEDRA
formula in two dimensions, so that the distance between points (x1 , y1 , z1 )
and (x2 , y2 , z2 ) is simply
p
d = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 .
(14.1)
Applying this
√ to the triangle in Figure 14.4, we see that the other two sides
have length 3, so that the cosine law for triangles results in
22 =
√
2
3 +
√
√ √
2
3 − 2 3 3 cos E,
or
1
cos E = ,
3
as expected.
Finding the dihedral angle is only a little trickier. Recall that the dihedral angle is the angle between adjacent faces of a Platonic solid. This angle
may be obtained as follows: slice the Platonic solid with a plane through the
midpoint of an edge and perpendicular to that edge. This creates an angle
which also passes through the centers of the two faces which meet in that
edge. So in general, the dihedral angle of a Platonic solid is always part of
a triangle whose vertices are the midpoint of a given edge and the centers
of the two faces which meet at that edge.
So let’s consider the edge given in Figure 14.4. Its midpoint is found
by averaging the respective coordinates (as in two dimensions), resulting
in (1, 0, 1). It is the edge where the top and front faces of the cube meet.
Since the positive z-axis goes through the center of the top face, its center is
(0, 0, 1), and likewise, the center of the front face is given by (1, 0, 0). Thus,
the triangle as shown in Figure 14.5 is formed. Since we have coordinates
for the three vertices of the triangle, we may find the lengths of the three
sides using (14.1) and use the cosine law for triangles to find cos D = 0, as
expected.
(0, 0, 1)
D
(1, 0, 1)
(1, 0, 0)
Figure 14.5
14.3. THE TETRAHEDRON AND OCTAHEDRON
217
In the future, we will simply jump from a diagram such as Figure 14.5 to
the final result, letting the reader fill in the necessary algebraic details. We
note that the reader familiar with vector geometry and the inner product
(or dot product) may see another way to calculate cos D from Figure 14.5,
and should feel free to do so.
14.3
The Tetrahedron and Octahedron
The reader who has completed Exercise 6(b) of Chapter 2 or studied Figure
9.4 may recall that the tetrahedron may be inscribed in a cube in two ways.
For the less thorough reader, we may see this fact by choosing a vertex of
the cube. Three square faces meet at this vertex, so that three diagonals
of these squares meet at this vertex as well. The other three ends of these
diagonals, together with the chosen vertex, form the vertices of a regular
tetrahedron. It might be instructive to build a cube at this point and draw
in the appropriate diagonals to see this.
The four remaining vertices of the cube also form a regular tetrahedron
which interpenetrates the first. The region common to these two tetrahedra
is an octahedron (see Exercise 6(b) of Chapter 2 again) whose vertices are
the centers of the faces of the cube.
For specificity, choose one of these tetrahedra; say, begin with the vertex
(1, 1, 1). The opposite ends of the diagonals meeting at this vertex have two
coordinates changed, resulting in (1, −1, −1), (−1, 1, −1), and (−1, −1, 1).
On the tetrahedron, each of these vertices is adjacent to the others, always
differing in exactly two coordinates.
The edge angle may be determined as before using a triangle with vertices
(0, 0, 0), (1, 1, 1), and (1, −1, −1). In other words, simply choose an edge of
the tetrahedron as a side of the triangle, and use the origin as the third
vertex. This results in cos E = − 31 .
To find the dihedral angle of the tetrahedron, recall that we may form
a triangle from the midpoint of an edge and the centers of the two faces
meeting at that edge.
So begin by choosing an edge of the tetrahedron, say the edge with ends
(1, 1, 1) and (1, −1, −1). The midpoint of this edge is (1, 0, 0). To find the
centers of the incident faces, we may use the following result: the center of
a regular polygon is found by averaging the coordinates of its vertices.
Since one face incident at this edge has vertices (1, 1, 1), (1, −1, −1), and
218
CHAPTER 14. COORDINATES OF POLYHEDRA
(−1, 1, −1), its center is given by
1
1
1
1 1 1
.
(1 + 1 + (−1)), (1 + (−1) + 1), (1 + (−1) + (−1)) =
, ,−
3
3
3
3 3 3
The center of the other incident face with vertices (1, 1, 1), (1, −1, −1), and
(−1, −1, 1) has coordinates ( 31 , − 13 , 13 ). This forms the triangle in Figure
14.6. Using the technique described earlier, we find that cos D = 31 .
In discussing the tetrahedron, we remarked that the centers of the faces
of the cube are the vertices of an octahedron. This should not be surprising
since the cube and octahedron are dual to each other (see Figure 9.5). The
octahedron in this figure may be “shrunk” until its vertices coincide with
the centers of the faces of the cube.
(1, 0, 0)
D
( 13 , 13 , − 13 )
q
√
2 2
3
2
3
( 13 , − 13 , 31 )
Figure 14.6
Given these relationships, it is not difficult to find the edge and dihedral
angles of the octahedron using the methods described in this chapter. Details
are left to the Exercises.
14.4
The Dodecahedron
Finding coordinates for the vertices of a dodecahedron is a little trickier
business. As an aid, we consider a dodecahedron constructed around a cube
as in Figure 2.10. This immediately gives us eight of the twenty vertices of
the dodecahedron. Only twelve remain to be found.
We begin with one roof built on the top face of the cube described in
§14.2, as given in Figure 14.7. Note that points p and q in this Figure have
the same x-coordinate, 0, and the same z-coordinate, unknown at this time
and denoted by z.
14.4. THE DODECAHEDRON
219
To determine the y-coordinate, recall from §1.1 that the edges and diagonals of a regular pentagon have lengths in the ratio 1 : τ . Since the cube
has edges of length 2 (diagonals of the pentagons), the length of the sides of
the pentagons must be τ2 . Since the y-coordinates of p and q are opposite
due to symmetry, they must be − τ1 and τ1 , respectively.
q = 0, τ1 , z
0, − τ1 , z = p
(−1, 1, 1)
(−1, −1, 1)
(1, −1, 1)
(1, 1, 1)
Figure 14.7
How may z be determined? One way is to note that the distance from q
to the vertex (1, 1, 1) must be τ2 , the length of the edges of the dodecahedron.
In other words, using (14.1),
p
(0 − 1)2 + (τ −1 − 1)2 + (z − 1)2 =
2
.
τ
Solving this quadratic equation for (z − 1)2 using Table 1.1 as needed, we
have
1
(z − 1)2 = 2 − τ = 2 ,
τ
so that
1
1
1
z = 1 + = τ, z = 1 − = 2 .
τ
τ
τ
Since z > 1, we choose the solution z = τ , the other solution being extraneous.
Because of the symmetrical way that the dodecahedron is assembled by
placing roofs on the faces of a cube, the coordinates for the other vertices
may easily be found.
Since we have just found coordinates for an edge of the dodecahedron,
calculating the edge angle may be done as usual.
To find the dihedral angle of the dodecahedron, we simply take the appropriate cross section of Figure 14.7, shown in Figure 14.8(a) and redrawn
in Figure 14.8(b).
220
CHAPTER 14. COORDINATES OF POLYHEDRA
(0, 0, τ )
D5,3
(1, 0, 1)
(a)
(−1, 0, 1)
(b)
Figure 14.8
Given the coordinates of the vertices in Figure 14.7, it is not hard to calculate the midpoints of the appropriate segments to obtain the coordinates
given in Figure 14.8(b). Once these are known, it is not difficult to use the
cosine law to find D5,3 . Details are left to the Exercises.
14.5
The Icosahedron
There is more than one way to find coordinates for the vertices of an icosahedron. Two that we will consider exploit the fact that the icosahedron and
dodecahedron are dual to each other.
Perhaps the conceptually simpler method is to note that just as vertices
for an octahedron may be obtained by finding the centers of the faces of a
cube, so the vertices of an icosahedron may be found by finding the centers
of the faces of a dodecahedron. This method, though, is rather tedious; full
use of our geometrical savvy reveals an algebraically simpler approach.
As in Figure 9.6, note that edges of the dodecahedron and its dual icosahedron perpendicularly bisect each other in pairs. How can we exploit this
relationship?
A slight shift in perspective will aid us in our task. In the previous section, we considered the dodecahedron as being circumscribed about a cube
so that the vertices of the cube were on the surface of the dodecahedron.
By slightly expanding the cube, we may imagine the dodecahedron being
inscribed in a cube, so that one edge of each roof lies in a face of the cube,
as in Figure 14.9. This implies that the edges of the dual icosahedron perpendicular to these six dodecahedral edges (such as uv being perpendicular
to pq) also lie on the surface of the cube. Since the icosahedron has just
twelve vertices, determining coordinates of the ends of these six edges will
yield coordinates for the vertices of an icosahedron.
14.5. THE ICOSAHEDRON
221
v
p
q
u
a
w
Figure 14.9
The trick is to look at a triangular face of the icosahedron such as ∆uwa.
Regardless of how long the edge uv is, ∆uwa will always be equilateral. But
if the icosahedron is regular, the lengths of the sides of this equilateral
triangle must be the same as the length of the edge uv.
Now p, q, u, and v lie in the same plane parallel to the xy-plane. Since
p and q have z-coordinate τ , so must u and v. It is also easy to see that u
and v have y-coordinate 0. However, their x-coordinates are unknown; if u
had coordinates (x, 0, τ ), then v has coordinates (−x, 0, τ ).
In light of the symmetrical way the six icosahedral edges like in the faces
of the cube, it is evident that w has coordinates (τ, x, 0). So if [uv] = [uw],
we see by using the distance formula that
p
p
(x − (−x))2 + (0 − 0)2 + (τ − τ )2 = (x − τ )2 + (0 − x)2 + (τ − 0)2 .
Multiplying out and collecting like terms yields
x2 + τ x − τ 2 = (x − 1)(x + τ 2 ) = 0.
Since x must be positive, we select x = 1 as the solution, x = −τ 2 being extraneous. Then u has coordinates (1, 0, τ ), and considerations of symmetry
allow the coordinates of the other vertices to be easily found. Once this is
done, calculations of the edge and dihedral angles of the icosahedron may
be completed using the techniques employed earlier.
222
CHAPTER 14. COORDINATES OF POLYHEDRA
It is worth remarking that the icosahedron has edge length 2 while the
dual dodecahedron has edge length τ2 . These lengths are in the ratio τ : 1.
Examination of Figure 9.6 reveals that each pair of mutually bisecting edges
may be thought of as the diagonals of a rhombus on a rhombic triacontahedron. Thus, the lengths of the diagonals of the faces of a rhombic
triacontahedron are in the ratio τ : 1.
14.6
Truncated Platonic Solids
Recall that in §5.1, we found out how far we needed to truncate the Platonic
solids so that an Archimedean solid with regular faces was produced. Here,
we use that information to find coordinates for these polyhedra.
We know (see §5.1) that to partially truncate a Platonic solid with triangular faces, we must divide the edges into equal thirds. Our basic problem,
then, is this: if a point divides an edge in a given ratio, and the coordinates
of the ends of that edge are known, what are the coordinates of that point?
A simple way to look at the problem is to consider one coordinate at
a time. (That this may be done is a consequence of the face that our
coordinate lists form a linear (vector) space. The interested reader should
consult a linear algebra text for more details.) In Figure 14.10, a and b are
real numbers. How do we find a number c such that c divides the interval
between a and b into the ratio λ1 : λ2 ? (We implicitly assume that both λ1
and λ2 are strictly positive.)
a
λ1 c
λ2
b
Figure 14.10
We may rephrase this problem by seeking the number c that satisfies
λ1
c−a
=
.
b−c
λ2
Solving this equation for c yields
c=
λ2
λ1
a+
b.
λ1 + λ2
λ1 + λ2
(14.2)
As it turns out, this is valid in the three cases a < b, a = b, or a > b; that
is, for all values of a and b.
14.6. TRUNCATED PLATONIC SOLIDS
223
So if we wanted to truncate the tetrahedron described in §14.3, we would
need to find a point c dividing an edge in the ratio 1 : 2, as in Figure 14.11.
(1, −1, −1)
(1, 1, 1)
1
c
2
Figure 14.11
Using (14.2), we see that this point is given by
2
1
1 1
c = (1, 1, 1) + (1, −1, −1) = 1, ,
.
3
3
3 3
As suggested
√ by §5.1, the cube is truncated by dividing its edges in the
of the truncated cube divides an edge
ratio 1 : 2 : 1, so that a vertex
√
of the cube in the ratio 1 : 2 + 1. Similarly, since the dodecahedron
is truncated by dividing its edges in the ratio 1 : τ : 1, a vertex of the
truncated dodecahedron divides an edge of the dodecahedron in the ratio
1 : τ + 1. By using (14.2), then, vertices for all the partially truncated
Platonic solids may be found.
Finding coordinates for the vertices of the cuboctahedron and icosidodecahedron is a simple matter, as the vertices of these polyhedra are the
midpoints of the edges of the cube and the dodecahedron, respectively. Details are left to the Exercises.
At this point, the budding computer graphicist may be wondering if it
is necessary to calculate every vertex of a polyhedron in order to render it
graphically. For this seems to involve a lot of work! Rest assured that this
is not necessary. But more advanced tools are required for a more efficient
rendering – matrices and symmetry groups. These topics will be introduced
in the next chapter.
224
CHAPTER 14. COORDINATES OF POLYHEDRA
14.7
Exercises
1. If you are familiar with equations for lines and planes in three dimensions, find equations for the lines and planes which contain the edges
and faces, respectively, of the cube as suggested in §14.2.
2. If you are familiar with the inner product (dot product), use it to show
that the angle D in Figure 14.5 has measure 90◦ .
3. As suggested in §14.3, find the edge and dihedral angles of the octahedron using the techniques described in this chapter.
4. As suggested in §14.4, find the edge and dihedral angles of the dodecahedron as described in that section.
5. As suggested at the beginning of §14.5, find a vertex of an icosahedron
by finding the center of a face of the dodecahedron. How does your
result differ from the results at the end of §14.5 for the vertices of a
regular icosahedron? Why?
6. As suggested in §14.5, find the edge and dihedral angles of the icosahedron using the techniques described in this chapter.
7. Find coordinates for the vertices of one or more of the partially truncated Platonic solids as described in §14.6.
8. Recall that the vertices of a cuboctahedron are the midpoints of the
edges of a cube or octahedron. Find coordinates for the vertices of a
cuboctahedron.
9. Recall that the vertices of an icosidodecahedron are the midpoints of
the edges of a dodecahedron or icosahedron. Find coordinates for the
vertices of an icosidodecahedron.
10. It is evident that a rhombicuboctahedron may be inscribed in a cube.
A view from the top (as above the cap shown in Figure 6.7) looks like
Figure 14.12.
Now assume the sides of the central square in Figure 14.12 have length
2, so that the vertices of this “top” square are (−1, −1, z), (−1, 1, z),
(1, 1, z), and (1, −1, z), where z is half the length of the edges of the
circumscribing cube. Observing that the length of the edge of the
circumscribing cube is also the width of the octagon in Figure 14.12,
find coordinates for the vertices of a rhombicuboctahedron.
14.7. EXERCISES
225
Figure 14.12
11. Thinking along the lines of the previous exercise, use Figure 6.10 to
determine coordinates for the vertices of a rhombitruncated cuboctahedron.
226
CHAPTER 14. COORDINATES OF POLYHEDRA
Chapter 15
Matrices and Symmetry
Groups
15.1
Introduction
As suggested in the previous chapter, efficient rendering of polyhedral images
requires more than a knowledge of coordinates for the vertices of polyhedra.
Familiarity with matrices and symmetry groups is extremely useful.
Matrices are typically dicussed in an undergraduate course in linear algebra, while symmetry groups are usually presented in an abstract algebra
course. Only the barest introduction to these topics can be given in a single
chapter. The interested reader should be cautioned, though, that many textbooks in linear algebra approach the topic of matrices from a purely algebraic
perspective. In such texts, motivation for studying matrices is the solution
of simultaneous linear equations in several variables. The reader should look
for a text, such as Banchoff and Wermer’s Linear Algebra Through Geometry
(ISBN 0-387-97586-1), which considers matrices from a geometric viewpoint.
15.2
Definitions and Notations
Let R3 denote the set of all possible coordinate lists consisting of three real
numbers. Recall that in the last chapter, care was taken to distinguish a
point from its coordinates. Indeed, two coordinate systems may be in use
at the same time, such as when considering an equation for a plane curve in
both Cartesian and polar coordinates.
In this discussion, the only coordinate system under consideration is the
Cartesian coordinate system in three dimensions. So rather than say that
227
228
CHAPTER 15. MATRICES AND SYMMETRY GROUPS
the point p has coordinates (1, −1, −1), we may simply refer to the point
(1, −1, −1).
Before discussing matrices in particular, we must look at functions on
R3 in general. For example, consider the function A : R3 → R3 defined by
A((x, y, z)) = (x, y, 0).
This function takes a point and changes its z-coordinate to 0. Although
defined algebraically, this function also has a geometric interpretation. A
projects a point onto the xy-plane. In other words, if a point p is moved
towards the xy-plane along a path parallel to the z-axis, it would eventually
intersect the xy-plane at the point A(p).
Many other functions on R3 are possible, such as:
B((x, y, z)) = (x2 , y 2 , z 2 ),
(15.1)
C((x, y, z)) = (x + y, y − z, x),
(15.2)
D((x, y, z)) = (0, 0, 0),
(15.3)
y
E((x, y, z)) = (cos x, sin z, e ).
(15.4)
In a moment, we will look at those functions which may be represented
by a matrix.
A linear combination of the variables x, y, and z is an expression of
the form
ax + by + cz;
that is, the variables are simply multiplied by numbers and then added
together. Some of the numbers may be negative or zero, so that
y−z
is also a linear combination of x, y, and z.
However,
x2 + 4y − πz
is not, since the x is squared. The variables may only be multiplied by
numbers, and may not appear as arguments to any other function.
We say that a function F is linear if given a point p, the coordinates
of F(p) are linear combinations of the coordinates of p. Thus, the functions
A, C, and D described earlier are linear, while B and E are not.
Note that 0 = 0x + 0y + 0z is a linear combination of x, y, and z, but
that 1 is not. For no choice of real numbers a, b, and c does the equation
ax + by + cz = 1
15.2. DEFINITIONS AND NOTATIONS
229
hold for all values of x, y, and z.
A linear function F on R3 is often represented by a 3 × 3 matrix whose
first row consists of the coefficients of x, y, and z, respectively, in the first
coordinate of F((x, y, z)), whose second row consists of the coefficients of x,
y, and z, respectively, in the second coordinate of F((x, y, z)), and whose
third row is likewise defined. With this notation in mind, the function C
described above may be represented as


1 1 0




C = 0 1 −1 .


1 0 0
When using matrix notation, it is
point in a column, as in
  
x
1 1
  
  
C y  = 0 1
  
z
1 0
customary to write the coordinates of a
  

x
x+y
  

  

−1 y  =  y − z  .
  

0
z
x
0
Recall that this is a special case of an operation often referred to as matrix
multiplication. In the general case, we have

  

a11 a12 a13
x
a11 x + a12 y + a13 z

  


  

=
(15.5)
a21 a22 a23  y  a21 x + a22 y + a23 z  .

  

a31 a32 a33
z
a31 x + a32 y + a33 z
The linear functions A and

1


A = 0

0
D above are represented by the matrices



0 0
0 0 0






1 0 , D = 0 0 0 .



0 0
0 0 0
The matrix I defined by


1 0 0




I = 0 1 0


0 0 1
230
CHAPTER 15. MATRICES AND SYMMETRY GROUPS
is called the identity matrix; it has the property that for any point p,
I(p) = p.
The purist may remark that a matrix simply represents a function and
is not a function itself. As it turns out, if the underlying coordinate system
changes, the matrix representing a linear function also changes. But since
we will not be deviating from the use of our three-dimensional Cartesian
coordinate system, we will not need to notationally distinguish between a
linear function and its matrix.
15.3
Matrices and Symmetry
Certain types of matrices are especially
We begin with an example.
Consider the matrix

0 1


R = 0 0

1 0
relevant to a study of polyhedra.

0


1 .

0
Now consider the effect of R on the vertices of the cube described in §14.2:

   
   
   
0 1 0
1
−1
−1
1
−1
−1

   
   
   

   
   
   
0 0 1 −1 = −1 , R  1  = −1 , R −1 =  1  .

   
   
   
1 0 0
−1
1
−1
−1
1
−1
By examining Figure 14.3, it is apparent that R is a clockwise rotation
through 120◦ about an axis through the center of the cube and the vertex
(1, 1, 1). Since the axis also passes through (−1, −1, −1), the vertices (1, 1, 1)
and (−1, −1, −1) are not moved by R:
   
   
1
1
−1
−1
   
   
   
   
R 1 = 1 , R −1 = −1 .
   
   
1
1
−1
−1
We also see that
   
−1
1
   
   
R 1  =  1 ,
   
1
−1
 
−1
   
   
R −1 =  1  ,
   
1
1

1


1


1

   
   
R  1  = −1 .
   
−1
1
15.3. MATRICES AND SYMMETRY
231
Thus R takes each vertex of the cube into another, not necessarily distinct,
vertex of the cube.
The rotation R is said to be a symmetry of the cube in that if we begin
with the cube as described in §14.2 and apply the linear function R, the
resulting cube is indistinguishable from the original in that it occupies the
same region of space.
Another way to say this is that if we begin with a list of the eight vertices of the cube and apply R to each one, we get the same eight vertices,
although usually in a different order. Thus, R is said to be a permutation
of the vertices of the cube. It turns out that every symmetry of the cube
produces a corresponding permutation of its vertices. However, not every
permutation of the cube’s vertices yields a symmetry of the cube; for example, a permutation which merely switches the ends of one edge of the cube
and leaves the others intact does not yield a symmetry of the cube. For a
thorough discussion of the theory of permutations, consult a textbook on
abstract algebra.
Another symmetry of the cube is given by


0 −1 0




S = 1 0 0 .


0 0 1
By examining the effect of S on the vertices of the cube, we see that S
corresponds to a 90◦ counterclockwise rotation around the x-axis (as viewed
from the positive z-axis).
The reader might pause for a few moments and try to discover some other
symmetries of the cube and their corresponding matrices. For the impatient
reader, we will answer the following question: how many symmetries of the
cube are there?
Let us begin by counting symmetries such as R described above; that is, a
rotation around an axis which contains a diagonal of the cube. If we imagine
holding the opposite vertices of a diagonal of a cube with our thumb and
middle finger, we may physically rotate the cube in the manner described
by R. Performing this rotation again, we obtain a 240◦ rotation about the
axis, or the symmetry R ◦ R = R2 . (It is customary, when working with
matrices, to denote repeated composition by the appropriate exponent and
call the composition of matrices matrix multiplication.) If we perform
this rotation again, a 360◦ rotation about the axis is obtained; but this does
not in fact move the cube at all, since every vertex ends up exactly where
232
CHAPTER 15. MATRICES AND SYMMETRY GROUPS
it started. This may be described by the relationship
R ◦ R ◦ R = R3 = I.
Thus, each diagonal of the cube generates two symmetries of the cube. Since
there are four diagonals, we have found eight symmetries of the cube so far.
Technically, the identity, I, which does not move the cube at all, is
considered a symmetry of the cube. We will count it, however, only at the
end. If we counted it as being generated by a rotation about an axis, it
would be overcounted several times.
We may realize the rotation S described above by holding a cube so
that our thumb and middle finger touch the centers of two opposite faces
of the cube. If S describes a 90◦ counterclockwise rotation about the axis
passing through these facial centers, S ◦ S = S2 describes a 180◦ rotation
about this axis, and S3 described a 270◦ rotation. Of course S4 = I, since a
360◦ rotation is obtained. As we are not counting I, three rotations are thus
produced. Since three rotations are produced by each of the three pairs of
opposite faces, nine more rotations bring the count to 17.
By holding a cube by the midpoints of a pair of opposite edges, a new
rotation is generated. As this must be a 180◦ rotation, only one rotation
is generated by each of the six pairs of opposite edges. These six rotations
bring the count to 23.
Then, as promised earlier, we count the identity, I, bringing the count
up to 24.
15.4
Finding a Matrix
Now that we have enumerated the 24 rotations of the cube, the question
naturally arises: since the rotation R in §15.3 was represented by a matrix,
can we find matrix representations for all rotations of the cube?
Of course the reader must, by now, know the answer to such a leading
question. To explain the technique, we consider an example.
Let M be the matrix


3 −2




M =  9 5 6 .


−8 7 12
4
15.4. FINDING A MATRIX
233
Using (15.5), the reader may easily verify that
   
4
1
   
   
M 0 =  9  ,
   
−8
0
   
3
0
   
   
M 1 = 5 ,
   
7
0
   
−2
0
   
   
M 0 =  6  .
   
12
1
But the results of these operations are just the columns of M!
So if we reverse this procedure, we arrive at the following: if R is a
rotation (or any other linear function), the three columns of the matrix
reprentations for R are given by
 
1
 
 
R 0 ,
 
0
 
0
 
 
R 1 ,
 
0
 
0
 
 
R 0 .
 
1
Let us see how this works with an example. Consider the clockwise 90◦
rotation about the x-axis. That is, your viewpoint is from the positive xaxis, and you’re turning the cube a one-quarter rotation to your right. If we
call this rotation R, what is the matrix for R?
We first examine R((1, 0, 0)). Now (1, 0, 0) is just the center of the front
face of the cube. When we rotate the cube, this point does not move as it
is on the axis of rotation. Thus,
   
1
1
   
   
R 0 = 0 .
   
0
0
What happens to (0, 1, 0)? This point is the center of the right face
(looking from our viewpoint). When we rotate, this point is moved to the
center of the bottom face of the cube. Thus,
   
0
0
   
   
R 1 =  0  .
   
−1
0
234
CHAPTER 15. MATRICES AND SYMMETRY GROUPS
Finally, we look at (0, 0, 1), the center of the top face of the cube. After
rotation, this point becomes the center of the right face of the cube, so that
   
0
0
   
   
R 0 = 1 .
   
0
1
Combining our observations, we see that the matrix for R is given by


1 0 0




R = 0 0 1 .


0 −1 0
(15.6)
Thus, a matrix representation may be found for any symmetry of the
cube in the manner just described. We’ll continue working on this task in
the next few sections.
15.5
Reflections
Not all symmetries of the cube are rotations. Consider for a moment the
xy-plane as a mirror in which we may reflect the cube. In other words, we
reflect the top face to the bottom, and vice versa, through the plane z = 0.
This is easy to describe algebraically as well, for it becomes a simple matter
of changing the sign of the z-coordinate. The plane z = 0 is called a plane
of symmetry because it divides the cube into two identical pieces, each a
mirror image of the other. Using (15.5), it is easy to see that the matrix
which changes the sign of the z-coordinate is given by


1 0 0




Sz = 0 1 0  .


0 0 −1
(15.7)
Of course, the xz-plane and the yz-plane are also planes of symmetry for
analogous reasons. Algebraically, these change the signs of the y-coordinate
15.5. REFLECTIONS
235
and x-coordinate, respectively, and have matrix representations


1 0 0




Sy = 0 −1 0 ,


0 0 1

−1 0 0




Sx =  0 1 0 ,


0 0 1

respectively.
Does the cube have other planes of symmetry? (The reader may wish
to pause here and consider the question before going further.) Imagine a
plane passing through a pair of opposite edges of the cube. Such a plane
also divides the cube into two identical pieces, each the mirror image of the
other. Since there are twelve edges on the cube in six opposite pairs, this
line of thought provides six more planes of symmetry of the cube, and thus
six additional reflections.
So the cube has nine planes of symmetry, each generating a reflection.
But there are other ways to generate reflections of the cube. For example,
select any of the 24 rotations described earlier, and consider the symmetry
Sz R = Sz ◦ R.
Recall the convention of writing matrix composition as “multiplication,” and
also recall that for any point p,
Sz ◦ R(p) = Sz (R(p)),
so that function application is read “right-to-left.” This symmetry first
rotates the cube as described by R, and then reflects it in the xy-plane as
described by Sz . Thus Sz R leaves the cube occupying the same region of
space as it originally did, and hence is a symmetry of the cube. Since each
rotation R generates a different reflection Sz R, the 24 reflections of the cube
may be produced in this way.
To find the matrix representation for a reflection, recall that it is sufficient to determine the action of the reflection on the points (1, 0, 0), (0, 1, 0),
and (0, 0, 1), and use the results as the columns for its matrix. Another way
to produce the matrix is by using a generalization of (15.5), which is a
236
CHAPTER 15. MATRICES AND SYMMETRY GROUPS
formula for matrix multiplication:



x1 x2 x3
a11 a12 a13






a21 a22 a23   y1 y2 y3  =



z 1 z2 z3
a31 a32 a33

a11 x1 + a12 y1 + a13 z1 a11 x2 + a12 y2 + a13 z2 a11 x3 + a12 y3 + a13 z3





a21 x1 + a22 y1 + a23 z1 a21 x2 + a22 y2 + a23 z2 a21 x3 + a22 y3 + a23 z3  .


a31 x1 + a32 y1 + a33 z1 a31 x2 + a32 y2 + a33 z2 a31 x3 + a32 y3 + a33 z3
(15.8)
It is readily seen that to multiply matrices M1 and M2 , take the columns of
M2 , find their images under M1 using (15.5), and use the results as columns
for M1 M2 . The reader should check that if R is given by (15.6) and Sz is
given by (15.7), the matrix Sz R is given by


 

1 0 0
1 0 0
1 0 0


 



 

Sz R = 0 1 0  0 0 1 = 0 0 1 .


 

0 0 −1
0 −1 0
0 1 0
15.6
Direct and Opposite Symmetries
To date, we have generated 24 rotations of the cube along with 24 reflections
of the cube. The set of rotations is sometimes called the rotation group of
the cube or the octahedral group (since it is also the rotation group for the
octahedron), while the set including the rotations and reflections is called
the symmetry group of the cube. The term “group” here means more
than simply a collection of objects; the reader interested in group theory
should consult an abstract algebra text.
Rotations are called direct symmtries, while reflections are called opposite symmetries. We just learned how to generate them, but now ask a
more subtle question: given a symmetry of cube, such as


0 −1 0




S = 0 0 1 ,


1 0 0
15.6. DIRECT AND OPPOSITE SYMMETRIES
237
how can we determine if it is direct or opposite?
One way is to puzzle out the geometry of the symmetry by looking at its
effect on the points (1, 0, 0), (0, 1, 0), and (0, 0, 1). There is also an algebraic
method, taken from linear algebra, involving the determinant of a matrix,
defined by


a11 a12 a13




det a21 a22 a23  = a11 a22 a33 + a12 a23 a31 + a13 a21 a32


a31 a32 a33
− a13 a22 a31 − a11 a23 a32 − a12 a21 a33 . (15.9)
At first glance, this formula looks rather strange until we remember we are
isolating one definition from a broad mathematical topic (linear algebra) as
an aid in performing a certain calculation. The interested reader should
consult a relevant textbook.
There are many ways to calculate the determinant, each yielding the
same result, of course. One popular mnemonic device for calculating determinante of 3 × 3 matrices is as follows. First, copy the first two columns of
the matrix to its right, as in Figure 15.1. Take the products of terms along
the “up” arrows, add them, then subtract the products of terms along the
“down” arrows.
a11 a12 a13 a11 a12
b11 b12 b13 b11 b12
c11 c12 c13 c11 c12
Figure 15.1
Try this with the symmetry S given above. It turns out that all products
are 0 except one, making the calculation det S = −1 somewhat easier. And
now the reason for introducing the determinant in the first place: if S is a
rotation, then det S = 1, while if S is a reflection, det S = −1.
Another nice property of the determinant is that it is multiplicative. In
other words, if M1 and M2 are two matrices, then
det(M1 M2 ) = (det M1 )(det M2 ).
(15.10)
One consequence of this result is that the product of two opposite symmetries is direct. (It would be instructive to take a moment to try and see
238
CHAPTER 15. MATRICES AND SYMMETRY GROUPS
this now.) To see this, let S1 and S2 be two reflections; that is, opposite
symmetries. Then using (15.10),
det(S1 S2 ) = (det S1 )(det S2 ) = (−1) · (−1) = 1.
But since det(S1 S2 ) = 1, then S1 S2 must be a rotation (a direct symmetry).
Of course this presupposes that if S1 and S2 are reflections, then S1 S2
is a symmetry of the cube. But the product of any two symmetries is
always another symmetry; this is one reason the symmetries form a group.
If a symmetry leaves a cube occupying the same region of space, then any
product of symmetries also leaves a cube occupying the same region of space,
and hence is also a symmetry of the cube.
15.7
A Final Look
As the reader has undoubtedly surmised, there is much more that can be
said about symmetries of the cube. We content ourselves with a summary
description of the matrices representing symmetries of the cube.
Note that the matrices considered thus far have the property that in
every row and column there are two zeroes, along with a +1 or −1. This is
true of all symmtries of the cube. Considering matrices exclusively with +1
terms, we get the following six, where the first three are direct symmetries
and the latter three are opposite symmetries.

 
 
 
 
 

1 0 0
0 1 0
0 0 1
1 0 0
0 0 1
0 1 0

 
 
 
 
 


 
 
 
 
 

0 1 0 , 0 0 1 , 1 0 0 , 0 0 1 , 0 1 0 , 1 0 0 .

 
 
 
 
 

0 0 1
1 0 0
0 1 0
0 1 0
1 0 0
0 0 1
Now we consider changing some of the +1 terms to −1. Take the example of the first matrix listed above, the identity matrix. With three nonzero
positions, there are 23 possible ways (including I itself) to fill these positions with either +1 or −1. Similarly, there are 8 symmetries of the cube
associated with each of the other five matrices obtained by taking all possible changes of signs of the 1’s in those matrices. Thus, the 8 × 6 = 48
symmetries of the cube are produced.
This enumeration is nice because it provides a purely algebraic way of
describing the symmetries of the cube. But this is just what the budding
computer graphicist needs, in addition to a working knowledge of coordinates for the vertices of polyhedra, in order to enter the sublime world of
polyhedral graphics.
15.8. THE DIRECT SYMMETRIES
15.8
239
The Direct Symmetries
For reference, a complete list of the 24 direct symmetries is given below.
Multiplying each of these symmetries by any given reflection will produce
the 24 opposite symmetries.
Let R1,1,1 be that symmetry which rotates the cube 120◦ clockwise
around the axis through (1, 1, 1) and (−1, −1, −1). Let R1,y,1 = R−1,y,−1
be that symmetry whose axis of rotation joins the midpoints of (1, 1, 1) and
(1, −1, 1), and (−1, 1, −1) and (−1, −1, −1). Then the 24 direct symmetries
are:

1. R1,1,1

0 1 0




= 0 0 1: Clockwise 120◦ rotation about the axis.


1 0 0

2. R21,1,1

0 0 1




= 1 0 0: Counterclockwise 120◦ rotation about the axis.


0 1 0
3. R1,1,−1


0 0 −1




= 1 0
0 : Clockwise 120◦ rotation about the axis.


0 −1 0

0
1
0





4. R21,1,−1 =  0 0 −1: Counterclockwise 120◦ rotation about the


−1 0 0
axis.

1




= −1 0 0: Clockwise 120◦ rotation about the axis.


0 −1 0

5. R1,−1,1
0
0
240
CHAPTER 15. MATRICES AND SYMMETRY GROUPS

0 −1


6. R21,−1,1 = 0

1
axis.
0



−1: Counterclockwise 120◦ rotation about the

0
0
0

0 0 −1




= −1 0 0 : Clockwise 120◦ rotation about the axis.


0 1 0

7. R−1,1,1

−1 0
0


8. R2−1,1,1 =  0

−1
axis.
0
0



1: Counterclockwise 120◦ rotation about the

0


1 0 0




9. Rx = 0 0 1: Clockwise 90◦ rotation about the x-axis.


0 −1 0

1
0
0





10. R2x = 0 −1 0 : 180◦ rotation about the x-axis.


0 0 −1

1 0
0





11. R3x = 0 0 −1: Counterclockwise 90◦ rotation about the x-axis.


0 1 0

0 0 −1


12. Ry = 0 1

1 0



0 : Clockwise 90◦ rotation about the y-axis.

0
15.8. THE DIRECT SYMMETRIES
13.
14.
15.
16.
17.
18.
19.
20.
241


−1 0 0




R2y =  0 1 0 : 180◦ rotation about the y-axis.


0 0 −1


0 0 1




R3y =  0 1 0: Counterclockwise 90◦ rotation about the y-axis.


−1 0 0


0 1 0




Rz = −1 0 0: Clockwise 90◦ rotation about the z-axis.


0 0 1


−1 0 0




2
Rz =  0 −1 0: 180◦ rotation about the z-axis.


0
0 1


0 −1 0




R3z = 1 0 0: Counterclockwise 90◦ rotation about the z-axis.


0 0 1


−1 0 0




Rx,1,1 =  0 0 1: 180◦ rotation about the axis.


0 1 0


−1 0
0




Rx,1,−1 =  0
0 −1: 180◦ rotation about the axis.


0 −1 0


0 0 1




R1,y,1 = 0 −1 0: 180◦ rotation about the axis.


1 0 0
242
CHAPTER 15. MATRICES AND SYMMETRY GROUPS

0
−1
0


21. R1,y,−1 =  0 −1

−1 0

0 1
0



0 : 180◦ rotation about the axis.

0





22. R1,1,z = 1 0 0 : 180◦ rotation about the axis.


0 0 −1

0
−1


23. R1,−1,z = −1

0
0
0
0



0 : 180◦ rotation about the axis.

−1


1 0 0




24. I = 0 1 0: The identity.


0 0 1
15.9
Exercises
1. Decide which of the following functions from R3 to R3 are linear. For
those which are linear, give a matrix representing the function.
(a) A((x, y, z)) = (1, 0, 0),
(b) B((x, y, z)) = (y, z, x),
(c) C((x, y, z)) = (x2 , y 2 , z 2 ),
(d) D((x, y, z)) = (3x − y, x + y − z, z − y),
(e) E((x, y, z)) = (x − 1, y + z, z − 3),
(f) F((x, y, z)) = (−y, −x, −z).
2. Of the functions described in the previous exercise, which are symmetries of the cube? Rotations? Reflections?
15.9. EXERCISES
243
3. Perform the following operations involving matrices:
(a)
 
3
3 2 −1
 

 

 1 0 1  2 ,
 

6
−1 4 5

(b)

 
5

 

 
 3 y 7 1 ,

 
−1 0 2
0
x
0 1
(c)

1


0

−1
−1
3
1


3




1  0 0 5 ,


−2
1 −3 2
2
1
0
(d)



0 −1
0 0 0






−1 0 0  4 −9 3 .



0 0 8
0 0 0
1
4. Let M1 and M2 be given as follows:




−2 1 0
3 4
0








M1 =  0 1 1  , M2 = 1 0 −1 .




1 0 −1
1 −2 0
Find each of the following:
(a) M1 ((0, 0, 0)),
(b) M2 ((0, 1, 0)),
(c) M1 ((3, −1, 2)),
(d) M2 ((4, −5, 6)),
244
CHAPTER 15. MATRICES AND SYMMETRY GROUPS
(e) M1 M2 ,
(f) M2 M1 ,
(g) det M1 ,
(h) det M2 ,
(i) det M1 M2 ,
(j) det M2 M1 .
5. Describe the effects of the following symmetries on the cube in Figure
14.3.
(a)


1 0
0




0 −1 0  ,


0 0 −1
(b)


−1




 0 −1 0  ,


−1 0
0
0
0
(c)


−1 0 0




 0 0 1 ,


0 1 0
(d)


0 1 0




−1 0 0 .


0 0 1
6. Recall that the octahedron and cube are dual polyhedra (see Chapter
9). With this in mind, write a convincing argument that the symmetry
group of the octahedron is the same as the symmetry group of the cube.
7. Find all symmetries of the cube in Figure 14.3 with the property that
the center of the front face is taken to the center of the top face.
15.9. EXERCISES
245
8. Find all symmetries of the cube in Figure 14.3 with the property that
the vertex (1, 1, 1) is taken to the vertex (1, −1, −1).
9. Find all symmetries of the cube in Figure 14.3 with the property that
the vertex (1, 1, 1) remains fixed and the vertex (−1, 1, 1) is taken to
the vertex (1, 1, −1).
10. A symmetry T of the cube interchanges the positive x-axis and positive
y-axis, and also satisfies T((1, 1, 1)) = (1, 1, −1). Find a matrix for T.
11. It M is a matrix, there may be a matrix N with the property
MN = NM = I.
In this case, N is called the inverse of M and is denoted by M−1 .
The matrix Sz in (15.7) satisfies
Sz ◦ Sz = I,
so Sz is its own inverse; i.e., S−1
= Sz . This makes sense, since
z
applying Sz twice undoes the effect of Sz : Sz reflects about a plane,
then applying Sz again reflects back.
(a) Find the inverse of the matrix R given in (15.6).
(b) Find the inverse of the matrix R given by


0 0 −1




R = 0 1 0  .


1 0 0
12. Consider the tetrahedron with vertices (1, 1, 1), (1, −1, −1), (−1, 1, −1),
and (−1, −1, 1) which is inscribed in the cube in Figure 14.3. A moment’s thought reveals that any symmetry of the tetrahedron must be
a symmetry of the cube. However, a symmetry of the cube need not
be a symmetry of the tetrahedron (for example, a 90◦ rotation about
the center of a square face takes the tetrahedron into the other regular
tetrahedron inscribable in the cube).
In the language of abstract algebra, the symmetries of the tetrahedron
form a subgroup of the symmetries of the cube. Find the symmetries
of the tetrahedron. Keeping in mind the discussion in the previous
paragraph, give a reason for the number of symmetries of the tetrahedron. How many are direct? How many are opposite?
246
CHAPTER 15. MATRICES AND SYMMETRY GROUPS
Chapter 16
Graph Theory and Polyhedra
16.1
Introduction and Motivation
If asked to draw a cube, how might you do it? You might produce one of
the drawings shown in Figure 16.1
(a)
(b)
Figure 16.1
Figure 16.1(a) is a typical perspective drawing, while Figure 16.1(b) is
drawn “face on.” The former has the advantage that it looks somewhat
like a real cube, while the latter has the feature that no line of the drawing
crosses any of the others.
Yet we recognize both as cubes. Each includes the eight vertices of a
cube as well as the twelve edges. Moreover, three edges are incident at each
vertex in the drawings.
247
248
CHAPTER 16. GRAPH THEORY AND POLYHEDRA
These properties of the drawings of a cube are not sufficient, however,
to describe the adjacency of the vertices of a cube. Each of the graphs in
Figure 16.2 has eight vertices and twelve edges, with three edges meeting
at each vertex. Yet one can be “untangled” so that the graph of a cube is
obtained, and one cannot. Can the reader decide which one yields a cube?
6
5
6
5
7
4
7
4
8
3
8
3
1
2
1
(a)
2
(b)
Figure 16.2
This suggests the following question: besides having eight vertices and
twelve edges, with three edges incident at each vertex, what other properties
must a graph possess so that it corresponds exactly to the adjacency of
vertices on a cube?
Such questions belong to a fascinating branch of mathematics known as
graph theory. While some mathematicians devote their entire professional
lives to its study, we can provide only the barest introduction here.
In studying polyhedra from a graph-theoretical perspective, we restrict
our attention to one particular aspect of polyhedra; namely, the adjacency of
their vertices. Features such as the size and shape of faces are not relevant.
For example, the graph of vertex adjacency for a rectangular prism is
also given in Figure 16.1. Since vertex adjacency is the only relevant feature under discussion, edge lengths on the polyhedron are irrelevant. As a
result, Figure 16.1 represents the vertex adjacency for a parallelepiped or
the frustum of a square pyramid.
Like any other branch of mathematics, graph theory has its own “language” of definitions and concepts which, over time, have been relevant to its
particular application. Before going further, we must significantly enlarge
our graph-theoretical vocabulary.
16.2. BASIC DEFINITIONS
16.2
249
Basic Definitions
A graph G is essentially a drawing of points, or vertices, connected by
lines, or edges. Usually a carefully drawn picture is sufficient to represent
a graph. But as seen in Figure 16.1, there may be more than one way to
“draw” what is essentially the same graph. To avoid such bias, it is possible
to define a graph abstractly without a drawing by specifying a set V(G) of
its vertices, a set E(G) of its edges, and a function ψ, called the incidence
function, which specifies the ends of any particular edge. We denote the
number of vertices and edges on a graph by V and E, respectively, to be
consistent with the notations introduced in Chapter 2.
For example, consider a graph with V = 6 and E = 12 given by
V(G) = {1, 2, 3, 4, 5, 6},
E(G) = {e1 , e2 , e3 , ..., e12 }
ψ(e1 ) = 12,
ψ(e2 ) = 23,
ψ(e3 ) = 31,
ψ(e4 ) = 45,
ψ(e5 ) = 56,
ψ(e6 ) = 64,
ψ(e7 ) = 14,
ψ(e8 ) = 42,
ψ(e10 ) = 53,
ψ(e11 ) = 36,
ψ(e12 ) = 61.
ψ(e9 ) = 25,
There are many ways to draw this graph; three are given below.
4
2
1
1
2
3
4
5
6
5
4
6
3
(a)
(c)
1
2
3
6
5
(b)
Figure 16.3
Is there a reason to prefer one of the drawings to the others? To emphasize that this graph is planar; that is, it can be drawn on a sheet of paper
(a plane) without any of its edges crossing, one might draw the graph as in
250
CHAPTER 16. GRAPH THEORY AND POLYHEDRA
Figure 16.3(b). To emphasize that this graph is regular; that is, the same
number of edges are incident at each vertex, one might draw the graph as
in Figure 16.3(a).
The point to be made is that a drawing of a graph is only a convenient
representation. When there is ambiguity, or perhaps when necessary for
proving a complex theorem, a graph can be specified in terms of V(G),
E(G), and ψ as described above. For our purposes, it will usually suffice to
just give a drawing of a graph.
Note that the definition of a graph does not preclude two vertices being
joined by more than one edge, as in Figure 16.4.
v1
v2
e
v3
v4
Figure 16.4
Here, two distinct edges join vertices v1 and v3 . Also, an edge may begin and
end at the same vertex, such as edge e in Figure 16.4. Such an edge is called
a loop. Graphs with these features do not usually arise when representing
adjacency of vertices on a polyhedron. When a graph has no loops and
no two vertices are joined by more than one edge, the graph is said to be
simple. Most of our discussion will focus on simple graphs.
The degree of a vertex v ∈ V(G), denoted by d(v), is the number of
edge ends incident at v. In the graph shown in Figure 16.4, d(v1 ) = 3, and
d(v4 ) = 3 as well since the loop e contributes two ends to v4 .
A basic theorem results from counting edge ends. On the one hand, we
may create the sum
X
d(v),
v∈V(G)
which indicates we sum the degrees of all the vertices of G. But since each
16.2. BASIC DEFINITIONS
251
edge contributes two ends to this sum, the result must be 2E, so that
X
d(v) = 2E
(16.1)
v∈V(G)
for any graph G.
It may happen, as does rather frequently with graphs derived from polyhedra, that every vertex has the same degree. Graphs with this property
are called regular graphs. If q is the common degree of the vertices of a
regular graph, the graph is said to be q–regular. In this case, d(v) = q for
all v ∈ V(G), so that (16.1) becomes
qV = 2E.
(16.2)
Compare this equation with (P30 ) of §2.3.
A complete graph is a simple graph where each vertex is connected to
every other. A complete graph with n vertices is denoted by Kn (see Figure
16.5). Note that Kn is (n − 1)–regular.
(a)
(b)
(c)
Figure 16.5
A bipartite graph is a graph whose vertices may be partitioned into
two sets, V1 (G) and V2 (G), such that each edge has one end in V1 (G) and
the other in V2 (G). Such a graph is usually drawn so that the partition of
vertices is clear, as in Figure 16.6.
Figure 16.6
252
CHAPTER 16. GRAPH THEORY AND POLYHEDRA
While this may seem like a lot of new definitions and concepts, they are
all relevant to graphs derived from polyhedra. In the next section, we’ll see
how they apply to graphs derived from the Platonic solids.
16.3
The Platonic Graphs
—The Tetrahedron
As promised, we now undertake a survey of the graphs of the vertex
adjacencies of the Platonic solids. The simplest is the tetrahedron, the only
Platonic solid with the property that each vertex is adjacent to each of the
other vertices. But this means that its graph is just the complete graph on
four vertices, shown in Figure 16.5(a).
—The Cube and Octahedron
As we have seen earlier, the graph of a cube is a 3–regular simple graph
with eight vertices. However, we have also seen (see Figure 16.2) that not
every 3–regular, simple graph with eight vertices is the graph of a cube. So
is there a way to distinguish which such graph corresponds to a cube?
An affirmative answer to this question lies with the observation that the
graph of a cube is bipartite. We may see this in several ways. First, examine
the graph in Figure 16.1(b) and find V1 (G) and V2 (G) as required. (The
reader is encouraged to do so now.) Second, as suggested by Exercise 6(b) of
Chapter 2, a regular tetrahedron may be inscribed in a cube in two ways so
that the edges of the tetrahedron are diagonals of the faces of the cube. One
then observes that the ends of each edge of the cube are vertices of different
tetrahedra, so that the required partition of vertices may be produced by
taking the vertices of the two different tetrahedra.
Finally, we may look at the coordinates for the vertices of a cube as
desribed in §14.2. Call a vertex even if the number of coordinates which
are −1 is even, such as (−1, 1, −1) or (1, 1, 1). Analogously, call a vertex
odd if the number of coordinates which are −1 is odd, such as (1, 1, −1) or
(−1, −1, −1). Then observe that each edge of the cube joins an even vertex
and an odd vertex. Thus, the set of even vertices and the set of odd vertices
form the required partition of vertices.
This is now enough. If a graph with eight vertices is simple, 3–regular,
and bipartite, then it is in fact a graph of the vertex adjacency of a cube.
In essence, then, there is only one such graph, unique up to a relabelling of
vertices and edges.
A proof of this assertion follows. It is included as it is not difficult
16.3. THE PLATONIC GRAPHS
253
and illustrates how many of the concepts defined earlier may be used in
combination.
Theorem: There is essentially one simple, 3–regular, bipartite graph with eight vertices. It is the graph describing the
vertex adjacency of the cube.
To see this, note that the graph must have 12 edges. This follows from
X
2E =
d(v) = 8 · 3,
v∈V(G)
since each of the 8 vertices has degree 3.
Since the graph is 3–regular, bipartite, and has 12 edges, the vertices
must be partitioned into two groups of four so that a total of 12 = 4 · 3
edges end in each partition of vertices. Thus our graph looks like Figure
16.7, where one vertex is arbitrarily joined to three of the others.
1
2
3
4
5
6
7
8
Figure 16.7
Due to the symmetry of the graph drawn thus far, there are two alternatives for vertex 6: either join 6 to vertices 1, 2, and 3, or join 6 to vertex
4 and two of the vertices 1, 2, and 3. These alternatives are drawn in Figure
16.8.
1
2
3
4
1
2
3
4
5
6
7
8
5
6
7
8
(a)
(b)
Figure 16.8
254
CHAPTER 16. GRAPH THEORY AND POLYHEDRA
We now observe that it is impossible for Figure 16.8(a) to occur. For
since the graph is 3–regular and bipartite, vertex 4 must be joined to three
of the vertices 5, 6, 7, and 8. Since the graph is simple, these three vertices
must all be different. But vertices 5 and 6 are already ends of three edges,
leaving only two vertices, 7 and 8, available. Thus, there is no way to
complete the graph drawn in Figure 16.8(a).
Now since vertices 5 and 6 are already ends of three edges, vertices 3
and 4 must each be joined to vertices 7 and 8. This leaves just two ways to
consistently complete the graph in Figure 16.8(b): either join vertices 1 and
7, and 2 and 8, or join vertices 1 and 8, and 2 and 7. Both alternatives are
shown in Figure 16.9.
1
2
3
4
1
2
3
4
5
6
7
8
5
6
7
8
(a)
(b)
Figure 16.9
Now redraw Figure 16.9(b) by switching the locations of vertices 7 and
8, but maintaining adjacencies of the vertices. The result is Figure 16.10.
1
2
3
4
5
6
8
7
Figure 16.10
But Figure 16.10 looks exactly like Figure 16.9(a)! This means that
Figures 16.9(a) and (b) are essentially the same graph, since one may be
obtained from the other by simply relabelling some of the vertices (in this
case, switching vertices 7 and 8). Such graphs are said to be isomorphic.
Moreover, by redrawing Figure 16.9(a); that is, by putting its vertices at
different locations on paper while maintaining vertex adjacency, the familiar
Figure 16.11 is obtained.
16.3. THE PLATONIC GRAPHS
255
6
2
4
8
7
3
1
5
Figure 16.11
This demonstrates that if we adhere to the conditions that our graph
is simple, 3–regular, bipartite, and has eight vertices, we must eventually
create a graph such as Figure 16.11, a representation of the cube. In other
words, the Theorem is proved.
–The Icosahedron and Dodecahedron
A graph of the vertex adjacency of the icosahedron must have twelve
vertices, thirty edges, and be 5–regular since five edges meet at each vertex
of the icosahedron. There are many graphs with these properties, so there
is no simple theorem analogous to our recent result about the cube. This
graph is shown in Figure 16.12.
Figure 16.12
256
CHAPTER 16. GRAPH THEORY AND POLYHEDRA
Likewise, there is no simple theorem characterizing the graph of vertex
adjacency of the dodecahedron. It must, however, have 20 vertices, 30 edges,
and be 3–regular. This graph is shown in Figure 16.13.
Figure 16.13
16.4
Exercises
1. Show that there is no 3–regular graph with seven vertices.
2. Decide for which positive integers q and V it is possible to create a
q–regular graph on V vertices. It may be helpful to think in terms of
an algorithm for creating such a graph.
When is it possible to create a simple, q–regular graph on V vertices?
A simple, bipartite, q–regular graph on V vertices? Again, thinking
in terms of an algorithm may be fruitful.
3. The graphs below show the vertex adjacency for various convex polyhedra. Redraw the graphs so that the polyhedra are readily apparent.
For, example, Figure 16.2 may be redrawn to yield a cube as shown
in Figure 16.1. (Hint: It may be helpful to use Euler’s formula to
determine the number of faces on the polyhedron.)
16.4. EXERCISES
(a)
(b)
(c)
257
258
CHAPTER 16. GRAPH THEORY AND POLYHEDRA
(d)
4. Prove the following statement, using logic similar to that used in proving the theorem in §16.3: There is essentially one 4–regular simple
graph on six vertices: it is the graph of the vertex adjacency of the
octahedron.
5. (a) Show that the graph of the vertex adjacency of a hexagonal prism
is simple, 3–regular, and bipartite.
(b) By giving a counterexample, show that a simple, 3–regular bipartitite graph with twelve vertices need not necessarily correspond
to the graph of the vertex adjacency of a hexagonal prism.
6. A graph is said to have a perfect matching if there is a set of edges
M with the following properties:
(a) no two edges in M are incident; that is, no two edges in M have
a vertex in common, and
(b) every vertex of the graph is the end of some edge in M.
For each of the Platonic graphs, find a perfect matching.
7. A graph is said to be tripartite if the vertices may be partitioned into
three sets V1 , V2 , and V3 such that the ends of each edge lie in two
different sets, and any pair of sets contains the ends of at least one
edge.
(a) Find a simple, 5–regular, tripartite graph with twelve vertices.
(b) Show that the graph of the vertex adjacency of the icosahedron
(see Figure 16.12) is not tripartite.
Appendix A
Basic Constructions
A.1
Finding the perpendicular bisector and midpoint of a segment
This construction finds the perpendicular bisector and the midpoint of a
given segment.
p
q
(a)
p
q
(b)
p
q
(c)
Figure A.1
Let the segment pq be given. Open the compass wider than half the
length of the segment (otherwise the construction will not work). With
compass point at p, draw arcs above and below the segment, as in (a).
Repeat with the compass point at q, as in (b). Finally, join the intersections
of the pairs of arcs, as in (c). This line is the perpendicular bisector of pq,
so that it intersects the segment pq at its midpoint.
259
260
APPENDIX A. BASIC CONSTRUCTIONS
A.2
Constructing a perpendicular line
p
p
p
(a)
(b)
(c)
Figure A.2
Let p be given on a line, as in (a). With the compass point at p, draw
arcs on either side of p intersecting the line, as in (b). The points where the
arcs intersect the line are ends of a line segment. The perpendicular bisector
(see above) of this segment will pass through p, as in (c).
A.3
Bisecting an angle
p
p
(a)
p
(b)
(c)
Figure A.3
Let the angle with vertex p be given. With compass point at p, draw
an arc which passes through both sides of the angle, as in (a). With the
compass point at the points of intersection of this arc and the angle, draw
arcs which meet in the interior of the angle, as in (b). (The compass may
need to be adjusted at this step.) Join the intersection of these arcs and p,
as in (c). This line bisects the given angle.
A.4. CONSTRUCTING AN EQUILATERAL TRIANGLE
A.4
261
Constructing an equilateral triangle
q
p
q
p
(a)
q
p
(b)
(c)
Figure A.4
Open the compass to the desired length of the sides of the triangle, and
draw a line. With the compass point at some point on the line, say p, mark
off a point q on the same line, as in (a). With the compass set to the same
length, create intersecting arcs by placing the compass point at p, then at
q, as in (b). Complete the equilateral triangle as in (c).
A.5
Copying an angle
q
p
original angle
p
q
p
(a)
(b)
q
p
(c)
Figure A.5
Draw a line and a point p on the line as in (a). With the compass point
at the vertex of the original angle, draw an arc passing through both sides
of the angle. With this same compass setting, draw an arc with compass
point at p, intersecting the line at q as in (b). Measure the width of the arc
on the original angle. With this compass setting, draw an arc with compass
point at q intersecting the arc drawn in (b). Draw a line through p and this
intersection as in (c). This produces a copy of the original angle.
262
A.6
APPENDIX A. BASIC CONSTRUCTIONS
Trisecting a segment
p
q
p
q
p
t
u
v
t
s
s
r
(a)
q
(b)
r
(c)
Figure A.6
To trisect segment pq, draw a ray from p as in (a). With the compass
set not too wide and the compass point at p, mark off t on the ray drawn in
(a), as in (b). With compass point at t, mark off s, and with compass point
at s, mark off r. Note that pr is divided into equal thirds.
Draw rq. Copy ∠prq at t (see above), thereby creating similar triangles.
This will create pu, which is therefore one-third of the segment pq, as in
(c). With compass set to the length [pu] and compass point at u, create uv,
finishing the construction.
To divide a segment into, say, five congruent parts, simply mark off five
points at step (b) rather than three. This method may be extended to divide
a segment into any number of congruent parts.