# Affine Fullerene C60 in a GS-Quasigroup

Authors: Vladimir Volenec; Zdenka Kolar-Begović; Ružica Kolar-Šuper
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

It will be shown that the affine fullerene C60, which is defined as an affine image of buckminsterfullerene C60, can be obtained only by means of the golden section. The concept of the affine fullerene C60 will be constructed in a general GS-quasigroup using the statements about the relationships between affine regular pentagons and affine regular hexagons. The geometrical interpretation of all discovered relations in a general GS-quasigroup will be given in the GS-quasigroup C
(
1
/
2
(
1
+
5
)
).


---

## Body

## 1. Introduction

The fullerenes are closed carbon-cage molecules containing only pentagonal and hexagonal rings.C
60 is the first fullerene that was theoretically conceived and experimentally obtained. The geometrical structure of C
60 is a truncated icosahedron with a carbon atom at the corners of each hexagon and a bond along each edge (Figure 1). The sixty-carbon cluster with the geometry of a truncated icosahedron is named buckminsterfullerene [1, 2].Figure 1The affine regular icosahedron is defined as an affine image of a regular icosahedron. Let the affine regular icosahedron be given with the pairs of opposite verticesa, a
′; b, b
′; c, c
′; d, d
′; e, e
′; f, f
′ as in Figure 2. Let us divide each edge of this icosahedron into three equal parts and then omit two lateral parts. On each of the twenty faces of the icosahedron on the sides the three segments in their middle parts are left. Let us connect the adjacent ends of these segments, so that an affine regular hexagon is formed on each side of an icosahedron, and an affine regular pentagon appears in the neighborhood of each vertex of icosahedron (Figure 3).Figure 2Figure 3The obtained polyhedron consists of twelve affine regular pentagons and twenty affine regular hexagons. It is an affine version of buckminsterfullereneC
60 which will be calledaffine fullerene
C
60. It is presented in Figure 3, from where it is obvious how the labels of vertices of that polyhedron are chosen, starting from the labels of vertices of the affine regular icosahedron.We will prove that the complete affine fullereneC
60 can be presented only by means of the golden section. The concept of a GS-quasigroup will be used in this consideration.


## 2. GS-Quasigroup

A quasigroup(
Q
,
·
) is said to be a golden section quasigroup or shortly a GS-quasigroup [3] if it satisfies the (mutually equivalent) identities
(1)
a
(
a
b
·
c
)
·
c
=
b
,
(2)
a
·
(
a
·
b
c
)
c
=
b
,
and the identity of idempotence
(3)
a
a
=
a
.
GS-quasigroups are medial quasigroups; that is, the identity
(4)
a
b
·
c
d
=
a
c
·
b
d
is valid [4].As a consequence of the identity of mediality, the considered GS-quasigroup(
Q
,
·
) satisfies the identities of elasticity and left and right distributivity; that is, we have these identities:
(5)
a
b
·
a
=
a
·
b
a
,
(6)
a
·
b
c
=
a
b
·
a
c
,
(7)
a
b
·
c
=
a
c
·
b
c
.
Further, the identities
(8)
a
(
a
b
·
b
)
=
b
,
(9)
(
b
·
b
a
)
a
=
b
,
(10)
a
(
a
b
·
c
)
=
b
·
b
c
,
(11)
(
c
·
b
a
)
a
=
c
b
·
b
,
(12)
a
(
a
·
b
c
)
=
b
(
b
·
a
c
)
,
(13)
(
c
b
·
a
)
a
=
(
c
a
·
b
)
b
and equivalencies
(14)
a
b
=
c
⟺
a
=
c
·
c
b
,
(15)
a
b
=
c
⟺
b
=
a
c
·
c
also hold.LetC be the set of points of the Euclidean plane. For any two different points a
,
b we define a
b
=
c if the point b divides the pair a
,
c in the ratio of the golden section. In [3], it is proved that (
C
,
·
) is a GS-quasigroup. We will denote that quasigroup by C
(
(
1
/
2
)
(
1
+
5
)
) because we have c
=
(
1
/
2
)
(
1
+
5
) if a
=
0 and b
=
1. The figures in this quasigroup C
(
(
1
/
2
)
(
1
+
5
)
) can be used for illustration of “geometrical” concepts and relations in any GS-quasigroup.


## 3. Affine Regular Pentagons and Hexagons in GS-Quasigroups

From now on, let(
Q
,
·
) be any GS-quasigroup. The elements of the set Q are said to bepoints.In each medial quasigroup, the concept of a parallelogram can be introduced by means of two auxiliary points. In [5], it is proved that the points a
,
b
,
c
,
d are the vertices of aparallelogram denoted by Par
(
a
,
b
,
c
,
d
), if and only if there are two points p and q such that p
a
=
q
b and p
d
=
q
c. It is also shown that if the statement Par
(
a
,
b
,
c
,
d
) holds, then the equalities p
a
=
q
b and p
d
=
q
c are equivalent. In a general GS-quasigroup, the notation of a parallelogram can be characterized by the equivalency Par
(
a
,
b
,
c
,
d
)
⇔
a
·
b
(
c
a
·
a
)
=
d (Figure 4).Figure 4In [3], some properties of the quaternary relation Par on the set Q are proved. We will mention only the property which will be used afterwards.Lemma 1.
From Par(
a
,
b
,
c
,
d
) and Par(
c
,
d
,
e
,
f
) there follows Par(
a
,
b
,
f
,
e
).We will say thatb is themidpoint of the pair of points a
,
c and we write M
(
a
,
b
,
c
) if and only if Par
(
a
,
b
,
c
,
b
) holds. The statement M
(
a
,
b
,
c
) holds if and only if c
=
b
a
·
b [3].The concept of the affine regular hexagon [6] in a GS-quasigroup is defined in the following way. We will say that (
a
1
,
a
2
,
a
3
,
a
4
,
a
5
,
a
6
) is anaffine regular hexagon with the vertices a
1
,
a
2
,
a
3
,
a
4
,
a
5
,
a
6 and the center o and we write ARH
o
(
a
1
,
a
2
,
a
3
,
a
4
,
a
5
,
a
6
) if the statements Par(
o
,
a
i
-
1
,
a
i
,
a
i
+
1
) hold (
i
=
1,2,3,4
,
5,6
), where indexes are taken modulo 6 (Figure 5). The following statement can be proved [6].Figure 5Lemma 2.
An affine regular hexagon is uniquely determined by any three consecutive vertices.The pointso
,
a
1
,
a
2
,
a
3
,
a
4 determine the figure which will be denoted by the symbol HARH
o
(
a
1
,
a
2
,
a
3
,
a
4
), “half” of the affine regular hexagon with the center o (Figure 5).The following results [6] will be very useful.Lemma 3.
Letn
∈
N, n
≥
3. If the statements H
A
R
H
c
12
(
b
1
,
a
1
,
a
2
,
b
2
), H
A
R
H
c
23
(
b
2
,
a
2
,
a
3
,
b
3
)
,
…, and H
A
R
H
c
n
-
1
,
n
(
b
n
-
1
,
a
n
-
1
,
a
n
,
b
n
) are valid, then there exists a unique point c
n
1 so that the statement H
A
R
H
c
n
1
(
b
n
,
a
n
,
a
1
,
b
1
) is valid too. (The case for n
=
5 is illustrated in Figure 6.)Figure 6Lemma3 implies the following statement.Lemma 4.
Letn
∈
N, n
≥
3. If the statements A
R
H
c
12
(
b
1
,
a
1
,
a
2
,
b
2
,
d
21
,
d
12
), A
R
H
c
23
(
b
2
,
a
2
,
a
3,b
3
,
d
32
,
d
23
)
,
…
,
A
R
H
c
n
-
1
,
n
(
b
n
-
1
,
a
n
-
1
,
a
n
,
b
n
,
d
n
,
n
-
1
,
d
n
-
1
,
n
) are valid, then there exist unique points c
n
1
,
d
n
1
,
d
1
n so that the statement A
R
H
c
n
1
(
b
n
,
a
n
,
a
1
,
b
1
,
d
1
n
,
d
n
1
) is valid, too.The pointsa
,
b
,
c
,
d successively are said to be the vertices of thegolden section trapezoid [7] denoted by GST(
a
,
b
,
c
,
d
) if the identity a
·
a
b
=
d
·
d
c holds (Figure 7). It can be proved that the following equivalency GST
(
a
,
b
,
c
,
d
)
⇔
c
=
a
(
d
b
·
b
) holds. The following statement is also valid.Figure 7Lemma 5.
Any of the three statementsG
S
T
(
a
,
b
,
c
,
d
), G
S
T
(
b
,
e
,
f
,
c
), and a
e
=
d
f is a consequence of the two remaining statements.In [8], it is proved that any two of the five statements
(16)
GST
(
a
,
b
,
c
,
d
)
,
GST
(
b
,
c
,
d
,
e
)
,
GST
(
c
,
d
,
e
,
a
)
,
GST
(
d
,
e
,
a
,
b
)
,
GST
(
e
,
a
,
b
,
c
)
imply the remaining statements.The pointsa
,
b
,
c
,
d
,
e successively are said to be the vertices of theaffine regular pentagon [8] denoted by ARP
(
a
,
b
,
c
,
d
,
e
) if any two (and then all five) of the five statements (16) are valid (Figure 7).Lemma 6.
An affine regular pentagon is uniquely determined by any three of its vertices.Now, we are going to study the relationships between the previously defined geometrical concepts in a general GS-quasigroup.Lemma 7.
If the statements Par(
d
1
,
o
,
b
2
,
a
1
), Par(
d
2
,
o
,
b
1
,
a
2
) are valid, then (i)
the statementsd
1
b
1
=
d
2
b
2 and GST(
d
1
,
a
1
,
a
2
,
d
2
) are equivalent (Figure 8);
(ii)
the statementsd
1
b
1
=
d
2
b
2 and GST(
a
1
,
b
1
,
b
2
,
a
2
) are equivalent (Figure 8).Figure 8Proof.
(i) We have the equalities(17)
a
1
=
d
1
·
o
(
b
2
d
1
·
d
1
)
,
a
2
=
d
2
·
o
(
b
1
d
2
·
d
2
)
,
and we have to prove the equivalency of the equalities
(18)
d
1
·
d
1
a
1
=
d
2
·
d
2
a
2
,
d
1
b
1
=
d
2
b
2
.
However, we get
(19)
a
1
=
d
1
·
o
(
b
2
d
1
·
d
1
)
=
(
6
)
d
1
o
·
d
1
(
b
2
d
1
·
d
1
)
=
(
5
)
d
1
o
·
(
d
1
·
b
2
d
1
)
d
1
=
(
11
)
d
1
o
·
(
d
1
b
2
·
b
2
)
=
(
4
)
(
d
1
·
d
1
b
2
)
·
o
b
2
=
(
10
)
o
(
o
d
1
·
b
2
)
·
o
b
2
=
(
6
)
o
·
(
o
d
1
·
b
2
)
b
2
=
(
13
)
o
·
(
o
b
2
·
d
1
)
d
1
and thus we get
(20)
d
1
·
d
1
a
1
=
d
1
·
d
1
[
o
·
(
o
b
2
·
d
1
)
d
1
]
=
(
6
)
(
d
1
·
d
1
o
)
·
d
1
[
d
1
·
(
o
b
2
·
d
1
)
d
1
]
=
(
5
)
,
(
2
)
(
d
1
·
d
1
o
)
·
o
b
2
=
(
4
)
d
1
o
·
(
d
1
o
·
b
2
)
and analogously d
2
·
d
2
a
2
=
d
2
o
·
(
d
2
o
·
b
1
). Because of that, it is necessary to prove the equivalency of the equality d
1
o
·
(
d
1
o
·
b
2
)
=
d
2
o
·
(
d
2
o
·
b
1
) and the equality b
2
=
(
d
2
·
d
1
b
1
)
·
d
1
b
1 which is, according to (15), equivalent to d
1
b
1
=
d
2
b
2. As we get
(21)
d
2
o
·
(
d
2
o
·
b
1
)
=
(
8
)
d
2
o
·
[
d
2
o
·
d
2
(
d
2
b
1
·
b
1
)
]
=
(
6
)
d
2
[
o
·
o
(
d
2
b
1
·
b
1
)
]
=
(
2
)
,
(
5
)
[
d
1
·
d
1
(
d
2
d
1
·
d
1
)
]
[
o
·
o
(
d
2
b
1
·
b
1
)
]
=
(
4
)
d
1
o
·
[
d
1
o
·
(
d
2
d
1
·
d
2
b
1
)
(
d
1
b
1
)
]
=
(
6
)
d
1
o
·
[
d
1
o
·
(
d
2
·
d
1
b
1
)
(
d
1
b
1
)
]
,
these should be equivalent equalities:
(22)
d
1
o
·
(
d
1
o
·
b
2
)
=
d
1
o
·
[
d
1
o
·
(
d
2
·
d
1
b
1
)
(
d
1
b
1
)
]
,
b
2
=
(
d
2
·
d
1
b
1
)
·
d
1
b
1
,
which is obvious.
(ii) Firstly, let us prove that from GST(
a
1
,
b
1
,
b
2
,
a
2
) there follows d
1
b
1
=
d
2
b
2.
We have the equalities(23)
d
1
=
a
1
·
b
2
(
o
a
1
·
a
1
)
,
d
2
=
a
2
·
b
1
(
o
a
2
·
a
2
)
,
b
2
=
a
1
(
a
2
b
1
·
b
1
)
.
Therefore we get
(24)
d
1
b
1
=
[
a
1
·
b
2
(
o
a
1
·
a
1
)
]
b
1
=
a
1
[
a
1
(
a
2
b
1
·
b
1
)
·
(
o
a
1
·
a
1
)
]
·
b
1
=
(
6
)
[
a
1
·
a
1
(
a
2
b
1
·
b
1
)
]
[
a
1
(
o
a
1
·
a
1
)
]
·
b
1
=
(
12
)
[
(
a
2
b
1
)
(
a
2
b
1
·
a
1
b
1
)
·
a
1
(
o
a
1
·
a
1
)
]
b
1
=
(
7
)
[
(
a
2
·
a
2
a
1
)
b
1
·
a
1
(
o
a
1
·
a
1
)
]
b
1
=
(
4
)
[
(
a
2
·
a
2
a
1
)
a
1
·
b
1
(
o
a
1
·
a
1
)
]
b
1
=
(
9
)
[
a
2
·
b
1
(
o
a
1
·
a
1
)
]
b
1
=
(
6
)
[
a
2
b
1
·
a
2
(
o
a
1
·
a
1
)
]
b
1
=
(
7
)
(
a
2
b
1
·
b
1
)
[
a
2
(
o
a
1
·
a
1
)
·
b
1
]
,
d
2
b
2
=
[
a
2
·
b
1
(
o
a
2
·
a
2
)
]
·
a
1
(
a
2
b
1
·
b
1
)
=
(
6
)
[
a
2
b
1
·
a
2
(
o
a
2
·
a
2
)
]
·
a
1
(
a
2
b
1
·
b
1
)
=
(
4
)
(
a
2
b
1
·
a
1
)
[
a
2
(
o
a
2
·
a
2
)
·
(
a
2
b
1
·
b
1
)
]
=
(
7
)
(
a
2
b
1
·
a
1
)
[
a
2
(
a
2
b
1
·
b
1
)
·
(
o
a
2
·
a
2
)
(
a
2
b
1
·
b
1
)
]
=
(
8
)
(
a
2
b
1
·
a
1
)
[
b
1
·
(
o
a
2
·
a
2
)
(
a
2
b
1
·
b
1
)
]
=
(
4
)
(
a
2
b
1
·
b
1
)
[
a
1
·
(
o
a
2
·
a
2
)
(
a
2
b
1
·
b
1
)
]
,
so it is necessary to prove the equality
(25)
a
2
(
o
a
1
·
a
1
)
·
b
1
=
a
1
·
(
o
a
2
·
a
2
)
(
a
2
b
1
·
b
1
)
.
Really, we have
(26)
a
1
·
(
o
a
2
·
a
2
)
(
a
2
b
1
·
b
1
)
=
(
9
)
,
(
11
)
(
a
1
·
a
1
b
1
)
b
1
·
[
(
o
·
a
2
b
1
)
b
1
·
(
a
2
b
1
·
b
1
)
]
=
(
7
)
[
(
a
1
·
a
1
b
1
)
·
(
o
·
a
2
b
1
)
(
a
2
b
1
)
]
b
1
=
(
4
)
[
a
1
(
o
·
a
2
b
1
)
·
(
a
1
b
1
·
a
2
b
1
)
]
b
1
=
(
7
)
[
a
1
(
o
·
a
2
b
1
)
·
(
a
1
a
2
·
b
1
)
]
b
1
=
(
4
)
[
(
a
1
·
a
1
a
2
)
·
(
o
·
a
2
b
1
)
b
1
]
b
1
=
(
11
)
(
a
1
·
a
1
a
2
)
(
o
a
2
·
a
2
)
·
b
1
=
(
10
)
[
a
2
(
a
2
a
1
·
a
2
)
·
(
o
a
2
·
a
2
)
]
b
1
=
(
5
)
[
(
a
2
·
a
2
a
1
)
a
2
·
(
o
a
2
·
a
2
)
]
b
1
=
(
7
)
[
(
a
2
·
a
2
a
1
)
(
o
a
2
)
·
a
2
]
b
1
=
(
4
)
[
(
a
2
o
)
(
a
2
a
1
·
a
2
)
·
a
2
]
b
1
=
(
5
)
[
(
a
2
o
)
(
a
2
·
a
1
a
2
)
·
a
2
]
b
1
=
(
6
)
[
a
2
(
o
·
a
1
a
2
)
·
a
2
]
b
1
=
(
5
)
[
a
2
·
(
o
·
a
1
a
2
)
a
2
]
b
1
=
(
11
)
a
2
(
o
a
1
·
a
1
)
·
b
1
.
Now, we are going to prove that d
1
b
1
=
d
2
b
2 implies GST
(
a
1
,
b
1
,
b
2
,
a
2
). According to (i), from the hypotheses of (ii), there follows the statement GST
(
d
1
,
a
1
,
a
2
,
d
2
) and, from this statement and the equality d
1
b
1
=
d
2
b
2, according to Lemma 5, there follows the statement GST
(
a
1
,
b
1
,
b
2
,
a
2
).Lemma 8.
If the statementsH
A
R
H
o
(
c
1
,
b
1
,
b
2
,
c
2
), H
A
R
H
d
1
(
a
1
,
b
1
,
c
1
,
e
1
), and H
A
R
H
d
2
(
a
2
,
b
2
,
c
2
,
e
2
) are valid, then the statement G
S
T
(
a
1
,
b
1
,
b
2
,
a
2
) and equality d
1
b
1
=
d
2
b
2 are equivalent (Figure 9).Figure 9Proof.
The assumptions of the lemma imply the statements Par(
o
,
b
2
,
b
1
,
c
1
), Par(
o
,
b
1
,
b
2
,
c
2
), Par(
b
1
,
c
1
,
d
1
,
a
1
), and Par(
b
2
,
c
2
,
d
2
,
a
2
), and then, according to Lemma 1, parallelograms Par(
o
,
b
2
,
a
1
,
d
1
), Par(
o
,
b
1
,
a
2
,
d
2
) follow from the first and the third, and the second and the fourth parallelogram, respectively. Owing to these last statements, according to Lemma 7(ii), statements GST(
a
1
,
b
1
,
b
2
,
a
2
) and d
1
b
1
=
d
2
b
2 are equivalent.Lemma 9.
With the assumptionA
R
H
(
a
b
,
b
a
,
b
c
,
c
b
,
c
a
,
a
c
), the statement f
′
a
b
=
e
′
a
c follows from the equalities d
′
b
c
=
f
′
b
a, d
′
c
b
=
e
′
c
a (Figure 10).Figure 10Proof.
Supposing that a more precise statementARH
o
(
a
b
,
b
a
,
b
c
,
c
b
,
c
a
,
a
c
) is valid, so the statements Par(
c
b
,
c
a
,
o
,
b
c
) and Par(
b
a
,
o
,
a
c
,
a
b
) are valid. From the statements Par(
c
b
,
c
a
,
o
,
b
c
), d
′
c
b
=
e
′
c
a there follows d
′
b
c
=
e
′
o, which together with d
′
b
c
=
f
′
b
a gives the equality f
′
b
a
=
e
′
o, and this statement and the statement Par(
b
a
,
o
,
a
c
,
a
b
) imply the equality f
′
a
b
=
e
′
a
c.Theorem 10.
The statementA
R
P
(
a
b
,
a
f
′
,
a
d
,
a
e
′
,
a
c
) follows from the statements A
R
H
(
a
b
,
b
a
,
b
c
,
c
b
,
c
a
,
a
c
), A
R
H
(
b
c
,
b
d
′
,
d
b
′
,
d
c
′
,
c
d
′
,
c
b
), A
R
P
(
b
c
,
b
d
′
,
b
e
,
b
f
′
,
b
a
), A
R
P
(
c
b
,
c
d
′
,
c
f
,
c
e
′
,
c
a
), A
R
H
(
b
a
,
b
f
′
,
f
b
′
,
f
a
′
,
a
f
′
,
a
b
), and A
R
H
(
c
a
,
c
e
′
,
e
c
′
,
e
a
′
,
a
e
′
,
a
c
) (Figure 11).Figure 11Proof.
It is sufficient to prove that the statement GST(
a
f
′
,
a
b
,
a
c
,
a
e
′
) follows from the statements ARH
o
(
a
b
,
b
a
,
b
c
,
c
b
,
c
a
,
a
c
), HARH
d
′
(
b
d
′
,
b
c
,
c
b
,
c
d
′
), GST(
b
d
′
,
b
c
,
b
a
,
b
f
′
), GST(
c
d
′
,
c
b
,
c
a
,
c
e
′
), HARH
f
′
(
b
f
′
,
b
a
,
a
b
,
a
f
′
), and HARH
e
′
(
c
e
′
,
c
a
,
a
c
,
a
e
′
). Firstly, according to Lemma 8, the equality d
′
b
c
=
f
′
b
a follows from the statements HARH
o
(
c
b
,
b
c
,
b
a
,
a
b
), HARH
d
′
(
b
d
′
,
b
c
,
c
b
,
c
d
′
), HARH
f
′
(
b
f
′
,
b
a
,
a
b
,
a
f
′
), and GST(
b
d
′
,
b
c
,
b
a
,
b
f
′
). Owing to the same lemma, the equality d
′
c
b
=
e
′
c
a follows from the statements HARH
o
(
b
c
,
c
b
,
c
a
,
a
c
), HARH
d
′
(
c
d
′
,
c
b
,
b
c
,
b
d
′
), HARH
e
′
(
c
e
′
,
c
a
,
a
c
,
a
e
′
), and GST(
c
d
′
,
c
b
,
c
a
,
c
e
′
).
Now, all requirements of Lemma9 are satisfied and the equality f
′
a
b
=
e
′
a
c follows accordingly.
Finally, according to Lemma8, the statements HARH
o
(
b
a
,
a
b
,
a
c
,
c
a
), HARH
f
′
(
a
f
′
,
a
b
,
b
a
,
b
f
′
), and HARH
e
′
(
a
e
′
,
a
c
,
c
a
,
c
e
′
) and equality f
′
a
b
=
e
′
a
c imply the statement GST(
a
f
′
,
a
b,a
c
,
a
e
′
).


## 4. Construction of an Affine Fullerene C60 in a GS-Quasigroup

In this section, we are going to construct an affine fullereneC
60 in a general GS-quasigroup by means of the previously discovered statements about affine regular pentagons and hexagons in a general GS-quasigroup.Theorem 11.
An affine fullereneC
60 can be constructed in each GS-quasigroup.Proof.
For the sake of clarity, each step of the proof is precisely presented in figures in the GS-quasigroupC
(
(
1
/
2
)
(
1
+
5
)
) and each sequence of the proof of the theorem can be followed on the Schlegel diagram (Figure 12).
Let us start with the four given pointsb
a
,
b
c
,
c
b
,
b
d
′. The affine regular hexagons ARH(
a
b
,
b
a
,
b
c
,
c
b
,
c
a
,
a
c
) and ARH(
b
c
,
b
d
′
,
d
b
′
,
d
c
′
,
c
d
′
,
c
b
) can be constructed according to Lemma 2.
Owing to Lemma6, affine regular pentagons ARP(
b
c
,
b
d
′
,
b
e
,
b
f
′
,
b
a
) and ARP(
c
b
,
c
d
′
,
c
f
,
c
e
′
,
c
a
) can be obtained. If we apply Lemma 2 again, we can get ARH(
b
a
,
b
f
′
,
f
b
′
,
f
a
′
,
c
f
′
,
a
b
) and ARH(
c
a
,
c
e
′
,
e
c
′
,
e
a
′
,
a
e
′
,
a
c
).
According to Theorem10, the existence of these obtained affine regular hexagons and affine regular pentagons around an affine regular hexagon will result in the existence of the affine regular pentagon ARP(
a
b
,
a
f
′
,
a
d
,
a
e
′
,
a
c
) (Figure 13).
According to Lemma2, the already obtained points a
d
,
a
f
′
,
f
a
′ uniquely determine ARH
(
d
a
,
a
d
,
a
f
′
,
f
a
′
,
f
d
′
,
d
f
′
) and then, because of Lemma 4, the statements ARH
(
d
a
,
a
d
,
a
f
′
,
f
a
′
,
f
d
′
,
d
f
′
), ARH
(
f
a
′
,
a
f
′
,
a
b
,
b
a
,
b
f
′
,
f
b
′
), ARH
(
b
a
,
a
b
,
a
c
,
c
a
,
c
b
,
b
c
), and ARH
(
c
a
,
a
c
,
a
e
′
,
e
a
′
,
e
c
′
,
c
e
′
) imply the statement ARH (
e
a
′
,
a
e
′
,
a
d
,
d
a
,
d
e
′
,
e
d
′
). In the same way, the statements ARH
(
e
b
,
b
e
,
b
d
′
,
d
b
′
,
d
e
′
,
e
d
′
), ARH
(
d
b
′
,
b
d
′
,
b
c
,
c
b
,
c
d
′
,
d
c
′
), ARH(
c
b
,
b
c
,
b
a
,
a
b
,
a
c
,
c
a
), and ARH(
a
b
,
b
a
,
b
f
′
,
f
b
′
,
f
a
′
,
a
f
′
) imply ARH
(
f
b
′
,
b
f
′
,
b
e
,
e
b
,
e
f
′
,
f
e
′
) and, analogously, the statements ARH
(
f
c
,
c
f
,
c
e
′
,
e
c
′
,
e
f
′
,
f
e
′
), ARH
(
e
c
′
,
c
e
′
,
c
a
,
a
c
,
a
e
′
,
e
a
′
), ARH(
a
c
,
c
a
,
c
b
,
b
c
,
b
a
,
a
b
), and ARH(
b
c
,
c
b
,
c
d
′
,
d
c
′
,
d
b
′
,
b
d
′
) imply ARH
(
d
c
′
,
c
d
′
,
c
f
,
f
c
,
f
d
′
,
d
f
′
).
This consideration is presented in Figure14.
These two precisely described procedures will also be used later. It will also be denoted which figure is used for the geometrical presentation of the obtained implications in the GS-quasigroupC
(
(
1
/
2
)
(
1
+
5
)
).
Now, according to Theorem10, the statements ARH
(
d
b
′
,
b
d
′
,
b
c
,
c
b
,
c
d
′
,
d
c
′
), ARH
(
b
c
,
b
a
,
a
b
,
a
c
,
c
a
,
c
b
), ARP
(
b
c
,
b
a
,
b
f
′
,
b
e
,
b
d
′
), ARP
(
c
b
,
c
a
,
c
e
′
,
c
f
,
c
d
′
), ARH
(
b
d
′
,
b
e
,
e
b
,
e
d
′
,
d
e
′
,
d
b
′
), and ARH
(
c
d
′
,
c
f
,
f
c
,
f
d
′
,
d
f
′
,
d
c
′
) imply ARP(
d
b
′
,
d
e
′
,
d
a
′
′
,
d
f
′
,
d
c
′
). The statements ARP
(
e
c
′
,
e
f
′
,
e
b
′
′
,
e
d
′
,
e
a
′
) and ARP
(
f
a
′
,
f
d
′
,
f
c
′
′
,
f
e
′
,
f
b
′
) can be obtained similarly (Figure 15).
Thanks to Lemma4, the statements ARH
(
e
d
′
,
d
e
′
,
d
b
′
,
b
d
′
,
b
e
,
e
b
), ARH
(
b
d
′
,
d
b
′,d
c
′
,
c
d
′
,
c
b
,
b
c
), ARH(
c
d
′
,
d
c
′
,
d
f
′
,
f
d
′
,
f
c
,
c
f
), and ARH(
f
d
′
,
d
f
′
,
d
a
′
′
,
a
d
′
′
,
a
f
′
,
f
a
′
) imply ARH (
a
d
′
′
,
d
a
′
′
,
d
e
′
,
e
d
′
,
e
a
′
,
a
e
′
). We can find the affine regular hexagons ARH
(
b
e
′
′
,
e
b
′
′
,
e
f
′
,
f
e
′
,
f
b
′
,
b
f
′
) and ARH(
c
f
′
′
,
f
c
′
′
,
f
d
′
,
d
f
′
,
d
c
′
,
c
d
′
) in the same way (Figure 16).
If we apply Theorem10, we will discover three new affine regular pentagons (Figure 17). The statement ARP(
d
f
′
,
d
c
′
,
d
b
′
,
d
e
′
,
d
a
) follows from ARH(
d
f
′
,
f
d
′
,
f
a
′
,
a
f
′
,
a
d
,
d
a
), ARH(
f
a
′
,
f
b
′
,
b
f
′
,
b
a
,
a
b
,
a
f
′
), ARP(
f
a
′
,
f
b
′
,
f
e
′
,
f
c
′
′
,
f
d
′
), ARP(
a
f
′
,
a
b
,
a
c
,
a
e
′
,
a
d
), ARH
(
f
d
′
,
f
c
′
′
,
c
f
′
′
,
c
d
′
,
d
c
′
,
d
f
′
), and ARH(
a
d
,
a
e
′
,
e
a
′
,
e
d
′
,
d
e
′
,
d
a
), and similarly we get ARP(
e
d
′,e
a
′
,
e
c
′
,
e
f
′
,
e
b
) and ARP(
f
e
′
,
f
b
′
,
f
a
′
,
f
d
′
,
f
c
).
Now, we are in a position to use Lemma4 whose application gives the statements about three new affine regular hexagons (Figure 18). The statements ARH(
b
d
′
,
d
b
′
,
d
e
′
,
e
d
′
,
e
b
′
′
,
b
e
′
′
), ARH(
e
d
′
,
d
e
′
,
d
a
,
a
d
,
a
e
′
,
e
a
′
), ARH(
a
d
,
d
a
,
d
f
′
,
f
d
′
,
f
a
′
,
a
f
′
), and ARH(
f
d
′
,
d
f
′
,
d
c
′
,
c
d
′
,
c
f
′
′
,
f
c
′
′
) imply ARH(
c
d
′
,
d
c
′
,
d
b
′
,
b
d
′
,
b
c
′
′
,
c
b
′
′
). Similarly, we can get the affine regular hexagons ARH(
a
e
′
,
e
a
′
,
e
c
′
,
c
e
′
,
c
a
′
′
,
a
c
′
′
) and ARH(
b
f
′
,
f
b
′
,
f
a
′
,
a
f
′
,
a
b
′
′
,
b
a
′
′
).
If we apply Theorem10, we will obtain the three new affine regular pentagons ARP(
a
e
′
,
a
c
′
′
,
a
b
′
′
,
a
f
′
,
a
d
′
′
), ARP(
b
f
′
,
b
a
′
′
,
b
c
′
′
,
b
d
′
,
b
e
′
′
), and ARP(
c
d
′
,
c
b
′
′
,
c
a
′
′
,
c
e
′
,
c
f
′
′
) (Figure 19).
Finally, the application of Lemma3 will allow us to close the complete structure. We have to prove that the points b
a
′
′
,
a
b
′
′
,
a
c
′
′
,
c
a
′
′
,
c
b
′
′
,
b
c
′
′ are the vertices of an affine regular hexagon.
By applying Lemma3 we get that the statements HARH(
c
a
′
′
,
a
c
′
′
,
a
e
′
,
e
a
′
), HARH(
e
a
′
,
a
e
′
,
a
d
′
′
,
d
a
′
′
), HARH(
d
a
′
′
,
a
d
′
′
,
a
f
′
,
f
a
′
), and HARH(
f
a
′
,
a
f
′
,
a
b
′
′
,
b
a
′
′
) imply HARH(
b
a
′
′
,
a
b
′
′
,
a
c
′
′
,
c
a
′
′
).
Analogously, we have that HARH(
a
b
′
′
,
b
a
′
′
,
b
f
′
,
f
b
′
), HARH(
f
b
′
,
b
f
′
,
b
e
′
′
,
e
b
′
′
), HARH(
e
b
′
′
,
b
e
′
′
,
b
d
′
,
d
b
′
), and HARH(
d
b
′
,
b
d
′
,
b
c
′
′
,
c
b
′
′
) imply HARH(
c
b
′
′
,
b
c
′
′
,
b
a
′
′
,
a
b
′
′
), and HARH(
b
c
′
′
,
c
b
′
′
,
c
d
′
,
d
c
′
), HARH(
d
c
′
,
c
d
′
,
c
f
′
′
,
f
c
′
′
), HARH(
f
c
′
′
,
c
f
′
′
,
c
e
′
,
e
c
′
), and HARH(
e
c
′
,
c
e
′
,
c
a
′
′
,
a
c
′
′
) imply HARH(
a
c
′
′
,
c
a
′
′
,
c
b
′
′
,
b
c
′
′
).
These obtained three halves of affine regular hexagons HARH(
b
a
′
′
,
a
b
′
′
,
a
c
′
′
,
c
a
′
′
), HARH(
c
b
′
′
,
b
c
′
′
,
b
a
′
′
,
a
b
′
′
), and HARH(
a
c
′
′
,
c
a
′
′
,
c
b
′
′
,
b
c
′
′
) determine the affine regular hexagon ARH(
b
a
′
′
,
a
b
′
′
,
a
c
′
′
,
c
a
′
′
,
c
b
′
′
,
b
c
′
′
) (Figure 19). This completes the proof of the theorem.Figure 12Figure 13Figure 14Figure 15Figure 16Figure 17Figure 18Figure 19Thus, we have proved that any affine fullereneC
60 can be obtained only by applying the golden section.



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*Source: 950103-2014-07-07.xml*