1. Introduction
The close connection between the cubic structure and algebraic structures on cubic curves was studied in [
1,
2,
3], and this correlation was further examined in [
4]. The cubic structure was defined in [
4]. Let
Q be a nonempty set whose elements are called
points, and let
be a ternary relation on
Q. Such a relation and the ordered pair
will be called a
cubic relation and a
cubic structure, respectively, if the following properties are satisfied:
- C1.
For any two points , there is a unique point such that (i.e., ).
- C2.
The relation is totally symmetric (i.e., implies , , , , and ).
- C3.
, , , , and imply , which can be clearly written in the form of the following table:
In [
4], numerous examples of cubic structures are presented, with one notable example in which
Q is the set of all non-singular points of a cubic curve in the plane. In this context, the notation
signifies that points
a,
b, and
c∈
Q are collinear. Therefore, in a general cubic structure
, if
holds true, then we will also say that points
a,
b, and
c form a
line. If this statement is not valid, then we say that
is a
triangle.
The concept of tangentials of points was introduced in [
5]. The point
is said to be the
tangential of the point
a if
holds true. If
is the tangential of the point
a, then we will also say that the point
a is an antecedent of the point
. It is obvious that every point has one and only one tangential
. The tangential
of the tangential
of a point
a is called its
second tangential. We will always denote the tangential and the second tangential of any point
x as
and
. The validity of
implies the validity of
. Two distinct points having the same tangential are called
corresponding points. All points such that any two of them are corresponding points (i.e., have the same tangential) are said to be
associated. The maximal number of associated points always equals
for some
, and the number
m is called the
rank of the observed cubic structure. In the case of a cubic structure where collinear triples of non-singular points are observed on a cubic curve in the complex plane, ranks 0, 1, or 2 appear, depending on whether the cubic has a spike, an ordinary double point, or is without singular points. We will mention here some of the results from [
5] in the form of several lemmas.
Lemma 1. Let and be corresponding points with the common tangential , let o be any point, and let and be points such that and . Then, and are corresponding points with the common tangential such that . In addition, there is a point c such that and , and o and c are corresponding points.
Lemma 2. If ; ; ; and , then ; ; and are pairs of corresponding points, and the tangentials , , and satisfy .
Lemma 3. If , , , and are associated points with the common tangential , then there exist points p, q, and r such that , , , , , and , and points , p, q, and r are associated.
Lemma 4. Suppose holds, where , and are mutually different points. All different antecedents of points , , and in a cubic structure of rank 2 can be denoted by , , , and ; , , , and ; and , , , and such that the following hold:
Points a, b, c, d, e, and f are said to form a quadrilateral if there exist lines , , , and , and we say that the points from each pair of points ; ; and are opposite. Lemma 2 actually asserts that pairs of opposite vertices of a quadrilateral are corresponding.
We will say that
a,
d;
b,
e; and
c,
f are pairs of opposite vertices of a
complete quadrilateral if there exist lines
,
,
, and
. According to [
5] (Theorem 3.5), pairs of opposite vertices have common tangentials which are collinear points.
The motivation for this paper is drawn from classical books [
6,
7,
8,
9] which extensively covered the properties of cubics. Additionally, a wealth of various research papers on this subject exist, although we will not provide a detailed list.
2. Some Properties of Tangentials in a General Cubic Structure
The following theorem allows the construction of the tangential of a point when the tangential of another given point is known:
Theorem 1. Let a point a and its tangential be given. For each point p, let q, r, s, and t be points such that , , , and hold. Then, point t is the tangential of point p.
Proof. This statement follows from the table
Theorem 2. Let q and r be fixed points, and let , , , and be points such that , , and . Then, point s such that is also a fixed point.
Proof. Let
be the tangential of
q. From the table
we obtain
. Let
,
,
, and
be points such that
,
, and
. Then, the statement
, which should be proven, follows from the table
Theorem 3. If p, q, r, and s are points such that , and hold, and u and v are points such that and hold, then for the tangential of point a, the statement holds.
Proof. This statement follows from the table
Theorem 4. Let , , , and be points such that there is a point which satisfies and , and let t be any point. If points , , , and satisfy , , , and , then there is a point satisfying and .
Proof. If
is the tangential of point
t, and
is the point such that
, then from the tables
t | a1 | a2 | | t | c1 | c2 |
t | b1 | b2 | | t | d1 | d2 |
t′ | e1 | e2 | | t′ | e1 | e2 |
we obtain
and
. □
Theorem 5. Let points and have the common tangential , and let b be any point. If c and d are points such that and hold, then there is a point e satisfying and .
Proof. Let point
e be such that
. Then,
follows from the table
Theorem 6. In a cubic structure, there are as many triangles with vertices a, b, and c whose sides pass through the three given points d, e, and f of this cubic structure as there are antecedents of each point (i.e., in a structure of rank m, there are such triangles).
Proof. Let
g be the point such that
, and let
h be the point satisfying
. If
a is some antecedent point of the point
h (i.e., if
h is the tangential
of the point
a), then
a is one of the required points. Indeed, if
b and
c are points such that
and
, then we obtain
from the table
The number of solutions is equal to the number of antecedents of point h. □
Theorem 7. If , , and , where , then the tableimplies . If points , , and have the common tangential, then points , , and have the common tangential, as do points , , and . Proof. Here,
and
imply
and
, and since
and
also hold true, the tables
| | | | | | |
| | | | | | |
| | | | | | |
imply
and
(i.e.,
and
, respectively). In the same way, the second claim follows from the first statement by cyclically replacing the indices
. □
Theorem 8. Let points , , and be tangentials of points a, b, and c. Then, and imply .
Proof. This statement follows from the table
The next theorem is the converse of Lemma 2.
Theorem 9. If and hold, and if points a and b have the common tangential , then there is a point f satisfying and (i.e., there exists the quadrilateral ).
Proof. Let
f be a point such that
. From the table
we obtain
.
Theorem 10. Let a, b, c, d, e, and f be points such that , , and . The following four statements are equivalent: , , , and .
Proof. Because of the symmetry of our ternary relation, it suffices to prove that
if and only if
. If
, then
follows from the table
Conversely, if
, then
follows from the table
The following theorem is the converse of Lemma 3.
Theorem 11. If there exist lines , , , , , and , then points a, b, c, and d have the common tangential t, and points p, q, r, and t have the common tangential.
Proof. If
t is the tangential of point
a, then the fact that point
t is the common tangential of points
b,
c, and
d follows from the tables
q | d | b | , | | p | d | c | , | | p | c | d | . |
c | r | b | b | r | c | b | q | d |
a | a | t | a | a | t | a | a | t |
According to the table
the point
is the tangential of
p. The proof for points
q and
r is similar. □
Theorem 12. Let , , and hold. Points a, b, and c have the common tangential if and only if there exists a point p such that , , and .
Proof. Let
, and let
p be a point where
. Statements
and
follow from the tables
Conversely, if there is a point
p such that
,
, and
, and if
q is the tangential of the point
a, then from the tables
we find that
q is also the tangential of
b and
c. □
Theorem 13. If , , and , then there exists a point p satisfying and .
Proof. The proof follows by applying the table
Theorem 14. If , , and , then and are equivalent statements.
Proof. Suppose
holds. Since
implies
, we find that
(i.e.,
). Due to Theorem 13, there is a point
p such that
and
, which imply
(i.e.,
holds true). Conversely, suppose
holds. From the table
we find that
. As
implies
, we obtain
. □
Following [
10], a trio of points
is called a
triad if for points
d,
e, and
f such that
,
, and
, the statement
also holds true (i.e., there is a quadrilateral
in which the triad
is a triangle). We will call this quadrilateral the
circumscribed quadrilateral and line
the
complementary line of the triad
. Obviously, the quadrilateral
is also circumscribed to three other triads
,
, and
, to which the lines
,
, and
are complementary. Due to Lemma 1, the pairs of points
;
; and
have the common tangentials
,
, and
belonging to one line.
Theorem 15. If is the circumscribed quadrilateral of the triad , and if g, h, and i are points such that , , and hold, then these points also form the triad whose complementary line is the line , where , , and are the tangentials of points d, e, and f.
Proof. We already have
,
,
, and
, and the last relation implies
. From the tables
b | e | h | , | | a | d | g | , | and | a | d | g | , |
c | f | i | c | f | i | b | e | h |
d | d | d′ | e | e | e′ | f | f | f′ |
we obtain
,
, and
which, together with
, yields the statement of the theorem. □
Theorem 16. Let , , and be given points, let be any cyclic permutation of , and let points and be defined in such a way that and hold. Then, holds for .
Proof. The statement follows using the table
3. Properties of the Tangentials in the Cubic Structures of Ranks 0 and 1
Theorem 17. In a cubic structure of rank 0, the antecedents of three collinear points are also collinear.
Proof. Let a, b, and c be antecedents of points , , and , respectively, and let hold true. Suppose that there is a point d such that . This implies , and thus . Since each point has only one antecedent, we conclude that . □
Theorem 18. In a cubic structure of rank 1, let and be two different points having the common tangential a, and let b be a point different from a which has the common tangential with a. Then, holds true.
Proof. Let
c be a point such that
, and let
be the tangential of
a. From the table
it follows that point
c has the tangential
. If
, then we would have
and
. This would lead to the contradiction
. Consequently,
. □
Theorem 19. If points a and b have the tangentials and , respectively, and if point is such that , then there exists a point c to which point is tangential and which satisfies .
Proof. Let c be a point such that . This implies , and the point is uniquely determined. □
Theorem 20. In a cubic structure of rank 1, let hold true, and let ; ; and be pairs of different points with common tangentials , , and , respectively. Then, the indices of these points can be chosen such that , , , and (i.e., such that is a quadrilateral).
Proof. Let us choose arbitrary labeled points with tangentials and . Due to Theorem 19, one of the points with the tangential lies on the same line with points and . Let us label this point with and the other point with . We therefore have . Then, because of axiom C1, and cannot be valid. According to Theorem 19, and must hold. From any of these two statements, due to C1, it follows that cannot be valid, and according to Theorem 19, must hold. □
4. The Properties of Tangentials in Cubic Structures of Rank 2
Theorem 21. If non-collinear points a, b, and c in a cubic structure of rank 2 have collinear tangentials , , and , respectively, then is a triad.
Proof. Each of the points , , and has four antecedents , , and . According to Lemma 4, the indices of these points can be chosen such that, among others, , , , and hold, where points , , and are the original points a, b, and c. □
Theorem 22. Let the triad and the line be given. If g, h, and i are points such that , , and , then is also a triad.
Proof. Here,
holds, and the existence of four lines in the theorem implies the existence of the lines
,
,
, and
. From the table
we obtain
. Points
g,
h, and
i are not collinear, because otherwise
would follow from the table
which is not true. Due to Theorem 20,
is a triad. □
Theorem 23. Let a, b, and c be three non-collinear points in a cubic structure of rank 2. Then, there are four triples of the form such that , , and hold true.
Proof. Let
g be a point such that
, and additionally, let
h be a point such that
. If we assume that
,
, and
hold, then from the table
it follows that point
h is necessarily the tangential of
d. Therefore, for point
d, we have to take one of the four antecedents of
h. Let us take one such point
d, and then let
e and
f be points such that
and
hold. Then
follows from the table
In the previous inference, if points a, b, and c are collinear, then , and thus . Therefore, point a is one of the antecedents of point h, and any of the other three can be taken as point d. In case , we obtain and . Theorem 16 solves the problem of finding a triangle whose “sides” pass through the given points.
Theorem 24. If , , , and are different points with the common tangential , then implies , where is the tangential of .
Proof. According to Lemma 2, there is a point p such that , , and points p an have the common tangential . Let t be a point such that . As , , , and imply , we obtain that t is the tangential of p (i.e., ). □
Let a, b, c, and d be associated points with the common tangential p. According to Lemma 3, there are points d, e, and f such that , , , , , and . Points p, e, f, and g are associated with the common tangential . According to the same lemma, there are points , , and such that , , , , , and . Points , , , and are associated with the common tangential .
Theorem 25. Using the same labeling and the results from above, there exist points q, r, s, and t such that , , , , , , , , , , , , , , , and hold true, and points q, r, s, and t are associated.
Proof. Let point
q be such that
. Then,
,
, and
follow from the tables
f | d | b | , | | e | d | c | , | and | e | c | d | . |
f | g | e1 | e | g | f1 | e | f | g1 |
p′ | a | q | p′ | a | q | p′ | a | q |
Similarly, implies , , and , and from , we obtain , , and , and implies , , and . Finally, switching to the tangentials, from , , , and , we obtain , , , and (i.e., ). □
5. Hesse Configuration
The configuration
of points and lines obtained in Lemma 4 is called the
Hesse configuration. In this configuration, we have the line
, nine lines of the forms
,
, and
, and six lines of the form
, where
is any permutation of
. The lines of this configuration can be divided into 4 quadruplets of lines, each of which contains all 12 points of the configuration:
, | , | , | , |
, | , | , | , |
, | , | , | , |
, | , | , | . |
Moreover, because of the following two theorems, the points of the Hesse configuration lie with some other points on some more lines.
Theorem 26. Using the notation from Lemma 4, there are points , , and such that there are 18 lines , , , , , and , where , , and .
Proof. Let
,
, and
be points such that
,
, and
hold. From the table
we obtain
. Cyclically permuting letters
a,
b, and
c in the previous table proves
, and repeating it proves
. □
Theorem 27. For each permutation of , for the points from Theorem 22 there is a line .
Proof. The proof follows by applying the table
Theorem 28. Let be the common tangential of mutually different points , , , and in a cubic structure of rank 2. If o is any point, and if , , , and are points such that , then , , , and are mutually different points with the common tangential such that , where is the tangential of point o. In addition, there are mutually different points c, d, and e which are different from point o such that points c, d, e, and o have the common tangential and , , , , , , , , , , , and hold true.
Proof. Let and be different points with the common tangential , let o be any point, and let and be points such that and . Then, according to Lemma 1, points and are different and have the common tangential such that , where is the tangential of o. Aside from that, there is a point c such that and , and points o and c have the common tangential . Points o and c are different because otherwise, we would have and , where . In a cubic structure of rank 2, each point has four different antecedent points. Let , , , and be different points with the common tangential . If o is any point, and , , , and are points such that , then points , , , and are mutually different. Due to the previous facts, points , , , and have the common tangential such that , and there are points c, d, and e such that , , , , , and and which have the tangential . Points c, d, and e are mutually different because, for example, would imply , which contradicts , where , and they are also different from point o. There is also point f with tangential and different from o such that and . It must coincide with one of points c, d, or e. Due to and , it can be neither c nor d, and thus (i.e., and hold true). Similarly, one can show that , , , and . □
If we now rename points
o,
c,
d, and
e as
,
,
, and
, respectively, then we have 16 lines of the form
, where the indices
are such that either all three are different or all three are equal to zero. Otherwise, one index is equal to zero, and the other two are equal and different from zero. We find the
Table 1 from which one can see which point
lies on the same line with some point
and some point
. We once again obtain the Hesse configuration
from Lemma 4.
Theorem 29. The following quadrilaterals exist in the Hesse configuration shown in the previous table: Proof. The proof is obvious when referring to the above table. For example, the last statement is a consequence of the existence of lines , , , and . □
6. The de Vries Configuration
In [
11,
12], it is claimed (and in [
13], it is proven) that for an elliptic cubic, there are only two non-isomorphic configurations
in which there are three disjoint quadruples of points such that no two points from a particular quadruple are on one of the 16 lines of the configuration. One of these configurations is the Hesse configuration, and the other can be called the
de Vries configuration. Both configurations were also observed in [
14,
15,
16]. All five authors used the properties of the ambient space (i.e., the properties of the projective plane in which the cube is embedded). In this paper, we will observe the de Vries configuration in any cubic structure by means of that structure (i.e., using only axioms C1– C3). The observed cubic structure should be of at least rank 1.
We start from three non-collinear points
,
, and
. Let
and
be points such that
and
, and let
and
be two different points with the common tangential
. Let
,
,
, and
be points such that
,
,
, and
hold, and let
be the point such that
. Then, by using the tables
a01 | a0 | a1 | , | | a01 | a0 | a1 | , | | a01 | a3 | a3 | . |
a2 | c0 | b3 | a3 | c0 | b2 | a2 | b0 | c2 |
a2 | b0 | c2 | a3 | b0 | c3 | a2 | c3 | b1 |
we obtain
,
, and
. If
and
are points such that
and
, then from the tables
b1 | c3 | a2 | , | | b1 | c2 | a3 |
c0 | a3 | b2 | c0 | a2 | b3 |
a1 | b0 | c1 | a1 | b0 | c1 |
we have
and
. Finally, from the tables
c0 | b0 | a0 | , | | c0 | b0 | a0 | , | | c0 | b0 | a0 | . |
a3 | c2 | b1 | a2 | c1 | b2 | a1 | c2 | b3 |
b2 | a2 | c1 | b3 | a1 | c2 | b1 | a2 | c3 |
we obtain
,
, and
. Therefore, we proved the following theorem:
Theorem 30. There exists a configuration of points , , and , where , with the corresponding Table 2. The configuration from Theorem 30 is the de Vries configuration. To create it, we begin such that points and have the common tangential satisfying . However, such a property appears several times more in the configuration (i.e., the following holds true):
Theorem 31. In the configuration from Theorem 30, pairs of points , ; , ; , ; , ; , ; and , have common tangentials , , , , , and , respectively, such that , , , , , and hold true.
Proof. We will prove all statements using only
Table 2 and statements
,
,
,
,
, and
(i.e., independent from the construction method of the observed configuration). Statements
,
,
, and
can be derived from the tables
b0 | c0 | a0 | , | | b0 | c1 | a1 | , | | b0 | c2 | a2 | , | | b0 | c3 | a3 | . |
c2 | b2 | a0 | c2 | b3 | a1 | c0 | b3 | a2 | c0 | b2 | a3 |
a2 | a3 | a23 | a2 | a3 | a23 | a0 | a1 | a01 | a0 | a1 | a01 |
We obtain
,
,
, and
from the tables
c0 | a0 | b0 | , | | c0 | a1 | b1 | , | | c0 | a3 | b2 | , | | c0 | a2 | b3 | . |
a3 | c3 | b0 | a3 | c2 | b1 | a0 | c2 | b2 | a0 | c3 | b3 |
b2 | b3 | b23 | b2 | b3 | b23 | b0 | b1 | b01 | b0 | b1 | b01 |
Then,
,
,
, and
follow from the tables
a0 | b0 | c0 | , | | a0 | b1 | c1 | , | | a0 | b2 | c2 | , | | a0 | b3 | c3 | . |
b2 | a3 | c0 | b2 | a3 | c1 | b0 | a2 | c2 | b0 | a3 | c3 |
c2 | c3 | c23 | c2 | c3 | c23 | c0 | c1 | c01 | c0 | c1 | c01 |
□
Theorem 32. Under the conditions of Theorem 30, there exist complete quadrilaterals , , , , and .
Proof. Using
Table 2, it is easy to check the existence of quadruples of the required lines. For instance, for the last quadrilateral, we have the lines
,
,
, and
. □
From
Table 2, we see that the lines
pass alternately through points
and
, while the lines
pass alternately through points
and
. In fact, we find two Steiner octagons with fundamental points
,
and
,
, and all 16 lines of the configuration were used. However, it is also possible to form two Steiner octagons from all 16 lines of configuration with fundamental points
,
and
,
, 2 with fundamental points
,
and
,
, 2 with fundamental points
,
and
,
, 2 with fundamental points
,
and
,
, and 2 with fundamental points
,
and
,
. In the two observed octagons, opposing vertices have the common tangential, and the same holds for the other Steiner octagons.
Using the following theorems, we will show how to associate yet another de Vries configuration to the one from Theorem 30.
Theorem 33. There exists a complete quadrilateral .
Proof. From the tables
a3 | a2 | a23 | , | | a3 | a2 | a23 | , | | b3 | b2 | b23 | , | | c3 | c2 | c23 |
b3 | b2 | b23 | b1 | b0 | b01 | c1 | c0 | c01 | a1 | a0 | a01 |
c1 | c1 | c23 | c2 | c2 | c01 | a3 | a3 | a01 | b2 | b2 | b01 |
we obtain the lines
,
,
, and
, which proves the statement. □
Since the opposite vertices of the quadrilaterals have common tangentials, we have the following:
Corollary 1. The pairs of points , ; , ; and , have common tangentials.
Theorem 34. There exist points , , , , , and such that there exist lines , , , , , , , , , , , and . Furthermore, there exist quadrilaterals , , and . Finally, the pairs of points , ; , ; and , have common tangentials.
Proof. Let
,
,
,
,
, and
be such that there are lines
,
,
,
,
, and
. The existence of the remaining six lines can be inferred from the following tables:
c0 | b1 | a1 | , | | b0 | c1 | a1 | , | | a0 | c1 | b1 | , |
b0 | c3 | a3 | c0 | b3 | a2 | c0 | a2 | b3 |
a0 | a2 | x2 | a0 | a3 | x3 | b0 | b2 | y2 |
|
a0 | c1 | b1 | , | | a0 | b1 | c1 | , | | a0 | b1 | c1 | , |
c0 | a3 | b2 | b0 | a3 | c3 | b0 | a2 | c2 |
b0 | b3 | y3 | c0 | c2 | z2 | c0 | c3 | z3 |
The existence of these 12 lines proves the existence of the three mentioned quadrilaterals, and the last statement is an immediate consequence of [
5] (Theorem 3.4). □
Theorem 35. There exists a de Vries configuration of points , , , , , , , , , , , and with the corresponding Table 3. Proof. Except for the lines from Theorem 33, one should prove the existence of yet another 12 lines. But this follows from the following tables:
a0 | a1 | a01 | , | | a0 | a1 | a01 | , | | a2 | a3 | a23 | , | | a2 | a3 | a23 | , |
b0 | b2 | y2 | b0 | b3 | y3 | b0 | b2 | y2 | b0 | b3 | y3 |
c0 | c3 | z3 | c0 | c2 | z2 | c2 | c0 | z2 | c2 | c1 | z3 |
|
a0 | a2 | x2 | , | | a0 | a2 | x2 | , | | a0 | a3 | x3 | , | | a0 | a3 | x3 | , |
b0 | b1 | b01 | b2 | b3 | b23 | b0 | b1 | b01 | b2 | b3 | b23 |
c0 | c3 | z3 | c2 | c0 | z2 | c0 | c2 | z2 | c2 | c1 | z3 |
|
a0 | a2 | x2 | , | | a0 | a2 | x2 | , | | a0 | a3 | x3 | , | | a0 | a3 | x3 | , |
b0 | b2 | y2 | b2 | b1 | y3 | b2 | b0 | y2 | b0 | b3 | y3 |
c0 | c1 | c01 | c2 | c3 | c23 | c2 | c3 | c23 | c0 | c1 | c01 |
Comparing
Table 3 in the text of this theorem with
Table 2 in Theorem 30 reveals that they represent the same configuration. □
Using
Table 2, we proved Theorems 31 and 32 concerning the existence of some common tangentials and quadrilaterals. Similarly, by employing
Table 3, we can prove analogous theorems about the existence of common tangentials and quadrilaterals in this second de Vries configuration.