Tiling a Plane with Semi-Regular Equilateral Polygons with 2m-Sides ()
1. Introduction
The problem of tiling a plane is an ancient one, which even the mathematicians of ancient Egypt, Greece, Persia, China and other old civilizations were familiar with. Tiling comes down to dividing a plane into polygons that would completely cover it without having any overlaps or gaps, following certain regularity, depending on the type, shape and arrangement of the polygons [1]. So, with tiling, the goal is to divide the plane into polygons that would only have common sides and vertices. Then, for the polygons that have one common side it is said that they are the adjacent polygons, and the point of the plane in which the vertices of the adjacent polygons meet is named the node of that partition of the plane. The node is called regular if all angles of the polygons meeting at it are equal. The two nodes are considered to be equal if the number of angles that meet at it is the same. The problem of tiling comes down to determining all possible divisions of a plane with the polygons:
1) The division of a plane with regular polygons, or when all the polygons and all the nodes are mutually equal. Such tiling is called a regular one.
2) The division of a plane in a way that several types of regular polygons meet at a node. Such tiling a plane is called an Archimedes one, or a semi-regular one.
3) Special cases of tiling a plane.
The first two cases have been largely researched, and there can be found more about them in [1] and a catalog of tiling can be seen in [2] [3] [4].
We are interested in special cases, and that is tiling a plane with semi-regular equilateral polygons, i.e. when equilateral semi-regular polygons meet at a node. Prior to the analysis of the problem of tiling a plane with semi-regular polygons, let us mention some basic points of view and theorems that are valid for tiling a plane with regular polygons.
Theorem 1. The only proper tiling is possible with equilateral triangles, squares and regular hexagons, and in such a way that six, four and three of them meet at a single node.
Proof: Since the sum of angles at each node is 2π and the value of the interior angles of regular polygon
, if
regular polygons meet at the vertex, then it follows that:
.
From here, after rearranging and solving the equation, we get the following equation
from which, after solving by k, we find that
.
From the condition
it follows that
. On the basis of this, it is found that for
, the value is
, and for
it is
, while for
, it is
[1].
2. Semi-Regular Equilateral Polygons and Formulation of a Problem
1) Polygon
or closed polygonal line is the union along
and is write shortly
,
. Points
are vertices, and lines
are sides of polygon
. The angles on the inside of a polygon formed by each pair of adjacent sides are angles of the polygon.
2) Given polygon
with vertices
,
, lines of which
polygonal diagonals if indices are not consecutive natural numbers, that is
. We can draw
diagonals from each vertex of the polygon with n number of vertices.
3) Exterior angle of the polygon
with vertex
is the angle
with one side
, and vertex
, and the other one is extension of the side
through vertex
.
4) The interior angle of the polygon
with vertex
is the angle
for which
. That is the angle with one
, side, and the other side
. Sum of all interior angles of the polygon is defined by equation
. In which k is number of turning around the polygon in certain direction.
5) A regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length). Regular polygon with n sides of b length is marked as
. The formula for interior angles
of the regular polygon
with n sides is
. A non-convex regular polygon is a regular star polygon. For more about polygons in [4] [5] [6].
6) Polygon that is either equiangular or equilateral is called semi-regular polygon. Equilateral polygon with different angles within those sides are called equilateral semi regular polygons, whereas polygons that are equiangular and with sides different in length are called equiangular sem regular polygons. For more about polygons in [1] [2] [3].
7) If we construct a polygon
with
sides,
,
with vertices
over each side of the convex polygon
,
,
with vertices
, that is
, we get new polygon with
(Figure 1) marked as
.
Here are the most important elements and terms related to constructed polygons:
· Polygon
with vertices
,
constructed over each side
of polygon
with which it has one side in common is called edge polygon for polygon
.
·
are the sides of polygon
.
·
are diagonals
of the polygon
drawn from the top
and that implies
.
· Angles
are interior angles of vertices
of the polygon
. and are denoted as
. Interior angles
of the polygon of the vertices
are denoted as
.
Figure 1. Convex semi regular polygon
with
sides constructed above the regular polygon
.
· Polygon
of the side a constructed over the side b of the polygon
is isosceles, with
equal sides, is denoted as
.
·
denotes the angle between its two consecutive diagonals drawn from the vertices
for which it is true
.
· Regular polygon
is called corresponding regular polygons of the semi-regular polygon
.
· If the isosceles polygon
is constructed over each side of the b regular polygon
with n sides, then the constructed polygon with
of equal sides is called equilateral semi-regular polygon which is denoted as
.
· Interior angles of a semi-regular polygon at odd vertices are marked with
, and those at even vertices are marked with
(Figure 1).
· To a semi-regular equilateral polygon
with
with equal sides, there can be “inscribed” regular n-side polygons: by joining odd vertices,
or even vertices
. [5].
To analyze the metric properties of regular polygons, it is sufficient for us to know one basic element, the length of a side, while for the semi-regular polygons this is not sufficient [5] Therefore, in addition to side a of a semi-regular polygon, for the analysis of the metric properties we will use another element of it, and that is the angle between side a of the semi-regular polygon and side b of its “inscribed” regular polygon, which we mark with δ, i.e.
(Figure 2) [5] [6] [7].
To show that a semi-regular equilateral 2n-side polygon is given by side a and angle δ we write:
.
Figure 2. Basic elements of equilateral semi-regular polygon
of side
and angle δ.
If
is the interior angle of the “inscribed” regular polygon
, then
gives interior angles at odd vertices, and
gives the ones at even vertices of the semi-regular polygon
of a side a, where
marks the angle between the sides of polygons
and
(Figure 2). Here, we consider that a regular polygon with
sides (segment) is “inscribed” to a semi-regular equilateral quadrilateral (rhombus) [5] [6] [7] [8].
Let us mark semi-regular polygon
with
sides as determined with the characteristic elements
and interior angles
, with
.
Here are some more results on the interior angles of semi-regular polygon
, with
sides, which we need, and the proof of which can be seen in papers [5].
Theorem 2. A semi-regular equilateral convex polygon
, with
sides and characteristic angle
has the folowing:
interior angles along the vertices of the “inscribed” regular polygon
, with side b, all equal to angle α and the following applies
(1)
interior angles, along the vertices of isosceles polygons
, with m equal arms constructed over each side of a regular polygon
, as a common side. These interior angles are equal to angle β and the following applies:
(2)
The dependence of convexity of a semi-regular equilateral polygon
, with
sides from the value of angle δ is expressed by the following theorem:
Theorem 3. Semi-regular, equilateral polygon
with
sides is convex if the following is valid for angle δ:
;
, (3)
Note that for
, a convex semi-regular polygon
becomes regular, and since this is not the subject of our research here, these values of angle δ are excluded from further consideration. The proof of this theorem can be seen in [6] [7] [8].
3. Main Results
3.1. Tiling a Plane with Semi-Regular Polygons of the Same Type
Tiling a plane with equilateral convex semi-regular polygons differs from tiling a plane with regular ones, and it belongs to a special group of tiling. Based on the characteristics of the semi-regular equilateral polygons, the following types of tiling a plane with semi-regular polygons can be differentiated:
A) Tiling a plane with semi-regular polygons when the equal number of semi-regular polygons of the same type meet at each node;
B) Tiling a plane with semi-regular polygons when semi-regular equilateral polygons of different types and equal sides meet at one node;
C) Tiling a plane with semi-regular polygons when semi-regular equilateral polygons of different types and different sides meet at one node.
In this paper we shall consider the cases of tiling a plane as stated under section (A).
Let us consider tiling a plane if tiling is performed with semi-regular equilateral polygons.
If it is possible to perform tiling with one type of semi-regular polygons, than due to the existence of the two types of the interior angles: the angles along the vertices of the inscribed regular polygon, equal to angle
,
and angles along the vertices of the edge polygons, equal to angle
, the following types of nodes can be differentiated [9]:
1) Nodes at which the vertices meet, to which the interior angles equal to the angle α correspond;
2) Nodes at which the vertices meet, to which the interior angles equal to angle β correspond;
3) Nodes at which the vertices meet, to which the interior angles equal to angles α and β correspond.
Let us assume that it is possible to perform tiling with semi-regular equilateral convex polygons of the same type, constructed in the manner as described above, with the characteristic elements
and interior angles
.
Then in that case there is any non-negative integers t, s which are not simultaneously equal to zero, such that at one node, there are t vertices meeting, to which the interior angles equal to angle α correspond, and/or s vertices, to which the interior angles equal to angle β correspond.
Based on the value of interior angles α and β, and the fact that in that case, the sum of angles at each node is equal to 2π, the stated conditions may be written down in the form of the following equation
(4)
In this way, the problem of tiling a plane with one type of semi-regular convex equilateral polygons can be described as the solution of Equations (3.1). A semi-regular equilateral polygon is convex for
when the polygon is regular.
Based on this, for different values of
and δ we also have different Diophantine equations. Here we analyze the case when characteristic angle
and l is any natural number greater than one,
. (A case when
was dealt with in paper [9] ). For this value of angle δ, the Equation (4) has the following form:
After rearranging and solving, the last equation can be written in the following form:
(5)
representing the Diophantine equation in which the unknown parameters are
.
Should we denote the coefficients respectively with:
(6)
the equation can be written in a simpler form
.
For different values of
Equation (5) has different forms. Let us analyze the possibilities of tiling a plane with semi-regular polygons with
, i.e. with those with which the inscribed regular polygon has
sides closed segment.
The Diophantine equation for tiling a plane with semi-regular equilateral polygons with
sides reads
(7)
where
and unknowns
are the non-negative integers, i.e.
The following theorem applies:
Theorem 4. Let
be an equilateral semi-regular polygon with
sides and let
be its characteristic angle. Tiling a plane with semi-regular equilateral polygon
can be performed only if
or
, that is, the Diophantine Equation (7) has a non-negative integer solutions only for
or
.
Proof: To show the theorem’s claim, let us note that the problem of tiling a plane with a semi-regular equilateral polygon with 2m sides is equivalent to determining the set of all the solutions of the corresponding Diophantine Equation (7). Any pair
of the non-negative integers that are not simultaneously equal to zero and that meet the given equation represent its solution.
Let us show that the Diophantine Equation (7) has solutions only when
or
. Further, note that for a given value of angle
for the interior angles of a semi-regular equilateral polygon equal to angle α the following relation applies:
, and for angles equal to angle β the following applies:
.
Note that the solution of Equation (7) represents a pair of non-negative integers (t, s) that depend on the value of natural number
and that the following cases can be distinguished:
Case 1: If
, Equation (7) has the following form:
(8)
From here we find that one solution, for
of Equation (7) is a pair
, which does not depend on the choice of natural numbers
.
Case 2: If
the Equation (7) reads:
From here we find that
And from here, we find that s is a natural number, if and only if
is a natural number, or if
.
If it were that:
, then one of these cases could arise:
1)
.
2)
because
.
Let us consider both cases.
1) If it were that:
, then it would be that
, which is impossible, because by assumption
and
.
If it where that:
then it would be that
were from it would follow that
. Since
then it means that
or
, whence we get that
if
or
.
- If
, then
, and then it is that
,
. A pair of (0, 4) is another solution of Diophantine Equation (7).
- if
, then
, which is contrary to the assumption that
, and, in this case, the Equation (7) has no solution.
2) If the case is that
, then
. Hence, it follows that l is a natural number greater than 1, if
or
, that is, if
or
.
- If it were that
, then it would be that
. Based on that, it would be that
, so pair
is another solution of the corresponding Diophantine Equation (7)
- and the case when
is not possible, because then it would be that
, which is contrary to the assumption that
.
Case 3: Let
. Let us transform Equation (7) as follows:
(9)
From the last equation
it follows that s is a natural number, if one of the cases occur:
1)
or
2)
.
Let us consider each of the cases.
1) If
, then
and
is a natural number only for
and
.
In case when
, then
, so the pair of
is also the solution of Equation (7).
When
then
, and
, thus another pair of non-negative integers
has been determined
which is the solution of Equation (7).
2) Let us also consider case 2). Suppose that
is defined by Equation (8) and that
. Then from the assumption that
, it follows that
i.e. then the integer is
.
In relation to the constraint of the value of integer p, let us observe the following cases:
1)
, then
, We have already considered this case;
2)
then from equation
it follows that
, and from this we get that
. Since from the assumption that
, it follows that it must be that
.
For this value of number, it is
and
. So the pair of
is the solution of Diophantine Equation (7) for every natural number
.
1) If
, then from equation
we get that
, or
. It follows from this equation that t is a natural number, if and only if
and
are natural numbers, that is, if
and
.
From the requirement
, it follows that it must be that
or
.
For
it applies that
and then it is
and
. Furthermore, from the requirement that
, it follows that it must be that:
, because
by assumption. It follows then that
. Now, for
from
, it follows that
if
and
, then
.
Based on that, for
we have determined another solution of Diophantine Equation (7), and that is
.
If
, then from the condition that
and equation
, it follows that
is a natural number only if p is an even number, i.e.
. Furthermore, since
, by assumption, it follows that
. From this inequation we then get that
. Since
is only for
, which is contrary to the assumption that
. Thus, there is no
for which
, and the consequence of this is that for
, in this case, there is no solution for Diophantine Equation (7).
We have, thus, shown that the Diophantine Equation (7) has a solution only if
or
and that the set of solutions of Equation (7) is:
,
.
3.2. Tiling a Plane with Semi-Regular Equilateral Quadrilaterals
If in Equation (7) we put that
, we get the following equation:
(10)
Equation (10) represents the corresponding Diophantine equation for the problem of tiling a plane with equilateral semi-regular quadrilaterals with a characteristic angle
and interior angles which are equal to angle
or angle
.
Note that for
the interior angles of the quadrilateral are equal, and they are
, so it is regular, and it is not subject to our investigation here.
Therefore, we shall further consider the case when
, i.e. when the interior angles of an equilateral quadrilateral are different. The theorem on nodes in tiling a plane with semi-regular equilateral quadrilateral holds.
Theorem 5. In relation to the selected value
, when tiling a plane with a semi-regular equilateral quadrilateral, only the following nodes can appear:
, and in this case node (0, 3) appears only in the case of tiling a plane with a semi-regular quadrilateral whose characteristic angle is
, and interior angles are
and
.
Proof. The problem of determining nodes is equivalent to the problem of finding all non-negative solutions of the corresponding Diophantine Equation (10). The following cases are possible:
1) If
, then
, so, obviously, pair
is the one solution of Equation (10) for all
. That is, when tiling a plane with a semi-regular equilateral quadrilateral, a node
appears. That is, 2l vertices of semi-regular quadrilaterals which have the internal angles equal to angle
are joined at one node.
2) If
, then Equation (10) has the following form
. The solution of equation
can be written in the form of
, from which it follows that
is a non-negative integer if, and only if
or
. If it were that
, value
(when the quadrilateral is regular) and value
do not meet the assumption that
. Only value
meets the assumption, and then
, and it follows that
. It follows that node
appears only when tiling a plane with a semi-regular equilateral quadrilateral with a characteristic angle
, and interior angles
and
.
3) Let
still be natural numbers. If we write Equation (10) in the form of
, it follows that s is a natural number, if
. Since
, if:
a)
, that is, if
, then it still is that
. Thus, pair
is the solution of Equation (10). Hence, it follows that when tiling a plane with a semi-regular equilateral quadrilateral, a node (2, 2) appears at
which two vertices join, with an interior angle equal to angle
and two vertices whith corresponding interior angles equal to angle β for each choice of natural number
.
b) If it were that
, and then it would be that
and then
, and
, so pair
is another solution of Diophantine Equation (10).
Thus, when tiling a plane with semi-regular equilateral quadrilateral, there is a node
at which the following vertices meet: vertex
with an interior angle equal to angle
and another vertex with a corresponding interior angle which is equal to angle
.
c) If it were that
, and then it would be that
, and then
. From the requirement that
, it follows that
. Since by assumption
, it follows that the case is not possible.
d) If
, then it would be
and
. Since by assumption
are natural numbers, the following conjunction must hold:
and
. Hence we find that the conjunction is valid only when
, i.e.
if
. Thus, for
it is
and
. This pair (4, 1) is obtained from the previous pair
if
. For
, we have that
and
. We got a pair, as in case a).
We have, thus, determined the set of all solutions of Equation (10),
, i.e. all nodes that can appear when tiling a plane with semi-regular equilateral quadrilaterals, with a characteristic angle
and interior angles equal to angle
and angle
.
A graphical presentation of the corresponding Diophantine equation with nodes when tiling a plane with semi-regular quadrilaterals is shown in Figure 3.
Table 1 also lists the basic values of semi-regular quadrilaterals with the corresponding Diophantine equation, as an example of tiling a plane with semi-regular equilateral quadrilaterals for various values
.
Let us now consider tiling a plane with some of the semi-regular equilateral quadrilaterals.
We noted that for
the characteristic angle is
, and that then the
Figure 3. Position of nodes when tiling a plane with a semi-regular equilateral quadrilateral on the graph of the corresponding Diophantine equation.
Table 1. Diophantine equations with a set of solutions for tiling a plane with semi-regular equilateral quadrilaterals with characteristic values of angle δ, and interior angles, for various values of parameter
.
1—quadrilateral is regular.
interior angles of the quadrilateral are equal;
and
so the observed quadrilateral is regular (square). Tiling a plane with these quadrilaterals has been previously considered [2].
For
, the corresponding Diophantine equation is
. The set of solutions of this equation is:
, and characteristic angle
, while the values of the interior angles are
,
respectively.
Let us consider the example of tiling a plane when the following nodes appear: (6, 0) and (0, 3) (Figure 4) and nodes (0, 3), (4, 1) and (2, 2) (Figure 5).
For
when tiling a plane with a semi-regular quadrilateral with a characteristic angle
and one node,
, at which two quadrilaterals meet, with vertices to which two interior angles equal to angle
correspond, and two quadrilaterals with vertices to which interior angles equal to angle
correspond (Figure 6).
Figure 4. A fragment of tiling a plane with a semi-regular quadrangle with
and with nodes (6, 0) and (0, 3).
Figure 5. Tiling a plane with semi-regular equilateral quadrilaterals with
, and nodes (0, 3) and (4, 1).
Figure 6. A fragment of tiling a plane with a semi-regular quadrilateral with
and with node (2, 2).
A case of tiling a plane with a semi-regular quadrilateral in which there is a node (8.0) and a node (2.2) is shown in Figure 7, while tiling a plane with a semi-regular equilateral quadrilateral with nodes (5.1) and (2.2) is shown in Figure 8.
3.3. Tiling a Plane with Semi-Regular Equilateral Hexagons
If, in Equation (7) which corresponds to the problem of tiling a plane with semi-regular equilateral polygons with 2m-sides,
with
and unknowns t, s are non-negative integers, i.e.
, we insert that
, we get the following equation:
(11)
Equation (11) represents Diophantine equation for the problem of tiling a plane with equilateral semi-regular hexagons with characteristic angle
and interior angles that are equal to angle
or angle
. Each pair
of the non-negative integers that are not simultaneously equal to zero and that meet Equation (11) is the solution of the equation.
Figure 7. Tiling a plane with semi-regular quadrilaterals with node (8.0) and node (2.2).
Figure 8. Tiling a plane with semi-regular quadrilateral when with
and nodes (5, 1) and (2, 2).
The theorem holds.
Theorem 6. In relation to the selected value
when tiling a plane with a semi-regular equilateral hexagon, only nodes
can appear for all values of the characteristic angle
and interior angles
and
.
Proof: The problem of determining nodes is equivalent to the problem of finding all non-negative solutions of Diophantine Equation (11). In doing so, let us consider the following cases:
Case 1: If
, the equation has the following form
, whence we find that there is a pair
, which represents the solution of Equation (11) for all
.
From this, it follows that when tiling a plane with a semi-regular equilateral hexagon, there appears node (2l) at which two vertices meet with an interior angle equal to angle
and for each choice of a natural number
.
Case 2: If
, then Equation (11) reads
. Let us examine for which values of
it is that
and it is a natural number. Note that
and that
is valid only when
is divided by 2. This is possible only when
or
. If it were that
than it would be that
or
and these values do not meet the requirement that
. If it were that
then it could be that
or
, and also none of these values meets the requirement that
. Thus, we conclude that there is no pair (t, s) in which
, and which is the solution of Equation (11) and which meets the requirement
, and there is no node (0, s) as well, when tiling a plane with semi-regular equilateral hexagon.
Case 3: Let us assume that
. Then from equation
we get that
. Note that
only if s is an even number.
Let
. Let us determine the values for natural number p.
Since
for
then from the assumption that
and
it follows that it must be that
, which is possible if
, which is equivalent to the requirement that
, or the requirement that
. Hence, from the requirement that
it follows that
only when
. For this value
is
, and
, for all values of
.
We have, thus, determined another solution of Diophantine Equation (11).
That is, when tiling a plane with a semi-regular equilateral hexagon, a node (1, 2) appears at which one vertex of a semi-regular equilateral hexagon meets, with a corresponding interior angle equal to angle α, and two vertices of a semi-regular equilateral hexagon, with a corresponding interior angle equal to angle β for each selection of natural number
.
We have, thus, shown that the set of all solutions of Diophantine Equation (11) is
, i.e. that only nodes
can occur when tiling a plane with a semi-regular equilateral hexagon, for all values of
.
Examples of tiling a plane with semi-regular equilateral hexagons with corresponding Diophantine equations are shown in Table 2, and graphically in Figure 9 and Figure 10.
Figure 9. Tiling a plane with a semi-regular equilateral hexagon, with nodes (4, 0) and (1, 2).
Figure 10. A case of tiling a plane with a semi-regular equilateral hexagon with nodes (16, 0) and (1, 2).
Table 2. Diophantine equations with a set of solutions for tiling a plane with a semi-regular equilateral hexagon, and with characteristic values of angle δ, and interior angles, for various values of parameter
.
4. Conclusions
The paper dealt with the possibility of tiling the Euclidean plane with convex semi-regular equilateral polygons. The research was conducted by observing a set of solutions for the corresponding Diophantine equation of the following form:
, where t, s are the nonnegative integers that are not simultaneously equal to zero, and
, are the interior angles of a semi-regular equilateral polygon PN.
It has been shown that each solution of this equation represents one node and it shows how many semi-regular equilateral polygons with the corresponding interior angles meet at that node. It has also been shown that of all semi-regular equilateral polygons with 2m-sides, a plane may be tiled only with semi-regular quadrilaterals and semi-regular hexagons. Graphically presented cases are just some of the possible ones that depend on the value of the characteristic angle.