dirichlet tessellations of a plane by congruent...

NEW ZEALAND JOURNAL OF MATHEMATICSVolume 39 (2009), 79-101

DIRICHLET TESSELLATIONS OF A PLANEBY CONGRUENT QUADRANGLES

Yusuke Takeo and Toshiaki Adachi

(Received February 2007)

Abstract. In this paper we give a necessary and sufficient condition for some

tessellations of a plane R2 by congruent convex quadrangles to be Dirichlet.

1. Introduction

A family R ={Rj

}of convex closed polygons is said to be a convex tessellation

(or a convex tiling) of a plane R2 if it satisfies the following properties:(i) R2 =

⋃j Rj ,

(ii) if Rj ∩Rk 6= ∅, then Rj ∩Rk is either an edge of each of Rj and Rk or a vertexof each of them.

For given a set P ={

Pj

}of points on R2, we define a region Rj as a set of points

whose nearest point in P is Pj , that is Rj =⋂

k 6=j

{x ∈ R2

∣∣ d(x,Pj) 5 d(x,Pk)}

.Then we see the family

{Rj

}of these regions turns to be a convex tessellation, which

is called a Volonoi diagram associated with a set of sites P. Volonoi diagrams arequite important tessellations which are used in wide area of science. A convextessellation R of a plane is said to be Dirichlet if we can choose a point PR ∈ R foreach region R ∈ R so that R be a Volonoi diagram associated with

{PR

∣∣ R ∈ R}.When we study convex tessellations of a plane, it is an interesting problem to

give a condition that they are Dirichlet. For a vertex of a region of a convextessellation, it is said to be n–valent if there are exactly n edges emanating fromthis vertex. In other words, a vertex is n–valent if it is an intersection of exactlyn regions. When each vertex is odd valent, a necessary and sufficient conditionthat a convex tessellation to be Dirichlet is given in [1]. This condition is closelyrelated to a condition that divisions of circles are “Dirichlet”. If a vertex of aDirichlet tessellation is odd–valent, then the alternative sum of angles formed byedges emanating this vertex is positive. But when a tessellation contains even–valent vertices, the situation is not so simple.

If all regions of a tessellation R are rectangles, we can reduce our problem to aproblem on divisions of lines, and can obtain a condition that R to be Dirichlet.But in general, it seems not so easy to get a condition for convex tessellationscontaining even–valent vertices. In this paper we restrict ourselves on tessellations

2000 Mathematics Subject Classification Primary 52C20, Secondary 05B45.Key words and phrases: Dirichlet tessellation, quadrangles, sites.The second author is partially supported by Grant–in–Aid for Scientific Research (C) (No.

17540072), Japan Society for the Promotion of Sciences.

80 YUSUKE TAKEO and TOSHIAKI ADACHI

by congruent convex quadrangles which are also “locally congruent”, and study acondition that they are Dirichlet.

2. Tessellations by Trapezoids

In this paper we study tessellations all of whose regions are congruent to eachother. Among such tessellations those tessellations by congruent rectangles are thesimplest examples. When R be a tessellation by congruent rectangles we take anarbitrary point P0 in the interior of a region R0 ∈ R. If R ∈ R and R0 are adjacent ,that is R∩R0 6= ∅, we choose a point PR ∈ R so that it is the reflection point of P0

with respect to the line containing R0 ∩R. We can easily see that we can continuesuch an operation to all rectangles. Thus we find R is Dirichlet and

{PR

∣∣ R ∈ R}is a set of sites.

Figure 1. Tessellation by Congruent Rectangles

Next to tessellations by congruent rectangles we can consider tessellations bycongruent parallelograms and tessellations by congruent trapezoids. In this sectionwe study these tessellations. A tessellation R of a plane R2 by congruent convexquadrangles is said to be a tessellation of type (2, 2) if edges of each R ∈ R areclassified into two types, which are called point–type and line–type, and satisfies thefollowing conditions:

(i) If two edges are adjacent, they are different type.

(ii) If two quadrangles R,R′ ∈ R are adjacent, then the edge R∩R′ being regardedas an edge of R and that being regarded as an edge of R′ are of the same type.

(iii) If two quadrangles R,R′ ∈ R are adjacent by the edge R ∩ R′ of point–type,then they are symmetric with respect to the midpoint of this edge.

(iv) If two quadrangles R,R′ ∈ R are adjacent by the edge R ∩ R′ of line–type,then they are symmetric with respect to the line containing this edge.

Lemma 1. If a tessellation R of a plane is of type (2, 2), then each of its regionis congruent to a trapezoid.

Proof. Suppose each region of R is congruent to a quadrangle R0 = �ABCD ∈ Rwhich is not a trapezoid. We may suppose its angles satisfy ∠A + ∠B > π and∠B + ∠C 6= π.

When the edge AB is of point–type, hence the edge AD is of line–type, we takea quadrangle R′0 = �AB′C′D which is the reflection of R0 with respect to the lineAD. We see the edge AB′ is of point–type. Let R−1 be a quadrangle which issymmetric to R0 with respect to the midpoint of the edge AB, and let R′−1 be aquadrangle which is symmetric to R′0 with respect to the midpoint of the edge AB′.

DIRICHLET TESSELLATIONS OF A PLANE 81

Since both of the edges AB and AB′ are of point–type, we see R−1 6= R′−1. Thusby the assumption ∠A + ∠B > π, we find the interior of the set R−1 ∩ R′−1 is notempty, which is a contradiction.

Next we consider the case that the edge AB is of line–type. As ∠B +∠C 6= π byour assumption, we find either ∠B + ∠C > π or ∠A + ∠D > π holds. By applyingthe above argument to the edge BC or to the edge AD, which are of point–type,we find a contradiction. Thus we get our conclusion. �

Lemma 2. If R is a tessellation of type (2, 2), then parallel edges of each regionare of line–type

Proof. When each region is congruent to a parallelogram, our assertion is trivial.So we suppose each region of R is congruent to a trapezoid �ABCD whose edgesAD and BC are parallel and whose angles satisfy ∠B + ∠C 6= π. If the edge AB isof line–type, then we find a contradiction by the same argument as in the proof ofLemma 1, hence get our conclusion. �

By these lemmas tessellations of type (2, 2) are like the following figures. Thusone can see that such a tessellation is a generalization of tessellations by rectangles.

A

B C

D(L)

(P)(P)

(L)

A

B C

D(L)

(P)(P)

(L)

Figure 2. Tessellation of Type (2, 2)

We now study when a tessellation of type (2, 2) is Dirichlet.

Theorem 1. Let R be a tessellation of type (2, 2) each of whose regions are congru-ent to a trapezoid �ABCD. We suppose edges AD, BC are parallel and the lengthb of BC is not shorter than that of AD. If we denote by h the distance betweenthese parallel edges, then R is Dirichlet if and only if this trapezoid is a rectangleor it satisfies ∠B < π/2, ∠C < π/2 and

2b < h

(1

sin∠B cos∠B+

1sin∠C cos∠C

). (2.1)

Proof. When the trapezoid K = �ABCD is a rectangle, it is obvious that Ris Dirichlet. We suppose K satisfies the conditions in the assertion. We takeperpendicular bisectors `1, `2 for edges AB and DC, respectively. Denoting by Qi

the crossing point of `i and the line BC, we find the length of the segment BQ1

is h/(2 sin∠B cos∠B) and the length of the segment Q2C is h/(2 sin∠C cos∠C).Thus the condition (2.1) guarantees that `1, `2 meet in the interior of K. We denoteby P ∈ K the crossing point of `1 and `2, and take a point PR ∈ R for each region


R ∈ R which corresponds to P. If R,R′ ∈ R are adjacent by the edges of line–type, it is clear that the line containing R ∩R′ is the perpendicular bisector of thesegment PRPR′ . If R,R′ ∈ R are adjacent by edges of point–type, as PR,PR′ lie onthe perpendicular bisector of the edge R ∩R′, we also find that the line containingR ∩R′ is the perpendicular bisector of the segment PRPR′ . Thus {PR | R ∈ R} isthe set of sites for R, hence R is a Dirichlet tessellation.

On the other hand, we study the case that R is a Dirichlet tessellation. As thelength of AD is not longer than that of BC, we may suppose ∠B 5 π/2. If wesuppose ∠C > π/2, we consider a band obtained by regions in R two of whichare adjacent by edges of point–type. We label each trapezoid in this band as Ri

is adjacent with Ri−1 and Ri+1 like in the Figure 3. We take a point P0 in theinterior of R0. We choose a point P1 ∈ R1 so that the line containing R0∩R1 is theperpendicular bisector of the segment P0P1. If P1 remains in the interior of R1, as∠C > π/2, we see the distance between the point P1 and the base line containingthe edge BC is shorter than the distance between the point P0 and the base line.In this case we choose a point P2 ∈ R2 so that the line containing R1 ∩ R2 is theperpendicular bisector of the segment P1P2. Since the vertical edges in the bandare classified into two classes, in view of distances of these points from the baseline, we find this operation stops in finite steps. This shows that the tessellation ofthis band does not have sites as a Volonoi diagram, which is a contradiction. Thuswe find ∠B 5 π/2 and ∠C 5 π/2.

When ∠B = ∠C = π/2 we see K is a rectangle. When ∠B < π/2, ∠C = π/2 or∠B = π/2, ∠C < π/2, along the same lines as above we find a band obtained by Kdoes not have sites. Hence we have ∠B < π/2, ∠C < π/2 if K is not a rectangle.

Finally we shall show that the trapezoid satisfies the inequality (2.1). Let�BAC′D′ and �A′B′DC be trapezoid which are symmetric to K in R2 with respectto the midpoints of the edges AB and CD, respectively. Since ∠C′AB = ∠BAQ1,we find the intersection of �ABCD and the reflection of �BAC′D′ with respect tothe line AB is the triangle 4Q1AB. Similarly, we see the intersection of �ABCDand the reflection of �A′B′DC with respect to the line CD is the triangle 4Q2CD.Thus if R is Dirichlet, then a site in each region should be in the interior of thedomain corresponding to M = 4Q1AB ∩4Q2CD ∩K. Since the interior of M isnot empty if and only if the inequality (2.1) holds and since P ∈ M if M 6= ∅, weget our conclusion. �

RR R R R3210-1

P P P0 1 2

R R R RR-1 0 1 2 3

Figure 3. A Band of Congruent Trapezoids

Remark 1. Since the length of the edge AD is b−h(cot∠B+cot∠C), the trapezoid�ABCD of a Dirichlet tessellation of type (2, 2) in Theorem 1 satisfies

h(cot∠B + cot∠C) < b and ∠B + ∠C > π/2.


A tessellation R of a plane R2 by congruent convex quadrangles is said to be atessellation of type (3, 1) if each R ∈ R has three edges of point–type and an edgeof line–type, and satisfies the following conditions:

(i) If two quadrangles R,R′ ∈ R are adjacent, then the edge R∩R′ being regardedas an edge of R and that being regarded as an edge of R′ are of the same type.

(ii) If two quadrangles R,R′ ∈ R are adjacent by the edge R ∩ R′ of point–type,then they are symmetric with respect to the midpoint of this edge.

(iii) If two quadrangles R,R′ ∈ R are adjacent by the edge R ∩ R′ of line–type,then they are symmetric with respect to the line containing this edge.

Lemma 3. If R is a tessellation of type (3, 1), then each of its region is congruentto a trapezoid. If an edge is of line–type, it is one of parallel edges.

Proof. We take a region R0 = �ABCD ∈ R. We may suppose the edge AD is ofline–type, hence AB,DC are of point-type. We consider the quadrangle �AB′C′Dwhich is the reflection of R0 with respect to the line AD. As R is of type (3, 1),the edges AB′,DC′ are of point-type. If we suppose the vertex D is 3–valent in R,as the edge CD is of point–type, we find ∠C + 2∠D = 2π and ∠B = ∠D. Theseleads us to ∠A = 0, which is a contradiction. Applying the same argument on thevertex A, we find the valences of both of vertices A,D are greater than 3 in R. Wehence obtain 2∠A + 2∠B 5 2π and 2∠C + 2∠D 5 2π. Thus we find equalities holdin these inequalities, and AD,BC are parallel. We hence get the conclusion. �

A D

B C

D A

P

PP

P

* *

*#

(L)

(P)

(P)

(P)

A D

B C

(L) (L)

(L)

(P) (P)

(P)

(P)

(P)

(P)


We shall show that tessellations of type (3, 1) are essentially not Dirichlet.

Theorem 2. Let R be a tessellation of type (3, 1) each of whose regions are con-gruent to a trapezoid �ABCD. We suppose edges AD,BC are parallel. If R isDirichlet, then this trapezoid satisfies ∠B = ∠C (> π/4) and hence R can beregarded of type (2, 2).

Proof. We take a region R0 = �ABCD ∈ R. We may suppose the length of ADis not longer than that of BC. Applying the same argument as in the proof ofTheorem 1 to a band, we find ∠B 5 π/2, ∠C 5 π/2. In order to show our theoremwe suppose ∠B > ∠C.


We consider the case BC is of point–type. We take a region R? = A?CBD?,which is symmetric to R0 with respect to the midpoint of the edge BC. We choosea point P in the interior of R0, and take points P?, P∗ which are reflections of Pwith respect to the lines BC, CD, respectively. We also take a point P] which is areflection of P? with respect to the line CA?. Since ∠C < ∠B (= ∠BCA?), we findP∗ and P] are not reflections each other with respect to the line BC

(see Figure 4

),

which is a contradiction.For the case AD is of point–type, by the same argument we find a contradiction.

Thus we obtain that ∠B = ∠C (5 π/2). �

Here we consider where the site exists in each region of a Dirichlet tessellationof type (3, 1). We take a band like the Figure 5, and label each region so that Ri isadjacent with Ri+1 and Ri−1 in this band. We denote R0 = �ABCD where edgesAD, BC are parallel and ∠B = ∠C 5 π/2. We denote by Q1, Q2 the intersectionof the line BC and the perpendicular bisectors of edges AB, CD, respectively. Aswe see in the proof of Theorem 1, considering three regions R−1, R0, R1 we see thesite of R0 should lies in the interior of the set M = 4Q1AB ∩Q2CD ∩R0.

A D

B CQQ2 1

P0

1R

R1

'

-1R

A D

B C

P0

1

1

0-1

12 QQ

R RR

R'

Figure 5. 2∠A–Rotation((−2∠C)–rotation

)Let Pi ∈ Ri be the intersection of the perpendicular bisectors of the edges

Ri−1 ∩ Ri and Ri ∩ Ri+1. We denote by R′i+1 the reflection of Ri+1 with respectto the line containing Ri ∩ Ri+1. Since ∠B = ∠C, if we rotate Ri for 2∠A anti–clockwisely centered at Pi, we find it coincides with R′i+1, though their faces aredifferent.

Let ϕ : R2 → R2 be a 2∠A–rotation centered at P0. We call two pointedsets (Gi,Pi), i = 1, 2, in a plane R2 congruent if there is a motion f of R2 withf(G1) = G2 and f(P1) = P2. We denote by h the distance between two paralleledges AD, BC, and by b the length of the base edge BC, which is not shorter thanthe length of the edge AD. For a Dirichlet tessellation R of type (3, 1) we findby Theorems 1, 2 and Remark 1 that its fundamental isosceles trapezoid �ABCDsatisfies −2h cot∠A < b < −2h/ sin 2∠A, in particular, it satisfies (π/2 <)∠A <3π/4. For such positive h, b,∠A, we set

r(b, h,∠A) =

{(2h+ b sin 2∠A)/(4 sin2∠A), if h/ sin∠A ≤ b < −2h/ sin 2∠A,(b+ h cot∠A)/(2 sin∠A), if −2h cot∠A < b < h/ sin∠A.

We can then conclude the following:

Theorem 3. Let R be a Dirichlet tessellation of type (3, 1).


(1) If each region is a rectangle, all the point of the interior of given region can bea site of this region.

(2) If each region is not a rectangle, a site of a region R0 ∈ R lies in the interiorof the set M =

⋂∞i=−∞ ϕi(M) =

⋂∞i=−∞ ϕi(R0). On the contrary every point

in the interior of this set can be a site of R0 for R.

(3) When ∠A = (p/q)π for some relatively prime positive integers p, q (2p >

q, 3q > 4p), if we take a set of sites{

Pi

}for a band

{Ri

}, then these pairs

(Ri, Pi) satisfies the following:

(a) (Ri, Pi) is congruent to (Ri+q, Pi+q) when q is even,

(b) (Ri, Pi) is congruent to (Ri+2q, Pi+2q) when q is odd.

(4) In this case that ∠A = (p/q)π with relatively prime positive integers p, q (2p >q, 3q > 4p), the set M is as follows:

(a) When p is even, it is a q-regular polygon. The length of each edge of thispolygon is 2r(b, h,∠A) tan

(π/q

).

(b) When p is odd and one of the following conditions holds

(i)1− cos(π/q) cos∠A

sin∠A(cos(π/q)− cos∠A

)h ≤ b < −2h/ sin 2∠A,

(ii) −2h cot∠A < b ≤ cos(π/q)− cos∠Asin∠A

(1− cos(π/q) cos∠A

)h,

it is also a q–regular polygon. The length of each edge of this polygon is2r(b, h,∠A) tan

(π/q

).

(c) When p is odd and

cos(π/q)− cos∠Asin∠A


)h < b <1− cos(π/q) cos∠A

sin∠A(

cos(π/q)− cos∠A)h

holds, then it is a 2q–polygon. All the distances between P0 and verticesof this polygon M are the same. It is

12{

(h2 + b2 sin2∠A)(1 + cos2∠A− 2 cos(π/q) cos∠A

)− 2bh sin∠A

(cos(π/q)(1 + cos2∠A)− 2 cos∠A

)}1/2

×(

sin(π/q))−1(sin∠A)−2.

This polygon has two kinds of edges which appear alternatively. Theselengths are

b sin∠A(1− cos(π/q) cos∠A

)− h(

cos(π/q)− cos∠A)

sin(π/q) sin2∠A.

andh(1− cos(π/q) cos∠A

)− b sin∠A

(cos(π/q)− cos∠A

)sin(π/q) sin2∠A

(5) When ∠A is not in the case of (3), the set M is a ball centered at P0 of radiusr(b, h,∠A).


Proof. The first three assertions are obvious (see Figure 5). In order to see otherassertions, for R0 = �ABCD ∈ R we denote by H the foot of perpendicular of theline BC passing through P0, and by H1,H2 the mid points of the edges AB andCD, respectively. We put θ = ∠B = π −∠A. By Theorems 1, 2 and Remark 1, wesee 2h cot θ < b < 2h/ sin 2θ.

We need to study the behaviors of edges AB,BC,CD under the action of 2∠A–rotation ϕ centered at P0. We first compare the length P0H of P0H and the lengthsP0H1 = P0H2 of P0H1 and P0H2. As we have ∠HP0Q1 = π/2− ∠P0Q1H = θ, wefind

P0H = HQ1 cot θ =(BQ1 − BH

)cot θ =

(h

sin 2θ− b

2

)cot θ,

P0H1 = H1Q1 − P0Q1 =h

2 cos θ−(

h

sin 2θ− b

2

)1

sin θ=

b

2 sin θ− h cot θ

2 sin θ.

Thus we obtain that P0H ≤ P0H1 when h/ sin θ ≤ b < 2h/ sin 2θ, and P0H >P0H1 when 2h cot θ < b < h/ sin θ. If ∠A/π is not rational, we see the segmentϕj(edge AB) is not parallel to the edge AB and the segment ϕj(edge BC) is notparallel to the edge BC for an arbitrary integer j. Thus we obtain the fifth assertion.

Next we consider the case ∠A = (p/q)π. Generally, if a line ` does not passthrough P0, the distinct q lines `, ϕ(`), . . . , ϕq−1(`) form a q-regular polygon whoselengths of edges are 2 tan(π/q)–times of the distance between P0 and `. We regardthis regular polygon as a union of q isosceles triangles whose lengths of isosceles are1/ cos(π/q)–times of the distance between P0 and ` (see Figure 6).

We here consider the behaviors of lines AB,BC and CD under the action ofϕ and study the intersection M′ of q–regular polygons obtained by each of theselines. As the angle between two vectors

−→AB and

−→CD is 2∠A, we see ϕ(line AB) =

line CD. On the other hand, as the angle between two vectors−→AB and

−→BC is ∠A,

we see the angle between two vectors ϕj(−→AB) and

−→BC is (2j − 1)∠A. Therefore,

when p is even hence q is odd, we find that two lines ϕ(q+1)/2(line AB) and BCare parallel and P0 lies on the same side with respect to these lines. Thus weonly need to compare P0H and P0H1. But when q is even hence p is odd, linesϕj(line AB) (j = 1, . . . , q − 1) are not parallel to the line BC. And when p andq are odd, though ϕ(q+1)/2(line AB) and BC are parallel, P0 lies on the oppositeside with respect to these lines. Thus, in these two cases we have to compareP0H, P0H/ cos(π/q), P0H1 and P0H1/ cos(π/q).

When p is even and P0H ≤ P0H1, or when p is odd and P0H/ cos(π/q) ≤ P0H1,the setM′ is a q–regular polygon formed by lines BC, ϕ(BC), . . . , ϕq−1(BC) whoseedges have lengths 2P0H tan(π/q). When p is even and P0H1 ≤ P0H, or when p isodd and P0H1/ cos(π/q) ≤ P0H, the setM′ is a q–regular polygon formed by linesAB, ϕ(AB), . . . , ϕq−1(AB) whose edges have lengths 2P0H1 tan(π/q). When p isodd and P0H1 cos(π/q) < P0H < P0H1/ cos(π/q), the set M′ is the intersection oftwo q–regular polygons formed by BC, . . . , ϕq−1(BC) and AB, . . . , ϕq−1(AB). Weset F, G, K like Figure 7. In this Figure H′1 is ϕj0(H1) for some j0, the line FK is theline BC, the line KG is ϕj0(line AB), and ∠FP0H = ∠H′1P0H = ∠GP0H′1 = π/q.


P0

Figure 6

P

HHF

K

G

0

1'

Figure 7

Since we haveP0H = P0K cos∠HP0K,

P0H′1 = P0K cos∠H′1P0K = P0H cos(π/q) + P0K sin(π/q) sin∠HP0K,

we obtain

P0K2

sin2(π/q) = P0K2

sin2(π/q)(cos2∠HP0K + sin2∠HP0K

)= P0H

2sin2(π/q) +

(P0H1 − P0H cos(π/q)

)2= P0H

2+ P0H1

2 − 2 P0H× P0H1 cos(π/q).

We also find thatHK sin(π/q) = P0K sin(π/q) sin∠HP0K = P0H1 − P0H cos(π/q),

H′1K sin(π/q) = P0H− P0H1 cos(π/q).

The lengths of edges of the 2q–polygon are 2HK and 2H′1K, and distances betweenP0 and its vertices are P0K. Here if we substitute the lengths of P0H and P0H1,we find the conditions P0H1 cos(π/q) < P0H and P0H < P0H1/ cos(π/q) turn to

cos(π/q) + cos θsin θ

(1 + cos(π/q) cos θ

)h < b and b <1− cos(π/q) cos θ

sin θ(

cos(π/q) + cos θ)h,

and obtain

4P0K2

sin2(π/q) sin4 θ = (h2 + b2 sin2 θ)(1 + cos2 θ + 2 cos(π/q) cos θ

)− 2bh sin θ

(2 cos θ + cos(π/q)(1 + cos2 θ)

),

2HK sin(π/q) sin2 θ = b sin θ(1 + cos(π/q) cos θ

)− h(

cos(π/q) + cos θ),

2H′1K sin(π/q) sin2 θ = h(1 + cos(π/q) cos θ

)− b sin θ

(cos(π/q) + cos θ

).

In the last stage, we have to show that the domainM′ coincides with the domainM obtained by the segments AB, BC and CD under the action of ϕ. We draw aline which is parallel to the line BC and passes through P0. We denote by T thecrossing point of this line and the segment AB. Since ∠H1P0H = ∠A, we find

∠H1P0T = ∠A− π/2 = (2p− q)π/(2q),which is not less than π/q except the case q = 2p− 1. Hence if q 6= 2p− 1, we haveBH1 > P0H1 tan(π/q).


We next consider the case q = 2p− 1. We take a point S on the line AB so thatit satisfies the following conditions:

(i) Two points B and S lie on the same side with respect to H1,

(ii) ∠H1P0S = π/q.When q = 3 and p = 2, it is obvious that the point S lies on the segment BH1 ifand only if P0H1 ≤ P0H. When q ≥ 5, we have

SP0 cos∠SP0H = SP0 sin(π/(2q)

)= P0H1

sin(π/(2q)

)cos(π/q)

< P0H1 cos(π/q).

Hence we find that SP0 cos∠SP0H < P0H if P0H1 ≤ P0H/ cos(π/q). Thus S lieson the segment BH1 under this assumption. In any case, we may consider the lineAB instead of the segment AB. We can say the same thing for segments BC andCD. We hence obtain the fourth assertion. �

Remark 2. The proof of Theorem 3 shows the position of M in a fundamentaltrapezoid R0 of a Dirichlet tessellation R of type (3, 1).(1) When ∠A = (p/q)π with even p, the position of M in R0 is as follows:

(a) In the case h/ sin∠A < b < −2h/ sin 2∠A, the edge BC contains an edgeof M, and M∩AB = ∅, M∩ CD = ∅;

(b) In the case b = h/ sin∠A, each of edges AB, BC, CD contains an edgeof M;

(c) In the case −2h cot∠A < b < h/ sin∠A, each of the edges AB and CDcontains an edge of M, and M∩ BC = ∅.

(2) When ∠A = (p/q)π with odd p, the position of M in R0 is as follows:(a) In the case

1− cos(π/q) cos∠Asin∠A


)h < b < − 2hsin 2∠A

,

the edge BC contains an edge of M, and M∩AB = ∅, M∩ CD = ∅;(b) In the case(

1− cos(π/q) cos∠A)h = sin∠A


)b,

the edge BC contains an edge of M and each of the edges AB, CDcontains a vertex of M;

(c) In the case

cos(π/q)− cos∠Asin∠A


)h < b <1− cos(π/q) cos∠A

sin∠A(

cos(π/q)− cos∠A)h,

each of edges AB, BC, CD contains an edge of M;(d) In the case(

cos(π/q)− cos∠A)h = sin∠A


)b,

each of the edges AB, CD contains an edge of M and the edge BCcontains a vertex of M;


(e) In the case

−2h cot∠A < b <cos(π/q)− cos∠A

sin∠A(1− cos(π/q) cos∠A

)h,each of the edges AB,CD contains an edge of M and M∩ BC = ∅.

(3) When ∠A/π is not rational, it satisfies the following:(a) In the case h/ sin∠A < b < −2h/ sin 2∠A, the edge BC contains a point

of M, and M∩AB = ∅, M∩ CD = ∅;(b) In the case b = h/ sin∠A, each of edges AB,BC,CD contains a point ofM;

(c) In the case −2h cot∠A < b < h/ sin∠A, each of the edges AB and CDcontains a point of M, and M∩ BC = ∅.

In all the cases, we have M∩AD = ∅.

Remark 3. In Theorem 3 (4), (5) and Remark 2, we find some cases do not occurcorresponding to the angle ∠A.(1) When ∠A > (2/3)π, the case −2h cot∠A < b < h/ sin∠A does not occur.

(2) When ∠A = (p/q)π(< (2/3)π

)with an odd integer p and the inequality

2 cos(π/q)(cos∠A)2 − cos∠A ≥ cos(π/q) holds, the case −2h cot∠A < b ≤(cos(π/q)− cos∠A

)h/{

sin∠A(1− cos(π/q) cos∠A

)}does not occur.

(3) When ∠A = (p/q)π(< (2/3)π

)with an odd integer p and the inequality

2(cos∠A)2 − cos(π/q) cos∠A ≥ 1 holds, the case −2h cot∠A < b ≤(1 −

cos(π/q) cos∠A)h/{

sin∠A(

cos(π/q)− cos∠A)}

does not occur.

We give some figures of bands and of the domains of sites for Dirichlet tessella-tions of type (3, 1).

Example 1. (1) When ∠A = 2π/3, we see the set M is a regular triangle. Anedge ofM is contained in the base edge BC and other edges of a fundamentaltrapezoid do not have an intersection withM. The point (Ri, Pi) is congruentto (Ri+6, Pi+6) for arbitrary i (see Figure 8).

(2) When ∠A = 4π/7, we see the set M is a 7-regular polygon and (Ri, Pi) iscongruent to (Ri+14, Pi+14) (see Figures 9, 10, 11, 12).

R0R R R R R R R65431 2-1

Figure 8. Sites and the Region of Sites for a Tessellation of Type(3, 1) with ∠A = 2π/3


A D

B C

Figure 9. The Region of Sites for a Tessellation of Type (3, 1)with ∠A = 4π/7 and P0H < P0H1.

A D

B C

Figure 10. The Region of Sites for a Tessellation of Type (3, 1)with ∠A = 4π/7 and P0H1 < P0H.

A D

B C

Figure 11. The Region of Sites for a Tessellation of Type (3, 1)with ∠A = 4π/7 and P0H = P0H1.


RR R R R R R R R R R R R R R R0 1-1 2 3 4 5 6 7 8 9 1 1 11 1 140 2 3

Figure 12. Sites for a Tessellation of Type (3, 1) with ∠A = 4π/7.

Example 2.

(1) When ∠A = 3π/5, we see the set M is either a 5–regular polygon or a 10–polygon. The pointed set (Ri, Pi) is congruent to (Ri+10, Pi+10)

(see Figures

13, 14).

(2) When ∠A = 5π/7, as we have 2 cos2(2π/7) + cos(π/7) cos(2π/7) > 1, we seethe set M is a 7-regular polygon (see Figure 15). The pointed set (Ri, Pi) iscongruent to (Ri+14, Pi+14).

(3) When ∠A = 5π/8, we see the set M is either an 8–regular polygon or a 16–polygon. The pointed set (Ri, Pi) is congruent to (Ri+8, Pi+8)

(see Figures

17, 18).

(4) When ∠A = 7π/10, as we have 2 cos2(3π/10) + cos(π/10) cos(3π/10) > 1, wesee the setM is a 10–regular polygon

(see Figure 16

). The pointed set (Ri, Pi)

is congruent to (Ri+10, Pi+10).

A D

CB

A D

CB

A D

CB

Figure 13. Three Types of Regions of Sites for Tessellations ofType (3, 1) with ∠A = 3π/5.

RR R R R R R R R R10987654321RR 0-1

Figure 14. Sites for a Tessellation of Type (3, 1) with ∠A = 3π/5


A D

B C

Figure 15. The Re-gion of Sites for a Tes-sellation of Type (3, 1)with ∠A = 5π/7

A D

B C

Figure 16. The Re-gion of Sites for Tes-sellations of Type (3, 1)with ∠A = 3π/10.

A

B

D

C

A D

B C

A D

C D

Figure 17. Three Types of Regions of Sites for Tessellations ofType (3, 1) with ∠A = 5π/8.

RR R R R R R R R R0-1 1 2 3 4 5 6 7 8

Figure 18. Sites for a Tessellation of Type (3, 1) with ∠A = 5π/8

Next we study the existence region of sites for general Dirichlet tessellations oftype (2, 2). We take a band like the Figure 20 and put R0 = �ABCD, where edgesAD, BC are parallel and ∠B < π/2, ∠C < π/2. We employ the notations givenabove for a band in a Dirichlet tessellation of type (3, 1). We denote by R′i+1∪R′i+2

the reflection of Ri+1 ∪Ri+2 with respect to the line containing Ri ∩Ri+1, and byR′′i+2 the reflection of R′i+2 with respect to the line containing R′i+1 ∩ R′i+2 (seeFigure 19). Through this operation the intersection Pi+2 of the perpendicularbisectors is mapped onto Pi. Moreover we can see that the image of Ri through a2(∠A + ∠D)-rotation centered at Pi coincides with R′′i+2.


Let M denote the set 4Q1AB∩4Q2CD∩R0. We take the set M1 in R1 whichcorresponds to M , denote by M ′1 its reflection with respect to the line containingR0 ∩ R1, and set N = M ∩M ′1 (see Figure 20). Let ψ00, ψ01 : R2 → R2 be thereflection with respect to perpendicular bisector of the edge BC and the 2∠D-rotation centered at P0, respectively. If we define a motion ψ0 by ψ0 = ψ01 ◦ ψ00,we see R′1 = ψ0

(R0

), M ′1 = ψ0(M). Considering four regions R−1, R0, R1, R2, we

see a site of R0 should lies in the interior of N .

D

CB

A

R

RR

R

R

R

R1

1

0

2

2

3

3

'''

"

"

Figure 19. 2(∠A + ∠D)–Rotation

If we denote by ψ : R2 → R2 the 2(∠A +∠D)–rotation centered at P0, we find thefollowing:

Proposition 1. Let R be a Dirichlet tessellation of type (2, 2) whose regions arenot rectangles.(1) A site of a region R0 ∈ R lies in the interior of the set N =

⋂∞i=−∞ ψi(N) =⋂∞

i=−∞ ψi(R0 ∩R′1

).

(2) On the contrary, every point in the interior of N can be a site of R0 for R.

(3) When ∠A +∠D = (p/q)π for some relatively prime positive integers p, q (p >q, 2p < 3q), if we take a set of sites

{Pi

}for a band

{Ri

}, then (Ri, Pi) is

congruent to (Ri+2q, Pi+2q).

We give some figures of bands and of the regions of sites for Dirichlet tessellationsof type (2, 2).

Example 3.

(1) When ∠A = 17π/24, ∠D = 15π/24, we have ∠A + ∠D = 4π/3. In this casewe see (Ri, Pi) is congruent to (Ri+6, Pi+6) (see Figure 20).

(2) When ∠A = 13π/20, ∠D = 3π/4, we have ∠A + ∠D = 7π/5. In this case wesee (Ri, Pi) is congruent to (Ri+10, Pi+10) (see Figures 21, 22).

(3) When ∠A = 5π/8, ∠D = 13π/24, we have ∠A + ∠D = 7π/6. In this case wesee (Ri, Pi) is congruent to (Ri+12, Pi+12)

(see Figures 23, 24

).


A D

B CQQ

P

12

0

A D

B C

RR R R R R R R0 1-1 2 3 4 5 6

Figure 20. Tessellation of Type (2, 2) with ∠A = 17π/24 and∠D = 15π/24

R R R R R R R R R R RR 0 011 12 3 4 5 6 7 8 9-

Figure 21. Sites for a Tessellation of Type (2, 2) with ∠A =13π/20 and ∠D = 3π/4

A tessellation R of a plane R2 by congruent convex quadrangles is said to be atessellation of type (1, 3) if each R ∈ R has three edges of line–type and an edge ofpoint-type, and satisfies the following conditions:

(i) Every vertex is not 3-valent in R.

(ii) If two quadrangles R, R′ ∈ R are adjacent, then the edge R∩R′ being regardedas an edge of R and that being regarded as an edge of R′ are of the same type.

(iii) If two quadrangles R,R′ ∈ R are adjacent by the edge R ∩ R′ of point–type,then they are symmetric with respect to the midpoint of this edge.

(iv) If two quadrangles R,R′ ∈ R are adjacent by the edge R ∩ R′ of line–type,then they are symmetric with respect to the line containing this edge.


A D

BC

Figure 22. The Re-gion of Sites for a Es-sellation of Type (2, 2)with ∠A = 13π/20 and∠D = 3π/4

Figure 23. The Re-gion of Sites for a Tes-sellation of Type (2, 2)with ∠A = 5π/8 and∠D = 13π/24

RR R R R R R R R R R R R R0 1-1 2 3 4 5 6 7 8 9 10 11 12

Figure 24. Sites for a Tessellation of Type (2, 2) with ∠A = 5π/8and ∠D = 13π/24

Lemma 4. If R is a tessellation of type (1, 3), then each of its region is congruentto a trapezoid. It has an edge of line–type jammed by right angles. If it is not arectangle, its parallel two edges are also of line–type.

Proof. We take a region R0 = �ABCD ∈ R. If the edge AB is of point–type, thenthe edge AD is of line–type. We consider a quadrangle R′0 = �AB′C′D which is thereflection of R0 with respect to the line AD and a quadrangle R−1 = �ABD†C†

which is symmetric to R0 with respect to the midpoint of the edge AB. As the edgeAC† should be of line–type, we take a quadrangle R′−1 = �AB′′D′′C† which is thereflection of R−1 with respect to the line AC†. Since the vertex A is not 3–valentin R, we see R′0 6= R′−1, hence the interior of R′0 ∩R′−1 should be empty. Thus weobtain ∠A + ∠B 5 π.

We now show that R0 is a trapezoid. We suppose it is not a trapezoid. In thiscase we may suppose ∠C + ∠D > π and ∠D > π/2. The above argument shows


that the edge CD is of line–type. If we suppose the edge AD is of line–type, weconsider the quadrangle R′0 = �AB′C′D. Since the vertex D is not 3–valent, whenthe edge DC′ is also of line–type, it should hold that 4∠D 5 2π, and when DC′ isof point–type, it should hold that 3∠D +∠C 5 2π. These are contradiction, hencethe edge AD is of point type. Thus we have ∠A +∠D < π as we suppose R0 is nota trapezoid. We hence find ∠B +∠C > π. We apply the argument on the edge CDto the edge BC of line–type. Since the edges AB and CD are of line–type, we find∠B ≤ π/2, ∠C ≤ π/2, which is a contradiction by the same way. Thus we find R0

is a trapezoid.Next we study the angles of R0. We may suppose the edges AD,BC are parallel,

the edge CD is of line–type, and ∠D = π/2. When AD is of line–type, aboveargument shows that ∠D = π/2 and hence ∠C = π/2. When AD is of point–type,considering our tessellation at the vertex D we see ∠A + ∠D 5 π. Thus we find∠A 5 π/2 hence ∠B = π/2. Since the edges AB, BC are of line–type in this case,applying the above argument at the vertex B, we get ∠B = π/2 = ∠A.

We finally study parallel edges of R0 when it is not a rectangle. We may supposeAD, BC are parallel and ∠C = ∠D = π/2, ∠A > π/2. If we suppose AD is of point–type, considering at the vertex D we see 2∠A+2∠D 5 2π, which is a contradiction.If we suppose BC is of point-type, we take the trapezoid R? = �A?CBD? whichis symmetric to R0 with respect to the midpoint of the edge BC. Since A?C,A?D?

should be of line-type, considering at the vertex A? we see either 4∠A 5 2π or3∠A+∠B 5 2π holds, which are also contradiction. Thus we get our conclusion. �

A D

B C

(L)

(L)

(L)

(P)


Theorem 4. Every tessellation R of type (1, 3) is Dirichlet.

Proof. We suppose each region of R is congruent to a trapezoid K = �ABCDwhose edges AD,BC are parallel and of line-type, and whose angles satisfy ∠C =∠D = π/2. We take a point P which is in the interior of K and which lies on theperpendicular bisector for the edge AB. If we take a point PR ∈ R for each regionR ∈ R which corresponds to P, we can easily check that

{PR

∣∣ R ∈ R} is the setof sites for R by just the same way as in the proof of Theorem 1. �

Remark 4. Let R be a tessellation of R2 of type (1, 3) each of whose region iscongruent to a trapezoid R0 = �ABCD ∈ R like the Figure 25. We suppose R0 isnot a rectangle. We denote by Q the intersection of the perpendicular bisector `for the edge AB and the line BC. One can easily see the following:


(1) A site of R0 lies in the interior of 4QAB ∩R0

(see Figure 26

).

(2) Every point in the interior of 4QAB ∩R0 can be a site of R0 for R.Moreover, if we denote by (R,PR) a pair of region and its site as a Volonoi diagraminduced by R, we find the following:(1) If PR0 lies on `, all such pointed regions are congruent each other.

(2) If it does not lie on `, there are two congruence classes.

(3) If (R,PR) and (R′,PR′) are not congruent and R,R′ are adjacent to eachother, then they are adjacent by edges of point-type.

QC

DA

B Q

A D

B C

Figure 26. The Regions of Sites for Tessellations of Type (1, 3)

We shall call a tessellation R of a plane R2 by congruent convex quadranglesa tessellation of type (0, 4) if it satisfies the following condition: If two regionsR,R′ ∈ R are adjacent, they are symmetric with respect to the line containing theedge R ∩R′. We now suppose R = �ABCD is a region of a tessellation R of type(0, 4). It is clear that the vertex V (= A,B,C,D) is nV–valent (nV ≥ 3) if and onlyif the angle at V satisfies ∠V = 2π/nV. Hence the valencies of vertices only dependon the quadrangle. Moreover, if the vertex A is of odd–valent, then we see thelength of edges containing A satisfy AB = AD and the angles at adjacent verticessatisfy ∠B = ∠D.

Lemma 5. If R is a tessellation of type (0, 4), then each of its region is one ofthe following:

(i) a rectangle,

(ii) a rhombus formed by two congruent regular triangles,

(iii) a quadrangle formed by two congruent rectangular triangles whose other anglesare π/6, π/3.

Proof. We note that the angle at a vertex of R = �ABCD ∈ R is greater thanπ/2 if and only if it is 3–valent. Therefore, if all vertices of R are even valent, wesee R is a rectangle.

We consider the case that two adjacent vertices, say A, B, are odd valent. Thenwe have ∠A = ∠C, ∠B = ∠D and hence 2/nA+2/nB = 1. As we have 2(nA+nB) =nAnB, it is a contradiction because the right hand side is odd.

Next we study the case that one vertex, say A, is odd valent and the other areeven valent. Since the angle at an even valent vertex is not greater than π/2, we seeA is 3–valent. Hence it holds that 2/nB + 1/nC = 2/3. As we have 2(nB − 3)nC =


3nB and nC ≥ 4, we find 3nB ≥ 8(nB − 3). As nB ≥ 4, this lead us to nB = 4and nC = 6. Since AB = AD, we find R is formed by two congruent rectangulartriangles whose other angles are π/6, π/3 (see Figure 27).

Finally we study the case two vertices which are not adjacent, say A, C, are oddvalent and other vertices are even valent. In this case we have 1/nA+2/nB+1/nC =1. If we denote as nB = 2m, we see m(nA +nC) = (m−1)nAnC. As nA +nC is evenand nAnC is odd, we see m is odd, hence m ≥ 3. On the other hand, as we have{(m− 1)nA −m}nC = mnA and nC ≥ 3, we see mnA ≥ 3{(m− 1)nA −m}, whichis equivalent to 3m ≥ (2m − 3)nA. Since nA ≥ 3, we have 3m ≥ 3(2m − 3) andhence m = 3. Thus we have 3(nA +nC) = 2nAnC. Again as nC ≥ 3 we find nA = 3and hence nC = 3. Therefore R is formed by two congruent regular triangles (seeFigure 28). �

Figure 27. Rhombus Type Figure 28.Non–Trapezoid Type

As we can easily see in Figures 27, 28, we obtain the following.

Theorem 5. Every tessellation of a plane of type (0, 4) is Dirichlet.

We here also study a tessellation R of a plane R2 by congruent convex quadran-gles each of whose regions have two edges of line–type and two edges of point–typeand satisfies the following conditions:

(i) Every vertex is 4-valent in R.

(ii) Same type edges are adjacent.

(iii) If two quadrangles R,R′ ∈ R are adjacent, then the edge R∩R′ being regardedas an edge of R and that being regarded as an edge of R′ are of the same type.

(iv) If two quadrangles R,R′ ∈ R are adjacent by edges of point-type, then theyare symmetric with respect to the midpoint of the edge R ∩R′.

(v) If two quadrangles R,R′ ∈ R are adjacent by edges of line–type, then they aresymmetric with respect to the line containing the edge R ∩R′.

Lemma 6. If R is a tessellation of a plane mentioned as above, then each regionis a rectangle.

Proof. We take a region R0 = �ABCD ∈ R. We may suppose AD, CD are ofline–type. We consider quadrangles R′0 = �AB′C′D, R∗ = �A∗B∗CD which arethe reflections of R0 with respect to the lines AD and CD, respectively.


If one of the edges DC′, DA∗ is of line–type, we have ∠D = π/2 by the firstcondition. When both of these edges are of line–type, as the edges AB, BC, AB′ andCB∗ are of point–type, we find by the first condition that ∠A +∠B = ∠B +∠C =π. Thus we see R0 is a rectangle. When the edge DC′ is of point–type, we see∠C = π/2. Since the edges BC, CB∗ are of point–type, we have ∠B +∠C = π andR0 is a rectangle.

If both of these edges DC′, DA∗ are of point–type, by the first condition we see3∠D + ∠A = 2π and ∠A = ∠C. By the second condition we find the edge AB′ isof line–type. Since the edge AB is of point-type, we get 3∠A + ∠B = 2π, hence∠A = ∠D. Thus we see R0 is a rectangle also in this case. �

We should note that there exist many other tessellations of a plane by congruenttrapezoids which are not of types mentioned in this section. For example, thetessellation by congruent trapezoids with right angle which is given in Figure 29is Dirichlet. The tessellation by congruent trapezoids given as Figure 30 and thetessellation by congruent rhombus given as Figure 31 are not Dirichlet.

Figure 29 Figure 30

A B

A' B'

A B" "

Figure 31

3. Tessellation by Quadrangles Inscribed About Circles

In the previous section except a tessellation in Lemma 5 we studied tessellationsgiven by congruent trapezoids. In this section we study tessellations by congruentconvex quadrangles which are not trapezoid. A tessellation R of a plane R2 bycongruent convex quadrangles is said to be a tessellation of type (4, 0) if two adjacentquadrangles R,R′ ∈ R are symmetric with respect to the midpoint of the edgeR ∩R′. It is clear that each vertex is 4–valent in R. A tessellation of type (4, 0) islike the following Figure 32.

A

B C

DA

B CD



In order to study Dirichlet tessellations of type (4, 0), we here recall a localproperty of Dirichlet tessellations of a plane. Let R be a convex tessellation of aplane. At each vertex x of this tessellation we denote by θ1, θ2, . . . , θn the anglesaround x formed by edges emanating from x. Here n is the valency of x and we settheir indices so that they increase in the anti–clockwise direction.

When R is Dirichlet, we take the set of sites{

PR}. If R,R′ are adjacent, theline containing R ∩ R′ is the perpendicular bisector of the segment PRPR′ . Thusone can easily see the following (c.f.[1]):

Lemma 7. If R is a Dirichle tessellation of a plane, then at each vertex x thealternative sum of angles around x satisfies

n∑i=1

(−1)i−1θi =

{> 0, when x is odd valent,= 0, when x is even valent.

By use of this lemma we find a necessary and sufficient condition that tessellationof type (4, 0) is Dirichlet.

Theorem 6. A tessellation R of type (4, 0) is Dirichlet if and only if each regionR ∈ R is inscribed about some circle whose center lies in the interior of R.

Proof. We suppose each region R is inscribed about some circle whose center PR

lies in the interior of R. Since all regions are congruent each other, those circleshave the same radius. Therefore, if R,R′ are adjacent, denoting the edge R ∩ R′by AB we see the quadrangle APRBPR′ is a rhombus. Hence the line containingthe edge R ∩R′ is the perpendicular bisector of the segment PRPR′ . We find R isDirichlet.

Next we suppose R is Dirichlet. We take a region R0 = �ABCD ∈ R. ByLemma 7 and Figure 32, we see ∠A − ∠B + ∠C − ∠D = 0. Thus we obtain∠A + ∠C = ∠B + ∠D = π and find that R0 is inscribed about a circle.

We now study the center T of the circumscribed circle of R0. If we supposeit does not lie in the interior of R0, all vertices of R0 lie on a half-circle of thiscircumscribed circle. Therefore, if the edge BC is the nearest edge from T amongedges of R0, then ∠B < π/2, ∠C < π/2. Let Q1,Q2 be the intersection of theline BC and the perpendicular bisectors of edges AB,CD, respectively. Since thesebisectors meet at T, we find the interior of the set4Q1AB∩4Q2CD is an empty set.This is a contradiction, because the site of R0 as a region of Dirichlet tessellationlies in this interior. We hence obtain that T lies in the interior of R0. �

We here give another Dirichlet tessellation of a plane which is formed by con-gruent quadrangles and all of whose vertices are 4–valent (see Figure 33). It isconstructed by regular rectangles and by right crosses. This does not satisfy ourtype–conditions, but each quadrangles are inscribed about circles. Each quadranglehas two right angles. Its two edges which jam a right angle have the same length.

At the end of this paper, we should note that there are many other Dirichlettessellations by congruent convex quadrangles containing non–4–valent vertices.The tessellation given in Figure 35 is constructed by regular triangles and by rightthree forked edges.


A

B C

D

Figure 33. AB = AD, ∠C = π/2. Its site lies inside the gray domain.

Figure 34Figure 35

References

[1] P. Ash and E.D. Bolker, Recognizing Dirichlet tessellations, Geom. Dedicata19 (1985), 175–206.

[2] G.I. Grima and A. Marquez, Computational Geometry on Surfaces, KluwerAcad. Pbul. 2001, Dordrecht, Boston and London.

[3] B. Grunbaum and G.C. Shephard, Tilings with congruent tiles, Bull. A.M.S.3 (1980), 951–973.

Yusuke Takeo

Division of Mathematics and Mathematical ScienceDepartment of Computer Science and Engineering

Graduate School of Engineering, Nagoya Institute

of TechnologyGokiso

Nagoya, 466-8555

JAPAN

[email protected]

Toshiaki AdachiDepartment of MathematicsNagoya Institute of TechnologyNagoya 466-8555

JAPAN

[email protected]

dirichlet tessellations of a plane by congruent...

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