Characterization of the Two-Dimensional Five-Fold Translative Tiles
Qi Yang Chuanming Zong
math.MG 52C22


Characterization of the Two-Dimensional

Five-Fold Translative Tiles

Qi Yang and Chuanming Zong




12.5cm
Abstract. In 1885, Fedorov discovered that a convex domain can form a lattice tiling of the Euclidean plane if and only if it is a parallelogram or a centrally symmetric hexagon. It is known that there is no other convex domain which can form a two-, three- or four-fold translative tiling in the Euclidean plane, but there are centrally symmetric convex octagons and decagons which can form five-fold translative tilings. This paper characterizes all the convex domains which can form five-fold translative tilings of the Euclidean plane, which consist of two classes of octagons and one class of decagons.


2010 Mathematics Subject Classification: 52C20, 52C22, 05B45, 52C17, 51M20




1. Introduction

In 1885, Fedorov <cit.> proved that a convex domain can form a lattice tiling in the plane if and only if it is a parallelogram or a centrally symmetric hexagon; a convex body can form a lattice tiling in the space if and only if it is a parallelotope, an hexagonal prism, a rhombic dodecahedron, an elongated dodecahedron, or a truncated octahedron. As a generalized inverse problem of Fedorov's discovery, in 1900 Hilbert <cit.> listed the following question in the second part of his 18th problem: Whether polyhedra also exist which do not appear as fundamental regions of groups of motions, by means of which nevertheless by a suitable juxtaposition of congruent copies a complete filling up of all space is possible. Try to verify Hilbert's problem in the plane, in 1917 Bieberbach suggested Reinhardt (see <cit.>) to determine all the two-dimensional convex tiles. However, to complete the list turns out to be challenging and dramatic. Over the years, the list has been successively extended by Reinhardt, Kershner, James, Rice, Stein, Mann, McLoud-Mann and Von Derau (see <cit.>), its completeness has been mistakenly announced several times! In 2017, M. Rao <cit.> announced a completeness proof based on computer checks.

Let K be a convex body with (relative) interior int(K) and (relative) boundary ∂ (K), and let X be a discrete set, both in 𝔼^n. We call K+X a translative tiling of 𝔼^n and call K a translative tile if K+X=𝔼^n and the translates int(K)+ x_i are pairwise disjoint. In other words, if K+X is both a packing and a covering in 𝔼^n. In particular, we call K+Λ a lattice tiling of 𝔼^n and call K a lattice tile if Λ is an n-dimensional lattice. Apparently, a translative tile must be a convex polytope. Usually, a lattice tile is called a parallelohedron.

As one can predict that to determine the parallelohedra in higher dimensions is complicated. Through the works of Delone <cit.>, Štogrin <cit.> and Engel <cit.>, we know that there are exact 52 combinatorially different types of parallelohedra in 𝔼^4. A computer classification for the five-dimensional parallelohedra was announced by Dutour Sikirić, Garber, Schürmann and Waldmann <cit.> only in 2015.

Let Λ be an n-dimensional lattice. The Dirichlet-Voronoi cell of Λ is defined by

    C={ x:  x∈𝔼^n,  x,  o≤ x, Λ},

where X, Y denotes the Euclidean distance between X and Y. Clearly, C+Λ is a lattice tiling and the Dirichlet-Voronoi cell C is a parallelohedron. In 1908, Voronoi <cit.> made a conjecture that every parallelohedron is a linear transformation image of the Dirichlet-Voronoi cell of a suitable lattice. In 𝔼^2, 𝔼^3 and 𝔼^4, this conjecture was confirmed by Delone <cit.> in 1929. In higher dimensions, it is still open.

To characterize the translative tiles is another fascinating problem. At the first glance, translative tilings should be more complicated than lattice tilings. However, the dramatic story had a happy end! It was shown by Minkowski <cit.> in 1897 that every translative tile must be centrally symmetric. In 1954, Venkov <cit.> proved that every translative tile must be a lattice tile (parallelohedron) (see <cit.> for generalizations). Later, a new proof for this beautiful result was independently discovered by McMullen <cit.>.

Let X be a discrete multiset in 𝔼^n and let k be a positive integer. We call K+X a k-fold translative tiling of 𝔼^n and call K a translative k-tile if every point x∈𝔼^n belongs to at least k translates of K in K+X and every point x∈𝔼^n belongs to at most k translates of int(K) in int(K)+X. In other words, K+X is both a k-fold packing and a k-fold covering in 𝔼^n. In particular, we call K+Λ a k-fold lattice tiling of 𝔼^n and call K a lattice k-tile if Λ is an n-dimensional lattice. Apparently, a translative k-tile must be a convex polytope. In fact, similar to Minkowski's characterization, it was shown by Gravin, Robins and Shiryaev <cit.> that a translative k-tile must be a centrally symmetric polytope with centrally symmetric facets.

Multiple tilings was first investigated by Furtwängler <cit.> in 1936 as a generalization of Minkowski's conjecture on cube tilings. Let C denote the n-dimensional unit cube. Furtwängler made a conjecture that every k-fold lattice tiling C+Λ has twin cubes. In other words, every multiple lattice tiling C+Λ has two cubes sharing a whole facet. In the same paper, he proved the two- and three-dimensional cases. Unfortunately, when n≥ 4, this beautiful conjecture was disproved by Hajós <cit.> in 1941. In 1979, Robinson <cit.> determined all the integer pairs { n,k} for which Furtwängler's conjecture is false. We refer to Zong <cit.> for an introduction account and a detailed account on this fascinating problem, respectively, to pages 82-84 of Gruber and Lekkerkerker <cit.> for some generalizations.

Let P denote an n-dimensional centrally symmetric convex polytope, let τ (P) be the smallest integer k such that P can form a k-fold translative tiling in 𝔼^n, and let τ^* (P) be the smallest integer k such that P can form a k-fold lattice tiling in 𝔼^n. For convenience, we define τ (P)=∞ if P can not form translative tiling of any multiplicity. Clearly, for every centrally symmetric convex polytope we have

    τ (P)≤τ^*(P).


In 1994, Bolle <cit.> proved that every centrally symmetric lattice polygon is a lattice multiple tile. However, little is known about the multiplicity. Let Λ denote the two-dimensional integer lattice, and let P_8 denote the octagon with vertices (1,0), (2,0), (3,1), (3,2), (2,3), (1,3), (0,2) and (0,1). As a particular example of Bolle's theorem, it was discovered by Gravin, Robins and Shiryaev <cit.> that P_8+Λ is a seven-fold lattice tiling of 𝔼^2.

In 2000, Kolountzakis <cit.> proved that, if D is a two-dimensional convex domain which is not a parallelogram and D+X is a multiple tiling in 𝔼^2, then X must be a finite union of translated two-dimensional lattices. In 2013, a similar result in 𝔼^3 was discovered by Gravin, Kolountzakis, Robins and Shiryaev <cit.>.


In 2017, Yang and Zong <cit.> studied the multiplicity of the multiple translative tilings by proving the following results: Besides parallelograms and centrally symmetric hexagons, there is no other convex domain which can form a two-, three- or four-fold translative tiling in the Euclidean plane. However, there are particular octagons and decagons which can form five-fold translative tilings. Meanwhile, Zong <cit.> characterized all the two-dimensional five-fold lattice tiles.

This paper characterizes all the two-dimensional five-fold translative tiles by proving the following theorem.

Theorem 1. A convex domain can form a five-fold translative tiling of the Euclidean plane if and only if it is a parallelogram, a centrally symmetric hexagon, a centrally symmetric octagon (under a suitable affine linear transformation) with vertices
v_1=( 3 2-5α 4, -2), v_2=( -1 2-5α 4, -2), v_3=( α 4-3 2, 0), v_4=( α 4-3 2, 1), v_5=- v_1, v_6=- v_2, v_7=- v_3 and v_8=- v_4, where 0<α <2 3, or with vertices v_1=(2-β, -3), v_2=(-β, -3), v_3=(-2, -1), v_4=(-2, 1), v_5=- v_1, v_6=- v_2, v_7=- v_3 and v_8=- v_4, where 0<β≤ 1, or a centrally symmetric decagon with 𝐮_1=(0,1), 𝐮_2=(1,1), 𝐮_3=(3 2, 1 2), 𝐮_4=(3 2,0), 𝐮_5=(1,-1 2), 𝐮_6=-𝐮_1, 𝐮_7=-𝐮_2, 𝐮_8=-𝐮_3, 𝐮_9=-𝐮_4 and 𝐮_10=-𝐮_5 as the middle points of its edges.


Remark 1. It was shown by Zong <cit.> that a convex domain can form a five-fold lattice tiling of the Euclidean plane if and only if it is a parallelogram, a centrally symmetric hexagon, a centrally symmetric octagon (under a suitable affine linear transformation) with vertices v_1=(-α , -3 2), v_2=(1-α , -3 2), v_3=(1+α , -1 2), v_4=(1-α , 1 2), v_5=- v_1, v_6=- v_2, v_7=- v_3 and v_8=- v_4, where 0<α <1 4, or with vertices
v_1=(β , -2), v_2=(1+β , -2), v_3=(1-β , 0), v_4=(β , 1), v_5=- v_1, v_6=- v_2, v_7=- v_3, v_8=- v_4, where 1 4<β <1 3, or a centrally symmetric decagon (under a suitable affine linear transformation) with u_1=(0, 1), u_2=(1, 1), u_3=(3 2, 1 2), u_4=(3 2, 0), u_5=( 1,-1 2), u_6=- u_1, u_7=- u_2, u_8=- u_3, u_9=- u_4 and u_10=- u_5 as the middle points of its edges. In fact, all the two-dimensional convex five-fold translative tiles are five-fold lattice tiles. They take different representations just for the proof purpose.




2. Preparation

Let P_2m denote a centrally symmetric convex 2m-gon centered at the origin, let v_1, v_2, …, v_2m be the 2m vertices of P_2m enumerated in the clock order, and let G_1, G_2, …, G_2m be the 2m edges, where G_i is ended by v_i and v_i+1. For convenience, we write

    V={ v_1,  v_2, …,  v_2m}

and

    Γ={G_1, G_2, …, G_2m}.


Assume that P_2m+X is a τ (P_2m)-fold translative tiling in 𝔼^2, where X={ x_1,  x_2,  x_3, …} is a discrete multiset with x_1= o. Now, let us observe the local structure of P_2m+X at the vertices v∈ V+X.

Let X^ v denote the subset of X consisting of all points x_i such that

    v∈∂ (P_2m)+ x_i.

Since P_2m+X is a multiple tiling, the set X^ v can be divided into disjoint subsets X^ v_1, X^ v_2, … , X^ v_t such that the translates in P_2m+X^ v_j can be re-enumerated as P_2m+ x^j_1, P_2m+ x^j_2, …, P_2m+ x^j_s_j satisfying the following conditions:

1. v∈∂ (P_2m)+ x^j_i holds for all i=1, 2, …, s_j.

2. Let ∠^j_i denote the inner angle of P_2m+ x^j_i at v with two half-line edges L^j_i,1 and L^j_i,2, where L^j_i,1, x^j_i- v and L^j_i,2 are in clock order. Then, the inner angles join properly as

    L^j_i,2=L^j_i+1,1

holds for all i=1, 2, … , s_j, where L^j_s_j+1,1=L^j_1,1.

For convenience, we call such a sequence P_2m+ x^j_1, P_2m+ x^j_2, …, P_2m+ x^j_s_j an adjacent wheel at v. It is easy to see that

    ∑_i=1^s_j∠^j_i =2w_j·π

hold for positive integers w_j. Then we define

    ϕ ( v)=∑_j=1^tw_j= 12π∑_j=1^t∑_i=1^s_j∠^j_i

and

    φ ( v)=♯{ x_i:  x_i∈ X,  v∈ int(P_2m)+ x_i}.


Clearly, if P_2m+X is a τ (P_2m)-fold translative tiling of 𝔼^2, then

    τ (P_2m)= φ ( v)+ϕ ( v) (1)

holds for all v∈ V+X.

To prove Theorem 1, we need the following known results.

Lemma 1 (Yang and Zong <cit.>). Assume that P_2m is a centrally symmetric convex 2m-gon centered at the origin and P_2m+X is a τ(P_2m)-fold translative tiling of the plane, where m≥4. If 𝐯∈ V+X is a vertex and G∈Γ+X is an edge with 𝐯 as one of its two ends, then there are at least ⌈(m-3)/2⌉ different translates P_2m+𝐱_i satisfying both

    𝐯∈∂(P_2m)+𝐱_i

and

    G\{𝐯}⊂ int(P_2m)+𝐱_i.



Lemma 2 (Yang and Zong <cit.>). Assume that P_2m is a centrally symmetric convex 2m-gon centered at the origin, P_2m+X is a translative multiple tiling of the plane, and 𝐯∈ V+X. Then we have

    ϕ ( v)=κ·m-1 2+ℓ·1 2,

where κ is a positive integer and ℓ is the number of the edges in Γ+X which take v as an interior point.


Lemma 3 (Yang and Zong <cit.>). When m≥ 6, we have

    τ (P_2m)≥ 6.



Lemma 4 (Bolle <cit.>). A convex polygon is a k-fold lattice tile for a lattice Λ and some positive integer k if and only if the following conditions are satisfied:

1. It is centrally symmetric.

2. When it is centered at the origin, in the relative interior of each edge G there is a point of 1 2Λ.

3. If the midpoint of G is not in 1 2Λ then G is a lattice vector of Λ.





3. Proof of Theorem 1

Lemma 5. A centrally symmetric convex octagon P_8 is a five-fold translative tile if and only if it is, under an affine linear transformation, one with vertices v_1=( 3 2-5α 4, -2), v_2=( -1 2-5α 4, -2), v_3=( α 4-3 2, 0), v_4=( α 4-3 2, 1), v_5=- v_1, v_6=- v_2, v_7=- v_3 and v_8=- v_4, where 0<α <2 3, or one with vertices
v_1=(2-β, -3), v_2=(-β, -3), v_3=(-2, -1), v_4=(-2, 1), v_5=- v_1, v_6=- v_2, v_7=- v_3 and v_8=- v_4, where 0<β≤ 1.

Proof. Suppose that X is a discrete subset of 𝔼^2 and P_8+X is a five-fold translative tiling of the plane. First of all, it follows from Lemma 1 that

    φ ( v)≥⌈4-3 2⌉ =1(2)

holds for all v∈ V+X.  On the other hand, by Lemma 2 we have

    ϕ ( v)=κ·3 2+ℓ·1 2,(3)

where κ is a positive integer and ℓ is a nonnegative integer. In fact ℓ is the number of the edges which take v as an interior point. Thus, to prove the lemma, it is sufficient to deal with the following four cases:

Case 1. ϕ(𝐯)≥ 5 holds for a  vertex 𝐯∈ V+X. It follows by (1) and (2) that

    τ(P_8)=φ(𝐯)+ϕ(𝐯)≥ 6,(4)

which contradicts the assumption that P_8+X is a five-fold translative tiling of the plane.

Case 2. ϕ(𝐯)=4 holds for a  vertex 𝐯∈ V+X. It follows by (3) that ℓ≠ 0 and therefore 𝐯∈ int(G) holds for some G∈Γ+X. Assume that v^*_1 and v^*_2 are the two ends of G. Applying Lemma 1 to { v^*_1, G} and { v^*_2, G}, respectively, one can deduce that

    φ(𝐯)≥ 2

and therefore

    τ(P_8)=φ(𝐯)+ϕ(𝐯)≥ 6,(5)

which contradicts the assumption.

Case 3. ϕ(𝐯)=3 holds for a  vertex 𝐯∈ V+X. Then (3) has and only has two groups of solutions
{κ , ℓ}={ 1, 3} or {2, 0}.

Subcase 3.1. {κ , ℓ}={ 1, 3}. Then, there are three edges G'_1, G'_2 and G'_3 in Γ +X satisfying

    v∈ int(G'_i),   i=1, 2, 3.

Next, we study the multiplicity by considering the relative positions of these edges.


Subcase 3.1.1. G'_1=G'_2=G'_3. Assume that v^*_1 and v^*_2 are the two ends of G'_1. Then X^ v^*_1 has two identical points. By computing the angle sum of all the adjacent wheels at v^*_1 it can be deduced that

    ϕ(𝐯_1^*)≥4.

Then, by Case 1 and Case 2 we get

    τ(P_8)=ϕ(𝐯_1^*)+φ(𝐯_1^*)≥6,(6)

which contradicts the assumption.

Subcase 3.1.2. G'_2=G'_3 and G'_1∦ G'_2. Then there are two adjacent wheels at 𝐯, one
 has five translates P_8+ x_1, P_8+ x_2, …, P_8+ x_5 and the other has two translates P_8+ x'_1 and P_8+ x'_2, as shown by Figure 1.



By re-enumeration we may assume that ∠_1, ∠_2, ∠_3 and ∠_4 are inner angles of P_8 and ∠_5=π, as shown by Figure 1. Guaranteed by linear transformation, we assume that the two edges G_1 and G_3 of P_8 are horizontal and vertical, respectively. Suppose that G_1'=G_1+𝐱_5. Let 𝐯_1^* and 𝐯_2^* be the two ends of G_1', let L denote the straight line determined by 𝐯_1^* and 𝐯_2^*, let G_3^* denote the edge of P_8+𝐱_4 lying on L with two ends 𝐯 and 𝐯_3^*, and let G_4^* denote the edge of P_8+𝐱_1 lying on L with two ends 𝐯 and 𝐯_4^*.

By Lemma 1, there is a 𝐱^*_3∈ X^𝐯_1^* such that 𝐯_3^*∈ int(P_8)+𝐱^*_3. Clearly, by the convexity of P_8, both 𝐯_3^* and 𝐯_1^* belong to int(P_8)+𝐱'_1. Thus, we have
𝐱^*_3≠𝐱'_1. Meanwhile, since both 𝐯_3^* and 𝐯_1^* belong to int(P_8)+𝐱_2, we have 𝐱_2≠𝐱^*_3 and therefore

    φ(𝐯_3^*)≥3.(7)

Then, the only chance to keep P_8+X a five-fold tiling is ϕ(𝐯_3^*)=2. Similarly, one can deduce

    ϕ(𝐯_1^*)=ϕ(𝐯_2^*)=ϕ(𝐯_3^*)=ϕ(𝐯_4^*)=2.(8)


By (3) it is easy to see that the local configuration of P_8+X^ v is essentially unique when ϕ ( v)=2. In other words, it is determined by the one that v is not its vertex. Consequently, the set X has four points y_1, y_2, y_3 and y_4 satisfying

    𝐯_1^*=𝐯_4+𝐲_1,  𝐯∈ int(P_8)+𝐲_1,(9)


    𝐯_2^*=𝐯_7+𝐲_2,  𝐯∈ int(P_8)+𝐲_2,(10)


    𝐯_3^*=𝐯_3+𝐲_3,  𝐯∈ int(P_8)+𝐲_3(11)

and

    𝐯_4^*=𝐯_8+𝐲_4,  𝐯∈ int(P_8)+𝐲_4.(12)


Clearly, by the convexity of P_8 we have 𝐲_1≠ 𝐲_2, 𝐲_1≠𝐲_3 and 𝐲_2≠𝐲_4. For convenience, we write 𝐯_i=(x_i,y_i). If 𝐲_2=𝐲_3, then by (10) and (11) we have

    y_3=y_7.(13)

If 𝐲_1=𝐲_4, then by (9) and (12) we get

    y_4=y_8.(14)

However, it is obvious that (13) and (14) can not hold simultaneously. Therefore, we still get

    φ(𝐯)≥3

and therefore

    τ(P_8)=φ(𝐯)+ϕ(𝐯)≥6,(15)

which contradicts the assumption.



Subcase 3.1.3. G'_1≠ G'_2 and G'_1∥ G'_2. Let 𝐯_1^* and 𝐯_2^* be the two ends of G'_1, and let 𝐯_3^* and 𝐯_4^* be the two ends of G'_2. Without loss of generality, we suppose that 𝐯_3^* is between 𝐯_1^* and 𝐯_2^*, as shown by Figure 2. By Lemma 1, X has three points y_1, y_2 and y_3 satisfying both

    v^*_i∈∂ (P_8)+ y_i,   i=1, 2, 3

and

    v∈ int(P_8)+ y_i,   i=1, 2, 3.

By the convexity of P_8 it is easy to see that these three points are pairwise distinct. Then, we get

    φ(𝐯)≥3

and therefore

    τ(P_8)=ϕ(𝐯)+φ(𝐯)≥6,(16)

which contradicts the assumption.

Subcase 3.1.4. G'_1∦ G'_2, G'_1∦ G'_3 and G'_2∦ G'_3. By studying the angle sum at 𝐯, it can be deduced that P_8+X^𝐯 is an adjacent wheel of seven translates.  Suppose that 𝐱_2∈ X^𝐯 and  G'_1 is an edge of P_8+𝐱_2. Since G'_1, G'_2 and G'_3 are mutually non-collinear, X^ v has two points
𝐱_1 and 𝐱_3 such that 𝐯 is a common vertex of both P_8+𝐱_1 and  P_8+𝐱_3, and P_8+𝐱_2 joins both P_8+𝐱_1 and P_8+𝐱_3 at non-singleton parts of G'_1, respectively. Let 𝐯^*_1 and 𝐯^*_2 be the two ends of G'_1, let L denote the straight line determined by 𝐯^*_1 and 𝐯^*_2, let G^*_1 denote the edge of P_8+𝐱_1 lying on L with ends 𝐯 and 𝐯^*_3, and let G^*_2 denote the edge of P_8+𝐱_3 lying on L with ends 𝐯 and 𝐯^*_4, as shown in Figure 3.



By studying the corresponding angles of the adjacent wheel at v, it is easy to see that P_8+X^𝐯 has exact two translates which contain both v^*_1 and v^*_3 as interior points. On the other hand, by Lemma 1, P_8+X^ v^*_3 has at least one more
translate which contains v^*_1 as an interior point. Thus, we have

    φ ( v^*_1)≥ 3.

Then, the only chance to keep P_8+X a five-fold tiling is ϕ(𝐯_1^*)=2. Similarly, one can deduce

    ϕ(𝐯_1^*)=ϕ(𝐯_2^*)=ϕ(𝐯_3^*)=ϕ(𝐯_4^*)=2.(17)

By repeating the argument between (8) and (15), it can be deduced that

    τ(P_8)=φ(𝐯)+ϕ(𝐯)≥6,(18)

which contradicts the assumption.


Subcase 3.2. {κ, ℓ}={2,0} holds at every vertex v∈ V+X. Then P_8+ X^𝐯 is an adjacent wheel of eight translates P_8+𝐱_1, P_8+𝐱_2, … , P_8+𝐱_8, as shown in Figure 4. Let 𝐯_i^* be the second vertex of P_8+ x_i connecting to 𝐯 by an edge. Since ϕ ( v)=3, every v^*_i is an interior point of exact two of these eight translates. Consequently, for every v^*_i, there are two different translates P_8+ y_i
and P_8+ y'_i in P_8+X^ v^*_i both contain v as an interior point.



On the other hand, it can be easily deduced that there is only one point x∈ X such that both v^*_1 and v^*_2 belong to
∂ (P_8)+ x and v∈ int(P_8)+ x. It is v^*_2- v+ x_1. Therefore, at least one of the two points y_2 and y'_2 is different from both y_1 and y'_1. Then, we get

    φ(𝐯)≥3

and

    τ(P_8)=φ(𝐯)+ϕ(𝐯)≥6,(19)

which contradicts the assumption.


As a conclusion of the previous cases, if P_8+X is a five-fold translative tiling, then ϕ(𝐯)=2 must hold at some 𝐯∈ V+X.


Case 4. ϕ(𝐯)=2 holds for a vertex 𝐯∈ V+X. It follows by (3) that ϕ ( v)=2 holds if and only if κ=1 and ℓ =1. In other words, P_8+X^𝐯 is an adjacent wheel of five translates. By re-enumeration we may assume that ∠_1, ∠_2, ∠_3 and ∠_4 are inner angles of P_8 and ∠_5=π, as shown  by Figure 5. Guaranteed by linear transformation, we assume that the edges G_1 and G_3 of P_8 are horizontal and vertical, respectively.



Let G_1^* denote the edge of P_8+𝐱_5 such that 𝐯∈ int(G_1^*) with two ends 𝐯_1^* and 𝐯_2^*, let L denote the straight line determined by 𝐯_1^* and 𝐯_2^*, let G_3^* denote the edge of P_8+𝐱_4 lying on L with ends 𝐯 and 𝐯_3^*, and let G_4^* denote the edge of  P_8+𝐱_1 lying on L with ends 𝐯 and 𝐯_4^*. If  ϕ(𝐯_1^*)=3 or  ϕ(𝐯_2^*)=3, by Subcase 3.1 we have τ(P_8)≥6, which contradicts the assumption that P_8+X is a five-fold translative tiling. Thus we have ϕ(𝐯_1^*)=2 and  ϕ(𝐯_2^*)=2. Similarly, we have

    ϕ(𝐯_1^*)=ϕ(𝐯_2^*)=ϕ(𝐯_3^*)=ϕ(𝐯_4^*)=2.(20)


Since the configuration of P_8+X^ v is essentially unique if ϕ ( v)=2, by considering the wheel structures at v, v^*_1, v^*_2,
v^*_3 and v^*_4, there are four points y_1, y_2, y_3 and y_4 in X satisfying

    𝐯_1^*=𝐯_4+𝐲_1,  𝐯∈ int(P_8)+𝐲_1,(21)


    𝐯_2^*=𝐯_7+𝐲_2,  𝐯∈ int(P_8)+𝐲_2,(22)


    𝐯_3^*=𝐯_3+𝐲_3,  𝐯∈ int(P_8)+𝐲_3(23)

and

    𝐯_4^*=𝐯_8+𝐲_4,  𝐯∈ int(P_8)+𝐲_4.(24)



By the convexity of P_8 it follows that 𝐲_1≠𝐲_2, 𝐲_1≠𝐲_3 and 𝐲_2≠𝐲_4. For convenience, we write 𝐯_i=(x_i,y_i). If 𝐲_1=𝐲_4, then by (21) and (24) we have

    y_4=y_8.(25)

If 𝐲_2=𝐲_3, then by (22) and (23) we have

    y_3=y_7.(26)

It is obvious that (25) and (26) can not hold simultaneously. Therefore, we have either y_1≠ y_4 or y_2≠ y_3.

On the other hand, since ϕ ( v)=2, the three inequalities 𝐲_3≠𝐲_4, 𝐲_2≠𝐲_3 and 𝐲_1≠𝐲_4 can not hold simultaneously. Otherwise, it can be deduced that

    φ( v)≥ 4

and therefore

    τ(P_8)=φ(𝐯)+ϕ(𝐯)≥ 6,(27)

which contradicts the assumption. Since 𝐲_1=𝐲_4 and 𝐲_2=𝐲_3 are symmetric, it is sufficient to deal with two subcases.

Subcase 4.1. 𝐲_2=𝐲_3.  Let 𝐯'_1 and 𝐯'_2 be the two vertices of P_8+𝐱_2 that adjacent to 𝐯, as shown in Figure 6.  By convexity it is easy to see that 𝐯'_1∈ int(P_8)+𝐲_4. Since 𝐲_2=𝐲_3, we have y_3=y_7. Then 𝐯'_1 is an interior point of P_8+𝐲_2 as well. Thus we get φ(𝐯'_1)≥ 3. If ϕ(𝐯'_1)≥ 3, then we have

    τ (P_8)=ϕ(𝐯'_1)+φ(𝐯'_1)≥ 6,(28)

which contradicts the assumption. Thus we must have ϕ(𝐯'_1)=2. By lemma 1, there is a point 𝐲_5∈ X^𝐯'_1 such that 𝐯∈ int(P_8)+𝐲_5.



Subcase 4.1.1. 𝐯'_1 is a vertex of P_8+𝐲_5. If 𝐯'_1 is a vertex of P_8+𝐲_1, as shown by Figure 6, then one can deduce that 𝐯'_2 is a vertex of P_8+𝐲_1. On the other hand, it follows by (24) that v'_2 is a vertex of P_8+ y_4 as well. Therefore, we have ϕ ( v'_2)≠ 2 and thus

    ϕ(𝐯'_2)≥ 3.(29)

Let 𝐯'_3 denote the vertex 𝐯_2+𝐲_1 of P_8+𝐲_1, as shown in Figure 6. By lemma 1, there is a point 𝐳∈ X^𝐯'_3 such that 𝐯'_2∈ int(P_8)+𝐳. Then it can be deduced that 𝐯'_3∈ int(P_8)+𝐱_4 and thus 𝐳≠𝐱_4. Since y_3=y_7, it can be shown that 𝐯'_3∉ P_8+𝐲_2 and therefore 𝐳≠𝐲_2. In addition, we have

    𝐯'_2∈(  int(P_8)+𝐱_4)∩(  int(P_8)+𝐲_2).

Thus we have

    φ(𝐯'_2)≥3(30)

and consequently

    τ(P_8)=φ(𝐯'_2)+ϕ(𝐯'_2)≥6,(31)

which contradicts the assumption.

If 𝐯'_1 is not a vertex of P_8+𝐲_1, remembering the subcase assumption, then we have 𝐲_1≠𝐲_5. In fact, in this case all
y_1, y_3, y_4 and y_5 are pairwise distinct. Thus we have

    φ(𝐯)≥ 4

and

    τ(P_8)=φ(𝐯)+ϕ ( v)≥ 6,(32)

which contradicts the assumption.

Subcase 4.1.2. 𝐯'_1 is an interior point of an edge of P_8+𝐲_5. It follows from the convexity of P_8 that 𝐯'_1 is an interior point of both P_8+𝐲_4 and P_8+𝐲_3. Therefore we have 𝐲_5∉{𝐲_3,𝐲_4}. If 𝐲_5≠𝐲_1, then all y_1, y_3, y_4 and y_5 are pairwise distinct. Thus we have

    φ(𝐯)≥4

and

    τ(P_8)=φ(𝐯)+ϕ ( v)≥6,(33)

which contradicts the assumption.

Thus, to avoid contradiction, we must have 𝐲_5=𝐲_1. Notice that 𝐯'_1 is an interior point of P_8+𝐱_5, and P_8+𝐲_1 has only two edges G_4+𝐲_1 and G_5+𝐲_1 which contain interior points of P_8+𝐱_5. Since ϕ(𝐯'_1)=2, by studying the structure of the adjacent wheel at 𝐯'_1, one can deduce that 𝐯'_1 must be an interior point of G_5+𝐲_1. Then we have

    y_5-y_4=y_4-y_3(34)

and

    y_3-y_2=2 (y_4-y_3).(35)



Let 𝐯_5^* and 𝐯_6^* be the two ends of  G_5+𝐲_1. Suppose that 𝐯_5^* is on the left of 𝐯'_1. By Lemma 1, there is a point 𝐲_6∈ X^𝐯_5^* such that 𝐯'_1∈ int(P_8) +𝐲_6.

It is obvious that 𝐯_5^* is an interior point of both P_8+𝐱_5 and P_8+𝐲_2. Thus we have 𝐲_6∉{𝐲_2,𝐱_5}. If 𝐯_5^* is not lying on the boundary of P_8+𝐲_4, then we have 𝐲_4≠𝐲_6. Consequently, all y_2, y_4, y_6 and x_5 are pairwise distinct. Thus we have

    φ(𝐯'_1)≥4

and

    τ(P_8)=φ(𝐯'_1)+ϕ ( v'_1)≥6,(36)

which contradicts the assumption.

To avoid the contradiction, the point 𝐯_5^* must belong to the boundary of P_8+𝐲_4. Furthermore, since the y-coordinate of 𝐯_5^* is equal to the y-coordinates of both 𝐯'_1 and 𝐯_3+𝐲_4, the point 𝐯_5^*
must be the vertex 𝐯_3+𝐲_4 of P_8+𝐲_4.

Let v denote the x-coordinate of v, and let w_1, w_2 and w_3 denote the x-coordinates of v_3+ y_4, v^*_1 and
v^*_5, respectively. First, by computing the x-coordinate of v^*_4 in two ways we get

    w_1+(x_7-x_6)+(x_6-x_5)+(x_5-x_4)=v+(x_6-x_5)

and thus

    w_1 =v-(x_7-x_6)-(x_5-x_4).(37)

On the other hand, since y_2= y_3, by computing the distance between v^*_3 and v^*_4 in two ways we get

    (x_7-x_6)+(x_6-x_5)+(x_5-x_4)+v-w_2=2(x_6-x_5)

and thus

    w_2=v+(x_7-x_6)-(x_6-x_5)+(x_5-x_4).

Since v^*_5 is the left vertex of G_5+ y_1, we get

    w_3=w_2+(x_5-x_4)=v+(x_7-x_6)-(x_6-x_5)+2(x_5-x_4).(38)

Then, v^*_5= v_3+ y_4 implies w_1=w_3 and

    2(x_7-x_6)+3(x_5-x_4)=x_6-x_5.(39)


In conclusion, recalling (34) and (35), a centrally symmetric octagon with G_1 horizontal, G_3 vertical and y_2= y_3 is a five-fold translative tile only if

    {[               y_5-y_4=y_4-y_3,;            y_3-y_2=2(y_4-y_3),; x_6-x_5=2(x_7-x_6)+3(x_5-x_4).                                ].(40)

Guaranteed by linear transformations, by choosing y_4-y_3=1, x_6-x_5=2 and x_5-x_4=α and keeping the symmetry in mind, one can deduce that the candidates are the octagons D_8(α) with vertices v_1=( 3 2-5α 4, -2), v_2=( -1 2-5α 4, -2), v_3=( α 4-3 2, 0), v_4=( α 4-3 2, 1), v_5=- v_1, v_6=- v_2, v_7=- v_3 and v_8=- v_4, where 0<α <2 3.

Let Λ (α) denote the lattice generated by 𝐮_1=(2,0) and 𝐮_2=(1+α 2, 1). It can be easily verified by Lemma 4 that D_8(α)+Λ (α) is indeed a five-fold tiling of 𝔼^2.

Subcase 4.2. 𝐲_3=𝐲_4. Then by (23) and (24) we have

    y_3=y_8(41)

and

    x_8-x_3=2(x_1-x_2).(42)




By Lemma 1, there is 𝐲_5∈ X^𝐯'_1 such that 𝐯∈ int(P_8)+𝐲_5. Since y_3=y_8, by convexity we have 𝐯'_1∈ int(P_8)+𝐲_3 and then 𝐲_5≠𝐲_3.

If ϕ(𝐯'_1)=3 and 𝐯'_1∈ int(G) holds for some G∈Γ+X, by Subcase 3.1 we have

    τ(P_8)=φ ( v'_1)+ϕ ( v'_1)≥6,(43)

which contradicts the assumption. Thus we have either ϕ(𝐯'_1)=2 or ϕ(𝐯'_1)=3 and 𝐯'_1 is a vertex of P_8+𝐱 for all 𝐱∈ X^𝐯'_1.

If both 𝐲_5≠𝐲_1 and 𝐲_5≠𝐲_2 hold simultaneously, then we have

    φ(𝐯)≥ 4

and therefore

    τ(P_8)=φ ( v)+ϕ ( v) ≥6,(44)

which contradicts the assumption. So we must have either 𝐲_5=𝐲_1 or 𝐲_5=𝐲_2.

Suppose that 𝐲_5=𝐲_1. If 𝐯'_1 is a vertex of P_8+𝐲_1, then we have

    y_5-y_4=y_4-y_3.(45)

If 𝐯'_1 is an interior point of an edge of P_8+𝐲_1, then by Subcase 3.1 we have ϕ(𝐯'_1)=2. By studying the structure of the adjacent wheel at 𝐯'_1, one can deduce that 𝐯'_1 must be an interior point of G_5+𝐲_1. Since G_5+𝐲_1 is horizontal, we also obtain (45).

In conclusion, recalling (41) and (42), a centrally symmetric octagon with G_1 horizontal, G_3 vertical and y_3= y_4 is a five-fold translative tile only if

    {[            y_3=y_8,;    y_5-y_4=y_4-y_3,; x_8-x_3=2(x_1-x_2).                     ].(46)

Guaranteed by linear transformation, by choosing y_4-y_3=2, x_1-x_2=2 and x_6=β and keeping symmetry in mind, one can deduce that the candidates are the octagons D_8(β) with vertices v_1=(2-β, -3), v_2=(-β, -3), v_3=(-2, -1), v_4=(-2, 1), v_5=- v_1, v_6=- v_2, v_7=- v_3 and v_8=- v_4, where 0<β≤ 1.

Let Λ (β ) denote the lattice generated by 𝐮_1=(2,0) and 𝐮_2=(1+β 2, 1). 
It can be easily verified by Lemma 4 that D_8(β)+Λ (β) is indeed a five-fold tiling of 𝔼^2.

Lemma 5 is  proved. 

Lemma 6. Let P_10 be a centrally symmetric decagon centered at the origin and let X be a discrete multiset of 𝔼^2. If P_10+X is a five-fold translative tiling of 𝔼^2, then it is  a five-fold lattice tiling of 𝔼^2.

Proof. Let v_1, v_2, …, v_10 denote the ten vertices of P_10 enumerated in the clock order, let G_i denote the edge with ends v_i and v_i+1, and let u_i denote the middle point of G_i. Suppose that X is a discrete subset of 𝔼^2 and P_10+X is a five-fold translative tiling of the plane. First of all, it follows from Lemma 1 that

    φ ( v)≥⌈5-3 2⌉ =1(47)

holds for every v∈ V+X.  On the other hand, by Lemma 2 we have

    ϕ ( v)=κ· 2+ℓ·1 2,(48)

where κ is a positive integer and ℓ is the number of the edges which contain v as a relative interior point.

Now we proceed to show that ϕ(𝐯)=2 holds for every vertex 𝐯∈ V+X by dealing with the following two cases.

Case 1. ℓ =0 holds for all vertices 𝐯∈ V+X. Then it follows by (48) that ϕ(𝐯) can take only two values, two or four.

If ϕ(𝐯)=4, the local arrangements P_10+X^ v can be divided into two adjacent wheels, each contains five translates.
Suppose that P_10+ x_1, P_10+ x_2, …, P_10+ x_5 is such a wheel at v and 𝐯=𝐯_k+𝐱_1. Then, the wheel can be determined by P_10+𝐱_1 explicitly as following:

    𝐯=𝐯_k+4+𝐱_2,   G_k+4+ x_2=G_k-1+𝐱_1,


    𝐯=𝐯_k+8+𝐱_3,   G_k+8+ x_3=G_k+3+𝐱_2,


    𝐯=𝐯_k+2+𝐱_4,   G_k+2+ x_4=G_k+7+𝐱_3,


    𝐯=𝐯_k+6+𝐱_5,   G_k+6+ x_5=G_k+1+𝐱_4,

where 𝐯_10+i=𝐯_i and G_10+i=G_i.



Without loss of generality, as shown by Figure 8, we take 𝐯=𝐯_1+𝐱_1, 𝐯_1^*=𝐯_2+𝐱_1 and 𝐯_2^*=𝐯_10+𝐱_1. By Lemma 1, for each v^*_i there is a y_i∈ X^ v^*_i such that

    v∈ int(P_10)+ y_i.

In fact, by the previous analysis, we have y_1= v^*_1- v_4 and therefore by convexity

    v^*_2∈ int(P_10)+ y_1.

Thus, the two points y_1 and y_2 are different. Then we have

    φ ( v)≥ 2

and

    τ (P_10)=φ ( v)+ϕ ( v)≥ 6,(49)

which contradicts the assumption.

This means that, in this case ϕ(𝐯)=2 must hold for all v∈ V+X.


Case 2. ℓ≠0 holds at a vertex 𝐯∈ V+X. In other words, there is an edge G∈Γ +X such that v∈ int(G). Clearly, by (48) we have ϕ ( v)≥ 3.

Suppose that 𝐯_1^* and 𝐯_2^* are the two ends of G. By Lemma 1, there are two different points 𝐱_1∈ X^𝐯_1^* and 𝐱_2∈ X^𝐯_2^* such that

    𝐯∈( int(P_10)+𝐱_1)∩( int(P_10)+𝐱_2).

Then we have φ(𝐯)≥2. If ϕ ( v)≥ 4, one can deduce that

    τ (P_10)=φ(𝐯)+ϕ(𝐯)≥6,(50)

which contradicts the assumption.

If ϕ(𝐯)=3, by (48) one can deduce that P_10+X^𝐯 consists of seven translates P_10+𝐱_1, P_10+𝐱_2, …, P_10+𝐱_7, and there is another G'∈Γ+X which contains 𝐯 as an interior point. Suppose that G is an edge of P_10+𝐱_6 and G' has two ends 𝐯_5^* and 𝐯_6^*. We deal with three subcases.

Subcase 2.1. G'||G and G'≠ G. Without loss of generality, we assume that 𝐯_5^* is between 𝐯_1^* and 𝐯_2^*. Then, by lemma 1 we have 𝐲_i∈ X^𝐯_i^* such that

    𝐯∈ int(P_10)+𝐲_i,    i=1, 2, 5.

It is obvious that y_1, y_2 and y_5 are pairwise distinct. Thus, we have
φ(𝐯)≥3 and therefore

    τ(P_10)=φ(𝐯)+ϕ(𝐯)≥6,(51)

which contradicts the assumption.

Subcase 2.2. G'= G. Then P_10+X^𝐯 can be divided into two adjacent wheels, as shown by Figure 9.



Let P_10+𝐱_6 and P_10+𝐱_7 be the two translates that contain G as a common edge. Without loss of generality, suppose that G=G_6+𝐱_7 and 𝐯=𝐯_7+𝐱_1, as shown in Figure 9. Let L be the straight line determined by 𝐯_1^* and 𝐯_2^*, let G_1^* be the edge of P_10+𝐱_1 lying on L with ends 𝐯 and 𝐯_3^*, and let G_2^* be the edge of P_10+𝐱_1 with ends 𝐯 and 𝐯_4^*.

It is easy to see that ϕ(𝐯_1^*)≥ 3 since v^*_1 is an interior point of G^*_1. In fact, to avoid contradiction, we must have ϕ(𝐯_1^*)=3 and the adjacent wheel at 𝐯_1^* can be divided into two adjacent wheels. Since 𝐯_1^*=𝐯_6+𝐱_7, by Lemma 1 and the structure of the adjacent wheel that consists of five translates, we have three points 𝐲_1, 𝐲_2, 𝐲_3∈ X^𝐯_1^* such that

    𝐯_1^*=𝐯_8+𝐲_1,  𝐯_3^*∈ int(P_10)+𝐲_1,(52)


    𝐯_1^*=𝐯_10+𝐲_2,  𝐯_3^*∈ int(P_10)+𝐲_2(53)

and

    𝐯_1^*=𝐯_4+𝐲_3,  𝐯∈ int(P_10)+𝐲_3.(54)


Clearly, we also have 𝐯_3^*∈ int(P_10)+𝐱_4. Since 𝐯_1^*∈ int(P_10)+𝐱_4, thus we have 𝐱_4∉{𝐲_1,𝐲_2}, φ(𝐯_3^*)≥3 and ϕ(𝐯_3^*)=2. By Lemma 1 and the structure of the adjacent wheel with five translates, there is a point y_4∈ X^𝐯_3^* such that

    𝐯_3^*=𝐯_4+𝐲_4,  𝐯∈ int(P_10)+𝐲_4.(55)

Furthermore, by Lemma 1 we have 𝐲_5∈ X^𝐯_4^* such that 𝐯∈ int(P_10)+𝐲_5.
By (54), (55) and convexity we have

    𝐯_4^*∈( int(P_10)+𝐲_3)∩( int(P_10)+𝐲_4),
 𝐲_5∉{𝐲_3,𝐲_4}, φ(𝐯)≥3 and thus

    τ(P_10)=φ(𝐯)+ϕ(𝐯)≥6,(56)

which contradicts the assumption.

Subcase 2.3. G'∦ G. Suppose that G is an edge of P_10+𝐱_6 with ends 𝐯^*_1 and 𝐯^*_2 which contains v as an interior point. Since G'∦ G, there is a translates P_10+𝐱' in X^𝐯 that meets P_10+𝐱_6 at a non-singleton part of G. Let L be the line determined by  𝐯^*_1 and 𝐯^*_2. Let G_1^* be the edge of P_10+𝐱' lying on L with ends 𝐯^*_3 and 𝐯. Since 𝐯^*_1∈ int(G_1^*), we have φ ( v^*_1)≥ 2 and therefore ϕ(𝐯^*_1)≤ 3. On the other hand, since ℓ≠ 0 at v^*_1, we must have ϕ ( v^*_1)≥ 3. Thus we get

    ϕ ( v^*_1)=3.(57)


Since G'∦ G, then P_10+X^𝐯 cannot be divided into smaller adjacent wheels. There are exact two corresponding inner angles of two translates in P_10+X^𝐯 at 𝐯 that are divided into two positive measure parts by G^*_1. Thus there are exact two translates in P_10+X^𝐯 that contain both 𝐯^*_3 and 𝐯^*_1 as  interior points. By Lemma 1, there is a translate P_10+𝐲 in P_10+X^𝐯_3^* that contains 𝐯^*_1 as an interior point and therefore φ(𝐯^*_1)≥3. Then, by (57) we get

    τ(P_10)=φ(𝐯^*_1)+ϕ(𝐯^*_1)≥6,(58)

which contradicts the assumption.

As a conclusion, we have proved that ϕ(𝐯)=2 must hold for all vertices 𝐯∈ V+X if P_10+X is a five-fold translative tiling.

Let P_10 be a centrally symmetric convex decagon centered at the origin with vertices 𝐯_1, 𝐯_2, …, 𝐯_10 enumerated in the anti-clock order. Let G_i denote the edge with ends 𝐯_i and 𝐯_i+1 and let u_i denote the middle point of G_i. Then, we define

    {[          a_1  = u_1- u_6,;          a_2  = u_2- u_7,;          a_3  = u_3- u_8,;          a_4  = u_4- u_9,;          a_5 = u_5- u_10. ].

According to Lemma 8 of Zong <cit.>, we have

    a_1- a_2+ a_3- a_4+ a_5= o.(59)


Assume that x_1= o∈ X. Since ϕ(𝐯)=2 holds for every vertex 𝐯∈ V+X, by studying the structure of the adjacent wheel at 𝐯 we have

    ∑_i=1^5 z_i𝐚_i∈ X,    z_i∈ℤ.

For convenience, we define

    Λ ={∑ z_i𝐚_i:  z_i∈ℤ, i=1,2,…,5}.(60)


Since ϕ(𝐯)=2 hold for all vertices, we have φ(𝐯)=3 for every vertex v as well. Suppose that the adjacent wheel at 𝐯_1 is P_10+𝐱_i, i=1, 2, …, 5. Let 𝐯_i^* be the common vertex of P_10+𝐱_i and P_10+𝐱_i+1 other than v_1 as shown by Figure 10, where 𝐱_6=𝐱_1 and 𝐱_1= o. By Lemma 1, we have 𝐲_i∈ X^𝐯_i^* such that 𝐯_1∈ int(P_10)+𝐲_i. In fact, it can be explicitly deduced that

    {[                            𝐲_1   = v^*_1- v_4= a_1- a_4+ a_5,;                            𝐲_2 = v^*_2- v_10=2 a_1- a_2+ a_5,;                            𝐲_3 = v^*_3- v_6=2 a_1-2 a_2+ a_3,;                            𝐲_4  = v^*_4- v_2=- a_2+ a_3- a_5,;                            𝐲_5  = v^*_5- v_8=- a_1+ a_2- a_5. ].(61)





By (61) and symmetry it can be shown that 𝐲_i≠𝐲_i+1, where 𝐲_6=𝐲_1. For example, if y_1= y_2 (as shown in Figure 10), then by symmetry we will get that (P_10+ x_2)∩ (P_10+ y_1) is a parallelogram and y_1= v^*_1- v_3, which contradicts the first equation of (61). Thus, any triple of {𝐲_1,𝐲_2,…,𝐲_5} cannot be identical. Restricted by φ(𝐯)=3, these points have to satisfy one of the following five groups of conditions: (i). 𝐲_1=𝐲_3 and 𝐲_2=𝐲_4; (ii). 𝐲_1=𝐲_3 and 𝐲_2=𝐲_5; (iii). 𝐲_1=𝐲_4 and 𝐲_2=𝐲_5; (iv). 𝐲_1=𝐲_4 and 𝐲_3=𝐲_5 and (v). 𝐲_2=𝐲_4 and 𝐲_3=𝐲_5.

Case (i). 𝐲_1=𝐲_3 and 𝐲_2=𝐲_4. Then, by (61) and (59) we get

    {[  a_1- a_4+ a_5=2 a_1-2 a_2+ a_3,; 2 a_1- a_2+ a_5=- a_2+ a_3- a_5,;      a_1- a_2+ a_3- a_4+ a_5= o,                                  ].(62)



    {[ 2 a_2-2 a_4+ a_5= a_1+( a_3- a_4),;       a_4-2 a_5=2 a_1-( a_3- a_4),;          a_2- a_5= a_1+( a_3- a_4)                                    ].

and therefore

    {[                a_1= a_1,; a_2=-2 a_1+4( a_3- a_4),; a_3=-4 a_1+6( a_3- a_4),; a_4=-4 a_1+5( a_3- a_4),; a_5=-3 a_1+3( a_3- a_4),                          ].(63)

which means that Λ is a lattice with a basis { a_1,  a_3- a_4}.
Furthermore, since u_i=1 2 a_i∈1 2Λ, it follows by Lemma 4 that
P_10+Λ is indeed a five-fold lattice tiling.


Case (ii). 𝐲_1=𝐲_3 and 𝐲_2=𝐲_5. Then, by (61) and (59) we have

    {[  a_1- a_4+ a_5=2 a_1-2 a_2+ a_3,; 2 a_1- a_2+ a_5=- a_1+ a_2- a_5,;      a_1- a_2+ a_3- a_4+ a_5= o,                                  ].(64)



    {[ a_1+ a_4+ a_5=- a_3+2( a_2+ a_5),;         3 a_1+4 a_5=2( a_2+ a_5),;  a_1- a_4+2 a_5=- a_3+( a_2+ a_5)                                   ].

and therefore

    {[  a_1=8 a_3-6( a_2+ a_5),;  a_2=6 a_3-4( a_2+ a_5),;                a_3= a_3,; a_4=-3 a_3+3( a_2+ a_5),; a_5=-6 a_3+5( a_2+ a_5),                          ].(65)

which means that Λ is a lattice with a basis { a_3,  a_2+ a_5}.
Furthermore, since u_i=1 2 a_i∈1 2Λ, it follows by Lemma 4 that
P_10+Λ is indeed a five-fold lattice tiling.


Case (iii). 𝐲_1=𝐲_4 and 𝐲_2=𝐲_5. Then, by (61) and (59) we get

    {[   a_1- a_4+ a_5=- a_2+ a_3- a_5,; 2 a_1- a_2+ a_5=- a_1+ a_2- a_5,;       a_1- a_2+ a_3- a_4+ a_5= o                                  ].(66)



    {[ 2 a_2- a_3+2 a_5=-( a_1- a_2)+ a_4,;            a_2+2 a_5=-3( a_1- a_2),;          a_3+ a_5=-( a_1- a_2)+ a_4                                     ].

and therefore

    {[  a_1=4 a_4+6( a_1- a_2),;  a_2=4 a_4+5( a_1- a_2),;  a_3=3 a_4+3( a_1- a_2),;                a_4= a_4,; a_5=-2 a_4-4( a_1- a_2),                          ].(67)

which means that Λ is a lattice with a basis { a_4,  a_1- a_2}.
Furthermore, since u_i=1 2 a_i∈1 2Λ, it follows by Lemma 4 that
P_10+Λ is indeed a five-fold lattice tiling.


Case (iv). 𝐲_1=𝐲_4 and 𝐲_3=𝐲_5. Then, by (61) and (59) we have

    {[    a_1- a_4+ a_5=- a_2+ a_3- a_5,; 2 a_1-2 a_2+ a_3=- a_1+ a_2- a_5,;       a_1- a_2+ a_3- a_4+ a_5= o,                                   ].(68)



    {[ a_2- a_4+ a_5= a_3-( a_1+ a_5),;  3 a_2+2 a_5= a_3+3( a_1+ a_5),;       a_2+ a_4= a_3+( a_1+ a_5)                                 ].

and therefore

    {[  a_1=4 a_3-5( a_1+ a_5),;  a_2=3 a_3-3( a_1+ a_5),;                a_3= a_3,; a_4=-2 a_3+4( a_1+ a_5),; a_5=-4 a_3+6( a_1+ a_5),                          ].(69)

which means that Λ is a lattice with a basis { a_3,  a_1+ a_5}.
Furthermore, since u_i=1 2 a_i∈1 2Λ, it follows by Lemma 4 that
P_10+Λ is indeed a five-fold lattice tiling.

Case (v). 𝐲_2=𝐲_4 and 𝐲_3=𝐲_5. Then, by (61) and (59) we have

    {[  2 a_1- a_2+ a_5=- a_2+ a_3- a_5,; 2 a_1-2 a_2+ a_3=- a_1+ a_2- a_5,;       a_1- a_2+ a_3- a_4+ a_5= o,                                   ].(70)



    {[                  2 a_1- a_3=-2 a_5,; 3 a_1+ a_3-3 a_4=3( a_2- a_4)- a_5,;     a_1+ a_3-2 a_4=( a_2- a_4)- a_5                                     ].

and therefore

    {[ a_1=3 a_5+3( a_2- a_4),; a_2=6 a_5+5( a_2- a_4),; a_3=8 a_5+6( a_2- a_4),; a_4=6 a_5+4( a_2- a_4),;               a_5= a_5,                         ].(71)

which means that Λ is a lattice with a basis { a_5,  a_2- a_4}.
Furthermore, since u_i=1 2 a_i∈1 2Λ, it follows by Lemma 4 that
P_10+Λ is indeed a five-fold lattice tiling.

As a conclusion of these five cases, Lemma 6 is proved. 


Lemma 7 (Zong <cit.>). A convex decagon can form a five-fold lattice tiling of the Euclidean plane if and only if, under a
suitable affine linear transformation, it takes u_1=(0,2), u_2=(2,2), u_3=(3,1), u_4=(3,0), u_5=(2,-1), u_6=- u_1, u_7=- u_2, u_8=- u_3, u_9=- u_4 and u_10=- u_5 as the middle points of its edges.


Remark 2 (Zong <cit.>). Let W denote the quadrilateral with vertices w_1=(-1,2), w_2=(-1,3 2), w_3=(-4 3, 4 3) and w_4=(-3 2,3 2).  A centrally symmetric convex decagon with u_1=(0,2), u_2=(2,2), u_3=(3,1), u_4=(3,0), u_5=(2,-1), u_6=- u_1, u_7=- u_2, u_8=- u_3, u_9=- u_4 and u_10=- u_5 as the middle points of its edges if and only if one of its vertices is an interior point of W.

Proof of Theorem 1. Theorem 1 follows from Lemma 3, Lemma 5, Lemma 6 and Lemma 7. 




Acknowledgements. For helpful comments and suggestions, the authors are grateful to Professor S. Robins and Professor G. M. Ziegler. This work is supported by 973 Program 2013CB834201.


amsplain

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Qi Yang, School of Mathematical Sciences, Peking University, Beijing 100871, China

Corresponding author:

Chuanming Zong, Center for Applied Mathematics, Tianjin University, Tianjin 300072, China

Email: cmzong@math.pku.edu.cn



