Tilings of Parallelograms with Similar Right Triangles

Transcription

1 Discrete Comput Geom (2013) 50: DOI /s Tilings of Parallelograms with Similar Right Triangles Zhanjun Su Chan Yin Xiaobing Ma Ying Li Received: 1 October 2012 / Revised: 29 April 2013 / Accepted: 3 June 2013 / Published online: 18 June 2013 Springer Science+Business Media New York 2013 Abstract We say that a triangle T tiles the polygon A if A can be decomposed into finitely many non-overlapping triangles similar to T. A tiling is called regular if there are two angles of the triangles, say α and β, such that at each vertex V of the tiling the number of triangles having V as a vertex and having angle α at V is the same as the number of triangles having angle β at V. Otherwise the tiling is called irregular. Let P(δ) be a parallelogram with acute angle δ. In this paper we prove that if the parallelogram P(δ) is tiled with similar triangles of angles (α,β,π/2), then (α, β) = (δ, π/2 δ) or (α, β) = (δ/2,π/2 δ/2), and if the tiling is regular, then only the first case can occur. Keywords Parallelogram Regular and irregular tiling Right triangle We say that a triangle T tiles the polygon A if A can be decomposed into finitely many non-overlapping triangles similar to T. In fact, it is in general very difficult to determine the set of similar triangles which tile a given polygon. Previous papers [1 4,6] have looked at tiling polygons with similar triangles. In such case one starts with a collection of triangles and are allowed to scale the triangles when placing them. Recently, Laczkovich [5] has looked at tiling polygons, particularly regular triangles, squares, rectangles, and regular n-gons with n large enough, with congruent triangles. Clearly, the number of distinct non-similar triangles T such that parallelogram P can be dissected into finitely many triangles similar to T is infinite. In this paper we look at tiling parallelograms with similar right triangles. Let us recall some definitions from [4]. Let A be a polygon with vertices V 1,...,V N, and suppose that A is decomposed into non-overlapping similar triangles 1,..., t Z. Su (B) C. Yin X. Ma Y. Li College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang , People s Republic of China suzj888@163.com

2 470 Discrete Comput Geom (2013) 50: of angles α, β, γ.letv 1,...,V m (m N) be an enumeration of the vertices of the triangles 1,..., t. For each i = 1,...,m we denote by p i (resp. q i and r i )the number of those triangles j whose angle at the vertex V i is α (resp. β and γ ). If i N and the angle of A at the vertex V i is δ i, then If i > N then we have either or p i α + q i β + r i γ = δ i. (1) p i α + q i β + r i γ = 2π (2) p i α + q i β + r i γ = π. (3) Namely, (2) holds if V i is in the interior of A and whenever V i is on the boundary of a triangle j then necessarily V i is a vertex of j. In the other cases (3) holds. It is clear that the coefficients p i, q i, r i must satisfy p i = q i = r i = t. (4) The tiling will be called regular if one of the following statements is true: p i = q i for every i = 1,...,m; p i = r i for every i = 1,...,m; q i = r i for every i = 1,...,m. Otherwise the tiling is called irregular. Let P(δ) be a parallelogram with acute angle δ. We first consider the regular tilings of parallelograms with similar right triangles and get the following theorem. Theorem 1 If parallelogram P(δ) has a regular tiling with similar right triangles of angles (α,β,π/2), then (α, β) = (δ, π/2 δ). Proof Suppose that P(δ) can be regularly tiled with similar right triangles of angles (α,β,π/2). If at each vertex of the tiling the number of angles α is the same as that of β, then for the angle δ at a vertex of P(δ) we let δ = p(α + β) + r π/2, where p, r are integers. But 0 <δ<π/2, which is impossible. Therefore by symmetry we may assume that at each vertex of the tiling the number of angles β is the same as that of right angles π/2. In the following we shall consider Eqs. (1) (3). We first consider Eq. (1) and rewrite it as pα+r(β+π/2) = δ or π δ, where p andr are integers. If pα+r(β+π/2) = δ, then by 0 <δ<π/2 we get δ = pα. This implies that only the angle α occurs at the vertex with angle δ.ifpα + r(β + π/2) = π δ, then by π/2 <π δ<πwe have r = 0 or 1. When r = 1, we obtain p = 0, and thus π δ = β + π/2. So δ = α and (α, β) = (δ, π/2 δ), the result is true. When

3 Discrete Comput Geom (2013) 50: Fig. 1 Regular tiling r = 0, we have π δ = pα. Therefore only the angle α occurs at the vertex with angle π δ. Secondly, we consider Eq. (2) and rewrite it as pα + r(β + π/2) = 2π. Clearly, r 3. If r = 0, then only α occurs. If r = 1, then each of β and π/2 occurs once and α occurs p times with p > 1. If r = 2, then p = 2 and each of α, β, π/2 occurs twice. If r = 3, then p = 0 and β = π/6, and thus (α, β) = (π/3,π/6).byδ = pα, we get p = 1 and (α, β) = (δ, π/2 δ), the result is true. Thirdly, we consider Eq. (3) and denote it by pα +r(β +π/2) = π. Clearly, r 1. If r = 1, then p = 1, and thus each of the angles α, β, π/2 occurs once at the vertices with angle π.ifr = 0, then π = pα. From the discussions above we conclude that if (α, β) = (δ, π/2 δ) then at the vertices with angles π or 2π the number of angle α occurs is not less than that of β and at vertices with angles δ or π δ only angle α occurs, which contradicts Eq. (4), and the proof is complete. Figure 1 shows that the right triangle (δ, π/2 δ, π/2) can regularly tile the parallelogram P(δ). In order to consider the irregular tiling of parallelograms with similar right triangles, we introduce the following lemma. Lemma 2 The parallelogram P(3π/10) cannot be tiled with similar right triangles of angles (π/10, 2π/5,π/2). Proof Suppose that P(3π/10) can be tiled with similar right triangles of angles (π/10, 2π/5,π/2). Then by Theorem 2 of [2], we may assume that the coordinates of each vertex of P(3π/10) and of the triangles of the tiling belong to the field Q(cot(π/10)). Now, by Lemma 8 of[2], we have Q(cot(π/10)) Q(ζ ), where ζ = e (π/20)i. Also, since 3 is prime to 20, there is an automorphism φ of Q(ζ ) such that φ ( cot a 10 π) = cot 3a 10 π for every a not divisible by 10. Consider the map (x, y) = (φ(x), φ(y)) (x, y Q(ζ )). This map defines a collineation that maps every triangle of the tiling into a triangle of angles (3π/10,π/5,π/2).Also, does not change the orientation of these triangles (see pp of [2]). Now maps P(3π/10) into a parallelogram of angles 9π/10

4 472 Discrete Comput Geom (2013) 50: and π/10. Therefore, the conjugate tiling induced by gives a tiling of parallelogram of angles 9π/10 and π/10 with triangles of angles (3π/10,π/5,π/2). But this is clearly impossible, since the angles of the triangles are greater than π/10. This proves the lemma. Theorem 3 If the parallelogram P(δ) has an irregular tiling with similar right triangles of angles (α,β,π/2), then (α, β) = (δ, π/2 δ) or (α, β) = (δ/2,π/2 δ/2). Proof Suppose that P(δ) has an irregular tiling with similar right triangles of angles (α,β,π/2). Ifα = β, then (α,β,π/2) = (π/4,π/4,π/2) and this is the case of (δ, π/2 δ, π/2). Therefore, we may assume that α>β, and thus π/4 <α<π/2. Since the tiling is irregular, there is an equation pα +qβ +r π/2 = τ with p > q and τ {δ, π δ, π, 2π}, where p, q, r are integers. That is, (p q)α = τ (q +r) π/2. We need to consider the following cases. Case 1: τ = δ. In this case we have (p q)α = δ (q + r) π/2. Since p > q and δ<π/2, we get q + r = 0, and thus pα = δ. Since α>π/4, this implies p = 1, α = δ, and (α, β) = (δ, π/2 δ). Case 2: τ = π δ. Suppose (p q)α = π δ (q + r) π/2. This implies q + r 1. If q + r = 0, then pα = π δ. Since π/4 <α<π/2, we get p = 2or p = 3. If p = 2, then α = π/2 δ/2 and we have (α, β) = (π/2 δ/2,δ/2). If p = 3, then α = (π δ)/3 and β = π/6 + δ/3. Since α>β, this implies δ<π/4. Thus the angle δ cannot be tiled with angles α. Therefore, we have δ = nβ. Thus n = 1orn = 2, which give δ = π/4 orδ = π, both are impossible. If q + r = 1, then (p q)α = π/2 δ. Since α>π/4, this implies α = π/2 δ and (α, β) = (π/2 δ, δ). Case 3: τ = π. In this case we have (p q)α = π (q + r) π/2. This implies q +r 1. If q +r = 0, then pα = π. Since π/4 <α<π/2, we get p = 3, α = π/3, and (α, β) = (π/3,π/6). Ifδ = π/3, then δ = α, and thus (α, β) = (δ, π/2 δ). If δ = n π/6, then n = 1orn = 2. When n = 2, this is the above case. When n = 1, then δ = π/6, and thus (α, β) = (π/2 δ, δ). If q + r = 1, then (p q)α = π/2. From π/4 <α<π/2we know that this case is impossible. Case 4: τ = 2π. In this case we have (p q)α = 2π (q + r) π/2. Since p > q and π/4 <α<π/2, we have q + r 3. If q + r = 3, then (p q)α = π/2. From π/4 <α<π/2weknow this is impossible. If q + r = 2, then (p q)α = π. Since π/4 <α<π/2, we get α = π/3, which is the result in Case 3. If q + r = 1, then (p q)α = 3π/2. This implies α = 3π/8 orα = 3π/10. When α = 3π/8, then (α, β) = (3π/8,π/8).Ifδ = 3π/8orδ = π/8, our result is true. Otherwise, we have δ = m π/8. It is clear that m = 1, 3. If m = 2, then δ = π/4, and this is the case of (α, β) = (π/2 δ/2,δ/2).ifm 4, then δ = m π/8 π/2, a contradiction. When α = 3π/10, then (α, β) = (3π/10,π/5). Ifδ = 3π/10, or δ = π/5, our result is true. Otherwise, we have δ = n π/5. It is clear that n = 1. If n = 2, then δ = 2π/5, and this is the case of (π/2 δ/2,δ/2).ifn 3, then δ = n π/5 >π/2, a contradiction.

5 Discrete Comput Geom (2013) 50: Fig. 2 Irregular tiling If q + r = 0, then pα = 2π. Since π/4 <α<π/2, we have α = 2π/5, α = π/3, or α = 2π/7. When α = 2π/7, then (α, β) = (2π/7, 3π/14). Ifδ = 2π/7, or δ = 3π/14, then the result is true. If δ = 2π/7 and δ = 3π/14, then by 0 <δ<π/2we may assume that δ = m 3π/14. If m = 2, then δ = 3π/7, and this is the case of (π/2 δ/2,δ/2). If m 3, then δ = m 3π/14 >π/2, a contradiction. When α = π/3, this is a result of Case 3. When α = 2π/5, then (α, β) = (2π/5,π/10).Ifδ = 2π/5orδ = π/10, the result is true. Otherwise, we have δ = n π10.ifn = 2, this is the case of (π/2 δ/2,δ/2). If n = 3, then δ = 3π/10, and by Lemma 2 we conclude a contradiction. If n 5, then δ = n π/10 π/2, a contradiction. The proof is complete. Figure 2 shows that the right triangles (δ, π/2 δ, π/2) and (δ/2,π/2 δ/2,π/2) can irregularly tile the parallelogram P(δ), where the numbers 1, 2, 3 in the left figure denote the angles δ, π/2 δ, π/2, respectively; the numbers 1, 2, 3 in the right figure denote the angles δ/2,π/2 δ/2,π/2, respectively. Remark 4 Since the above proofs never used that P(δ) is a parallelogram, only that two of its angles equal δ and two equal π δ, for the trapezoid T (δ) with acute angle δ, we have the same results as that of parallelogram P(δ). Acknowledgments The authors wish to thank the anonymous referees for their careful reading and helpful suggestions which we accepted gratefully. Su s research was partially supported by National Natural Science Foundation of China ( ) and NSF of Hebei Province (A ). References 1. Butler, S., Chnug, F., Graham, R., Laczkovich, M.: Tilings of polygons with lattice triangles. Discrete Comput. Geom. 44, (2010) 2. Laczkovich, M.: Tilings of polygons with similar triangles. Combinatorica 10, (1990) 3. Laczkovich, M.: Tilings of triangles. Discrete Math. 140, (1995) 4. Laczkovich, M.: Tilings of polygons with similar triangles II. Discrete Comput. Geom. 19, (1998) 5. Laczkovich, M.: Tilings of convex polygons with congruent triangles. Discrete Comput. Geom. 48, (2012) 6. Szegedy, B.: Tilings of the square with similar right triangles. Combinatorica 21, (2001)