J. Appl. Math. & Computing Vol. 23(2007), No. 1 - 2, pp. 321 - 327
Website: http://jamc.net
TILINGS OF PARALLELOGRAMS WITH SIMILAR TRIANGLES ZHANJUN SU∗ AND REN DING
Abstract. We say that a triangle ∆ tiles the polygon P if P can be decomposed into finitely many non-overlapping triangles similar to ∆. Let P be a parallelogram with angles δ and π − δ (0 < δ ≤ π/2) and let ∆ be a triangle with angles α, β, γ (α ≤ β ≤ γ). We prove that if ∆ tiles P then either δ ∈ {α, β, γ, π −γ, π −2γ} or dimLP =dimL∆ . We also prove that for every parallelogram P, and for every integer n (where n ≥ 2, n 6= 3) there is a triangle ∆ so that n similar copies of ∆ tile P. AMS Mathematical Subject Classification : 52C20. Key words and phrases : Parallelogram, similar triangle, linear space, tiling.
We say that a triangle ∆ tiles the polygon P if P can be decomposed into finitely many non-overlapping triangles similar to ∆. In [8] Szegedy considered the tilings of the square with similar right triangles and in [5] Laczkovich discussed the tilings of rectangles with similar triangles. In the present paper we consider the tilings of parallelograms with similar triangles. Let P be a parallelogram with angles δ and π − δ (0 < δ ≤ π2 ) and let LP = {q1 δ + q2 π | q1 , q2 ∈ Q}, where LP is a linear space over the rational field Q. Let ∆ be a triangle with angles α, β, γ (α ≤ β ≤ γ) and let L∆ = {q1 α + q2 β + q3 γ | qi ∈ Q, i = 1, 2, 3}, where L∆ is a linear space over the rational field Q. Theorem 1. If ∆ tiles P, then either (i) δ ∈ {α, β, γ, π − γ, π − 2γ}, or Received August 4, 2005. Revised November 30, 2005. ∗ Corresponding author. This research was supported by Hebei NSF A2005000144 and the Science Foundation of Hebei Normal University. c 2007 Korean Society for Computational &Applied Mathematics and Korean SIGCAM.
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(ii) dimLP =dimL∆ . Proof. Let P be decomposed into the triangles ∆1 , ∆2 , . . . , ∆N such that each ∆j is similar to ∆. Let V be a vertex of one of the triangles ∆j and let {∆j1 , ∆j2 , . . . , ∆jk } be the set of triangles having V as a vertex. Let θjs denote the angle of ∆js at V (s = 1, 2, . . . , k). If V is a vertex of P with angle δ, then we have k X
θjs = δ.
(1)
s=1
If V is a vertex of P with angle π − δ, then k X
θjs = π − δ.
(2)
s=1
If V is not a vertex of P, then either k X
θjs = 2π
(3)
θjs = π.
(4)
s=1
or k X s=1
In fact, (3) holds if V is in the interior of P, and V is necessarily a vertex of ∆j whenever V is on the boundary of a triangle ∆j ; in all other cases we have (4). Since 3γ ≥ 2γ + β ≥ 2γ + α ≥ α + β + γ = π, γ + 2β ≥ α + β + γ = π and 0 < δ ≤ π2 , each of the equations (1) and (2) must take the form of one of the following equations: 2γ = π − δ, β + γ = π − δ, kα + γ = δ or π − δ(k ≥ 0) and kα + mβ = δ or π − δ(k, m ≥ 0). If any of the equations 2γ = π−δ, β+γ = π−δ, γ = δ, γ = π−δ, α+γ = π−δ, α = δ, β = δ, α + β = δ, α + β = π − δ occurs among the equations (1) and (2), then (i) is true. Therefore we may assume that each of the equations (1) and (2) is of the form ki α + γ = π − δ, (ki ≥ 2)
(5)
ki α + mi β = δ or π − δ, (max{ki , mi } ≥ 2).
(6)
or
Now we suppose that neither of (i) and (ii) is true. Since dimLP ≤ 2 and dimL∆ ≤ 3, we need consider the following cases: dimL∆ = 3; dimLP = 1 and dimL∆ = 2; dimLP = 2 and dimL∆ = 1.
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First we consider the case when dimL∆ = 3. Then α, β, γ are linearly independent over Q. Hence, if pα + qβ + rγ = vπ
(7)
holds, where p, q, r, v are integers, then p = q = r = v. Consequently, the left hand sides of the equations (4) and (3) are of the form α+β +γ and 2α+2β +2γ, respectively. Since the left hand sides of the equations (1), (2), (3), and (4) contain N α0 s, β 0 s and γ 0 s (as they involve all the angles of the triangles ∆1 , ∆2 , . . . , ∆N ), this implies that the left hand sides of the equations (1) and (2) contain the same number of α0 s, β 0 s and γ 0 s. However, each of the equations (5) contains more α0 s than γ 0 s. Hence the total number of α0 s in the equations (1) and (2) is greater than the number of γ 0 s unless both are zero. In this case, however, the number of β 0 s will be different. This shows that dimL∆ = 3 is impossible. Next we suppose that dimLP = 1 and dimL∆ = 2. Then δ is a rational multiple of π. We distinguish between two cases. Case 1. Equations of the form (5) do occur among the equations (1) and (2). Then the total number of α0 s is strictly greater than the number of γ 0 s in the equations (1) and (2) and hence there must be an equation (7) among the equations (3) or (4) such that p < r (and v = 1 or v = 2). Subtracting rα + rβ + rγ = rπ from (7) we obtain (p − r)α + (q − r)β = (v − r)π.
(8)
Let kα + γ = π − δ, k ≥ 2 be an equation of the form (5). Subtracting α + β + γ = π we obtain (k − 1)α − β = −δ.
(9)
If (8) and (9) determine α and β then, as δ is a rational multiple of π, α and β are also rational multiples of π. This implies dimL∆ = 1 which is not the case. Therefore the equations (8) and (9) are not independent, that is, there is a number c such that c(k − 1) = p − r, −c = q − r, −cδ = (v − r)π. Since p < r and k ≥ 2, we have c < 0, q > r, v > r. Hence by v ≤ 2 and p < r, we obtain p = 0, r = 1, v = 2. This gives π = (v − r)π = −cδ = δ/(k − 1) and (k − 1)π = δ, which is impossible. Case 2. Each of the equations (1) and (2) is of the form (6). Let kα + mβ = δ
(10)
be one of these equations. Then either k ≥ max{m, 2} or m ≥ max{k, 2}. Suppose the former (the latter can be considered similarly). Then following the argument of the Case 1 we find an equation of the form (7) with p < r. If p, q, r are all positive then we can subtract α + β + γ = π from (7) so that we may assume min{p, q} = 0. As in the case 1, we find that the equations (10) and (8) can not be independent and hence there is a number c such that ck = p − r, cm = q − r, cδ = (v − r)π.
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Then c < 0 and, by m ≤ k, we obtain p ≤ q and v < r. This gives, by min{p, q} = 0, p = 0. We also have (r − v)π = −cδ = rδ/k < rπ/k, and k(r − v) < r. Since v = 1 or 2, v < r, k ≥ 2, this implies k = 2, v = 2, r = 3. Thus c = −3/2 and δ = 2π/3, contradicting the assumption of 0 < δ ≤ π/2. Last we suppose that dimLP = 2 and dimL∆ = 1. Then α, β, γ are rational multiples of π. Let p1 α + q1 β + r1 γ = δ
(11)
p2 α + q2 β + r2 γ = π − δ
(12)
and be two equations among the equations (1) and (2). Since α, β, γ are rational multiples of π, by the equations (11) and (12), δ, π − δ are also rational multiples of π. This implies dimLP = 1, which is not the case. Theorem 2. In Theorem 1, if dimLP =dimL∆ = 2 and γ = pα + qβ, p, q ∈ Q, then p and q are integers. Proof. As proved in Theorem 1, we need consider the equations (5) and (6), that is, the following equations. kα + γ = π − δ, (k ≥ 2); kα + mβ = δ, (max{k, m} ≥ 2); kα + mβ = π − δ, (max{k, m} ≥ 2). We distinguish between two cases. Case 1. First we consider the equations kα + γ = π − δ, (k ≥ 2), 0 0 k α + mβ = δ, (max{k , m} ≥ 2). Since α + β + γ = π, we have
(1 − k)α + β = δ, 0 k α + mβ = δ. 0
0
Because dimL∆ = 2, we obtain that m = 1, 1 − k = k . Since k ≥ 2, k ≤ 0 and 0 by m = 1, we have max{k , m} = 1, which is impossible. Case 2. Next we consider the equations kα + mβ = δ, (max{k, m} ≥ 2), 0 0 0 0 k α + m β = π − δ, (max{k , m } ≥ 2). Since γ = pα + qβ, we obtain that kα + mβ = δ, 0 0 (1 + p − k )α + (1 + q − m )β = δ. 0
0
Because dimL∆ = 2, we obtain k = 1 + p − k and m = 1 + q − m , that is, 0 0 0 0 p = k + k − 1 and q = m + m − 1. So p and q are integers, for k, m, k , m are integers.
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Figure 1 We know that it is difficult to determine the number n of triangles that tile the given polygon P. Soifer’s book [7] contains several results concerning tilings of triangles. He proves, e.g., that every triangle T can be cut into any number n of similar triangles, except n = 2, 3, 5. In [2] Golomb conjectures that if the pieces must be congruent to each other and similar to the triangle T , then the number of pieces is necessarily n2 , 3n2 or n2 + k 2 , which is proved in [6]. We consider only the case where the polygon P is a parallelogram. Theorem 3. For every parallelogram P, and for every integer n (where n ≥ 2, n 6= 3) there is a triangle ∆ so that n similar copies of ∆ tile P. Proof. If n = 2, then join the opposite vertices of the parallelogram P, and we obtain 2 congruent triangles. If n = 2k (k ≥ 2), then we can decompose P into k smaller congruent parallelograms by k − 1 lines parallel to one side of P and decompose each of the small parallelograms into two congruent triangles and in this way we decompose P into 2k congruent triangles. Now let n = 2k + 1 (k ≥ 2). It is obvious that any triangle T can be decomposed into 2k triangles similar to T as shown in Figure 1. So, for any parallelogram P we decompose it first into two congruent triangles by a diagonal, then decompose one of the two congruent triangles into 2k small triangles as mentioned above and by doing so we decompose P into n = 2k + 1 similar triangles. Now let us consider the following problem: Can we decompose a given parallelogram into 3 similar triangles? We have the following result. Theorem 4. A parallelogram P can be decomposed into 3 similar triangles if and only if P satisfies one of the following two conditions: (1) The parallelogram P is a rectangle; (2) A diagonal of P is perpendicular to a side of P. Proof. If the condition (1) or (2) is fulfilled, obviously the parallelogram P can be decomposed into 3 similar triangles (see Figure 2).
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Now suppose that P can be decomposed into 3 similar triangles by 2 segments. We need consider the following cases. Without loss of generality we suppose the parallelogram P= ABCD (whose vertices are A, B, C and D counterclockwise), and ∠A ≤ ∠B. Case 1. If a line crosses adjacent sides of the parallelogram P without passing any vertex of P, then the line cuts P into a triangle and a pentagon. In any case, a pentagon cannot be decomposed into 2 similar triangles, so we can’t obtain 3 similar triangles in this case. Case 2. If a line passes through a vertex and a non-incident side of the parallelogram P, then the line divides P into a triangle and a quadrangle, so we need only to consider diagonals of the quadrangle. Subcase 1. Take a point E in BC, consider the triangles ABE, ADE, CDE. If ∆ABE ∼ ∆CDE, since ∠B > ∠CDE and ∠B 6= ∠CED, then ∠B = ∠C = π/2, by ∠EAD < π/2 and ∠EDA < π/2 we have ∠AED = π/2 (also see Figure 2). So ABCD is a rectangle and we obtain 3 similar triangles ABE, ADE, CDE, which is the case (1) of Theorem 4. Subcase 2. Take a point E in BC, consider the triangles ABD, DBE, DEC. If 4DBE ∼ 4DEC, then ∠BED = ∠CED = π/2. Notice 4ABD ∼ 4DEC, ∠A = ∠C, ∠ABD > ∠EDC, therefore ∠ABD = ∠DEC = π/2. So AB ⊥ DB, DE ⊥ BC. In this case, we obtain 3 similar triangles 4ABD, 4BED and 4DEC, which is the case (2) of Theorem 4. Subcase 3. Take a point E in AB, consider the triangles AED, BED, CDB. If 4AED ∼ 4BED, then ∠AED = ∠BED = π/2. Because 4AED ∼ 4CBD, ∠A = ∠C, ∠AED > ∠CDB. Therefore ∠AED = ∠CBD = π/2, therefore DE ⊥ AB, CB ⊥ DB, we also obtain three similar triangles, which is the case (2) of Theorem 4. Subcase 4. Take a point E in DC, consider the triangles AED, AEC, ABC. If 4AED ∼ 4AEC, then AE ⊥ CD, this is a contradiction. Case 3. If a segment is exactly the diagonal of the parallelogram, then we have the following 2 cases. Take a point E in AC. 4BEA ∼ 4BEC implies BE ⊥ AC, 4ADC ∼ 4BEA implies ∠D = π/2, so quadrangle ABCD is a rectangle, and we obtain three similar triangles. Take a point E in BD. 4AEB ∼ 4AED implies AE ⊥ BD, because 4AEB ∼ 4BCD, so ∠C = π/2, hence quadrangle ABCD is a rectangle, we also obtain three similar triangles. The proof is complete. For similar decompositions using congruent rather than similar triangles see [1].
References 1. V. G. Boltyanskii, Equivalent and Equidecomposable Figures, Topics in Mathematics Series, D. C. Heath and Company 1963.
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Figure 2 2. S. W. Golomb, Replicating figures in the plane, Math Gaz 48 (1964). 3. A. Hanaki and I. Miyamoto, Classification of association schemes with 18 and 19 vertices, J. Appl. Math. & Computing(old:KJCAM) 5 (1998), 633-641. 4. Taenam Kim, Dukhwan Oh, and Funki Lim, A new parallel algorithm for rooting a tree, J. Appl. Math. & Computing(old:KJCAM) 5 (1998), 427-432. 5. M. Laczkovich, Tilings of polygons with similar triangles, Combinatorica 10 (1990), 281306. 6. S. L. Snover, C. Waiveris and J. K. Williams, Rep-tiling, Discrete Math. 91 (1991). 7. A. Soifer, How Does One Cut a Triangle ? Center for Excellence in Mathematical Education Colorado Springs 1990. 8. B. Szegedy, Tilings of the square with similar right triangles, Combinatorica 21(1) (2001), 139-144. Zhanjun Su received his BSc, MSc and Ph.D. from Hebei Normal University under the direction of Prof. Ren Ding. His research interests focus on discrete geometry, convex geometry and combinatorial geometry. College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang 050016, P. R. China e-mail:
[email protected] Ren Ding is a professor of mathematics, supervising Ph.D. programs at Hebei Normal University. His research interests focus on discrete geometry, convex geometry and combinatorial geometry. College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang 050016, P. R. China e-mail:
[email protected]
