Janusz Januszewski TRANSLATIVE PACKING OF UNIT SQUARES INTO EQUILATERAL TRIANGLES

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1 DEMONSTRATIO MATHEMATICA Vol. XLVIII No 015 Janusz Januszewski TRANSLATIVE PACKING OF UNIT SQUARES INTO EQUILATERAL TRIANGLES Communicated by Z. Pasternak-Winiarski Abstract. Every collection of n (arbitrary-oriented) unit squares can be packed translatively into any equilateral triangle of side length.755 n. Let the coordinate system in the Euclidean plane be given. For 0 α i ă π{, denote by Spα i q a square in the plane with sides of unit length and with an angle between the x-axis and a side of Spα i q equal to α i. Furthermore, let T psq be an equilateral triangle with sides of length s and with one side parallel to the x-axis. A collection of unit squares admits a translative packing into a set C if there are mutually disjoint translated copies of the members of the collection contained in C. The question of packing of unit squares into squares or triangles (with the possibility of rotations) is a well-known problem (see [1], [], [4] and [9]). Some upper bounds concerning translative packing (without the possibility of rotations) of unit squares into a square are given in [6]. Covering problems are discussed in [7]. In this note, we propose the question of translative packing of squares into an equilateral triangle. Denote by t n the smallest number t such that any collection of n unit squares Spα 1 q,..., Spα n q admits a translative packing into T ptq. The problem is to find t n for n 1,,,.... Claim 1. t 1 ` a{ «.. Proof. Let λ 1 ` a{. Obviously, Spπ{4q cannot be packed translatively into T pλ 1 ɛq for any ɛ ą 0 (see Fig. 1, left). The squares Spπ{1q as well as Sp5π{1q cannot be packed translatively either. Consequently, t 1 λ Mathematics Subject Classification: 5C15. Key words and phrases: packing, translative packing, triangle. DOI: /dema c Copyright by Faculty of Mathematics and Information Science, Warsaw University of Technology Download Date 1/1/17 6:58 AM

2 Translative packing of unit squares 45 Fig. 1 Let s α be the smallest number such that Spαq can be packed translatively into T ps α q. At least one angle between a side of Spαq and a side of T ps α q is between π{6 and π{. Without loss of generality, we can assume that π 6 α π. By Fig. 1 (right) we see that ˆ s α 1 ` pcos α ` sin αq. It is easy to check that this value is not greater than ˆ ˆ 1 ` cos π 4 ` sin π ˆ 1 ` 4 This implies that t 1 λ 1. λ 1. Claim. t ˆ cosrarctanp ` 1qs ` ` sinrarctanp ` 1qs «.59. Proof. Put γ arctanp ` 1q, let Spαq and Spβq be unit squares and let σ ˆ cos γ ` ` sin γ d d 0 8 ` ` 6 9 «.59. Part I of the proof. First we show that Spαq and Spβq can be packed translatively into T pσq, i.e., that t σ. We can assume that 0 β α ă π{. We say that a square S is packed along a side of a triangle T provided S Ă T and a vertex of S belongs to this side. Case 1: α π{, β π{6 and α ` β π{. Let T pλ 5 q be a triangle with vertices p0, 0q, pλ 5, 0q, p 1 λ 5, (see Fig., left) cos β sin α λ 5 ` sin β ˆ ` 1 ` cos β ` cos α ˆ ` 1 ` tan α λ 5q, where sin α. Download Date 1/1/17 6:58 AM

3 454 J. Januszewski Since where apx a, y a q, bpx b, y b q, x a λ 5 Fig. y b y a px b x a q tan α, sin β ` cos β cos β, y a cos β, x b sin α ` cos α ` sin α, y b sin α, it follows that Spαq and Spβq can be packed translatively into T pλ 5 q. It can be shown that λ 5 is maximal at α ` β π{. Consequently, replacing β by π{ α we obtain By λ 5 f 1 pαq cos α ` 1 ` f 1 1pαq sin α ` 1 ` sin α. cos α, we have λ 5 f 1 pγq σ. This implies that Spαq and Spβq can be packed translatively into T pσq. Case : α π{, β π{6 and α ` β ă π{. If α 1 π{ α and β 1 π{ β, then α 1 π{6, β 1 π{ and α 1 ` β 1 π pα ` βq ą π{ (see Fig., right). By Case 1, we conclude that Spαq and Spβq can be packed translatively into T pσq. Case : α π{, β π{. Let T pλ 6 q be a triangle with vertices p0, 0q, pλ 6, 0q, p 1 λ 6, (see Fig., left) ˆ sin α ` cos α sin β sin β ` cos β λ 6 ` ` sin β ` 1 ` tan α λ 6q, where sin α. Download Date 1/1/17 6:58 AM

4 Translative packing of unit squares 455 Since where cpx c, y c q, dpx d, y d q, x c Fig. y d y c px d x c q tan α, sin β ` cos β ` sin β, y c sin β, x d λ 6 sin α sin α, y d sin α ` cos α, it follows that Spαq, Spβq can be packed translatively into T pλ 6 q. By Bλ 6 Bα 0 and α β, replacing α by β, we obtain λ 6 f pβq cos β sin β ` cos β ˆ ` ` sin β. It can be shown that f 1 pβq ă 0 provided π{ β π{. This implies that λ 6 f ˆπ 1 ` 4 «.09 ă σ. Case 4: α π{6, β π{6. If α 1 π{ α and β 1 π{ β, then α 1 π{ and β 1 π{. By Case, we conclude that Spαq and Spβq can be packed translatively into T pσq. Case 5: α π{, π{6 ă β ă π{. Observe that angles between a side of Spαq and Spβq and the left-hand side of T pσq are equal to α π ` π α 5 6 π α ą π and β π β ă π 6, respectively (see Fig., right and Fig. 4, left). By Cases 1 and, we deduce that the squares can be packed translatively into T pσq along the left-hand side of this triangle. Case 6: π{6 ă α ă π{, π{6 ă β ă π{. Observe that angles between a side of Spαq and Spβq and the left-hand side of T pσq are equal to α π{ α ă π{6 and β π{ β ă π{6, respectively. By Case 4, we deduce that the squares can be packed translatively into T pσq. Download Date 1/1/17 6:58 AM

5 456 J. Januszewski Fig. 4 Case 7: π{6 ă α ă π{, β π{6. The angles between a side of Spαq and Spβq and the right-hand side of T pσq are equal to α π ` π α α π 6 ă π 6 (see β β π 6 in Fig. 4, left) and β π ` β ą π (see β in Fig., right), respectively. By Cases 1 and, we deduce that the squares can be packed translatively along the right-hand side of T pσq. Part II of the proof. We show that Spγq and Spδq, where γ arctanp ` 1q and δ π γ arctan 1, cannot be packed translatively into T pσ ɛq for any ɛ ą 0, i.e., that t σ. It is sufficient to consider only packings along sides of T pσq (see Fig. 5 and Fig. 6). The angles between a side of Spγq and Spδq and the left-hand side of T pσq are equal to γ 5 6 π γ and δ π δ, respectively. The angles between a side of Spγq and Spδq and the right-hand side of T pσq are equal to γ γ π 6 and δ π ` δ, respectively. Since γ δ and δ γ, it suffices to consider four possibilities of packings (packings along the right-hand side of T pσq give the same results as packings along the left-hand side). Let ɛ ą 0. By Case 1, we conclude that Spγq and Spδq cannot be packed translatively along the base of T pσ ɛq so that Spγq is to the left of Spδq (see Fig. 5, left). Let ˆ λ 7 sin δ ` cos δ ` cos δ «.6499 ą σ. By Fig. 5 (right), we see that Spγq and Spδq cannot be packed translatively along the base of T pλ 7 ɛq so that Spγq is to the right of Spδq. Let λ 8 cos δ sin γ tan γ ˆ ` 1 ` ˆ ` 1 ` sin δ ` cos γ sin γ ` cos γ «.94 ą σ. By Fig. 6 (left), we see that Spγq and Spδq cannot be packed translatively along the left-hand side of T pλ 8 ɛq so that Spγq is lower than Spδq. Download Date 1/1/17 6:58 AM

6 Translative packing of unit squares 457 Fig. 5 Finally, let λ 9 cos γ sin δ tan γ ` cos γ ` ˆ 1 ` Fig. 6 ˆ ` sin δ ` 1 ` cos δ sin γ «.495 ą σ. By Fig. 6 (right), we see that Spγq and Spδq cannot be packed translatively along the left-hand side of T pλ 9 ɛq so that Spγq is higher than Spδq. Results presented in Claims, 4 and 5 will be applied in the proof of the Theorem. Claim. ˆ ˆ t 1 ` cos arctan ˆ ` sin arctan «.756. Proof. Put ϑ arctan, and let Q be a deltoid with vertices p0, 0q, px p, 0q, px p {, p1 ` {qx p q, p0, y q q, Download Date 1/1/17 6:58 AM

7 458 J. Januszewski Fig. 7 where y q `1 ` { cos ϑ ` sin ϑ « and x p y q { «1.078 (see Fig. 7, left). Since any triangle T py q q can be divided into three deltoids congruent to Q (see Fig. 4, right), it follows that to prove Claim it suffices to show that any unit square can be packed translatively into Q. Let Spαq be a unit square. There is no loss of generality in assuming that π{6 α π{ (the straight line going through the points p and q is the axis of symmetry of Q). A simple calculation shows that y h x h ` y q, where x h cos α, y h sin α ` cos α. Moreover, y g ą px g x p q, where x g cos α ` sin α, y g sin α. This implies that Spαq can be packed translatively into Q. Obviously, T ptq can be divided into four equilateral triangles of side lengths 1 t (see Fig. 7, right). Claim 4. t 4 t 1 p ` a{q «4.46. Claim 5. t 7 ` 4 6 « Proof. Melissen (see [8]) found the densest packing of seven equal circles into an equilateral triangle. We apply a part of its arrangement. We pack five circles of radius r { as in Fig. 8 (left). The length of the segment uv equals rp ` q ` 6. Since any unit square can be packed translatively into any circle of radius {, by Claim 1 we deduce that t 7 ` 6 ` t 1 ` 4 6. Download Date 1/1/17 6:58 AM

8 Translative packing of unit squares 459 Theorem. Let n be a positive integer. Then ˆd 10 4 d 11 ` 6 n. t n ` Furthermore, the equality holds only for n. Proof. Put ξ By Claims 1 4 we know that and that d 10 4 d 11 ` 6 ` 4 « t n ă ξ n for n P t1,, 4u, t ξ. Now assume that 5 n 15. Denote by µ n, the smallest number s such that n circles of unit radius can be packed into T psq. The problem of minimizing the side of an equilateral triangle into which n congruent circles can be packed is a well-known question. The values of µ n are known, among others, for n ă 16 (see [8]). We know that: µ 5 µ 6 ă 7.47, µ 8 µ 9 µ 10 ă 9.47, µ 11 µ 1 ă 10.9, µ 1 µ 14 µ 15 ă Each unit square is contained in a circle of radius {. Consequently, n unit squares can be packed translatively into T p µ n {q. It is easy to verify that t n µ n ă.7 n for n P t5,..., 15uzt7u. Unfortunately, µ 7 ` 4 and µ 7 { ą ξ 7. The inequality t 7 ă.7 7 follows from Claim 5. Obviously, T ptq can be divided into four equilateral triangles of side lengths 1 t (see Fig. 7, right). Consequently, t 16 t 4 ă Finally assume that n 17. There exists an integer m such that pm 1qm ă n mpm ` 1q. Download Date 1/1/17 6:58 AM

9 460 J. Januszewski Fig. 8 By Fig. 8 (right), we deduce that mpm`1q circles of radius { can be packed into T p pm 1q ` 6q. Since it follows that Thus pm 1qm ` 1 n, m 1 ` 8n 7. t n ˆ1 ` 8n 7 1 ` 6. It is easy to check that the value on the right side of this inequality is smaller than ξ n provided n 17. Remark. Arguing as in the proof of Theorem of [6] or Theorem 7 of [5], we can show that t lim n nñ8 4n 1. References [1] H. T. Croft. K. J. Falconer, R. K. Guy, Unsolved Problems in Geometry, Springer Verlag, Berlin, 1991, [] G. Fejes-Tóth, Packing and covering, in: Handbook of Discrete and Computational Geometry, J. E. Goodman, J. O Rourke, eds., 1997, [] E. Friedman, Packing unit squares in squares: a survey and new results, Electron. J. Combin., Dynamic Survey 7, 000. [4] F. Göbel, Geometrical packing and covering problems, in: Packing and Covering in Combinatorics, A. Schrijver, ed., Math. Centrum Tracts 106 (1979), [5] H. Groemer, Covering and packing properties of bounded sequences of convex sets, Mathematica 9 (198), [6] J. Januszewski, Translative packing of unit squares into squares, International Journal of Mathematics and Mathematical Sciences, Volume 01 (01), Article ID 6101, 7 pages, doi: /01/6101. Download Date 1/1/17 6:58 AM

10 Translative packing of unit squares 461 [7] J. Januszewski, Translative covering by unit squares, Demonstratio Math. 46() (01), [8] H. Melissen, Densest packings of congruent circles in an equilateral triangle, Amer. Math. Monthly 100(10) (199), [9] N. Oler, A finite packing problem, Canad. Math. Bull. 4 (1961), J. Januszewski INSTITUTE OF MATHEMATICS AND PHYSICS UNIVERSITY OF TECHNOLOGY AND LIFE SCIENCES Al. Prof. S. Kaliskiego BYDGOSZCZ, POLAND januszew@utp.edu.pl Received September, 01. Download Date 1/1/17 6:58 AM