International Mathematics and Mathematical Sciences Volume 01, Article ID 61301, 7 pages doi:10.1155/01/61301 Research Article Translative Packing of Unit Squares into Squares Janusz Januszewski Institute of Mathematics and Physics, University of Technology and Life Sciences, Kaliskiego 7, 85-789 Bydgoszcz, Poland Correspondence should be addressed to Janusz Januszewski, januszew@utp.edu.pl Received 8 March 01; Revised 4 May 01; Accepted 4 May 01 Academic Editor: Naseer Shahzad Copyright q 01 Janusz Januszewski. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Every collection of n arbitrary-oriented unit squares admits a translative packing into any square of side length.5 n. 1. Introduction Let i be a positive integer, let 0 α i <π/, and let a rectangular coordinate system in the plane be given. One of the coordinate system s axes is called x-axis. Denote by S α i a square in the plane with sides of unit length and with the angle between the x-axis and a side of S α i equal to α i. Furthermore, by I s denote a square with side length s and with sides parallel to the coordinate axes. We say that a collection of n unit squares S α 1,...,S α n admits a packing intoasetc if there are rigid motions σ 1,...,σ n such that the squares σ i S α i are subsets of C and that they have mutually disjoint interiors. A packing is translative if only translations are allowed as the rigid motions. For example, two unit squares can be packed into I, but they cannot be packed into I ɛ for any ɛ>0. Three and four unit squares can be packed into I as well see Figure 1. Obviously, two, three, or four squares S 0 can be translatively packed into I. If either α 1 / 0orα / 0, then two squares S α 1 and S α cannot be translatively packed into I. The reason is that for every α / 0, the interior of any square S α translatively packed into I covers the center of I see Figure 1. The problem of packing of unit squares into squares with possibility of rigid motions is a well-known problem e.g., see 1 3. The best packings are known for several values of n. Furthermore, for many values of n, there are good packings that seem to be optimal. In this paper, we propose the problem of translative packing of squares. Denote by s n the smallest number s such that any collection of n unit squares S α 1,...,S α n admits
International Mathematics and Mathematical Sciences Figure 1 Q 3 λ 1 Q Q 1 5 λ Figure a translative packing into I s. The problem is to find s n for n 1,, 3,... Obviously, s n > n. By 4, Theorem 7, we deduce that lim n s n / n 1. We show that s n.5 n. 1.1. Packing into Squares Example.1. We have s 1. Each unit square can be translatively packed into I,butit is impossible to translatively pack S π/4 into I ɛ for any ɛ>0. Example.. We have s 5 see 5. Here, we only recall that two squares: S arctan 1/ and S arctan cannot be translatively packed into I 5 ɛ for any ɛ>0 see Figure. Example.3. We have s 4. Four squares S π/4 admit a translative packing into I see Figure 3, where / λ 3 /. In Figure 3 and Figure 4, we illustrate the cases when λ andλ /, respectively. By these three pictures, we conclude that four squares S π/4 cannot be translatively packed into I ɛ, for any ɛ>0. Consequently, s 4. On the other hand, four circles of radius / can be packed into I see Figure 4. Since any square S α i can be translatively packed into a circle of radius /, it follows that s 4. Lemma.4 see 5. Every unit square can be translatively packed into any isosceles right triangle with legs of length 5.
International Mathematics and Mathematical Sciences 3 λ Figure 3 Figure 4 Theorem.5. If n 3, thens n 10 5 / 3 n. Proof. Let S α 1,S α, and S α 3 be unit squares and put λ 1 1 ( 10 5 )..1 Three congruent quadrangles Q 1,Q, and Q 3, presented in Figure, of side lengths λ 1, 5,.5, and λ λ 1 5, are contained in I λ 1. Since the length of the diagonal of S α i is smaller than.5, by Lemma.4 we deduce that S α i can be translatively packed into Q i for i 1,, 3. Consequently, the squares S α 1,S α, and S α 3 can be translatively packed into I s and 10 5 s 3 λ 1 3.. 3 Now assume that 4 n 16. Denote by m n the smallest number s such that n circles of unit radius can be packed into I s. The problem of minimizing the side of a square into which n congruent circles can be packed is a well-known question. The values of m n are known, among others, for n 16 see Table..1 in 6 or 7. We know that m 4 4, m 5 < 4.83, m 6 < 5.33, m 7 < 5.74, m 8 <m 9 6, m 10 < 6.75, m 11 < 7.03, m 1 <m 13 < 7.47, m 14 <m 15 <m 16 8..3
4 International Mathematics and Mathematical Sciences Figure 5 Since each unit square is contained in a circle of radius /, it follows that n unit squares can be translatively packed into I m n /. Itiseasytoverifythat s n 10 5 m n < n,.4 3 for n 4, 5,...,16. Finally, assume that n>16. We use two lattice arrangements of circles. There exists an integer m 4 such that either m <n m m or m m<n m 1..5 Obviously, m circles of radius / can be packed into I m see Figure 5. Moreover, m m circles of radius / can be packed into I m / see Figure 5 ; it is easy to check that provided m 4. If m 1 n m m, then s n m ( ) 3m < m,.6 n 10 5 1 < n..7 3 If m m 1 n m 1, then s n m 1. Since m m 1 n, it follows that m 1/ 4n 3 1/. Thus, s n ( 1 ) 1 10 5 4n 3 < n..8 3 By Theorem.5, Examples.1 and., we conclude the following result.
International Mathematics and Mathematical Sciences 5 Corollary.6. Every collectionof n unit squares admits a translative packing into any square of area.5 n, that is, s n.5 n. Furthermore, s 5. For n 1980, the following upper bound is better than the bound presented in Theorem.5. Lemma.7. Let n be a positive integer, and let k be the greatest integer not over 4 n.then s n ( 1 π ) [ ] 1 k n k k 4k..9 Proof. Assume that S α 1,...,S α n is a collection of n unit squares. Let k be the greatest integer not over 4 n,letη π/k, and put ζ ( 1 η )[ ] 1 k n k 4k..10 For each i {1,...,n}, there exists j {1,,...,k} such that ( j 1 ) η αi <jη..11 Put ϕ α i j 1 η. We have 0 ϕ<η. Moreover, let λ 3 and λ 4 denote the lengths of the segments presented in Figure 6. Since λ 3 λ 4 cos ϕ sin ϕ 1 ϕ<1 η,.1 it follows that S α i is contained in a square P i with side length 1 η and with the angle between the x-axis and a side of P i equal to j 1 η. We say that P i is a j-square. To prove Lemma.7,itsuffices to show that P 1,...,P n can be translatively packed into I ζ. Denote by A j the total area of the j-squares, for j 1,...,k. Obviously, k A j n ( 1 η ). j 1.13 Put h j A j ζ ( 1 η ) ( 1 η ),.14 for j 1,...,k. Observe that k h j A 1 A k ζ ( 1 η ) k ( 1 η ). j 1.15
6 International Mathematics and Mathematical Sciences 1 λ 3 λ 4 ϕ Figure 6 Thus, k n ( 1 η ) h j ζ ( 1 η ) k ( 1 η ). j 1.16 The equation n ( 1 η ) x ( 1 η ) k ( 1 η ) x.17 is equivalent to x x ( 1 η ) k 1 8k n ( 1 η ) 0..18 It is easy to verify that x ζ is a solution of this equation. Consequently, k n ( 1 η ) h j ζ ( 1 η ) k ( 1 η ) ζ. j 1.19 We divide I ζ into k rectangles R 1,...,R k, where R j is a rectangle of side lengths ζ and h j. Since the diagonal of each P i equals 1 η and [ A j ζ ( 1 η )][ h j ( 1 η )],.0 it follows that all j-squares admit a translative packing into R j for j 1,...,k see Figure 6. Hence, P 1,...,P n, and consequently S α 1,...,S α n can be translatively packed into I ζ. Thisimpliesthats n ζ. Theorem.8. Let n be a positive integer. Then s n ( π ) n 4 n O 1,.1 as n. Proof. Let k be the greatest integer not over 4 n. Since n k 4k < n k n n< n 1,.
International Mathematics and Mathematical Sciences 7 by Lemma.7, it follows that, for n>1, s n < ( ) π [ ( 1 1 4 ) ] n n 1 4 n n ( π ) 4 ( )( π ) n 1 1 π π 4. n 1.3 References 1 H. T. Croft, K. J. Falconer, and R. K. Guy, Unsolved Problems in Geometry, Springer, New York, NY, USA, 1991. E. Friedman, Packing unit squares in squares: a survey and new results, The Electronics Combinatorics # DS7, 009. 3 F. Göbel, Geometrical packing and covering problems, in Packing and Covering in Combinatorics, A. Schrijver, Ed., vol. 106 of Mathematical Centre tracts, pp. 179 199, 1979. 4 H. Groemer, Covering and packing properties of bounded sequences of convex sets, Mathematika, vol. 9, no. 1, pp. 18 31, 198. 5 J. Januszewski, Translative packing two squares in the unit square, Demonstratio Mathematica, vol. 36, no. 3, pp. 757 76, 003. 6 G. Fejes Tóth, Packing and covering, in Handbook of Discrete and Computational Geometry, J.E.Goodman and J. O Rourke, Eds., pp. 19 41, CRC, Boca Raton, Fla, USA, 1997. 7 http://www.packomania.com/, updated by E. Sprecht.
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