Packing 10 or 11 Unit Squares in a Square

Let s(n) be the side of the smallest square into which it is possible pack n unit squares. We show that s(10) = 3 + √ 1 2 ≈ 3.707 and that s(11) ≥ 2 + 2 √ 4 5 ≈ 3.789. We also show that an optimal packing of 11 unit squares with orientations limited to 0◦ or 45◦ has side 2+2 √ 8 9 ≈ 3.886. These results prove Martin Gardner’s conjecture that n = 11 is the first case in which an optimal result requires a non-45◦ packing. Let s(n) be the side of the smallest square into which it is possible to pack n unit squares. It is known that s(1) = 1, s(2) = s(3) = s(4) = 2, s(5) = 2 + √ 1 2 , and that s(6) = s(7) = s(8) = s(9) = 3. For larger n, proofs of exact values of s(n) have been published only for n = 14, 15, 24, 35, and when n is a square. The first published proof that s(6) = 3 is by Kearney and Shiu [3] and the other results are reported in Erich Friedman’s dynamic survey [1]. We prove here that s(10) = 3+ √ 1 2 ≈ 3.707 (Theorem 1) and that s(11) ≥ 2+2 √ 4 5 ≈ 3.789 (Theorem 2). The 10-square packings in Figure 1 are optimal. The most efficient known packing of 11 squares, shown in Figure 2 and due to Walter Trump, has side about 3.8772 and includes unit squares tilted at about 40.182◦. s = 3 + √ 1 2 ≈ 3.707 Figure 1: Best packings of 10 squares the electronic journal of combinatorics 10 (2003), #R8 1