Tuyển tập các báo cáo nghiên cứu khoa học về toán học trên tạp chí toán học quốc tế đề tài: On a covering problem for equilateral triangles. | On a covering problem for equilateral triangles Adrian Dumitrescu Minghui Jiang y Department of Computer Science University of Wisconsin-Milwaukee Milwaukee WI 53201-0784 USA Email ad@ Department of Computer Science Utah State University Logan UT 84322-4205 USA Email mjiang@ Submitted July 22 2006 Accepted Feb 26 2008 Published Feb 29 2008 Mathematics Subject Classification 52C15 Abstract Let T be a unit equilateral triangle and Tt . Tn be n equilateral triangles that cover T and satisfy the following two conditions i Ti has side length ti 0 tị 1 ii Ti is placed with each side parallel to a side of T. We prove a conjecture of Zhang and Fan asserting that any covering that meets the above two conditions i and ii satisfies 52n t ti 2. We also show that this bound cannot be improved. 1 Introduction Inspired by an old problem of Erdos about packing smaller squares in a unit square 2 3 4 Zhang and Fan 7 have recently considered the following covering problem for the equilateral triangle. Let T be a unit equilateral triangle and T Tt . Tng be a set of n equilateral triangles that cover T and satisfy the following two conditions i Ti has side length ti 0 ti 1 ii T is placed with each side parallel to a side of T. All triangles are viewed as closed sets. An example of such a covering with 5 triangles is shown in Fig. 1 a . Define U n inf T covering T n n X ti. i t Since the triangles in the covering are smaller than T each triangle Ti can cover at most one vertex of T so the condition n 3 is necessary. Recently Zhang and Fan showed the following upper bounds on U n U n 3 n for even n 4 U n 4 nêã for odd Supported in part by NSF CAREER grant CCF-0444188. Supported in part by NSF grant DBI-0743670. THE ELECTRONIC JOURNAL OF COMBINATORICS 15 2008 R37 1 n 7. In particular U 4 2 follows. They also found that U 3 2 and U 5 9 4. It should be noted here that the inequality U k 2 for some k 3 does not imply for instance U k 1 2 so in particular having U 3 2 does
