The closest packing of convex two-dimensional domains
KeywordsCommon Point Close Packing Lattice Packing Convex Cover Common Boundary Point
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- 1.Acta Sci. Math. (Szeged), 12 (1950), 62–67, see Theorem I and the remarks on page 66.Google Scholar
- 2.A convex polygon having at most six sides will be called a hexagon.Google Scholar
- 3.My first result, Theorem 2, was obtained in 1947, and was described in seminars in London, Cambridge, Bristol and Princeton in the years 1948–49; its most important consequence was announeed in a paper byJ. H. H. Chalk and myself (J. L. M. S., 23 (1948), 178–187 (179)). Detailed proofs of the results were given in the version of the present paper originally submitted to Acta mathematica.Google Scholar
- 1.This inequality is not difficult to prove. By continuity considerations it suffices to prove the inequality in the case whenK is strictly convex. If one considers a lattice Λ with determinantd(K) giving a lattice packing of strictly convex setsK, it follows from the well know theory ofMinkowski (Diophantische Approximationen (Teubner, Berlin 1947), § 4, or seeK. Mahler, Proc. London Math. Soc. (2) 49 (1946), 158) that each setK+x withx in Λ has a boundary point in common with the boundaries of just six of the other sets of this form. If tac-lines are drawn toK through these six points of contact, care being taken to ensure that opposite tac-lines are parallel, they bound an open hexagonH cireumscribingK, and no two of the hexagonsH+x withx in Λ have common points. Thus we see thath(K)≤a(H)≤d(A)=d(K).Google Scholar
- 2.Abh. Math. Sem. Hamb. Univ., 10 (1934), 216–230 or seeK. Mahler, Proc. K. Ned. Akad. v. Wet. (Amsterdam), 50 (1947), 692–703.Google Scholar
- 1.An open setK is said to be strictly convex if it is such that, for every pair of distinct pointsa andb on the boundary ofK, every inner pointc of the line segment joininga tob is inK.Google Scholar
- 1.W. Blaschke,Kreis und Kugel (Leipzig, 1916), § 18.Google Scholar
© Almqvist & Wiksells Boktryckeri 1951