A Note on Rectangle Covering with Congruent Disks

In this note we prove that, if Sn is the greatest area of a rectangle which can be covered with n unit disks, then 2 ≤ Sn/n < 3 √ 3/2, and these are the best constants; moreover, for ∆(n) := (3 √ 3/2)n − Sn, we have 0.727384 < lim inf ∆(n)/ √ n < 2.121321 and 0.727384 < lim sup∆(n)/ √ n < 4.165064. The problem of covering sets in the plane with figures of prescribed shape has been extensively studied in literature–even though the dual packing problem received comparatively much more attention–both from the theoretical and computational viewpoint, also in virtue of its practical applications. In this note we study the extreme values for the area of a rectangle covered by a fixed number of congruent disks. Our aim is here to give precise bounds for the maximum value of this area. Let then Sn be the greatest area of a rectangle which can be covered with n closed disks of unit radius. Here we prove the following two facts. Theorem 1. For every n ∈ N, 2n ≤ Sn < 3 √ 3 2 n. These are the best possible constants: minn∈N Sn/n = 2 and lim supn→∞ Sn/n = 3 √ 3/2. Define moreover ∆(n) := 3 √ 3 2 n− Sn, α := lim inf n→∞ ∆(n) √ n , β := lim sup n→∞ ∆(n) √ n . Then one has Theorem 2. 0.727384 . . . ≤ α ≤ 2.121320 . . . 0.727384 . . . ≤ β ≤ 4.165063 . . . First, let C1, . . . , Cn be the circles covering a rectangle (that we treat as fixed) and O1, . . . , On their centers, and recall that the Voronoi cell Vori of the circle Ci is the set of points Q inside the rectangle such that the distance of Q from Oj is greater or equal than its distance from Oi for all j 6= i. Proof of Theorem 1. The leftmost inequality is trivial. Just take a rectangle built by juxtaposing n squares, each inscribed in a circle as in Figure 1: each square has area 2, hence the rectangle has area 2n. The constant 2 is the best possible one because the largest rectangle with fixed circumcircle is the square, that is we have equality for n = 1. 2010 Mathematics Subject Classification. 52C15, 05B40. 1