On a conjecture of L. Fejes Tóth and J. Molnár about circle coverings of the plane

Tóth and Molnár (Math Nachr 18:235–243, 1958) formulated the conjecture that for a given homogeneity q the thinnest covering of the Euclidean plane by arbitrary circles is greater or equal a function S ( q ) . Florian (Rend Semin Mat Univ Padova 31:77–86, 1961) proved that if the covering consists of only two kinds of circles then the conjecture is true supposed that S ( q ) ≤ S ( 1 / q ) what can be easily veriﬁed by a computer. In this paper we consider the general case of circles with arbitrary radii from an interval of the reals. We set up two further functions M 0 ( q ) and M 1 ( q ) and prove that the conjecture is true if S ( q ) is less than or equal to S ( 1 / q ) , M 0 ( q ) and M 1 ( q ) . As in the case of two kinds of circles this can be readily conﬁrmed by computer calculations. (For q ≥ 0 . 6 we even do not need the function M 1 ( q ) for computer aided comparisons.) Moreover, we obtain Florian’s result in a shorter different way.


Introduction
A circle covering K(q) of homogeneity q of the Euclidean plane is a countable set of closed circular discs C i with radii r i such that every point of the plane belongs to at least one circle of K(q) and q := inf(r i /r j ), i, j = 1, 2, . . .. Given q with 0 ≤ q ≤ 1 it is of interest to determine the density D(q) of a thinnest covering of the plane. (For a formal definition of a thinnest covering see, for example, the classical book [8].) There have been several attempts to determine D(q) or accurate bounds of it. For upper bounds of D(q) cf. [2,[5][6][7][8], for lower bounds [1,3,4,9,10]. The most outstanding open conjecture concerning a lower bound (considered by G. Blind in 1974 as "hopeless to prove analytically" (cf. [1])) had been proposed by Tóth and Molnár (first published 1958, [10]). It states that for 0 < q ≤ 1 Dedicated to the memory of August Florian.
with equality for q = 1 and, in the limit, for q = 0. Geometrically S(q) is based on the following setting. The centres O 1 , O 2 , O 3 of three closed circular discs C 1 , C 2 , C 3 with radii r 1 , r 2 , r 3 which have exactly one common point (but no common inner point) form a triangle T with angles α, β, γ (see Fig. 1; to choose a scale b is set to be 1).
Supposing that no circle cuts into the opposite side of T the quotient of the area covered by C 1 , C 2 , C 3 within T and the area ΔT of T is given by That no circle cuts into the opposite side of T can be neglected if it turns out in the course of determining the minimum of the function δ(q) that this assumption is satisfied by itself. L. Fejes Tóth and J. Molnár had shown in [10] that min δ(q) ≤ D(q) and assumed that min δ(q) = S(q).
In this paper we consider a representation of δ(q) together with its adhering boundary conditions derived by the author in [2]. In this representation δ(q) is expressed as a function of q, β, r 1 and γ (see Fig. 1). We prove that for given q and β δ(q, β, r 1 , γ ) has a unique stationary point at which δ(q, β, r 1 , γ ) assumes its minimum δ 0 (q, β). However, this minimum is not always assumed within the limits of the domain B defined by all boundary conditions. If the minimum is attained within B computer calculations show that M 0 (q) := min β δ 0 (q, β) > S(q). As for the boundary values of B we prove that in order to find minδ(q) only two further functions remain which have to be compared to S(q) by computer aid. First, if q < 0.6 and |O 1 , O 3 | = r 1 + r 3 , the minimum of the covering density will yield a function M 1 (q) which has to be taken into account. Second, if 0 < q ≤ 1 and two radii of the circles C 1 , C 2 , C 3 coincide we will prove that minδ(q) = S(q) or minδ(q) = S(1/q), this way recovering a result by Florian [3], but in a totally different (and shorter) way. As already pointed out in [3] computer calculations establish that S(q) ≤ S(1/q). Showing by computer aid that also M 1 (q) > S(q) for 0 < q < 0.6 we therefore obtain that S(q) is the smallest of the four functions M 0 (q), M 1 (q), S(1/q) and S(q) for any homogeneity q, this way confirming the conjecture of Tóth and Molnár.

ı(q) for constant q andŘ
eferring to the parameters specified in Fig. 1 we assume that r 1 ≥ r 2 ≥ r 3 and r 3 /r 1 = q, that is r 3 = r 1 q. Because the circles C 1 , C 2 , C 3 have no common inner point we can suppose that their common intersection point lies within the triangle T . In view of r 1 ≥ r 2 ≥ r 3 we can further conclude that α ≤ π/2 and 1/(1 + q) ≤ r 1 ≤ 1/(1 − q). (2.1) As derived in [2] δ(q) can be expressed by the equation: 2) The following notations will turn out to be very useful: With these notations (2.2) reads as As for boundary conditions apart from (2.1) it was shown in [2] that the following inequalities have to be satisfied: with the first inequality representing r 2 ≤ r 1 and the second to r 3 ≤ r 2 .
We now agree to keep the parameters q and β fixed with 0 < q < 1 and 0 < β < π.
As for variables, we pick γ and R, so that we consider δ as a function δ(γ, R) subject to the boundary conditions (2.1) and (2.5). Condition (2.1) then assumes the form In a more suggestive form the boundary condition (2.5) can be written as The domain of δ(γ, R) defined by (2.6) and (2.7) will be referred to by B. If we only consider the rectangle defined by (2.6) we will refer to it as A. -A typical shape of B is illustrated by Fig. 2 (for which q = 3/5 and β = 0.628). Lemma 2.1 G is a positive, strongly convex function of γ , γ ∈ (0, π − β). It assumes its minimium for γ = γ 0 which is the unique solution of the equation , further the first, second and third derivatives of g(x): ≥ 0 for x ≤ π/2 the first derivative of g is increasing so that γ 0 ≤ π/2 and γ 0 + β ≤ π/2 cannot hold together, from which we get that in any case γ 0 ≥ π/2 − β.

Theorem 2.2
The function δ(γ, R) has exactly one stationary point (γ 0 , R 0 ) ∈ A at which it assumes its minimum within the domain A.
. Solving this equation with respect to R we get a contradiction. So we choose the second root R 2 for R. By taking into account γ 0 and setting R 0 := R 2 we thus obtain (2.11) According to the properties of γ 0 stated in Lemma 2.1 this shows that (γ 0 , R 0 ) ∈ A.
According to Theorem 4.1 and its proof in [2] min δ(q) can only occur within triangles T with α ≤ γ or α = β or r 2 = r 3 (the latter leading to case (f)). For discussing the cases (a) and (b) we will assume these conditions.
As for the cases (c) and (d), since our goal is at last to find minimal values on the boundaries of B in respect to all angles β, we will make use of the assumption within conjecture (1.1) that no circle must cut into the opposite side of the triangle T . However, in case (d) this will alter the upper bound of R.
The next section will be devoted to the cases (c) and (d).

Two radii in a line
If r 1 = 1/(1 + q) and no circle cuts into an opposite side of T , r 1 and r 3 (= r 1 q 2 ) form the segment of a straight line, and the third radius is orthogonal to this line (see Fig. 4). We will establish a representation for δ(q) anew on the basis of Fig. 4. From this drawing we can read off readily: The area ΔT of the triangle T is given by 2ΔT = r 2 tan(α) = r 2 (1 + q) = q tan(γ ), arctan(q) ≤ α ≤ π/4 and, looking for the smallest possible and the largest β, we get π/4 + arctan(q) ≤ β ≤ π/4 + arctan(1/q) which is equivalent to arctan q ≤ α ≤ π/4.

Remark 3.3
As one can see by numerical calculations the minimum ofδ 1 (q, α) is assumed for α = arctan q, hence one would obtain Next we will discuss case (d) altered by the assumption that no circle cuts into an opposite side of the triangle T . Then due to r 1 being as long as possible the situation illustrated in Fig. 5 occurs.

Theorem 3.4 min β δ 2 (q, β) ≥ S(q).
Next we will discuss the cases (e) and (f) of the boundary conditions stated above, yet without the assumption that no circle must cut into an opposite side of the triangle T , since looking for minima this fact will turn out by itself.
For 0 < q < 0.6 we have not only to take into account the minima of δ 0 (q, β), δ 12 (q, β) and δ 23 (q, β) but also the minimum ofδ 1 (q, α) which gives rise to M 1 (q) (Definition 3.2). With this in mind we can formulate the following Theorem 5.1 If M 0 (q) and S(1/q) are less than or equal to S(q) for 0.6 ≤ q ≤ 1 -this can be verified by computer calculations -the conjecture of L. Fejes Tóth and J. Molnár can be considered as confirmed.
If M 0 (q), M 1 (q) and S(1/q) are less than or equal to S(q) for 0 < q < 0.6 -this can also be shown by computer calculations -the correctness of the conjecture ensues.