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

Abstract

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 $S(q)\le S(1/q)$ 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 $q\ge 0.6$ 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.

1 Introduction

A circle covering $\mathbf{K(q)}$ 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 $\mathbf{K(q)}$ and $q:=\inf (r_i/r_j),i,j=1,2,\ldots$. Given q with $0\le q\le 1$ 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 $0<q\le 1$

$$\begin{aligned} D(q)\ge S(q):=\min _{0<x\le 1}\frac{\pi -2(1-q^2)\arctan (\sqrt{1-x^2}/(x+q))}{2\sqrt{1-x^2}(x+q)} \end{aligned}$$

(1.1)

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 $\alpha ,\beta ,\gamma$ (see Fig. 1; to choose a scale b is set to be 1).

Fig. 1

Triangle T

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 $\varDelta T$ of T is given by

$$\begin{aligned} \delta (q):=\frac{\alpha r_1^2+\beta r_2^2+\gamma r_3^2}{2\varDelta T}. \end{aligned}$$

(1.2)

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 $\delta (q)$ that this assumption is satisfied by itself.

L. Fejes Tóth and J. Molnár had shown in [10] that $\min \delta (q)\le D(q)$ and assumed that $\min \delta (q)=S(q)$.

In this paper we consider a representation of $\delta (q)$ together with its adhering boundary conditions derived by the author in [2]. In this representation $\delta (q)$ is expressed as a function of $q,\beta ,r_1$ and $\gamma$ (see Fig. 1). We prove that for given q and $\beta$$\delta (q,\beta ,r_1,\gamma )$ has a unique stationary point at which $\delta (q,\beta ,r_1,\gamma )$ assumes its minimum $\delta _0(q,\beta )$. 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 $M_0(q):= \min _{\beta }\delta _0(q,\beta ) > S(q)$. As for the boundary values of B we prove that in order to find min$\delta (q)$ 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 $0<q\le 1$ and two radii of the circles $C_1, C_2, C_3$ coincide we will prove that min$\delta (q)=S(q)$ or min$\delta (q)= S(1/q)$, 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 $S(q)\le S(1/q)$. 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.

2 $\delta (q)$ for constant q and $\beta$

Referring to the parameters specified in Fig. 1 we assume that $r_1\ge r_2\ge r_3$ and $r_3/r_1=q$, that is $r_3=r_1q$. 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 $r_1\ge r_2\ge r_3$ we can further conclude that

$$\begin{aligned} \alpha \le \pi /2 \quad \mathrm {and} \quad 1/(1+q)\le r_1\le 1/(1-q). \end{aligned}$$

(2.1)

As derived in [2] $\delta (q)$ can be expressed by the equation:

$$\begin{aligned} \delta (q)= & {} r_1^2[(\pi -\gamma (1-q^2))\cot \gamma -(\pi -(\beta +\gamma )(1-q^2))\cot (\beta +\gamma )] \nonumber \\&+\,\beta \cot \beta -\beta \sqrt{4r_1^2-(r_1^2(1-q^2)+1)^2}, \end{aligned}$$

(2.2)

The following notations will turn out to be very useful:

$$\begin{aligned}&R:=r_1^2, p:=1-q^2, G:=(\pi -\gamma p)\cot \gamma -(\pi -(\beta +\gamma )p)\cot (\beta +\gamma ),\nonumber \\&\quad W:=\sqrt{4R-(Rp+1)^2}; \end{aligned}$$

(2.3)

due to $1/(1+q)\le r_1\le 1/(1-q)$ the function W is well defined.

With these notations (2.2) reads as

$$\begin{aligned} \delta (q)=R G+\beta \cot (\beta ) - \beta W. \end{aligned}$$

(2.4)

As for boundary conditions apart from (2.1) it was shown in [2] that the following inequalities have to be satisfied:

$$\begin{aligned} Rp\cot (\beta +\gamma )\le W-\cot \beta \le Rp\cot \gamma \end{aligned}$$

(2.5)

with the first inequality representing $r_2\le r_1$ and the second to $r_3\le r_2$.

We now agree to keep the parameters q and $\beta$ fixed with $0<q<1$ and $0<\beta <\pi$.

As for variables, we pick $\gamma$ and R, so that we consider $\delta$ as a function $\delta (\gamma ,R)$ subject to the boundary conditions (2.1) and (2.5). Condition (2.1) then assumes the form

$$\begin{aligned} \pi /2-\beta \le \gamma <\pi -\beta \ , \ \ \gamma > 0 \quad \mathrm {and} \quad 1/(1+q)^2\le R\le 1/(1-q)^2. \end{aligned}$$

(2.6)

In a more suggestive form the boundary condition (2.5) can be written as

$$\begin{aligned} \gamma \le \mathrm {arccot}\left( \frac{W-\cot \beta }{Rp}\right) \le \beta +\gamma . \end{aligned}$$

(2.7)

The domain of $\delta (\gamma ,R)$ 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 $\beta =0.628$).

Fig. 2

Domain A and $B\subset A$

Lemma 2.1

G is a positive, strongly convex function of $\gamma$, $\gamma \in (0,\pi -\beta )$. It assumes its minimium for $\gamma = \gamma _0$ which is the unique solution of the equation

$$\begin{aligned} p\cot \gamma _0+\frac{\pi -\gamma _0p}{\sin ^2\gamma _0}= p\cot (\beta +\gamma _0)+\frac{\pi -(\beta +\gamma _0)p}{\sin ^2(\beta +\gamma _0)}. \end{aligned}$$

(2.8)

Moreover, $\gamma _0\ge \pi /2-\beta$.

Proof

From $\delta (q)=RG+\beta \cot \beta -\beta W\ge 1$ and $\beta \cot \beta < 1$ we infer $G> \beta W/R\ge 0$.

Next we consider the function $g(x):=(\pi -xp)\cot (x)$ for $0< x < \pi )$ which constitutes $G=g(\gamma )-g(\beta +\gamma )$, further the first, second and third derivatives of g(x):

$$\begin{aligned} g'(x)= & {} -p\cot (x)-\frac{\pi -xp}{\sin ^2(x)},\ g''(x)=\frac{2}{\sin ^2(x)}\left( p+(\pi -xp)\cot (x)\right) \end{aligned}$$

(2.9)

$$\begin{aligned} g'''(x)= & {} \frac{-2}{\sin ^4(x)}\left( (2+\cos (2x))(\pi -px)+\frac{3p}{2}\sin (2x)\right) . \end{aligned}$$

(2.10)

$g'''(x)\le 0$: If $x\le \pi /2$ this is obvious. For $\pi /2<x<\pi$ it suffices to show that the function $f(x):=(2+\cos (2x)(\pi -x)+3/2\sin (2x)\ge 0$.

From $f'(x)=-2((\pi -x)\sin (2x)+1-\cos (2x))=-2\sin (2x)((\pi -x)-\tan (\pi -x))\le 0$ we obtain that f is decreasing and therefore, because of $f(\pi /2)=\pi /2$ and $f(\pi )=0$ we can conclude that $f\ge 0$.

Due to $g'''\le 0$ the second derivative $g''$ is decreasing, hence $G''(\gamma )=g''(\gamma )-g''(\beta +\gamma ) > 0$, i.e, G is strongly convex.

Equation (2.8) follows from $G'(\gamma )=g'(\gamma )-g'(\beta +\gamma )=0$. Because $g''(x)\ge 0$ for $x\le \pi /2$ the first derivative of g is increasing so that $\gamma _0\le \pi /2$ and $\gamma _0+\beta \le \pi /2$ cannot hold together, from which we get that in any case $\gamma _0\ge \pi /2-\beta$. $\square$

Theorem 2.2

The function $\delta (\gamma ,R)$ has exactly one stationary point $(\gamma _0,R_0)\in A$ at which it assumes its minimum within the domain A.

Proof

$\frac{\partial \delta }{\partial \gamma } = 0$ implies Eq. (2.8) from which we implicitly obtain the unique point $\gamma _0$ and $G_0:=G(\gamma _0)$.

From $\frac{\partial \delta }{\partial R} = 0$ we infer $G=\beta \frac{2-(Rp+1)p}{W}$. Solving this equation with respect to R we get

$$\begin{aligned} R_{1/2}=\frac{1}{p^2}\left( 1+q^2\pm 2q\frac{G}{\sqrt{p^2\beta ^2+G^2}}\right) . \end{aligned}$$

$R=R_1$ implies

$$\begin{aligned} 0<G=\frac{\beta }{W}\left( 2-p-(1+q^2+2q\frac{G}{\sqrt{p^2\beta ^2+G^2}})\right) =\frac{\beta }{W}\left( \frac{-2qG}{\sqrt{p^2\beta ^2+G^2}}\right) <0, \end{aligned}$$

a contradiction. So we choose the second root $R_2$ for R. By taking into account $\gamma _0$ and setting $R_0:=R_2$ we thus obtain

$$\begin{aligned} R_0=\frac{1}{p^2}\left( 1+q^2-2q\frac{G_0}{\sqrt{p^2\beta ^2+G_0^2}}\right) . \end{aligned}$$

(2.11)

$p^2=(1-q^2)^2=(1-q)^2(1+q)^2$, hence

$$\begin{aligned} \frac{1}{(1+q)^2}= & {} \frac{1-2q+q^2}{p^2}\le \frac{1}{p^2}\left( 1+q^2-2q\frac{G_0}{\sqrt{p^2\beta ^2+G_0^2}}\right) =R_0\\\le & {} \frac{1}{p^2}\left( 1+q^2+2q\frac{G}{\sqrt{p^2\beta ^2+G_0^2}}\right) \le \frac{1}{(1-q)^2}. \end{aligned}$$

According to the properties of $\gamma _0$ stated in Lemma 2.1 this shows that $(\gamma _0,R_0)\in A$.

$\frac{\partial ^2\delta }{\partial \gamma ^2}(\gamma _0,R_0)=R_0\frac{\partial ^2 G}{\partial \gamma ^2}>0$ by Lemma 2.1. Moreover, $\frac{\partial ^2\delta }{\partial \gamma \partial R}(\gamma _0,R_0) = \frac{\partial ^2\delta }{\partial R \partial \gamma }(\gamma _0,R_0) =\frac{\partial G}{\partial \gamma }(\gamma _0)=0$.

$$\begin{aligned} \frac{\partial ^2\delta }{\partial R^2} = \dots = \frac{\beta }{W}\left( p^2+\left( \frac{2-(Rp+1)p}{W}\right) ^2\right) >0. \end{aligned}$$

Therefore the Hesse determinant $H(\gamma _0,R_0)=R_0 \frac{\partial ^2 G}{\partial \gamma ^2}(\gamma _0,R_0)\frac{\partial ^2 G}{\partial R^2}(\gamma _0,R_0) >0$ which confirms that $\delta (\gamma ,R)$ has a relative minimum in its unique stationary point $(\gamma _0,R_0)$. Because $\delta (\gamma ,R)$ is continous we have found its absolute minimum $\delta _0(q,\beta )$ on A. $\square$

In Fig. 3 the function $\delta (\gamma .R)$ is plotted for $q=0.4$ and $\beta = \pi /3$.

Fig. 3

$\delta (\gamma ,R)$

Substituting $R_0$ in W one obtains

$$\begin{aligned} W_0=\frac{2q\beta }{\sqrt{p^2\beta ^2+G_0^2}}, \end{aligned}$$

(2.12)

and for $\delta _0(q,\beta )=\delta (\gamma _0,R_0)=R_0G_0+\beta (\cot \beta -W_0)$ a short calculation shows that

$$\begin{aligned} \delta _0(q,\beta )=\frac{1+q^2}{p^2}G_0-\frac{2q}{p^2}\sqrt{\beta ^2p^2+G_0^2}+\beta \cot \beta . \end{aligned}$$

(2.13)

Now we add the boundary conditions (2.7) to the conditions (2.6) calling forth two possibilities, namely $(\gamma _0,R_0)\in B$ and $(\gamma _0,R_0)\notin B$.

If $(\gamma _0,R_0)$ satisfies condition (2.6) then the minimum $\delta (\gamma _0,R_0)$ will be assumed within B in which case we introduce the following function:

Definition 2.3

Let $M_0(q):=\min _{\beta }\delta _0(q,\beta )$ for $\gamma _0 \le \mathrm {arccot}(\frac{W_0-\cot (\beta )}{R_0p})\le \beta +\gamma _0$, besides the assumption that $\gamma _0$ and $W_0$ can both be expressed by $G_0$ according to (2.11) and (2.12), respectively.

If $(\gamma _0,R_0)\notin B$, we have to check the following boundaries of B, what we will do aside the assumption that $(\gamma _0,R_0)\notin B$:

• (a) $\gamma =\pi /2-\beta$, if $\beta \le \pi /2$ and $\gamma =0$ for $\beta >\pi /2$,

• (b) $\gamma =\pi -\beta$,

• (c) $R=1/(1+q)^2$,

• (d) $R=1/(1-q)^2$,

• (e) $Rp\cot (\beta +\gamma )=W-\cot \beta$, which stands for $r_1=r_2$ and

• (f) $Rp\cot \gamma =W-\cot \beta$, which represents the case $r_3=r_1q=r_2$.

According to Theorem 4.1 and its proof in [2] min $\delta (q)$ can only occur within triangles T with $\alpha \le \gamma$ or $\alpha =\beta$ or $r_2=r_3$ (the latter leading to case (f)). For discussing the cases (a) and (b) we will assume these conditions.

Using $\rightarrow$ to indicate that a variable approaches to a point and also for a function to converge to a limit, we thus obtain

(a) If $\gamma =\pi /2-\beta$ for $\beta \le \pi /2$ then $\alpha \le \gamma$ or $\alpha =\beta$ implies $\alpha =\gamma =\pi /2$ or $\alpha =\beta = \pi /2$, respectively. If both, $\alpha$ and $\gamma$, approach $\pi /2$, then $R \rightarrow \infty$ which because of $R\le 1/(1-q)^2$ means that $q\rightarrow 1$. Therefore we can conclude that $\delta (q)\ge$ min $\delta (1)$. If $\alpha$ and $\beta$ approach $\pi /2$ then $\gamma \rightarrow 0$ and by Eq. (2.2) we obtain $\delta (q) \rightarrow \infty$.

If  $\pi /2<\beta <\pi$ then $\gamma =0$ implies $\alpha =0$ for $\alpha \le \gamma$ from which we can infer $\delta (q)\rightarrow \infty$, and with respect to $\alpha =\beta$ the outcome will be the same.

(b) $\gamma =\pi -\beta$, i.e. $\alpha \rightarrow 0$:

$$\begin{aligned} \delta (q)={\lim }_{\alpha \rightarrow 0}(R[(\pi -(\pi -\beta )p)\cot (\beta )+(\pi -\pi p)\cot \alpha ]+\beta \cot \beta -\beta W\rightarrow +\infty . \end{aligned}$$

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 $\beta$, 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).

Fig. 4

$r_1=1/(1+q)$

3 Two radii in a line

If $r_1=1/(1+q)$ and no circle cuts into an opposite side of T, $r_1$ and $r_3\; (=r_1q^2)$ 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 $\delta (q)$ anew on the basis of Fig. 4. From this drawing we can read off readily:

The area $\varDelta T$ of the triangle T is given by $2\varDelta T=r_2 \; \tan (\alpha )=r_2(1+q)=q\tan (\gamma )$, $\arctan (q)\le \alpha \le \pi /4$ and, looking for the smallest possible and the largest $\beta$, we get $\pi /4+\arctan (q)\le \beta \le \pi /4+\arctan (1/q)$ which is equivalent to $\arctan q\le \alpha \le \pi /4$.

We therefore obtain

$$\begin{aligned} \delta (q,\beta )= & {} \frac{1}{r_2}\left( \frac{\alpha }{(1+q)^2} +\beta \, r_2^2 +\frac{\gamma \, q^2}{(1+q)^2}\right) \end{aligned}$$

(3.1)

$$\begin{aligned}= & {} \frac{1}{r_2}\left( \alpha \, r_2^2 \cot ^2\alpha +\beta \, r_2^2 + \gamma \, q^2 r_2^2\cot ^2\alpha \right) \nonumber \\= & {} \frac{\tan \alpha }{1+q} \left( (\alpha +\gamma \,q^2)\cot ^2\alpha +\pi -\alpha -\gamma \right) \nonumber \\= & {} \frac{1}{1+q}\left( \alpha (\cot \alpha -\tan \alpha )+\pi \,\tan \alpha +\gamma \,(q^2\cot \alpha -\tan \alpha )\right) \nonumber \\= & {} \frac{1}{1+q} \left( 2\alpha \,\cot (2\alpha )+\pi \,\tan \alpha +\arctan \left( \frac{1}{q}\tan \alpha \right) (q^2\cot \alpha -\tan \alpha )\right) \nonumber \\=: & {} \bar{\delta }_1(q,\alpha ). \end{aligned}$$

(3.2)

Theorem 3.1

If $q\ge 0.6$ then $\delta (q,\beta )\ge 1.219...>S(1)$ for every $\beta$.

Proof

As shown above $\delta (q,\beta )=\bar{\delta }_1(q,\alpha )$. Observing that $1/r_2=(1+q)\cot \alpha \ge 1+q$ and $\tan \alpha \ge q$, substituting $\gamma$ within Eq. (3.1) by $\pi -\beta -\alpha$ we obtain

$$\begin{aligned} \bar{\delta }_1(q,\alpha )= & {} \frac{1}{r_2}\frac{\alpha (1-q^2)}{(1+q)^2}+\beta \,r_2+\frac{(\pi -\beta )q^2}{(1+q)^2\,r_2}\ge \frac{\alpha \,(1-q)}{r_2(1+q)} +\beta \frac{\tan \alpha }{1+q}+\frac{(\pi -\beta )q^2}{(1+q)}\\&\ge \alpha \,\cot (\alpha )(1-q)+\beta \frac{\tan \alpha }{1+q}+\frac{\pi \,q^2}{1+q}-\frac{\beta \,q^2}{1+q}\\&\ge \alpha \,\cot (\alpha )(1-q)+\frac{\beta \,(q-q^2)}{1+q}+\frac{\pi \,q^2}{1+q}. \end{aligned}$$

Since $\alpha \,\cot \alpha$ is a decreasing function and $\alpha \le \pi /4$ we further obtain

$$\begin{aligned} \bar{\delta }_1(q,\alpha )\ge \frac{\pi }{4}(1-q)+ \frac{\beta \,q}{1+q}(1-q)+\frac{\pi \,q^2}{1+q}=:\bar{\epsilon }_1(q,\beta ). \end{aligned}$$

$\bar{\epsilon }_1(q,\beta )$ is an increasing function of $\beta$ from which we can infer that

$$\begin{aligned} \bar{\delta }_1(q,\alpha )\ge \frac{\pi }{4}(1-q)+\frac{q(1-q)}{1+q}\left( \frac{\pi }{4}+\arctan q\right) +\frac{\pi \,q^2}{1+q}=:\tilde{\delta }_1(q). \end{aligned}$$

$\tilde{\delta }_1(q)$ is an increasing function of q (easily checked for $q\ge 0.5$), and $\tilde{\delta }_1(0.6)>1.219>S(1)$.$\square$

Definition 3.2

For $0<q<0.6$ and $\bar{\delta }_1(q,\alpha )$ according to (3.2) we set $M_1(q):=\min _{\alpha }\bar{\delta }_1(q,\alpha )$ for $\arctan (q)\le \alpha \le \pi /4$.

Remark 3.3

As one can see by numerical calculations the minimum of $\bar{\delta }_1(q,\alpha )$ is assumed for $\alpha =\arctan q$, hence one would obtain

$$\begin{aligned} M_1(q)= \frac{1}{1+q}\left( \pi \,q+\frac{1-q^2}{q}\arctan q\right) . \end{aligned}$$

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.

Fig. 5

$r_1=1/\sqrt{1+q^2}$

From Fig. 5 one immediately realizes substituting $r_1$ with r and $a=rq+r_2$ that $r_2=r\cot \beta$, $a=r(q+\cot \beta )$, $2\varDelta T=ar$, $\sin \gamma =r=1/\sqrt{1+q^2}$, $\gamma =\mathrm {arccot}(q)$ and the range of $\beta$ is $\pi /4\le \beta \le \mathrm {arccot}(q)$.

Representing $\delta (q)$ as a function of q and $\beta$, referred to as $\delta _2(q,\beta )$, we therefore obtain:

$$\begin{aligned} \delta _2(q,\beta )= & {} \frac{1}{ar}(\alpha r^2+\beta r^2\cot ^2\beta +\gamma r^2q^2)=\frac{r}{a}(\pi -\beta -\gamma +\beta \cot ^2\beta +\gamma q^2)\\= & {} \frac{1}{q+\cot \beta }(\pi -\beta +\beta \cot ^2\beta -(1-q^2)\mathrm {arccot}(q)). \end{aligned}$$

As one can easily check by means of the first partial derivative of $\delta _2(q,\beta )$, $\delta _2(q,\beta )$ is a decreasing function of $\beta$ for $\pi /4\le \beta \le \pi /2$. Setting $\beta =\mathrm {arccot}(q)$ and considering the function $s(q,x):= \frac{\pi -2(1-q^2)\arctan (\sqrt{1-x^2}/(x+q))}{2\sqrt{1-x^2}(x+q)}$ which constitutes S(q) (see Eq. (1.1)), one gets

$$\begin{aligned} \min _{\beta }\delta _2(q,\beta )=\frac{1}{2q}(\pi -2(1-q^2)\mathrm {arccot}(q))=s(q,0)\ge S(q). \end{aligned}$$

This way we have proven.

Theorem 3.4

$\min _{\beta }\delta _2(q,\beta )\ge S(q)$.

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.

4 Coverings with only two kind of circles

Theorem 4.1

If $r_1=r_2$ then $\min \delta (q)=S(q)$, and if $r_2=r_3$ then $\min \delta (q)=S(1/q)$.

Proof

According to (2.5) $r_1=r_2$ means $Rp\cot (\beta +\gamma ) = W-\cot (\beta )$, a quadratic equation in R which entails the solution ([2],(10))

$$\begin{aligned} R_{12}:=\frac{\sin \alpha }{p^2\sin \beta }\left( 2\sin \alpha \sin \beta -p\cos \gamma -\sqrt{(2\sin \alpha \sin \beta -p\cos \gamma )^2-p^2}\right) . \end{aligned}$$

Setting $R=R_{12}$ in $\delta (q)=RG+\beta \cot (\beta )-\beta W$ yields

$$\begin{aligned} \delta _{12}(q,\beta ,\gamma ):=\frac{\pi -\gamma p}{p^2\sin \gamma }\left( q^2\cos \gamma -\cos (2\beta +\gamma )-\sqrt{(q^2\cos \gamma -\cos (2\beta +\gamma )^2-p^2}\right) . \end{aligned}$$

Keeping $\gamma$ fixed, within a few calculation steps one obtains that

$$\begin{aligned} \frac{\partial \delta _{12}(q,\beta ,\gamma )}{\partial \beta }=-2\frac{\sin (2\beta +\gamma )}{\sqrt{(q^2\cos \gamma -\cos (2\beta +\gamma )^2-p^2}}\,\delta _{12}(q,\beta ,\gamma ). \end{aligned}$$

This shows: If $2\beta +\gamma \le \pi$ then $\delta _{12}(q,\beta ,\gamma )$ is decreasing in $\beta$ and if $2\beta +\gamma \ge \pi$ then $\delta _{12}(q,\beta ,\gamma )$ is increasing, hence $\delta _{12}(q,\beta ,\gamma )$ is minimal for $2\beta +\gamma =\pi$ which means $\alpha =\beta$. If this is the case then one can find by some elaborate but elementary calculations (cf. [2],(14)) that

$$\begin{aligned} \delta _{12}(q,\beta ,\pi -2\gamma )= & {} \dots =\frac{\pi q^2+2\beta (1-q^2)}{(1-q^2)^2\sin 2\beta }\left( q\sin \beta -\sqrt{1-q^2\cos ^2\beta }\right) ^2\nonumber \\=: & {} \delta _{12}(q,\beta ) \end{aligned}$$

(4.1)

Fig. 6

Geometric interpretation of s(2, x)

If $r_2=r_3$ then (as already derived in [2]) $W-\cot (\beta )=Rp\cot \gamma$ yields

$$\begin{aligned} R_{23}:= \frac{\sin \gamma }{p^2\sin \beta }\left( 2\sin \beta \sin \gamma -p\cos (\beta -\gamma )-\sqrt{(2\sin \beta \sin \gamma -p\cos (\beta -\gamma ))^2-p^2}\right) , \end{aligned}$$

which, substituted for R within $\delta (q)=RG+\beta \cot (\beta )-\beta W$ and writing $\pi -\alpha -\gamma$ instead of $\beta$ entails

$$\begin{aligned} \delta _{23}(q,\alpha ,\gamma ):=\frac{\pi q^2+\alpha p}{p^2\sin \alpha }\left( \cos \alpha -q^2\cos (\alpha +2\gamma )-\sqrt{(\cos \alpha -q^2\cos (\alpha +2\gamma ))^2-p^2}\right) . \end{aligned}$$

One can easily deduce that

$$\begin{aligned} \frac{\partial \delta _{23}(q,\alpha ,\gamma )}{\partial \gamma } =-2\frac{q^2\sin (\alpha +2\gamma )}{\sqrt{(\cos \alpha -q^2\cos (\alpha +2\gamma )^2-p^2}}\delta _{23}(q,\alpha ,\gamma ), \end{aligned}$$

and ascertain that $\delta _{23}(q,\alpha ,\gamma )$ is minimal for $\alpha +2\gamma =\pi$, i.e. $\beta = \gamma$. In this case (as already carried out in [2]) elementary calculations lead to

$$\begin{aligned} \delta _{2,3}(q,\alpha ,(\pi -\alpha )/2)= \dots =\frac{\pi -2\beta (1-q^2)}{(1-q^2)^2\sin 2\beta }\left( \sin \beta -\sqrt{q^2-\cos ^2\beta }\right) ^2=:\delta _{23}(q,\beta ).\nonumber \\ \end{aligned}$$

(4.2)

Fig. 7

$S(q),M_0(q),S_1(q),M_1(q)$

As also conveyed in [2], interpreting geometrically x and q within the function $s(q,x):= \frac{\pi -2(1-q^2)\arctan (\sqrt{1-x^2}/(x+q))}{2\sqrt{1-x^2}(x+q)}$ constituting S(q), straightforward calculations link $\beta$ to x (see Fig. 6 for $\alpha =\beta$): One obtains $\min _{0<\beta <\pi /2} \delta _{12}(q,\beta )=S(q)$ and $\min _{0<\beta <\pi /2}\delta _{2,3}(q,\beta )=S(1/q)$. $\square$

5 Conclusions

We have come so far to know that for all $q\ge 0.6$$\min \delta (q)$ is to be found among the minima of the densities $\delta _0(q,\beta )$ (Eq. 2.13), $\delta _{12}(q,\beta )$ (Eq. 4.1) and $\delta _{23}(q,\beta )$ (Eq. 4.2) described by the functions $M_0(q)$ (Definition 2.3), S(q) and S(1 / q) (see Eq.(1.1)).

For $0<q<0.6$ we have not only to take into account the minima of $\delta _0(q,\beta )$, $\delta _{12}(q,\beta )$ and $\delta _{23}(q,\beta )$ but also the minimum of $\bar{\delta }_1(q,\alpha )$ 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 $0.6\le q\le 1$ — 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.

The functions $M_0(q)$, $M_1(q)$, S(q) and S(1 / q) are visualized in Fig. 7.