Thinnest Covering of a Circle by Eight, Nine, or Ten Congruent Circles

Transcription

1 Combinatorial Computational Geometry MSRI Publications Volume 5, 005 Thinnest Covering of a Circle by Eight, Nine, or Ten Congruent Circles Abstract. Let r n be the maximum radius of a circular disc that can be covered by n closed unit circles. We show that r n = + cos(π/()) for n = 8, n = 9, n = 0.. Introduction What is the maximum radius r n of a circular disk which can be covered by n closed unit circles? The determination of r n for n 4 is an easy task: we have r = r =, r 3 = / 3 r 4 =. The problem of finding r 5 has been motivated by a game popular on fairs around the turn of the twentieth century [Neville 95; Ball Coxeter 987, pages 97 99]. The goal of the game was to cover a circular space painted on a cloth by five smaller circles equal to each other. The difficulty consisted in the restriction that an on-line algorithm had to be used, that is no circle was allowed to be moved once it had been placed. Neville [95] conjectured that r 5 = this has been verified by K. Bezdek [979; 983] who also determined the value of r 6 = The proofs of these cases are complicated. The case n = 7 is again easy. We have r 7 = if 7 unit circles cover a circle C 7 of radius, then one of them is concentric with C 7 while the centers of the other circles lie in the vertices of a regular hexagon of side 3 concentric with C 7. In his thesis Dénes Nagy [975] claimed without proof that r n = + cos ( π/() ) for n = 8 n = 9 that, as for n = 7, the best arrangement has ()-fold rotational symmetry. He conjectured the same for n = 0. Krotoszyński [993] claimed to have proved this even for n. Unfortunately, his proof contains some errors. In fact, Melissen Schuur (see [Melissen 997]) gave a counter example for n =. Part of this work was done at the Mathematical Sciences Research Institute at Berkeley, CA, where the author was participating in a program on Discrete Computational Geometry. The research was also supported by OTKA grants T 0300, T , T

2 36 In this note we settle the cases n = 8, n = 9, n = 0. Theorem. Let r n be the maximum radius of a circular disk which can be covered by n unit circles. Then we have for n = 8, 9, 0 r n = + cos π n. Moreover, if for n = 8, n = 9, or n = 0, n unit circles cover a circle C n of radius r n, then one of them is concentric with C n the centers of the other circles are situated in the vertices of a regular ()-gon at a distance sin ( π/() ) from the center of C n. The analogous problems of the thinnest covering of a square an equilateral triangle with a given number of equal circles, as well as the dual problem concerning the densest packing of a given number of equal circles in a circle, a square or an equilateral triangle have been investigated intensively. A comprehensive account can be found in [Melissen 997]. Generally, given a compact set C in a metric space, one can consider the problems of the densest packing of n balls in C the thinnest covering of C with n balls. In lack of similarity the problems are formulated in a dual form. Let r C (n) be the maximum number with the property that n balls of radius r C (n) can be packed in C let R C (n) be the minimum number with the property that n balls of radius r C (n) can cover C. The basic task is, of course, to design effective algorithms determining the values of r C (n) R C (n), as well as the corresponding arrangements. So far only the case of r C (n) for C a square has been solved, by an algorithm devised by Peikert [994]; see also [Peikert et al. 99]. Exact solutions are generally known only for small values of n. The only exception is the problem of densest packing of circles in an equilateral triangle. When C is an equilateral triangle, r C (n) is known for all n of the form k(k + )/, the triangular numbers; see [Groemer 960; Oler 96]. If C is an equilateral triangle with side-length, we have for such triangular numbers r C (n) = /(k + 3 ). The optimal arrangement is given by the regular triangular lattice. Many conjectured best arrangements of circles, both for packing for covering, have been constructed using different heuristic algorithms. The examples show that optimal arrangements quite often contain freely movable circles. This raises the following questions. Does there exist a compact set C for which for infinitely many n an optimal packing of (covering with) n congruent circles contains a freely movable circle? Does there exist a C for which there is no n at all such that an optimal packing of (covering with) n congruent circles contains a freely movable circle? Is there a constant c, possibly depending on C but independent of n such that the number of freely movable circles in an optimal arrangement is at most c?

3 COVERING A CIRCLE WITH EIGHT, NINE, OR TEN CIRCLES 363 The densest packing of n = k(k + )/ circles in an equilateral triangle shows another interesting phenomenon. According a conjecture of Erdős Oler [Croft et al. 99, page 48] if n is a triangular number, then r C (n) = r C (n ), that is, the optimal arrangement for n circles is obtained by removing one circle from the optimal arrangement of n circles. The conjecture has been confirmed for n = 6 n = 0 [Melissen 997]. There is a similar situation on the sphere: it is known (see [Rankin 955], for example) that if C = S d, the d-dimensional sphere, then r C (d + 3) = r C (d + 4) =... = r C (d + ). This suggests the following question. For a given compact set C, are there natural numbers k = k(c) K = K(C) such that r C (n) > r C (n + k) R C (n) > R C (n + K) for every n? I conjecture that the answer is yes if C = S d also if C is a compact convex set in Euclidean or spherical space, but I would not be surprised if the answer turned out to be no for general compact sets, or even for convex bodies in hyperbolic geometry.. Proof of the Theorem For the proof we modify the argument used by Schütte [955] for the determination of the thinnest covering of the sphere by 5 7 congruent caps. Clearly, it suffices to show the second statement of the theorem, from that it follows immediately that no circle of radius greater than r n can be covered by n unit circles (n = 8, n = 9, o). The proof of the three cases are similar, however the case n = 0 is more complicated. We shall leave two of the more involved discussions for n = 0 to Section 3. In the treatment of all three cases the functions f r (α) F r (α) defined for 0 α π by f r (α) = arcsin sin(α/) r ( F r (α) = r arcsin sin(α/) ( r sin arcsin sin(α/) r )) + sin α play an important role. Here r > is not a variable but a parameter. The geometric meaning of f r (α) F r (α) is the following: Let C be a circle of radius r centered at o let C be a unit circle with center õ C such that bdc bd C intersect, say in a b. If ]aõb = α, then f(α) = ]aob F(α) is the area of the domain bounded by the segments õa, õb the arc ab of bdc. We have f r (α) = cos(α/) ( r sin (α/) ), f / r (α) = (r )sin(α/) ( r sin (α/) ), 3/

4 364 F r (α) = F r (α) = ( sin α sin α sin(α/) ( r sin (α/) ) + cosα /, ) ( cosα ( sin (α/) r ) r cos (α/) (r sin (α/)) 3/ + cos α Hence it is easily seen that f r (α) is concave strictly increasing for 0 α π. The concavity of F r (α) needs some calculation. To check it, we have to show that ( cos α sin α r) r cos α + cos α ( r sin α ) 3/ > 0. Introducing the abbreviations s = sin(α/) c = cos(α/), we have cosα ( s r ) r c + c ( r s ) 3/ > cos α ( s r ) r c + c ( r s ) = ( c ) ( c r ) r c + c ( r + c ) = (r )( c)( + 3c) + c 3 ( c) > 0. Let C n be a circle of radius r n centered at o let C 0,...,C be closed unit circles with centers o 0,...,o covering C n. We assume that for a circle C i, i = 0,...,, for which C i bdc n? the centers of C i C n lie on the same side of the radical axis of the circles C i C n. Otherwise we reflect C i in this radical axis still get a covering of C n. Let C 0,...,C be unit circles in the position described in the theorem, that is so that C 0 is concentric with C n the centers ō,...,ō of C,...,C are situated in the vertices of a regular ()-gon at distance sin ( π/() ) from o. We are going to show that the two arrangements of circles are congruent. The following lemma claims that the two arrangements of circles {C i } n i=0 {C i } n i=0 have the same topological structure. Lemma. Exactly one of the circles {C i } i=0 is contained in intc n. Moreover, no three of the circles intersecting bdc n can have a common point. Since r n >, there is a circle, say C 0, which is contained in intc n. Observe that an arc of bdc n which is covered by a unit circle spans at o an angle not greater than arcsin(/r n ). Since 6arcsin r 8 = < π 7arcsin r 9 = < π, it follows that if n = 8 or n = 9, then C i bdc n? for i =,...,n. We also observe that three unit circles with a common point cannot cover a part of bdc n whose angle spanned at o exceeds arcsin(/r n ). Since 4arcsin r 8 + arcsin r 8 = < π ).

5 COVERING A CIRCLE WITH EIGHT, NINE, OR TEN CIRCLES 365 5arcsin r 9 + arcsin r 9 = < π, it follows that for n = 8 9 no three of the circles C i, i =,...,n, can have a common point. This argument breaks down when n = 0. We have 7arcsin = < π, showing that at most two of the circles {C i } 9 i=0 are contained in int C 0; however 8arcsin = > π, so we cannot exclude in this way that two of the circles {C i } 9 i=0 are contained in intc 0. Also the proof that no three of the circles intersecting bdc n can have a common point requires a different argument. Melissen [997, pp. 08 ] proved the Lemma for n = 0 using an argument based on the investigation of distances. In Section 3 we repeat Melissen s argument for the proof of the first part of the Lemma give an alternative proof for the second statement, using estimations of areas. Let D i, i = 0,...,n, be the Dirichlet cell of C i with respect to C n. From the considerations above it follows that each vertex in the cell complex of the Dirichlet cells is trihedral, D 0 is an ()-gon, while for i =,...,, D i is a curved quadrilateral bounded by three line segments an arc of bdc n. Thus the cell complex of the cells D i is isomorphic to the cell complex of the Dirichlet cells D i of the circles C i. We introduce some notations. We describe them for the circles C i. The same symbols with a bar will be used for the corresponding objects quantities for the circles C i (see Figure ). Let the vertices of D 0 be p,...,p let the vertices of Dirichlet cells on bdc n be q,...,q. We write p n = p, q n = q assume, as we may without loss of generality, that the notation is chosen so that the vertices of D i are p i, p i+, q i+, q i for i =,...,n. We write α i = ]q i o i q i+, β i = ]p i o i q i, γ i = ]p i+ o i q i+, δ i = ]p i o i p i+, ε i = ]p i o 0 p i+. We note that the assumption that o o i lie on the same side of the radical axis of C n C i implies that α i π for i =,...,n. It is easy to check that ᾱ i = 6π n, βi = γ i = (n 5)π n, δ i = ε i = π n.

6 366 Figure. Using the relation ᾱ i = 6π/() one can verify that ( 6π ) f = π n n. We dissect C n into the triangles T i = p i o 0 p i+, T i = p i o i p i+, T i = p i o i q i, T + i = p i+ o i q i+ into the regions R i bounded by the segments o i q i, o i q i+ the arc q i q i+ of bdc n, for i =,...,. We shall estimate the total area of these domains show that it is less than the area of C n unless the arrangement of the circles C 0,...,C is congruent to that of the circles C 0,...,C. Observe that the triangles T i T i are congruent, so that We have δ i = ε i = π = δ i = ε i () (β i + γ i ) = (n )π (α i + δ i ) = (n )π α i. () Hence we get ( π = ]q i oq i+ f rn (α i ) (n )f α ) i rn. n ( π ) α i (n )fr n = (n ) 6π n n = ᾱ i. (3) Now we are in the position to estimate the total area of the parts of C n. Using Jensen s inequality we get

7 COVERING A CIRCLE WITH EIGHT, NINE, OR TEN CIRCLES 367 (a(t i ) + a(ti )) sin ε i (n )sin (a(t i ) + a(t + i )) (sin β i + sin γ i ) (n )sin ε i n, (4) (β i + γ i ), (5) n ( a(r i ) F rn (α i ) (n )F α ) i rn. (6) n In view of () we have Write (n )sin ε i = sin ε i = n α = α i n. (a(t i ) + a(t i )). Then we have, in view of (), (β i + γ i ) (n )π = α; n n hence, by (5) (6), ( a(t i ) + a(t + i ) + a(r i) ) ( ( (n )π ) ) (n ) sin α + F rn (α). n The function ( ) (n )π sin α + F rn (α) n is concave for 0 α π, as it can be checked numerically, decreasing for α = 6π/(). Therefore it is decreasing for 6π/() α π. Observing that 6π n = ᾱ i we deduce that (n )π n ( a(t i ) + a(t + i ) + a(r i) ) (n ) 6π (n 0)π = = n n β i = γ i, ( ( )) (n 0)π 6π sin + F rn n n = (sin β i + sin γ i ) + F rn (ᾱ i ) ( = a(t i ) + a(t + i ) + a(r i) ). (7)

8 368 Adding inequalities (4) (7) we get a(c n ) = (a(t i ) + a(ti ) + a(t i ) + a(t + i ) + a(r i)) (a(t i ) + a(t i ) + a(t i ) + a(t + i ) + a(r i)) = a(c n ). Therefore we have equality in all of the inequalities (5) (7). This can only occur if the arrangements of the circles C 0,...,C C 0,...,C are congruent. 3. Proof of the Lemma for n = 0 Let C 0,...,C 9 be closed unit circles covering the circle C 0 of radius. As in the previous section, we denote the center of C i, i = 0,...,9 by o i the center of C 0 by o. We shall follow the argument of Melissen to show that no eight of the circles can cover bd C 0. Suppose that bdc 0 7 i=0 C i. Since the angular measure of an arc of bdc 0 covered by a unit circle is at most arcsin(/ ) 7arcsin = < π, no proper subset of the circles C i, i = 0,...,7 covers bd C 0, hence no three of the arcs C i bdc 0, 0 i 7 intersect. This property defines a cyclic order of the arcs C i bdc 0. We assume that the notation is chosen so that this cyclic order coincides with the order of the indices, that is C 0 C bdc 0?,...,C 6 C 7 bdc 0?,C 7 C 0 bdc 0?. We choose points q,...,q 8 from the sets C 0 C bdc 0,...,C 7 C 0 bdc 0, respectively. Recall that the maximum angular measure of an arc of bdc 0 covered by three unit circles with a common point is arcsin(/ ). Since arcsin + 5arcsin = < π, no three of the circles C i, i = 0,...,7 have a common point. Let p i = bdc i C i intc 0 for i =,...,7, p 8 = C 7 C 0 intc 0 (see Figure ). The main observation of Melissen is that the points p i, i =,...,8, cannot be covered by two circles. This follows easily from the following result: Proposition. We have for i =,...,8, with p i = p i+8. p i p i+3 > p i p i+4 >

9 COVERING A CIRCLE WITH EIGHT, NINE, OR TEN CIRCLES 369 C i+ q i+ C i q i+ q i= q i+8 p i+ p i p i+ q i+3 o q i+7 p i+3 q i+6 q i+4 q i+5 Figure. Indeed, the first inequality readily implies that one of the circles C 8 C 9 contains all points p i with an odd subscript the other contains all points with an even subscript. This, however, contradicts the second inequality. In order to estimate the distances between the points p i we need a lower bound for the distance op i. Let h(ϑ) be the minimum distance between o a point of intersection of the boundaries of two unit circles that cover an arc of angular measure ϑ from bdc 0, with 0 ϑ 4arcsin(/ ). It is easy to see that this minimum distance is achieved in the symmetric position when each of the unit circles cover an arc of angular measure ϑ from bdc 0. Using some trigonometry we calculate that h(ϑ) = cos ϑ r 0 sin ϑ 4 cos ϑ 4. Writing s = sin ϑ 4 we have ( s r 0 s cos 4) ϑ h (ϑ) = r 0 s h (ϑ) = ( r 0 s)3/ cos ϑ + ( + r 0 6r 0 s + 4r4 0 s )cos ϑ 4 8( r 0 s ) 3/. It immediately follows that h(ϑ) is increasing. Observing that we get + r 0 6r 0 s + 4r4 0 s > r h (ϑ) > cos ϑ + (r )cos ϑ 4 8 = 4 cos ϑ 4 + (r )cos ϑ

10 370 The minimum of the right side is 8( r 0 5 4) > 0, showing that h(ϑ) is convex. Now we are in the position to estimate the distances between the points p i. Write ψ = ]p i op i+3 ξ = ]p i op i+4. Then we have, on the one h, ψ = π ]p i+3 oq i+4 6 ]q i+j oq i+j+ ]q i+7 op i j=4 π arcsin 6arcsin = ψ min = > π ξ = π ]p i+4 oq i+5 6 ]q i+j oq i+j+ ]q i+7 op i j=45 π arcsin 4arcsin = ξ min = > π, on the other h, ψ = ]p i oq i+ + ]q i+ oq i+ + ]q i+ op i+3 arcsin + arcsin = < π ξ = ]p i oq i+ + ]q i+ oq i+ + ]q i+ op i+3 + ]p i+3 op i+4 arcsin + 4arcsin = π ξ min. By the law of cosines we get p i p i+3 = op i + op i+3 op i op i+3 cosψ, p i p i+4 = op i + op i+4 op i op i+4 cosξ. Let ϑ, ϑ, ϑ 3 be the angular measure of the arc of bdc 0 covered by the pair of circles C i, C i, C i+, C i+3, C i+3, C i+4, respectively. As the triangles p i op i+3 p i op i+4 are obtuse, we get lower bounds for p i p i+3 p i p i+4 if we substitute for op i, op i+3, op i+4 their minimum values for cosψ cos ξ their maximum values: We have, for j =, 3, p i p i+3 h (ϑ ) + h (ϑ ) h(ϑ )h(ϑ )cos ψ min, p i p i+4 h (ϑ ) + h (ϑ 3 ) h(ϑ )h(ϑ 3 )cos ξ min. ϑ + ϑ j π 8arcsin. Since h(ϑ) is increasing convex, therefore, ( h(ϑ ) + h(ϑ j ) h π 4arcsin ).

11 COVERING A CIRCLE WITH EIGHT, NINE, OR TEN CIRCLES 37 The functions h + h h h cosψ min h + h h h cosξ min are homogeneous of degree one in the variables h h, thus, they are convex. They are also increasing in both variables. Therefore p i p i+3 h (π 4arcsin(/ ))( cosψ min ) = = h(π 4arcsin(/ ))sin ψ min = > p i p i+4 h (π 4arcsin(/ ))( cos ξ min ) = = h(π 4arcsin(/ ))sin ξ min = >. This completes the proof of the Proposition at the same time the proof of the first part of the Lemma. It remains to show the second part of the Lemma, namely that if nine of the circles C 0,...,C 9 intersect bdc 0, then no three of them can have a common point. This part of the Lemma can be settled by estimating areas in a similar way as we did in the previous section. Suppose that C 0 bdc 0 =? C i bdc 0? for i =,...,9. We shall scrutinize the cell complex formed by the Dirichlet cells D i of the circles C i, 0 i 9, with respect to C 0. We may assume that D i bdc 0? for i =,...,9, otherwise bdc 0 is covered by eight circles, which we already excluded. Without loss of generality we may suppose that the arcs D i bdc 0, i =,...,9, are situated on bdc 0 in their natural cyclic order. We shall exclude the possibility that three of the Dirichlet cells D,...,D 9 intersect. We note that three circles can intersect without their corresponding Dirichlet cells having a common point, however the case when no three of the cells D,...,D 9 intersect has been already discussed in the previous section. Observe that D i D i±j =? for i =,...,9, j = 3, 4. (8) Otherwise the circles C i,c i±...,c i±j cover from bdc 0 an arc whose angular measure is at most arcsin(/ ), while the angular measure of the arc covered by the other 9 j 5 circles cannot exceed 0arcsin(/ ). Since arcsin(/ ) + 5arcsin(/ ) = < π, this is impossible. Suppose that three of the cells D,...,D 9 intersect. In view of (8) they must belong to consecutive indices. Assume, say, that D D D 3?. If no further triple of the cells D,...,D 9 intersect, then the cells are arranged as depicted on the left side of Figure 3, where the Dirchlet cells are drawn by broken lines. We shall refer to this situation as Case. If there is another intersecting triple, say D i, D i+, D i+, among the cells D,...,D 9, then {,,3} {i,i+,i+}?. Otherwise the total angular

12 37 q 4 o 4 q 3 o 3 o o q q o 9 p p p 3 q 5 p p q o 4 3 o q q 6 p o 9 o o 0 0 q 5 o 4 q 4 o 3 q 3 o q Figure 3. measure of the arcs of bd C 0 covered by the circles is at most 4arcsin(/ ) + 6arcsin(/ ) = < π. D cannot be the member of another intersecting triple of cells. For, if D, D 3, D 4 or D 9, D, D intersect, then D D 4? or D 9 D 3?, which is impossible by (8). Thus, the only cidates for another triple of intersecting cells are {D 3,D 4,D 5 } {D 8,D 9,D }. As these triples are disjoint, only one of them can have a nonempty intersection. Hence, the other case we have to investigate is that besides D, D, D 3, say D 3, D 4, D 5 have a common point. This is Case which is represented on the right side of Figure 3. As before, we denote by o i, i =,...,0, the center of C i, by q i, i =,...,9, the vertices of Dirichlet cells on bdc 0 choosing the notation so that q i is common to D i D i. We denote the vertices of D 0 in their consecutive order for the two cases by p, p 3, p 4,..., p 9 p, p 3, p 5,..., p 9, respectively, so that p q is the side common to D D 9 (see Figure 3). We divide C 0 into the following regions: (i) the 6-gon P = p o p 3 o 3 p 4 o 4 p 5 o 5 p 6 o 6 p 7 o 7 p 8 o 8 p 9 o 9, the pentagon P = q o p 3 o 3 q 3, the segment S cut off from C 0 by the chord q q 3, the quadrilaterals Q i = o i q i+ o i+ p i+, 3 i 9, the regions R i, i 9, i, bounded by the segments o i q i, o i q i+ the arc q i q i+ of bdc n in Case ; (ii) the 4-gon P = p o p 3 o 3 p 5 o 5 p 6 o 6 p 7 o 7 p 8 o 8 p 9 o 9, the pentagons P = q o p 3 o 3 q 3 P 3 = q 4 o 3 p 5 o 5 q 5, the two segments S S cut off from C 0 by the chords q q 3 q 4 q 5, respectively, the quadrilaterals Q i = o i q i+ o i+ p i+, 5 i 9, the regions R i, i 9, i, 4, bounded by the segments o i q i, o i q i+ the arc q i q i+ of bdc n in Case.

13 COVERING A CIRCLE WITH EIGHT, NINE, OR TEN CIRCLES 373 As we saw in the previous section, P can be dissected into pairs of congruent triangles one half of the triangles making up the cell D 0. Hence we get in Case a(p ) 8sin π = (9) 4 a(p ) 7sin π 7 = (9 ) in Case. The length of four sides of the pentagon P (P 3 ) is bounded above by, while the length of the fifth side is at most. The area of such a pentagon cannot exceed the area of a pentagon with four sides of length one side of length inscribed into a circle. The radius r of the circle is determined implicitly by the equation 4arcsin r + arcsin r = π. r 0 =.0736 is an upper bound for r ( r0 sin arcsin + sin π arcsin ) r 0 = r 0 is an upper bound for the area of the pentagon. Thus, we have, in Case, a(p ) (0) a(p ) + a(p 3 ) (0 ) The area of S (S ) cannot exceed ( r0 arcsin ) sin arcsin = , the area of a segment of C 0 cut off by a chord of length. Hence Using the rough estimate a(q i ) we get a(s ) () a(s ) + a(s ) ( ) 9 a(q i ) 7 () i=3 9 a(q i ) 5, ( ) i=5 respectively. We estimate the total area of the regions R i using the method developed in the previous section. Let x = ]q oq 3 in Case x = ]q oq 3 + ]q 4 oq 5 in

14 374 Case. Then we have x arcsin(/ ) x 4arcsin(/ ), respectively. Writing α i = ]q i o i q i+ we have π x = ]q i oq i+ ( ) f r0 (α i ) 8f r0 α i 8 π x = hence 7 8 i 9, i = i 9, i =, 4 i 9, i = i 9, i =,4 ]q i oq i+ i 9, i = i 9, i =,4 f r0 (α i ) 8f r0 ( 8 i 9, i i 9, i,4 ( π x ) ( π arcsin ) α i fr 0 fr 8 0 = α i ), ( π x ) ( π 4arcsin ) α i fr 0 fr 7 0 = , 7 respectively. Observing that F r0 (α) is decreasing for α.9 we get for the total area of the regions R i the estimate a(r i ) ( ) F r0 (α i ) 8F r0 α i 8 i 9, i = i 9, i = i 9, i = ( π arcsin )) 8F r0 (f r 0 = (3) 4 in Case i 9, i =,4 ( ) a(r i ) F r0 (α i ) 7F r0 α i 7 i 9, i =,4 i 9, i =,4 ( π 4arcsin )) 7F r0 (f r 0 = (3 ) 7 in Case. From inequalities (9) (3) we conclude that a(c 0 ) = a(p ) + a(p ) + a(s ) + 9 a(q i ) + i= < 0.4 < a(c 0 ) i 9, i = a(r i )

15 COVERING A CIRCLE WITH EIGHT, NINE, OR TEN CIRCLES 375 in Case, it follows from (9 ) (3 ) that a(c 0 ) = a(p ) + a(p ) + a(p 3 ) + a(s ) + a(s ) + 0 < 0.4 < a(c 0 ) 9 a(q i ) + a(r i ) i=5 i 9, i,4 in Case, yielding in both cases a contradiction. This completes the proof of the Lemma. References [Ball Coxeter 987] W. W. R. Ball H. S. M. Coxeter, Mathematical recreations essays, Dover Publications Inc., New York, 987. Original edition, 89. [Bezdek 979] K. Bezdek, Optimal covering of circles, Thesis, Budapest, 979. In Hungarian. [Bezdek 983] K. Bezdek, Über einige Kreisüberdeckungen, Beiträge Algebra Geom. no. 4 (983), 7 3. [Croft et al. 99] H. T. Croft, K. J. Falconer, R. K. Guy, Unsolved problems in geometry, Problem Books in Mathematics, Springer, New York, 99. [Groemer 960] H. Groemer, Über die Einlagerung von Kreisen in einen konvexen Bereich, Math. Z 73 (960), [Krotoszyński 993] S. Krotoszyński, Covering a disk with smaller disks, Studia Sci. Math. Hungar. 8:3-4 (993), [Melissen 997] H. Melissen, Packing covering with circles, Ph.D. Thesis, 997. [Nagy 975] D. Nagy, Coverings their applications, Thesis, Budapest, 975. In Hungarian. [Neville 95] E. H. Neville, On the solutions of numerical functional equations, illustrated by an account of a popular puzzle its solution, Proc. London Math. Soc. () 4 (95), [Oler 96] N. Oler, A finite packing problem, Canad. Math. Bull. 4 (96), [Peikert 994] R. Peikert, Dichteste Packungen von gleichen Kreisen in einem Quadrat, Elem. Math. 49: (994), 6 6. [Peikert et al. 99] R. Peikert, D. Würtz, M. Monagan, C. de Groot, Packing circles in a square: a review new results, pp in System modelling optimization (Zurich, 99), Lecture Notes in Control Inform. Sci. 80, Springer, Berlin, 99. [Rankin 955] R. A. Rankin, The closest packing of spherical caps in n dimensions, Proc. Glasgow Math. Assoc. (955), [Schütte 955] K. Schütte, Überdeckungen der Kugel mit höchstens acht Kreisen, Math. Ann. 9 (955), 8 86.

16 376 Gábor Fejes Tóth Alfréd Rényi Institute of Mathematics Hungarian Academy of Sciences P.O.Box 7 H-364 Budapest Hungary gfejes@renyi.hu