Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II: Cases of N = 10, 11, and 12

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1 Original Paper Forma,, 57 75, 007 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II: Cases of N = 0,, and Teruhisa SUGIMOTO * and Masaharu TANEMURA, The Institute of Statistical Mathematics, Minami-Azabu, Minato-ku, Tokyo , Japan Department of Statistical Science, The Graduate University for Advanced Studies, Minami-Azabu, Minato-ku, Tokyo , Japan * address: sugimoto@ismacjp (Received August 6, 007; Accepted October 3, 007) Keyword: Spherical Cap, Sphere, Packing, Covering, Tammes Problem Abstract Let C i (i =,, N) be the i-th open spherical cap of angular radius r and let M i be its center under the condition that none of the spherical caps contains the center of another one in its interior We consider the upper bound, r N, (not the lower bound!) of r of the case in which the whole spherical surface of a unit sphere is completely covered with N congruent open spherical caps under the condition, sequentially for i =,, i N, that M i is set on the perimeter of C i, and that each area of the set ( ν= C ν ) C i becomes maximum In this paper, for N = 0,, and,, we found out that the solutions of our sequential covering and the solutions of the Tammes problem were strictly correspondent Especially, we succeeded to obtain the exact closed form of r 0 for N = 0 Introduction The circle on the surface of a sphere is called a spherical cap Among the problems of packing on the spherical surface, the closest packing of congruent spherical caps is the most famous, and is usually known as the Tammes problem (TAMMES, 930) So far, the mathematically proved solution of Tammes problem were given for N =,,, and 4 (SCHÜTTE and VAN DER WAERDEN, 95; DANZER, 963; FEJES TÓTH, 97) The exact closed form of solution in the cases for N =,, 9,,, and 4 are known, but in the case for N = 0, the solution was only known in the rage [54479, 54480] by DANZER (963) In our study, the condition that none of spherical caps contains the center of another one in its interior is called Minkowski condition (SUGIMOTO and TANEMURA, 003, 004, 006) Let C i be the i-th open spherical cap of angular radius r and let M i be its center under the Minkowski condition (i =,, N) Then, our problem is as follows; the whole spherical surface of a unit sphere is completely covered with N congruent open spherical caps under the condition, sequentially for i =,, N, that M i is set on the perimeter of C i, and that each area of the set ( i ν = C ν ) C i becomes maximum That is, our problem is to calculate the upper bound of r for our sequential covering In our previous paper 57

2 58 T SUGIMOTO and M TANEMURA (SUGIMOTO and TANEMURA, 006; hereafter, we refer it as I), we calculated the upper bounds for N =,, 9 Here, we define a half-cap as the spherical cap whose angular radius is r/ and which is concentric with the original cap If the centers of N half-caps are placed on the positions of the centers of C i (i =,, N), we get the packing with N congruent half-caps Then, for N =,, 9, we found that the upper bound of our problem with N congruent spherical caps and the value of angular diameter of the Tammes problem are equivalent, and that the location of centers of our problem is correspondent to that of the Tammes problem (SUGIMOTO and TANEMURA, 003, 004, 006) Further, it should be said that our method is a systematic and a different approach to the Tammes problem from the works by SCHÜTTE and VAN DER WAERDEN (95), etc In this paper, we calculate the upper bounds of r for N = 0,, and theoretically by using our sequential covering procedure As a result, for N = 0,, and, we found out that the solutions of our sequential covering and the solutions of the Tammes problem were strictly correspondent as the cases of N =,, 9 Especially, the exact closed form for N = 0 is obtained (see Eq (0)) Further, we presented a systematic method which is different from the approach of DANZER (963) Preparations Overlapping area and union W i Throughout this paper we assume that the center of the unit sphere is the origin O = (0, 0, 0) and represent the surface of this unit sphere by the symbol S In the following, open spherical caps are simply written as spherical caps unless otherwise stated We define the geodesic arc between an arbitrary pair of points T = (x, y, z ) and T = (x, y, z ) on S as the inferior arc of the great circle determined by T and T Then the spherical distance between T and T is defined by the length of geodesic arc of this pair of points, and we denote d s (T, T ) = cos (x x + y y + z z ) as the spherical distance between T and T In order to solve our problem, we present the overlapping area of congruent spherical caps under the Minkowski condition Assume C i and C j be two congruent spherical caps, of angular radius r, which are mutually overlapping under the Minkowski condition, and let A ij = A(C i C j ) be the overlapping area where A(X) is the area of X When r d s (M i, M j ) r, the overlapping area of C i and C j is given by A ij hij r cos cos = cos cos r 4cos sin r cos hij r h ij + π () tan sin where h ij = cos (cos r/cos (d s (M i, M j )/)) is the spherical distance between cross points of C i and C j It is obvious that A ij is a monotone increasing function of h ij when r is fixed (SUGIMOTO and TANEMURA, 003, 006) Let N be the number of spherical caps when the whole spherical surface S is completely covered And, let C i be the perimeter of C i (i =,, N) Then we define: Wi = Wi Ci max A( Wi Ci), i, K, N ; W C Mi Ci = =

3 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 59 In other words, W i is the union of W i and C i satisfying the condition that the area A(W i C i ) is maximum with the restriction M i C i Hereafter, we call the case of () that i i ν = C ν is in an extreme state We are always necessary to examine whether ν = C ν is in an extreme state in our sequential covering procedure Then, we calculate the area A(W i C i ) by using () and the area formula of spherical triangle Note that, in order to simplify calculation, we often use the fact that A(W i C i ) is maximum with the restriction M i C i is the same as that A((W i C i ) c ) is maximum with the restriction M i C i Upper bounds r N and r N We define r N as the upper bound of angular radius r for the case in which N congruent open spherical caps completely cover the whole spherical surface S under the condition, sequentially for i =,, N, that M i is set on C i, and that each area of the set W i C i becomes maximum Next, we define another upper bound of r, r N, such that the set N ν = C ν which contains W N cannot cover S under the Minkowski condition Then r N should be equal to the spherical distance of the largest interval in the uncovered region (W N ) c of S It is because, when the angular radius r is equal to r N, the set N ν = C ν which contains W N can cover S except for a finite number of points or a line segment under our sequential covering Therefore, N ν = C ν is in an extreme state if and only if at least one endpoint of the interval, which has the above mentioned spherical distance r N, comes on the perimeter C N Further, when there are two or more uncovered points, the spherical distance of any pair of these uncovered points is less or equal to r N since the largest interval is assumed to be r N Then, we can put the center M N of C N at one of the uncoverd points At this moment, we see that N ν = C ν which contains W N covers S without any gap Then, we notice the fact that r N is equal to r N In this paper, we calculate each solution for N = 0,, and using above fact Refer to I for detailed explanations of r N and r N 3 Results 3 N = 0 According to the considerations and results for N = 7, 8, and 9 of I, the angular radius for N = 0 should be shorter than r 9 = cos (/3) 3096 rad and should be larger than tan 075 rad Therefore, let us consider the case for N = 0 under the assumption tan r r 9 Even if our answer for N = 0 is obtained by this assumption tan r, we will also consider the range of r < tan later If two spherical caps C a (the coordinates of the center: (a, a, a 3 )) and C b (the coordinates of the center: (b, b, b 3 )) that satisfy M b C a intersect, we will have the coordinates (x, y, z) of cross points of the perimeters C a and C b by solving the following simultaneous equations: ax + ay + az 3 = cos r, bx + by + bz 3 = cos r, x + y + z = () 3

4 60 T SUGIMOTO and M TANEMURA Note that there are two cross points of C a and C b When (a, a, a 3 ) = (0, 0, ) and (b, b, b 3 ) = (sin r, 0, cos r), we get z = cos r, sin r x cos r z = cos r, x + y + z = ( 4) When K = (x, y, z ) and K = (x, y, z ) are the solutions of simultaneous Eq (4), the coordinates of cross points of C a and C b are as follows: ( ) + cos r cos r cos r cos r ( x, y, z)=,, cos r, 5 sin r sin r + cos r cos r cos r cos r ( x, y, z)=,, cos r 6 sin r sin r First, as in the cases of N = 7, 8, and 9 in I, the centers M, M, and M 3 are placed at (0, 0, ), K, and (sin r, 0, cos r), respectively At this time, from Theorem in I, the set 3 ν = C ν is in an extreme state when the centers M, M, and M 3 satisfy the relations d s (M, M ) = d s (M, M 3 ) = d s (M, M 3 ) = r Then, let K 3 be the one of the cross points of the perimeters C and C, and let it be outside C 3 In addition, let K 4 be one of the cross points of C and C 3, and let it not be the south pole (0, 0, ) The explicit expressions of cross points K 3 and K 4 are given in Appendix Then, from the cases of N = 7, 8, and 9 in I, we assume that the allocation of points K 4 for M 4 satisfies the condition that ν = C ν is in an extreme state It is because, in the range tan r < π/, our computations show that the area A(W 3 C 4 ) is maximum when M 4 is put at K as we will see below Let us place the center M 4 at K 4 and move it to K along the arc K 4 K of C 3 We calculate the area A(W 3 C 4 ) against the moving point M 4 numerically for several fixed values of r among tan r < π/ The results are shown in Fig In this figure, the horizontal axis is the position of M 4 on the arc K 4 K of C 3 and the vertical axis is the area A(W 3 C 4 ) In our computation, the arc K 4 K is divided into 00 equal intervals and the area A(W 3 C 4 ) is calculated on 0 end points of the intervals Hereafter, the similar computations are performed for determination of centers of spherical caps (see Figs, 4, and 7) As to the two values of r in Fig, we find that this curve of A(W 3 C 4 ) is symmetrical at the center of the arc K 4 K (it is evident from the spherical symmetry) and A(W 3 C 4 ) is maximum at both ends The same fact as above would hold for every values of r in the range tan 4 r < π/ Therefore, we expect that ν = C ν is in an extreme state if and only if M 4 is put at K or K 4 for tan r < π/ To make sure, we shall check that 4 these points K and K 4 satisfy the condition that ν = C ν is in an extreme state after obtaining the exact values of the angular radius r at the last paragraph in this subsection

5 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 6 Fig The curve of A(W 3 C 4 ) when M 4 is moved on the arc K 4 K of C 3 Here, as in the cases of N = 7, 8, and 9 in I, the arc K 4 K is divided into 00 equal intervals and the area A(W 3 C 4 ) is calculated on 0 end points of the intervals Note that the curve of r 075 corresponds to the case of r = tan The values of r of other curves are described in the text Therefore, we choose M 4 on the point K Then, let K 5 = (x 5, y 5, z 5 ) be one of the cross points of perimeters C 4 and C, and let it be outside C 3 Further, let K 6 = (x 6, y 6, z 6 ) be one of the cross points of C 4 and C 3, and let it not be the south pole (0, 0, ) The explicit expressions of cross points K 5 and K 6 are given in Appendix Next we calculate the area A(W 4 C 5 ) when M 5 is put at a certain point on the arc K 6 K 5 of C 4 and is moved on the arc Since the fact that A(W i C i ) is maximum with the restriction M i C i is the same as that A((W i C i ) c ) is maximum with the restriction M i C i, in order to simplify calculation, we calculate A((W 4 C 5 ) c ) against the moving point M 5 numerically for several fixed values of r among tan r < π/ Here, the computation is performed as in the determination of M 4 Figure (a) shows the results In this figure, the horizontal axis is the position of M 5 on the arc K 6 K 5 of C 4 and the vertical axis is the area A((W 4 C 5 ) c ) As a result, for the two values of r in Fig (a), the curve of A((W 4 C 5 ) c ) is symmetrical at the center of the arc K 6 K 5 (it is evident from the spherical symmetry) and A((W 4 C 5 ) c ) is maximum when M 5 is placed on K 6 or K 5 The same fact as above would hold for every values of r in the range tan r < π/ Therefore, we expect 5 that ν = C ν is in an extreme state if and only if M 5 is put at K 5 or K 6 for tan r < π/ To make sure, we shall check whether the point K 5 and K 6 such points after obtaining the exact values of the angular radius r at the last paragraph in this subsection like the case of M 4 We choose M 5 on the point K 5 as in the cases of N = 7, 8, and 9 in I Then, let K 7 = (x 7, y 7, z 7 ) be one of the cross points of the perimeters C 5 and C 4, and let it be outside of C Similarly, let K 8 = (x 8, y 8, z 8 ) be one of the cross points of C 5 and C, and let it be outside of C The exact coordinates of K 7 and K 8 are also given in Appendix When each M, M, M 3, M 4, and M 5 is placed at (0, 0, ), K, (sin r, 0, cos r), K, and K 5, respectively, the shape of the uncovered region (W 5 ) c is formed as a kite K 8 K 4 K 6 K 7 on the unit sphere (see Fig 3) We note the sides of the kite K 8 K 4 K 6 K 7 are not geodesic arcs

6 6 T SUGIMOTO and M TANEMURA Fig (a) The curve of A((W 4 C 5 ) c ) when M 5 is moved on the arc K 6 K 5 of C 4 (b) The curve of A((W 5 C 6 ) c ) when M 6 is moved on the arc K 7 K 8 of C 5 The similar computation method as in Fig is taken See the legend in Fig but are perimeters of spherical caps From the configuration of the vertices K 8, K 4, K 6, and K 7 of the kite K 8 K 4 K 6 K 7, for the range tan r < π/, the relations of spherical distance between each vertices always hold as follows: d K, K d K, K, s d K, K d K, K, s ( 8 4)= s( 8 7) ( 6 4)= s( 6 7) < < < < d K, K d K, K d K, K, s 6 4 s 8 4 s 6 8 d K, K d K, K d K, K s 6 4 s 4 7 s 6 8 ( 7)

7 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 63 Fig 3 The kite K 8 K 4 K 6 K 7 on unit sphere In addition, for tan r < cos (( )/7), there hold the following relations of spherical distance between each vertices of the kite K 8 K 4 K 6 K 7 d s (K 6, K 4 ) r < d s (K 4, K 7 ) < d s (K 8, K 4 ) (8) Note that we find numerically that these relations (7) and (8) hold by using mathematical software In passing, cos (( )/7) rad is equal to the upper bound r 8 for N = 8 (SUGIMOTO and TANEMURA, 003, 004, 006) Now, we want to place five centers of C 6, C 7, C 8, C 9, and C 0 in the uncovered kite K 8 K 4 K 6 K 7 under the Minkowski condition In this kite, we need to consider five points which keep spherical distance r with each other Here, we search the position of M 6 which satisfies the condition that A((W 5 C 6 ) c ) is 6 maximum with the restriction M 6 C 5 (ie ν = C ν is in an extreme state) From the restriction M 6 C 5, M 6 is put at a certain point on the arc K 7 K 8 of C 5 and, during M 6 is moved on the arc K 7 K 8 of C 5, we calculate A((W 5 C 6 ) c ) against the moving point M 6 numerically for several fixed values of r among tan r < r 7 = cos ( (4/ 3 ) cos(7π/8)) rad Figure (b) shows the graph of computational results In this figure, the horizontal axis is the position of M 6 on the arc K 7 K 8 of C 5 and the vertical axis is the area A((W 5 C 6 ) c ) As a result, we find that A((W 5 C 6 ) c ) is maximum when M 6 is 6 put on K 8 Therefore, we expect that ν = C ν is in an extreme state if and only if M 6 is put at K 8 in the range tan r < r 7 We shall check that K 8 is such point after obtaining the exact values of the angular radius r as in the cases of M 4 and M 5 We choose here M 6 on the point K 8 Let K 9 = (x 9, y 9, z 9 ) be one of the cross points of the perimeters C 6 and C, and let it be outside of C Similarly, let K 0 = (x 0, y 0, z 0 ) be one of the cross points of C 6 and C 5, and let it be outside of C The exact coordinates of K 9 and K 0 are also given in Appendix Here, let us assume two points K C 3, and K C 4 In addition, we suppose K 9, K 0, K, and K satisfy the relations

8 64 T SUGIMOTO and M TANEMURA = = = = = d K, K d K, K d K, K s 8 9 s 8 0 s 9 0 d K, K d K, K d K, K s 9 s 0 s ( 9) Then, we see that K 8 K 9 K 0 is a spherical equilateral triangle and that K 9 K K K 0 is a spherical square (see Fig 3) If such an arrangement of vertices K 9, K 0, K, and K is actually possible, the upper bound r 0 ( r 9 ) is considered to be equal to a side-length (eg, the spherical distance between the points K 9 and K ) of the spherical square K 9 K K K 0 Then, if M 7, M 8, and M 9 are placed on the points K 9, K 0, and K, respectively, the surface 9 S is covered by the set ν = C ν except for the point K That is, our sequential covering is realized At this time, K and K are just cross points of C 7 and C 3, and C 8 and C 4, 7 8 respectively Note that the points K 9 and K 0 are the positions such that ν = C ν and ν = C ν are in an extreme state, respectively However, it is not checked yet that K 9 and K 0 are such positions We shall check these facts after obtaining the exact values of the angular radius r and the coordinates of K and K From the above consideration that our r is equal to a side-length of spherical square K 9 K K K 0, the equation r = d s (K 9, K ), for instance, should be satisfied Then, we get the exact value of r which satisfies this equation Therefore, to begin with, we calculate the coordinates of the points K 9 and K which satisfy (9) Since the point K 9 is a cross point of the perimeters C 6 and C, the coordinates of K 9 are calculable by using (3), (5), and the coordinates of K 8 in Appendix like the case of N = 9 in I The result is given in Appendix Next, we need to calculate the coordinates of K without using the coordinates of K 9 Otherwise, we cannot get an equation of r in a closed form So we pay attention to the spherical distance = d s (K 8, K ) Then, by applying the spherical cosine theorem to the spherical isosceles triangle K 8 K 9 K of legs d s (K 8, K 9 ) = d s (K 9, K ) = r, cos may be written as follows: + 3 3cos r+ cos r cos r cos r cos r cos r cos l = + cos r It is because the inner angle at K 9 of the spherical isosceles triangle K 8 K 9 K is the sum of the interior angles of spherical equilateral triangle K 8 K 9 K 0 and spherical square K 9 K K K 0 In this connection, the inner angle of spherical equilateral triangle of side-length r is cos cos r, cos r + and the inner angle of spherical square of side-length r is cos cos r cos r +

9 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 65 As a result, we can obtain the coordinates of K = (x, y, z ) by using simultaneous equations d s (K, M 3 ) = r, d s (K 8, K ) =, and x + y + z = Here, M 3 = (sin r, 0, cos r) and refer to Appendix for the coordinates of K 8 The explicit coordinates of cross point K are shown in Appendix From d s (K i, K j ) = cos (x i x j + y i y j + z i z j ) and the coordinates of K 9 and K in Appendix, we get the equation of the following type r = d s (K 9 (r), K (r)) Then, the equation is solved against r by using mathematical software The form of the solution and its value is obtained as r = r = tan cos tan L rad 0 9 ( 0) By using (0), we find numerically at an arbitrary precision that the relation (9) is attained as we expected Note that we get also another solution r 9753 rad (the exact equation for this value is omitted since it is complicated) in the range of tan r cos (/3) 3096 rad when r = d s (K 9, K ) is solved against r But, for r 9753 rad, we find d s (K, K ) < r numerically Namely, K and K do not satisfy the relation (9) Therefore, when M 9 is put at K, C 9 will cover K Thus, we can exclude this answer r 9753 rad as the solution of inadequacy As we have noted, we check here whether the allocations of the points K, K 5, K 8, K 9, 4 and K 0 for M 4, M 5, M 6, M 7, and M 8, respectively, satisfy the condition that ν = C ν, ν = C ν, ν = C ν, ν = C ν, and ν = C ν are in an extreme state, each, by using the value of (0) First, from the considerations for determinations of M i mentioned above, we checked numerically that the area A(W i C i ) (or A((W i C i ) c )) are maximum with the restriction M i C i ( i ν = C ν are in an extreme state) when M 4, M 5, and M 6 are put at K, K 5, and K 8, respectively Then, these results are graphically presented by the curve of r 5448 corresponding to the value of (0) in Figs and Therefore, for N = 0, our choice of M 4, M 5, and M 6 are justified Next, we examine the position of M 7 where A((W 6 C 7 ) c ) is maximum Then, we calculate A((W 6 C 7 ) c ) against the moving point M 7 on the arc K 0 K 9 of C 6 numerically for several fixed values of r among tan r < r 8 = cos (( )/7) rad As a result, the curve of A((W 6 C 7 ) c ) is symmetrical at the center of the arc K 0 K 9 (it is evident from the shape of uncovered region (W 6 ) c ) and A((W 6 C 7 ) c ) is maximum at both end for the two values of r in Fig 4(a) The same fact as above would hold for every values of r in the range tan r < r 8 Figure 4(a) illustrates the result of computation for M 7 In this figure, the horizontal axis is the position of M 7 on the arc K 0 K 9 of C 6 and the vertical axis is A((W 6 C 7 ) c ) Then, the result is graphically presented by the curve of r 5448 corresponding to the value of (0) in Fig 4(a)

10 66 T SUGIMOTO and M TANEMURA Fig 4 (a) The curve of A((W 6 C 7 ) c ) when M 7 is moved on the arc K 0 K 9 of C 6 (b) The curve of A((W 7 C 8 ) c ) when M 8 is moved on the arc K 0 K of C 7 The similar computation method as in Fig is taken See the legend in Fig 7 Therefore, for the value of (0), we check numerically that ν = C ν are in an extreme state if and only if M 7 is put at K 9 or K 0 In this paper, we choose M 7 on the point K 9 Then, in order to find the optimal position of M 8, let us place M 8 at K 0 and move it to K along the arc K 0 K of C 7 Therefore, we calculate A((W 7 C 8 ) c ) against the moving point M 8 on the arc K 0 K When r is equal to (0), we find numerically that A((W 7 C 8 ) c ) is maximum with the restriction M 8 C 7 ( 8 ν = C ν is in an extreme state) if and only if M 8 is put at K 0 The curve of r 5448 in Fig 4(b) presents this result In Fig 4(b), the horizontal axis is the position of M 8 on the arc K 0 K of C 7 and the vertical axis is A((W 7 C 8 ) c ) Hence, for N = 0, our choices for M 7 and M 8 are justified As mentioned above, when eight spherical caps whose angular radius is the value of (0) are placed on S according to our sequential covering, the uncovered region (W 8 ) c is a

11 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 67 Fig 5 (a) Our sequential covering for N = 0 (b) Our solution of Tammes problem for N = 0 Both viewpoints are (0, 0, 0) In this example, the coordinates of the centers are respectively (0, 0, ), (06335, , ), (09458,0, ), (06335, , ), ( 0769, , ), ( , 05768, 05593), ( 03883, 07836, ), ( , , ), ( , 07490, ), and ( , 0395, ) quadrangle on S Then, from Corollary of Theorem in I and the relations for four vertices of (W 8 ) c, we find that d s (K, K ) is equal to the spherical distance of the largest interval in the uncovered region (W 8 ) c Therefore, r 9 is equal to (0) In addition, we find that 9 ν = C ν is in an extreme state if and only if M 9 is put on the point K C 8 Therefore, we choose M 9 on K, and then K is the unique uncovered point on S For N = 0, we initially assumed that r should be in the range (tan, cos (/3)] Then, as a result of the investigation, our upper bound r 0, (0), has fallen within the range (tan, r 9 ] However, one might suspect that the fact is due to the assumption So, if r is in the range (0, tan ], we examine whether W 9 is able to cover S except for finite points When r is assumed to be equal to tan, we find that the set W 9 leave an uncovered region on S For its detail, refer to the consideration of Subsec 3 Furthermore, for 0 < r < tan, the uncovered region would become still bigger Hence, the upper bounds r 0 cannot be in the range (0, tan ] as in the cases of N = 8 and 9 in I Thus, we note that our initial assumption tan < r r 9 is also confirmed (r = tan is excluded from the above consideration) As a result of consideration above, the set W 9 covers S except for K Therefore, when M 0 is put at the point K, the whole of S is covered by 0 ν = C ν which contains W 9 (see Fig 5(a)) Thus, our consideration that our r is equal to the upper bound r 0 (a side-length of the spherical square K 9 K K K 0 which satisfies (9)) is confirmed and (0) is certainly the upper bound for N = 0 3 N = and It is expected that the solution r for N = should not be larger than r 0 Then, we assume r r r 0 Further, we assume tan r Hence, the relations (7) and (8) hold because of the assumption tan r r r 0 Then, we can use the same configuration of the first five spherical caps of the case N = 0 When the fifth spherical cap C 5 is put on the sphere, in the same way as the foregoing subsection, a quadrilateral kite K 8 K 4 K 6 K 7 on the sphere might be formed as the uncovered region Further, because of the condition that

12 68 T SUGIMOTO and M TANEMURA Fig 6 The kite K 8 K 4 K 6 K 7, the spherical equilateral triangle K 8 K 9 K 0, and the spherical regular pentagon K 9 K K 6 K K 0 6 ν = C ν is in an extreme state, the center M 6 should be placed again on the point K 8 Then, let K 9 be one of the cross points of the perimeters C 6 and C, and let it be outside C Similarly, let K 0 be one of the cross points of C 6 and C 5, and let it be outside C At this time, the uncovered region (W 6 ) c is reduced to the pentagon K 9 K 4 K 6 K 7 K 0 which is bounded by perimeters of spherical caps Before considering the case of N =, we look back upon the cases of N = 9 and 0 For N = 9, we considered two spherical equilateral triangles K 8 K 9 K 0 and K 9 K 6 K 0 in the kite K 8 K 4 K 6 K 7 on the sphere At that time, we placed the centers of four caps on each vertices of two spherical equilateral triangles and obtained the upper bound r 9 by using the relation that the angular radius r is equal to a side-length of the spherical equilateral triangle (SUGIMOTO and TANEMURA, 003, 006) For N = 0, in Subsec 3, we considered the spherical equilateral triangle K 8 K 9 K 0 and the spherical square K 9 K K K 0 (see Fig 3) Then, the centers of five caps are put on each vertices and the upper bound r 0 is obtained by using the relation that r is equal to a side-length of the spherical square K 9 K K K 0 Now, for N =, we assume the spherical equilateral triangle K 8 K 9 K 0 and the spherical regular pentagon of side-length r on the kite K 8 K 4 K 6 K 7, in order to place six more caps under the Minkowski condition Therefore, here, let us assume K 9, K 0, K C or C 3, and K C 4 or C 5 satisfy = = = = d ( K, K )= d ( K, K )= d ( K, K ) d K, K d K, K d K, K d K, K s 8 9 s 8 0 s 9 0 s 9 s 6 s 6 s 0 ( ) If the arrangement of vertices K 9, K 0, K, and K are actually possible, we obtain the upper bound r ( r 0 ) by using the relation that the angular radius r is equal to a side-length of the spherical regular pentagon K 9 K K 6 K K 0 as the cases of N = 9 and 0 (see Fig 6) Then, for example, when M i (i = 6,, 0) are placed on the points K 8, K 9, K 0, K, and K 6, i respectively, we guess that ν = C ν is in an extreme state and that K is an uncovered point on S Therefore, finally, M can be placed at K However, it is not checked yet that K 9, i K 0, K, and K 6 satisfy such a condition that ν = C ν is in an extreme state for i = 7,, 0, respectively We shall check this fact after obtaining the exact values of the angular radius

13 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 69 r and the coordinates of K and K Now, by applying the spherical cosine and sine theorems, the inner angle of spherical regular pentagon of side-length r is given as follows: cos r 5 cos ( cos r + ) Here, we pay attention to the spherical isosceles triangle K 9 K K 6 whose legs satisfy d s (K, K 9 ) = d s (K, K 6 ) = r Then, by applying the spherical cosine theorem to this isosceles triangle K 9 K K 6, we have ds( K6, K9)= cos cos r+ cos r cos r 5 ( ) Note that the left hand side of () is presented as the function of r by using the coordinates of K 6 and K 9 Refer to Appendix for the explicit coordinates of K 6 and K 9 Equation () is solved against r by using mathematical software As a result, the value of r is obtained as r = tan (3) Therefore, the side-length of spherical regular pentagon K 9 K K 6 K K 0 which satisfies the relation () is tan Here, from the relation (8), let us consider the special case r = d s (K 6, K 4 ) When r = d s (K 6, K 4 ) is solved against r by using mathematical software, we get again the solution r = tan Therefore, r = d s (K 6, K 4 ) = d s (K 6, K 7 ) = tan from the relation (7) On the other hand, from the fact that the side-length of spherical regular pentagon K 9 K K 6 K K 0 which satisfies () is tan, d s (K 6, K ) = d s (K 6, K ) = tan Thus, we find the result as follows: K K 4 and K K 7 if and only if r = tan It follows that the spherical regular pentagon K 9 K K 6 K K 0 which satisfies () is identical with the spherical regular pentagon K 9 K 4 K 6 K 7 K 0 of side-length tan We note that the vertex K 8 of kite K 8 K 4 K 6 K 7 is on the perimeter of the first spherical cap C then In addition, we have shown that tan is equal to the upper bound of the range of angular diameter for the kissing number k = 5 (SUGIMOTO and TANEMURA, 003, 006) Therefore, as a result, we find that the spherical regular pentagon K 9 K 4 K 6 K 7 K 0 satisfies the relation as follows: d s (K 8, K 9 ) = d s (K 8, K 0 ) = d s (K 9, K 0 ) = d s (K 9, K 4 ) = d s (K 4, K 6 ) = d s (K 6, K 7 ) = d s (K 7, K 0 ) = d s (P, K 9 ) (4) = d s (P, K 4 ) = d s (P, K 6 ) = d s (P, K 7 ) = d s (P, K 0 ) = tan,

14 70 T SUGIMOTO and M TANEMURA Fig 7 The curve of A((W 8 C 9 ) c ) when M 9 is moved on the arc K 7 P of C 8 The similar computation method as in Fig 3 is taken See the legend in Fig 3 where P is the center point of spherical regular pentagon K 9 K 4 K 6 K 7 K 0 From the above considerations, for tan r r 0, we see that the maximal spherical regular pentagon of side-length r on the uncovered region (W 6 ) c is coincident with the spherical regular pentagon K 9 K 4 K 6 K 7 K 0 which satisfies the relation (4) Then, for N =, let us examine whether our sequential covering is actually possible when the angular radius r of caps is equal to tan 075 First, we find numerically that the allocations of points K, K 5, K 8, and K 9 for M 4, M 5, M 6, and M 7, respectively, satisfy the condition that each A(W i C i ) (or A((W i C i ) c )) is maximum with restriction M i C i ( i ν = C ν are in an extreme state) for i = 4, 5, 6 and 7 As mentioned above, the results are indicated by the curve of r 075 rad corresponding to r = tan in Figs,, and 4(a) Then, we see that K 0 is the cross point of C 6, C 5, and C 7 Furthermore, K 4 is the cross point of C, C 3, and C 7 Therefore, from the restriction M 8 C 7, M 8 is put at a certain point on the arc K 0 K 4 of C 7 Then, we move M 8 on the arc K 0 K 4 of C 7 and search for the position of M 8 where the area A((W 7 C 8 ) c ) is maximum For r = tan, we 8 checked numerically that ν = C ν are in an extreme state if and only if M 8 is put at K 0 or K 4 The fact is graphically presented by the curve of r 075 in Fig 4(b) Note that the arc K 0 K in Fig 4(b) turns out the arc K 0 K 4 since K K 4 for r = tan Thus, in this paper, M 8 is put at K 0 Then, the uncovered region (W 8 ) c is the quadrangle K 4 K 6 K 7 P on S Next, for r = tan, let us place M 9 at K 7 and move it to P along the arc K 7 P of C 8, and then we search for the position of M 9 where the area A((W 8 C 9 ) c ) is maximum As a result, we 9 find that ν = C ν is in an extreme state when M 9 is put at K 7 The curve of r 075 in Fig 7 indicates this fact In this figure, the horizontal axis is the position of M 9 on the arc K 7 P of C 8 and the vertical axis is the area A((W 8 C 9 ) c ) Therefore, we put M 9 at K 7 Then, K 6 and P are the cross points on C 9 and the uncovered triangle K 4 K 6 P on S is the equilateral triangle of side-length tan Hence, when M 0 is put on the point K 6 C 9, we see that

15 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 7 Fig 8 (a) Our sequential covering for N = (b) Our solution of Tammes problem for N = Both viewpoints are (0, 0, 0) In this example, the coordinates of the centers are respectively (0, 0, ), (07639, , 0447), (089443,0, 0447), (07639, , 0447), ( 0736, 05573, 0447), ( 0736, 05573, 0447), ( 07639, , 0447), ( ,0, 0447), ( 07639, , 0447), (0736, 05573, 0447), (0736, 05573, 0447), and (0, 0, ) 0 ν = C ν is in an extreme state automatically and that S is covered by the set W 0 except for two points K 4 (or K ) and P (the center point of spherical regular pentagon K 9 K 4 K 6 K 7 K 0 ) from the relation (4) At this time, we see that P is the cross point of perimeters C 7, C 8, and C 9 Furthermore, as a result of calculation by using the coordinates of K 9, K 0, and K 7 for r = tan, P is just the north pole (0, 0, ) certainly Therefore, if M is put on K 4 C 0, ν = C ν is in an extreme state and P is a unique uncovered point on S Then, in order to cover whole of S, we have to put one more cap at P In other words, when r = tan, according to our sequential covering, we can put twelve caps on S under Minkowski covering Therefore, M and M are put on K 4 and P, respectively, then ν = C ν which contains W covers the whole of S Namely, we are able to see that d s (P, K 4 ) = tan = r Thus, we consider that tan is the upper bound r for N = As a result, we find that the positions of caps of our sequential covering for N = correspond to the regular icosahedral vertices (see Fig 8(a)) Therefore, if all spherical caps of our sequential covering for N = are replaced by half-caps, all of those half-caps contact with other five half-caps and there is no space for those half-caps to move Then, how about r? From the results of N = 0 and, it becomes obvious that our initial assumption r = tan r r 0 is indispensable At the time the center M 6 is placed on the point K 8, we had to place five more caps of the angular radius r on the uncovered region (W 6 ) c under the Minkowski covering Then, from the above considerations, we see that the maximal spherical regular pentagon of side-length r (r r r 0 ) on (W 6 ) c is identical with the spherical regular pentagon K 9 K 4 K 6 K 7 K 0 which satisfies the relation (4) So, we conclude that r is equal to r = tan and use the same configuration of the first eleven spherical caps of the case N = However, when r = tan, our sequential covering for N = covers S except for the point P But, if only the eleventh cap C is replaced by a closed cap, our eleven caps can completely cover the whole of S under the Minkowski condition Thus, for N =, we need a special care as above

16 7 T SUGIMOTO and M TANEMURA Fig 9 (a) Our solution of Tammes problem for N = 0 (b) Our solution of Tammes problem for N = Note that the spheres are drawn with the wireframe 4 Conclusion From the results of Sec 3, if all of those spherical caps are replaced by half-caps, the results of our problem for N = 0,, and are coincident with those of the Tammes problem (see Figs 5, 8, and 9) Especially, we obtained in this paper the exact closed form of r 0 for N = 0 (see (0)), whereas Danzer have obtained the range [54479, 54480] of angular diameter for N = 0 (DANZER, 963) Further, Danzer have solved the Tammes problem for N = 0 and through the consideration on irreducible graphs obtained by connecting those points, among N points, whose spherical distance is exactly the minimal distance Then he needed the independent considerations for N = 0 and, respectively However, we presented a systematic method which is different from the approach of Danzer about the Tammes problem Namely, as shown in Sec 3, our method is able to obtain a solution for N by using the results for the case N or N In addition, in this study, we have considered the Tammes (packing) problem from the standpoint of sequential covering The advantages of our approach are that we only need to observe uncovered region in the process of packing and that this uncovered region decreases step by step as the packing proceeds A set of spherical caps is said to be a Minkowski set if none of its elements contains in its interior the center of another In our study, the condition of Minkowski set of centers is called a Minkowski condition, and the covering which satisfies the Minkowski condition is called a Minkowski covering On the other hand, replacing each spherical cap in a Minkowski set of spherical caps with a concentric caps of radius half as big as the original, FEJES TÓTH (999) called a Minkowski packing (ie, the Minkowski packing is our packing of half-caps) and demonstrated the upper bounds for densities of a Minkowski packing and a Minkowski set Note that their densities are sharp for N = 3, 4, 6, and, and are asymptotically sharp for great values of N Then, we checked that our solutions for N = 3, 4, 6, and are coincident with his results In I, we defined the efficiency of covering on spherical surface Our efficiency is the reciprocal of the density of a Minkowski set

17 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 73 Therefore, our solutions for N = 3, 4, 6, and are the worst efficient covering under the Minkowski condition It is interesting to point out that our method presented in this paper will be also useful for providing a conjectured value of optimal angular radius for any N by using the already known value for N This will be done as follows: (i) first we set the angular radius of a spherical cap by using the known exact or approximate value for N, namely, we put r = ˆr N where ˆr N is the exact or approximate value of r for N ; (ii) then we put N spherical caps with radius r through our method of sequential covering; (iii) in this stage, it is possible that the Minkowski condition will be broken especially when the N-th spherical cap is placed If this is true, decrease the size of r so far as the Minkowski condition is satisfied and go to Step (ii) Otherwise, check if the whole of S except for a point or a line segment is covered by N spherical caps If it holds, the procedure ends, else increase r by a prescribed value r then go to Step (ii) We will be able to estimate the value of an optimal angular radius for any N by this procedure Throughout our paper, we used the mathematical software, Maple 8 and 95, which is capable of manipulating complicate algebraic expressions exactly and is also useful for numerical computations The authors would like to thank Prof L Danzer, Univ of Dortmund, and Prof H Maehara, Univ of Ryukyu, for their helpful comments This research was supported by the Ministry of Education, Science, Sports and Culture, Grantin-Aid for JSPS Fellows, 5 83, and Young Scientists (B), Appendix: The Coordinates of K i (i = 3, 4, 5, 6, 7, 8, 9, and ) In the following, r (tan r < π/) is the angular radius The coordinates of K i (i = 3, 4, 5, 6, 7, and 8) are shown for reader s convenience, although indicated in I K 3 = (x 3, y 3, z 3 ): K 4 = (x 4, y 4, z 4 ): sin r cos r cos r cos rsin r cos r+,, cos r ( cos r + ) ( cos r + ) cos rsin r cos r+ cos r + K 5 = (x 5, y 5, z 5 ): cos rsin r cos r+ 4cos r cos r,, cos r + cos r + sin r cos r cos r cos rsin r cos r+,, cos r ( cos r + ) ( cos r + )

18 74 T SUGIMOTO and M TANEMURA K 6 = (x 6, y 6, z 6 ): cos rsin r cos r+ cos r + K 7 = (x 7, y 7, z 7 ): cos rsin r cos r+ 4cos r cos r,, cos r + cos r + ( ) cos rsin r cos r+ cos r cos r + 4cos r cos r cos r + + cos rsin r 3cos r+ cos r,, cos r K 8 = (x 8, y 8, z 8 ): ( + ) cos rsin r cos r cos r 9cos r cos r cos r cos r cos r+ 5cos r+ cos r 3 9cos r cos r cos r+ cos rsin r cos r cos r, 9cos r cos r cos r K 9 = (x 9, y 9, z 9 ): 3 sin r cos r 5cos r cos r+ 4sin rcos r cos r+ 3, 3 9cos r cos r cos r+ 9cos r cos r cos r+ 3 cos r cos r+ cos r cos r 3 3 9cos r cos r cos r+ K = (x, y, z ) are shown in cos (see Subsec 3) and the coordinates of K 8 due to their lengthy expressions (( ) x = x cos lcos r+ y sin r+ z sin r+ z cos lsin r+ x z cos r cos r (( + cos l z cos lcos r+ x cos lsin rcos r+ x z sin rcos r+ z z cos r ( / ) + y8 y8 cos r) y8 cos r) 8 y8 ( + z8 + x8 cos r+ x8z8sin rcos r z8cos r),

19 Packing and Minkowski Covering of Congruent Spherical Caps on a Sphere, II 75 y cos l + z = y 8 8 x8 + z8tan r x y 8, z = + tan r x REFERENCES DANZER, L (963) Endliche Punktmengen auf der -Sphäre mit möglichst großen Minimalabstand, Universität Göttingen (Finite Point-Set on S with Minimum Distance as Large as Possible, Discrete Mathematics, 60, 3 66 (986)) FEJES TÓTH, L (97) Lagerungen in der Ebene, auf der Kugel und im Raum, nd ed, Springer-Verlag, Heidelberg FEJES TÓTH, L (999) Minkowski circle packings on the sphere, Discrete & Computational Geometry,, 6 66 SCHÜTTE, K, and VAN DER WAERDEN, B L (95) Auf welcher Kugel haben 5, 6, 7, 8, oder 9 Punkte mit Mindestabstand Eins Platz?, Mathematische Annalen, 3, 96 4 SUGIMOTO, T and TANEMURA, M (003) Packing and Minkowski covering of congruent spherical caps on a sphere, Research Memorandum, ISM, 90 SUGIMOTO, T and TANEMURA, M (004) Packing and covering of congruent spherical caps on a sphere, Symmetry: Art and Science - 004, ISIS-Symmetry, SUGIMOTO, T and TANEMURA, M (006) Packing and Minkowski covering of congruent spherical caps on a sphere for N =,, 9, Forma,, 97 5 TAMMES, P M L (930) On the origin of number and arrangement of the places of exit on the surface of pollen grains, Recueil de Travaux Botaniques Néerlandais, 7, 84