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Kepler's Spheres and Rubik's Cube Kepler'sSpheres and Rubik'sCube JAMES G. PROPP Universityof Maryland College Park,MD 20742 "How manyspheres of radiusr may simultaneouslybe tangentto a fixedsphere of the same radius?" This questiongoes back at least to the year 1694, when it was consideredby Isaac Newtonand his contemporaryDavid Gregory.Johann Kepler had already shown in 1611 [8] that twelve outer spherescould be arrangedaround a centralsphere, and now Gregoryclaimed that a thirteenthcould be added. Newton disagreed,but neitherman provedhis claim,and it was not until1874 thata proof was found,substantiating Newton's conjecture. (R. Hoppe's originalproof is described in [2]; morerecently, proofs have been foundby Giinter[7] and by Schiitteand van der Waerden [10]. Perhapsthe mostelegant proof known is the one givenby John Leech [9]. For a historyof the problem,see [6].) It's worthnoting that the 4-dimensionalversion of thisproblem is stillunsolved; no one knows how to arrangemore than twenty-fourhyperspheres around a central hypersphere,but neitherhas it been shownthat a twenty-fifthcannot be added. In higherdimensions the situationis even less well understood,with the remarkable exception of dimensions8 and 24, for which the maximumcontact numbersare preciselyknown [1]. The sourceof difficultyin the originalGregory-Newton problem (and the reasonit took 180 yearsto be solved)is thatthere is almostroom for a thirteenthsphere; the twelve spheres of Kepler can be pushed and pulled in all sortsof ways, and it's crediblethat some sortof fiddling could create a space big enoughto accommodatean extrasphere. This note will deal witha relatedquestion, in the spiritof Erno Rubik:if we label the twelve spheresand roll themover the surfaceof the innersphere at will, what permutationsare achievable?The answeris surprising,and the proofrequires only the rudimentsof analyticgeometry and group theory.One fringebenefit of this enterpriseis that it leads to a naturalcoordinatization of the verticesof the regular icosahedron. An equivalentformulation of the problemis gottenby consideringonly the centers of the spheres:we imaginetwelve vertices, free to move on a sphereof fixedradius R = 2r about a fixedorigin but subjectto theconstraint that no twovertices may ever be closer togetherthan R. We will findit convenientto switchback and forth betweenthe two formulations.For definiteness,put r = 1, 1 = 1. To begin, we must prescribean initialconfiguration for the outer spheres.One possibilitythat springsto mindis to arrangethe twelvevertices to forma regular icosahedron.As we'll see, anytwo distinctvertices of the icosahedroninscribed in the unit sphereare at least /2 ~-2/ I~C5 1. 0 1 231 This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:54:42 PM All use subject to JSTOR Terms and Conditions 232 MATHEMATICS MAGAZINE unitsapart, so not onlyis thisconfiguration possible, but no two of the outerspheres touch (see FIGURE1). It is clear thatany rigidmotion of the icosahedronabout its centercan be achievedby a jointmotion of the spheres;and so, since the rotational symmetriesof the icosahedroncomprise the alternatinggroup A5 [5, pp. 49-50], the number of icosahedralarrangements of the spheres accessible fromthis starting configurationis at least IA51= 60. FIGURE 1 But thereis anotherway of placingthe twelvespheres around one, and moreover, thisarrangement extends to a sphere-packingof three-space.Let the latticeA consist of all points(a/X;, b/xl2, c/l42) witha, b, c integersand a + b + c even. Then no two pointsof A are closerthan 1, and so the pointsof A mayserve as the centersof non-overlappingspheres of radius 2 (see FIGURE 2). This is the cubic close-packing discoveredby Kepler.Each vertexhas twelvenearest neighbors, and in particularthe - xV XX -- rCFIGURE2 /~7 This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:54:42 PM All use subject to JSTOR Terms and Conditions VOL. 61, NO. 4, OCTOBER 1988 233 neighborsof (0, 0,0) are (+11/2,? 11/2,?) (1C+ , O,+ 11/2) (O, ? 11/2, ? 112) . These verticesform a cuboctahedron(see FIGURE 3(a)). Its rotationalsymmetry group,like thatof the cube and the octahedron,is S4; one mayconcretely picture the S4 action as permutingthe fourpairs of antipodaltriangular faces. Hence at least IS4 -24 cuboctahedralarrangements are accessiblefrom the initialposition. The S4 action on the vertices,like the A5 action,performs 'only even permutationson the twelvevertices/spheres. D FIGURE 3(a) J Now comes a pleasant surprise:the two arrangementsof the twelve spheres (icosahedraland cuboctahedral)may be deformedinto one another!This meansthat in our Rubik-likemanipulations, we can avail ourselvesof both the A5 and S4 symmetries. To deformthe cuboctahedroninto the icosahedron,move each pointalong a great circle(as indicatedin FIGURE 3(a) forselected points) so thateach squareface of the cuboctahedronbecomes a pairof triangularfaces of the icosahedron. For instance,the points A-(- ,o, fl and c= (,o, A) both move toward(0, 0, 1), whilethe point B= (,-0 , ?) moves toward (0, - 1,0). (This is a variationon the constructiongiven in [5, pp. 51-53].) We mayparametrize this joint motion with a variablet, whoseinitial value is This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:54:42 PM All use subject to JSTOR Terms and Conditions 234 MATHEMATICS MAGAZINE A(t) = (-cos t,O,sint) B(t) = (O, -sin t,cos t) C(t) = (cos t,O,sin t) and so on. The distancebetween A(t) and B(t) is -cos2t + sin2t+ (sin t - cos t 2-sin 2t, whichincreases from 1 to F as t runsfrom x/4 to ff/2.The distancebetween A(t) and C(t) is 2 cos t, which decreasesfrom F2 to 0 as t approaches7T/2. Let us interveneand stop the processwhen the lengthsof AB and AC are equal, whichoccurs at someunique time t =0, and write A' =A(O), B' B(O), etc. The symmetryof the procedurenow guaranteesthat B'C' = A'B'- B'F' - F'A', so thattriangles A'B'C' and A'B'F' are congruentequilateral triangles. In fact, the vertices A', .., L' formtwenty such triangleson the sphereof radiusR, and thisimplies that they are arrangedicosahe- drally,with labelings as shownin FIGURE3(b). A , FCf G 11 FIGURE 3(b) (By equating (A'B')2 = 2 -sin2O and (A'C )2 = 4cos20 = 2 + 2cos2O one obtains -1 ~~~2 cos2O =- and sin 2 = C C5 so thatthe commonlength of the segmentsis A2- 2/U This is the side-lengthof an icosahedronof circumradius 1. Also,the distancebetween A' and F' is 2 sin0, whose ratioto A'C' = A'B' = 2 cos 0 is tan0; thismay easilybe This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:54:42 PM All use subject to JSTOR Terms and Conditions VOL. 61, NO. 4, OCTOBER 1988 235 shownto equal the goldenratio (1 + VU)/2.Hence A'I'K'C' is a goldenrectangle, as is any rectangleformed by two diagonalsof the icosahedron.) Since it does not matterwhether our initialconfiguration is icosahedral or cubocta- hedral,we may as well assumethe latter.Let G be the groupof permutationsof the twelvespheres that can be achievedby the operationsof rotatingthe cuboctahedron, switchingbetween cuboctahedraland icosahedralformation, and rotatingthe icosa- hedron,in all combinations.(We don't claim that G containsall the achievable permutations.)Note thatpoints which were antipodalat the startremain antipodal throughout.Also notethat since G containscopies of both S4 and A5, its orderis at least 120. If we rotatethe labeled icosahedronof FIGURE 3(b) by 72 degreesclockwise about its axis A'K', the pointsare permutedby (A')(B'C'D'E'F')(G'H'L'I'J')(K') whichcorresponds to the permutation y = (A)(BCDEF)(GHLIJ)(K) of the verticesof the cuboctahedron.On the otherhand, if we rotatethe cuboc- tahedronby 90 degreesclockwise about the verticalaxis, the pointsare permutedby a = (ABCD)(EFGH)(IJKL). Since each of thesepermutations belongs to G, so does theirproduct a-y= (ABDFC)(EG)(HIKLJ). But (ay)5 = the two-cycle(E G). Hence, in general,any two antipodalpoints may be exchangedwithout affecting the finalposition of the otherten. The precedingobservation tells us thatthe group G is a wreathproduct of theform 26.H, where H is a groupacting on pairs of antipodalpoints. (For an explanationof thisnotation, see [3].) Thatis, everyelement of G is uniquelydetermined by itsaction on the six principaldiagonals of the polyhedron,with a choice of orientation;if you like, thinkof H as a permutationgroup acting on six coins,and 26.H as the set of actions one gets by allowingnot onlypermutations taken from H but also arbitrary flips.Let A, B, etc., denotethe diagonalsAK, BL, etc. Then the image of -yunder the naturalhomomorphism from G to H is y =(A)(BC DEF and similarly,the imageof a is a = (A B C D F). To findgenerators of the groupH, we need to knowthe generatorsof G. a and y aren'tenough, but we'll onlyneed one more.The S4 subgroupof G is generatedby a = (ABCD)(EFGH)(IJKL) (90 degreerotation about the axis throughface A B CD) and /3= (ABF)(CJE)(DGI)(HKL) (120 degree rotationabout the axis throughface ABF), while the A5 subgroupis generatedby This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:54:42 PM All use subject to JSTOR Terms and Conditions 236 MATHEMATICS MAGAZINE y= (A)(BCDEF)(GHLIJ)(K) (72 degreerotation about axis A'K') and 8 = (ABF)(CJE)(DGI)(HKL) = (120 degree rotationabout the axis throughface A'B'F'). The images of those elementsin H are a = (A BCD)(EF) : A BF)(C DE ) 'y= (A) BC DEF ) and theymust generate H. Since all threepermutations are even, H mustbe a subgroupof the alternating group A6, and in fact H is A6 itself.We prove this by successivelyexamining stabilizersubgroups: 1. The action of H on A, B, C, D, E, F is clearlytransitive, so all of its 1-point stabilizersubgroups are isomorphic,and have orderJH1/6. 2. y (which belongsto the stabilizerof A) acts transitivelyon B, C, D, E, F, so thatall 2-pointstabilizer subgroups are isomorphic,and have order{H1/6 5.
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