1 4 ANALYTICAL TREATMENT OF THE POLYTOPES KEGULAKLY 



the simplex of coordinates and in three regular hexagons in planes 

 parallel to the plane x k = 0, x- = 0, x 6 = containing the opposite 

 face A i A 2 A^ of that simplex. As these faces are perpendicular to 

 each other according to the theorem of art 5, we find a regular 

 prismotope (3 ; 6), occurring twenty times. 



d). For x x -\- x 2 = 7 we have to combine each of the possibilities 

 a\ =4, x 2 = 3 and a? 4 = 3, œ 2 = 4 with x 2 , x k , œ bi x 6 = (2110). 

 So we find fifteen prisms P co . 



e). The equation ^ = 4 gives œ. 2i œ 39 œ\, œ 5 , œ 6 == (32110). So 

 here we find six e i e 3 S(b). 



So the limiting poly topes are 



6^/9(5), 15P f0 , 20 (3; 6), 15P C0 , 6^ 3 £(5) ; 



their number is 62. Of these the Qe ± e 2 S(b) are of polytope im- 

 port, the 15 P l0 of polyhedron import, the 20 (3 ; 6) of face import, 

 the 15 P co of edge import and the 6^^/9(5) of vertex import. 



In order to find the number of the limiting polyhedra we enu- 

 merate the limiting polyhedra of the limiting polytopes. 



a). The determination of the limiting polyhedra of the six poly- 

 topes (32100) runs parallel to the investigation of (32110) in the 

 preceding article, as we remarked already in the last footnote. 

 So we find in the same order of succession and with the import 

 with respect to the simplex S (5) of its space: btO of polyhedron 

 import, 10jP 3 of edge import and 5 tT of vertex import, i. e. twenty 

 limiting polyhedra. 



b'). The prism P t0 is limited by two tO and fourteen prisms, viz. 

 six P 4 (or cubes) and eight P 6 , i.e. by sixteen polyhedra. 



c). The prismotope (3 ; 6) is limited by nine prisms, six P s and 

 three i J 6 . 



ct). The prism P co is limited by two CO and fourteen prisms, 

 viz. six P\ (or cubes) and eight P s , i. e. by sixteen polyhedra. 



e). The polytope (32110) of the preceding article is limited by 

 thirty polyhedra. 



So the number of polyhedra is 



\ (6 X 20 + 15 X 16 + 20 X 9 -f 15 X1G + 6X 3 °) = 480 - 



According to the law of Euler the number of faces is 



1080 + 480 -f 2 — (360 -f- 62)= 1140. 



So the resulting symbol of characteristic numbers is (360, 1080, 

 1140, 480, 62). 



► 



9. The direct method alluded to in the beginning of art. 6 rests 



