DERIVED FROM THE REGULAR POLYTOPES. 55 



Finally, in the set with <? 4 , two cells — belonging to different groups 

 — cannot have a vertex in common ; so here each net symbol 

 represents only 4 of the system of vertices. 



We now indicate schematically how we can determine all the 

 constituents of the different nets of C m . To that end we have 



1°. to deduce from the preceding developments the net symbols 

 necessary in every case, 



2°. to calculate the coordinates of the centres of the different 

 constituents, by multiplying the coordinates of a vertex, of the 

 midpoint of an edge , of the centre of a face and of the centre 

 of a limiting body of [2, 0, 0, 0] by the extension number, 



3°. to determine the vertices contained in the net symbols, lying 

 at the same minimum distance from these centres. 



As we shall have to consider the "extended" vertex, midpoint of 

 edge, centre of face, or centre of limiting body mentioned sub 2° 

 as new origin of parallel axes of coordinates in order to be able 

 to obtain the simplest representation of the sets of vertices menti- 

 oned sub 3° we will denote this extended point henceforth by 0' . 



Of each of the three sets we will treat some examples, of 

 the first e l e 2 iV((7 16 ) and ce i e 2 N(C i6 ), of the second e 2 e z N(C iQ ) 

 and ce i e 3 N( C iQ ) , of the third e x e k A\ C i6 ) , e x e 2 e 3 e k N(C ie ) and 

 ce x e 2 e 3 e± A67 16 . Afterwards we will put on record the coordinate 

 svmbols of all the constituents in Table VII. 



81. Case e l e. 2 JHir(C 16 ). Net symbol 



[12 «, + 6, 12a 2 + 4, 12^ + 2, 12fl 4 -f-0], S«i even. 



Here the constituent of polytope import is [6, 4, 2, 0] = e x e 2 C 16 . 

 There are no constituents of body and face import as the opera- 

 tions e k and e z do not present themselves. So we have only to 

 determine the polytopes of edge and vertex import. 



Edge gap prism. By extension the centre 1, 1, 0, of the edge 

 (2, 0) of [2, 0, 0, 0] becomes 6, 6, 0, 0. By putting in the net 

 symbol a i = 0, (i = 1, 2, 3, 4), we find among others the vertices 

 (6,4) [2,0] and by putting a A =a 2 =\, tf 3 = tf 4 = 0, and taking 

 the movable digits 6 , 4 with the negative sign we find also the 

 vertices (6, 8) [2, 0]; with respect to the new axes with the point 

 6, 6, 0, as new origin O' these two groups of vertices can be 

 represented together by the symbol [2,0] [2,0]. So we find a 

 measure polytope C 6 which is to be interpreted here as a prism on 

 a cube, F c . 



