DERIVED FROM THE REGULAR POLTTOPES. 1 1 



(n = 5) is half the sum of the numbers of the limiting bodies of 

 the limiting polytopes. 



7. We now treat at full length the example (32110) mentioned 

 above. 



The number of vertices is 5! divided by 2! i. e. 120:2 — 60. 

 The number of edges passing through each vertex is five, for in 



sTTTo 



each of the five brackets indicates two coordinates with difference 

 unity the interchanging of which furnishes a new vertex joined by 

 an edge to the "pattern vertex", i. e. to the point with the coor- 



(50 V 5 

 dinates 3, 2, 1, 1, 0. So the number of edges is — —~ — =150. 



In the case of this polytope we have to consider successively the 

 equations. 



a) . . . x\ -\- x 2 -f- x 3 -f- x k = 7 or x b = , 



b) ... a? 4 -j- a? 2 -f- a?j = 6 or x tv -j- x b = 1 , 



c) . . . x^ x 2 ■—— o or Xo ■ x 4 Xr. ==z iC , 



d) . . . w ± = 3 or x 2 -\- x s -f- x k -\- x- ö = 4. 



Each of the two equations placed on the same line is a conse- 

 quence of the other; for in treating the polytope (32110) we have 

 to suppose that the true coordinate values of any point have been 

 increased by such a common amount as to make the sum of the 

 coordinates equal to seven. 



a). Both the equations x ± -\- x 2 -\- x 3 -\- x^ = 7 and x 5 = give 

 a? 4 , œ l3 x 3 , x k = (3211), i. e. for the Coordinates x ly x 2 , x 3 , x k we can 

 take any permutation of 3, 2, 1 , I. But if we subtract a unit from all 

 the coordinates and write x b = — land x i} x 2 , x 3 , œ\ = (2100), whereby 

 the constant sum 7 of the coordinates is changed into 2, we see 

 (compare the table in art. 3) that the obtained form is a tT 1 ). 

 This tT presents itself five times, as the subscript 5 in x b = 

 may be any of the ^ve numbers 1,2,3,4,5. 



b). In the case x ± -j- x 2 -\- x 3 = 6 which implies #? 4 -j- x 5 = 1 we 

 have to combine the two systems œ u x 2 , x 3 = (32 1 ) and a? 4 , x b = (10); 

 i. e. we have to combine the system at it w 2 , x s = (321) with each 

 of the two possibilities #? 4 = 1 , x b = and a? 4 = , a? 5 = 1 giving 



x ) That (2100) is a polyhedron with 12 vertices, 18 edges, 8 faces limited by 4 

 regular hexagons and 4 equilateral triangles is immediately found by treating the 

 symbol (2100) in the manner taught here. 



