DERIVED FROM THE REGULAR POLYTOPES. 79 



The spaces 8^ represented by + x x + œ$ = 5 -J- 3 give 2 2 . (5) 2 = 

 40 limits (53) -J- [3 1 1] of body import. 



The spaces # 4 represented by S + œ^ = 13 give 2 4 limits (53311) 

 of vertex import, where (5 3 3 1 1) = (4 2 2 0). *) 



The spaces 8 k represented by 2 + x { = 11 give 2 4 limits (5331 — 1 ) 

 introduced by the alternate truncation. 



So the limiting polytopes are 



10 e ± <7 46 + 40 P tT + 16 e 2 S(5) + 16 é?^ /S(5), 



i. e. 82 in toto. 



Now from the list of limiting bodies 



e ± C i6 ...80,16 tT 



P tT ...2tT, 4P 3 , 4 P 6 



<? 2 tf(5) ... 5 00, 10 P 3 , 5 O 



^^£(5) ...btT, 10 P 6 , 10 P 3 , 56'0 



of the four different limiting polytopes we can deduce that our 

 polytope is limited by 



1(10 X 8 + 16 X 5)0, £(10 X 16 + 40X 2 + 1G X 5) ^ 7 , 



1(16X5 + 16X5)00, 1(40X4 + 16X10 + 16X10)^3, 



1(40X4 + 16X10) A 



i. e. by 720 polyhedra, viz. 



80 O, 160 tT, 80 00, 240 P 3 ; 160 P 6 . 



Now finally, according to Euler's rule, the number of faces is 

 1840. So the result is 



(480, 1680, 1840, 720, 82). 



This example shows that the method explained is sufficient for 

 8 5 , as far as the characteristic numbers themselves are concerned. 

 But if we want to extend our knowledge of these hmpd. — in 

 relation with the difficulty of realising their lopsided form — by 

 determining the numbers of the different hinds of limits the method 

 is insufficient even in 8 5 and has to be completed, in one sense 

 or other, with respect to the different kinds of edges and of faces. 

 We shall see that the direct method, which will be explained in 

 the next article, furnishes this complement at least expense. 



95. Here once more the direct method in view is based on the 



') This (42200) with edges 2J/2 is similar to (21100) with edges J/2, i.e. to e 2 S(5). 

 Likewise (5331-1) leads by (64420) to (32210) or (32110), i.e. to e 1 e z S (5). 



