AND ENUMERATION OF POLYEDRA. 65 



The system to be completed by four more triplets is 

 Aab Abe Acd Ade Aef Afa 

 fab bed de ef fb bd, 



C must have one duad of consecutive letters : either Cde, 

 or Ce/, must be written : which can only be followed by Cdb, 

 Cebi bef, or Ceb, Cfb, bde, respectively, as is easily verified. 

 But either of these sets of four repeats b four times, which has 

 thrice occurred before, thus introducing a heptagonal face, 

 which is here inadmissible. 



The faces here determined would be 

 A C a b e d ef 



■(A) 



e a ej -v 



7 3 5 5 4 f ^ 



b e d ef \ 



7 3 4 5 6 -^ 



6 3 3 



A C a b e d ef 



6 3 3 



which are two octaedra on a heptagonal base 6, the one the 

 reflected image of the other. 



We take then a and d for our triangles, opposite faces 

 about the hexagonal base. Looking now at 

 Aab Abe Aed Ade Aef Afa, 

 fab be ede ef fb ce, 



we see that it is of no consequence whether we take Cbc or 

 C^for the triplet containing a duad of consecutive letters, 

 because all is symmetrical about the opposite faces. 

 Taking the former, we must write either 



Cbc CbfCfcfce, or Cbc Cce Ceb bef 

 which give the faces 



AC abed ef, AC abcdef (B) 



63356346 63365364 



which are merely reflected images of each other. Further, 

 if we take a for base in the figure (6) of the last article, and 

 d for crown, we find that (6) is equally represented by 

 a d (C c ef A 6 \, 



63X3653 6 4/ 



CcefAb being the faces which lie in order about a, as is 

 evident from the triplets aCc, ace, aef, a/A, aAb, abC: 

 therefore (6) and (B) are the same octaedron. 

 K 



