AND ENUMERATION OF POLTEDRA. 53 



6. Returning to consider our n-edron P, having a 

 (w — l)-gonal base and only triedral summits, and having 

 only two triangular faces, abc &ndjkl, we have determined of 

 its 2n — 4 summits the following n-fl, 

 Aab, Abc, Acd, Ade Ajk Akl Aim, Ama, abc, jkl. 



The edges of the w-edron are 3« — 6 (Theo. II.). Of 

 these, n — I are sides of the (w — l)-gonal base, n — 1 others 

 pass through the angles of the base, and n — 4 never meet the 

 base, but form the broken line which connects the vertices of 

 the two triangular faces. If we pass along this line from the 

 triangular face k to the similar face b, we shall traverse all 

 those edges of the intervening faces, /, m, and a, which do 

 not meet the base. These n — 4 edges are X in the face /, fi in 

 m, and a in a, such that X4-ft-|-a=w — 4. If X=:l, / has but 

 one edge that does not meet the base, and as the duad jl, in 

 the triplet jkl, must appear a second time, that one edge is 

 jl, which must meet the edge Im, passing through the base 

 angle Aim. In this case jlm is one of the triplets of the 

 system. But if X=2, jlm is not a triplet. 



^ 7. To clear the matter, let n=14. 



X=2, /i=5, a=w— 4— 2— 5=3. 

 In the triplets to be formed, we have to dispose of the duads 



cd, de, ef,fg, gh, hi, ij, Im, ma, ac, jl, 

 which have been only once employed, and the whole system 

 of triplets must exhibit no duad oftener or seldomer than 

 twice. 



One edge of the broken line in / is^/; the second must be 

 il, meeting it in ijl, for i is the face next to j remote from 

 jkl ; and the duads il and Im must occur a second time : this 

 determines the triplets 



lij and Imi. 

 Next we have 5 edges of the broken line on the face m, of 

 which one is mi, because that duad must appear again, and 

 tlie others are in order mi, mh, mg, mf, me. It is necessary 



