202 
EEV. T. P. ETRE^V TAX ON AETOPOLAE POLTEDEA. 
OE. Therefore jS is in O', and is e±l. Now o(e±l) being convanescible after the 
evanescence of oe, must before that be in a face not tiiangular. Of such faces S has 
only E, O, and H the polar of But oo' is in E ; therefore o{e+l) is in H. 
The conditions, therefore, that the triangle oe{e-±l) shall contain a second leading 
system, are, that O' shall be the triangle e{e±l)f, having the triaces e±l and/,, that oe 
shall be an edge distinct from o;?, and that o(e+l) shall be an edge of H, no triangle. 
XXIII. The signatures of the pyramid about e are 
...(G-2) (G-1) G (G+1)... 
... : (^+1) ^ {e-l) : ... 
Avhere the colon is the point/. As the only summits of the figm’e S not in the con- 
tour of Q are o and o\ e has the edge oe, or de, or both. The edges of [e, e-\-\,f) (the 
triangle O') are O'(G-l) and 0'(G-2), and of {e, e-l,/) are O'G and 0'(G+1); of 
Avhich none is O'E unless G — 1=:E, or G — 2=E; or G=E, or G-{-l=E. 
If the first or third be true, e is nodal. 
If G — 2=E, G — 1=E+1, and e-\-l is nodal. 
If G-1-1=E, G=:E— 1, and e— 1 is nodal. 
Therefore O'E is never an edge of (e, e+1,/), unless one summit of this triangle is 
nodal ; and OE is always an edge of S, if no summit of O' is nodal. 
^Vhen no summit of O' is nodal, o(e+l) is always an edge of a quadrilateral. For 
in the triangle (e, e-f-lr/) neither 0'(G — 1) nor 0'(G — 2) is 0'(E-|-1), because e+1 not 
being nodal, E-fl is neither G — 1 nor G — 2. As then 0'(E+1) is no edge, o'{e-\-l) 
is none, and o(e-\-l) is in G — 2, whose summits are e+l,/, e+2, o. 
And in the triangle (e, e — 1,/) neither O'G nor O'(G-i-l) is 0'(E — 1) ; wherefore 
o(e— 1) is in G-]-!, Avhose summits are e— 1,/, e — 2, o. 
Consequently, when no summit of the triangle O' is nodal, S has two leadmg systems, 
and maybe a repetition of some other of the \{r.{r — 2) — 2+2,_,} S above (XXI.) con- 
structed. 
XXIV. Let e be nodal, and G=E. The triangle {e, 6—1,/) has the edge O'E; but 
the triangle {e, 6+1, /) has not the edge O'E. In the latter case OE is an edge of S ; 
but not in the former, unless 6 be a pentace containing ef, eo and eo', with two sides of Q. 
But G=E has only the summits e, e — 1, o, o'; therefore e is no pentace, and OE is not 
an edge of S in the former case. Hence Avhen e is nodal, and G=E, one line ef^ can 
be draAvn to make one triangle O' or {e, 6—1,/) such that O'E is an edge of S, and OE 
is not; and another, [e, 6+i,/) such that OE is an edge and O'E is not. 
"When O'E is no edge, 0'(G — 1) and 0'(G — 2) are neither of them 0'(E+1) in 
(6, 6+1, /); and neither O'G nor 0'(G+1) is 0'(E — 1) in [e, e — 1, /). Therefore 
0(E+1) is an edge in one case, and 0(E — 1) in the other; and o(6+l) is in (G — 2), 
whose summits are 6+1, / 6+2, o in the former case, and o(6 — 1) is in G+1, whose 
summits are e — 1,/, 6—2, o, in the latter. Hence 6+1 is a triace in O'. 
The like conclusion folloAvs from supposing e nodal, and G— 1=:E, that one generator 
