406 
ON CUBIC CONES AND CURVES. 
[351 
The tangent planes, or, say the tangents at the inflexions in piano, for instance, 
at the inflexions 1, 2, 3, form a trilateral, and we have thus corresponding to each of 
the three planes a trilateral formed by the tangents at the inflexions on such plane; 
and there are of course four systems, each of three trilaterals formed by the tangents 
at the nine inflexions. 
13. I say that if S = 0, then in one of the four systems the trilaterals become 
each of them a line, that is, the tangents at the nine inflexions meet by threes in three 
lines. 
14. This may be shown by means of the before-mentioned canonical form 
a? + y % + z s + Glxyz = 0 
of the equation of a cubic cone, for then the notation of the inflexions being in 
accordance with the foregoing scheme, the coordinates may be taken to be 
(1) x — 0, y + z = 0, (4) x = 0, y + coz = 0, (7) x = 0, y + co-z = 0, 
(2) y = 0, z + x = 0, (5) y = 0, z + cox — 0, (8) y = 0, z + orx — 0, 
(3) z — 0, x + y = 0, (6) z = 0, x+coy = 1, (9) z = 0, x + co 2 y = 0, 
where co denotes an imaginary cubic root of unity, and the equations of the tangents are 
(1) 
— 2 lx + y + z = 0, 
(4) 
— 2 lx + coy + arz = 0, 
(7) 
— 2 lx + Ci) 2 y 4- COZ = 0, 
(2) 
x — 2ly + z= 0, 
(5) 
o' 
II 
3 
+ 
CM 
1 
3 
(8) 
(ox — 2ly + (o~z = 0, 
(3) 
x + y — 2 Iz — 0, 
(6) 
COX + 0) 2 y — 2 Iz = 0, 
(9) 
(o 2 x + coy — 2Z^ = 0. 
15. The value of S is = —Z + Z 4 , and for each of the values 1, 0, &>, to 2 , of Z, which 
give >8=0, we have a system of the nine tangents meeting by threes in three lines, 
viz. the systems are 
= 1, 
123, 
456, 
789, 
= 0, 
147, 
258, 
369, 
= (O , 
159, 
267, 
348, 
= <w 2 , 
168, 
249, 
357. 
16. It is proper to notice that starting with the systems in question, or what is 
the same thing, with a single set of each system, say the sets 123, 147, 159, 168, we 
obtain as the condition to be satisfied by Z, the equation 
Z (4Z 3 - 3Z - 1) (4Z 3 - 3loo + 1) (4Z 3 - 3Zco 2 + 1) = 0, 
or, as it may otherwise be written, 
Z (21— l) 2 (Z — 1) (2Zco — l) 2 {loo — 1) (2Zeo 2 — 1) 2 (Z&> 2 — 1) = 0, that is, (- l + Z 4 ) (1 + 8Z 3 ) 2 = 0 ; 
it is clear that a factor has dropped out, and that the true form of the condition is 
(— Z + Z 4 ) (1 + 8Z 3 ) 3 = 0 ;