308 BROOKS — ORTHIC CURVES. [May 20, 



If we regard r as a parameter, we may say that 



(x — a) (x — i8) {x — r) = r(x—a) (^ ~J) {x — y) 



is the equation of the pencil of orthic cubics which have the triad 

 a^y. It will be convenient to give a pencil of this sort some name ; 

 let us refer to it as a central pencil, noting for our justification that 

 the centroid of the triad is the centre of every curve of the pencil. 



If there were any real point other than a, /S, or y, on two curves of 

 this pencil, it would map into a real point of the 2;-plane, not the 

 origin, which would be on two of the lines through the origin. As 

 this is manifestly impossible, it follows that : 



Tivo orthic cubics which have a triad in common have no other real 

 intersection. 



Now we know that two cubic curves intersect in nine points, and 

 that if the curves given by the equation 



^x — a) {x — I3)(x — y) = r{x — a)(x — ft) {x — y) 



really constitute a pencil, there must be six imaginary points whose 

 coordinates satisfy this equation whatever the value of r. Let us 

 form the following table of coordinates. The real intersections are 



Xt^ = a, Xj ^ a, 



X2 — ft, X.^ r- ft, 



•^3 = r> ^^3 = r- 



It is evident that each of the following points : 

 x1 = a, ^4 = J, 

 X, = a, x, = y, 

 Xe = ft, Xfi'= a, 



x-j = ft, x-j = y^ 

 Xg = y, Xg = a, 



•^9 ^^ Ty ^9 ^^^^ ft, 



satisfies the equation, independently of t. These points, the six 

 imaginary intersections of the pencil, are the antipoints^ obtained 

 ^Cayley, Collected Mathematical Papers, Volume VI, p. 499. 



