ON A CASE OF THE INVOLUTION AF+BG+ CH=0, WHERE 
A, B, C, F, G, H ARE TERNARY QUADRICS. 
[From the Messenger of Mathematics, vol. xxn. (1893), pp. 182—186.] 
We have here the six conics 
A= 0, B = 0, (7 = 0, F= 0, (7 = 0, H= 0; 
the curves AF= 0 and BG = 0 are quartics intersecting in 16 points, and if 8 of 
these lie in a conic H = 0, then the remaining 8 will be in a conic (7=0. I take 
the first set of eight points to be 1, 2, 3, 4, 5, 6, 7, 8; the quartics J.F= 0 and 
BG = 0 each pass through these eight points; and I assume for the moment 
A = 1234, F— 5678 ; R=1256, (7 = 3478, 
viz. that A = 0 is a conic through the points 1, 2, 3, 4, and similarly for F, G, B. 
Here H — 0 is a conic through the points 1, 2, 3, 4, 5, 6, 7, 8, or attending only to 
the last four points it is a conic through 5, 6, 7, 8; we have therefore a linear 
relation between F, G, H, and supposing the implicit constant factors to be properly 
determined, this may be taken to be F+G + H = 0; the identity AF+ BG + CH = 0 
thus becomes F (A — C) + G (B — G) = 0. We have thus F a numerical multiple of 
B — G, and by a proper determination of the implicit factor we may make this relation 
to be F = B—G; the last equation then gives G = G—A, and from the equation 
F + G + H = 0, we have H = A — B; the six functions thus are 
A, B — C, or if we please, A— D, B — C, 
B, C-A 
G, A-B 
B-D, G-A, 
C -D, A-B, 
where D is an arbitrary quadric function. The solution 
(A-B) (B - C) + (B-D) (C- A) + (C-D)(A - B) = 0 
of the involution is an obvious and trivial one.