622] 
CONNECTED WITH MALFATTI S PROBLEM. 
549 
It is very remarkable that, introducing the foregoing relation K = 0, there is not 
in the solution any indication that the intersection has become a conic and two points, 
but the solution gives eight determinate points, viz. the before-mentioned points 
P, Pi, Q, Qi, P, Pi, and I, J. 
To develope the solution, remark that, in virtue of the relation in question, we have 
s/BG=±F, \ZCA = ±G, *JAB = ±H, 
where the signs must be such that the product is = FGH (viz. they must be all 
positive, or else one positive and the other two negative); for, taking the product to 
be + FGH, the equations give 
0 = ABC-FGH, 
that is, 
0 = A {BG- F 2 )-F(GH- AF), = K(Aa — Ff), = KV, 
which is true in virtue of the relation K = 0. Taking the signs all positive, we have 
for op, y 2 , z 2 , yz, zx, xy, the values f 2 , g 2 , h 2 , gh, hf, fg, viz. we have thus the points 
(f 9> h )> (~f ~9> ~ h ), 
which are the points I, J. Taking the signs one positive and the other two negative, 
say VBG = F, VCA = — G, VAB = — H, we find for x\ y' 2 , z 2 , yz, zx, xy the values 
be, — , gh, bg, ch, viz. we have thus the points 
0 c 
[■Jbo, P 1 , 
bg\ 
l-Obo, -t, 
_ b c J ) 
v ’ Obc 
\/bcP 
\ Vèc 
Pbcf 
which are the points P, P 1 ; and the other two combinations of sign give of course 
the points Q, Qj and P, Pj respectively. 
If the coefficients (a, b, c, f, g, h), instead of the foregoing relation K= 0, satisfy 
the relation 
abc — a/ 2 — bg 2 — ch 2 — 2fgh = 0, say K' — 0, 
the quadric surfaces intersect in 8 points, the coordinates of which are given by the 
general formulae: but the expressions assume a very simple form. Writing for shortness 
F’ = gh + cif, G' = hf+bg, H' = fg + ch, 
then, in virtue of the assumed relation, 
\/BC=±F', a/CA = ±G\ VAB = ±H', 
where the signs are such that the product of the three terms is positive, viz. they 
must be all positive, or else one positive and the other two negative. For, assuming 
it to be so, we have