Replacing now x by 1/x, y by 1/y, z by \/z, we Bee that
the point of contact of Xx + fiy + vz = and the quartic,
i. e. any inflexion of the quartic, lies on

2 Ax 2 (gy + hz) + 2 By 2 (hz +fx) + 2 Cz 2 (fx + gy)

+ (4 ahc — 4>fgh — A) xyz = 0.
Of the twelve intersections of this cubic with the quartic
six lie on the conic T = which touches the cubic at each
vertex of the triangle of reference. Therefore the remaining six
intersections of the cubic and quartic, namely, the inflexions
of the quartic, lie on a conic (Ch. XII, § 7). We can verify
that

(afghA + bfghB + cfghC+2bcghF+2cahfG + 2abfgH) K
+ M{2Ax 2 (gy + hz) + 2By 2 (hz +fa) + 2Gz 2 (fx + gy)

+ (4<abc — ifgh — A) xyz]
= T{2 A (gG + hff)x 2 + . .. + ...
_ (4,fF 2 + 2gh A + Saf A) yz-. ..-...}.
The contents of the last brackets { } equated to zero give
the equation of the conic through the inflexions, which is
readily seen to be the same as that given above. This proof
is due to Richmond and Stuart, Proc. London Math. Soc, II.
i (1903), p. 130. »

The remaining intersections of the quartic with the
inflexions-conic are those intersections of the quartic with
the line M = which do not lie on the conic T = ; i. e. the
intersections of the line with the conic U = 0.
