I gave a general solution to the following problem:-Given a cubic curve, to find tioo inscribed triangles ABC and DEF such that, if P be any point on the cubic curve, the two pencils of lines P(A, B, C) and P(D, E, F) shall be apolar for all positions of P on the curve. In the former paper, the solution given was unsymmetrical. I have since then found a symmetrical solution, which I now proceed to describe. 2. It easily follows from the above paper that, if the chord DEF of the given cubic

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... touch its Cayleyan, and if ABC be an inscribed apolar triangle, the necessary and sufficient condition that A, B, C and D, E, F be co-apolar triads with respect to the given cubic curve is that the common intersection G of the tangents at D, E, F be the pole of the line DEF with respect to the triangle ABC. For the apolar locus of ABC and DEF passes through A, B, C and touches DG, EG, FG at D, E, F respectively. Hence the given cubic and the apolar locus have the following nine-points in common, A, and must therefore coincide unless A, B, C be collinear, which is contrary to the hypothesis. 8. It will be convenient for our purpose to regard the given cubic curve as the projection from a point on itself of a skew quartic formed by the intersection of two quadrics. Any one chord of the quartic defines a unique quadric of the above system, and we shall call chords, which are non-intersecting generators of the same quadric, chords of the same system.