The twenty-seven lines upon the cubic surface ... by Archibald Henderson.

94 ON SOME CONFIGURATIONS ASSOCIATED WITH THE

interpreted as a planographic representation of the projection of the
lines upon a cubic surface having only one conical node, from the
nodal point, upon an arbitrary plane into the projection of the
"mystic hexagram."
It is perhaps worthy of remark that the six lines 1, 2, 3, 4, 5, 6 lie
upon the quadric cone having D for its vertex, and for its base the
conic section (an hyperbola) having for its equations
12 (2 + y2 + 2) - 5 (6yz- 8zx + 5xy) = 0, w = 0.
The intersections of this cone by the planes x = 0, y 0, z = 0, respectively, have for their equations:
= 0, (2y -) (y-2z) = 0;
y = 0, (x + 3z) (3x + z) = 0;
z = 0, (3x + 4y) (4x + 3y) = 0.
48. A Deduction from Cayley's Theorem on the Pascalian Configuratioln.
The theorem of Cayley* mentioned in ~ 45, together with the
argument in ~ 46, leads to the well-known conclusion:
Given any two triads of planes a, b, c; f, g, h; then it is possible
to find four points O0, 02, 0, O4 such that the polar plane of any one
of these points with respect to one triheder is identical with its polar
plane with respect to the other triheder. Considering any one of
the points, say O0, then it is possible to draw six lines through 0,
whose positions are defined as follows:
'line 1 meets the lines af, bg, ch,
2,,,,,, ag, bh, cf,
) 4,,,,,, aj; bh, c g,,,,,,, ag, bf, ch,,, 6,,,,,, h, bg, cf.
Then these six lines together with the nine lines af, ag, ah; bf, bg,
bh; cf, cg, ch determine a cubic surface upon which they lie, for which
the point 0, is the only conical point.
This conclusion may be more generally phrased as follows:
Through the nine lines of mutual intersection of two triheders can
be drawn four cubic surfaces, each possessing only one conical point

* Coll. Math. Papers, Vol. vI. pp. 129-134.