MATHEMATICS: H. S. WHITE 
465 
fewest possible points, seven. It is this: // seven points on a twisted 
cubic be joined, two mid two, by twenty-one lines, then any seven planes that 
contain these twenty -one lines will osculate a second cubic curve. 
The proof to be given is analytic, consisting in a repeated application 
of Pascal's theorem concerning a hexagon inscribed in a conic, adapted 
to the six projecting lines from a seventh point by means of Clebsch's 
so called Randerungs-princip. 
Pascal's theorem may be expressed in notation readily understood, 
for six points a, b, c, d, e,/of a conic (coordinates (ai, a2, az) etc.) by the 
vanishing of the determinant 
((ab),(de))^ {{bc),{ef)), {. (cd) , (fa) ) , 
{{ab),{de)), 
( {ab), (de) )3 
or any one of the 59 equivalent forms. This is reducible to the equa- 
tion, in three-rowed determinants of point-coordinates, 
(abe) (cde) (adf) (bcf) = (abf) (cdf) (ade) (bee). (1) 
(See Encyklopddie der mathematischen Wissenschaften, 32, §7.) 
Clebsch's Randerungs-princip applied to this equation gives the 
condition that six points in space {ai, ai, az, a^, etc., or briefly a, b, c, d, 
e,f, shall be projected from a seventh point (gi, g2, gz gi) by six genera- 
tors of a quadric cone, viz., 
(abeg) (cdeg) (adfg) (bcfg) = (abfg) (cdfg) (adeg) (bceg), (2) 
If the seven points are on a twisted cubic, this relation may be written 
in 7 • 15 • 3 different ways, according to the choice of the seventh 
point, g, the two points e and /, and the two ways of forming pairs from 
the remaining four points a, b, c, d. Let equation (2) denote all of this 
type. 
Seven planes, 1, 2, 3, 4, 5, 6, 7 may be determined in 30 different ways 
so as to contain the 21 lines that join pairs of the points a, b, c, d, e,f,g. 
Any one such choice constitutes a triad system on those seven letters, 
since its seven sets of three points in a plane must contain every pair 
of points once and only once. Fix one set thus, planes, 1, 2, 3, 4, 5, 6, 7 
containing the points ade, afg, bdf, beg, cdg, cef, abc, respectively. These 
seven planes, in order to osculate a twisted cubic, must satisfy condi- 
tions on planes precisely dual to the 315 conditions of type (2) on points. 
On account of their similarity we need to verify only one. This is a 
slightly tedious reckoning, which we give in extenso. 
The question shall be: Does the plane 7 intersect the six planes 1, 2, 3, 
4, 5, 6 in six tangents to a conic? Is the following condition verified? 
(1257) (3457) (1467) (2367) = (1267) (3467) (1457) (2357). (3) 
