510 PROCEEDINGS OF THE AMERICAN ACADEMY. 



joins this (m — 2) -tuple point to any point of C m never contains more 

 than one other point of C m ; nevertheless during its revolution around 

 the line, it contains successively all the other points of the curve. If we 

 take an ordinary point of C m as vertex and construct a cone with C m as 

 base, this cone will be of the (m — l)st order. This cone contains the 

 line joining the (m — 2)-tuple point to the vertex as an (m — 2)-tuple 

 edge, and can therefore have no double edges. The curve can therefore 

 have no apparent double points when viewed from an ordinary point of 

 the curve. We therefore have h = m — 2;* thus a twisted curve of 

 order m that has an (m — 2)-tuple point has m — 2 apparent double 

 points. 



2. Let C m be a twisted curve of order m having an (m — 2)-tuple 

 point at A (where 4 < m). This curve will lie on a cone of order two, 

 say K 2 , whose vertex is A ; for any plane through the point A can con- 

 tain only two other points of O m , the lines joining these points to A 

 being generators of a quadric cone. Not more than two tangents to G m 

 at A can lie in one plane, for the plane would then meet the curve in 

 more than m points, which is only possible if the plane contains the 

 curve. Every unicursal curve of order m with an actual (m — 2)-tuple 

 point thus lies on a quadric cone. We can show that C m can be cut out 

 of this cone K 2 by a monoid. Clearly C m is the complete or partial 

 intersection of K 2 with some surface of order fx, say aS^, having a £-tuple 

 point at A. Now m — 2 < 2 k, since 2 k is the multiplicity of A on the 

 complete intersection of S* and K 2 . It is evident that k is the order of 

 the cone that contains the m — 2 tangents to C m at A that lie on K 2 , 

 provided m - 2 ^ \ k (k + 3) - \ {k - 2) (k + 1) - 1, where 2 ^ k, 

 i. e. provided in — 2 < 2 k, where 2 < k. We can thus always take k = 



or — - — , according as m is even or odd. The surface *S M with a 



£-iuple point at A is determined by \ O + 1) (/< + 2) (ji + 3) — ^ k 

 (k + 1) (k + 2) — 1 points. In order not to have £ M break up into K 2 

 and a component of order \n — 2 having a (k — 2)-tuple point at A, we 

 must take J (ja - 1) (i (ji + 1) - I (k — 2) (k — 1) k of these points 

 off K 2 . There are ju (p + 2) — k 2 points remaining that can be taken 

 arbitrarily, but in such a way that <S M shall contain C m . S^ meets C m 

 in k (m — 2) points at A, and will therefore contain C m entirely if it 

 contains m \i — (tn — 2) k + 1 additional points of C m . S^ can therefore 

 be determined so as to contain C m if 



* Salmon's Geometry of Three Dimensions (1882), No 330, Example 2. 



