74 KANSAS UNIVERSITY SCIENCE BULLETIN. 



Let us now try to construct a cubic surface having three binodes 

 that is the surface (3B S ). Consider a tetrahedron SABC, and draw 

 the lines AT/, AT/' in the plane SAB, BT /' , in the plane SBC, 

 CTi" in the plane SCA. We can construct a twisted cubic K s { 

 going through A, B, C and tangent to AT/, BT /', CT /". If we 

 take then a point arbitrarily on Bl Ave may remark that /?| is 

 perfectly determined by the conditions that it pass through 0, A, 

 B, C, be tangent to A T / and to the planes SBC and S( A in B and I '. 

 For let E be the point where AT / meets SB. Then the conic pro- 

 jection of B 2 on the plane SBC must go through \x>, A B, C, E and be 

 tangent to SC, uj being the projection of on the plane SBC, while 

 the conic projection of R 2 on the plane SAB must go through 

 A, B, u/and be tangent to SB and AE. R% is thus perfectly deter- 

 mined as the intersection of the projecting cones from two of its 

 points. 



7. Suppose now that R\ and h% be tangent in D, and let DT 

 be their common tangent. Then the two projective pencils of the 

 tangent planes of 2 i and 2 2 , have DT for common axis. The double 

 planes of the projectivity in question will be the tangent planes to 

 S 3 in D, and their intersection DT will lie on the surface. D will 

 therefore be a binode B 4 that is such that the intersection of the bi- 

 planes be on the surface. This is all clearly shown when the contact 

 of the two curves is considered as the limit of two common points 

 innnitely near each other, and this consideration shows also that 

 the point B 4 is equivalent — as far as the class of the surface goes — 

 to two ordinary nodes. There is no difficulty therefore in constructing 

 the surface B 4 , having a binode in D. We give ourselves DT and 0, 

 and have to construct two twisted cubics going through 0, D and 

 tangent to DT. By subjecting the two curves to the condition of 

 having one or two more common points, which can be chosen arbit- 

 rarily we obtain the surfaces (C 2 + B 4) and (2C 2 + B 4). 



We can easily obtain the lines on the surface B 4 by direct con- 

 siderations ( 7 ). There are two common bisecants of Bf and B 2 

 through D, each of them counting for two, since there is contact of 



the two curves in D, while DT is also on >S 3. We thus have a 1, a 5, 



therefore R\ and R% have a bisecant a 6 not going through D. 

 Suppose that a%=z a 3 , a 4 EE a 5 . Then the two biplanes are (DT, a 2 ) 

 and (DT, a 4 ), and if they are met by a 6 in E and F, the lines DE and 

 DP are also on S 3 . Let now P be an arbitrary plane through DT, 

 it meets hi, h'2 and a 6 in points G 1, G 2 and H, and TIG 2 meet re- 

 spectively DT' in points A' 1, A' 2 , which are seen at once to be in 



7. Cailey, A Memoir on Cubic Surfaces. Phil. Transacts. 1869. p. 231. Schlaefii, 

 Ibid. 1863. p. 193. 



