668 Proceedings of the Royal Irish Academy. 



surface S, of which and 0' are also generators. This quadric must 

 pass through the four double points. 



Let ;S' be the quadric surface which contains all the points in the 

 second system corresponding to the points of S regarded as the first 

 system. Then 0' will lie on 8', and the rest of the intersection of S 

 and S' will be a twisted cubic C, which passes through the four double 

 points. 



Take any point P on C, and draw any plane through P. Then 

 every ray of the first system of the pencil through P in this plane 

 will have as its correspondent in the second system the ray in some 

 other plane pencil Z. One, at least, of the rays in the pencil L will 

 cut the cubic C. Call this ray X' , and draw its correspondent Xin 

 the first system passing through P. 



We thus haye a pair of corresponding rays X and X', each of 

 which intersects the twisted cubic C. 



Draw pairs of corresponding planes through X and X'. The locus 

 of their intersection will be a quadric S", which also contains the 

 four double points. 



S" and C, being of the second and the third order respectively, will 

 intersect in six points. Two of these are on X and X', and are thus 

 distinguished. The four remaining intersections will be the required 

 double points, and thus the problem has been solved. 



