J17 



Fig. 5. 

 in ^ departs from A above a in the direction AC. If this branch 

 formed at A a finite angle with AC, then every line inside this 

 angle would be tangent at A in every plane except /?, and this is 

 obviously at variance with the results already obtained. For the com- 

 plete demonstration it is necessary to know that the linesegment 

 ACn cannot have points in common with F^, converging towards yl. 

 Now this is obvious if we remember that when the lines c«„ converge 

 towards a, the point A„ on AR converges towards A, and that the 

 point A only counts double on a. 



The possibility might be put forward that the branches in the 

 converging planes (Sn have only A as limiting set in ^. Then how- 

 ever, it is unavoidable that the converging planes show ovals, con- 

 tracting from above towards A. Now all these ovals would cross 

 AB, hence A would be limiting point of F^ on AB and the entire 

 line AB would belong to F\ a possibility excluded at the outset. 



Between the branches AP and ^4aS the surface F^ was assumed 

 to be situated above u, hence the part of F^ coimecting AS and 

 AR lies below u. Now if the planes /?i, j^^, ^^ . . . converge towards 

 /? from the other side and if we consider the front halves of these 

 planes (fig. 5), it may be shown in exactly the same way that in 

 i^ a branch departs from A below <( in the direction AD. 



Taking these results together, it is found that ^ is point of inflexion 

 in [3 with a for taugent. 



Before passing on to § 4 we shall prove the following theorem : 

 It is impo.s.fibh that a point of intersection A counts double on a 



