742 



through c. In any such plane however two brandies arrive at A 

 from above n, one on the set which joins AP to AQ and the other 

 on the set by which AR is connected with A^. Now if A were 

 cusp in a plane through c then, considering the branches arrive 

 from above h, at the utmost one plane could pass through h in 

 which A is isolated above a and this contradicts the above results. 

 Hence A must be double point in every plane through c: a contra- 

 diction. 



It now remains to consider the case that the oval in a has a 

 for tangent at A. We shall consider separately the following possi- 

 bilities : 



I. There exists a semi plane through h above i<. in which A is 

 not isolated. 



II. No such semiplane exists. 



I. A is not isolated above a in a plane J through h. Then in 

 the semiplane of fS above « two branches depart from A, because 

 A is cusp or double point in every plane not containing a and line 

 h has a second point B in common with F*. From the way in 

 which the branches meeting at ^ in « are connected, follows that 

 in every plane through /; two branches arrive from below «, hence 

 A is ordinary double point in (S. Here, there is no danger of a line 

 as we assumed that no second line of F' intersects a. If (f revolves 

 round b then in one of the two directions A will remain ordinary 

 double point till 6 coincides with a. 



From this it follows that the semiplanes of ^ and y in which the 

 cuspidal branches depart from A, are situated on the same side of 

 ff, let us say below 6. In 6 we now choose a line d through A 

 separated from h by the tangents at A. The same reasoning used 

 before shows that A would have to be double point in every plane 

 through d. Only for the plane through a and d a slight alteration 

 is required, which however is selfevident. The impossibility of 

 assumption I has thus been proved. 



II. A is isolated in every semiplane through h above «. In q^^v^ 

 plane through A not containing a the point A is double point or 

 cusp, hence in every plane through h (=|= «) A i& cusp and all the 

 branches arrive at A from below «. It follows that A must be cusp 

 in every plane except «, all branches arriving at A from below a. 

 This however is only possible if the cuspidal tangents form one 

 plane h through a, which plane has nothing but line a in common 



with F*. Tiet a sequence of planes 8j, f, all passing through 



a converge towards p. In each of these an oval passes through A. 



Now suppose the oval in fn crosses the line a at A. Then four 



