159 
the co-ordinates of a triple point have to satisfy the six equations: 
aAyy + BBei + yCu + Du + el = 9, 
in which Aj, etc. represent derivatives according to x, and a. 
The number of points has to be found, for which 
Ashe Ae Az Ane By 
BNA ANB 
Cuero Ce Ge C=O. 
De) DD Dac We 
| #, E,, EB. EE, | 
| 
According to a well-known rule we find for this 
(5°—4? + 37—2? + 1?) (n—2)?. 
There are therefore 15(n—2)? curves c, with a triple point S’). 
In such a point the nodal curves have the same tangents d,d’. 
Any straight line passing through S is to be considered as a cuspidal 
tangent c. 
The null-system therefore has 15(n—2)? singular points. 
9. I now take three points P, Q, R, arbitrarily, and consider (ef. § 7) 
the curves (D)pg and (D)pr. To begin with they have the point P 
in common; for there is a dr, which has P as a node, and PQ as 
tangent and a de, for which one of the tangents lies along PR. 
Those curves have further in common the (52—9) points D, for 
which QR is one of the tangents d. Another group of common 
points consists of the singular points S. 
Let U be one of the still remaining intersections. There is in 
that case a dr with tangents U P and UQ, and also a dr with 
tangents UP and UR. From this it ensues that all é with node 
U have the straight line U P as tangent, consequently belong to a 
pencil in which the tangents d,d’ form a parabolic involution. 
The double rays of this involution have then coincided in U P, 
and U is cusp for only one cuspidal c. If such a point is called 
unicuspidal point, it follows from (5n—8)?’—1—(5n—9)—15(n— 2)? 
that (10n?—25n + 12) unicuspidal curves send their tangent through 
P. The cuspidal tangents of the unicuspidal points envelop a curve 
of class (10n?—25n + 12). 
10. In any point C of the straight line a I draw the two null- 
1) If nm =3, and the system has 5 base-points, the 15 triple points are easy to 
indicate. One of them e.g. is the intersection of B,B, with Bs By. 
