802 
points P and P’ of A,A, there exists therefore a correspondence 
(2,1) with 3 coincidences, from which follows: 
On any straight line joining two of the five base points A,,...., A, 
of S,* lie three points P, so that the straight lines joining P to the 
three remaining base points, are parts of degenerations belonging to 
the same pencil. 
5. We can arrive at the same results in an entirely different way, 
where at the same time the relation between the points P appears. 
With a view to this we shall first prove an auxiliary proposition. 
We start from a net of cubics with base points P, A,,...., A, 
and suppose the degenerations formed by PA,,...., PA, with 
completing conics to belong to the same pencil. We know that 
through P there pass two more straight lines which together with 
two conics through A,,...., A, form also degenerations of the net. 
Let us take PA,A, for triangle of coordinates and let us put 
PA, =p, tt, + pt, =(pe)=90, PA, red a te (qe) =F 
A,A,=a,2,+0,0,+4,2,=(ar7)=—0, B,B,=b,2,+6,¢,4b,0,=(b2)=0, 
The conics through A,, B,, A,, B,, A,, B, and through A,, B,, 
A,, B,, A,, B, belong both to the pencil (px) (ga) + 4 (aa) (b2) = 0. 
For the former conic 4 must be chosen such that it passes through 
A,, for the latter such that it passes through A,. Hence 4 must be 
resp. equal to —p,g,:a,6, and —p,g,:a,6,. The former conic is 
completed to a degeneration by «, =O, the latter by a, = 0. 
The straight line (av) =O belongs to a conic through P, A,, A,, 
and has therefore the equation c,v,v, + c,a,7, + ¢,v,c, = 0. By these 
three curves the net: 
) 2, }a,b, (pe) (qe)-p‚g,(0e) (be) + A4, ja, be (pe) (Ge)—Pags (22) (be); 
+ A, (az) (c,2,0,+¢,2,0,+¢,c,2,) = 0 
is defined. 
By assuming #,=rx, and by putting the condition that these 
straight lines be parts of degenerations in the net, we find through 
the elimination of A,,4,,4, and through division by p, + p‚r and 
q, + 9,7” the equation 
b, : ¢,+ 6,7 ee 
a,(6,+5, r) Te b, (4,44, r) ’ a,C,7r ai (c, +¢, r) (a, +a, r) ae 
This equaticn defines therefore the two straight lines m and n 
which pass through P and are parts of degenerations. 
In the net is a curve which has a double point in P. For this 
A, = Sh A. =e pass hs and the nodal tangents are found 
P3929, ss), 
out of: 
0. 
