( 237 ) 
C3 intersects K? in 6 points; these 6 points are at the same 
time the points of contact on K? of the common tangents of K? 
and Cé, From this we can conclude that for these points the end- 
points of the chords of intersection coincide; so in these points 
conics of pencil (1 2 3 4) will touch &?. If we imagine one of these 
points 7, to be given, a conic of the pencil passes through this 
point, and as it touches A’*, the corresponding pole falls in 7; 
from this we conclude that: 
The 6 points of intersection 7, ... 7} of C® and K? are 6 of 
the points of coincidence of both involutions; so we have still to 
account for 6 other points of coincidence. Let us call one of these 
points Ajo, then the conic through Aj, will meet K? in A,, Ay, A3, A4 
and will pass through 434, the pole of 4, A,. Hence these six 
points can be divided into three pairs of points (Aj, 434) (By, Bs4) 
(Cig C34). 
By the way we remark that the obtained result is in accordance 
with the fact, that six conics of a pencil touch an arbitrary conic. 
9. The found three pairs of points determine the three conics 
which will solve the problem. We however add a second deduction, 
