PROBLEMS AND SOLUTIONS. 
595 
[485 
485] 
' are two 
[y G and 
ïame two 
rawing a 
neets the 
conic A, 
C in the 
ic surface 
V, N, N'. 
,nd which 
plane in 
acludes a 
A which 
he points 
ough the 
a of the 
and the 
point on 
c passing 
e surface 
l applies, 
my point 
lies A, B 
tics A, B 
0 passing 
the same 
i point a, 
msecutive 
•ough the 
nil touch 
or (what 
fence the 
; entirely 
B, B', Y, 
ve points 
shown as 
before that the surface will include the quadric surface 2. But there still remains 
for consideration the case where 0 is a conical point on the surface, and I do not 
at present see how this is to be treated. 
I remark that, taking three lines xx, yy', zz avhich meet in a point 0, then if 
a surface be such that every section through xx r includes a conic, every section through 
yy' includes a conic, and every section through zz includes a conic; and if besides 
the two points, say X, X\ on the conics through the line xx are ordinary points on 
the surface, then the surface will include a quadric surface. In fact, if the surface 
has at each of the points X, X' an ordinary tangent plane, then the conics through 
xx', and (as conics of the series) the two conics B, C all of them touch the two 
tangent planes; hence, constructing as before the quadric surface 2, this also touches 
the two tangent planes: and taking through xx a plane meeting the conic A in the 
points L, L', the section of the surface includes a conic which touches the section of 
the quadric surface 2 at the points X, X', and besides meets it in the points L, L'; 
such conic coincides therefore with the section of the quadric surface 2; that is, every 
section of the surface through the line xx' includes the conic which is the section of 
the quadric surface 2; and the surface thus includes as part of itself the quadric sur 
face 2. 
[Vol. x., July to December, 1868, pp. 17—19.] 
2609. (Proposed by Professor Cayley.)—Given three conics passing through the 
same four points; and on the first a point A, on the second a point B, and on the 
third a point G. It is required to find, on the first a point A', on the second a 
point B’, and on the third a point G', such that the intersections of the lines 
A'B' and AC, A'G' and AB, lie on the first conic; 
B'C' and BA, B'A' and BC, lie on the second conic ; 
C'A' and CB, C'B' and CA, lie on the third conic. 
Solution by the Proposer. 
Let the six intersections in question be called a, a!; ¡3, /3' ’■> y, J > respectively; 
then BC intersects second conic in /3', third conic in y; GA intersects third conic in <y\ 
first conic in a; AB intersects first conic in a', second conic in /3; and we have 
A' the intersection of ot/3', ya', 
B' the intersection of /3y', a{3', 
G' the intersection of ya', /3y'; 
and it has to be shown that the points A', B', C' so determined lie—A' on the first 
conic, B! on the second conic, C' on the third conic. 
75—2