548 
PROBLEMS AND SOLUTIONS. 
[485 
respectively, then the number is = (M, m) (JS T , n)(P, p) (Q, q) (R, r) {1, 2, 4, 4, 2, 1}, that 
is, 1 MNPQR + 2 "EMNPQ r + &c. The number is not, I believe, known in any other 
of the problems. In particular, (Prob. 7) we do not as yet know the number of the 
conics which touch a given curve at five points. It would be interesting to obtain this 
number; but (judging from the analogous question of finding the double tangents of a 
curve) the problem is probably a very difficult one. 
[Vol. V. p. 37.] 
1857. (Proposed by Professor Cayley.)—If for shortness we put 
P = oc? + y 3 + z 3 , Q — yz 2 + y 2 z + zap + z 2 x + xy 2 + x 2 y, R = xyz, 
P 0 = a 3 +b 3 + c 3 , Q 0 = be 1 + b 2 c + ca 2 + c 2 a + ab 2 + a?b, R 0 = abc; 
then (a, ¡3, 7) being arbitrary, show that the cubic curves 
P, 
Pn, 
ß, 
Q, 
Qo> 
7 
R 
Rn 
= 0 pass all 
of them through the same nine points, lying six of them upon a conic and three of 
them upon a line; and find the equations of the conic and line, and the coordinates 
of the nine points of intersection; find also the values of (a : /3 : 7) in order that 
the cubic curve may break up into the conic and line. 
[Yol. v. p. 37.] 
1730. (Proposed by Professor Cayley.)—Show that (I) the condition in order that 
the roots k 1} k 2 , k 3 of the equation 
7& 3 + (“ 9 ~ i a + i/3 + f 7) № + (— g — f a — l /3 + £7) k - a = 0 (A) 
may be connected by a relation of the form 
k 3 (k x - k 2 ) - (k 2 - k 3 ) = 0, (1) 
and (II) the result of the elimination of a, b, c from the equations 
a 2 (b + c) = — 2a, (2) 
b 2 (c + a) = 2/3, (3) 
c 2 (a + &)=-27, (4) 
(b — c)(c — ci)(a — b)= — 4g, (5) 
are each 
4 (/3 — 7) (7 — cl) (a — (3)g 3 + 4 (— 2a 3 /3 + 42a 2 /3 2 — Z'ZoPfty) g 2 
+ (£ “ 7) (7 ~ «) ( a - /3) g + 2 (/3 - 7) 2 (7 - a) 2 (a - /3) 2 = 0. (B)