610 
PROBLEMS AND SOLUTIONS. 
[383 
may be the two tangents to the oval,) then we obtain a system of six points, (part 
of the system of eighteen points) ; viz., the coordinates of the six points are of the 
form (a, b, c), (b, c, a), (c, a, b), (a, c, b), (b, a, c), (c, b, a) and therefore the six points 
are in a conic. 
6. To verify this, if we take y — 0x for the equation of a tangent from the point 
(# = 0, y = 0), the equation (1 + 0 s ) ie s + 6Wx 2 z + z 3 = 0 must have a pair of equal roots, 
1 q_ 0 s 
giving for 6 the equation (1 + 0 3 ) 2 + S2l 3 6 3 — 0; and we then find z = x, that is, 
6 being determined by the foregoing equation, the coordinates of the point of contact 
1 -f- 0 ' 
are x : y : z — 1 : 6 : £iq~’ r00 ^ s °f th e equation in 6 are of the form 
$i, 6<>, 0 S , \ , and assuming that the curve has an oval, there are two real 
V\ Vo Vo 
roots 0 lt 
1 
V 
Hence, writing x : y : z = 1 : 6 X : — 
1 + 6i 
U0, 
a : b : c, the substitution -3- 
0i 
for 0, gives x : y : z = b : a : c, that is, the coordinates of the points of contact of 
the tangents to the oval, from the point (x = 0, y = 0) are (a, b, c) and (b, a, c) 
respectively; and writing successively (y, z, x) and (z, x, y) in place of (x, y, z), the 
coordinates for the tangents from (y = 0, z = 0) are (b, c, a), (c, b, a); and those for 
the tangents from (z = 0, x = 0) are (c, a, b) and (a, c, b); so that the coordinates of 
the six points of contact are a system of the form in question. 
[Vol. iv. p. 107.] 
1812. (Proposed by Professor Cayley.)—Find the envelope of a series of circles 
which touch a given straight line and have their centres in the circumference of a 
given circle. {See Quest. 1771.} 
[Vol. iv. pp. 108, 109.] 
1816. (Proposed by R. Ball, M.A.)—Express the roots of the equation 
(1ae — 4bel + 3c 2 ) («¿c 4 + 4bx 3 + 6cx- + 4dx + e)- — 3 {(ac — b 2 ) x A + 2 (ad — be) a? 
+ (ae + 2 bd — 3c 2 ) x 2 + 2 (be — cd) x + (ce — d 2 )} 2 = 0, 
in terms of the roots a, ß, y, 8 of x* + 4bai 3 + 6cæ 2 + 4dx + e = 0. 
Solution by Professor Cayley. 
Writing 
U = (a, b, c, d, e\x, y) 4 — a (x — ay) (x - (3y) (x — 7y) (x — 8y), 
H — (ac — b 2 , ^ (ad — be), ^ (ae + 2bd — 3c 2 ), ^ (be — cd), ce — d 2 ][x, y) 4 , 
I =ae — 4 bd + 3c 2 ;