146 
ON THE PORISM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE, 
[129 
or, what is the same thing, 
x 2 : y 2 : z 2 = y x z x (bn 2 + cm 1 — 2/mn) 
: z x x x (cl 1 + an 2 — 2gnl) 
: x x y x (am 2 + bn- — 21dm). 
It is not necessary for the present purpose, but it may be as well to give the 
corresponding solution of the problem: 
Given that one of the tangents through the point (£, 77, £) to the conic 
(a, b, c, f g, K$x, y, z) 2 = 0 
is the line l x x + m x y + n x z = 0; to find the equation to the other tangent. 
Let Lx + m. 2 y + n 2 z — 0 be the other tangent, then 
(a,... ££, 77, £) 2 . (a,... Jx, y, z) 2 - {(a ... #£, 77, g$x, y, z)} 2 
= (kx + m x y + n x z) (l.pc + m 2 y + mz) 
to a constant factor pres. Assume successively y = 0, z = 0; z = 0, x = 0 ; x = 0, y = 0 ; 
then we have 
L : m 2 : n 2 = m x n x [a (a,... $£, 77, £) 2 ~ ( a £ + il V + g£f} 
: n x l x {b (a,... ££, 77, £) 2 - (Jig + bg +/£) 2 } 
: l x m x [c (a, ... $£, 77, £) 2 ~ (g£ +fv + c£) 2 }; 
or, as they may be more simply written, 
l 2 : m 2 : n 2 = m x n x (m? + <S»f + 2$vO 
: n, 2, <Mf‘ + ar - 
Returning now to the solution of the first problem, I shall for the sake of 
simplicity consider the formulas obtained by taking for the equation of the conic, 
ax 2 + ¡3y 2 + 7 z 2 = 0. 
We see, therefore, that if this conic be intersected by the line lx + my + nz = 0 in 
the points (x x , y x , z x ) and (x 2 , y 2 , z 2 ), then 
x 2 : y 2 : z 2 = y x z x (ym 2 + an 2 ) 
: z x x x (an 2 + (3l 2 ) 
: x x y x (¡3l 2 + am 2 ).