596 
PROBLEMS AND SOLUTIONS. 
[485 
Taking x = 0, y = 0, z = 0 for the equations of the sides of the triangle ABG, the 
equations of the three conics may be taken to be U = 0, V = 0, W = 0, where the 
functions U, V, W are such that identically U + V + W = 0; and then observing that 
C 
the conics pass through the points (y = 0, z = 0), (z = 0, x = 0), (x= 0, y — 0), respectively, 
we see that the equations may be taken to be 
( 0, - b, c, f lt g x , Jh\x, y, z) 2 = 0, 
(a, 0, - c, / a , g 2 , K\x, y, zf = 0, 
(-a, b, 0, /, g 3 , h 3 \x, y, z) 2 = 0, 
where 
fi +f2 +/3 = 0, g x + g. 2 4- g 3 =0, h x + h. 2 + h 3 = 0. 
The coordinates of the points a, /3, y, a, ¡3y are at once found to be 
a, 
( 
0 , 
0, 
- 2^1) ; 
a', ( b , 
2/q, 
0) 
 
(- 
2L, 
a, 
0 ); 
ß\ ( 0, 
c , 
2/ 2 ) 
7» 
( 
0, 
-2/3, 
b ); 
7> (2^3> 
0, 
a ) 
and hence the equations of /3y', ya, a/3' are 
¡3y ; ax + 2h 2 y — 2g 3 z = 0, 
ya' ; — 2hjx + by + 2f 3 z = 0, 
a/3'; 2g x x - 2f 2 y + cz = 0. 
Hence for the point A', which is the intersection of ya', a/3', coordinates are 
be + 4/, / 3 , 4/3 g x + 2ch x , U x f, -2bg 1 ; 
and A' will be on the first conic if only 
(0, - b, c, /, g x , hffbc + 4/ 2 / 3 , 4f 3 g x + 2ch x , 4h x f - 2bg x f = 0, 
viz. this equation is 
- b ( 16f 3 2 g x 2 + 1 QfsgAc + 4h x 2 c 2 ) 
+ 0 ( I6V/2 2 -1 Qf^gAb + 4<7 1 2 6 2 ) 
+ 2/i( IQgAfifs- tyifsb +8h 1 2 f 2 c-4 ! g 1 h 1 bc) 
+ 29i (+ 16/q/ 2 2 / 3 - 8g x f 2 f 3 b + 4 hf'.bc - 2g 1 b 2 c ) 
+ 2 Jh (+ 16gf,f. 2 + 8 h x f 2 f 3 c + *g x f 3 bc + 2 h x bc 2 ) = 0,