114] 
STEINERS EXTENSION OF MALFATTl’s PROBLEM. 
61 
Eliminating w between the first and second equations and between the second and 
third equations, 
V - 2be [ax + Q- n y + ^z)+cy + bz = 0, 
2a ' 2a 
'r + b + Ta Z J + ^ 
0; 
and from these equations (cy — bz)~ — 0, or the point of contact lies in the section 
cy-bz = 0. It follows that the equations of the tactors are 
-2a(aa~\ / -2bc)x + (c + a)y+(b + a)z+ (2aa- '26c) w = 0, 
(c + b) x - 2/3 (b& - V - 2ca) y + ( a + b)z + (2b/3 - V - 2ca) w = 0, 
(b + c) x + (a + c) y — 2y (cy — V — 2ab) z + (2cy — V — 2a&) w = 0, 
where a, b, c still remain to be determined. 
Now the separators pass through the point of intersection of the determinators; 
the equations of these give for the point in question, 
x : y : z : w = (2j3y + l)(-a + /3+y + 2a/3y) 
: (2ya +1)( a- /3 + y + 2a(3y) 
: (2a/3 + 1)( a + /3 - y + 2a/3y) 
: 4a 2 /3 2 y 2 — 1 + a- + /3 2 + y 2 ; 
and the values of a, b, c are therefore 
a : b : c = (2/3y+ 1) (— a + /3 + y 4- 2a(3y) 
: (2ya + 1)( a — (3 + y + 2a/3y) 
: (2a/3 + 1) ( a + /3 - y + 2a/3y), 
which are to be substituted for a, b, c in the equations of the separators and tactors 
respectively. 
Now proceeding to find the bisectors, let \x +/zy + vz +pw = 0 be the equation 
of a section touching the determinators, 
¿*-£» + ¿* + »=0. + + w = 
and suppose, as before, A 2 = X. 2 + y? + v 2 — 2yv — 2v\ — 2\y — 2p-; the conditions of con 
tact are 
±0A = £X-(/8 + j|)/A+£i'-2 p, 
yA = yX, + yy — (y + v — 2p,