STEINER’S EXTENSION OF MAEFATTI’S PROBLEM. 
261 
+ x/I(N/a® - ©) (yA"+/f^“+ a") 
+/{-9'/sc+ii5v/ca+©yaB-ajf-(®iE)-ajr)) 
=/(^'a'+ V +/'') (i''"” +/f*" + »") 
+yg(v/ac-®){sA”+yp."+«'') 
+v/g(v/gE-©)(/A'+v+/''') 
jc®( - ) = (yac-®) (yai - ©) {/ Yz+ya(Y+z) -/) . 
What, however, is really required=*^, is the value of 0(+) ; to find this, 
Jc0(+) = Z>c’0( — )+2 &cK 
=(yac-®)(v/al-®){/Yz+ya(Y+z)+/} 
+2JcK - 2/(^/lC-®) ), 
the second line of which is 
2(x/ic-®)(ygs-®){^(ygc+®)(N/gi+»)-/} 
iv 
=2(s/gc-®)(y3l-®)yg«, 
where 
|^(\X C) ; 
and consequently 
k<D(+) = (yiC-(a)(x/Sl-l){/YZ+x/ifY+Z)+/+2^x/I}, 
a reduction, which on account of its peculiarity, I have thought right to work out 
in full. 
The condition of contact is 
<i>(+)=vV"=^(ya«-®)(yis-B)yj+v.yT+z^. 
* It may be shown without difficulty that the ( — ) sign would imply that the sections touching z=0, x—O 
and .r=0, y — 0 were sections touching x=0 at the same point. By taking the (— ) sign in each equation we 
should have the solution of the problem ” to determine three sections of a surface of the second order, the two 
sections of each pair touching one of three given sections at the same point,” which is not without interest ; 
the solution may be completed without any difficulty. 
