ON STEINERS SURFACE.
3
[556
sitions, (1) resting with
edges horizontal, or say
tetrahedron and being
le nodal lines form a
vertical. And I proceed
'f the surface.
¡o be the perpendicular
the distance from the
)riginally homogeneous;
lue, the equation will
section of the tetra-
lich is to an edge of
triangle, if X, Y, Z
opposite side of the
a face of the tetra-
r , Z; we consequently
be the length of an edge,
rpendicular from a summit
ts of opposite edges, then
at right angles thereto, be
quadrilateral) ; viz. in each
= $s; meeting in the centre
556]
where the coordinates X, Y, Z are the perpendicular distances from the sides of the
triangle which is the section of the tetrahedron. To simplify, I write
h — \ 2q + ly
that is,
2 \-h
q ~2h-2\’
the equation then is
VX + VF+V^+V(22+1)(X+F + Z) = 0;
or, proceeding to rationalize, we have first
q(X+ Y+Z) = YYZ + \fZX+*JXY,
and thence
q 2 (A + 1 + Zf — (YZ + ZX + X Y) = 2*JXYZ (y*X + \J Y 4- \/Z);
and finally
{q>(X+Y+Zy-YZ-ZX-XY}* = 4<(2q + l)XYZ(X+ Y + Z).
This is a quartic curve, having for double tangents the four lines X = 0, Y= 0, Z= 0,
X + Y+Z= 0, the last of these being the line infinity touching the curve in two
imaginary points, since obviously the whole real curve lies within the triangle. This
is as it should be: the double tangents are the intersections of the plane w = \ by
the singular planes of the surface.
To find the points of contact, writing for instance Z= 0, the equation becomes
that is,
whence
tf(X+Yy--XY = 0,
X 2 +( 2 ~ \)XY+ F 2 = 0;
giving the two points of contact equi-distant from the centre; these are imaginary if
q > but otherwise real, which agrees with what follows. (See the Table afterwards
referred to.)
The nodal lines of the surface are (x-y= 0, z-w = 0), (y -z = 0, x-w = 0),
(z — x = 0, y w = 0). Considering the first of these, we have for its intersection with
the plane w = A,
X = Y, (X+Y+Z),= (2q + 1) (X + Y + Z),
and the last equation gives
that is,
Z = (2q + l)(2X + Z),
0 = (2q + l)X+qZ,
1—2
