[556 
556] 
ON STEINERS SURFACE. 
5 
q : 2q + 1; and taking 
es — q, —q, 2^ + 1 will 
have the three nodes 
;asy to verify that this 
b Z) = 0, 
in the four real points 
observe that, writing 
uation may be written 
B- = 0, 
nary conics. The fore- 
il point of the surface. 
Y + Z), 
point X— Y=Z, for a 
also 
Then we have 
and the equation is 
that is, 
or finally 
2A = (/3 — 7> 2 + (7 — a) 2 + (a — /3) 2 , 
il = (ß-ry)(cy- a )(a- /3). 
X + Y+Z = 3 u, 
YZ + ZX + XY = Su 2 — A, 
XYZ= u 3 -uA + £l, 
{9 u 2 — 9 (Su 2 — A)} 2 — 324ui (u 3 — uA + H) = 0, 
(— 2 u 2 + A) 2 — 4 u (u 3 — u A + il) = 0, 
A 2 — 4 uCl = 0, 
where the lowest terms in /3 — <y, 7 — a, a — /3 are of the order 3, and the theorem 
is thus proved. The case in question, q = — ^ or \ = ^h, is where the plane passes 
through the centre of the tetrahedron. 
When q = % = 
2 A-h 
2k-2\’ 
or A = §h, the equation is 
(X 2 +Y 2 + Z 2 -2YZ- 2 ZX - 2 XY) 2 = 128X7£(X + Y+Z). 
Here each of the lines X = 0, Y = 0, Z = 0 is an osculating tangent having with the 
curve a 4-pointic intersection. 
When 
q = 0 = 
2 A — h 
2 h - 2A ’ 
or A = \h, the equation is 
that is, 
(YZ+ZX + XY) 2 -4XYZ(X+Y+Z) = 0, 
Y 2 Z 2 + ^ 2 X 2 + X 2 Y 2 - 2XYZ (X + Y+ Z) = 0; 
viz. each angle of the triangle is here a cusp. 
When q = — or A = 0, the curve is 
{X 2 + Y 2 + Z 2 — 2 (YZ + ZX + XY)} 2 = 0, 
viz. the plane is here the base of the tetrahedron, and the section is the inscribed 
circle taken twice. 
For tracing the curves, it is convenient to find the intersections with the lines 
Y — Z = 0, Z — X = 0, X—Y— 0 drawn from the centre of the triangle to the vertices ; 
each of these lines passes through a node, and therefore besides meets the curve in 
two points. Writing, for instance, Y = X, the equation becomes 
[q 2 (2X + Z) 2 - 2XZ - X 2 } 2 - 4 (2q + 1) XX(2X +Z) = 0 ; 
[qZ + (2q + 1) X} 2 [q 2 Z 2 + (4q 2 -2q-*)XZ + (4q 2 - 4>q + 1) X 2 } = 0, 
viz. this is