f f :x 
ON STEINERS SURFACE. 
[556 
We may imagine the tetrahedron placed in two different positions, (1) resting with 
one of its faces on the horizontal plane, (2) with two opposite edges horizontal, or say 
with the horizontal plane passing through the centre of the tetrahedron and being 
parallel to two opposite edges; or, what is the same thing, the nodal lines form a 
system of rectangular axes, one of them, say that of z, being vertical. And I proceed 
to consider, in the two cases respectively, the horizontal sections of the surface. 
In the first case, the coordinates x, y, z, w may be taken to be the perpendicular 
distances of a point from the faces of the tetrahedron, w being the distance from the 
base. We have*, if the altitude be h, 
x + y + z + w = h] 
an equation which may be used to homogenize any equation not originally homogeneous; 
thus, for the plane w = A, of altitude A, we have 
w 
^(x + y + z + w), 
or, what is the same thing, 
The equation of the surface is 
w = ( x + v + z )- 
\/x + Vy + \/z + \/w = 0, 
and if we herein consider w as having the last-mentioned value, the equation will 
belong to the section by the plane w — A. I remark that the section of the tetra 
hedron, by this plane, is an equilateral triangle, the side of which is to an edge of 
the tetrahedron as h— A : h. For a point in the plane of the triangle, if X, Y, Z 
are the perpendiculars on the sides, then 
X + Y + Z = P, 
(if for a moment P is the perpendicular from a vertex on the opposite side of the 
triangle, viz. we have P = P> if p be the perpendicular for a face of the tetra 
hedron). And it is clear that x, y, z are proportional to X, Y y Z; we consequently 
have, for the equation of the section, 
*JX Y Z -\- 
\/ 
h- A 
(X+Y+Z) = 0, 
* I take the opportunity of remarking that in a regular tetrahedron, if s be the length of an edge, 
p the perpendicular from a summit on an edge (or altitude of a face), h the perpendicular from a summit 
on a face (or altitude of the tetrahedron), and q the distance between the mid-points of opposite edges, then 
S =j2 k ' p= 2ji h ’ «=2*- 
The tetrahedron can, by means of planes through the mid-points of the edges at right angles thereto, be 
divided into four hexahedral figures (8 summits, 6 faces, 12 edges, each face a quadrilateral) ; viz. in each 
such figure there are, meeting in a summit of the tetrahedron, three edges, each = A s ; meeting in the centre 
three edges, each and six other edges, each =ip. 
556] 
where the 
triangle wi 
that is, 
the equati 
or, proceed 
and thence 
and finally 
This is a 
X+Y+Z 
imaginary 
is as it s 
the singul 
To fin 
that is, 
whence 
giving the 
q>%, but 
referred to 
The 
(z — x — 0, 
the plane 
and the 
that is,