596 
PROBLEMS AND SOLUTIONS. 
[383 
With respect to the construction of the four planes, 
x+y — z — w = 0, x—y+z—w= 0, x — y — z+w = 0, x+y+z+w= 0, 
it is to be observed that if through any edge of the tetrahedron, for instance the 
edge x — 0, y = 0, we draw the plane x — y — 0 through the point 0, then the harmonic 
of this in regard to the planes x = 0, y = 0 is the plane x + y = 0; we have thus six 
planes, one through each edge of the tetrahedron, viz., these are y + z = 0, z + x = 0, 
x + y = 0, x+w = 0, y + iu= 0, z + w= 0; the six planes being the faces of a hexahedron, 
which is such that the vertices of the tetrahedron are four of the eight vertices of 
the hexahedron. The pairs of opposite faces of the hexahedron meet in three lines 
lying in the plane x + y + z + w = 0, and consequently forming a triangle such that 
through each side of the triangle there pass two opposite faces of the hexahedron; 
the planes x + y — z — w = 0, x—y + z — w = 0, x — y— z + iv = 0 are the harmonics of the 
plane x + y+ z + iu = 0 in respect of the pairs of opposite faces of the hexahedron; 
viz., the plane x + y — z — w = 0 is the harmonic of the plane x + y + z + w = 0 in respect 
to the planes x + y = 0, z-tw = 0', and the like for the other two planes x — y + z — w= 0 
and x—y—z+w= 0 respectively. 
[Vol. iv. July to December, 1865, pp. 17, 18.] 
1710. (Proposed by Professor Cayley.)—Trace the curve y 4 - 2y 2 zx — z 4 = 0, where 
the coordinates are such that x + y + z = 0 is the line infinity. 
Solution by the Proposer. 
We have x = 
; or writing y = 8z, then x = z > that is 
x : y : z = 8 4 — 1 : 28 3 : 2fP. 
Hence, we see that y, z are indefinitely small in comparison of x, 
if 8 = oc, and then x : y : z= 8 4 : 28 s : 28 2 , that is y 2 = 2zx; 
or, if 8 = 0, and then x : y : z = — 1 : 28 3 : 2d' 2 , that is z 3 = — 2y' 2 x; 
so that in the neighbourhood of the point (y = 0, z = 0) there are two branches coin 
ciding with the parabola y 2 = 2zx and with the semicubical parabola z 3 = — 2y 2 xz, 
respectively. 
To find the points at infinity we have x + y+z = 0, that is 8 4 + 2d 3 + 28 2 — 1 = 
(8 + 1)(8 3 + 8 2 + 8 — 1) = 0; and observing that the equation 8 3 + 8 2 + 8— 1 — 0 has one 
real root, say 8 = k, if k be the real root of the equation k 3 + k 2 + k — 1 = 0 (k= •505 
nearly),—there are two real points at infinity, viz., corresponding to 8 = — 1, we have 
the point (0, —1, 1), and corresponding to 8 = k the point (— 1— k, k, 1). 
The equation of the tangent at a point (a, /3, 7) is 
x (— ß 2 <y) + y (2ß 3 — 2nßy) + z (— aß 2 — 2j 3 ) = 0,