485] 
PROBLEMS AND SOLUTIONS. 
549 
[Vol. v. pp. 38, 39.] 
1834. (Proposed by Professor Cayley.)—1. It is required to find on a given cubic 
curve three points A, B, C, such that, writing x = 0, y — 0, z — 0 for the equations of 
the lines BG, CA, AB respectively, the cubic curve may be transformable into itself by 
the inverse substitution (ax -1 , /3y -1 , yz -1 ) in place of x, y, z respectively, a, /3, y being 
disposable constants. 
2. In the cubic curve ax (y 2 + z 2 ) + by (z 2 + x 2 ) + cz (x 2 + y 2 ) + 2Ixyz — 0 the inverse 
points (x, y, z) and (x -1 , y -1 , z -1 ) are corresponding points (that is, the tangents at 
these two points meet on the curve). 
Solution by the Proposer, S. Roberts, M.A., and others. 
Since the points A, B, G are on the curve, the equation is of the form 
fy 2 z + gz 2 x + lix 2 y + iyz 2 + jzx 2 + hxy 2 + 2 Ixyz = 0 ; 
hence this equation must be equivalent to 
My , VT a , ha? P , iPY 2 , ¿Y a ' 2 , ka P 2 , Zlo-fiy _ 0 
yz 
or, 
p 
xy 
' 1 „2 
yr 
ZX-‘ 
xy* 
xyz 
j -g y' 2 z + k — z 2 x + i~ x 2 y + h -yz 2 +f^zx 2 +g ~ xy 2 + 2lxyz = 0, 
which will be the case if 
f=j 
. a 
P 
g=k-, h = i-, i-h-, j=f~, k = g r ~ 
y y a y J J a ü (3 
This implies fgh = ijk; and if this condition be satisfied, then a : /3 : y can be deter 
mined, viz. we have a : ¡3 : y — if : ij : hf, which satisfy the remaining equations, so 
that the only condition is fgh = ijk. 
Writing in the equation of the curve x~0, we find fy 2 z + iyz 2 = 0, that is, the line 
x = 0 meets the curve in the points (x = 0, y = 0), (x = 0, z = 0), and (x =0, fy + iz = 0). 
We have thus on the curve the three points 
(x = 0, fy + iz = 0), (y = 0, gz +jx = 0), (z = 0, hx + ky = 0), 
and in virtue of the assumed relation fgh = ijk, these three points lie in a line. 
Hence the points A, B, G must be such that BG, GA, AB respectively meet the curve 
in points A', B', G', which three points lie in a line; that is, we have a quadrilateral 
whereof the six angles A, B, C, A', B’, G' all lie on the curve. It is well known 
that the opposite angles A and A', B and B', G and G' must be corresponding points, 
that is, points the tangents at which meet on the curve. And conversely taking A, G 
any two points on the curve, A' a corresponding point to A (any one of the four