485] 
PROBLEMS AND SOLUTIONS. 
573 
that is, the equation of the twelfth order breaks up into four equations each of the 
third order. The geometrical theory may also be further developed. In fact, assuming 
on each of the three lines respectively a certain sense as positive (and thus isolating 
a set of three solutions) the construction is, on the three lines, from the points A', B', C' 
respectively, measure off the distances A’A = B'B — C'C = s. Then the points A, B, C 
form on the three lines respectively three homographic series; that is, the lines 
BG, CA, AB are respectively generating lines of three hyperboloids, viz. hyperboloids 
which pass respectively through the second and third lines, the third and first lines, 
and the first and second lines. Taking the given point 0 as the centre of projection, 
and projecting the whole figure on any plane whatever, the projections of the lines 
BG are the tangents of a conic which is the projection of the visible contour of the 
hyperboloid generated by the lines BG\ and the like for the lines GA and AB. 
Hence in the projection, or plane figure, we have a triangle whereof the sides A', B', C' 
are the projections of the three given lines respectively; inscribed in this triangle we 
have a variable triangle ABC, such that the side 
BG envelopes a conic, say (M), which touches B' and G', 
CA envelopes a conic, say (B), which touches C' and A', 
AB envelopes a conic, say (G), which touches A' and B'. 
The conics (H)(7?)((7) have three common tangents, say L, M, N\ the conics 
(B) and (G) having besides the common tangent A', 
(C) and (M) having besides the common tangent B', 
(M) and (B) having besides the common tangent G', 
so that the common tangents of the conics (B) and (C), (C) and (A), (H) and (B) are 
the lines A', B', C' each once, and the lines L, M, IV each three times. In the entire 
series of triangles ABC there are three triangles which degenerate into the lines L, M, N 
respectively, these being in fact the projections of the triangles ABG of the solid 
figure which lie in a plane with 0. Or, what is the same thing, the planes of the 
required triangles ABC of the solid figure are the planes through 0 and the three 
lines L, M, and IV, respectively. 
[Vol. vii. pp. 34—36.] 
1993. (Proposed by T. Cotterill, M.A.)—If P is a point on a circle, in which 
A and B are fixed points on a diameter at equal distances from its centre, the curve 
envelope of lines cutting harmonically the two circles whose centres are A and B and 
radii AP, BP respectively, is independent of the position of P on the circle. 
Solution by Professor Cayley. 
1. More generally, the problem may be thus stated: If two conics touch at I, J 
the lines 01, OJ respectively; if P be a variable point on the first conic, and OAB