592 
PROBLEMS AND SOLUTIONS. 
[383 
10. Referring to the foregoing figure, if instead of the point 1 we take on the 
line x, a point 1', and construct therewith the hexagon 1'2'3'4'5'6' ; then if a, a' be the 
(foci or) sibi-conjugate points of the range 1, 4, 1', 4' on the line x\ ¡3, /3' the sibi- 
conjugate points of the range 2, 5, 2', 5' on the line y ; and y, 7' the sibi-conjugate 
points of the range 3, 6, 3', 6' on the line z ; the points in question form two triangles 
af3y, a'j3'y', such that for each triangle the angles lie in the given lines and the sides 
pass through the given points. This is an elegant geometrical construction for the 
problem of the in-and-circumscribed triangle, in the particular case where the given 
points A, B, G lie in the given lines x, y, z, respectively. 
11. The points 1, 2, 3, 4, 5, 6, A, B, G constitute a system of 9 points which lie 
in 9 lines of 3 each. The points a, /3, 7, a', /3', y, A. B, G constitute a radically 
distinct system of 9 points lying in 9 lines of 3 each; viz., in the former system 
there are 3 sets of 3 lines which contain all the 9 points; in the latter system 
there is only the set of lines Aaa', B/3/3', Gyy which contains all the nine points. The 
last-mentioned system may be constructed as follows: The points /3, /3' and 7, 7' are 
arbitrary: A is the intersection of the lines /3y and ¡3'y'; and then joining A with 
the point of intersection of the lines /3y and /3'y we have a an arbitrary point 
on the joining line ; the lines ay and /3/3' meet in the point B, the lines a/3 and yy 
in the point C; the lines Cj3' and By will then meet in a point a' on the line Aa; 
and we have thus the figure of the nine points a, /3, 7, a, /3', y, A, B, G. 
[Vol. ill. pp. 78, 79.] 
1667. (Proposed by Professor Sylvester.) 
Show that the discriminant of the form 
ax 5 + b\x A y + c\ 2 x?y 3 + cy 3 x 2 y 3 + byxy 4 + ay 5 
will be a rational integral function of the quantities a, b, c, Xy, X 5 + y 5 , and of the 
second degree only in respect to the last of them. 
■ 
In general 
Solution by Professor Cayley. 
Disct. (a, b, c, d, e, f'$\x + yy, X'x 4- y yf — (\y — X'y) 20 Disct. (a, b, c, d, e, f\x, y) 5 . 
Hence first, if (X, y, X', y) = (0, 1, 1, 0), then 
Disct. (a, b, c, d, e, f\y, x) 5 = Disct. (a, b, c, d, e, f\x, y) 5 ; 
and secondly, if co be an imaginary fifth root of unity and (X, y, X', y) = (eo, 0, 0, 1), 
then 
Disct. (a, b, c, d, e, f\wx, yf — Disct. (a, b, c, d, e, f\x, yf.