485] 
PROBLEMS AND SOLUTIONS. 
581 
Writing them A = (12 ‘' 3 * 5) 2 
1(12, 345) 
12 
(and therefore A + /a = 1) we find (A, /a) 
by substituting Aa a + /A0£ 2 , A^ + /a/3 2 , Ayj + yy- 2 , A + fi for x, y, z, 1 in the equation 
x, y, z , 1 =0 
a 3> ßa> 73> 1 
®4> Pl> 0^45 1 
a 5> ßö) y5> 1 
of the plane 345 ; we have thus 
A (1345) + /a (2345) = 0, 
A : ¡X = (2345) : - (1345) = (2345) : (3451), 
(12, 345)2 : 1(12, 345) = (2345) : (3451), 
that is 
whence 
that is 
= (2345) : (3451), 
or completing by symmetry 
1111 
1 • \ = (2345) : (3451) : (4512) : (5123) : (1234), 
which is the theorem for the case n = 5. The general case depends, it is clear, upon 
similar reasoning in a {n — 2)dimensional geometry; leading to the conception in this 
geometry of a figure of (71 — 1) points such that the line joining any two of them is 
at right angles to the line joining any other two of them. 
[Yol. vu. p. 106.] 
2331. (Proposed by Professor Cayley.)—Show that it is possible to find (X, Y, Z) 
linear functions of the trilinear coordinates (oc, y, z) such that the equations xX = yY=zZ 
may determine four given points. 
[Vol. viil, July to December, 1867, p. 26.] 
2321. (Proposed by Professor Cayley.)—Given a conic, to find four points such 
that all the conics through the four points may have their centres in the given conic. 
[Vol. viil p. 36.] 
2371. (Proposed by Professor Cayley.)—(4). If P, Q be two points taken at 
random within the triangle ABC, what is the chance that the points A, B, P, Q may 
form a convex quadrangle ?