383] 
PROBLEMS AND SOLUTIONS. 
571 
the equation of the polar is ^ + - = 0 ; the equation may also be written 
{a.h x -\-fih y -\-<y§ z y xyz= 0, and it thus appears that the line just defined as the polar is 
in fact the second or line polar of the point in regard to the three lines BG, CA, AB 
considered as forming a cubic curve.} 
[Vol. II. July to December 1864, pp. 6—9.] 
1505. (Proposed by Professor Cayley.)—If P, Q, 1, 2, 3, 4 be points on a conic, 
then the four points PI, Q2 ; P2, Ql ; P3, Q4 ; P4, Q3 lie on a conic passing through 
the points P and Q. 
Solution by the Proposer. 
This is an immediate consequence of the theorem of the anharmonic property of 
the points of a conic. For if (PI, P2, P3, P4) denote the anharmonic ratio of the 
lines PI, P2, P3, P4, and so in other cases; then 
(P2, PI, P4, P3) = (PI, P2, P3, P4) = (Q1, Q2, Q3, Q4); 
(P2, PI, P4, P3) = (Q1, Q2, QS, Q4), 
which proves the theorem. 
In particular, if P, Q are the circular points at infinity, then the conic is a circle. 
Moreover the points PI, Q2 ; P2, Q1 are the antifocal points of 1, 2; viz., calling these 
1', 2', then 12 and 1'2' are lines at right angles to each other, having a common 
centre 0, but such that l'2' = i. 12, (i = V—l, as usual); or, what is the same thing, 
01 = 02 = i. 01' = i. 02'. And the theorem is as follows: viz., if 1, 2, 3, 4 are points 
on a circle, and 
V, 2' are the antifocal points of 1, 2, 
3' 4' 3 4 
» 5> 5, ”, “» 
then 1', 2', 3', 4' are points on a circle. 
As an a posteriori proof, take the centre of the given circle as origin, so that 
( a i, A), ( a 2, AX («3, AX ( a 4, Pi) being the coordinates of 1, 2, 3, 4, and the radius 
being taken as unity, we have 
ttl 2 + ft 2 = a 2 2 + A 2 = «s 2 + /S 3 2 = «4 2 + A 2 = 1 • 
Suppose for a moment that x, y are the coordinates of the antifocal points of 1, 2; we 
have 
00 - «1 ± i (y - Pi) = 0, X - a 2 + i (y - A) = 0, 
that is 
x + iy = + ip 
x — iy = a 2 — ip 2 , 
72—2