558 
PROBLEMS AND SOLUTIONS. 
[485 
being two conditions for the determination of the position of the point G ; these give 
CG, GH as linear functions of CF; the distances GF, CG, CH of the point G from 
the points F, G, H in the line FGH are connected by a quadratic equation, and hence 
substituting for GG, GH their values in terms of CF, we have a quadratic equation 
for GF ; as the given conditions are satisfied when G coincides with A or with B, 
the roots of this equation are CF=AF and GF = BF. But if CF=AF, then the 
linear relations give CG=AG and GH = AH, that is, G is a point opposite to A in 
regard to the line FGH. And similarly if CF=BF, then G is a point opposite to 
B in regard to the line FGH. But G being opposite to A or B, the fourth concyclic 
focus D will be opposite to B or A ; that is, the pairs A, B and G, D of concyclic 
foci lie symmetrically on opposite sides of the line FGH ; this of course implies that 
the four points lie on a circle. 
5. Taking Y — 0 as the equation of the line FGH, x 2 + y 2 — 1 = 0 as the equation 
of the circle through the four points A, B, G, D, then these lie on a proper cubic 
(x 2 + y 2 + 1 ) x + lx 2 + ny 2 = 0 
(not passing through the points F, G, H) and the four foci are given as the inter 
sections with the circle x 2 + y 2 — 1 = 0 of the pair of lines 
x 2 — 2 nx — nl = 0. 
But if we attempt to describe with the same four foci a cubic 
(x 2 + y 2 + 1) y + l'x 2 4- 2m'xy + ny 2 = 0, 
then the foci are given as the intersections with the circle x 2 + y 2 — 1=0 of the conic 
y 2 4- 2m'x — 21'y + m' 2 — n'l' = 0. 
In order that these may coincide with the points (A, B, G, D) we must have 
(x 2 — 2 nx — nl) + (y 2 + 2 m'x — 21' y + m’ 2 — n'l') = x 2 + y 2 — 1; 
that is 
m' — n, V = 0, — nl + n 2 — n'l' — — 1. 
The last equation is n'l' = n 2 4-1 — nl, which, assuming that nl is not equal to n 2 +1, 
{in this case the cubic {x 2 + y 2 + 1) x + lx 2 + my 2 = 0 would reduce itself to the line 
and conic (x + n) (x 2 + y 2 + a ^j = 0}, since V — 0, gives n' = oo , and therefore the cubic 
(x 2 + y 2 +1) y + l'x 2 + 2m'xy + n'y 2 = 0, 
reduces itself to y 2 = 0, that is, the cubic in question reduces itself to the line 
FGH twice repeated, and the line infinity. 
6. The conclusion is that F, G, H being given points on a line, and A and B 
being any other given points, there is not any proper cubic passing through F, G, H 
and having A, B for concyclic foci : and the primd facie objection to the truth of 
the theorem is thus removed.