396 
PROFESSOR CAYLEY ON THE PROBLEM OF 
and consequently 
<px—e B multiplied into 
X\2x 2 -Gx+4) 
+X (— 6# 2 +18#+L) 
+ Ax 2 -{-lx +a|, 
where the constants L, l , X have to be determined. The number of triangles vanishes 
when the curve is a line or a conic, that is <px = 0 for x=A, X=0, £=0, and for x=X=2, 
£=Q: we thus have 
0= 4 + 1, 
0=40+2L+2Z+6*. 
Moreover, the data being sibireciprocal, the result must be so likewise ; we must there- 
fore have L = 1. We thus obtain L=/=x= — 4; so that finally 
<px= { X 2 (2# 2 — 6.r+ 4) -f X( — Gx 2 + 1 8# — 4) -f- Ax 2 — Ax— 4£ } eB. 
Second process, by correspondence: form a=c=l) = F=x. We have 
c-x--x!+2{f-<p-<p')=0; 
also from the special consideration that the points D, F are given as the intersections of 
the curve x, by the first polar of the point e, which first polar does not pass through a, 
we have 
(f— f- f')+e(d- 8— i>')=0, 
and by the consideration that c, D are given as intersections, c a double intersection, of 
the curve with the first polar of the point c, which first polar does not pass through a, 
d-h-V +2(c-y— y') = 0, 
whence 
g-yc—x!=~M c -r— v 1 ) 
and 
c — y — 7 '=BA, 
so that this is 
g— %,— %'= — 4BcA 
Also 
so that 
= -4Be(-2X-2x+2+?). 
X =B(x-l)(X-2)e(X-l)(x-2) 
x ’=(X-2)e(X-l)(*-2)B(x-l), = x , 
g = Bi? multiplied into 
2(X-l)(X-2)(ar-lXa:-2)-4(-2X-2af+2 + g), 
viz. this is 
Be{X 2 (2x 2 -6x+4)+X(-6x 2 -F18x-4)-F4x 2 -4x-4?}. 
