240 ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. [514 
Moreover, the data being sibireciprocal, the result must be so likewise; we must 
therefore have L = l. We thus obtain L = l = \ = — 4; so that finally 
<fix = {X 2 (2# 2 — 6# + 4) + X (— 6# 2 + 18# — 4) + 4# 2 — 4# — 4£} eB. 
Second process, by correspondence: form a = c = D = F=x. We have 
c -%-%' + 2 (f-0-0') = O; 
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 
(i — (f> — ft) + e (d — 8 — S') = 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 — S — 8' + 2 (c — 7 — <y') — 0, 
whence 
and 
so that this is 
Also 
so that 
g-%- = (c-7-7) 
c — 7 — 7' = BA, 
g-%-% ,= - 45eA 
= - 4Be (- 2X — 2# + 2 + £). 
% = B (x - 1) (X - 2) e (X -1) (x - 2), 
X =(X -2) e(X-l)(x-2) B(x-1), = x , 
g = Be multiplied into 
2 (X — 1) (X - 2) (# — 1) (# — 2) — 4 (— 2X - 2# + 2 + £), 
viz. this is 
Be {X 2 (2# 2 — 6# + 4) + X (— 6# 2 + 18# — 4) + 4# 2 — 4# — 4|}. 
Third process: form c — e — F=B = x. 
g =% + %'-Red., 
X = X (# — 2) X) (# — 1) (X — 2) a, 
X = X{x — 2)D{x— 1) (X — 2) a, = x , 
X + = aD • 2X (X - 2) (# - 1) (# - 2). 
The first-mode reduction is here 
aD [(X- 2) X + (X - 4) 28 + (X - 3) 3/c + k] ;