238 
ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. [514 
and thence 
Moreover 
and 
g-%-%'= DF(c-y-y). 
X = (X-2)(x- 3) D(x- l)F(x-\), 
x ' = F(x-l)B(x-l)(X-2)(x-l), = %, 
X + X' = VF(X~ 2)2(x-3)(x-l)\ 
c — 7 — y = 2r + (X — 3) k — 2 (X — 2) (x — 3), 
as is easily obtained, but see also post, No. 29; hence 
g = BF multiplied into 
(X-2).2(>-3)0-l) 2 
+ (X- 2).- 2(>-3) 
+ 2t + (X — 3) k ; 
but I reject the term DX.(X—3)/e as in fact giving a heterotypic solution; I do 
not go into the explanation of this. And then substituting for 2t its value, we have 
g = I)F multiplied into 
(X-2).2x{x-l){x-2) 
+ X 2 - X + 8x - 3|, 
where the second factor is 
= X 2 + X (2* 3 - lCte 2 + 12x - 1) - 4# 3 + 20^ - 16x - 3£, 
which is the foregoing result. 
Case 40. B = D = F=e = x. By reciprocation of 39. 
No. = {x* + x (2X 3 - 10X 2 + 12X - 1) - 4X 3 + 20X 2 - 16X - 3£} ac. 
Case 41. c = e—D = F=x. 
X =Bx(X-2)(x -3)(X-3) a, 
x ' = X (x - 2) (X - 3) (x - 3) Ba, 
g = (x-3)(X-3)aB{x(X-2)+X(x-2)\, 
= 2 (x — 3) (X — 3) (xX — x— X) aB. 
Case 42. a — c = D — F = x. 
Functional Process; the curve is supposed to be the aggregate of two curves, say 
a = c = D = F=x+x.