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.