THE IX- AXE-CIRCUMSCRIBED TRIANGLE. 
397 
Third process : form c=e=F =B=x. 
g=X + x'- Red -> 
x =X(^-2)D(^-l)(X-2>, 
% f — X(^— 2)D(^— 1)(X— 2)«, = % , 
3C + % '=«D.2X(X-2)(®-l)(jr-2). 
The first-mode reduction is here Fig* 4. 
«D[(X-2)X+(X-4)2i+(X-3)3*+*]; 
where the last term aDz arises from the tangents 
cBa and eFa, each coinciding with a cuspidal tan- 
gent, as shown in the figure. 
The second-mode reduction is 
= al) . X(x— 2)(x—3), 
so that the two reductions together are 
=«D{(X-2)X-f(X-4)2H(X-3)3^-h«+X(^-2)(o;-3)}, 
viz. this is 
= aD { (X - 2)X + (X - 4)(25 + 3*) + 4* + X(a: - 2)(ar - 3) } ; 
or substituting for 2l-Foz and » the values x 2 — x—X and — 3X-fi§ respectively, and 
reducing, it is 
ciF) {X(2x 2 — Gx— 4) — 4x 2 + 4x~F 4§ } . 
Hence subtracting from % + %, written in the form 
dD { X 2 (2x 2 — Gx +4)+X(- 4x 2 + 1 2 sc - 8) } , 
the result is 
= «D { X 2 (2a 2 — Gx + 4) + X( - Gx 2 + 1 8a’- 4) + 4x 2 -4x- 4? } . 
On account of the symmetry we must divide by 2. 
Case 43. cl—g—g—x , B=X)=F=v/. 
Suppose for a moment that the angle a is a free point ; the locus of a is a curve the 
order of which is obtained from Case 28, by writing c=e~x, B = D = F=:?/; the locus 
in question meets a curve order a in {2Y(Y — 1)(Y — 2)-\-y}x(x— l)a points; wherefore 
the order of the locus is 
= {2Y(Y— 1)(Y— 2) 4-^ }#(#—!), 
and this locus meets the curve a=c=e=x in a number of points 
= {2Y(Y-l)(Y-2)+y}^-l), 
viz. this is the number of positions of the angle a ; but several of these belong to special 
forms of the triangle aBcDeF, giving heterotypic solutions, which are to be rejected ; the 
required number is thus 
{2Y(Y— 1)(Y— 2) +y}a?(x- 1 ) - Red. 
