514] ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. 
235 
Second process. Taking the form 
C — I) = e = x, B = F-y\ 
here 
No - = X + X ~ Red -> 
X=Yx(X- 2) 0 - 3) Ya, = y, 
and 
X + X ~ 2F 2 x — 3) (X — 2) a. 
There is a first-mode reduction, 
aY {2 r + 28 (Z - 4) + 3* (Z - 3)}, 
viz. this is 
aY { Z 2 — Z + 8« — 3£ 
+ (Z — 4) (« 2 — x + 8Z — 3£) 
+ (Z — 3) ( — 9Z + 3£)}, 
which is 
= «7 [X (x 2 — « — 6) — 4« 2 -f 12«}; 
and a second-mode reduction 
= aYX (« —2)(x— 3). 
Hence the two together are 
= aY[X (2« 2 — 6«) — 4« 2 + 12«} 
= 2F« (« — 3) (Z — 2) a, 
whence the result is 
= 2 (F 2 — F) « (« — 3) (Z — 2) a, 
which agrees with that obtained above. 
On account of the symmetry we must divide by 2. 
Case 30. e = D = F — x, a = c = y. By reciprocation of 29. 
No. =2Z (X-3)(x-2)y(y-l)B. 
On account of the symmetry we must divide by 2. 
Case 31. c — e = D — x, a — B = y. 
x = (F- 2)«(Z - 2) («-3) Fy, x ' = Fx ( X - 2) (« - 3) F(y-2), 
g=x(x-3)(X-2)F{(Y-2)y+Y(y-2)} 
= 2x (x — 3) (Z — 2){yY — y—Y)F. 
Case 32. F =B = a — x, D = e = y. By reciprocation of 31. 
No. = 2Z(Z- 3)(x-2){yY-y- Y) c. 
30—2