244 
ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. [514 
and subtracting from the before-mentioned number 
2« 2 («-l) F(F-l)(F-2) 
+ & (at - 1) y, 
the required number of positions of the angle a is 
= 2x (x — 1) («—2) F (F — 1) (F — 2) 
+ p(»-l)(«-2)i/ + IF(F-l)(F-2). 
The number of triangles is on account of the symmetry equal to one-sixth of this 
number. 
Case 44. e — D = F = x, a = c = B = y. 
X = ( F — 2) (y — 3) X (« — 2) (X — 3) y, 
x ' = X( æ -2)(X-3) ÿ (F-2)(y-3)(= x ), 
g =2<*-2)X(X-8)<r-2)y(y-S): 
there is a division by 2 on account of the symmetry. 
Case 45. a = D — B — x, c = e = F — y. 
X = (X — 2)y (X — 1) (y — 1)(F— 2)x, 
X — F(y — 2) X (y — 1) (X — 1) (x — 2), 
g =(X — 1) (y — 1) [xy (X — 2) (F—2) + IF(x-2) (y — 2)} 
= 2 (X — 1) (y — 1) {XFxy — XY(x + y) — xy (X + F) + 2xy + 2XF}. 
Case 46. a = c = y, B — I) — F—e = x. By reciprocation of 47, 
No. = y (y - 1) {¿c 2 + a; (2X 3 - 10X 2 + 12X - 1) - 4X 3 + 20X 2 - 16X - 3^} : 
there is a division by 2 on account of the symmetry. 
Case 47. D = F—y, a — c = e = B = x. 
The functional process is exactly the same as for No. 39 (a = c = e = B = x), with 
only F(F—1) written instead of DF; hence 
No. = F(F— 1) {X 2 + X (2a?- KXz 2 + 12x - 1) - 4« 3 + 20^ - 16«- 3f} : 
there is a division by 2 on account of the symmetry. 
Case 48. a = c = JD = F= x, e = B = y. 
The functional process, writing a = c = D = F = x + x', would be precisely the same 
as for Case 42, with only the factor y F written instead of eB; and we have thus 
the like result, viz. 
No. = {X 2 (2# 2 — 6x + 4) + X (— 6x 2 + 18« — 4) + 4« 2 — 4« — 4£) y Y, 
which on account of the symmetry must be divided by 2.