234 ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. [514 
Case 26. a = c = x, B = D = y, e = F=z. 
X=Y(x-l)(Y-l)z(Z-2)x, x=Z{z-2) Yx (F- 1) (x -1), 
g=x (x-1) F(F-l) {z(Z-2) + Z(z-2)} 
= 2x(x-l)Y(Y-l)(zZ-z-Z). 
Case 27. a = c — e — x, B = F= y. By reciprocation of 28. 
No. = [2x(x — 1) {pc — 2) + X) F(F— 1)D, 
where each triangle is counted twice, so that the number is really one half of this. 
Case 28. B —D = F—x, c — e = y. 
Here 
g=X + X ~ Yed. 
x = Xy(X-l)(y-l)(X-l)a, % ' = Xy(X-l)(2/-l)(X-l)a(= % ), 
% + X = a V (y~ !) • 2X (X -1) 2 . 
The reductions are those of the first and second mode as explained above, with 
the variation that the curves c and e are here identical, c = e, and that the curve D 
is identical with the curves B = F. 
First-mode reduction is 
a (G + 28 + 3k) B(B- 1) 
(where S, tc refer to the curve c = e), which is 
= a c (c — 1) B (B — 1); 
that is, the reduction is = a y (y — 1) X (X — 1). 
Second-mode reduction is 
cl(2t + 3i)c{c— 1) 
(where t, l refer to the curve B = D = F), which is 
= a{B(B-l)-b}c(c-l)-, 
that is, the reduction is =ay(y — l) [X (X — 1) — x). 
Hence the two together are = a y (y — 1) {2X (X — 1) — x); and subtracting from 
X + x we h ave 
g = ay(y-l).{2X{X-l){X-2) + x}\ 
but on account of the symmetry each triangle is reckoned twice, and the number of 
triangles is = |g. 
Case 29. a = c = B =x, I) = F — y. 
x = (X-2)(x-3)Ye(Y-l)x, %'= Fe(F- l) x(X-2)(X -3) (=%), 
g = 2x (x — 3) (X — 2) F (F — 1) e.