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PROFESSOR CAYLEY ON THE PROBLEM OF 
Case 44. £=D=F=a, a—c=Y\=y. 
X =(Y-2)(y-S)X(®-2)(X-3 )y, 
x '=X^-2)(X-3) 3 ,(Y-2)(y-3)(= % ) ) 
g=2( i r-2)X(X-3)(Y-2My-3): 
there is a division by 2 on account of the symmetry. 
Case 45. a= D=B=a, c=e=F =y. 
X = (X— 2)_y(X— l)(_y— 1)(Y— 2)^r, 
^=Y(y-2)X(y-l)(X-l)(a;-2) f 
g=(X-lXy-l){^(X-2)(Y-2)+XY(a ? -2)(y-2)} 
= 2(X-l)(y-l){XYjry-XY(a?+y)-ay(X+Y)+2a3r+2XY}. 
Case 46. a—c—y , B=D==F =e=x. By reciprocation of 47. 
No.=y(y— 1){a 2 -Fa(2X 3 — 10X 2 +12X— 1)— 4X 3 +20X 2 — 16X— 3f } : 
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 
Y(Y — 1) written instead of DF ; hence 
No.=Y(Y— 1){X 2 +X(2 a 3 — 10a 2 + 12a— 1) — 4a 3 + 20a 2 — 16a— 3s} : 
there is a division by 2 on account of the symmetry. 
Case 48. a=c=D= F=a, e=B=y. 
The functional process, writing a=c=D=F=%-}-a/, would be precisely the same as 
for Case 42, with only the factor yY written instead of eB ; and we have thus the like 
result, viz. 
No. = { X 2 (2 a 2 — 6 a -f- 4) +X( — 6 a 2 + 1 8a — 4) + 4 a 2 — 4a — 4f }y Y, 
which on account of the symmetry must be divided by 2. 
Case 49. a=B=y, c=e=T)=F=x. 
% = (Y— 2)a(X— 2)(a— 3)(X— 3)y, 
% ,= X(a— 2)(X— 3)(a— 3)Y(y— 2), 
g= ( A-3)(X-3){Ay(X-2)(Y-2)+XY(A-2)(y-2)} 
=2(A-3)(X-3){AyXY-(A+y)XY-(X+Y)Ay+2Ay+2XY}. 
Case 50. c=e= B=D=F=a. 
Functional process; by taking the curve c—e=B=T) = ~F as the aggregate of two 
curves, say =a+^. The cases are 
