390 
PROFESSOR CAYLEY ON THE PROBLEM OF 
Case 23. o= B=x, c—T)=y, e=F=z. 
x =(X-2MY-2KZ-2K % '=Z(,-2)Y(^-2)X(^-2), 
g =xyz(X — 2)( Y— 2)(Z — 2)-\-XYZ(x—2)(y — 2)(z— 2) 
= 2 {xyzXYZ — xyz( YZ + ZX -|- X Y ) — XYZ(^+~^' + x y) 
+ 2xyz{X + Y + Z) + 2XYZ(* +y+z)- ixyz - 4XYZ } . 
Case 24. a=D=x, c=B=y, e=F=z. 
x =Y(y-2)Xz(Z-2)x, ^=Z(z-2)Xy(Y-2)x, 
g=xX{Y(Z-2)z(y-2) + Z(Y-2)y(z-2)} 
= 2xX {yzYZi -yz( Y+ Z) - YZ(^ + z) + 2yZ + 2zY] *. 
Case 25. a=c=x , I) = F =y, e = B=z. 
x =Z(x-l)Yz(Y-l)x, x '=Yz(Y—l)xZ(x—l)(= x ), 
g = 2x(x — 1) Y( Y — 1 )zZ. 
But we have here eB as an axis of symmetry, so that each triangle is counted twice, 
or the number of distinct triangles is ~^g. 
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(Y —l)(x—l), 
g=x(x-l)Y(Y-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. ={2ff(ff-l)(ff-2)+X}Y(Y-l)D, 
where each triangle is counted twice, so that the number is really one half of this. 
Case 28. B=D=F=#, c=e=y. 
Here g=x+x'“ Peed. 
x =X^(X-l)(y-l)(X-l)«, x =Xy(X—l)(y—l)(X—l')a(= x ), 
X + yJ=ay(y-l).2X(X-l)\ 
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 J) is identical 
with the curves B = F. 
First-mode reduction is 
a(C+2&+3*)B(B — 1) 
(where cl, z refer to the curve c=e), which is 
=a c(c— 1)B(B — 1) ; 
that is, the reduction is =ay(y — 1)X(X — 1). 
* Printed in the Table, erroneously, 2„rX{ . . . + 2yz + 2YZ} . 
