388 
PROFESSOR CAYLEY ON THE PROBLEM OF 
draw B tangents cBa to the curve B, and F tangents eFa to the curve F ; we have thus in 
respect of the given tangent of D, BF positions of «, or in all BDF positions of a which 
will ultimately coincide with the cusp ; that is, BDF infinitesimal triangles of which the 
angles «, c, e coincide together at the cusp ; and for all the cusps together ^BDF such 
triangles : this would be what is wanted ; the difficulty is that as (of the two intersections 
at the cusp) each in succession might be taken for c, and the other of them for e, it would 
seem that the foregoing number ;cBDF should be multiplied by 2. 
Case 7. B=D=F=a\ Here g=%-l-%; , — Bed. and 
X=Xc(X-l)c(X— 1)«, x '=X<X- 1MX-1)«(= 7j ); 
that is, 
% +*'=2X(X— 1 face. 
The reductions of the two modes are as above, with only the variation that in the 
present case D is the same curve with the two curves B = F. That of the first mode is 
=X(X — 1 )ace, and that of the second mode is (2 7-f- 3 i)ace, which is = {X(X— 1) — x}ace ; 
together they are ={2X(X — 1 )—x}ace, or subtracting, we have 
g={2X(X-l)(X-2)+a?}ff«?. 
Case 8. a=c= B=x. 
x =(X-2)(x-3)F>eFx, ^=FeDx{X-2)(x-Z)(= x ), 
g = 2x(x— 3)(X— 2)<?DF. 
Case 9. D=F=c=x. By reciprocation of 8. 
No. = 2X(X — 3)(x — 2)acB. 
Case 10. a—c=D=x. 
X=~B(x—l)(K—2)eFx, ^=FeX(x-2)F(x-l), 
S = 2(x-l)(Xx-X-x)eBF. 
Case 11. D=F=«=a’. By reciprocation of 10. 
No. =2(X— 1)(X#— X— #)ceB. 
Second process: form ci= B— D=a\ 
%= (X — 2)c(X — 1 )eFx, yJ=FeXc(X-l)(x-2\ 
giving the former result. 
Case 12. c=c=x, ci=F)—y. 
7 =BxY(x-l)Fy, ■ x !=FxY(x-l)By(= 7 J, 
g = 2x(x— lj^YBF. 
Case 13. F=B=.r, «— l)=y. By reciprocation of 12. 
No. =2X(X—l)Yyce. 
