38G 
PEOFESSOE CAYLEY ON THE PEOBLEM OE 
Case 3. D=F=a\ Reciprocation from 2 ; or else, second process, 
x =BcXe(X-l)a, yj=Xe(X-l)cBa, 
g = 2X(X— 1)B ace. 
Third process: form F=B=a. We have here g=Z J r% — Red. 
x =XcDeXa, y!=XeDcXa (= x ), 
%+%'=2X 2 D ace; 
and the reductions are those of the first and second mode, as explained ante, Nos. 11, 12, 
viz. each of these is = XD ace, and together they are = 2XD ace; whence the foregoing 
result. 
Case 4. a=D=x. 
x =BcXeFx, y!=YeXBx(= x l 
g =2XacFBF. 
Observe this is what the result for Case 1 becomes on writing therein a=T)=x, viz. the 
opposite curves a, D may become one and the same curve without any alteration in the 
form of the result. 
Case 5. a= B=a\ 
X=(X-2 )cBeFx, x'=FeDcX(a-2), 
where 
(X-2)a+X(a-2)=2(Xa-X-a) ; 
wherefore 
g=2(Xr— X— x)ceJyF. 
Case 6. a—c=e—x: perhaps most easily by reciprocation of Case 7 ; or 
Second process, functionally by taking the curve a~c—e to be the aggregate curve 
x-\-ad. The triangle «BcDcF is here in succession each of the eight triangles: 
x F»x Da F 
X , , X ,, X ,, 
x ,, xt „ X ,, 
X ,, X ,, X ,, 
x? Ba'Da'F 
/y»' rtfJ 
55 ^ 55 ^ 55 
syJ /y» /yJ 
55 xA , 55 O' 55 
X „ x ,, xf ,, 
where the two top triangles give <px and <px' respectively ; the remaining triangles all 
belong to Case 2, and those of the first column give each 2(x 2 — a)a'BDF, and those of 
the second column each 2(x' 2 — a'jaBDF. We have thus 
(p(x + x') — <px — <px' = { 6 (x 2 x' + xx' 2 ) — 1 2xx' } BDF. 
Hence obtaining a particular solution and adding the constants, we have 
<px= (2a 3 — 6a 2 + ax + /3X-f-y£)BDF ; 
