[514 
514] 
ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. 
237 
yes, say 
of the 
ase to 
which it belongs; thus x'XxDxF is B = c — e = x, which is Case 10, and so in the 
other instances. Observing that cases 10 and 14 occur each twice, we have thus 
(f>(x + x') — (fix — (fix' = DF multiplied into 
4 (x — 1) (Xx — X— x) x' + .. 
+ {2a;(a; —l)(a;—2 ) + X}X'+ .. 
+ 4<x (x — 1)(XV — X' — x) + . . 
+ 2x(x — 1) xX' + . . 
+ 2x (x — 3) (X — 2) x' + . . 
(10) x 2 
(6) 
(14) x 2 
(12) 
(8) 
where the (. .)’s refer to the like functions with the two sets of letters interchanged. 
Developing and collecting, this is 
(f)(x + x) — cf)X — (f>x' = DF multiplied into 
2XX' 
+ 2X (Sx 2 x' + 3aV 2 + x' 3 — 10aV — ox 2 + 6x') 
+ 2X' (oc 3 + Sx 2 x' + 3aV 2 — ox 2 — 10xx' + 6x) 
— 12 (x 2 x' + xx' 2 ) + 40aV, 
and thence 
<f)x= DF multiplied into 
X 2 
+ X (2x 3 - 10x 2 + 12a;) - LX 
— 4a; 3 + 20a; 2 —lx — Xf, 
where the constants L, l, X have to be determined. Now for a cubic curve the 
number of triangles vanishes; that is, we have <f>x= 0 in each of the three cases, 
x = S, X = 6, |=18, 
„ X = 4, |=12, 
„ X = 3, £ = 10, 
and we thus obtain the three equations 
0 = 108 - 6L - SI - 18X, 
0= 88 — 4Z — SI - 12X, 
0 = 81 — SL - SI - 10X, 
giving L — 1, ¿ = 16, X= 3. Whence, finally, 
<f)x = {X 2 + X (2a; 3 - 10a; 2 + 12« - 1) - 4a; 3 + 20a; 2 - 16a; - 3£} DF. 
Second process, by correspondence. We have 
g~X~X + ^( e - e - e') = 0, 
e — e — e' + D (c — y — y') = 0,