514] ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. 
239 
The enumeration is 
Case 
X 
BxXi 
X 
xBx'X'eX', 
(42) 
x' 
.xX 
X 
&c. 
(11) 
X 
.x'X 
.X 
(H) 
x' 
.x'X 
X 
(17) 
X 
.xX' 
X 
(10) 
/ 
X 
.xX' 
.X 
(19) 
X 
.x'X' 
X 
(21) 
X 
. x'X' 
* 
(10) 
whence 
<f>(x + x') — <\)X — <f>x' = eB multiplied into 
4 (X — 1) (Xx — X — x) x' + . . 
+ 2x(x-l)X'(X' -1) +.. 
+ 4 (x — 1) (Xx —x — x)X’ + . . 
+ 2xXx'X' + .. 
+ 2xxXX' - 2 (x + x) XX' - 2 (X + X') xx' + 4 (Xx' + X'x) + . . 
(11) x 2 
(17) 
(10) x 2 
(19) 
(21) 
where the (. .)’s refer to the like functions with the two sets of letters interchanged. 
Developing and collecting, we have 
<f)(x + x) — (f)X — (J)x' = eB multiplied into 
X 2 (W+ 2x'*-6x) 
+ XX (4æ 2 + 8xx + 4«' 2 -12x-12x'+8) 
+ X' 2 (2x 2 + 4txx — 6x) 
+ X (— 12xx' — 6x' 2 + 18«') 
+ X' (- Qx 2 -I2xx' + I8x) 
+ 8xx, 
and consequently 
cpx = eB multiplied into 
X 2 (2x 2 — Qx + 4) 
+ X (— 6x 2 + 18a? + L) 
+ 4# 2 + lx + Af, 
where the constants L, l, A have to be determined. The number of triangles vanishes 
when the curve is a line or a conic, that is <f)X = 0 for x= 1, X = 0, £ = 0, and for 
x = X = 2, g = 6 ; we thus have 
0= 4+1, 
0 = 40 + 2 L + 21 + 6 A.