[514 
letters 
514] 
ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. 
247 
for a 
first 
giving in the three cases respectively 
3J + 6Z+18\ = 1116, 
3Z + 4£ + 12\ = 736, 
31 + 3L + 10X = 588; 
and we have then l = — 8, L = 64, \ = 42, so that the required number is 
= x?( +1) 
+ «*( 2X 3 -14X 2 + 28X — 11) 
+ x (- 10X 3 + 70X 2 -116X- 8) 
+ 12X 3 - 76X 2 + 64X 
+ £ (— Qx — 4X + 42 ). 
As a verification, observe that for 1 a conic, x = X = 2, £ = 6, this is = 0. 
Second process, by correspondence : form c=e = B — D = F — x. 
We have 
g =% + %'- R ed., 
X = X(x-2)(X-3)(x-3)(X-3 )a, 
X = X (x— 2) (X — 3) (x — 3) (X — 3) a, = % , 
X + X = a int0 
2 (x - 2) (® - 3) X (X - 3) 2 . 
Fig. 8. 
There is a first-mode reduction, which is 
= a {28 (X - 4) (X - 5) + 3* (X - 3) (X - 4) + * (X - 3) + 2r (X - 3)},