[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)},