248 ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. [514 
where the term a. 2t (X — 3) arises, as shown in the figure, and a second-mode reduction, 
which is 
= a {2t (« — 4) (« — 5) + 3t (« — 3) (x — 4)}; 
and the two together are = a into 
(.X - 4) (X - 5) (« 2 - x + 8X - 3f) 
+ (X —3) (X — 4)( 9X +3f) 
+ (X-3)/ -3X + A 
\+ X 2 - X + 8« - 3f/ 
+ (« — 4) ( x— 5) (X 3 — X + 8# — 3£) 
+ ( x - 3) ( x - 4) ( - 9# + 3£); 
that is, = a into 
— « 3 
+ « 2 . 2X 2 — 10X + 11 
+ x. - 10X 2 + 26X + 8 
+ 4X 2 + 44X 
+ £ (6« + 4X — 42); 
and subtracting this from the foregoing value of ^ which is = a into 
« 2 ( 2X 3 — 12X 3 + 18X) 
+ x (- 10X 3 + 60X 2 - 90X) 
+ 12X 3 - 72X 2 + 108X, 
the result is as before. 
There is a division by 2 on account of the symmetry. 
Case 51. a = c = e = B = D — x. By reciprocation of 50, 
No. is = X 3 ( +1) 
+ X 2 ( 2« 3 -14« 3 + 28a?-11) 
+ X (-lO«^ 70« 2 -116«- 8) 
+ 12« 3 — 7 6« 2 + 64« 
+ £(- 6X — 4« + 42). 
There is a division by 2 on account of the symmetry.