60, BULLETIN OF THE 



This equation involves only y, the radius of the inscribed circle 

 and the given bisectors of the angles o and ^ : hence we may de- 

 termine 7 from it. Eq. (8) becomes after a somewhat laborious 

 reduction 



64(a' — ai3x/2 + /3'')y«-f.8N/2a/3(4a' — 3>/2a/3 +4i3^)y« 

 + a-/3^(2a' — ai3v/2-l- 2 ]3'0 y' — a* /3* = 0. (9) 



These three solutions just given all involve trigonometrical re- 

 lations and are therefore properly classed as trigonometric solu- 

 tions. They may all, however, be made independently of trigo- 

 nometry. In the following we shall give the algebraical solutions 

 corresponding to the first and second trigonometrical solutions 

 together with athird and entirely independent solution. 



ALGEBRAICAL SOLUTION. 



From Fig. 2 we have 



c: AD: : a:DC: : OB:OD: : 1 :n 

 c:BE::6:EC::0 A :0E ::1 :m; 



from which 



B = ^ - , D = -iiL- , A ^ " , E = °^ , 

 1 + w 1 -\-n 1-j-m 1-j-m 



AD = cn,CD=:ra?i,BE=c7)i, CE = 6 m. 



Now 



c. A D = O A-' -f O B.O D or c'n = (~~^— ) V « /. ^ )', 



\l+m/ ^ U +n/ 



c. B E = O B= + O A.O E or c^ m = ( ^ )V m( ^ Y; 



whence 



or 



m(l-»)^,^r. (l-m) 

 (1 -f m)'' (1 + 1,)^ '^ ^ ^ 



Again 6 = A D -f- C D = n(c + a) . •. a* -f n' (c + a)' = c* 

 and a = C E -f- B E = 6 771 -{- c r?i = mn {c -\- a) -f cm 



= c m (1 -f ?7) -{■ amn . *, - = 



a m (1 -f- n) 

 Equating these two expressions 



1 — 77?. n \ + n^ 1 — n , 1 



• m = and 7i = 



771 



1 — n \ -\- n l-f-771 



