CHAP. V] TRIANGLES WITH RATIONAL ANGLE-BISECTORS. 211 



Cunliffe 98 found a rational triangle ABC with rational values for the 

 sides, altitudes and angle-bisectors. Circumscribe the circle with the 

 rational diameter d. Let the perpendicular bisectors mD, nE, pF of the 

 sides meet the circle at D, E, F (see right-hand figure on p. 210). Set 

 a = AD = DB, b = AE, c = BF. Then Dm = a?/d, En = b 2 /d, Fp = c 2 /d, 



Am = 3 d 2 - a 2 , Cn = 3 d* - b 2 , Cp = - tf - c 2 . 

 d d d 



Since the chords a, 6, c subtend arcs whose sum is the semi-circle, they 

 serve to form with the diameter an inscribed quadrilateral with sides 



a = MP, c = PQ, b = QN, d = MN. Hence 



JVP 2 = d 2 - a 2 , MQ 2 = d 2 - b 2 , 

 and 



MP-NQ + MN-PQ = NP-MQ, cd = # - a 2 - d 2 W - ab. 



Hence if d 2 a 2 and d 2 b 2 are rational squares, c as well as a and b are 

 rational. By the inscribed quadrilateral ACBD, 



AB-DC = DB-AC + AD-BC; 



hence DC is rational. Thus the angle-bisector Dl = (DB) 2 /DC is rational. 

 A second solution employs the inscribed circle with radius r and center S, 

 lengths a, b, c of the tangents from A, B, C, and foot T of the perpendicular 

 from S to AB. Then AS 2 = AT 2 + ST* = a 2 + r 2 . To satisfy it in 

 integers, take a = 2mnr/(m 2 n 2 ). Similarly, satisfy BS 2 = b 2 + r 2 . It 

 is proved that CS 2 = c 2 + r 2 is a rational square by use of 



abc = r 2 (a + 6 + c). 



W. Wright and C. Gill" employed an isosceles triangle with the equal 

 sides CA and CB, altitude CD, and intersection of the angle-bisectors 

 AP and BQ. Set x = AD, a = AC + x = semi-perimeter. Then 



CD = Va 2 - 2ax 

 will have a rational value ap if x = %a(l p 2 ). Then 



OD = C = a - 2 AO = al - 



It follows from certain proportions that CP is rational, while AP involves 

 Vl + p 2 . Hence the problem is solved ifl + p 2 =D = (l gp) 2 , say, 

 which gives p = 2q/(q 2 1). Taking q = 3, 4, 5, 7, we get four isosceles 

 triangles with the same perimeter and having rational sides, areas and 

 angle-bisectors. 



S. Jones 100 found a triangle whose sides x, y, z and angle-bisectors are 

 rational. Let nx and ny be the segments of z made by the bisector of the 

 opposite angle; mx and mz those of y. Hence y = (1 -f ri)mx/(l mri), 

 z (1 + m)nx/(l mri). The square of the bisector of angle (x, y) is 



98 The Gentleman's Math. Companion, London, 5, No. 27, 1824, 344-9. 



99 Ibid., 5, No. 30, 1827, 588-9. 



100 The Gentleman's Diary, or Math. Repository, London, 1840, 33-5, Quest. 1400. 



