212 HISTORY OF THE THEORY OF NUMBERS. [CHAP, v 



xy(l n 2 ), which is a square if (1 n)m(l mri) = D = m 2 n 2 , say, 

 whence m = (1 n)/n. Then y = (1 n z )x/n 2 , z = x/n. Then the bi- 

 sectors of angles (x, z) and (y, z) are rational if 2n 2 n and 2n 2 + n are 

 squares. Equating the first to p 2 n 2 , we get n. Then 2n 2 -f- n = D if 

 4 ^2 _ rj _ (2 pg) 2 , which determines p. W. Rutherford called the 

 sides a, b, c; the square of the bisector AD of angle A equals 4bcs(s a)/ 

 (b + c) 2 , where s = (a + b + c)/2. Thus 6cs(s - a) = D. Similarly, 

 abs(s c) = D, acs(s 6) = D. Hence s(s a)(s 6)(s c) = D and 

 the area is rational. Thus the problem is that treated by Cunliffe. 98 



J. Davey 101 found a triangle ABC in which the sides, the angle-bisector 

 CD, the median CE, and the segments AE = EB and ED of the base are 

 all integers. Take 



AC=(m+l)p, BC=(m-l}p, AD=(m+l)q, 



Then AE = mq, ED = q, CD 2 = (m 2 - l)(p 2 - g 2 ). Take 



CD = (m 2 - l)(p - ff ), 



whence p = w 2 <?/(m 2 2). Then 

 C# 2 = (m 2 



Hence take 5m 2 4 to be the square of 5(m l)r/s 1, thus obtaining 

 m rationally. 



Feldhoff 102 treated 31 problems on triangles in which certain elements 

 (area, perimeter, side) are rational, are equal, or are squares. In the tri- 

 angle formed by the juxtaposition of two rational right triangles, the angle- 

 bisectors are rational if two expressions of the form x 2 + 1 are squares. 103 



Worpitzky 32 stated that, if the rational triangle with sides (1) has its 

 angle-bisectors rational, then p = /z 2 v 2 , q = 2ju*>, r = p 2 cr 2 , s = 2po-. 



D. Biddle 104 found special oblique triangles having integral values for 

 the sides, area, altitude from one vertex and bisector of the angle at that 

 vertex. Use is made of 3 right triangles with a common side. 



R. Chartres 105 and others found integral values for the sides and the 

 bisector g of the largest angle such that the perimeter equals mg. 



* P. Dolguin 106 gave examples, but no general solution, of the problem 

 to find all triangles whose area, bisectors, medians, etc. are all rational. 



101 The Lady's and Gentleman's Diary, London, 1842, 69. He noted that J. Holroyd's 



solution, 1841, 57-8, leads only to degenerate triangles whose base equals the difference 

 of the other sides. 



102 Einige Satze iiber das Rationale Dreieck, Progr., Osnabriick, 1860. 



IDS j? or [f 2rs = 2mn is the common side, so that the composite triangle has the sides 

 6 = m 2 + n 2 , c = r 2 + s 2 , a = m z n 2 + r 2 s 2 , then 



a + b + c = 2(r 2 + m 2 ), b + c - a = 2(s 2 + n 2 ). 



Hence the quantity under the radical in the expression for the bisector a (Fuss 95 ) is a 

 product of four sums of two squares and hence equals such a sum. In the expression for 

 the bisector occurs the square root of E = ac(a + b + c)(a + c 6) = 4(r 2 + s 2 ) 

 X (r 2 + TO 2 )o(r 2 n 2 ). Replacing s by mn/r in a, we get a = (r 2 n 2 ) (1 + m 2 /r 2 ). Hence 

 E is a sum of two squares. The product a/3y is rational since the area is rational. 



104 Math. Quest. Educ. Times, 57, 1892, 32. 



105 Ibid., 66, 1897, 102-3. 



108 Vest, opytn. Fiziki (Spaczinski's Bote), Odessa, 1903, No. 355, 145-157 (Russian). 



