CHAP. IV] RIGHT TRIANGLES OF EQUAL AREA. 173 



the second part of Frenicle's 52 Traite; this process has been summarized 

 by A. Cunningham. 63 



Fermat 64 stated that he could give five right triangles of equal area and 

 had a method to find as many as one pleases, whereas Diophantus, V, 8, 

 and Vieta, Zetetica, IV, 11, gave only three. 



J. de Billy 65 noted that the right triangle with the legs 3r, (x + 4)r 

 will have the same area 6 as (3, 4, 5) if f (a? + 4)r 2 = 6. Thus x + 4 and 

 9 + (x + 4) 2 must be ; squares, which is the case if x = - 6725600/2405601. 



John Kersey 66 discussed the problem to deduce a rational right triangle 

 with the same area as a given one, and stated many problems on areas. 



L. Euler 67 discussed the solution of 



pr(p 2 r 2 } = qs(q 2 s 2 ), 



noting the case p = 11, r = 35, q = 23, s = 33. Hence the right 

 triangles formed 7 from 11, 35 and 23, 33 have equal areas. 



Euler 68 noted that if we take q = p, p 2 = r 2 + rs + s 2 , we get 



P / 2 + 3</ 2 



2r + s = V4p 2 -3s 2 = 2p - sf/g if - = 



s 



Take p = f 2 + 3g 2 , s = 4fg. Hence the values x = f 2 + 3g 2 , y = 4fg or 

 3<7 2 / 2 2fg give the sides 2xy, x 1 y 2 , x 2 + y 2 of three right triangles 

 with the same area xy(x 1 y 2 ). 



Grttson 22 (pp. 109-114) and Young 134 of Ch. XIX discussed the deter- 

 mination of three right triangles of equal area. 



J. Collins 69 employed the three right triangles with the legs 



v 2 x 2 , 2vx; v 2 y 2 , 2vy; z 2 v 2 , 2zv. 



The first two are of equal area if v 2 = x 2 + xy + y z . Set v = x t. 

 Then x = (t 2 y 2 )/(y + 2). The first and third have equal areas if 

 tf = x 2 - xz + z 2 . Set v = x - - s. Then x= (s 2 -- z 2 )/(2s - z). To make 

 the values of x equal, take t = m + n, y = m n, s = p + q, z = p q. 

 Then wn/(3w + ri) = pq/(3p + q) determines m in terms of p, q, n. 



J. Cunliffe 70 treated the problem to find k rational right triangles of 

 equal areas. For k = 3, let m 2 d= n 2 , 2mn be the sides of one triangle. In 



mn(m 2 n 2 } = pq(p 2 q 2 ), 



set p = m + r, q = n r, and solve the resulting quadratic for r. Thus 

 4r = db VR 3(m n}, where R = m 2 + 14mn + n 2 . Set 



R = ( m + n + s) 2 , 



63 Math. Quest. Educ. Times, 72, 1900, 31-2. 



64 Oeuvres, II, 263, letter to Mersenne, Sept. 1, 1643. He had asked (p. 259) for four. 



65 Inventum Novum, I, 38, Oeuvres de Fermat, III, 348. In his Diophantus Geometra, 



Paris, 1660, 108, 121, de Billy treated the problems of Diophantus V, 8, VI, 3. 



66 The Elements of Algebra, London, Books 3 and 4, 1674, 94, 124-142. 



67 Nova Acta Acad. Petrop., 13, 1795 (1778), 45; Comm. Arith., II, 285. 



68 Opera postuma, 1, 1862, 250-2 (about 1781). 



69 The Gentleman's Math. Companion, 2, No. 11, 1808, 123. 



70 New Series of the Math. Repository (ed., Th. Leybourn), 3, II, 1814, 60. 



