CHAP. IV] PROBLEMS ON AREAS OF RIGHT TRIANGLES. 175 



Diophantus took (3, 4, 5) as one triangle and stated that it is easy to find 

 two triangles such that the product of the legs of one is 12 (or 3) times that 

 of the other, as (9, 40, 41), (8, 15, 17). C. G. Bachet 74 " chose an arbitrary 

 triangle (pi, 61, hi) and the two triangles formed 7 from 61, hi and p 1} hi, 

 obtaining s = pi/(2hi). Fermat 75 gave general rules for finding two right 

 triangles whose areas are in a given ratio r/s, where r > s, viz., form the 

 triangles from 2r d= s, r =F s and 2s d= r, r =F s; or from 6r, 2r s and 

 4 r _j_ S} 4r 2s; or from r + 4s, 2r 4s and 6s, r 2s. Thus to find 

 three right triangles whose areas are proportional to given numbers r, s, t, 

 such that r + t = 4s, r > t, form the triangles from r + 4s, 2r 4s; 

 6s, r 2s ; 4s + t, 4s 2t. The areas of the triangles formed from 49, 2 ; 

 47, 2; 48, 1 are themselves the sides of a right triangle. 76 



L. Euler 77 found ten types of pairs of right triangles whose areas 

 A = pq(p 2 q 2 } and B = rs(r 2 s 2 ) have a given ratio a : b. He equated 

 r and s to two of the numbers p, 2p, q, 2q, p db q. For example, r = p, 

 s = p q give p -}- q :2p q = a :b, whence p : q = a + b :2a b; 

 taking r = p = a + b, we get q = 2a b, s = 2b a. He gave (pp. 

 222-3) several methods to make A/B a square (cf. Euler 33 of Ch. XV, Euler 81 

 of Ch. XVI, Euler 18 - 19 of Ch. XVIII and Euler 253 of Ch. XXII). 



A. Holm 78 noted that the problem leads to a cubic curve with two given 

 rational points, whence the chord determines a third. 



OTHER PROBLEMS INVOLVING ONLY AREA. 



An anonymous 79 Greek manuscript, probably dating between Euclid 

 and Diophantus, found the sides of a right triangle with the area 5 by 

 seeking a product of 5 and a square 36, divisible by 6, such that the product 

 5-36 is the area of a right triangle with the sides 9, 40, 41, and reduced 

 them in the ratio 1 : 6, which shows a knowledge of the fact that the area of 

 a right triangle with integral sides is a multiple of 6 (L. Pisano, 14 Ch. XVI). 



Diophantus, VI, 3, required a right triangle whose area increased by a 

 given number g yields a square. Take g = 5 and denote the triangle by 

 (hx, px, bx) ; we are to choose x so that ^pbx* + 5 = n?x 2 . Let (h, p, b) 

 be formed from m, 1/m and take n = m + 2-5/m. Then %pb = m 2 1/m 2 . 

 When this is subtracted from n 2 , the difference shall be 5 times a square. 

 Hence 100m 2 + 505 = D, say (10m + 5) 2 . Thus m = 24/5, n = 413/60, 

 x = 24/53. F. Vieta (Zetetica, V, 9) took g = r 2 + s 2 , formed the triangle 

 from (r + s) 2 , (r s) 2 , and divided its sides by 2(r + s)(r s) 2 ; the area 

 is now 2rs(r 2 + s 2 )/(r s) 2 , which added to g yields the square of (r 2 + s 2 )/ 

 (r s) . C. G. Bachet 74a remarked that g need not be the sum of two squares 



740 Diophanti Alex. Arith. . . . Commentariis . . . Avctore C. G. Bacheto, 1621, 333. 



75 Oeuvres, I, 319; French transl., Ill, 259. Cf., II, 224-6. 



76 Other solutions, Oeuvres de Fermat, II, 93, 250, 277; Oeuvres de Descartes, II, 165. 



De Billy gave the triangles formed from 6, 1; 7, 6; 8, 1; Oeuvres de Fermat, IV, 1912, 

 139; Bull. Bibl. Storia Sc. Mat. Fis., 12, 1879, 517. 



77 Opera postuma, 1, 1862, 224-7 (about 1773). 



78 Proc. Edinburgh Math. Soc., 22, 1903-4, 48. 



79 With German transl. by J. L. Heiberg and comments by H. G. Zeuthen, Bibliotheca Math., 



(3), 8, 1907-8, 121-131. 



