CHAP. V] RATIONAL TRIANGLES. 201 



the sides x* + y\ (1 + y~)x, c = (1 + x)(y* x). Conversely, if x, y, 

 y 2 x are positive, these numbers are the sides of a triangle, of area cxy. 



E. Turriere 61 noted several methods to find Heron triangles. There is 

 an infinitude of Heron triangles with sides in arithmetical progression such 

 that no two are similar. He investigated Heron triangles in which the semi- 

 perimeter p and p a, p b, p c are all rational squares, and the 

 analogous problem for inscriptible quadrilaterals. He 62 found Heron tri- 

 angles in which the sum of the squares of two sides is a square. 



F. R. Scherrer 620 made use of the theory of complex integers a + bi to 

 obtain the coordinates of the vertices, of the centers of the circumscribed, 

 inscribed, escribed and Feuerbach circles, of the intersection of the alti- 

 tudes, etc., of primitive Heron triangles. Cf. Simerka. 21a 



M. Rignaux 63 stated the final formulas of Schubert. 47 

 E. T. Bell stated and W. Hoover 64 proved incompletely that if u = 2, 

 Ui = 4, , u n+2 = 4w n+ i u nt then u n 1, u n , u n + 1 are the consecu- 

 tive sides of a triangle with integral area, and all such triangles are given 

 by this method. 



PAIRS OF RATIONAL TRIANGLES. 



Frans van Schooten 5 found two isosceles rational triangles with equal 

 perimeters and equal areas. Divide each into halves and let the right 

 triangles be formed from a, b and k, d respectively. By the perimeters, 



2(a 2 + 6 2 ) + 2(2a6) = 2(k 2 + d 2 ) + 2(2kd), a + b = k + d. 

 Set k = a + x, d = b x. The equality of the areas requires 



2z 2 + 3(a - b)x + a 2 - 4ab + b 2 = 0, x = %(r + 36 - 3a), 

 where r 2 = a 2 + 6 2 + 14a6. Set r = a + b -f c. Thus 



c 2 + 2bc 

 a ~ 12b-2c' 



The general solution thus involves the parameters b, c. For b = 1, c = 3, 

 we get a = 5/2, x = 1/2. Multiply the sides by 4. We get the right 

 triangles (20, 21, 29) and (12, 35, 37). Their doubles have the perimeter 

 98 and area 420. 



J. H. Rahn 65 devoted 8 pages to this problem, and J. Pell 62 pages. 

 There is first given the above solution by van Schooten, attributed to 

 Descartes. 



Several 66 gave straightforward solutions to van Schooten's problem. 



J. Cunliffe 67 treated the problem to find two triangles with rational 

 altitudes and segments of sides and with equal perimeters and equal areas. 



61 L'enseignement math., 18, 1916, 95-110. 



62 Ibid., 19, 1917, 259-261. Cf. Euler 21 of Ch. IV. 



620 Zeitschrift Math. Naturw. Unterricht, 47, 1916, 513-30. 



63 L'intermediaire des math., 24, 1917, 86. 



64 Amer. Math. Monthly, 24, 1917, 295, 471. Cf. Hoppe. 29 



65 Algebra, Zurich, 1659. Engl. transl. by T. Brancker, augmented by D. P., London, 1668, 



131-192. 



66 The Gentleman's Math. Companion, London, 5, No. 26, 1823, 183-5. 



67 New Series of the Math. Repository (ed., Th. Leybourn), 2, 1809, II, 54-7. 



