CHAP. IV] PROBLEMS ON AREAS OF RIGHT TRIANGLES. 177 



Fermat asked that the sum of the hypotenuse and one leg, diminished 

 by the area, shall be 4; the answer (17/3, 15/3, 8/3) is given in the Inventum 

 Novum, III, 33 (Oeuvres de Fermat, III, 389). Bachet found a right 

 triangle whose area increased (or decreased) by the hypotenuse is 4. 



Diophantus VI, 13 relates to a right triangle (px, bx, hx) whose area 

 increased by either leg is a square. Let A = pb/2. From Ax* + bx = m?x 2 , 

 x = 6/(m 2 A). Then Ax* + px = D requires that 



pbm* + Ab(b - p) = D. 



As in VI, 12, we may choose (p, b, h) similar to (3, 4, 5) so that the greater 

 leg b, b p and p + A are all squares, say b p = ra 2 . The preceding 

 condition is thus satisfied. Fermat's method (Oeuvres, III, 267) yields an 

 infinitude of triangles not similar to (3, 4, 5). 



Diophantus, VI, 15 [17], gave a right triangle (Sx, 15x, I7x) whose 

 area diminished [increased] either by the hypotenuse or one leg is a square. 

 Fermat 83 required that on subtracting the area from the hypotenuse or 

 one leg each difference be a square. 



Diophantus, VI, 19 [20], required a right triangle the sum of whose area 

 and hypotenuse is a square [cube], and perimeter a cube [square]. His 

 solution and various related papers are considered in Ch. XX. 



Diophantus, VI, 21 [22], required a right triangle the sum of whose 

 area and one leg is a square [cube], and perimeter a cube [square]. Use 

 a triangle given by the rule of Pythagoras, 1 after dividing its sides by 

 a + 1. The perimeter 4a + 2 is to be a cube. By the other condition, 

 2a + 1 = D. But 8 is the only cube which is double a square. Hence 

 a = 3/2. 



Diophantus VI, 23 [24] 84 relates to a right triangle the sum of whose 

 area and perimeter is a cube [square], and perimeter a square [cube]. 

 Use a triangle given by the rule of Plato. 1 The perimeter p = 2a 2 + 2a 

 is a square for a = 2/(m 2 2). Then a(a z 1) + p and hence 2m is to be 

 a cube for 2 < m 2 < 4, which is the case when m = 27/16. 



Bhdscara 12 found a right triangle whose area equals the hypotenuse. 



C. G. Bachet, at the end of book VI of his edition of Diophantus, added 

 22 problems. In the first 13, we are given the perimeter, or hypotenuse or 

 area of a rational right triangle and seek the maximum or minimum of some 

 specified function of the sides. In 14-18, we seek the sides, given the sum 

 of the legs or perimeter p, or p and the area A, or p and the product of the 

 sides. In 19, p and p A are to be squares. In 21 and 22, we are given 

 p or A and the perpendicular from the right angle to the hypotenuse. 



J. de Billy 85 found a right triangle in which one leg, the sum of the legs, 

 and the excess of each leg over double the area are all squares. If x and 

 y = 1 x are the legs, the conditions are that y and x 2 + y 2 be squares, 

 as is true if x = 40/49. If we formulate the problem algebraically and then 



83 His solution is in Inventum Novum, I, 26, 40; Oeuvres, III, 341, 349. 



84 For VI, 24, see T. L. Heath, Diophantus, 1885, 236-7; 1910, 244-5; P. Mansion, Mathesis, 



(4), 4, 1914, 145-9. 



85 Inventum Novum, I, 52; Oeuvres de Fermat, III, 359. 



13 



