CHAP. IV] PKOBLEMS ON AREAS OF RIGHT TRIANGLES. 181 



G. R. Perkins 104 noted that the triangles (40, 30, 50), (45, 24, 51), 

 (48, 20, 52) have equal perimeters and areas 600, 540, 480 in arithmetical 

 progression. 



V. J. Knisely 105 found the same result as had Perkins, by taking as the 

 sides 



(p 2 + 2pq)a, (2pq + 2q*)a, (p 2 + 2pq + 2q^a, 

 (p 2 - <? 2 )6, 2pqb, (p 2 + q^b, 



(p 2 4g 2 )c, 4p#c, (p 2 + 4g 2 )c. 



The conditions reduce to 



(p + 2q)a = pb, (p + q)a = pc, 2(p - q}b = pa + 2pc 4qc. 



Substitute into the third the values of b, c given by the first two conditions; 

 we get p = 4#, whence b = 6a/4, c = 5a/4. For q = 1, p = a = 4, c = 5, 

 6 = 6, we get the answer cited. A. B. Evans gave a long discussion said 

 to give the complete solution; but his numerical example involves very 

 large numbers. 



E. Lucas proposed and Moret-Blanc 106 solved the problems to find a 

 right triangle such that the square of the hypotenuse increased or diminished 

 by the area (or by double the area) is a square. 



Lucas 107 showed that the method of descent leads to a complete solution 

 of the second (double area) of the last two problems. 



C. de Comberousse 108 discussed rational right triangles whose area and 

 perimeter are equal. Eliminating z between x 2 -\- y 2 = z 2 and 



x + y + z = xy/2, 



we get y = 4 + S/(x 4). Thus x 4 is a divisor of 8, and the only 

 solutions are (x, y, z) = (5, 12, 13), (6, 8, 10). 



A. Holm 109 discussed a problem including the cases of Diophantus VI, 

 6-11, and the additions by Bachet and Fermat: Find a rational right 

 triangle such that the sum of given multiples of the area and three sides 

 shall be a given number. Taking (x 2 l)/jy, 2x/y as the sides, the con- 

 dition is 



= e - 

 V y y y J 



The discriminant of this quadratic for y is a quartic function Q(x) in which 

 the coefficient of x 4 and the constant term are squares. There are many 

 known methods of making Q(x) a square. 



RIGHT TRIANGLES WHOSE LEGS DIFFER BY UNITY. 



A. Girard 109a gave fourteen such triangles in which the least leg is 3, 20, 

 119, 696, 4059, 23660, 137903, 803760,- , 31509019100. 



104 The Analyst, Des Moines, 1, 1874, 151-4. Cf. Calculator. 98 



105 Math. Quest. Educ. Times, 20, 1874, 81-3. 



106 Nouv. Ann. Math., (2), 14, 1875, 510; (2), 20, 1881, 155-160. 



107 Bull. Bibl. Storia Sc. Mat., 10, 1877, 291-3. 



108 Algebre sup4rieure, 1, 1887, 190-1. 



109 Proc. Edinburgh Math. Soc., 22, 1903-4, 45-8; Math. Quest. Educ. Times, (2), 10, 



1906, 47-8. 

 1090 L'arith. de Simon Stevin par A. Girard, 1625, 629; Oeuvres, 1634, 158, col. 1. 



