CHAP. V] RATIONAL TRIANGLES. 195 



C. Gill 13 found integral sides x, y, z of a triangle the four diameters of 

 whose inscribed and escribed circles are integral squares r 2 and R 2 , R\, R 2 2 . 

 Take x + y + z = a 2 , y + 2 - x = fc 2 , z + 2 - y = c 2 , a; + 2/ - 2 = d\ 

 Let the condition a 2 = b 2 + c 2 + d 2 be satisfied. We get x = (a 2 - 6 2 )/2, 

 y, 0. It is known that 4A = r 2 a 2 = R 2 b 2 = Rlc 2 = Rid 2 . 



C. L. A. Kunze 14 derived eight rational triangles from the two rational 

 right triangles (3, 4, 5) and (5, 12, 13) by reducing their sides in proper 

 ratios so that any chosen leg of one shall equal any chosen leg of the other 

 and then juxtaposing the resulting triangles either with or without over- 

 lapping. Schlomilch 36 noted that we may start with any two rational 

 right triangles. 



C. F. Gauss, 14a whose attention had been called to Curtius' 2a problem 

 by Schumacher, stated that the sides of every triangle such that each side 

 and the radius r of the circumscribed circle are integers are of the form 



4abfg(a 2 + 6 2 ), 4a6(/ + flO( 8 / - W, 4ab(a 2 / 2 + Vf), 



where a, 6, /, g are positive integers, while r =(a 2 + 6*)(fltf 4 + &VO- We 

 obtain Curtius' numbers by taking a = g = 1, b = 2, / = 10, and deleting 

 the common factor 8. Many writers 146 derived Gauss' formula. 



E. W. Grebe 15 tabulated for 46 rational triangles the 12 rational values 

 of the segments of the altitudes and the segments of the sides cut off by the 

 altitudes. 



Grebe 16 gave a table of 496 rational triangles, showing also the area, 

 perimeter, altitudes, and diameter of the circumscribed circle. He began 

 with 32 rational right triangles (4, 3, 5), -, (195, 28, 197) with small 

 ratios of sides, took each pair of these triangles and multiplied their sides 

 by such factors as produce two triangles whose larger legs are equal. By 

 juxtaposition he formed a rational acute triangle. 



To find a triangle with integral sides whose area and perimeter are 

 equal, B. Yates 17 took, in accord with (1), the sides to be pq(r 2 + s 2 )/n, 

 rs(p 2 + q z )/n, (ps + qr)(pr qs}/n. The latter multiplied by pqrs/n is the 

 area. Equating the area to the perimeter 2pr(ps + qr)fn, we get 



qs(pr qs) = 2n. 



Integral solutions are found when n = 1, 2, 8. Many solvers used the 

 segments I, m, n into which the sides a, b, c are divided at the points of 

 contact of the inscribed circle of radius r. Thus I + m = a, I + n = b, 

 m + n = c. If s is the semi-perimeter, rs = 2s, whence r = 2. But 

 rV = slmn. Hence 4(7 + m + n) = Imn. The least side exceeds 2r = 4. 

 Hence we may take I + m = 5, 6, and find integral solutions. 



13 The Gentleman's Math. Companion, London, 5, No. 29, 1826, 509-512. 



14 Lehrbuch der Geometric, Jena, 1842, 205. 



140 Briefwechsel zwischen C. F. Gauss and H. C. Schumacher (ed., C. A. F. Peters), Altona, 



5, 1863, 375; letter of Oct. 21, 1847. Quoted in Archiv Math. Phys., 44, 1865, 504-6. 

 "* Archiv Math. Phys., 45, 1866, 220-231. 

 16 Eine Gruppe von Aufgaben iiber das geradlinige Dreieck, Progr., Marburg, 1856. 



16 Zusammenstellung von Stxicken rationaler ebener Dreiecke, Halle, 1864, 248 pp. 



17 The Lady's and Gentleman's Diary, London, 1865, 49-50, Quest. 2019. 



