CHAP. V] RATIONAL PYRAMIDS; RATIONAL TRIHEDRAL ANGLES. 223 



The next problem is to find rational solutions of 



O s = 02 + h 2 _ 2g h cos a, 6 2 = h 2 +f 2 - 2hfcos p, c 2 = / 2 + g 1 - 2fg cos 7, 

 where the cosines are given rational numbers. Set 



a = \g + h, b = rf + h, c = vg + f. 

 Then 



^(l - x 2 ) = 2h(\ + cos a), /(I - /i 2 ) = 2h(p + cos 0), 



0(1 - ? 2 ) = 2/(*> + cos 7). 

 Hence g/f has the value 



1 M 2 X + cos a 2(v + cos 7) 

 g : : 1 - X 2 ' /* + cos j8 = 1 - ? 2 



If cos a = cos j(3 = cos 7 = 0, we have a rectangular tetrahedron and 

 the problem reduces to that treated by Euler 3 of Ch. XIX to find three 

 squares such that their sums by pairs are squares. This process of Euler 

 leads in the general problem to 



P 2 (l + cos 7) + 2p(cos a + cos ft) + 1 ~ cos 7 + 2 cos a cos ft 



4(p + cos /3) 



For example, let M = N = 0, P = Q = 3, M 1 = PI = 2, NI = - 1, 

 Qi = 3. Then 



2 - 1 p 2 + 2p + 2 



coso:==0, cos/3 ==-, cos7 : -3-, -X= 6 +4 - 



Thus /, gr, /i are proportional to (6p + 4)(p 2 2p 4)(5p 2 + 2p 2), 

 6(6p + 4)(p 2 - l)(p 2 + 2p + 2), 3(p 2 - l)(p 2 - 4p - 2)(p 2 + 8p + 6). 

 For p = 0, we remove the factor 4, and get / = 8, g = 12, ^ = 9, 

 a = 15, & = 7, c = 12, V = 96, in which the signs may be taken positive. 

 For p = - 2 we get/ = 112, g = 72, h = 135, a = 153, b = 103, c = 152, 

 V = 120960. 



To obtain a rational quadrilateral, set + 7 = a or 2?r a. For 

 example, for cos a = cos /3 = cos 7 = ^, we have 



/ = (7p 2 - 4)(p 2 - 4)(2p - 1), flr = 8(p 2 - l)(p 2 + 2)(2p - 1), 



A = p(p 2 - l)(p + 4)(p 2 - 12p + 8). 



Thus, for p = %, we have the rational quadrilateral ABCD, in which 

 AB = 138, BC = 192, CD = 168, DA = 127, AC = 283, D = 120. 



We obtain a simpler solution by taking X = ^. Thus, for cos a = 3/7, 

 cos j8 = 0, cos 7 = 2/7, ^ = 2, we have X = 3 and / = 6, g = 7, h = 8, 

 a = 9, 6 = 10, c = 11, F = 48. For cos a = cos 7 = i, cos = , 

 ? = 2, we get the rational quadrilateral AB = 48, BC = 57, CD = 73, 

 DA = 80, AC = 63, BZ> = 112. 



H. Schubert 47 (pp. 50-7, or Schubert, 88 92-104) employed a rational 

 polygon inscribed in a circle of radius r and center C. Draw a perpendicular 

 to its plane at C and of length h such that in the right triangle of legs h 

 and r the angle opposite h is a Heron angle 47 /z. Thus we have a rational 



