CHAP. V] RATIONAL QUADRILATERALS. 217 



From one inscriptible quadrilateral we get two others (but not with per- 

 pendicular diagonals) by permuting the sides. The area of each of the 

 three quadrilaterals is the product of the three distinct diagonals divided 

 by double the area of the circumscribed circle (A. Girard; proof by Grebe, 

 Manuel de G<k>m., 1831, 435). 



L. N. M. Carnot 132 noted that the segments of the diagonals of a quad- 

 rilateral are expressible rationally in terms of the sides and diagonals. 



E. E. Kummer 133 noted that Chasles unriddled the obscurity of Brah- 

 megupta without perceiving the method used by the latter, and expressed 

 Brahmegupta's theorem in the following form. If the four sides of a 

 quadrilateral, inscriptible in a circle, have the values 



(a 2 + 6 2 )(c 2 - <2 2 ), (a, - 6 2 )(c 2 + d 2 }, 2cd(a* + 6 2 ), 2a6(c 2 + d 2 ), 



where a, 6, c, d are rational, then both diagonals (perpendicular to each 

 other), the segments of them, the area of the quadrilateral and the diameter 

 of the circumscribed circle are all rational. 



Kummer showed how to obtain all rational 

 quadrilaterals. Let ABCD have rational sides and 

 diagonals. Then the segments a, ft, 7, 5 of the 

 diagonals are rational. For, by 



6 2 = a 2 + AC 2 - 2a-AC cos u, 



cosu is rational; likewise cosy and cos (u + v). 

 Hence sin u sin v is rational; also sin 2 u and there- A > 

 fore sin u/sm v. But 



a _ sin w d sin w ft a sin u 

 /3 sinu' d sin v ' d d sin v' 



Hence /3/5, 1 + /3/5 = BD/d, 8 and /3 are rational. Similarly, a and 7 are 

 rational. Next, c = cos w is rational, in view of 



(1) a 2 = a 2 + /3 2 - 2a(3c. 



Set c = mjn, where m, n are relatively prime. Without loss of generality, 

 we may assume that a, a, (3 are integers with no common factor. To treat 

 one of two analogous cases leading to like results, let n be odd. Then n 

 must divide a/3. Thus a = rai, = sft, n = rs, 



(2) a 2 = r 2 al + s 2 /3? - 2ma 1 j3 l . 



Now i, @i are relatively prime, since a common factor would divide a. 

 We may take ft odd. The product of (2) by r 2 may be given the form 



^1^2 = (n 2 - m 2 )/3?, Fi = ar + r 2 ^ - wft, F 2 = ar - r 2 i + raft. 



If FI and F 2 were both divisible by a prime factor p of ft, then 2r 2 ai and 

 hence ra\ would be divisible by p, likewise a by (2), whereas a, a, /3 do not 



182 G6om6trie de position, Paris, 1803, 391-3. 

 133 Jour, fur Math., 37, 1848, 1-20. 



