194 



HISTORY OF THE THEORY OF NUMBERS. 



[CHAP. V 



C. Gill 10 made a computation which (although not so stated) in effect 



consists in finding three right triangles A OF, BOF, COE (see the figure 



below) with integral sides such that OF = OE 

 = OD = r is the radius of the inscribed cir- 

 cle, but omitted the condition that the sum of 

 their angles at shall be two right angles. 

 If AF = m, BF = n, CE = s, this condition 

 is mns = r\m + n + s). Thus his solution 

 fails. 



A. Cook 11 gave the following solution. Draw 

 OR perpendicular to AO to meet AB at R. 

 [The sides of any rational right triangle are 

 proportional to r 2 a 2 , 2m.] Hence we may 



take 



AF = (r 2 - a 2 )/(2a), AO = (r 2 + a 2 )/(2a), BF = (r 2 - 1 



BO = (r 2 + 6 2 )/(26), OF = r. 



By similar triangles, 



AF:FO::FO:FR = 



2ar 2 



i"2 - - f\ 



AO : RO = 



r(r 2 + a 2 ) 



r-2 - 



Hence we have RB = .BF FR. Since angles 

 the same lettered triangles are similar. Hence 



and OCB are equal, 



BR:BO::RO 



d 



:: BO:BC = 



(r 2 + 6 2 ) 2 (r 2 - a 2 ) 



where d = (r 2 - a 2 )(r 2 - 6 2 ) - 4a&r 2 . Hence 



DC = BC - BD = 2r 2 {(r 2 - a 2 )6 + (r 2 - 6 2 )a}/d. 



We may assign any values to a, b and any value, exceeding a and 6, to r. 

 For o = 16, 6 = 18, r = 72, we get AF = 154, AO = 170, BF = 135, 

 50 = 153, OC = 120, CD = 96, A = 289, AC = 250, BC = 231. 

 Several 12 employed Heron's formula 



A 2 = (B + S + s)(B + S - s)( - 5 + )(- 5 + 5 + s) 



for the square of the area A of a triangle with sides 2B, 2S, 2s. T. Baker 

 wrote x, y, z for the last three factors of A 2 . Then A 2 = xyz(x + y + z). 

 Let A = (axz) 2 . We get x rationally. " A. B. L." took B = x y, S = x, 

 s = x + y; then 3x 2 - - 12?/ 2 = D, whence u 2 - 3v* = 1. C. Holt equated 

 the last three factors of A 2 to 4pV, (g 2 + r 2 - - p 2 ) 2 , 4pV; by addition, 

 B + S + s = (<? 2 + r 2 + p 2 ) 2 . J. Anderson equated the product of the 

 four factors to s 2 x 2 . Hence 



{B 2 - (S 2 + s 2 )} 2 = s 2 (4S 2 - z 2 ) = s 2 (2S - ?/) 2 , 

 say; hence we get S and then B z . 



10 Ladies' Diary, 1824, 43, Quest. 1416. 



11 Ladies' Diary, 1825, 34-5. 



12 The Gentleman's Math. Companion, London, 5, No. 27, 1824, 289-292. 



notations in Math. Mag., 2, 1898, 224-5. 



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