194 HISTORY OF THE THEORY OF NUMBERS, [CHAP, v

C. Gill10 made a computation which (although not so stated) in effect consists in finding three right triangles AOF, EOF, COE (see the figure

below) with integral sides such that OF = OE = OD = r is the radius of the inscribed circle, but omitted the condition that the sum of their angles at 0 shall be two right angles. If AF = m, BF = n, CE = s, this condition is mns = r2(m + n + s). Thus his solution fails.

A. Cook11 gave the following solution. Draw OR perpendicular to AO to meet AB at R. ^ [The sides of any rational right triangle are a proportional to r2 ± a2, 2ra.] Hence we may take AF = (r2 - a2)/(2a), AO = (r2 + a2)/(2a), BF = (r2 - 62)/(26),

BO = (r2 + &2)/(2&), OF = r. By similar triangles,

Hence we have RB = BF ~ FR. Since angles jB0# and OC5 are equal, the same lettered triangles are similar. Hence

where d = (r2 - a2)(r2 - 62) - 4dbr2. Hence

DC = EG - BD = 2r2{(r2 - a2)& + (r2 - 62)a}/d

We may assign any values to a, 6 and any value, exceeding a and 6, to r. For a = 16, 6 = 18, r = 72, we get AF = 154, AO = 170, BF = 135, BO = 153, OC = 120, CD = 96, A£ = 289, AC = 250, BC = 231. Several12 employed Heron's formula

A2 = (S + S + s)(B + £ - s)(.B - S + s)(- B + S + s)

for the square of the area A of a triangle with sides 25, 25, 2s. T. Baker wrote x, y, z for the last three factors of A2. Then A2 = 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 3.T2 - 12y2 = D, whence uz - 3^ = 1. C. Holt equated the last three factors of A2 to 4pY, (c?2 + r2 - p2)2> 4p2r2; by addition, B + S + s = (<f + r2 + p2)2. J. Anderson equated the product of the four factors to s2z2. Hence

{B2 - (S2 + s2) }2 = s2(4S2 - x2) = s2(2^f - y)2, say; hence we get S and then B2.

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

11 Ladies' Diary, 1825, 34r-5.

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

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