Sources page biographical material

6.BL. TAN^-1 ⅓ + TAN^-1 ½ = TAN^-1 1, ETC.
This problem is usually presented with three squares in a row with lines drawn from one corner to the opposite corners of the squares. New section. Similar formulae occur in finding series for π. See 6.A and my Chronology of π.
L. Euler. Introductio in Analysin Infinitorum. Bousquet, Lausanne, 1748. Vol. 1, chap. VIII, esp. § 142 (??NYS). = Introduction to the Analysis of the Infinite; trans. by John D. Blanton; Springer, NY, 1988-1990; Book I, chap. VIII: On transcendental quantities which arise from the circle, pp. 101-115, esp. § 142, pp. 114-115. Developing series to calculate π, he considers angles a, b such that a + b = π/4, then examines the formula for tan (a + b) and says: "If we let tan a = ½, then tan b = ⅓ .... In this way we calculate ... π, with much more ease than ... before." Conway & Guy give some more details.

Carroll. ?? -- see Lowry (1972) and Conway & Guy (1996).

Størmer. 1896. See Conway & Guy.

Gardner. SA (Feb 1970) = Circus, chap. 11, prob. 3. Says he received the geometric problem from Lyber Katz who had been given it when he was in 4th grade in Moscow.

C. W. Trigg. A three square geometry problem. JRM 4:2 (Apr 1971) 90 99. Quotes a letter from Katz, dating his 4th year as 1931-32. Trigg sketches 54 proofs of the result, some of which generalize.

H. V. Lowry. Note 3331: Formula for π/4. MG 56 (No. 397) (Oct 1972) 224-225. tan ¹ 1/a  = tan^-1 1/b + tan^-1 1/c implies a(b+c) = bc - 1, hence (b a)(c a)  =  a² + 1, whence all integral solutions can be determined. Conway & Guy say this was known to Lewis Carroll.

J. R. Goggins & G. B. Gordon. Note 3346: Formula for π/4 (see Note 3331, Oct 1972). MG 57 (No. 400) (Jun 1973) 134. Goggins gets π/4 = Σ_n=1 tan^-1 1/F_2n+1, where F_n is the n th Fibonacci number. [I think this formula was found by Lehmer some years before??] Gordon also mentions Eureka No. 35, p. 22, ??NYS, and finds recurrences giving tan^-1 1/p_n + tan^-1 1/q_n = tan^-1 1/r_n.

Douglas A. Quadling. Classroom note 304: The story of the three squares (continued). MG 58 (No. 405) (Oct 1974) 212 215. The problem was given in Classroom note 295 and many answers were received, including four proofs published by Roger North. Quadling cites Trigg and determines which proofs are new. Trigg writes that tan ¹ 1/F_2n+2 + tan^-1 1/F_2n+1 = tan^-1 1/F_2n, which is the basis of Goggins' formula.

Alan Fearnehough. On formulas for π involving inverse tangent functions and Prob. 23.7. MS 23:3 (1990/91) 65-67 & 95. Gives four basic theorems about inverse tangents leading to many different formulae for π/4. The problem gives a series using inverse cotangents.

John H. Conway & Richard K. Guy. The Book of Numbers. Copernicus (Springer-Verlag), NY, 1996. Pp. 241-248 discusses relationships among values of tan^-1 1/n which they denote as t_n and call Gregory's numbers. Euler knew t = t₂ + t₃, t₁ = 2 t₃ + t₇ and t₁ = 5 t₇ + 2 t₁₈ - 2 t₅₇ and used them to compute π to 20 places in an hour. They say Lewis Carroll noted that t_n = t_n+c + t_n+d if and only if cd = n² + 1. In 1896, Størmer related Gaussian integers to Gregory numbers and showed how to obtain a Gregory number as a sum of other Gregory numbers. From this it follows that the only two-term expressions for π/4 are t₂ + t₃, 2 t₂ - 4 t₇, 2 t₃ + t₇ and 4 t₅ - t₂₃₉. This is described in Conway & Guy, but they have a misprint of 8 for 18 at the bottom of p. 246.