Sources page biographical material

6.BJ. 3D DISSECTION PUZZLES
This will cover a number of cases which are not very mathematical. I will record just some early examples. See also 6.G (esp. 6.G.1), 6.N, 6.W (esp. 6.W.7), 6.AP for special cases. The predecessors of these puzzles seem to be the binomial and trinomial cubes showing (a+b)³ and (a+b+c)³, which I have placed in 6.G.1. Cube dissections with cuts at angles to the faces were common in the 19C Chinese puzzle chests, often in ivory. I have only just started to notice these. It is hard to distinguish items in this section from other burr puzzles, 6.W.7, and I have tried to avoid repetition, so one must also look at that section when looking at this section.
Catel. Kunst-Cabinet. 1790. Der Vexierwürfel, p. 11 & fig. 32 on plate II. Figure shows that there are some cuts at angles to the faces, so this is not an ordinary cube dissection, but is more like the 19C Chinese dissected cubes.

C. Baudenbecher. Sample book or catalogue from c1850s. Op. cit. in 6.W.7. One whole folio page shows about 20 types of wooden interlocking puzzles, including most of the types mentioned in this section and in 6.W.5 and 6.W.7. Until I get a picture, I can't be more specific.

Slocum. Compendium. Shows: Wonderful "Coffee Pot"; Magic "Apple"; Magic "Pear"; Extraordinary "Cube"; Magic "Tub" from Mr. Bland's Illustrated Catalogue of Extraordinary and Superior Conjuring Tricks, etc.; Joseph Bland, London, c1890. He shows further examples from 1915 onward.

Hoffmann. 1893. Chap. II, pp. 107-108 & 141-142 = Hoffmann-Hordern, pp. 106-107, with photo.

No. 37: The Fairy Tea-Table. Photo on p. 107 shows a German example, 1870-1895.

No. 38: The Mystery. Photo on p. 107 shows a German example, 1870-1895, with instructions.

Western Puzzle Works, 1926 Catalogue.

No. 5075. Unnamed -- Fairy Tea-Table.

Last page shows 20 Chinese Wood Block Puzzles, High Grade. These are unnamed, but the shapes include various burr-like objects, cube, spheres, egg, barrel, tankard, pear and apple.

P. M. Grundy. A three-dimensional jig-saw. Eureka 7 (Mar 1942) 8-10. Consider a 2 x 2 x 2 array of unit cubes. He suggests removing and/or adding lumps on the interior faces to make a jig-saw. He then considers lumps in the form of a isosceles right triangular prism with largest face being a unit square. He finds there are 25 such pieces, subject to the conditions that there are most two removals, and that when there are two, they must be parallel. He gives a graphical view of a 2 x 2 x 2 formed from such pieces which gives some necessary, but not sufficient, conditions for a set of eight such pieces to be able to make a cube. One solution is shown. [If one assumes there is just one removal and one addition, I find just four pieces, which form a 1 x 2 x 2 block. One could view this as a 3-dimensional matching puzzle, where the internal faces have to match both like a head to tail matching, but also with correct orientation. MacMahon thought of notching pieces as an alternative to colouring edges, but his pieces were two-dimensional.]

6.BK SUPERELLIPSE
New section.
Gardner. SA (Sep 1965) = Carnival, chap. 18. Describes how the problem arose in the design of Sergel's Square, Stockholm, in 1959. Addendum in Carnival gives the results given by Gridgeman, below. Also says road engineers used such curves with n = 2.2, called '2.2 ellipses', from the 1930s for bridge arches.

N. T. Gridgeman. Lamé ovals. MG 54 (No. 387) (Feb 1970) 31 37. Lamé (c1818) seems to be the first to consider (x/a)ⁿ + (y/b)ⁿ = 1. Hein's design in Stockholm uses a/b = 6/5 and n = 5/2. Gerald Robinson used a/b = 9/7 and n = 2.71828..., which he determined by a survey asking people which shape they liked most. Gridgeman studies curvature, area, perimeter, evolutes, etc.

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.

6.BM. DISSECT CIRCLE INTO TWO HOLLOW OVALS
Consider a circle of radius 2. Cut it by two perpendicular diameters and by the circle of radius 1 about the centre. Two of the outer pieces (quarters of the annulus) and two of the inner pieces (quadrants) make an oval shape, with a hollow in the middle. The problem often refers to making two oval stools and the hollows are handholds! After the references below, the problem appears in many later books.
Jackson. Rational Amusement. 1821. Geometrical Puzzles, no. 9, pp. 25 & 84 & plate I, fig. 6. Mentions handholes. Solution is well drawn.

Endless Amusement II. 1826? Mentions handholes. Solution is well drawn.

Crambrook. 1843. P. 4, no. 6: A Circle to form two Ovals. Check??

Magician's Own Book. 1857. Prob. 36: The cabinet maker's puzzle, pp. 277 & 300. Solution is a bit crudely drawn. = Book of 500 Puzzles, 1859, pp. 91 & 114. = Boy's Own Conjuring Book, 1860, prob. 35, pp. 240 & 265.

Family Friend (Dec 1858) 359. Practical puzzle -- 4. I don't have the answer.

The Secret Out. 1859. The Oval Puzzle, pp. 380-381. Asks to 'produce two perfect ovals.' Solution is a bit crudely drawn, as in Magician's Own Book, but the text and numbering of pieces is different.

Illustrated Boy's Own Treasury. 1860. Prob. 34, pp. 401 & 441. Same crude solution as Magician's Own Book, but with different text, neglecting to state that the stools have handholes in their centres.

Magician's Own Book (UK version). 1871. On p. 282, in the middle of an unrelated problem, is the solution diagram, very poorly drawn -- the pieces of the oval stools are shown as having curved edges almost as though they were circular arcs. There is no associated text.

Hanky Panky. 1872. The oval puzzle, p. 123. Same crude solution as Magician's Own Book, but different text, mentioning handholes.

Mittenzwey. 1880. Prob. 256, pp. 46 & 97; 1895?: 285, pp. 50 & 99; 1917: 285, pp. 45 & 94. The stools are very poorly drawn, with distinct wiggles in what should be straight lines.

Hoffmann. 1893. Chap. II, no. 32: The cabinet maker's puzzle, pp. 104 & 137 138 = Hoffmann-Hordern, p. 102. Mentions hand holes. Well drawn solution.

Benson. 1904. The cabinet maker's puzzle, p. 201. Mentions hand holes. Poor drawing.

6.BN. ROUND PEG IN SQUARE HOLE OR VICE VERSA
Wang Tao K'un. How to get on. Late 16C. Excerpted and translated in: Herbert A. Giles; Gems of Chinese Literature; 2nd ed. (in two vols., Kelly & Walsh, 1923), in one vol., Dover, 1965, p. 226. "... like square handles which you would thrust into the round sockets ..."

Sydney Smith. Sketches of Moral Philosophy. Lecture IX. 1824. "If you choose to represent the various parts in life by holes upon a table, of different shapes, -- some circular, some triangular, some square, some oblong, -- and the persons acting these parts by bits of wood of similar shapes, we shall generally find that the triangular person has got into the square hole, the oblong into the triangular, and a square person has squeezed himself into the round hole. The officer and the office, the doer and the thing done, seldom fit so exactly that we can say they were almost made for each other." Quoted in: John Bartlett; Familiar Quotations; 9th ed., Macmillan, London, 1902, p. 461 (without specifying the Lecture or date). Irving Wallace; The Square Pegs; (Hutchinson, 1958); New English Library, 1968; p. 11, gives the above quote and says it was given in a lecture by Smith at the Royal Institution in 1824. Bartlett gives a footnote reference: The right man to fill the right place -- Layard: Speech, Jan. 15, 1855. It is not clear to me whether Layard quoted Smith or simply expressed the same idea in prosaic terms . Partially quoted, from 'we shall ...' in The Oxford Dictionary of Quotations; 2nd ed. revised, 1970, p. 505, item 24. Similarly quoted in some other dictionaries of quotations.

I have located other quotations from 1837, 1867 and 1901.

William A. Bagley. Paradox Pie. Vawser & Wiles, London, nd [BMC gives 1944]. No. 17: Misfits, p. 18. "Which is the worst misfit, a square peg in a round hole or a round peg in a square hole?" Shows the round peg fits better. He notes that square holes are hard to make.

David Singmaster. On round pegs in square holes and square pegs in round holes. MM 37 (1964) 335 337. Reinvents the problem and considers it in n dimensions. The round peg fits better for n < 9. John L. Kelley pointed out that there must be a dimension between 8 and 9 where the two fit equally well. Herman P. Robinson kindly calculated this dimension for me in 1979, getting 8.13795....

David Singmaster. Letter: The problem of square pegs and round holes. ILEA Contact [London] (12 Sep 1980) 12. The two dimensional problem appears as a SMILE card which was attacked as 'daft' in an earlier letter. Here I defend the problem and indicate some extensions -- e.g. a circle fits better in a regular n gon than vice versa for all n.

6.BO. BUTTERFLY PROBLEM
I have generally avoided classical geometry of this sort, but Bankoff's paper deserves inclusion.
Leon Bankoff. The metamorphosis of the butterfly problem. MM 60 (1987) 195 210. Includes historical survey of different proofs. The name first appears in the title of the solution of Problem E571, AMM 51 (1944) 91 (??NYS). The problem first occurs in The Gentlemen's Diary (1815) 39 40 (??NYS).
6.BP. EARLY MATCHSTICK PUZZLES
There are too many matchstick puzzles to try to catalogue. Some of them occur in other sections, e.g. 6.AO.1. Here I only include a few very early other examples. At first I thought these would date from mid to late 19C when matches first started to become available, but the earliest examples refer to slips of paper or wood. The earliest mention of matches is in 1858.
Rational Recreations. 1824. Exer. 23, p. 132. Double a sheep pen by adding just two hurdles. I have just realised this is a kind of matchstick puzzle and I suspect there are other early examples of this. It is a little different than most matchstick puzzles in that one is usually not given an initial pattern, but must figure it out. (A hurdle is a kind of panel woven from sticks or reeds used by shepherds to make temporary pens.)

Family Friend 2 (1850) 148 & 179. Practical Puzzle -- No. V. = Illustrated Boy's Own Treasury, 1860, Prob. 46, pp. 404 & 443. "Cut seventeen slips of paper or wood of equal lengths, and place them on a table, to form six squares, as in the diagram. ...."

Magician's Own Book. 1857. Prob. 20: Three square puzzle, pp. 273 & 296. (I had 87 & 110 ??) Almost identical to Family Friend, with a few changes in wording and a different drawing, e.g. "Cut seventeen slips of cardboard of equal lengths, and place them on a table to form six squares, as in the diagram. ...."

The Sociable. 1858. Prob. 2: The magic square, pp. 286 & 301. "With seventeen pieces of wood (lucifer matches will answer the purpose, but be careful to remove the combustible ends, and see that they are all of the same length) make the following figure: [a 2 x 3 array of squares]", then remove 5 matches to leave three squares.

Book of 500 Puzzles. 1859.

Prob. 2: The magic square, pp. 4 & 19. As in The Sociable.

Prob. 20: Three square puzzle, pp. 87 & 110. Identical to Magician's Own Book.

Boy's Own Conjuring Book. 1860. Three-square puzzle, pp. 235 & 259. Identical to Magician's Own Book.

Elliott. Within Doors. Op. cit. in 6.V. 1872. Chap. 1, no. 5: The three squares, pp. 28 & 31. Almost identical to Magician's Own Book, prob. 20, with a slightly different diagram.

Mittenzwey. 1880. Prob. 156-171, 202-212, 240, 242, pp. 32-33, 37-38, 44 & 83-85, 90, 94; 1895?: 179-196, 227-237, 269, 271, pp. 37-38, 41-42, 48 & 85-87, 92, 96; 1917: 179 196, 227-237, 269, 271, pp. 33-34, 38-39, 44 & 82-84, 88, 92. This is the first puzzle book to have lots of matchstick problems, though he doesn't yet use the name, calling them 'Hölzchen' (= sticks). Several problems occur elsewhere, e.g. in 6.AO.1. The last problem is the same as in Jackson.

Sophus Tromholt. Streichholzspiele. Otto Spamer, Leipzig, 1889; 5th ed., 1892; 14th ed., Leipzig, 1909; slightly revised and with a Preface by Rüdiger Thiele, Zentralantiquariat der DDR, Leipzig, 1986; Hugendubel, 1986. [Christopher 1017 is Spamer, 1890. C&B give 1890. There is an edition by Ullstein, Frankfurt, 1990.] This is the earliest book I know which is entirely devoted to matchstick puzzles.

Gaston Tissandier. Jeux et Jouets du jeune age Choix de récréations amusantes & instructives. Ill. by Albert Tissandier. G. Masson, Paris, nd [c1890]. P. 40, no. 4-5: Le problème des allumettes. Make five squares with nine matches. Solution has four small squares and one large one.

Clark. Mental Nuts. 1897, no. 56. The toothpicks. Use 12 toothpicks to make a 2 x 2 array of squares. Move three picks to form three squares.

H. D. Northrop. Popular Pastimes. 1901. No. 1: The magic square, pp. 65 & 71. = The Sociable.

6.BQ. COVERING A DISC WITH DISCS
The general problem is too complex to be considered recreational. Here I will mainly deal with the carnival version where one tries to cover a circular spot with five discs. In practice, this is usually rigged by stretching the cloth.
Eric H. Neville. On the solution of numerical functional equations, illustrated by an account of a popular puzzle and of its solution. Proc. London Math. Soc. (2) 14 (1915) 308-326. Obtains several possible configurations, but says "actual trial is sufficient to convince" that one is clearly the best, namely the elongated pentagon with 2-fold symmetry. This leads to four trigonometric equations in four unknown angles which theoretically could be solved, but are difficult to solve even numerically. He develops a modification of Newton's method and applies it to the problem, obtaining the maximal ratio of spot radius to disc radius as 1.64091. Described by Gardner and Singleton.

Ball. MRE. 10th ed., 1922. Pp. 253-255: The five disc problem. Sketches Neville's results.

Will Blyth. More Paper Magic. C. Arthur Pearson, London, 1923. Cover the spot, pp. 66-67. "This old "fun of the fair" game has been the means of drawing many pennies from the pockets of frequenters of fairs." Says the best approach is an elongated pentagon which has only 2-fold symmetry.

William Fitch Cheney Jr, proposer; editorial comment. Problem E14. AMM 39 (1932) 606 & 42 (1935) 622. Poses the problem. Editor says no solution of this, or its equivalent, prob. 3574, was received, but cites Neville.

J. C. Cannell. Modern Conjuring for Amateurs. C. Arthur Pearson, London, nd [1930s?]. Cover the spot, pp. 132-134. Uses discs of diameter 1 5/8 in to cover a circle of diameter 2 1/2 in. This is a ratio of 20/13 = 1.538..., which should be fairly easy to cover?? His first disc has its edge passing through the centre of the circle. His covering pattern has bilateral symmetry, though the order of placing the last two discs seems backward.

Walter B. Gibson. The Bunco Book. (1946); reprinted by Citadel Press (Lyle Stuart Inc.), Secaucus, New Jersey, 1986. Spotting the spot, pp. 24-25. Also repeated in summary form, with some extra observations in: Open season on chumps, pp. 97-106, esp. pp. 102 103. The circles have diameter 5" and the discs "are slightly more than three inches in diameter." He assumes the covering works exactly when the discs have five-fold symmetry -- which implies the discs are 3.090... inches in diameter -- but that the operator stretches the cloth so the spot is unsymmetric and the player can hardly ever cover it -- though it still can be done if one plays a disc to cover the bulge. "There is scarcely one chance in a hundred that the spectator will start correctly ...." On p. 103, he adds that "in these progressive times" the bulge can be made in any direction and that shills are often employed to show that it can be done, though it is still difficult and the operator generally ignores small uncovered bits in the shills' play in order to make the game seem easy.

Martin Gardner. SA (Apr 1959)?? c= 2nd Book, chap. 13. Describes the five disc version as Spot-the-Spot. Cites Neville.

Colin R. J. Singleton. Letter: A carnival game -- covering disks with smaller disks. JRM 24:3 (1992) 185-186. Responding to a comment in JRM 24:1, he points out that the optimum placing of five discs does not have pentagonal symmetry but only bilateral. Five discs of radius 1 can then cover a disc of radius 1.642.., rather than 1.618..., which occurs when there is pentagonal symmetry. He cites Gardner and E. H. Neville. His 1.642 arises because Gardner had truncated the reciprocal ratio to three places.

6.BR. WHAT IS A GENERAL TRIANGLE?
David & Geralda Singmaster, proposers; Norman Miller, solver. Problem E1705 -- Skewness of a triangle. AMM 71:6 (1964) 680 & 72:6 (1965) 669. Assume a  b  c. Define the skewness of the triangle as S = max {a/b, b/c, c/a} x min {a/b, b/c, c/a}. What triangles have maximum and minimum skewness? Minimum is S = 1 for any isosceles triangle. Maximum occurs for the degenerate triangle with sides 1, φ, 1+φ, where 1 + φ = φ², so φ = (1 + 5)/2 is the golden mean.

Baruch Schwarz & Maxim Bruckheimer. Let ABC be any triangle. MTr 81 (Nov 1988) 640-642. Assume AB < AC < BC and A < 90^o. Drawing BC and putting A above it leads to a small curvilinear triangular region where A can be. Making A equidistant from the three boundaries leads to a triangle with sides proportional to 33, 7, 8 and with angles 44.5^o, 58.5^o, 77^o. The sides are roughly in the proportion 6 : 7 : 8.

Gontran Ervynck. Drawing a 'general' triangle. Mathematics Review (Nov 1991). ??NYS -- cited by Anon., below. Notes that if we take an acute triangle with angles as different as possible, then we get the triangle with angles 45^o, 60^o, 75^o.

Anon. [possibly the editor, Tom Butts]. What is a 'general' triangle? Mathematical Log 37:3 (Oct 1993) 1 & 6. Describes above two results and mentions Guy's article in 8.C. Gives an argument which would show the probability of an acute triangle is 0.

Anon. [probably the editor, Arthur Dodd]. A very scalene triangle. Plus 30 (Summer 1995) 18-19 & 23. (Content says this is repeated from a 1987 issue -- ??NYS.) Uses the same region as Schwarz & Bruckheimer, below. Looks for a point as far away from the boundaries as possible and takes the point which gives the angles 45^o, 60^o, 75^o.

In 1995?, I experimented with variations on the definition of skewness given in the first item above, but have not gotten much. However, taking a = 1, we have 1  b  c  b + 1. Plotting this in the b, c plane gives us a narrow strip extending to infinity. For generality, it would seem that we want c = b + ½, but there is no other obvious condition to select a central point in this region. As fairly random points, I have looked at the case where c = b², which gives b = (1 + 3)/2 = 1.366.., c = 1.866.. -- this triangle has angles about 31.47^o, 45.50^o, 103.03^o -- and at the case where b = 3/2, which gives a triangle with sides proportional to 2, 3, 4 with angles about 28.96^o, 46.57^o, 104.46^o. In Mar 1996, I realised that the portion of the strip corresponding to an acute triangle tends to 0 !!

I have now (Mar 1996) realised that the situation is not very symmetric. Taking c = 1, we have 0  a  b  1  a + b and plotting this in the a, b plane gives us a bounded triangle with vertices at (0, 1), (½, ½), (1, 1). There are various possible central points of this triangle. The centroid is at (1/2, 5/6), giving a triangle with sides 1/2, 5/6, 1 which is similar to 3, 5, 6, with angles 29.93^o, 56.25^o, 93.82^o. An alternative point in this region is the incentre, which is at (½, ½{-1 + 22}), giving a triangle similar to 1, -1 + 22, 2 with angles 29.85^o, 65.53^o, 84.62^o. The probability of an acute triangle in this context is 2 - π/2 = .429.
6.BS. FORM SIX COINS INTO A HEXAGON
O O O O O

Transform O O O into O O in three moves.

O O

New section -- there must be older versions.

Young World. c1960. P. 13: Ringing the change. He starts with the mirror image of the first array.

Robert Harbin. Party Lines. Op. cit. in 5.B.1. 1963. The ring of coins, p. 31. Says it is described by Gardner, but gives no details. Notes that if you show the trick to someone

O O O and then give the coins in the mirror image pattern shown at the left, he will

O O O not be able to do it.

Putnam. Puzzle Fun. 1978.

No. 12: Create a space, pp. 4 & 27. With four coins, create the pattern OO  

on the right. [Takes two moves from a rhombic starting pattern.] O O

No. 13: Create a space again, pp. 4 & 38. Standard hexagon problem.