Sources page biographical material

6.L. CROSSED LADDERS
Two ladders are placed across a street, each reaching from the base of the house on one side to the house on the other side.

The simple problem gives the heights a, b that the ladders reach on the walls. If the height of the crossing is c, we easily get 1/c = 1/a + 1/b. NOTATION -- this problem will be denoted by (a, b).

The more common and more complex problem is where the ladders have lengths a and b, the height of their crossing is c and one wants the width d of the street. If the heights of the ladder ends are x, y, this situation gives x²   y² = a²   b² and 1/x + 1/y = 1/c which leads to a quartic and there seems to be no simple solution. NOTATION -- this will be denoted (a, b, c).
Mahavira. 850. Chap. VII, v. 180-183, pp. 243-244. Gives the simple version with a modification -- each ladder reaches from the top of a pillar beyond the foot of the other pillar. The ladder from the top of pillar Y (of height y) extends by m beyond the foot of pillar X and the ladder from the top of pillar X (of height x) reaches n beyond the foot of pillar Y. The pillars are d apart. Similar triangles then yield: (d+m+n)/c = (d+n)/x + (d+m)/y and one can compute the various distances along the ground. He first does problems with m = n = 0, which are the simple version of the problem, but since d is given, he also asks for the distances on the ground.

v. 181. (16, 16) with d = 16.

v. 182. (36, 20) with d = 12.

v. 183. x, y, d, m, n = 12, 15, 4, 1, 4.

Bhaskara II. Lilavati. 1150. Chap. VI, v. 160. In Colebrooke, pp. 68 69. (10, 15). (= Bijaganita, ibid., chap. IV, v. 127, pp. 205 206.)

Fibonacci. 1202. Pp. 397 398 (S: 543-544) looks like a crossed ladders problem but is a simple right triangle problem.

Pacioli. Summa. 1494. Part II.

F. 56r, prob. 48. (4, 6).

F. 60r, prob. 64. (10, 15).

Hutton. A Course of Mathematics. 1798? Prob. VIII, 1833: 430; 1857: 508. A ladder 40 long in a roadway can reach 33 up one side and, from the same point, can reach 21 up the other side. How wide is the street? This is actually a simple right triangle problem.

Victor Katz reports that Hutton's problem, with values 60; 37, 23 appears in a notebook of Benjamin Banneker (1731-1806).

Loyd. Problem 48: A jubilee problem. Tit Bits 32 (21 Aug, 11 & 25 Sep 1897) 385, 439 & 475. Given heights of the ladder ends above ground and the width of the street, find the height of the intersection. However one wall is tilted and the drawing has it covered in decoration so one may interpret the tilt in the wrong way.

Jno. A. Hodge, proposer; G. B. M. Zerr, solver. Problem 131. SSM 8 (1908) 786 & 9 (1909) 174 175. (100, 80, 10).

W. V. N. Garretson, proposer; H. S. Uhler, solver. Problem 2836. AMM 27 (1920) & 29 (1922) 181. (40, 25, 15).

C. C. Camp, proposer; W. J. Patterson & O. Dunkel, solvers. Problem 3173. AMM 33 (1926) 104 & 34 (1927) 50 51. General solution.

Morris Savage, proposer; W. E. Batzler, solver. Problem 1194. SSM 31 (1931) 1000 & 32 (1932) 212. (100, 80, 10).

S. A. Anderson, proposer; Simon Vatriquant, solver. Problem E210. AMM 43 (1936) 242 & 642 643. General solution in integers.

C. R. Green, proposer; C. W. Trigg, solver. Problem 1498. SSM 37 (1937) 484 & 860 861. (40, 30, 15). Trigg cites Vatriquant for smallest integral case.

A. A. Bennett, proposer; W. E. Buker, solver. Problem E433. AMM 47 (1940) 487 & 48 (1941) 268 269. General solution in integers using four parameters.

J. S. Cromelin, proposer; Murray Barbour, solver. Problem E616 -- The three ladders. AMM 51 (1944) 231 & 592. Ladders of length 60 & 77 from one side. A ladder from the other side crosses them at heights 17 & 19. How long is the third ladder and how wide is the street?

Geoffrey Mott-Smith. Mathematical Puzzles for Beginners and Enthusiasts. (Blakiston, 1946); revised ed., Dover, 1954. Prob. 103: The extension ladder, pp. 58-59 & 176 178. Complex problem with three ladders.

Arthur Labbe, proposer; various solvers. Problem 25 -- The two ladders. Sep 1947 [date given in Graham's second book, cited at 1961]. In: L. A. Graham; Ingenious Mathematical Problems and Methods; Dover, 1959, pp. 18 & 116 118. (20, 30, 8).

M. Y. Woodbridge, proposer and solver. Problem 2116. SSM 48 (1948) 749 & 49 (1949) 244 245. (60, 40, 15). Asks for a trigonometric solution. Trigg provides a list of early references.

Robert C. Yates. The ladder problem. SSM 51 (1951) 400 401. Gives a graphical solution using hyperbolas.

G. A. Clarkson. Note 2522: The ladder problem. MG 39 (No. 328) (May 1955) 147 148. (20, 30, 10). Let A = (a²   b²) and set x = A sec t, y = A tan t. Then cos t + cot t = A and he gets a trigonometrical solution. Another method leads to factoring the quartic in terms of a constant k whose square satisfies a cubic.

L. A. Graham. The Surprise Attack in Mathematical Problems. Dover, 1968. Problem 6: Searchlight on crossed ladders, pp. 16-18. Says they reposed Labbe's Sep 1947 problem in Jun 1961. Solution by William M. Dennis which is the same trigonometric method as Clarkson.

H. E. Tester. Note 3036: The ladder problem. A solution in integers. MG 46 (No. 358) (Dec 1962) 313 314. A four parameter, incomplete, solution. He finds the example (119, 70, 30).

A. Sutcliffe. Complete solution of the ladder problem in integers. MG 47 (No. 360) (May 1963) 133 136. Three parameter solution. First few examples are: (119, 70, 30); (116, 100, 35); (105, 87, 35). Simpler than Anderson and Bennett/Buker.

Alan Sutcliffe, proposer; Gerald J. Janusz, solver. Problem 5323 -- Integral solutions of the ladder problem. AMM 72 (1965) 914 & 73 (1966) 1125-1127. Can the distance f between the tops of the ladders be integral? (80342, 74226, 18837) has x = 44758, y = 32526, d = 66720, f = 67832. This is not known to be the smallest example.

Anon. A window cleaner's problem. Mathematical Pie 51 (May 1967) 399. From a point in the road, a ladder can reach 30 ft up on one side and 40 ft up on the other side. If the two ladder positions are at right angles, how wide is the road?

J. W. Gubby. Note 60.3: Two chestnuts (re-roasted). MG 60 (No. 411) (Mar 1976) 64-65. 1.  Given heights of ladders as a, b, what is the height c of their intersection? Solution: 1/c = 1/a + 1/b or c = ab/(a+b). 2. The usual ladder problem -- he finds a quartic.

J. Jabłkowski. Note 61:11: The ladder problem solved by construction. MG 61 (No. 416) (Jun 1977) 138. Gives a 'neusis' construction. Cites Gubby.

Birtwistle. Calculator Puzzle Book. 1978. Prob. 83, A second ladder problem, pp. 58-59 & 115-118. (15, 20, 6). Uses xy as a variable to simplify the quartic for numerical solution and eventually gets 11.61.

See: Gardner, Circus, p. 266 & Schaaf for more references. ??follow up.

Liz Allen. Brain Sharpeners. Op. cit. in 5.B. 1991. The tangled ladders, pp. 43-44 & 116. (30, 20, 10). Gives answer 12.311857... with no explanation.
6.L.1. LADDER OVER BOX
A ladder of length L is placed to just clear a box of width w and height h at the base of a wall. How high does the ladder reach? Denote this by (w, h, L). Letting x be the horizontal distance of the foot and y be the vertical distance of the top of the ladder, measured from the foot of the wall, we get x² + y² = L² and (x w)(y h) = wh, which gives a quartic in general. But if w = h, then use of x + y as a variable reduces the quartic to a quadratic. In this case, the idea is old -- see e.g. Simpson.

The question of determining shortest ladder which can fit over a box of width w and height h is the same as determining the longest ladder which will pass from a corridor of width w into another corridor of width h. See Huntington below and section 6.AG.

Simpson. Algebra. 1745. Section XVIII, prob. XV, p. 250 (1790: prob. XIX, pp. 272-273). "The Side of the inscribed Square BEDF, and the Hypotenuse AC of a right-angled Triangle ABC being given; to determine the other two Sides of the Triangle AB and BC." Solves "by considering x + y as one Quantity".

Pearson. 1907. Part II, no. 102: Clearing the wall, p. 103. For (15, 12, 52), the ladder reaches 48.

D. John Baylis. The box and ladder problem. MTg 54 (1971) 24. (2, 2, 10). Finds the quartic which he solves by symmetry. Editorial note in MTg 57 (1971) 13 says several people wrote to say that use of similar triangles avoids the quartic.

Birtwistle. Math. Puzzles & Perplexities. 1971. The ladder and the box problem, pp. 44-45. = Birtwistle; Calculator Puzzle Book; 1978; Prob. 53: A ladder problem, pp. 37 & 96 98. (3, 3, 10). Solves by using x + y - 6 as a variable.

Monte Zerger. The "ladder problem". MM 60:4 (1987) 239 242. (4, 4, 16). Gives a trigonometric solution and a solution via two quadratics.

Oliver D. Anderson. Letter. MM 61:1 (1988) 63. In response to Zerger's article, he gives a simpler derivation.

Tom Heyes. The old box and ladder problem -- revisited. MiS 19:2 (Mar 1990) 42 43. Uses a graphic calculator to find roots graphically and then by iteration.

A. A. Huntington. More on ladders. M500 145 (Jul 1995) 2-5. Does usual problem, getting a quartic. Then finds the shortest ladder. [This turns out to be the same as the longest ladder one can get around a corner from corridors of widths w and h, so this problem is connected to 6.AG.]

David Singmaster. Integral solutions of the ladder over box problem. In preparation. Easily constructs all the primitive integral examples from primitive Pythagorean triples. E.g. for the case of a square box, i.e. w = h, if X, Y, Z is a primitive Pythagorean triple, then the corresponding primitive solution has w = h = XY, x = X (X + Y), y = Y (X + Y), L = Z (X + Y), and remarkably, x - h = X², y - w = Y².
6.M. SPIDER & FLY PROBLEMS
These involve finding the shortest distance over the surface of a cube or cylinder. I've just added the cylindrical form -- see Dudeney (1926), Perelman and Singmaster. The shortest route from a corner of a cube or cuboid to a diagonally opposite corner must date back several centuries, but I haven't seen any version before 1937! I don't know if other shapes have been done -- the regular (and other) polyhedra and the cone could be considered.

Two-dimensional problems are in 10.U.

Loyd. The Inquirer (May 1900). Gives the Cyclopedia problem. ??NYS -- stated in a letter from Shortz.

Dudeney. Problem 501 -- The spider and the fly. Weekly Dispatch (14 & 28 Jun 1903) both p. 16. 4 side version.

Dudeney. Breakfast table problems, No. 320 -- The spider and the fly. Daily Mail (18 & 21 Jan 1905) both p. 7. Same as the above problem.

Dudeney. Master of the breakfast table problem. Daily Mail (1 & 8 Feb 1905) both p. 7. Interview with Dudeney in which he gives the 5 side version.

Ball. MRE, 4th ed., 1905, p. 66. Gives the 5 side version, citing the Daily Mail of 1 Feb 1905. He says he heard a similar problem in 1903 -- presumably Dudeney's first version. In the 5th ed., 1911, p. 73, he attributes the problem to Dudeney.

Dudeney. CP. 1907. Prob. 75: The spider and the fly, pp. 121 122 & 221 222. 5 side version with discussion of various generalizations.

Dudeney. The world's best problems. 1908. Op. cit. in 2. P. 786 gives the five side version.

Sidney J. Miller. Some novel picture puzzles -- No. 6. Strand Mag. 41 (No. 243) (Mar 1911) 372 & 41 (No. 244) (Apr 1911) 506. Contest between two snails. Better method uses four sides, similar to Dudeney's version, but with different numbers.

Loyd. The electrical problem. Cyclopedia, 1914, pp. 219 & 368 (= MPSL2, prob. 149, pp. 106 & 169 = SLAHP: Wiring the hall, pp. 72 & 114). Same as Dudeney's first, four side, version. (In MPSL2, Gardner says Loyd has simplified Dudeney's 5 side problem. More likely(?) Loyd had only seen Dudeney's earlier 4 side problem.)

Dudeney. MP. 1926. Prob. 162: The fly and the honey, pp. 67 & 157. (= 536, prob. 325, pp. 112 & 313.) Cylindrical problem.

Perelman. FFF. 1934. The way of the fly. 1957: Prob. 68, pp. 111 112 & 117 118; 1979: Prob. 72, pp. 136 & 142 144. MCBF: Prob. 72, pp. 134 & 141-142. Cylindrical form, but with different numbers and arrangement than Dudeney's MP problem.

Haldeman-Julius. 1937. No. 34: The louse problem, pp. 6 & 22. Room 40 x 20 x 10 with louse at a corner wanting to go to a diagonally opposite corner. Problem sent in by J. R. Reed of Emmett, Idaho. Answer is 50!

M. Kraitchik. Mathematical Recreations, 1943, op. cit. in 4.A.2, chap. 1, prob. 7, pp. 17 21. Room with 8 equal routes from spider to fly. (Not in his Math. des Jeux.)

Sullivan. Unusual. 1943. Prob. 10: Why not fly? Find shortest route from a corner of a cube to the diagonally opposite corner.

William R. Ransom. One Hundred Mathematical Curiosities. J. Weston Walch, Portland, Maine, 1955. The spider problem, pp. 144 146. There are three types of path, covering 3, 4 and 5 sides. He determines their relative sizes as functions of the room dimensions.

Birtwistle. Math. Puzzles & Perplexities. 1971.

Round the cone, pp. 144 & 195. What is the shortest distance from a point P around a cone and back to P? Answer is "An ellipse", which doesn't seem to answer the question. If the cone has height H, radius R and P is l from the apex, then the slant height L is (R² + H²), the angle of the opened out cone is θ = 2πR/L and the required distance is 2l sin θ/2.

Spider circuit, pp. 144 & 198. Spider is at the midpoint of an edge of a cube. He wants to walk on each of the faces and return. What is his shortest route? Answer is "A regular hexagon. (This may be demonstrated by putting a rubber band around a cube.)"

David Singmaster. The spider spied her. Problem used as: More than one way to catch a fly, The Weekend Telegraph (2 Apr 1984). Spider inside a glass tube, open at both ends, goes directly toward a fly on the outside. When are there two equally short paths? Can there be more than two shortest routes?

Yoshiyuki Kotani has posed the following general and difficult problem. On an a x b x c cuboid, which two points are furthest apart, as measured by an ant on the surface? Dick Hess has done some work on this, but I believe that even the case of square cross-section is not fully resolved.

6.N. DISSECTION OF A 1 x 1 x 2 BLOCK TO A CUBE
W. F. Cheney, Jr., proposer; W. R. Ransom; A. H. Wheeler, solvers. Problem E4. AMM 39 (1932) 489; 40 (1933) 113-114 & 42 (1934) 509-510. Ransom finds a solution in 8 pieces; Wheeler in 7.

Harry Lindgren. Geometric Dissections. Van Nostrand, Princeton, 1964. Section 24.2, p. 120 gives a variant of Wheeler's solution.

Michael Goldberg. A duplication of the cube by dissection and a hinged linkage. MG 50 (No. 373) (Oct 1966) 304 305. Shows that a hinged version exists with 10 pieces. Hanegraaf, below, notes that there are actually 12 pieces here.

Anton Hanegraaf. The Delian Altar Dissection. Polyhedral Dissections, Elst, Netherlands, 1989. Surveys the problem, gives a 6 piece solution and a 7 piece hinged solution.

6.O. PASSING A CUBE THROUGH AN EQUAL OR SMALLER CUBE --

PRINCE RUPERT'S PROBLEM
The projection of a unit cube along a space diagonal is a regular hexagon of side 2/3. The largest square inscribable in this hexagon has edge 6 - 2 = 1.03527618. By passing the larger cube on a slant to the space diagonal, one can get the larger cube having edge 32/4 = 1.06066172.

There are two early attributions of this. Wallis attributes it to Prince Rupert, but Hennessy says Philip Ronayne of Cork invented it. I have discovered a possible connection. Prince Rupert of the Rhine (1619-1682), nephew of Charles I, was a major military figure of his time, becoming commander-in-chief of Charles I's armies in the 1640s. In 1648-1649, he was admiral of the King's fleet and was blockaded with 16 ships in Kinsale Harbor for 20 months. Kinsale is about 20km south of Cork.

Ronayne wrote an Algebra, of which only a second edition of 1727 is in the BL. Schrek has investigated the family histories and says Ronayne lived in the early 18C. This would seem to make him too young to have met Rupert. Perhaps Rupert invented the problem while in Kinsale and this was conveyed to Ronayne some years later. Does anyone know the dates of Ronayne or of the 1st ed (Schrek only located the BL example of the 2nd ed)? I cannot find anything on him in Wallis, May, Poggendorff, DNB, but Google has turned up a reference to a 1917 history of the family which Schrek cites, but I have not yet tried to find this.

Hennessy's article says a little about Daniel Voster and details are in Wallis's . His father, Elias (1682   >1728) wrote an Arithmetic, of which Wallis lists 30 editions. The BL lists one as late as 1829. The son, Daniel (1705   >1760) was a schoolmaster and instrument maker who edited later versions of his father's arithmetic. The 1750 History of Cork quoted by Hennessy says the author had seen the cubes with Daniel. Hennessy conjectures that his example was made specially, perhaps under the direction of a mathematician. It seems likely that Daniel knew Ronayne and made this example for him.

John Wallis. Perforatio cubi, alterum ipsi aequalem recipiens. (De Algebra Tractatus; 1685; Chap. 109) = Opera Mathematica, vol. II, Oxford, 1693, pp. 470 471, ??NYS. Cites Rupert as the source of the equal cube version. (Latin and English in Schrek.) Scriba, below, found an errata slip in Wallis's copy of his Algebra in the Bodleian. This corrects the calculations, but was published in the Opera, p. 695.

Ozanam Montucla. 1778. Percer un cube d'une ouverture, par laquelle peut passer un autre cube égal au premier. Prob. 30 & fig. 53, plate 7, 1778: 319-320; 1803: 315-316; 1814: 268-269. Prob. 29, 1840: 137. Equal cubes with diagonal movement.

J. H. van Swinden. Grondbeginsels der Meetkunde. 2nd ed., Amsterdam, 1816, pp. 512 513, ??NYS. German edition by C. F. A. Jacobi, as: Elemente der Geometrie, Friedrich Frommann, Jena, 1834, pp. 394-395. Cites Rupert and Wallis and gives a simple construction, saying Nieuwland has found the largest cube which can pass through a cube.

Peter Nieuwland. (Finding of maximum cube which passes through another). In: van Swinden, op. cit., pp. 608 610; van Swinden Jacobi, op. cit. above, pp. 542-544, gives Nieuwland's proof.

Cundy and Rollett, p. 158, give references to Zacharias (see below) and to Cantor, but Cantor only cites Hennessy.

H. Hennessy. Ronayne's cubes. Phil. Mag. (5) 39 (Jan Jun 1895) 183 187. Quotes, from Gibson's 'History of Cork', a passage taken from Smith's 'History of Cork', 1st ed., 1750, vol. 1, p. 172, saying that Philip Ronayne had invented this and that a Daniel Voster had made an example, which may be the example owned by Hennessy. He gives no reference to Rupert. He finds the dimensions.

F. Koch & I. Reisacher. Die Aufgabe, einen Würfel durch einen andern durchzuschieben. Archiv Math. Physik (3) 10 (1906) 335 336. Brief solution of Nieuwland's problem.

M. Zacharias. Elementargeometrie und elementare nicht-Euklidische Geometrie in synthetischer Behandlung. Encyklopädie der Mathematischen Wissenschaften. Band III, Teil 1, 2te Hälfte. Teubner, Leipzig, 1914-1931. Abt. 28: Maxima und Minima. Die isoperimetrische Aufgabe. Pp. 1133-1134. Attributes it to Prince Rupert, following van Swinden. Cites Wallis & Ronayne, via Cantor, and Nieuwland, via van Swinden.

U. Graf. Die Durchbohrung eines Würfels mit einem Würfel. Zeitschrift math. naturwiss. Unterricht 72 (1941) 117. Nice photos of a model made at the Technische Hochschule Danzig. Larger and better versions of the same photos can be found in: W. Lietzmann & U. Graf; Mathematik in Erziehung und Unterricht; Quelle & Meyer, Leipzig, 1941, vol. 2, plate 3, opp. p. 168, but I can't find any associated text for it.

W. A. Bagley. Puzzle Pie. Op. cit. in 5.D.5. 1944. No. 12: Curios [sic] cubes, p. 14. First says it can be done with equal cubes and then a larger can pass through a smaller. Claims that the larger cube can be about 1.1, but this is due to an error -- he thinks the hexagon has the same diameter as the cube itself.

H. D. Grossman, proposer; C. S. Ogilvy & F. Bagemihl, solvers. Problem E888 -- Passing a cube through a cube of same size. AMM 56 (1949) 632 ??NYS & 57 (1950) 339. Only considers cubes of the same size, though Bagemihl's solution permits a slightly larger cube. No references.

D. J. E. Schrek. Prince Rupert's problem and its extension by Pieter Nieuwland. SM 16 (1950) 73 80 & 261 267. Historical survey, discussing Rupert, Wallis, Ronayne, van Swinden & Nieuwland. Says Ronayne is early 18C.

M. Gardner. SA (Nov 1966) = Carnival, pp. 41 54. The largest square inscribable in a cube is the cross section of the maximal hole through which another cube can pass.

Christoph J. Scriba. Das Problem des Prinzen Ruprecht von der Pfalz. Praxis der Mathematik 10 (1968) 241-246. ??NYS -- described by Scriba in an email to HM Mailing List, 20 Aug 1999. Describes the correction to Wallis's work and considers the problem for the tetrahedron and octahedron.

6.P. GEOMETRICAL VANISHING
Gardner. MM&M. 1956. Chap. 7 & 8: Geometrical Vanishing -- Parts I & II, pp. 114 155. Best extensive discussion of the subject and its history.

Gardner. SA (Jan 1963) c= Magic Numbers, chap. 3. Discusses application to making an extra bill and Magic Numbers adds citations to several examples of people trying it and going to jail.

Gardner. Advertising premiums to beguile the mind: classics by Sam Loyd, master puzzle poser. SA (Nov 1971) = Wheels, Chap. 12. Discusses several forms.

S&B, p. 144, shows several versions.

6.P.1. PARADOXICAL DISSECTIONS OF THE CHESSBOARD BASED

ON FIBONACCI NUMBERS
Area 63 version: AWGL, Dexter, Escott, White, Loyd, Ahrens, Loyd Jr., Ransom.
(W. Leybourn. Pleasure with Profit. 1694. ?? I cannot recall the source of this reference and think it may be an error. I have examined the book and find nothing relevant in it.)

Loyd. Cyclopedia, 1914, pp. 288 & 378. 8 x 8 to 5 x 13 and to an area of 63. Asserts Loyd presented the first of these in 1858. Cf Loyd Jr, below.

O. Schlömilch. Ein geometrisches Paradoxon. Z. Math. Phys., 13 (1868) 162. 8 x 8 to 5 x 13. (This article is only signed Schl. Weaver, below, says this is Schlömilch, and this seems right as he was a co editor at the time. Coxeter (SM 19 (1953) 135 143) says it is V. Schlegel, apparently confusing it with the article below.) Doesn't give any explanation, leaving it as a student exercise.

F. J. Riecke. Op. cit. in 4.A.1. Vol. 3, 1873. Art. 16: Ein geometrisches Paradoxon. Quotes Schlömilch and explains the paradox.

G. H. Darwin. Messenger of Mathematics 6 (1877) 87. 8 x 8 to 5 x 13 and generalizations.

V. Schlegel. Verallgemeinerung eines geometrischen Paradoxons. Z. Math. Phys. 24 (1879) 123 128 & Plate I. 8 x 8 to 5 x 13 and generalizations.

Mittenzwey. 1880. Prob. 299, pp. 54 & 105; 1895?: 332, pp. 58 & 106-107; 1917: 332, pp. 53 & 101. 8 x 8 to 5 x 13. Clear explanation.

The Boy's Own Paper. No. 109, vol. III (12 Feb 1881) 327. A puzzle. 8 x 8 to 5 x 13 without answer.

Richard A. Proctor. Some puzzles. Knowledge 9 (Aug 1886) 305-306. "We suppose all the readers ... know this old puzzle." Describes and explains 8 x 8 to 5 x 13. Gives a different method of cutting so that each rectangle has half the error -- several typographical errors.

Richard A. Proctor. The sixty-four sixty-five puzzle. Knowledge 9 (Oct 1886) 360-361. Corrects the above and explains it in more detail.

Will Shortz has a puzzle trade card with the 8 x 8 to 5 x 13 version, c1889.

Ball. MRE, 1st ed., 1892, pp. 34 36. 8 x 8 to 5 x 13 and generalizations. Cites Darwin and describes the examples in Ozanam-Hutton (see Ozanam-Montucla in 6.P.2). In the 5th ed., 1911, p. 53, he changes the Darwin reference to Schlömilch. In the 7th ed., 1917, he only cites the Ozanam-Hutton examples.

Clark. Mental Nuts. 1897, no. 33; 1904, no. 41; 1916, no. 43. Four peculiar drawings. 8 x 8 to 5 x 13.

Carroll-Collingwood. 1899. Pp. 316-317 (Collins: 231 and/or 232 (lacking in my copy)) = Carroll-Wakeling II, prob. 7: A geometrical paradox, pp. 12 & 7. 8 x 8 to 5 x 13. Carroll may have stated this as early as 1888. Wakeling says the papers among which this was found on Carroll's death are now in the Parrish Collection at Princeton University and suggests Schlömilch as the earliest version.

AWGL (Paris). L'Echiquier Fantastique. c1900. Wooden puzzle of 8 x 8 to 5 x 13 and to area 63. ??NYS -- described in S&B, p. 144.

Walter Dexter. Some postcard puzzles. Boy's Own Paper (14 Dec 1901) 174 175. 8 x 8 to 5 x 13 and to area 63.

C. A. Laisant. Initiation Mathématique. Georg, Geneva & Hachette, Paris, 1906. Chap. 63: Un paradoxe: 64 = 65, pp. 150-152.

Wm. F. White. In the mazes of mathematics. A series of perplexing puzzles. III. Geometric puzzles. The Open Court 21 (1907) 241 244. Shows 8 x 8 to 5 x 13 and a two piece 11 x 13 to area 145.

E. B. Escott. Geometric puzzles. The Open Court 21 (1907) 502 505. Shows 8 x 8 to area 63 and discusses the connection with Fibonacci numbers.

William F. White. Op. cit. in 5.E. 1908. Geometric puzzles, pp. 109 117. Partly based on above two articles. Gives 8 x 8 to 5 x 13 and to area 63. Gives an extension which turns 12 x 12 into 8 x 18 and into area 144, but turns 23 x 23 into 16 x 33 and into area 145. Shows a puzzle of Loyd: three piece 8 x 8 into 7 x 9.

Dudeney. The world's best puzzles. Op. cit. in 2. 1908. 5 x 5 into four pieces that make a 3 x 8.

M. Adams. Indoor Games. 1912. Is 64 equal to 65? Pp. 345-346 with fig. on p. 344.

Loyd. Cyclopedia. 1914. See entry at 1858.

W. Ahrens. Mathematische Spiele. Teubner, Leipzig. 3rd ed., 1916, pp. 94 95 & 111 112. The 4th ed., 1919, and 5th ed., 1927, are identical with the 3rd ed., but on different pages: pp. 101 102 & pp. 118 119. Art. X. 65 = 64 = 63 gives 8 x 8 to 5 x 13 and to area 63. The area 63 case does not appear in the 2nd ed., 1911, which has Art. V. 64 = 65, pp. 107 & 118 119 and this material is not in the 1st ed. of 1907.

Tom Tit?? In Knott, 1918, but I can't find it in Tom Tit. No. 3: The square and the rectangle: 64 = 65!, pp. 15-16. Clearly explained.

Hummerston. Fun, Mirth & Mystery. 1924. A puzzling paradox, pp. 44 & 185. Usual 8 x 8 to 5 x 13, but he erases the chessboard lines except for the cells the cuts pass through, so one way has 16 cells, the other way has 17 cells. Reasonable explanation.

Collins. Book of Puzzles. 1927. A paradoxical puzzle, pp. 4-5. 8 x 8 to 5 x 13. Shades the unit cells that the lines pass through and sees that one way has 16 cells, the other way has 17 cells, but gives only a vague explanation.

Loyd Jr. SLAHP. 1928. A paradoxical puzzle, pp. 19 20 & 90. Gives 8 x 8 to 5 x 13. "I have discovered a companion piece ..." and gives the 8 x 8 to area 63 version. But cf AWGL, Dexter, etc. above.

W. Weaver. Lewis Carroll and a geometrical paradox. AMM 45 (1938) 234 236. Describes unpublished work in which Carroll obtained (in some way) the generalizations of the 8 x 8 to 5 x 13 in about 1890 1893. Weaver fills in the elementary missing arguments.

W. R. Ransom, proposer; H. W. Eves, solver. Problem E468. AMM 48 (1941) 266 & 49 (1942) 122 123. Generalization of the 8 x 8 to area 63 version.

W. A. Bagley. Puzzle Pie. Op. cit. in 5.D.5. 1944. No. 23: Summat for nowt?, pp. 27-28. 8 x 8 to 5 x 13, clearly drawn.

Warren Weaver. Lewis Carroll: Mathematician. Op. cit. in 1. 1956. Brief mention of 8 x 8 to 5 x 13. John B. Irwin's letter gives generalizations to other consecutive triples of Fibonacci numbers (though he doesn't call them that). Weaver's response cites his 1938 article, above.