Sources page biographical material

6.AV. CUTTING UP IN FEWEST CUTS
Mittenzwey. 1880. Prob. 191, pp. 36 & 88-89; 1895?: 216, pp. 40 & 91; 1917: 216, pp. 37 & 87. Cut a 2 x 4 into eight unit squares with three cuts. First cuts into two squares, then overlays them and then cuts both ways.

Perelman. FFF. 1934.

Sectioning a cube. Not in the 1957 ed. 1979: prob. 122, pp. 170-171 & 182. MCBF: prob. 122, pp. 171-172 & 186. How many cuts to cut a cube into 27 cubes?

More sectioning. Not in the 1957 ed. 1979: prob. 123, pp. 171 & 182-183. MCBF: prob. 123, pp. 172-173 & 186. How many cuts to cut a chessboard into 64 squares?

Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. Prob. k, pp. 39, 189 & 191. Cut a cube into 27 cubelets.
6.AW. DIVISION INTO CONGRUENT PIECES
Polyomino versions occur in 6.F.4.

Quadrisecting a square is 6.AR.

See also: 6.AS.1, 6.AT.6.a, 6.AY, 6.BG.

For solid problems, see: 6.G.3, 6.G.4, 6.AP, 6.AZ?, 6.BC.

See: Charades, Enigmas, and Riddles, 1862, in 6.AW.2 for a quadrisection with pieces not congruent to the original.
6.AW.1. MITRE PUZZLE
Take a square and cut from two corners to the centre to leave ¾ of the square. The problem is to quadrisect this into four congruent parts.
Charles Babbage. The Philosophy of Analysis -- unpublished collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more details. F. 4r is "Analysis of the Essay of Games". F. 4v has an entry "8½ a Prob of figure" followed by the L tromino. 8½ b is the same with a mitre and there are other dissection problems adjacent -- see 6.F.3, 6.F.4, 6.AQ, 6.AY -- so it seems clear that he knew this problem.

Jackson. Rational Amusement. 1821. Geometrical Puzzles.

No. 15, pp. 26 & 86 & plate II, fig. 11. 2 squares, one double the size of the other, to be cut into four pieces to make a mitre. Just cut each along the diagonal.

No. 18, pp. 27 & 87 & plate II, fig. 14. Six equal squares to form a mitre. Cut each diagonally. [Actually you only need to cut three of the squares.]

Endless Amusement II. 1826? Prob. 5, pp. 192-193. Mitre puzzle -- says the pieces are not precisely equal. = New Sphinx, c1840, pp. 131-132.

Magician's Own Book. 1857.

Prob. 12: The quarto puzzle, pp. 269 & 294. Solution is a bit crudely drawn, but the parts are numbered to make it clear how they are combined. = Illustrated Boy's Own Treasury, 1860, No. 41, pp. 403 & 442.

Prob. 28: Puzzle of the two fathers, pp. 275-276 & 298. One father has L tromino (see 6.F.4), the other has the mitre. Solution carefully drawn and shaded. c= Landells, Boy's Own Toy-Maker, 1858, pp. 148-149.

Book of 500 Puzzles. 1859.

Prob. 12: The quarto puzzle, pp. 83 & 108. Identical to Magician's Own Book.

Prob. 28: Puzzle of the two fathers, pp. 89-90 & 112. Identical to Magician's Own Book..

Boy's Own Conjuring Book. 1860.

Prob. 11: The quarto puzzle, pp. 231 & 257. Identical with Magician's Own Book.

Prob. 27: Puzzle of the two fathers, pp. 237 238 & 262. Identical to Magician's Own Book.

Hanky Panky. 1872. The one quarterless square, p. 132

Hoffmann. 1893. Chap. X, no. 29: The mitre puzzle, pp. 347 & 386 = Hoffmann-Hordern, pp. 243 & 247. Photo on p. 247 shows Enoch Morgan's Sons Sapolio Color-Puzzle. This says to arrange the blocks 'in four equal parts so that each part will be the same size color and shape.' It appears that the blocks are isosceles right triangles with legs equal to a quarter of the original square. There are 6 blue triangles, 6 yellow triangles and 12 red triangles. I think there were originally 6 red and 6 orange, but the colors have faded, and I think one wants each of the four parts having one color.

Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co., NY), nd [1895]. P. 5: The mitre puzzle. Similar to Hoffmann. No solution.

Loyd. Origin of a famous puzzle -- No. 19: The mitre puzzle. Tit Bits 31 (13 Feb & 6 Mar 1897) 363 & 419. Nearly 50 years ago someone told him to quadrisect ¾ of a square into congruent figures. The L tromino was intended, but young Loyd drew the mitre shape instead. He says it took him nearly a year to solve it. But see Dudeney's comments below.

Clark. Mental Nuts. 1904, no. 31. Dividing the land. Quadrisect an L tromino and a mitre.

Pearson. 1907. Part II, no. 87: Loyd's mitre puzzle, pp. 87 & 178.

Dudeney. The world's best puzzles. Op. cit. in 2. 1908. He says he has traced it back to 1835 (Loyd was born in 1841) and that "strictly speaking, it is impossible of solution, but I will give the answer that is always presented, and that seems to satisfy most people." See also the solution to AM, prob. 150, discussed in 6.AY. Can anyone say what the 1835 source might be -- a version of Endless Amusement??

Wehman. New Book of 200 Puzzles. 1908.

P. 43: Puzzle of the two fathers. c= Magician's Own Book, with cruder solution.

P. 47: The quarto puzzle. c= Magician's Own Book, without the numbering of parts.

Loyd. Cyclopedia. 1914. A tailor's problem, pp. 311 & 381. Quadrisect half of a mitre. This has a solution with each piece similar to to the half mitre.

Loyd Jr. SLAHP. 1928. Wrangling heirs, pp. 35 & 96. Divide mitre into 8 congruent parts -- uses the pattern of Loyd Sr.

Putnam. Puzzle Fun. 1978. No. 106: Divide the shape, pp. 16 & 39. "Divide the given shape into four pieces, such that each and every piece is the same area." This is much easier than the usual version. Put two half-size mitres on the bottom edge and two trapeziums are left.
6.AW.2. REP TILES
Here one is cutting a shape into congruent pieces similar to the original shape. Section 5.J is a version with similar but non congruent pieces.
Charades, Enigmas, and Riddles. 1862: prob. 34, pp. 137 & 143; 1865: prob. 578, pp. 109 & 156. Divide isosceles right triangle and a hatchet-shaped 10-omino into four congruent pieces. In the first case, they are congruent to the original, but in the second case, the pieces are right trapeziums of height 1 and bases 2 and 2½.

Dudeney. The puzzle realm. Cassell's Magazine ?? (May 1908) 713-716. No. 3: An easy dissection puzzle. Quadrisect a right trapezium in form of square plus half a square. = AM, 1917, prob. 146, pp. 35 & 170.

C. Dudley Langford. Note 1464: Uses of a geometric puzzle. MG 24 (No. 260) (Jul 1940) 209 211. Quadrisects: L tromino, P pentomino, right trapezium (trapezoid), L tetromino, isosceles trapezium, another right trapezium, two squares joined at a corner. Gives some 9 sections and asks several questions, including asking about 3 D versions.

R. Sibson. Note 1485: Comments on Note 1464. MG 24 (No. 262) (Dec 1940) 343. Says some of Langford's 4 sections also give 9 sections. Mentions some 3 D versions and 144 sections.

Howard D. Grossman. Fun with lattice points. 13. A geometric puzzle. SM 14 (1948) 157 159. Cites Langford & Sibson. Gives two 9-sections obtained from Langford's 4 sections as asserted by Sibson. Gives alternative 4- & 9-sections. Gives a method of generating infinitely many examples on both square and triangular lattices.

Gardner. SA (May 1963) = Unexpected, chap. 19. Says Golomb started considering rep tiles in 1962 and wrote three private reports on them (??NYS). Gardner describes the ideas in them.

Solomon W. Golomb. Replicating figures on the plane. MG 48 (No. 366) (Dec 1964) 403 412. Cites Langford and adds numerous examples. Defines rep k and shows all k can occur.

Roy O. Davis. Note 3151: Replicating boots. MG 50 (No. 372) (May 1966) 175.

Rochell Wilson Meyer. Mutession: a new tiling relationship among planar polygons. MTg 56 (1971) 24 27. A and B mutually tessellate if each tiles an enlargement of the other.
6.AW.3. DIVIDING A SQUARE INTO CONGRUENT PARTS
In the 1960s, a common trick was to give someone a number of quadrisection problems where the parts happen to be congruent to the original figure -- e.g. the quadrisection of the square or the L-tromino. Then ask her to divide a square into five congruent parts. She usually tries to use square pieces in some way and takes a long time to find the obvious answer. c1980, Des MacHale told me that it was a serious question as to whether there was any non-trivial dissection of a square into five or even three congruent pieces. Sometime later, I found a solution -- slice the square into 6 equal strips and say part A consists of the 1st and 4th strips, part B is the 2nd and 5th, part C is the 3rd and 6th. However this is not what was intended by the problem though it leads to other interesting questions. Since then I have heard that the problem has been 'solved' negatively several times on the backs of envelopes at conferences, but no proof seems to have ever appeared. Very little seems to be published on this, so I give what little I know. Much of this applies to rectangles as well as squares. QARCH is an occasional problem sheet issued by the Archimedeans, the Cambridge (UK) student mathematics society.
Gardner, in an article: My ten favorite brainteasers in Games (collected in Games Big Book of Games, 1984, pp. 130-131) says the dissection of the square into five congruent parts is one of his favorite problems. ??locate

David Singmaster. Problem 12. QARCH III (Aug 1980) 3. Asks if the trisection of the square is unique.

David Singmaster. Response to Problem 12 and Problem 21. QARCH V (Jan 1981) 2 & 4. Gives the trisection by using six strips and unconnected parts. In general, we can have an n-section by cutting the square into kn strips and grouping them regularly. For n = 2, k = 4, there is an irregular dissection by using the parts as strips 1, 4, 6, 7 and 2, 3, 5, 8. If p is an odd prime, are there any irregular p-sections?

John Smith, communicated in a letter from Richard Taylor, editor of QARCH, nd, early 1981? Smith found that if you slice a square into 9 strips, then the following parts are congruent, giving an irregular trisection. 1, 2, 6; 3, 7, 8; 4, 5, 9.

David Singmaster. Divisive difficulties. Nature 310 (No. 5977) (9 Aug 1984) 521 & (No. 5979) (23 Aug 1984) 710. Poses a series of problems, leading to the trisections of the cube. No solutions were received.

Angus Lavery asserts that he can trisect the cube, considered as a 3 x 3 x 3 array of indivisible unit cubes. I had sought for this and was unable to find such a trisection and had thought it impossible and I still haven't been able to do it, but Angus swears it can be done, with one piece being the mirror image of the other two.

George E. Martin. Polyominoes -- A Guide to Puzzles and Problems in Tiling. MAA Spectrum Series, MAA, 1991. Pp. 29-30. Fig. 3.9 shows a 5 x 9 rectangle divided into 15 L-trominoes. Shrinking the length 9 to 5 gives a dissection of the square into 15 congruent pieces which are shrunken L-trominoes. Prob. 3.10 (very hard) asks for a rectangle to be dissected into an odd number of congruent pieces which are neither rectangles nor shrunken L-trominoes. He doesn't give an explicit answer, but on p. 76 there are several rectangles filled with an odd number of L, P and Y pentominoes. One might argue that the L and P have the shape of some sort of shrunken L-tromino, but the Y-pentomino is certainly not. Prob. 3.11 (unsolved) asks if a rectangle can be dissected into three congruent pieces which are not rectangles. Prob. 3.12 is a technical generalization of this and hence is also unsolved.

On 19 Jun 1996, I proposed the trisection of the square and Lavery's problem on NOBNET. Michael Reid demonstrated that Lavery's problem has no solution and someone said Lavery had only conjectured it. Reid also cited the following two proofs that the square trisection is impossible.

I. N. Stewart & A. Wormstein. Polyominoes of order 3 do not exist. J. Combinatorial Theory A 61 (1992) 130-136. ??NYS -- Reid says they show that if a rectangle is dissected into three congruent polyominoes, then each is a rectangle.

S. J. Maltby. Trisecting a rectangle. J. Combinatorial Theory A 66 (1994) 40-52. ??NYS -- Reid says he proves the result of Stewart & Wormstein without assuming the pieces are polyominoes.

Martin Gardner. Six challenging dissection tasks. Quantum (May/Jun 1994). Reprinted, with postscript, in Workout, chap. 16. Trisecting the square into congruent parts is his first problem. Cites Stewart & Wormstein. Then asks if one can have three similar parts, with just two, or none, congruent. Then asks the same questions about the equilateral triangle. All except the first question are solved, but the solutions for the fifth and sixth are believed to be unique. In the postscript, Gardner says Rodolfo Kurchan and Andy Liu independently suggested the problems with four parts and cites Maltby.
6.AW.4. DIVIDING AN L-TROMINO INTO CONGRUENT PARTS
See also 6.F.4.
F. Göbel. Problem 1771: The L shape dissection problem. JRM 22:1 (1990) 64 65. The L tromino can be dissected into 2, 3, or 4 congruent parts. Can it be divided into 5 congruent parts?

Editorial comment -- The L-shaped dissection problem. JRM 23:1 (1991) 69-70. Refers to Gardner.

Comments and partial solution by Michael Beeler. JRM 24:1 (1992) 64-69.

Martin Gardner. Tiling the bent tromino with n congruent shapes. JRM 22:3 (1990) 185 191.

6.AX. THE PACKER'S SECRET
This requires placing 12 unit discs snugly into a circular dish of radius 1 + 2 3 = 4.464.
Tissandier. Récréations Scientifiques. 5th ed., 1888, Le secret d'un emballeur, pp. 227-229. Not in the 2nd ed. of 1881 nor the 3rd ed. of 1883. Illustration by Poyet. He shows the solution and how to get the pieces into that pattern. No dimensions given. = Popular Scientific Recreations; [c1890]; Supplement: The packer's secret, pp. 855 856.

Hoffmann. 1893. Chap. X, no. 48: The packer's secret, pp. 356 & 394 = Hoffmann-Hordern, p. 255. Says the problem is of French origin. Gives dimensions 3½ and ¾, giving a ratio of 14/3 = 4.667. "The whole are now securely wedged together ...." [I think this would be a bit loose.]

"Toymaker". The Japanese Tray and Blocks Puzzle. Work, No. 1447 (9 Dec 1916) 168. Says to make the dish of radius 7 and the discs of radius 1½, again giving a ratio of 14/3 = 4.667. Makes "a firm immovable job ...."
6.AY. DISSECT 3A x 2B TO MAKE 2A x 3B, ETC.
This is done by a 'staircase' cut. See 6.AS.
Pacioli. De Viribus. c1500. Ff. 189v - 191r. Part 2. LXXIX. Do(cumento). un tetragono saper lo longare con restregnerlo elargarlo con scortarlo (a tetragon knows lengthening and contraction, enlarging with shortening ??) = Peirani 250-252. Convert a 4 x 24 rectangle to a 3 x 32 using one cut into two pieces. Pacioli's

description is cryptic but seems to have two cuts, making d c

three pieces. There is a diagram at the bottom of f. 190v, badly k f e

redrawn on Peirani 458. Below this is a inserted note which Peirani

252 simply mentions as difficult to read, but can make sense. The g  

points are as laid out at the right. abcd is the original 4 x 24 h a o b

rectangle. g is one unit up from a and e is one unit down from c.

Cut from c to g and from e parallel to the base, meeting cg at f. Then move cdg to fkh and move fec to hag. Careful rereading of Pacioli now seems to show he is using a trick! He cuts from e to f to g. then turns over the upper piece and slides it along so that he can continue his cut from g to h, which is where f to c is now. This gives three pieces from a single cut! Pacioli clearly notes that the area is conserved.

Although not really in this topic, I have put it here as it seems to be a predecessor of this topic and of 6.P.2.

Cardan. De Rerum Varietate. 1557, ??NYS. = Opera Omnia, vol. III, p. 248 (misprinted 348 and with running head Lib. XII in the 1663 ed.). Liber XIII. Shows 2A x 3B to 3A x 2B and half of 3A x 4B to 4A x 3B and discusses the general process.

Kanchusen. Wakoku Chiekurabe. 1727. Pp. 11-12 & 26 27. 4A x 3B to 3A x 4B, with the latter being square. Solution asserts that any size of paper can be made into a square: 'fold lengthwise into an even number and fold the width into an odd number' -- cf Loyd (1914) & Dudeney (1926) below.

Minguet. 1733. Pp. 117-119 (1755: 81-82; 1822: 136-137; 1864: 114-115). 3 x 4 to 4 x 3. Shows a straight tetromino along one side moved to a perpendicular side so both shapes are 4 x 4.

Charles Babbage. The Philosophy of Analysis -- unpublished collection of MSS in the BM as Add. MS 37202, c1820. ??NX. See 4.B.1 for more details. F. 4r is "Analysis of the Essay of Games". F. 4v has an entry "8½ c Prob of figure" followed by a staircase piece. F. 145-146 show two pieces formed into both rectangles. There are other dissection problems adjacent on F. 4v -- see 6.F.3, 6.F.4, 6.AQ, 6.AW.1.

Jackson. Rational Amusement. 1821. Geometrical Puzzles.

No. 7, pp. 24 & 83-84 & plate I, fig. 4. 9 x 16 to 12 x 12.

No. 12, pp. 25 & 85 & plate I, fig. 9. 4 x 9 to 6 x 6.

No. 14, pp. 26 & 86 & plate I, fig. 10. 10 x 20 to 13 1/3 x 15.

Endless Amusement II. 1826?

Prob. 1, p. 188. 5A x 6B to 6A x 5B and to 4A x 7B with two A x B projections. The 6A x 5B looks to be square. = New Sphinx, c1840, p. 136.

Prob. 22, pp. 200-201. 16 x 9 to fill a 12 x 12 hole. Does it by cutting in four pieces -- one 12 x 9 and three 4 x 3. = New Sphinx, c1840, pp. 136-137.

Prob. 33, pp. 210-211. Take a rectangle of proportion 2 : 3 and cut it into two pieces to make a square. Uses cut from 4A x 5B to 5A x 4B, but if we make the rectangle 4 x 6, this makes A = 1, B = 6/5 and the 'square' is 5 x 24/5. = New Sphinx, c1840, p. 140.

Nuts to Crack II (1833), no. 124. 9 x 16 to fill a 12 x 12 hole using four pieces. = Endless Amusement II, prob. 22.

Young Man's Book. 1839. Pp. 241-242. Identical to Endless Amusement II, prob. 22.

Boy's Own Book. 1843 (Paris): 436 & 441, no. 6. 5A x 6B to 6A x 5B and to 4A x 7B with two A x B projections. The 6A x 5B looks to be square. = Boy's Treasury, 1844, pp. 425 & 429. = de Savigny, 1846, pp. 353 & 357, no. 5. Cf de Savigny, below.

de Savigny. Livre des Écoliers. 1846. P. 283: Faire d'une carte un carré. View a playing card as a 5A x 4B rectangle and make a staircase cut and shift to 4A x 5B, which will be nearly square. [When applied to a bridge card, 3.5 x 2.25 in, the result is 2.8 x 2.8125 in.]

Magician's Own Book. 1857. Prob. 2: The parallelogram, pp. 267 & 291. Identical to Boy's Own Book, 1843 (Paris).

The Sociable. 1858. Prob. 19: The perplexed carpenter, pp. 292 & 308. 2 x 12 to 3 x 8. = Book of 500 Puzzles, 1859, prob. 19, pp. 10 & 26. = The Secret Out, 1859, p. 392.

Book of 500 Puzzles. 1859.

Prob. 19: The perplexed carpenter, pp. 10 & 26. As in The Sociable.

Prob. 2: The parallelogram, pp. 81 & 105. Identical to Boy's Own Book, 1843 (Paris).

Charades, Enigmas, and Riddles. 1860: prob. 29, pp. 60 & 64; 1862: prob. 30, pp. 136 & 142; 1865: prob. 574, pp. 107 & 155. 16 x 9 to 12 x 12. All the solutions have an extraneous line in one figure.

Boy's Own Conjuring Book. 1860. Prob. 2: The parallelogram, pp. 229 & 254. Identical to Boy's Own Book, 1843 (Paris).

Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 584-4, pp. 286 & 404. Looks like 3 x 4 to 2 x 6. The rectangles are formed by trimming a quarter off a playing card. The diagrams are not very precise, but it seems that the card is supposed to be twice as long as wide. If we take the card as 4 x 8, then the problem is 3 x 8 to 4 x 6.

Hanky Panky. 1872. The parallelogram, p. 107. "A parallelogram, ..., may be cut into two pieces, by which two other figures can be formed." Shows 5A x 4B cut, but no other figures.

Mittenzwey. 1880. Prob. 253 & 255, pp. 45-46 & 96-97; 1895?: 282 & 284, pp. 49 & 98-99; 1917: 282 & 284, pp. 45 & 93-94. 4 x 9 to 6 x 6. 16 x 9 to 12 x 12.

Cassell's. 1881. The carpenter's puzzle, p. 89. = Manson, 1911, p. 133. 3 x 8 board to cover 2 x 12 area.

Richard A. Proctor. Some puzzles; Knowledge 9 (Aug 1886) 305-306 & Three puzzles; Knowledge 9 (Sep 1886) 336-337. Cut 4 x 3 to 3 x 4. Discusses general method for nA x (n+1)B to (n+1)A x nB and notes that the shape can be oblique as well as rectangular.

Lemon. 1890. Card board puzzle, no. 58, pp. 11 12 & 99. c= The parallelogram puzzle, no. 620, pp. 77 & 120 (= Sphinx, no. 706, pp. 92 & 121). Same as Boy's Own Book, 1843 (Paris). In the pictures, A seems to be equal to B.

Don Lemon. Everybody's Pocket Cyclopedia. Revised 8th ed., 1890. Op. cit. in 5.A. P. 136, no. 8. 9 x 15 to 12 x 12. No solution.

Tom Tit, vol. 2. 1892. Les figures superposables, pp. 149-150. 3 x 2 to 2 x 3.

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Chinese Geometrical Puzzles No. 1, pp. 108 & 111. Same as Boy's Own Book, 1843 (Paris).

Hoffmann. 1893.

Chap. III, no. 8: The extended square, pp. 91 & 124 125 = Hoffmann-Hordern, p. 80. As in Boy's Own Book, 1843 (Paris), but A is clearly not equal to B.

Chap. III, no. 31: The carpenter's puzzle -- no. 2, pp. 103 & 137 = Hoffmann-Hordern, p. 101. 12 x 36 to 18 x 24.

Clark. Mental Nuts. 1897, no. 27. The leaking ship. 12 x 12 to 9 x 16.

Benson. 1904.

The extended square, p. 190. 5A x 6B square, but the other two figures are 8A x 7B and 8A x 5B with two A x B projections.

The carpenter's puzzle (No. 1), pp. 190 191. = Hoffmann, p. 103.

Anon [possibly Dudeney??] Breakfast Table Problems No. 331: A carpenter's dilemma. Daily Mail (31 Jan & 1 Feb 1905) both p. 7. 16 x 9 to 12 x 12.

Pearson. 1907. Part II, no. 1: The carpenter's puzzle, pp. 1 2 & 185 186. 2 x 12 to 3 x 8.

Wehman. New Book of 200 Puzzles. 1908.

P. 6: The perplexed carpenter. 2 x 12 to 3 x 8. c= The Sociable.

P. 9: The carpenter's puzzle. "A plank was to be cut in two: the carpenter cut it half through on each side, and found he had two feet still to cut. How was it?" This is very vague and can only be recognised as a version of our present problem because the solution looks like cutting a 2 x 6 to make a 3 x 4.

P. 16: The parallelogram. Identical to Boy's Own Book, 1843 (Paris).

P. 17: Another parallelogram. Takes a parallelogram formed of a square and a half a square and intends to form a square. He cuts 5A x 4B and makes 4A x 5B. But for this to be a square, it must be 20 x 20 and then the original was 25 x 16, which is not quite in the given shape.

M. Adams. Indoor Games. 1912. A zigzag puzzle, p. 349, with figs. on 348. 5A x 6B square, but the other two figures are 5A x 4B and 7A x 4B with two A x B projections.

Loyd. Cyclopedia. 1914.

The smart Alec puzzle, pp. 27 & 342. (= MPSL1, prob. 93, pp. 90 91 & 153 154.) Cut a mitre into pieces which can form a square. He trims the corners and inserts them into the notch to produce a rectangle and then uses a staircase cut which he claims gives a square using only four pieces. Gardner points out the error, as carefully explained by Dudeney, below. Since Dudeney gives this correction 1n 1911, he must have seen it in an earlier Loyd publication, possibly OPM?

The carpenter's puzzle, pp. 51 & 345. Claims any rectangle can be staircase cut to make a square. Shows 9 x 4 to 6 x 6 and 25 x 16 to 20 x 20. Cf Kanchusen (1727) and Dudeney (1926).

Dudeney. Perplexities. Strand Magazine 41 (No. 246) (Jun 1911) 746 & 42 (No. 247) (Jul 1911) 108. No. 45: Dissecting a mitre. "I have seen an attempt, published in America, ..." Sketches Loyd's method and says it is wrong. "At present no solution has been found in four pieces, and one in five has not apparently been published."

Dudeney. AM. 1917. Prob. 150: Dissecting a mitre, pp. 35 36 & 170 171. He fully describes "an attempt, published in America", i.e. Loyd's method. If the original square has side 84, then Loyd's first step gives a 63 x 84 rectangle, but the staircase cut yields a 72 x 73½ rectangle, not a square. Dudeney gives a 5 piece solution and says "At present no solution has been found in four pieces, and I do not believe one possible."

Dudeney. MP. 1926. Prob. 115: The carpenter's puzzle, pp. 43 44 & 132 133. = 536, prob. 338, pp. 116 117 & 320 321. Shows 9 x 16 to 12 x 12. "But nobody has ever attempted to explain the general law of the thing. As a consequence, the notion seems to have got abroad that the method will apply to any rectangle where the proportion of length to breadth is within reasonable limits. This is not so, and I have had to expose some bad blunders in the case of published puzzles ...." He discusses the general principle and shows that an n step cut dissects n² x (n+1)² to a square of side n(n+1). Gardner adds a note referring to AM, prob. 150. Cf Kanchusen (1727) & Loyd (1914)

Stephen Leacock. Model Memoirs and Other Sketches from Simple to Serious. John Lane, The Bodley Head, 1939, p. 299. Mentions 12 x 12 to 9 x 16.

Harry Lindgren. Geometric Dissections. Van Nostrand, 1964. P. 28 discusses Loyd's mitre dissection problem and variations. He also thinks a four piece solution is impossible.