6.       GEOMETRIC RECREATIONS

 

          6.A.    PI

 

          This is too big a topic to cover completely.  The first items should be consulted for older material and the general history.  Then I include material of particular interest.  See also 6.BL which has some formulae which are used to compute  π.  I have compiled a separate file on the history of  π.

 

Augustus De Morgan.  A Budget of Paradoxes.  (1872);  2nd ed., edited by D. E. Smith, (1915), Books for Libraries Press, Freeport, NY, 1967.

J. W. Wrench Jr.  The evolution of extended decimal approximations to  π.  MTr 53 (Dec 1960) 644‑650.  Good survey with 55 references, including original sources.

Petr Beckmann.  A History of  π.  The Golem Press, Boulder, Colorado, (1970);  2nd ed., 1971.

Lam Lay-Yong & Ang Tian-Se.  Circle measurements in ancient China.  HM 13 (1986) 325‑340.  Good survey of the calculation of  π  in China.

Dario Castellanos.  The ubiquitous  π.  MM 61 (1988) 67-98  &  148-163.  Good survey of methods of computing  π. 

Joel Chan.  As easy as pi.  Math Horizons 1 (Winter 1993) 18-19.  Outlines some recent work on calculating π and gives several of the formulae used.

 

Aristophanes.  The Birds.  ‑414.  Lines 1001‑1005.  In:  SIHGM, vol. 1, pp. 308‑309.  Refers to 'circle-squarers', possibly referring to the geometer/astronomer Meton.

 

E. J. Goodwin.  Quadrature of the circle.  AMM 1 (1894) 246‑247.

House Bill No. 246, Indiana Legislature, 1897.  "A bill for an act introducing a new mathematical truth ..."  In Edington's paper (below), p. 207, and in several of the newspaper reports.

(Indianapolis) Journal (19 Jan 1897) 3.  Mentions the Bill in the list of bills introduced.

Die Quadratur des Zirkels.  Täglicher Telegraph (Indianapolis) (20 Jan 1897) ??.  Surveys attempts since -2000 and notes that Lindemann and Weierstrass have shown that the problem is impossible, like perpetual motion.

A man of 'genius'.  (Indianapolis) Sun (6 Feb 1897) ??.  An interview with Goodwin, who says: "The astronomers have all been wrong.  There's about 40,000,000 square miles on the surface of this earth that isn't here."  He says his results are revelations and gives several rules for the circle and the sphere.

Mathematical Bill passed.  (Indianapolis) Journal (6 Feb 1897) 5.  "This is the strangest bill that has ever passed an Indiana Assembly."  Gives whole text of the Bill.

Dr. Goodwin's theaorem (sic)  Resolution adopted by the House of Representatives.  (Indianapolis) News (6 Feb 1897) 4.  Gives whole text of the Bill. 

The Mathematical Bill  Fun-making in the Senate yesterday afternoon _ other action.  (Indianapolis) News (13 Feb 1897) 11.  "The Senators made bad puns about it, ...."  The Bill was indefinitely postponed.

House Bills in the Senate.  (Indianapolis) Sentinel (13 Feb 1897) 2.  Reports the Bill was killed.

(No heading??)  (Indianapolis) Journal (13 Feb 1897) 3, col. 4.  "... indefinitely postponed, as not being a subject fit for legislation."

Squaring the circle.  (Indianapolis) Sunday Journal (21 Feb 1897) 9.  Says Goodwin has solved all three classical impossible problems.  Says  π = 3.2, using the fact that  Ö2 = 10/7,  giving diagrams and a number of rules. 

                    My thanks to Underwood Dudley for locating and copying the above newspaper items.

C. A. Waldo.  What might have been.  Proc. Indiana Acad. Science 26 (1916) 445‑446.

W. E. Edington.  House Bill No. 246, Indiana State Legislature, 1897.  Ibid. 45 (1935) 206‑210.

A. T. Hallerberg.  House Bill No. 246 revisited.  Ibid. 84 (1975) 374‑399.

Manuel H. Greenblatt.  The 'legal' value of pi, and some related mathematical anomalies.  American Scientist 53 (Dec 1965) 427A‑434A.  On p. 427A he tries to interpret the bill and obtains three different values for  π.

David Singmaster.  The legal values of pi.  Math. Intell. 7:2 (1985) 69‑72.  Analyses Goodwin's article, Bill and other assertions to find 23 interpretable statements giving 9 different values of  π !

Underwood Dudley.  Mathematical Cranks.  MAA, 1992.  Legislating pi, pp. 192-197.

 

C. T. Heisel.  The Circle Squared Beyond Refutation.  Published by the author, 657 Bolivar Rd., Cleveland, Ohio, 1st ed., 1931, printed by S. J. Monck, Cleveland;  2nd ed., 1934, printed by Lezius‑Hiles Co., Cleveland, ??NX  +  Supplement: "Fundamental Truth", 1936, ??NX,  distributed by the author from 2142 Euclid Ave., Cleveland.  This is probably the most ambitious publication of a circle-squarer _ Heisel distributed copies all around the world.

 

Underwood Dudley.  πt:  1832-1879.  MM 35 (1962) 153-154.  He plots 45 values of  π  as a function of time over the period 1832-1879 and finds the least-squares straight line which fits the data, finding that  πt = 3.14281 + .0000056060 t,  for  t  measured in years AD.  Deduces that the Biblical value of 3 was a good approximation for the time and that Creation must have occurred when  πt = 0,  which was in  -560,615.

Underwood Dudley.  πt.  JRM 9 (1976-77) 178 & 180.  Extends his previous work to 50 values of  π  over 1826-1885, obtaining  πt = 4.59183 - .000773 t.  The fact that  πt is decreasing is worrying _ when  πt = 1,  all circles will collapse into straight lines and this will certainly be the end of the world, which is expected in 4646 on 9 Aug at 20:55:33 _ though this is only the expected time and there is considerable variation in this prediction.  [Actually, I get that this should be on 11 Aug.  However, it seems to me that circles will collapse once  πt = 2,  as then the circumference corresponds to going back and forth along the diameter.  This will occur when  t = 3352.949547,  i.e. in 3352, on 13 Dec at 14:01:54 _ much earlier than Dudley's prediction, so start getting ready now!]

 

          6.B.    STRAIGHT LINE LINKAGES

 

          See Yates for a good survey of the field.

 

James Watt.  UK Patent 1432 _ Certain New Improvements upon Fire and Steam Engines, and upon Machines worked or moved by the same.  Granted 28 Apr 1784;  complete specification 24 Aug 1784.  14pp + 1 plate.  Pp. 4-6 & Figures 7‑12 describe Watt's parallel motion.  Yates, below, p. 170 quotes one of Watt's letters:  "... though I am not over anxious after fame, yet I am more proud of the parallel motion than of any other invention I have ever made."

P. F. Sarrus.  Note sur la transformation des mouvements rectilignes alternatifs, en mouvements circulaires; et reciproquement.  C. R. Acad. Sci. Paris 36 (1853) 1036‑1038.  6 plate linkage.  The name should be Sarrus, but it is printed Sarrut on this and the following paper.

Poncelet.  Rapport sur une transformation nouvelle des mouvements rectilignes alternatifs en mouvements circulaires et reciproquement, par Sarrut.  Ibid., 36 (1853) 1125‑1127.

A. Peaucellier.  Lettre au rédacteur.  Nouvelles Annales de Math. (2) 3 (1864) 414‑415.  Poses the problem.

A. Mannheim.  Proces‑Verbaux des sceances des 20 et 27 Juillet 1867.  Bull. Soc. Philomathique de Paris (1867) 124‑126.  ??NYS.  Reports Peaucellier's invention.

Lippman Lipkin.  Fortschritte der Physik (1871) 40‑??  ??NYS

L. Lipkin.  Über eine genaue Gelenk‑Geradführung.  Bull. Acad. St. Pétersbourg [=? Akad. Nauk, St. Petersburg, Bull.] 16 (1871) 57‑60.  ??NYS

L. Lipkin.  Dispositif articulé pour la transformation rigoureuse du mouvement circulaire en mouvement rectiligne.  Revue Univers. des Mines et de la Métallurgie de Liége 30:4 (1871) 149‑150.  ??NYS.  (Now spelled Liège.)

A. Peaucellier.  Note sur un balancier articulé a mouvement rectiligne.  Journal de Physique 2 (1873) 388‑390.  (Partial English translation in Smith, Source Book, vol. 2, pp. 324‑325.) Says he communicated it to Soc. Philomath. in 1867 and that Lipkin has since also found it.  There is also an article in Nouv. Annales de Math. (2) 12 (1873) 71‑78 (or 73?), ??NYS.

E. Lemoine.  Note sur le losange articulé du Commandant du Génie Peaucellier, destiné a remplacer le parallélogramme de Watt.  J. de Physique 2 (1873) 130‑134.  Confirms that Mannheim presented Peaucellier's cell to Soc. Philomath. on 20 Jul 1867.  Develops the inversive geometry of the cell.

[J. J. Sylvester.]  Report of the Annual General Meeting of the London Math. Soc. on 13 Nov 1873.  Proc. London Math. Soc. 5 (1873) 4 & 141.  On p. 4 is:  "Mr. Sylvester then gave a description of a new instrument for converting circular into general rectilinear motion, and into motion in conics and higher plane curves, and was warmly applauded at the close of his address."  On p. 141 is an appendix saying that Sylvester spoke "On recent discoveries in mechanical conversion of motion" to a Friday Evening's Discourse at the Royal Institution on 23 Jan 1874.  It refers to a paper 20 pages long but is not clear if or where it was published.

H. Hart.  On certain conversions of motion.  Cambridge Messenger of Mathematics 4 (1874) 82‑88 and 116‑120 & Plate I.  Hart's 5 bar linkage.  Obtains some higher curves.

A. B. Kempe.  On some new linkages.  Messenger of Mathematics 4 (1875) 121‑124 & Plate I.  Kempe's linkages for reciprocating linear motion.

H. Hart.  On two models of parallel motions.  Proc. Camb. Phil. Soc. 3 (1876‑1880) 315‑318.  Hart's parallelogram (a 5 bar linkage) and a 6 bar one.

V. Liguine.  Liste des travaux sur les systèms articulés.  Bull. d. Sci. Math. 18 (or (2) 7) (1883) 145‑160.  ??NYS ‑ cited by Kanayama.  Archibald; Outline of the History of Mathematics, p. 99, says Linguine is entirely included in Kanayama.

Gardner D. Hiscox.  Mechanical Appliances  Mechanical Movements  and  Novelties of Construction.  A second volume to accompany his previous Mechanical Movements, Powers and Devices.  Norman W. Henley Publishing Co, NY, (1904), 2nd ed., 1910.  This is filled with many types of mechanisms.  Pp. 245-247 show five straight-line linkages and some related mechanisms.

(R. Kanayama).  (Bibliography on linkages.  Text in Japanese, but references in roman type.)  Tôhoku Math. J. 37 (1933) 294‑319.

R. C. Archibald.  Bibliography of the theory of linkages.  SM 2 (1933‑34) 293‑294.  Supplement to Kanayama.

Robert C. Yates.  Geometrical Tools.  (As:  Tools; Baton Rouge, 1941);  revised ed., Educational Publishers, St. Louis, 1949.  Pp. 82-101 & 168-191.  Gets up to outlining Kempe's proof that any algebraic curve can be drawn by a linkage.

R. H. Macmillan.  The freedom of linkages.  MG 34 (No. 307) (Feb 1960) 26‑37.  Good survey of the general theory of linkages.

Michael Goldberg.  Classroom Note 312:  A six‑plate linkage in three dimensions.  MG 58 (No. 406) (Dec 1974) 287‑289.

 

          6.C.    CURVES OF CONSTANT WIDTH

 

          Such curves play an essential role in some ways to drill a square hole, etc.

 

L. Euler.  Introductio in Analysin Infinitorum.  Bousquet, Lausanne, 1748.  Vol. 2, chap. XV, esp. § 355, p. 190 & Tab. XVII, fig. 71.  = Introduction to the Analysis of the Infinite; trans. by John D. Blanton; Springer, NY, 1988-1990; Book II, chap. XV: Concerning curves with one or several diameters, pp. 212-225, esp. § 355, p. 221 & fig. 71, p. 481.  This doesn't refer to constant width, but fig. 71 looks very like a Reuleaux triangle.

L. Euler.  De curvis triangularibus.  (Acta Acad. Petropol. 2 (1778(1781)) 3‑30)  = Opera Omnia (1) 28 (1955) 298‑321.  Discusses triangular versions.

M. E. Barbier.  Note sur le problème de l'aiguille et jeu du joint couvert.  J. Math. pures appl. (2) 5 (1860) 273‑286.  Mentions that  perimeter = π * width.

F. Reuleaux.  Theoretische Kinematik; Vieweg, Brauschweig, 1875.  Translated:  The Kinematics of Machinery.  Macmillan, 1876;  Dover, 1964.  Pp. 129‑147.

Gardner D. Hiscox.  Mechanical Appliances  Mechanical Movements  and  Novelties of Construction.  A second volume to accompany his previous Mechanical Movements, Powers and Devices.  Norman W. Henley Publishing Co, NY, (1904), 2nd ed., 1910. 

Item 642: Turning a square by circular motion, p. 247.  Plain face, with four pins forming a centred square, is turned by the lathe.  A triangular follower is against the face, so it is moved in and out as a pin moves against it.  This motion is conveyed by levers to the tool which moves in and out against the work which is driven by the same lathe.

Item 681: Geometrical boring and routing chuck, pp. 257-258.  Shows it can make rectangles, triangles, stars, etc.  No explanation of how it works.

Item 903A: Auger for boring square holes, pp. 353-354.  Uses two parallel rotating cutting wheels.

H. J. Watts.  US Patents 1,241,175‑7 _ Floating tool‑chuck;  Drill or boring member;  Floating tool‑chuck.  Applied 30 Nov 1915;  1 Nov 1916;  22 Nov 1916;  all patented 25 Sep 1917.  2 + 1,  2 + 1,  4 + 1  pp + pp diagrams.  Devices for drilling square holes based on the Reuleaux triangle.

T. Bonnesen & W. Fenchel.  Theorie der konvexen Körper.  Berlin, 1934;  reprinted by Chelsea, 1971.  Chap. 15: Körper konstanter Breite, pp. 127‑141.  Surveys such curves with references to the source material.

G. D. Chakerian & H. Groemer.  Convex bodies of constant width.  In:  Convexity and Its Applications; ed. by Peter M. Gruber & Jörg M. Wills; Birkhäuser, Boston, 1983.  Pp. 49‑96.  (??NYS _ cited in MM 60:3 (1987) 139.)  Bibliography of some 250 items since 1930.

 

          6.D.   FLEXAGONS

 

          These were discovered by Arthur H. Stone, an English graduate student at Princeton in 1939.  American paper was a bit wider than English and would not fit into his notebooks, so he trimmed the edge off and had a pile of long paper strips which he played with and discovered the basic flexagon.  Fellow graduate students Richard P. Feynman, Bryant Tuckerman and John W. Tukey joined in the investigation and developed a considerable theory.  One of their fathers was a patent attorney and they planned to patent the idea and began to draw up an application, but the exigencies of the 1940s led to its being put aside, though knowledge of it spread as mathematical folklore.  E.g. Tuckerman's father, Louis B. Tuckerman, lectured on it at the Westinghouse Science Talent Search in the mid 1950s.

          S&B, pp. 148‑149, show several versions.  Most square versions (tetraflexagons or magic books) don't fold very far and are really just extended versions of the Jacob's Ladder _ see 11.L

 

Martin Gardner.  Cherchez la Femme [magic trick].  Montandon Magic Co., Tulsa, Okla., 1946.  Reproduced in:  Martin Gardner Presents; Richard Kaufman and Alan Greenberg, 1993, pp. 361-363.  [In:  Martin Gardner Presents, p. 404, this is attributed to Gardner, but Gardner told me that Roger Montandon had the copyright _ ??  I have learned a little more about Gardner's early life _ he supported himself by inventing and selling magic tricks about this time, so it may be that Gardner devised the idea and sold it to Montandon.].  A hexatetraflexagon.

"Willane".  Willane's Wizardry.  Academy of Recorded Crafts, Arts and Sciences, Croydon, 1947.  A trick book, pp. 42-43.  Same hexatetraflexagon.

Sidney Melmore.  A single‑sided doubly collapsible tessellation.  MG 31 (No. 294) (1947) 106.  Forms a Möbius strip of three triangles and three rhombi.  He sees it has two distinct forms, but doesn't see the flexing property!!

Margaret Joseph.  Hexahexaflexagrams.  MTr 44 (Apr 1951) 247‑248.  No history.

William R. Ransom.  A six‑sided hexagon.  SSM 52 (1952) 94.  Shows how to number the 6 faces.  No history.

F. G. Maunsell.  Note 2449:  The flexagon and the hexahexaflexagram.  MG 38 (No. 325) (Sep 1954) 213‑214.  States that Joseph is first article in the field and that this is first description of the flexagon.  Gives inventors' names, but with Tulsey for Tukey.

R. E. Rogers & Leonard L. D'Andrea.  US Patent 2,883,195 _ Changeable Amusement Devices and the Like.  Applied 11 Feb 1955;  patented 21 Apr 1959.  2pp + 1p correction + 2pp diagrams.  Clearly shows the 9 and 18 triangle cases and notes that one can trim the triangles into hexagons so the resulting object looks like six small hexagons in a ring.

M. Gardner.  Hexa‑hexa‑flexagon  and  Cherchez la femme.  Hugard's MAGIC Monthly 13:9 (Feb 1956) 391.  Reproduced in his:  Encyclopedia of Impromptu Magic; Magic Inc., Chicago, 1978, pp. 439-442.  Describes hexahexa and the hexatetra of Gardner/Montandon & Willane.

M. Gardner.  SA (Dec 1956) = 1st Book, chap. 1.  His first article in SA!!

Joan Crampin.  Note 2672:  On note 2449.  MG 41 (No. 335) (Feb 1957) 55‑56.  Extends to a general case having  9n  triangles of  3n  colours.

C. O. Oakley & R. J. Wisner.  Flexagons.  AMM 64:3 (Mar 1957) 143‑154.

Donovan A. Johnson.  Paper Folding for the Mathematics Class.  NCTM, 1957, pp. 24-25: Hexaflexagons.  Describes the simplest case, citing Joseph.

Roger F. Wheeler.  The flexagon family.  MG 42 (No. 339) (Feb 1958) 1‑6.  Improved methods of folding and colouring.

M. Gardner.  SA (May 1958) = 2nd Book, chap. 2.  Tetraflexagons and flexatube.

P. B. Chapman.  Square flexagons.  MG 45 (1961) 192‑194.  Tetraflexagons.

Anthony S. Conrad & Daniel K. Hartline.  Flexagons.  TR 62-11, RIAS, (7212 Bellona Avenue, Baltimore 12, Maryland,) 1962, 376pp.  This began as a Science Fair project in 1956 and was then expanded into a long report.  The authors were students of Harold V. McIntosh who kindly sent me one of the remaining copies in 1996.  They discover how to make any chain of polygons into a flexagon, provided certain relations among angles are satisfied.  The bibliography includes almost all the preceeding items and adds the references to the Rogers & D'Andrea patent, some other patents (??NYS) and a number of ephemeral items:  Conrad produced an earlier RIAS report, TR 60-24, in 1960;  Allan Phillips wrote a mimeographed paper on hexaflexagons;  McIntosh wrote an unpublished paper on flexagons;  Mike Schlesinger wrote an unpublished paper on Tuckerman tree theory. 

Sidney H. Scott.  How to construct hexaflexagons.  RMM 12 (Dec 1962) 43‑49.

William R. Ransom.  Protean shapes with flexagons.  RMM 13 (Feb 1963) 35‑37.  Describes 3‑D shapes that can be formed.  c= Madachy, below.

Robert Harbin.  Party Lines.  Op. cit. in 5.B.1.  1963.  The magic book, pp. 124-125.  As in Gardner's Cherchez la Femme and Willane.

Pamela Liebeck.  The construction of flexagons.  MG 48 (No. 366) (Dec 1964) 397‑402.

Joseph S. Madachy.  Mathematics on Vacation.  (Scribners, NY, 1966);  c= Madachy's Mathematical Recreations.  Dover, 1979.  Other flexagon diversions, pp. 76‑81.  Describes 3‑D shapes that one can form.  Based on Ransom, RMM 13.

Lorraine Mottershead.  Investigations in Mathematics.  Blackwell, Oxford, 1985.  Pp. 66-75.  Describes various tetra- and hexa-flexagons.

Douglas A. Engel.  Hexaflexagon + HFG = slipagon!  JRM 25:3 (1993) 161-166.  Describes his slipagons, which are linked flexagons.

Robert E. Neale (154 Prospect Parkway, Burlington, Vermont, 05401, USA).  Self-designing tetraflexagons.  12pp document received in 1996 describing several ways of making tetraflexagons without having to tape or paste.  He starts with a creased square sheet, then makes some internal tears or cuts and then folds things through to miraculously obtain a flexagon!

 

          6.E.    FLEXATUBE

 

          This is the square cylindrical tube that can be inverted by folding.  It was also invented by Arthur H. Stone, c1939, cf 6.D.

 

J. Leech.  A deformation puzzle.  MG 39 (No. 330) (Dec 1955) 307.  Doesn't know source.  Says there are three solutions.

M. Gardner.  Flexa-tube puzzle.  Ibidem 7 (Sep 1956) 129.  Cites the inventors of the flexagons and the articles of Maunsell and Leech (but he doesn't have its details).  (I have a note that this came with attached sample, but the copy I have doesn't indicate such.)

T. S. Ransom.  Flexa-tube solution.  Ibidem 9 (Mar 1957) 174.

M. Gardner.  SA (May 1958) = 2nd Book, chap. 2.  Says Stone invented it and shows Ransom's solution.

H. Steinhaus.  Mathematical Snapshots.  Not in the Stechert, NY, 1938, ed. nor the OUP, NY, 1950 ed.  OUP, NY:  1960: pp. 189‑193 & 326;  1969 (1983): pp. 177-181 & 303.  Erroneous attribution to the Dowkers.  Shows a different solution than Ransom's.

John Fisher.  John Fisher's Magic Book.  Muller, London, 1968.  Homage to Houdini, pp. 152‑155.  Detailed diagrams of the solution, but no history.

Highland Games (2 Harpers Court, Dingwall, Ross-Shire, IV15 9HT) makes a version called Table Teaser, made in a strip with end pieces magnetic.  Pieces are coloured so to produce several folding and inverting problems other than the usual one.  Bought in 1995.

 

          6.F.    POLYOMINOES, ETC.

 

          See S&B, pp. 15‑18.  See 6.F.1, 6.F.3, 6.F.4 & 6.F.5 for early occurrences of polyominoes.

          NOTATION.  Each of the types of puzzle considered has a basic unit and pieces are formed from a number of these units joined edge to edge.  The notation  N: n1, n2, ....  denotes a puzzle with  N  pieces, of which  ni  pieces consist of  i  basic units.  If  ni are single digit numbers, the intervening commas and spaces will be omitted, but the digits will be grouped by fives, e.g.  15: 00382 11.

 

Polyiamonds:  Scrutchin;  John Bull;  Daily Sketch;  B. T.s Zig-Zag;  Daily Mail;  Miller (1960);  Guy (1960);  Reeve & Tyrell;  O'Beirne (2 & 9 Nov 1961);  Gardner (Dec 1964 & Jul 1965);  Torbijn;  Meeus;  Gardner (Aug 1975);  Guy (1996, 1999);  Knuth,

Polycubes:  Rawlings (1939);  Editor (1948);  Niemann (1948);  French (1948);  Editor (1948);  Niemann (1948);  Gardner (1958);  Besley (1962);  Gardner (1972)

Solid Pentominoes:  Nixon (1948);  Niemann (1948);  Gardner (1958);  Miller (1960);  Bouwkamp (1967, 1969, 1978)

Polyaboloes:  Hooper (1774);  Book of 500 Puzzles (1859);  J. M. Lester (1919);  O'Beirne (21 Dec 61  &  18 Jan 62)

Polyhexes:  Gardner (1967);  Te Riele & Winter

Polysticks:  Benjamin;  Barwell;  General Symmetrics;  Wiezorke & Haubrich;  Knuth;  Jelliss;

Polyrhombs or Rhombiominoes:  Lancaster (1918);  Jones (1992).

Polylambdas:  Roothart.

Polyspheres _ see Section 6.AZ.

 

                    GENERAL REFERENCES

 

G. P. Jellis.  Special Issue on Chessboard Dissections.  Chessics 28 (Winter 1986) 137‑152.  Discusses many problems and early work in Fairy Chess Review.

Branko Grünbaum & Geoffrey C. Shephard.  Tilings and Patterns.  Freeman, 1987.  Section 9.4: Polyiamonds, polyominoes and polyhexes, pp. 497-511.  Good outline of the field with a number of references otherwise unknown.

Michael Keller.  A polyform timeline.  World Game Review 9 (Dec 1989) 4-5.  This outlines the history of polyominoes and other polyshapes.  Keller and others refer to polyoboloes as polytans.

Rodolfo Marcelo Kurchan (Parana 960 5 "A", 1017 Buenos Aires, Argentina).  Puzzle Fun, starting with No. 1 (Oct 1994).  This is a magazine entirely devoted to polyomino and other polyform puzzles.  Many of the classic problems are extended in many ways here.  In No. 6 (Aug 1995) he presents a labelling of the 12 hexiamonds by the letters  A, C, H, I, J, M, O, P, S, V, X, Y,  which he obtained from Anton Hanegraaf.  I have never seen this before.

 

Hooper.  Rational Recreations.  Op. cit. in 4.A.1.  1774.  Vol. 1, recreation 23, pp. 64-66.  Considers figures formed of isosceles right triangles.  He has eight of these, coloured with eight colours, and uses some of them to form "chequers or regular four-sided figures, different either in form or colour".

Book of 500 Puzzles.  1859.  Triangular problem, pp. 74-75.  Identical to Hooper, dropping the last sentence.

Dudeney.  CP.  1907.  Prob. 74: The broken chessboard, pp. 119‑121 & 220‑221.  The 12 pentominoes and a  2 x 2.

A. Aubry.  Prob. 3224.  Interméd. Math 14 (1907) 122-124.  ??NYS _ cited by Grünbaum & Shephard who say Aubry has something of the idea or the term polyominoes.

G. Quijano.  Prob. 3430.  Interméd. Math 15 (1908) 195.  ??NYS _ cited by Grünbaum & Shephard, who say he first asked for the number of  n‑ominoes.

Thomas Scrutchin.  US Patent 895,114 _ Puzzle.  Applied 20 Feb 1908;  patented 4 Aug 1908.  2pp + 1p diagrams.  Mentioned in S&B, p. 18.  A polyiamond puzzle _ triangle of side 8, hence with 64 triangles, apparently cut into 10 pieces (my copy is rather faint _ replace??).

Thomas W. Lancaster.  US Patent 1,264,944 _ Puzzle.  Filed 7 May 1917;  patented 7 May 1918.  2pp + 1p diagrams.  For a general polyrhomb puzzle making a rhombus.  His diagram shows an  11 x 11  rhombus filled with  19  pieces formed from  4  to  10  rhombuses.

John Milner Lester.  US Patent 1,290,761 _ Game Apparatus.  Filed 6 Feb 1918;  patented 7 Jan 1919.  2pp + 3pp diagrams.  Fairly general assembly puzzle claims.  He specifically illustrates a polyomino puzzle and a polyabolo puzzle.  The first has a Greek cross of edge  3  (hence containing  45  unit cells) to be filled with polyominoes _  11: 01154.  The second has an  8-pointed star formed by superimposing two  4 x 4  squares.  This has area  20  and hence contains  40  isosceles right triangles of edge  1, which is the basic unit of this type of puzzle.  There are  11: 0128  pieces.

Blyth.  Match-Stick Magic.  1921.  Spots and squares, pp. 68-73.  He uses matchsticks broken in thirds, so it is easier to describe with units of one-third.  6 units,  4 doubles and  2 triples.  Some of the pieces have black bands or spots.  Object is to form polyomino shapes without pieces crossing, but every intersection must have a black spot.  19  polyomino shapes are given to construct, including  7  of the pentominoes, though some of the shapes are only connected at corners.

"John Bull" Star of Fortune Prize Puzzle.  1922.  This is a puzzle with  20  pieces, coloured red on one side, containing  6  through  13  triangles to be assembled into a star of David with  4  triangles along each edge (hence  12 x 16 = 192  triangles).  Made by Chad Valley.  Prize of £250 for a red star matching the key solution deposited at a bank;  £150 for solution closest to the key;  £100 for a solution with  10  red and  10  grey pieces, or as nearly as possible.  Closing date of competition is 27 Dec 1922.  Puzzle made by Chad Valley Co. as a promotional item for John Bull magazine, published by Odhams Press.  A copy is in the toy shop of the Buckleys Shop Museum, Battle, East Sussex, to whom I am indebted for the chance to examine the puzzle and a xerox of the puzzle, box and solution.

Daily Sketch Jig-Saw Puzzle.  By Chad Valley.  Card polyiamonds.  39: 0,0,1,5,6, 12,9,6,  with a path printed on one side, to assemble into a shape of  16  rows of  15  with four corners removed and so the printed sides form a continuous circuit.  In box with shaped bottom.  Instructions on inside cover and loose sheet to submit solution.  No dates given, but appears to be 1920s, though it is somewhat similar to the Daily Mail Crown Puzzle of 1953 _ cf below _ so it might be much later.

B. T.s Zig-Zag.  B.T. is a Copenhagen newspaper.  Polyiamond puzzle.  33: 0,0,1,2,5  6,7,2,2,4  1,1,1,  Some repetitions, so I only see 20 different shapes.  To be fit into an irregular frame.  Solution given on 23 Nov 1931, pp. 1-2.  (I have a photocopy of the form to fill in; an undated set of rules, apparently from the paper, saying the solutions must be received by 21 Nov; and the pages giving the solution; provided by Jan de Geus.)

Herbert D. Benjamin.  Problem 1597: A big cutting-out design _ and a prize offer.  Problemist Fairy Chess Supplement (later called Fairy Chess Review) 2:9 (Dec 1934) 92.  Finds the  35  hexominoes and asks if they form a  14 x 15  rectangle.  Cites Dudeney (Tribune (20 Dec 1906));  Loyd (OPM (Apr-Jul 1908)) (see 6.F.1);  Dudeney (CP, no. 74) (see above)  and some other chessboard dissections.  Jelliss says this is the first dissection problem in this journal.

F. Kadner.  Solution 1597.  Problemist Fairy Chess Supplement (later called Fairy Chess Review) 2:10 (Feb 1935) 104-105.  Shows the  35  hexominoes cannot tile a rectangle by two arguments, both essentially based on two colouring.  Gives some other results and some problems are given as 1679-1681 _ ??NYS.

William E. Lester.  Correction to 1597.  Problemist Fairy Chess Supplement (later called Fairy Chess Review) 2:11 (Apr 1935) 121.  Corrects an error in Kadner.  Finds a number of near-solutions.  Editor says Kadner insists the editor should take credit for the two-colouring form of the previous proof.

Frans Hansson, proposer & solver?.  Problem 1844.  Problemist Fairy Chess Supplement (later called Fairy Chess Review) 2:12 (Jun 1935) 128  &  2:13 (Aug 1935) 135.  Finds both  3 x 20  pentomino rectangles.

W. E. Lester & B. Zastrow, proposers.  Problem 1923.  Problemist Fairy Chess Supplement (later called Fairy Chess Review) 2:13 (Aug 1935) 138.  Take an  8 x 8  board and remove its corners.  Fill this with the  12  pentominoes.

H. D. Benjamin, proposer.  Problem 1924.  Problemist Fairy Chess Supplement (later called Fairy Chess Review) 2:13 (Aug 1935) 138.  Dissect an  8 x 8  into the  12  pentominoes and the I-tetromino.  Need solution _ ??NYS.

Thomas Rayner Dawson & William E. Lester.  A notation for dissection problems.  Fairy Chess Review 3:5 (Apr 1937) 46‑47.  Gives all  n‑ominoes up to  n = 6.  Describes the row at a time notation.  Shows the pentominoes and a  2 x 2  cover the chessboard with the  2 x 2  in any position.  Asserts there are  108  7‑ominoes and  368  8‑ominoes _ citing F. Douglas & W. E. L[ester] for the hexominoes and J. Niemann for the heptominoes.

H. D. Benjamin, proposer.  Problem 3228.  Fairy Chess Review 3:12 (Jun 1938) 129.  Dissect a  5 x 5  into the five tetrominoes and a pentomino so that the pentomino touches all the tetrominoes along an edge.  Asserts the solution is unique.  Refers to problems 3026‑3030 _ ??NYS.

H. D. Benjamin, proposer.  Problem 3229.  Fairy Chess Review 3:12 (Jun 1938) 129.  Dissect an  8 x 8  into the  12  pentominoes and a tetromino so that all pieces touch the edge of the board.  Asserts only one tetromino works.

T. R. D[awson], proposer.  Problems 3230-1.  Fairy Chess Review 3:12 (Jun 1938) 129.  Extends prob. 3229 to ask for solutions with  12  pieces on the edge, using two other tetrominoes.  Thinks it cannot be done with the remaining two tetrominoes.

Editorial note:  The colossal count.  Fairy Chess Review 3:12 (Jun 1938) 131.  Describes progress on enumerating  8-ominoes (four people get  368  but Niemann gets  369)  and  9‑ominoes (numbers vary from  1237  to  1285).  All workers are classifying them by the size of the smallest containing rectangle.

W. H. Rawlings, proposer.  Problem 3930.  Fairy Chess Review 4:3 (Nov 1939) 28.  How many pentacubes are there?  Ibid 4:4 (Feb 1940) 75,  reports that both  25  or  26  are claimed, but the editor has only seen  24.  Ibid 4:5 (Apr 1940) 85, reports that R. J. F[rench] has clearly shown there are  23  _ but this considers reflections as equal _ cf the 1948 editorial note.

R. J. French, proposer and solver.  Problem 4149.  Fairy Chess Review 4:3 (Nov 1939) 43  &  4:6 (Jun 1940) 93.  Asks for arrangement of the pentominoes with the largest hole and gives one with  127  squares in the hole.  (See:  G. P. Jellis; Comment on Problem 1277; JRM 22:1 (1990) 69.  This reviews various earlier solutions and comments on Problem 1277.)

J. Niemann.  Item 4154: "The colossal count".  Fairy Chess Review 4:3 (Nov 1939) 44-45.  Announces that there are  369  8-ominoes,  1285  9-ominoes and  4654  10-ominoes, but Keller and Jelliss note that he missed a  10‑omino which was not corrected until 1966.

H. D. Benjamin.  Unpublished notes.  ??NYS _ cited and briefly described in G. P. Jelliss; Prob. 48 _ Aztec tetrasticks; G&PJ 2 (No. 17) (Oct 1999) 320.  Jelliss says Benjamin studied polysticks, which he called 'lattice dissections' around 1946-1948 and that some results by him and T. R. Dawson were entered in W. Stead's notebooks but nothing is known to have been published.  For orders  1, 2, 3, 4,  there are  1, 2, 5, 16  polysticks.  Benjamin formed these into a  6 x 6  lattice square.  Jelliss then mentions Barwell's rediscovery of them and goes on to a new problem _ see Knuth, 1999.

D. Nixon, proposer and solver.  Problem 7560.  Fairy Chess Review 6:16 (Feb 1948) 12  &  6:17 (Apr 1948) 131.  Constructs  3 x 4 x 5  from solid pentominoes.

Editorial discussion: Space dissection.  Fairy Chess Review 6:18 (Jun 1948) 141-142.  Says that several people have verified the  23  pentacubes but that  6  of them have mirror images, making  29  if these are considered distinct.  Says F. Hansson has found  77  6‑cubes (these exclude mirror images and the  35  solid  6-ominoes).  Gives many problems using  n-cubes and/or solid polyominoes, which he calls flat  n-cubes _ some are corrected in 7:2 (Oct 1948) 16 (erroneously printed as 108).

J. Niemann.  The dissection count.  Item 7803.  Fairy Chess Review 7:1 (Aug 1948) 8 (erroneously printed as 100).  Reports on counting  n‑cubes.  Gets the following.

                  n =                             4                    5                    6                    7

          flat pieces                            5                  12                  35                 108

          non-flats                              2                  11                  77                 499

              TOTAL                          7                  23                 112                 607

          mirror images                      1                    6                  55                 416

            GRAND TOTAL               8                  29                 167               1023

R. J. French.  Space dissections.  Fairy Chess Review 7:2 (Oct 1948) 16 (erroneously printed as 108).  French writes that he and A. W. Baillie have corrected the number of  6-cubes to  35 + 77 + 54 = 166.  Baillie notes that every  6-cube lies in two layers _ i.e. has some width  £ 2  _ and asks for the result for  n‑cubes as prob. 7879.  [I suspect the answer is that  n £ 3k  implies that an  n-cube has some width  £ k.]  Editor adds some corrections to the discussion in 6:18.

Editorial note.  Fairy Chess Review 7:3 (Dec 1948) 23.  Niemann and Hansson confirm the number  166  given in 7:2.

Daily Mail Crown Puzzle.  Made by Chad Valley Co.  1953.  26 pieces, coloured on one side, to be fit into a crown shape.  11 are border pieces and easily placed.  The other 15 are polyiamonds:  15: 00112 24012 11.  Prize of £100 for solution plus best slogan, entries due on 8 Jun 1953.

S. W. Golomb.  Checkerboards and polyominoes.  AMM 61 (1954) 675‑682.  Mostly concerned with covering the  8 x 8  board with copies of polyominoes.  Shows one covering with the  12  pentominoes and the square tetromino.  Mentions that the idea can be extended to hexagons.  S&B, p. 18, and Gardner (Dec 1964) say he mentions triangles, but he doesn't.

Walter S. Stead.  Dissection.  Fairy Chess Review 9:1 (Dec 1954) 2‑4.  Gives many pentomino and hexomino patterns _ e.g. one of each pattern of  8 x 8  with a  2 x 2  square deleted.  "The possibilities of the  12  fives are not infinite but they will provide years of amusement."  Includes  3 x 20,  4 x 15,  5 x 12  and  6 x 10  rectangles.  No reference to Golomb.  In 1955, Stead uses the  108  heptominoes to make a  28 x 28  square with a symmetric hole of size  28  in the centre _ first printed as cover of Chessics 28 (1986).

Jules Pestieau.  US Patent 2,900,190 _ Scientific puzzle.  Filed 2 Jul 1956;  patented 18 Aug 1959.  2pp + 1p diagrams.  For the  12  pentominoes!  Diagram shows the  6 x 10  solution with two  5 x 6  rectangles and shows the two-piece non-symmetric equivalence of the  N  and  F  pieces.  Pieces have markings on one side which may be used _ i.e. pieces may not be turned over.  Mentions possibility of using  n-ominoes.

Gardner.  SA (Dec 1957) = 1st Book, chap. 13.  Exposits Golomb and Stead.  Gives number of  n-ominoes for  n = 1, ..., 7.  1st Book describes Scott's work.  Says a pentomino set called 'Hexed' was marketed in 1957.  (John Brillhart gave me and my housemates an example in 1960 _ it took us two weeks to find our first solution.)

Dana Scott.  Programming a Combinatorial Puzzle.  Technical Report No. 1, Dept. of Elec. Eng., Princeton Univ., 1958, 20pp.  Uses MANIAC to find  65  solutions for pentominoes on an  8 x 8  board with square  2 x 2  in the centre.  Notes that the  3 x 20  pentomino rectangle has just two solutions.  In 1999, Knuth notes that the total number of solutions with the  2 x 2  being anywhere does not seem to have ever been published and he finds  16146.

M. Gardner.  SA (Sep 1958) c= 2nd Book, chap. 6.  First general mention of solid pentominoes, pentacubes, tetracubes.  In the Addendum in 2nd Book, he says Theodore Katsanis of Seattle suggested the eight tetracubes and the  29  pentacubes in a letter to Gardner on 23 Sep 1957.  He also says that Julia Robinson and Charles W. Stephenson both suggested the solid pentominoes.

C. Dudley Langford.  Note 2793:  A conundrum for form VI.  MG 42 (No. 342) (Dec 1958) 287.  4  each of the  L,  N,  and  T (= Y)  tetrominoes make a  7 x 7  square with the centre missing.  Also nine pieces make a  6 x 6  square but this requires an even number of  Ts.

J. C. P. Miller.  Pentominoes.  Eureka 23 (1960) 13‑16.  Gives the Haselgroves' number of  2339  solutions for the  6 x 10  and says there are  2  solutions for the  3 x 20.  Says Lehmer suggests assembling  12  solid pentominoes into a  3 x 4 x 5  and van der Poel suggests assembling the  12  hexiamonds into a rhombus.

C. B. & Jenifer Haselgrove.  A computer program for pentominoes.  Ibid., 16‑18.  Outlines program which found the  2339  solutions for the  6 x 10.  It is usually said that they also found all solutions of the  3 x 20,  4 x 15  and  5 x 12,  but I don't see it mentioned here and in JRM 7:3 (1974) 257, it is reported that Jenifer (Haselgrove) Leech stated that only the  6 x 10  and  3 x 20  were done  in 1960, but that she did the  5 x 12  and  4 x 15  with a new program in c1966.  See Fairbairn, c1962, and Meeus, 1973.

Richard K. Guy.  Some mathematical recreations I  &  II.  Nabla [= Bull. Malayan Math. Soc.] 7 (Oct  &  Dec 1960) 97-106  &  144-153.  Considers handed polyominoes, i.e. polyominoes when reflections are not considered equivalent.  Notes that neither the  5  plain nor the  7  handed tetrominoes can form a rectangle.  The  10  chequered handed tetrominoes form  4 x 10  and  5 x 8  rectangles and he has several solutions of each.  There is no  2 x 20  rectangle.  Discusses MacMahon pieces _ cf. 5.H.2 _ and polyiamonds.  He uses the word 'hexiamond', but not 'polyiamond'.  He considers making a 'hexagon' from the 19 hexiamonds.  Part II considers solid problems and uses the term 'solid pentominoes'.

Solomon W. Golomb.  The general theory of polyominoes: part 2 _ Patterns and polyominoes.  RMM 5 (Oct 1961) 3-14.  ??NYR.

J. E. Reeve & J. A. Tyrrell.  Maestro puzzles.  MG 45 (No. 353) (Oct 1961) 97‑99.  Discusses hexiamond puzzles, using the  12  reversible pieces.  [The puzzle was marketed under the name 'Maestro' in the UK.]

T. H. O'Beirne.  Pell's equation in two popular problems.  New Scientist 12 (No. 258) (26 Oct 1961) 260‑261.

T. H. O'Beirne.  Pentominoes and hexiamonds.  New Scientist 12 (No. 259) (2 Nov 1961) 316‑317.  This is the first use of the word 'polyiamond'.  He considers the  19  one‑sided pieces.  He says he devised the pieces and R. K. Guy has already published many solutions in Nabla.  He asks for the number of ways the 18 one-sided pentominoes can fill a  9 x 10.  In 1999, Knuth found this would take several months.

T. H. O'Beirne.  Some hexiamond solutions:  and an introduction to a set of  25  remarkable points.  New Scientist 12 (No. 260) (9 Nov 1961) 378‑379.

Maurice J. Povah.  Letter.  MG 45 (No. 354) (Dec 1961) 342.  States Scott's result of  65  and the Haselgroves' result of  2339  (computed at Manchester).  Says he has over  7000  solutions for the  8 x 8  board using a  2 x 2.

T. H. O'Beirne.  For boys, men and heroes.  New Scientist 12 (No. 266) (21 Dec 1961) 751‑752.

T. H. O'Beirne.  Some tetrabolic difficulties.  New Scientist 13 (No. 270) (18 Jan 1962) 158‑159.  These two columns are the first mention of tetraboloes, so named by S. J. Collins.

R. A. Fairbairn.  Pentomino Problems: The  6 x 10,  5 x 12,  4 x 15,  and  3 x 20  Rectangles _ The Complete Drawings.  Unpublished MS, undated, but c1962, based the Haselgrove's work of 1960.  ??NYS _ cited by various authors, e.g.  Madachy (1969), Torbijn (1969), Meeus (1973).  Madachy says Fairbairn is from Willowdale, Ontario, and takes some examples from his drawings.  However, the dating is at variance with Jenifer Haselgrove's 1973 statement - cf Haselgrove, 1960.  Perhaps this MS is somewhat later??  Does anyone know where this MS is now?  Cf Meeus, 1973.

Serena Sutton Besley.  US Patent 3,065,970 _ Three dimensional puzzle.  Filed 6 Jul 1960, issued on 27 Nov 1962.  2pp + 4pp diagrams.  For the  29  pentacubes, with one piece duplicated giving a set of  30.  Klarner had already considered omitting the  1 x 1 x 5  and found that he could make two separate  2 x 5 x 7s.  Besley says the following can be made:  5 x 5 x 6,  3 x 5 x 10,  2 x 5 x 15,  2 x 3 x 25;  3 x 5 x 6,  3 x 3 x 10,  2 x 5 x 9,  2 x 3 x 15;  3 x 4 x 5,  2 x 5 x 6,  2 x 3 x 10  (where the latter three are made with the  12  solid pentominoes and the previous four are made with the  18  non-planar pentacubes) but detailed solutions are only given for the  5 x 5 x 6,  3 x 5 x 6,  3 x 4 x 5.  Mentions possibility of  n-cubes. 

M. Gardner.  Polyiamonds.  SA (Dec 1964) = 6th Book, chap. 18.  Exposits basic ideas and results for the 12 double sided hexiamonds.  Poses several problems which are answered by readers.  The six-pointed star using 8 pieces has a unique solution.  John G. Fletcher and Jenifer (Haselgrove) Leech both showed the  3 x 12  rhombus is impossible.  Fletcher found the  3 x 11  rhombus has  24  solutions, all omitting the 'bat'.  Leech found  155  solutions for the  6 x 6  rhombus and  74  solutions for the  4 x 9.  Mentions there are  160  9-iamonds, one with a hole. 

John G. Fletcher.  A program to solve the pentomino problem by the recursive use of macros.  Comm. ACM 8 (1965) 621-623.  ??NYS _ described by Knuth in 1999 who says that Fletcher found the 2339 solutions for the  6 x 10  in 10 minutes on an IBM 7094 and that the program remains the fastest known method for problems of placing the 12 pentominoes.

M. Gardner.  Op art.  SA (Jul 1965) = 6th Book, chap. 24.  Shows the 24 heptiamonds and discusses which will tile the plane.

Solomon W. Golomb.  Tiling with polyominoes.  J. Combinatorial Theory 1 (1966) 280-296.  ??NYS.  Extended by his 1970 paper.

T. R. Parkin.  1966.  ??NYS _ cited by Keller.  Finds  4655  10-ominoes.

M. Gardner.  SA (Jun 1967) = Magic Show, chap. 11.  First mention of polyhexes.

C. J. Bouwkamp.  Catalogue of Solutions of the Rectangular  3 x 4 x 5  Solid Pentomino Problem.  Dept. of Math., Technische Hogeschool Eindhoven, July 1967, reprinted 1981, 310pp.

C. J. Bouwkamp.  Packing a rectangular box with the twelve solid pentominoes.  J. Combinatorial Thy. 7 (1969) 278‑280.  He gives the numbers of solutions for rectangles as 'known'.

              2 x 3 x 10   can be packed in      12  ways, which are given.

              2 x 5 x   6   can be packed in    264  ways.

              3 x 4 x   5   can be packed in  3940  ways.  (See his 1967 report.)

T. R. Parkin,  L. J. Lander  &  D. R. Parkin.  Polyomino enumeration results.  Paper presented at the SIAM Fall Meeting, Santa Barbara, 1 Dec 1967.  ??NYS _ described by Madachy, 1969.  Gives numbers of  n-ominoes, with and without holes, up to  n = 15,  done two independent ways.

Joseph S. Madachy.  Pentominoes _ Some solved and unsolved problems.  JRM 2:3 (Jul 1969) 181-188.  Gives the numbers of Parkin, Lander & Parkin.  Shows various examples where a rectangle splits into two congruent halves.  Discusses various other problems, including Bouwkamp's  3 x 4 x 5  solid pentomino problem.  Bouwkamp reports that the final total of 3940 was completed on 16 Mar 1967 after about three years work using three different computers, but that a colleague's program would now do the whole search in about three hours.

P. J. Torbijn.  Polyiamonds.  JRM 2:4 (Oct 1969) 216-227.  Uses the double sided hexiamonds and heptiamonds.  A few years before, he found, by hand, that there are  156  ways to cover the  6 x 6  rhombus with the 12 hexiamonds and  74  ways for the  4 x 9,  but could find no way to cover the  3 x 12.  The previous year, John G. Fletcher confirmed these results with a computer and he displays all of these _ but this contradicts Gardner (Dec 64) _ ??  He gives several other problems and results, including using the 24 heptiamonds to form  7 x 12,  6 x 14,  4 x 21  and  3 x 28  rhombuses.

Solomon W. Golomb.  Tlling with sets of polyominoes.  J. Combinatorial Theory 9 (1970) 60‑71.  ??NYS.  Extends his 1966 paper.  Asks which heptominoes tile rectangles and says there are two undecided cases _ cf. Marlow, 1985.  Gardner (Aug 75) says Golomb shows that the problem of determining whether a given finite set of polyominoes will tile the plane is undecidable.

C. J. Boukamp & D. A. Klarner.  Packing a box with  Y-pentacubes.  JRM 3:1 (1970) 10-26.  Substantial discussion of packings with  Y‑pentominoes and  Y-pentacubes.  Smallest boxes are  5 x 10  and  2 x 5 x 6  and  3 x 4 x 5.

Fred Lunnon.  Counting polyominoes.  IN:  Computers in Number Theory, ed. by A. O. L. Atkin & B. J. Birch; Academic Press, 1971, pp. 347-372.  He gets up through 18‑ominoes, but the larger ones can have included holes.  The numbers for  n = 1, 2, ...,  are as follows:  1, 1, 2, 5, 12,   35, 108, 369, 1285, 4655,   17073, 63600, 238591, 901971, 3426576,   13079255, 50107911, 192622052.  These values have been quoted numerous times.

Fred Lunnon.  Counting hexagonal and triangular polyominoes.  IN:  Graph Theory and Computing, ed. by R. C. Read; Academic Press, 1972, pp. 87-100.  ??NYS _ cited by Grünbaum & Shephard. 

M. Gardner.  SA (Sep 1972).  c= Knotted, chap. 3.  Says the  8  tetracubes were made by E. S. Lowe Co. in Hong Kong and marketed as "Wit's End".  Says an MIT group found  1390  solutions for the  2 x 4 x 4  box packed with tetracubes.  He reports that several people found that there are  1023  heptacubes _ but see Niemann, 1948, above.  Klarner reports that the heptacubes fill a  2 x 6 x 83.

Jean Meeus.  Some polyomino and polyamond [sic] problems.  JRM 6:3 (1973) 215-220.  (Corrections in 7:3 (1974) 257.)  Considers ways to pack a  5 x n  rectangle with some  n  pentominoes.  A. Mank found the number of ways for  n = 2, 3, ..., 11  as follows, and the number for  n = 12  was already known: 

                    0,  7,  50,  107,  541,  1387,  3377,  5865,  6814,  4103,  1010.

          Says he drew out all the solutions for the area 60 rectangles in 1972 (cf Fairbairn, c1962).  Finds that  520  of the  6 x 10  rectangles can be divided into two congruent halves, sometimes in two different ways.  For  5 x 12,  there are  380;  for  4 x 15,  there are 94.  Gives some hexomino rectangles by either deleting a piece or duplicating one, and an 'almost  11 x 19'.  Says there are  46  solutions to the  3 x 30  with the 18 one-sided pentominoes and attributes this to Mrs (Haselgrove) Leech, but the correction indicates this was found by A. Mank.

Jenifer Haselgrove.  Packing a square with Y-pentominoes.  JRM 7:3 (1974) 229.  She finds and shows a way to pack 45 Y-pentominoes into a  15 x 15,  but is unsure if there are more solutions.  In 1999, Knuth found  212  solutions.  She also reports the impossibility of using the Y-pentominoes to fill various other rectangles.

S. W. Golomb.  Trademark for 'PENTOMINOES'.  US trademark 1,008,964 issued 15 Apr 1975;  published 21 Jan 1975 as SN 435,448.  (First use:  November 1953.)  [These appear in the Official Gazette of the United States Patent Office (later Patent and Trademark Office) in the Trademarks section.]

M. Gardner.  Tiling with polyominoes, polyiamonds and polyhexes.  SA (Aug 75) (with slightly different title) = Time Travel, chap. 14.  Gives a tiling crierion of Conway.  Describes Golomb's 1966 & 1970 results. 

C. J. Bouwkamp.  Catalogue of solutions of the rectangular  2 x 5 x 6  solid pentomino problem.  Proc. Koninklijke Nederlandse Akad. van Wetenschappen A81:2 (1978) 177‑186.  Presents the  264  solutions which were first found in Sep 1967.

H. Redelmeier.  Discrete Math. 36 (1981) 191‑203.  ??NYS _ described by Jelliss.  Obtains number of n‑ominoes for  n £ 24.

Karl Scherer.  Problem 1045: Heptomino tessellations.  JRM 14:1 (1981‑82) 64.                              XX  

          Says he has found that the heptomino at the right fills a  26 x 42  rectangle.                     XXXXX

          See Dahlke below.

David Ellard.  Poly-iamond enumeration.  MG 66 (No. 438) (Dec 1982) 310‑314.  For  n = 1, ..., 12,  he gets  1, 1, 1, 3, 4,   12, 24, 66, 160, 448,   1186, 3342  n-iamonds.  One of the  8-iamonds has a hole and there are many later cases with holes.

Anon.  31: Polyominoes.  QARCH 1:8 (June 1984) 11‑13.  [This is an occasional publication of The Archimedeans, the student maths society at Cambridge.]  Good survey of counting and asymptotics for the numbers of polyominoes, up to  n = 24,  polycubes, etc.  10 references.

T. W. Marlow.  Grid dissections.  Chessics 23 (Autumn, 1985) 78‑79.

                                X                                             XX

          Shows  XXXXX  fills a  23 x 24  and   XXXXX   fills a  19 x 28.

Herman J. J. te Riele & D. T. Winter.  The tetrahexes puzzle.  CWI Newsletter [Amsterdam] 10 (Mar 1986) 33‑39.  Says there are:  7  tetrahexes,  22  pentahexes,  82  hexahexes,  333  heptahexes,  1448  octahexes.  Studies patterns of  28  hexagons.  Shows the triangle cannot be constructed from the  7  tetrahexes and gives  48  symmetric patterns that can be made.

Karl A. Dahlke.  Science News 132:20 (14 Nov 1987) 310.  (??NYS _ cited in JRM 21:3 and

                                                                               XX

          22:1 and by Marlow below.)  Shows  XXXXX  fills a  21 x 26  rectangle.

                    The results of Scherer and Dahlke are printed in JRM 21:3 (1989) 221‑223 and Dahlke's solution is given by Marlow below.

Karl A. Dahlke.  J. Combinatorial Theory A51 (1989) 127‑128.  ??NYS _ cited in JRM 22:1.  Announces a  19 x 28  solution for the above heptomino problem, but the earlier  21 x 26  solution is printed by error.  The  19 x 28  solution is printed in JRM 22:1 (1990) 68‑69.

Tom Marlow.  Grid dissections.  G&PJ 12 (Sep/Dec 1989) 185.  Prints Dahlke's result.

Brian R. Barwell.  Polysticks.  JRM 22:3 (1990) 165-175.  Polysticks are formed of unit lengths on the square lattice.  There are:  1, 2, 5, 16, 55  polysticks formed with  1, 2, 3, 4, 5  unit lengths.  He forms  5 x 5  squares with one  4-stick omitted, but he permits pieces to cross.  He doesn't consider the triangular or hexagonal cases.  See also Blyth, 1921, for a related puzzle.  Cf Benjamin, above, and Wiezorke & Haubrich, below.

General Symmetrics (Douglas Engel) produced a version of polysticks, ©1991, with  4  3‑sticks and  3  4-sticks to make a  3 x 3  square array with no crossing of pieces.

Kate Jones, proposer;  P. J. Torbijn, Jacques Haubrich, solvers.  Problem 1961 _ Rhombiominoes.  JRM 24:2 (1992) 144-146  &  25:3 (1993) 223‑225.  A rhombiomino or polyrhomb is a polyomino formed using rhombi instead of squares.  There are  20  pentarhombs.  Fit them into a  10 x 10  rhombus.  Various other questions.  Haubrich found many solutions.  See Lancaster, 1918.

Bernard Wiezorke & Jacques Haubrich.  Dr. Dragon's polycons.  CFF 33 (Feb 1994) 6-7.  Polycons (for connections) are the same as the polysticks described by Barwell in 1990, above.  Authors describe a Taiwanese version on sale in late 1993, using  10  of the  4‑sticks suitably shortened so they fit into the grooves of a  4 x 4  board _ so crossings are not permitted.  (An  n x n  board has  n+1  lines of  n  edges in each direction.)  They fit  15  of the  4-sticks onto a  5 x 5  board and determine all solutions.

                    CFF 35 (Dec 1994) 4 gives a number of responses to the article.  Brain Barwell wrote that he devised them as a student at Oxford, c1970, but did not publish until 1990.  He expected someone to say it had been done before, but no one has done so.  He also considered using the triangular and hexagonal lattice.  He had just completed a program to consider fitting  15  of the 4-sticks onto a  5 x 5  board and found over  180,000  solutions, with slightly under half having no crossings, confirming the results of Wiezorke & Haubrich.

                    Dario Uri also wrote that he had invented the idea in 1984 and called them polilati (polyedges).  Giovanni Ravesi wrote about them in Contromossa (Nov 1984) 23 _ a defunct magazine.

Chris Roothart.  Polylambdas.  CFF 34 (Oct 1994) 26-28.  A lambda is a  30o-60o-90o  triangle.  These may be joined along corresponding legs, but not along hypotenuses.  For  n = 1, 2, 3, 4, 5,  there are  1, 4, 4, 11, 12  n-lambdas.  He gives some problems using various sets of these pieces.

Richard Guy.  Letters of 29 May and 13 Jun 1996.  He is interested in using the  19  one-sided hexiamonds.  Hexagonal rings of hexagons contain  1, 6, 12  hexagons, so the hexagon with three hexagons on a side has 19 hexagons.  If these hexagons are considered to comprise six equilateral triangles, we have a board with  19 x 6  triangles.  O'Beirne asked for the number of ways to fill this board with the one-sided hexiamonds.  Guy has collected over  4200  solutions.  A program by Marc Paulhus found  907  solutions in eight hours, from which it initially estimated that there are about  30,000  solutions.  The second letter gives the final results _ there are  124,518  solutions.  This is modulo the  12  symmetries of the hexagon.  In 1999, Knuth found  124,519  and Paulhus has rerun his program and found this number.

Hilarie Korman.  Pentominoes: A first player win.  IN: Games of No Chance; ed. by Richard Nowakowski; CUP, 1997??, ??NYS - described in  William Hartston; What mathematicians get up to; The Independent Long Weekend (29 Mar 1997) 2.  This studies the game proposed by Golomb _ players alternately place one of the pentominoes on the chess board, aligned with the squares and not overlapping the previous pieces, with the last one able to play being the winner.  She used a Sun IPC Sparcstation for five days, examining about  22 x 109  positions to show the game is a first player win.

Nob Yoshigahara found in 1994 that the smallest box which can be packed with W-pentacubes is  5 x 6 x 6.  In 1997, Yoshya (Wolf) Shindo found that one can pack the  6 x 10 x 10  with Z-pentacubes, but it is not known if this is the smallest such box.  These were the last unsolved problems as to whether a box could be packed with a planar pentacube (= solid pentomino).

Marcel Gillen  &  Georges Philippe.  Twinform  462 Puzzles in one.  Solutions for Gillen's puzzle exchange at 17IPP, 1997, 32pp + covers.  Take 6 of the pentominos and place them in a  7 x 5  rectangle, then place the other six to make the same shape on top of the first shape.  There are  462  (= BC(12,6)/2)  possible puzzles and all of them have solutions.  Taking  F, T, U, W, X, Z  for the first layer, there is just one solution; all other cases have multiple solutions, totalling  22,873  solutions, but only one solution for each case is given here.)

Richard K. Guy.  O'Beirne's hexiamond.  In:  The Mathemagican and Pied Puzzler; ed. by Elwyn Berlekamp & Tom Rodgers, A. K. Peters, Natick, Massachusetts, 1999, pp. 85‑96.  He relates that O'Beirne discovered the 19 one-sided hexiamonds in c1959 and found they would fill a hexagonal shape in Nov 1959 and in Jan 1960 he found a solution with the hexagonal piece in the centre.  He gives Paulhus's results (see Guy's letters of 1996), broken down in various ways.  He gives the number of double-sided (i.e. one can turn them over) and single-sided  n-iamonds for  n = 1, ..., 7.  Cf Ellard, 1982, for many more values for the double-sided case.

                              n        1   2   3   4   5    6    7

                    double          1   1   1   3   4   12   24

                    single    1   1   1   4   6   19   44

          In 1963, Conway and Mike Guy considered looking for 'symmetric' solutions for filling the hexagonal shape with the 19 one-sided hexiamonds.  A number of these are described.

Donald E. Knuth.  Dancing links.  25pp preprint of a talk given at Oxford in Sep 1999, sent by the author.  Available as:  http://www-cs-faculty.stanford.edu/~knuth/preprints.html .  In this he introduces a new technique for backtrack programming which runs faster (although it takes more storage) and is fairly easy to adapt to different problems.  In this approach, there is a symmetry between pieces and cells.  He applies it to several polyshape problems, obtaining new, or at least unknown, results.  He extends Scott's 1958 results to get  16146  ways to pack the  8 x 8  with the 12 pentominoes and the  2 x 2.  He describes Fletcher's 1965 work.  He extends Haselgrove's 1973 work and finds 212 ways to fit 15 Y-pentominoes in a  15 x 15.  Describes Torbijn's 1969 work and Paulhus' 1996 work on hexiamonds, correcting the latter's number to  124,519.  He then looks for the most symmetric solutions for filling the hexagonal shape with the 19 one-sided hexiamonds, in the sense discussed by Guy (1999).  He then considers the 18 one-sided pentominoes (cf Meeus (1973)) and tries the  9 x 10,  but finds it would take a few months on his computer (a 500 MHz Pentium III), so he's abandoned it for now.  He then considers polysticks, citing an actual puzzle version that I've not seen.  He adapts his program to them.  He considers the 'welded tetrasticks' which have internal junction points.  There are six of these and ten if they are taken as one-sided.  The ten can be placed in a  4 x 4 grid.  There are  15  unwelded, one-sided, tetrasticks, but they do not form a square, nor indeed any nice shape.  He considers all  25  one-sided tetrasticks and asks if they can be fit into what he calls an Aztec Diamond, which is the shape looking like a square tilted 45o on the square lattice.  The rows contain  1, 3, 5, 7, 9, 7, 5, 3, 1  cells.  He thinks an exhaustive search is beyond present computing power.

G. P. Jelliss.  Prob. 48 _ Aztec tetrasticks.  G&PJ 2 (No. 17) (Oct 1999) 320.  Jelliss first discusses Benjamin's work on polysticks (see at 1946-1948 above) and Barwell's rediscovery of them (see above).  He then describes Knuth's Dancing Links and gives the Aztec Diamond problem.  Jelliss has managed to get all but one of the polysticks into the shape, but feels it is impossible to get them all in.

 

          6.F.1. OTHER CHESSBOARD DISSECTIONS

 

          See S&B, pp. 12‑14.  See also 6.F.5 for dissections of uncoloured boards.

 

Jerry Slocum.  Compendium of Checkerboard Puzzles.  Published by the author, 1983.  Outlines the history and shows all manufactured versions known then to him:  33 types in 61 versions.  The first number in Slocum's numbers is the number of pieces. 

Jerry Slocum  &  Jacques Haubrich.  Compendium of Checkerboard Puzzles.  2nd ed., published by Slocum, 1993.  90 types in 161 versions, with a table of which pieces are in which puzzles, making it much easier to see if a given puzzle is in the list or not.  This gives many more pictures of the puzzle boxes and also gives the number of solutions for each puzzle and sometimes prints all of them.  The Slocum numbers are revised in the 2nd ed. and I use the 2nd ed. numbers below.  (There was a 3rd ed. in 1997, NYR.  Les Barton is working on an extended version)

 

Henry Luers.  US Patent 231,963 _ Game Apparatus or Sectional Checker Board.  Applied 7 Aug 1880;  patented 7 Sep 1880.  1p + 1p diagrams.  15: 01329.  Slocum 15.5.1.  Manufactured as: Sectional Checker Baord Puzzle, by Selchow & Righter.  Colour photo of the puzzle box cover is on the front cover of the 1st ed. of Slocum's booklet.  B&W photo is on p. 14 of S&B.

??  UK patent application 16,810.  1892.  Not granted, so never published.  ??It may be possible to see the application??  (Hordern has an example with this number on it, by Feltham & Co.  However, in the 2nd ed., the cover is reproduced and it looks like the number may be 16,310.)  14: 00149.  Slocum 14.20.1.  Manufactured as: The Chequers Puzzle, by Feltham & Co.

Hoffmann.  1893.  Chap. III, no. 16: The chequers puzzle, pp. 97‑98 & 129‑130.  14: 00149.  Slocum 14.20.1.  Says it is made by Messrs Feltham, who state it has over 50 solutions.  He gives two solutions.  (Photo in Hordern, p. 61.)

Dudeney.  Problem 517 _ Make a chessboard.  Weekly Dispatch (4  &  18 Oct 1903), both p. 10.  8: 00010 12111 001.  Slocum 8.3.1.

Benson.  1904.  The chequers puzzle, pp. 202‑203.  As in Hoffmann, with only one solution.

Dudeney.  The Tribune (20  &  24 Dec 1906) both p. 1.  ??NX  Dissecting a chessboard.  Dissect into maximum number of different pieces.  Gets 18: 2,1,4,10,0, 0,0,1.  Slocum 18.1, citing later(?) Loyd versions.

Loyd.  Sam Loyd's Puzzle Magazine (Apr-Jul 1908) _ ??NYS, reproduced in:  A. C. White; Sam Loyd and His Chess Problems; 1913, op. cit. in 1; no. 58, p. 52.  = Cyclopedia, 1914, pp. 221 & 368, 250 & 373.  = MPSL2, prob. 71, pp. 51 & 145.  = SLAHP: Dissecting the chessboard, pp. 19 & 87.  Cut into maximum number of different pieces _ as in Dudeney, 1906.

Burren Loughlin  &  L. L. Flood.  Bright-Wits  Prince of Mogador.  H. M. Caldwell Co., NY, 1909.  The rug, pp. 7-13 & 65.  14: 00149.  Not in Slocum.

Loyd.  A battle royal.  Cyclopedia, 1914, pp. 97 & 351 (= MPSL1, prob. 51, pp. 49 & 139).  Same as Dudeney's prob. 517 of 1903.

Dudeney.  AM. 1917.  Prob. 293: The Chinese chessboard, pp. 87 & 213‑214.  Same as Loyd, p. 221.

Western Puzzle Works, 1926 Catalogue.  No. 79: "Checker Board Puzzle, in 16 pieces", but the picture only shows 14 pieces.  14: 00149.  Picture doesn't show any colours, but assuming the standard colouring of a chess board, this is the same as Slocum 14.15.

John Edward Fransen.  US Patent 1,752,248 _ Educational Puzzle.  Applied 19 Apr 1929;  patented 25 Mar 1930.  1p + 1p diagrams.  'Cut thy life.'  11: 10101 43001.  Slocum 11.3.1.

Emil Huber-Stockar.  Patience de l'echiquier.  Comptes-Rendus du Premier Congrès International de Récréation Mathématique, Bruxelles, 1935.  Sphinx, Bruxelles, 1935, pp. 93-94.  15: 01329.  Slocum 15.5.  Says there must certainly be more than 1000 solutions.

Emil Huber-Stockar.  L'echiquier du diable.  Comptes-Rendus du Deuxième Congrès International de Récréation Mathématique, Paris, 1937.  Librairie du "Sphinx", Bruxelles, 1937, pp. 64-68.  Discusses how one solution can lead to many others by partial symmetries.  Shows several solutions containing about 40 altogether.  Note at end says he has now got  5275  solutions.  This article is reproduced in Sphinx 8 (1938) 36-41, but without the extra pages of diagrams.  At the end, a note says he has  5330  solutions.  Ibid, pp. 75-76 says he has got 5362 solutions and ibid. 91-92 says he has  5365.  By use of Bayes' theorem on the frequency of new solutions, he estimates  c5500  solutions.  Haubrich has found 6013.  He intended to produce a book of solutions, but he died in May 1939 [Sphinx 9 (1939) 97].

F. Hansson.  Sam Loyd's 18-piece dissection _ Art. 48 & probs. 4152‑4153.  Fairy Chess Review 4:3 (Nov 1939) 44.  Cites Loyd's Puzzles Magazine.  Asserts there are many millions of solutions!  He determines the number of chequered handed  n-ominoes for  n = 1, 2, ..., 8  is  2, 1, 4, 10, 36, 110, 392, 1371.  The first 17 pieces total 56 squares.  Considers 8 ways to dissect the board into 18 different pieces.  Problems ask for the number of ways to choose the pieces in each of these ways and for symmetrical solutions.  Solution in 4:6 (Jun 1940) 93-94 (??NX of p. 94) says there are a total of  3,309,579  ways to make the choices.

C. Dudley Langford.  Note 2864:  A chess‑board puzzle.  MG 43 (No. 345) (Oct 1959) 200.  15: 01248.  Example with the underside coloured with reversed colours.  Gives two solutions and says there is at least one more.  Not in Slocum.

B. D. Josephson.  EDSAC to the rescue.  Eureka 24 (1961) 10‑12 & 32.  Uses the EDSAC computer to find two solutions of a 12 piece chessboard dissection.  12: 00025 41.  Slocum 12.9.

Leonard J. Gordon.  Broken chessboards with unique solutions.  G&PJ 10 (1989) 152‑153.  Shows Dudeney's problem has four solutions.  Finds other colourings which give only one solution.  Notes some equivalences in Slocum.

 

          6.F.2. COVERING DELETED CHESSBOARD WITH DOMINOES

 

          See also 6.U.2.

          There is nothing on this in Murray. 

 

Pál Révész.  Op. cit. in 5.I.1.  1969.  On p. 22, he says this problem comes from John [von] Neumann, but gives no details.

Max Black.  Critical Thinking, op. cit. in 5.T.  1946 ed., pp. 142 & 394, ??NYS.  2nd ed., 1952, pp. 157 & 433.  He simply gives it as a problem, with no indication that he invented it.

H. D. Grossman.  Fun with lattice points: 14 _ A chessboard puzzle.  SM 14 (1948) 160.  (The problem is described with 'his clever solution' from M. Black, Critical Thinking, pp. 142 & 394.)

S. Golomb.  1954.  Op. cit. in 6.F.

M. Gardner.  The mutilated chessboard.  SA (Feb 1957) = 1st Book, pp. 24 & 28.

Gamow & Stern.  1958.  Domino game.  Pp. 87‑90.

Robert S. Raven, proposer;  Walter P. Targoff, solver.  Problem 85 _ Deleted checkerboard.  In:  L. A. Graham; Ingenious Mathematical Problems and Methods; Dover, 1959, pp. 52 & 227.

R. E. Gomory.  (Solution for deletion of any two squares of opposite colour.)  In:  M. Gardner, SA (Nov 1962) = Unexpected, pp. 186‑187.  Solution based on a rook's tour.  (I don't know if this was ever published elsewhere.)

Michael Holt.  What is the New Maths?  Anthony Blond, London, 1967.  Pp. 68 & 97.  Gives the  4 x 4  case as a problem, but doesn't mention that it works on other boards.  (I include this as I haven't seen earlier examples in the educational literature.)

David Singmaster.  Covering deleted chessboards with dominoes.  MM 48 (1975) 59‑66.  Optimum extension to  n‑dimensions.  For an  n-dimensional board, each dimension must be  ³ 2.  If the board has an even number of cells, then one can delete any  n-1  white cells and any  n-1  black cells and still cover the board with dominoes (i.e.  2 x 1 x 1 x ... x 1  blocks).  If the board has an odd number of cells, then let the corner cells be coloured black.  One can then delete any  n  black cells and any  n-1  white cells and still cover the board with dominoes.

I-Ping Chu & Richard Johnsonbaugh.  Tiling deficient boards with trominoes.  MM 59:1 (1986) 34-40.  (3,n) = 1  and  n ¹ 5  imply that an  n x n  board with one cell deleted can be covered with  L  trominoes.  Some  5 x 5  boards with one cell deleted can be tiled, but not all can.

 

          6.F.3. DISSECTING A CROSS INTO  Zs  AND  Ls

 

Minguét.  Engaños.  1733.  Pp. 119-121 (1755: 85-86; 1822: 138-139).  Cross into 5 pieces, similar to Les Amusemens, but one  Z  is longer and one  L  is shorter.  Diagram shows  8  L  and  Z  shaped pieces, but it is not clear what the next problem wants _ either a piece or a label is missing.  Says one can make different figures with the pieces.

Les Amusemens.  1749.  P. xxxi.  Cross into 3  Z  pentominoes and 2  L  pieces.  The  Ls  are quite long.

Catel.  Kunst-Cabinet.  1790.  Das mathematische Kreuz, p. 10 & fig. 27 on plate I.  As in Les Amusemens, with long  Ls.

Bestelmeier.  1801.  Item 274 _ Das mathematische Kreuz.  Cross into 6 pieces, but the picture has an erroneous extra line.  It should be the reversal of the picture in Catel, the same as in Les Amusemens.

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. 4 is "Analysis of the Essay of Games".  F. 4.v has the dissection of the cross into 3  Z  pentominos and two  L  pieces.

Manuel des Sorciers.  1825.  Pp. 204-205, art. 21.  ??NX.  Dissect a cross into three  Zs  and two  Ls.

Boy's Treasury.  1844.  Puzzles and paradoxes, no. 3, pp. 424-425 & 428.  As in Les Amusemens.

Family Friend 3 (1850) 330 & 351.  Practical puzzle, No. XXI.  As in Les Amusemens, with long  Ls.

Magician's Own Book.  1857.  Prob. 31: Another cross puzzle, pp. 276 & 299.  As in Les Amusemens.

Landells.  Boy's Own Toy-Maker.  1858.  P. 152.  As in Les Amusemens.

Book of 500 Puzzles.  1859.  Prob. 31: Another cross puzzle, pp. 90 & 113.  As in Les Amusemens.  = Magician's Own Book.

Indoor & Outdoor.  c1859.  Part II, p. 127, prob. 5: The puzzle of the cross.  As in Les Amusemens.

Illustrated Boy's Own Treasury.  1860.  Practical Puzzles, No. 24, pp. 399 & 439.  Identical to Magician's Own Book.

Boy's Own Conjuring book.  1860.  Prob. 30: Another cross puzzle, pp. 239 & 263.  = Magician's Own Book, 1857.

Leske.  Illustriertes Spielbuch für Mädchen.  1864? 

Prob. 584-2, pp. 286 & 404.  4  Z  pentominoes to make a (Greek) cross.  (Also entered in 6.F.5.)

Prob. 584-8, pp. 287 & 405.  3  Z  pentominoes,  L  tetromino and  L  pentomino to make a Greek cross.  Despite specifically asking for a Greek cross, the answer is a standard Latin cross with  height : width = 4 : 3.

Mittenzwey.  1879?  Prob. 198, pp. 35 & 84.  As in Les Amusemens.

Cassell's.  1881.  P. 93: The magic cross.  = Manson, 1911, p. 139.  Same as Les Amusemens.

S&B, p. 20, shows a 7 piece cross dissection into 3  Zs,  2  Ls  and 2 straights, from c1890.

Handy Book for Boys and Girls.  Showing How to Build and Construct All Kinds of Useful Things of Life.  Worthington, NY, 1892.  Pp. 320-321: The cross puzzle.  As in Les Amusemens.

Hoffmann.  1893.  Chap. III, no. 29: Another cross puzzle, pp. 103 & 136.  As in Les Amusemens.  (Photo in Hordern, p. 65.)

Benson.  1904.  The Latin cross puzzle, p. 200.  As in Hoffmann.

Wehman.  New Book of 200 Puzzles.  1908.  Another cross puzzle, p. 32.  Usual form.

S. Szabo.  US Patent 1,263,960 _ Puzzle.  Filed 20 Oct 1917;  patented 23 Apr 1918.  1p + 1p diagrams.  As in Les Amusemens, with long  Ls. 

 

          6.F.4. QUADRISECT AN  L‑TROMINO, ETC.

 

          See also 6.AW.1 & 4.

 

Minguét.  Engaños.  1733.  Pp. 114-115 (1755: 80; 1822: 133-134).  Quadrisect  L-tromino.

Alberti.  1747.  Art. 30: Modo di dividere uno squadro di carta e di legno in quattro squadri equali, p. ?? (131) & fig. 56, plate XVI, opp. p. 130.

Les Amusemens.  1749.  P. xxx.  L-tromino ("gnomon") into 4 congruent pieces.

Vyse.  Tutor's Guide.  1771?  Prob. 9, p. 317 & Key p. 358.  Refers to the land as a parallelogram though it is drawn rectangular.

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. 4 is "Analysis of the Essay of Games".  F. 4.v 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.AQ, 6.AW.1, 6.AY, so it seems clear that he knew this problem.

Jackson.  Rational Amusement.  1821.  Geometrical Puzzles, no. 3, pp. 23 & 83 & plate I, fig. 2.

Manuel des Sorciers.  1825.  Pp. 203-204, art. 20.  ??NX.  Quadrisect L-tromino.

Family Friend 2 (1850) 118 & 149.  Practical Puzzle _ No. IV.  Quadrisect L-tromino of land with four trees.

Family Friend 3 (1850) 150 & 181.  Practical puzzle, No. XV.  15/16  of a square with 10 trees to be divided equally.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Mechanical puzzles, no. 8, p. 179 (1868: 190).  Land in the shape of an  L-tromino to be cut into four congruent parts, each with a cherry tree.

Magician's Own Book.  1857.

Prob. 3: The divided garden, pp. 267 & 292.  15/16  of a square to be divided into five (congruent) parts, each with two trees.  The missing  1/16  is in the middle.

Prob. 22: Puzzle of the four tenants, pp. 273 & 296.  Same as Parlour Pastime, but with apple trees.  (= Illustrated Boy's Own Treasury, 1860, No. 10, pp. 397 & 437.) 

Prob. 28: Puzzle of the two fathers, pp. 275-276 & 298.  Each father wants to divide  3/4  of a square.  One has  L‑tromino, other has the mitre shape.  See 6.AW.1.

Landells.  Boy's Own Toy-Maker.  1858. 

P. 144.  = Magician's Own Book, prob. 3.

Pp. 148-149.  = Magician's Own Book, prob. 27.

Book of 500 Puzzles.  1859.

Prob. 3: The divided garden, pp. 81 & 106.  Identical to Magician's Own Book.

Prob. 22: Puzzle of the four tenants, pp. 87 & 110.  Identical to Magician's Own Book.

Prob. 28: Puzzle of the two fathers, pp. 89-90 & 112.  Identical to Magician's Own Book.  See also 6.AW.1.

Charades, Enigmas, and Riddles.  1859?: prob. 28, pp. 59 & 63;  1862?: prob. 573, pp. 107 & 154.  Quadrisect  L-tromino, attributed to Sir F. Thesiger.

Boy's Own Conjuring book.  1860.

Prob. 3: The divided garden, pp. 229 & 255.  Identical to Magician's Own Book.

Prob. 21: Puzzle of the four tenants, pp. 235 & 260.  Identical to Magician's Own Book.

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

Illustrated Boy's Own Treasury.  1860.  Prob. 21, pp. 399 & 439.  15/16  of a square to be divided into five (congruent) parts, each with two trees.  c= Magician's Own Book, prob. 3.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 175, p. 88.  L-tromino into four congruent pieces, each with two trees.  The problem is given in terms of the original square to be divided into five parts, where the father gets a quarter of the whole in the form of a square and the four sons get congruent pieces.

Hanky Panky.  1872.  The divided orchards, p. 130.  L‑tromino into 4 congruent pieces, each with two trees.

Boy's Own Book.  The divided garden.  1868: 675.  = Magician's Own Book, prob. 3.

Cassell's.  1881.  P. 90: The divided farm.  = Manson, 1911, pp. 136-137.  = Magician's Own Book, prob. 3.

Lemon.  1890.

The divided garden, no. 259, pp. 38 & 107.  = Magician's Own Book, prob. 3.

Geometrical puzzle, no. 413, pp. 55 & 113 (= Sphinx, no. 556, pp. 76 & 116).  Quadrisect  L-tromino.

Hoffmann.  1893.  Chap. X, no. 41: The divided farm, pp. 352‑353 & 391.  = Magician's Own Book, prob. 3.  [One of the trees is invisible in the problem!]

Loyd.  Origin of a famous puzzle _ No. 18: An ancient puzzle.  Tit‑Bits 31 (13 Feb  &  6 Mar 1897) 363  &  419.  Nearly 50 years ago he was told of the quadrisection of  3/4  of a square, but drew the mitre shape instead of the  L‑tromino.  See 6.AW.1.

Benson.  1904.  The farmer's puzzle, p. 196.  Quadrisect an  L‑tromino.

Wehman.  New Book of 200 Puzzles.  1908.

The divided garden, p. 17.  = Magician's Own Book, prob. 3

Puzzle of the two fathers, p. 43.  = Magician's Own Book, prob. 28.

Puzzle of the four tenants, p. 46.  = Magician's Own Book, prob. 22.

Dudeney.  Some much‑discussed puzzles.  Op. cit. in 2.  1908.  Land in shape of an  L‑tromino to be quadrisected.  He says this is supposed to have been invented by Lord Chelmsford (Sir F. Thesiger), who died in 1878 _ see Charades, Enigmas, and Riddles (1859?).  But cf Les Amusemens.

Adams.  Indoor Games.  1912.  The clever farmer, pp. 23‑25.  Dissect  L‑tromino into four congruent pieces.

Blyth.  Match-Stick Magic.  1921.  Dividing the inheritance, pp. 20-21.  Usual quadrisection of  L-tromino set out with matchsticks.

Collins.  Book of Puzzles.  1927.  The surveyor's puzzle, pp. 2-3.  Quadrisect  3/4  of a square, except the deleted  1/4  is in the centre, so we are quadrisecting a hollow square.

Depew.  Cokesbury Game Book.  1939.  A plot of ground, p. 227.  3/4 of                                       XX     

          a square to be quadrisected, but the shape is as shown at the right.                                       XXX  

                                                                                                                                              X   XX

                                                                                                                                              XXXX

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?

Rowan Barnes-Murphy.  Monstrous Mysteries.  Piccolo, 1982.  Apple-eating monsters, pp. 40 & 63.  Trisect into equal parts, the shape consisting of a  2 x 4  rectangle with a  1 x 1  square attached to one of the central squares of the long side.  [Actually, this can be done with the square attached to any of the squares, though if it as attached to the end of the long side, the resulting pieces are straight trominoes.]

 

          6.F.5. OTHER DISSECTIONS INTO POLYOMINOES

 

Catel.  Kunst-Cabinet.  1790.

Das Zakk- und Hakenspiel, p. 10 & fig. 11 on plate 1.  4  Z‑pentominoes and 4  L‑tetrominoes make a  6 x 6  square.

Die zwolf Winkelhaken, p. 11 & fig. 26 on plate 1.  8  L‑pentominoes and 4  L‑hexominoes make a  8 x 8  square.

Bestelmeier.  1801.  Item 61 _ Das Zakken und Hakkenspiel.  As in Catel, p. 10, but not as regularly drawn.  Text copies some of Catel.

Manuel des Sorciers.  1825.  Pp. 203-204, art. 20.  ??NX  Use four L-trominoes to make a  3 x 4  rectangle or a  4 x 4  square with four corners deleted.

Family Friend 3 (1850) 90 & 121.  Practical puzzle _ No. XIII.  4 x 4  square, with 12 trees in the corners, centres of sides and four at the centre of the square, to be divided into 4 congruent parts each with 3 trees.  Solution uses 4  L-trominoes.  The same problem is repeated as Puzzle 17 _ Twelve-hole puzzle in (1855) 339 with solution in (1856) 28.

Magician's Own Book.  1857.  Prob. 14: The square and circle puzzle, pp. 270 & 295.  Same as Family Friend.  = Book of 500 Puzzles, 1859, prob. 14, pp. 84 & 109.  = Boy's Own Conjuring book, 1860, prob. 13, pp. 231-232 & 257.  c= Illustrated Boy's Own Treasury, 1860, prob. 8, pp. 396 & 437.  c= Hanky Panky, 1872, A square of four pieces, p. 117.

Landells.  Boy's Own Toy-Maker.  1858.  Pp. 146-147.  Identical to Family Friend.

Leske.  Illustriertes Spielbuch für Mädchen.  1864? 

Prob. 584-2, pp. 286 & 404.  4  Z‑pentominoes to make a Greek cross.  (Also entered in 6.F.3.)

Prob. 584-3, pp. 286 & 404.  4  L-tetrominoes to make a square.

Prob. 584-5, pp. 286 & 404.  8  L‑pentominoes and 4  L‑hexominoes make a  8 x 8  square.  Same as Catel, but diagram is inverted.

Prob. 584-7, pp. 287 & 405.  4  Z‑pentominoes and 4  L‑tetrominoes make a  6 x 6  square.  Same as Catel, but diagram is inverted.

Mittenzwey.  1879? 

Prob. 211, pp. 36 & 87.  4 x 4  square into 4  L-tetrominoes. 

Prob. 212, pp. 36 & 87.  6 x 6  square, as in Catel, p. 10.

Prob. 240, pp. 39 & 88.  4 x 4  square, with 12 trees, as in Family Friend.

S&B, p. 20, shows a 7 piece cross dissection into  3  Zs,  2  Ls  and  2  straights, from c1890.

Tom Tit, vol 3.  1893.  Les quatre  Z  et des quatre  L,  pp. 181-182.  = K, No. 27: The four  Z's  and the four  L's,  pp. 70‑71.  = R&A, Squaring the  L's  and  Z's,  p. 102.  6 x 6  square as in Catel, p. 10.

Sphinx.  1895.  The Maltese cross, no. 181, pp. 28 & 103.  Make a Maltese cross (actually a Greek cross of five equal squares) from 4  P-pentominoes.  Also:  quadrisect a  P‑pentomino.

Arthur Mee's Children's Encyclopedia 'Wonder Box'.  The Children's Encyclopedia appeared in 1908, so this is probably 1908 or soon thereafter.  4  Z‑pentominoes and 4  L‑tetrominoes make a  6 x 6  square and a  4 x 9  rectangle.

Wehman.  New Book of 200 Puzzles.  1908.  The square and circle puzzle, p. 5.  = Family Friend.

Burren Loughlin  &  L. L. Flood.  Bright-Wits  Prince of Mogador.  H. M. Caldwell Co., NY, 1909.  The Zoltan's orchard, pp. 24-28 & 64.  = Family Friend.

A. Neely Hall.  Carpentry & Mechanics for Boys.  Lothrop, Lee & Shepard, Boston, nd [1918].  The square puzzle, pp. 20‑21.  7 x 7  square cut into 1 straight tromino, 1  L‑tetromino and 7  L‑hexominoes.

Collins.  Book of Puzzles.  1927.  The surveyor's puzzle, pp. 2-3.  Quadrisect  3/4  of a square, except the deleted  1/4  is in the centre, so we are quadrisecting a hollow square.

W. Leslie Prout.  Think Again.  Frederick Warne & Co., London, 1958.  All square, pp. 42 & 129.  Make a  6 x 6  square from the staircase hexomino, 2  Y-pentominoes, an  N‑tetromino, an  L-tetromino  and 3  T-tetrominoes.  None of the pieces is turned over in the solution, though this restriction is not stated.

 

          6.G.    SOMA CUBE

 

Piet Hein invented the Soma Cube in 1936.  (S&B, pp. 40‑41.)  ??Is there any patent??

M. Gardner.  SA (Sep 1958) = 2nd Book, Chap 6.

Richard K. Guy.  Loc. cit. in 5.H.2, 1960.  Pp. 150-151 discusses cubical solutions _  234  found so far.  He proposes the 'bath' shape _ a  5 x 3 x 2  cuboid with a  3 x 1 x 1  hole in the top layer.

P. Hein, et al.  Soma booklet.  Parker Bros., 1969, 56pp.  Asserts there are  240  simple solutions and  1,105,920  total solutions, found by J. H. Conway & M. J. T. Guy with a a computer (but cf Gardner, below) and by several others.  [There seem to be several versions of this booklet, of various sizes.]

Thomas V. Atwater, ed.  Soma Addict.  4 issues, 1970‑1971, produced by Parker Brothers.  (Gardner, below, says only three issues appeared.)  ??NYS _ can anyone provide a set or xeroxes??

M. Gardner.  SA (Sep 1972) c= Knotted, chap. 3.  States there are  240  solutions for the cube, obtained by many programs, but first found by J. H. Conway & M. J. T. Guy in 1962, who did not use a computer, but did it by hand "one wet afternoon".

SOMAP  ??NYS _ ??details.  (Schaaf III 52)

Winning Ways, 1982, II, 802‑803 gives the SOMAP.

Jon Brunvall et al.  The computerized Soma Cube.  Comp. & Maths. with Appl. 12B:1/2 (1986  [Special issues  1/2  &  3/4  were separately printed as:  I. Hargittai, ed.; Symmetry _ Unifying Human Understanding; Pergamon, 1986.] 113‑121.  They cite Gardner's 2nd Book which says the number of solutions is unknown and they use a computer to find them.

 

          6.G.1. OTHER CUBE DISSECTIONS

 

          See also 6.N, 6.U.2, 6.AY.1 and 6.BJ.  The predecessors of these puzzles seem to be the binomial and trinomial cubes showing  (a+b)3  and  (a+b+c)3.  I have an example of the latter from the late 19C.  Here I will consider only cuts parallel to the cube faces _ cubes with cuts at angles to the faces are in 6.BJ.  Most of the problems here involve several types of piece _ see 6.U.2 for packing with one kind of piece.

 

Catel.  Kunst-Cabinet.  1790.  Der algebraische Würfel, p. 6 & fig 50 on plate II.  Shows a binomial cube:  (a + b)3  =  a3 + 3a2b + 3ab2 + b3.

Bestelmeier.  1801.  Item 309 is a binomial cube, as in Catel.  "Ein zerschnittener Würfel, mit welchem die Entstehung eines Cubus, dessen Seiten in  2  ungleiche Theile  a + b  getheilet ist, gezeigt ist."

Hoffmann.  1893.  Chap. III, no. 39: The diabolical cube, pp. 108 & 142.  6: 0, 1, 1, 1, 1, 1, 1,  i.e. six pieces of volumes  2, 3, 4, 5, 6, 7.

J. G.‑Mikusi_ski.  French patent.  ??NYS _ cited by Steinhaus.

H. Steinhaus.  Mikusi_ski's Cube.  Mathematical Snapshots.  Not in Stechert, 1938, ed.  OUP, NY:  1950: pp. 140‑142 & 263;  1960, pp. 179‑181 & 326;  1969 (1983): pp. 168-169 & 303.

Jan Slothouber & William Graatsma.  Cubics.  Octopus Press, Deventer, Holland, 1970.  ??NYS.  3 x 3 x 3  into  3  1 x 1 x 1  and  6  1 x 2 x 2.

M. Gardner.  SA (Sep 1972) c= Knotted, chap. 3.  Discusses Hoffmann's Diabolical Cube and Mikusi_ski's cube.  Says he has 8 solutions for the first and that there are just  2  for the second.  The Addendum reports that Wade E. Philpott showed there are just  13  solutions of the Diabolical Cube.  Conway has confirmed this.  Gardner briefly describes the solutions.  Gardner also shows the Lesk Cube, designed by Lesk Kokay (Mathematical Digest [New Zealand] 58 (1978) ??NYS), which has at least  3  solutions.

D. A. Klarner.  Brick‑packing puzzles.  JRM 6 (1973) 112‑117.  Discusses Slothouber‑Graatsma;  5 x 5 x 5  into  3  1 x 1 x 3  and  29  1 x 2 x 2;  Conway's  5 x 5 x 5  into  3  1 x 1 x 3,  1  2 x 2 x 2,  1  1 x 2 x 2  and  13  1 x 2 x 4.

Leisure Dynamics, the US distributor of Impuzzables, a series of  6  3 x 3 x 3  cube dissections identified by colours, writes that they were invented by Robert Beck, Custom Concepts Inc., Minneapolis.  However, the Addendum to Gardner, above, says they were designed by Gerard D'Arcey.

Michael Keller.  Polycube update.  World Game Review 4 (Feb 1985) 13.  Reports results of computer searches for solutions.  Hoffmann's Diabolical Cube has  13;  Mikusinski's Cube has  2;  Soma Cube has  240;  Impuzzables:  White _ 1;  Red _ 1;  Green _ 16;  Blue _ 8;  Orange _ 30;  Yellow _ 1142.

Michael Keller.  Polyform update.  World Game Review 7 (Oct 1987) 10‑13.  Says that Nob Yoshigahara has solved a problem posed by O'Beirne:  How many ways can  9  L‑trominoes make a cube?  Answer is  111.  Gardner, Knotted, chap. 3, mentioned this.  Says there are solutions with  n  L‑trominoes and  9‑n  straight trominoes for  n ¹ 1 and there are  4  solutions for  n = 0.  Says the Lesk Cube has  4  solutions.  Says Naef's Gemini Puzzle was designed by Toshiaki Betsumiya.  It consists of the  10  ways to join two  1 x 2 x 2  blocks.

H. J. M. van Grol.  Rik's Cube Kit _ Solid Block Puzzles.  Analysis of all  3 x 3 x 3  unit solid block puzzles with non‑planar  4‑unit and  5‑unit shapes.  Published by the author, The Hague, 1989, 16pp.  There are  3  non‑planar tetracubes and  17  non‑planar pentacubes.  A  3 x 3 x 3  cube will require the  3  non‑planar tetracubes and  3  of the non‑planar pentacubes _ assuming no repeated pieces.  He finds  190  subsets which can form cubes, in  1  to  10  different ways.

 

          6.G.2. DISSECTION OF  63  INTO  33,  43  AND  53,  ETC.

 

H. W. Richmond.  Note 1672:  A geometrical problem.  MG 27 (No. 275) (Jul 1943) 142.  AND  Note 1704:  Solution of a geometrical problem (Note 1672).  MG 28 (No. 278) (Feb 1944) 31‑32.  Poses the problem of making such a dissection, then gives a solution in 12 pieces:  three  1 x 3 x 3;  4 x 4 x 4;  four 1 x 5 x 5;  1 x 4 x 4;  two  1 x 1 x 2  and a  V‑pentacube.

Anon. [= John Leech, according to Gardner, below].  Two dissection problems:  2.  Eureka 13 (1950) 6  &  14 (1951) 23.  Asks for such a dissection using at most 10 pieces.  Gives an 8 piece solution due to R. F. Wheeler.  [Cundy & Rollett; Mathematical Models; 2nd ed., pp. 203‑205, say Eureka is the first appearance they know of this problem.  See Gardner, below, for the identity of Leech.]

Richard K. Guy.  Loc. cit. in 5.H.2, 1960.  Mentions the 8 piece solution.

J. H. Cadwell.  Some dissection problems involving sums of cubes.  MG 48 (No. 366) (Dec 1964) 391‑396.  Notes an error in Cundy & Rollett's account of the Eureka problem.  Finds examples for  123 + 13 = 103 + 93  with 9 pieces  and  93 = 83 + 63 + 13  with 9 pieces.

J. H. Cadwell.  Note 3278:  A three‑way dissection based on Ramanujan's number.  MG 54 (No. 390) (Dec 1970) 385‑387.  7 x 13 x 19  to  103 + 93  and  123 + 13  using 12 pieces.

M. Gardner.  SA (Oct 1973) c= Knotted, chap. 16.  He says that the problem was posed by John Leech.  He gives Wheeler's initials as E. H. ??  He says that J. H. Thewlis found a simpler 8‑piece solution, further simplified by T. H. O'Beirne, which keeps the  4 x 4 x 4  cube intact.  This is shown in Gardner.  Gardner also shows an 8‑piece solution which keeps the  5 x 5 x 5  intact, due to E. J. Duffy, 1970.  O'Beirne showed that an 8‑piece dissection into blocks is impossible and found a 9‑block solution in 1971, also shown in Gardner.

Harry Lindgren.  Geometric Dissections.  Van Nostrand, Princeton, 1984.  Section 24.1, pp. 118‑120 gives Wheeler's solution and admires it.

Richard K. Guy, proposer;  editors & Charles H. Jepson [should be Jepsen], partial solvers.  Problem 1122.  CM 12 (1987) 50  &  13 (1987) 197‑198.  Asks for such dissections under various conditions, of which (b) is the form given in Eureka.  Eight pieces is minimal in one case and seems minimal in two other cases.  Eleven pieces is best known for the first case, where the pieces must be blocks, but this appears to be the problem solved by O'Beirne in 1971, reported in Gardner, above.

Charles H. Jepsen.  Additional comment on Problem 1122.  CM 14 (1988) 204‑206.  Gives a ten piece solution of the first case.

Chris Pile.  Cube dissection.  M500 134 (Aug 1993) 2-3.  He feels the  1 x 1 x 2  piece occurring in Cundy & Rollett is too small and he provides another solution with 8 pieces, the smallest of which contains 8 unit cubes.  Asks how uniform the piece sizes can be.

 

          6.G.3. DISSECTION OF A DIE INTO NINE  1 x 1 x 3

 

Hoffmann.  1893.  Chap. III, no. 17: The "Spots" puzzle, pp. 98‑99 & 130‑131.  Says it is made by Wolff & Son.

Benson.  1904.  The spots puzzle, pp. 203‑204.  As in Hoffmann.

Collins.  Book of Puzzles.  1927.  Pp. 131‑134: The dissected die puzzle.  The solution is different than Hoffmann's.

Rohrbough.  Puzzle Craft.  1932.  P. 21 shows a dissected die, but with no text.  The picture is the same as in Hoffmann's solution.

Slocum.  Compendium.  Shows Diabolical Dice from Johnson Smith catalogue, 1935.

Harold Cataquet.  The Spots puzzle revisited.  CFF 33 (Feb 1994) 20-21.  Brief discussion of two versions.

David Singmaster.  Comment on the "Spots" puzzle.  29 Sep 1994, 2pp.  Letter in response to the above.  I note that there is no standard pattern for a die other than the opposite sides adding to seven.  There are  23 = 8  ways to orient the spots forming  2, 3, and 6.  There are two handednesses, so there are 16 dice altogether.  (This was pointed out to me perhaps 10 years before by Richard Guy and Ray Bathke.  I have since collected examples of all 16 dice.)  However, Ray Bathke showed me Oriental dice with the two spots of the 2 placed horizontal or vertically rather than diagonally, giving another 16 dice (I have 5 types), making 32 dice in all.  A die can be dissected into  9  1 x 1 x 3  pieces in 6 ways if the layers have to alternate in direction, or in 21 ways in general.  I then pose a number of questions about such dissections.

 

          6.G.4. USE OF OTHER POLYHEDRAL PIECES

 

S&B.  1986.  P. 42 shows Stewart Coffin's 'Pyramid Puzzle' using pieces made from truncated octahedra and his 'Setting Hen' using pieces made from rhombic dodecahedra.  Coffin probably devised these in the 1960s _ perhaps his book has some details of the origins of these ideas.  ??check.

Mark Owen & Matthew Richards.  A song of six splats.  Eureka 47 (1987) 53‑58.  There are six ways to join three truncated octahedra.  For reasons unknown, these are called 3‑splats.  They give various shapes which can and which cannot be constructed from the six 3‑splats.

 

          6.H.   PICK'S THEOREM

 

Georg Pick.  Geometrisches zur Zahlenlehre.  Sitzungsberichte des deutschen naturwissenschaftlich‑medicinischen Vereines für Böhmen "Lotos" in Prag (NS) 19 (1899) 311‑319.  Pp. 311‑314 gives the proof, for an oblique lattice.  Pp. 318‑319 gives the extension to multiply connected and separated regions.  Rest relates to number theory.  [I have made a translation of the material on Pick's Theorem.]

Charles Howard Hinton.  The Fourth Dimension.  Swan Sonnenschein & Co., London, 1906.  Metageometry, pp. 46-60.  [This material is in Speculations on the Fourth Dimension, ed. by R. v. B. Rucker; Dover, 1980, pp. 130-141.  Rucker says the book was published in 1904, so my copy may be a reprint??]  In the beginning of this section, he draws quadrilateral shapes on the square lattice and determines the area by counting points, but he counts  I + E/2 + C/4,  which works for quadrilaterals but is not valid in general.

H. Steinhaus.  O mierzeniu pól p_askich.  Przegl_d Matematyczno‑Fizyczny 2 (1924) 24‑29.  Gives a version of Pick's theorem, but doesn't cite Pick.  (My thanks to A. M_kowski for an English summary of this.)

H. Steinhaus.  Mathematical Snapshots.  Stechert, NY, 1938, pp. 16-17 & 132.  OUP, NY:  1950: pp. 76‑77 & 260 (note 77);  1960: pp. 99‑100 & 324 (note 95);  1969 (1983): pp. 96‑97 & 301 (note 107).  In 1938 he simply notes the theorem and gives one example.  In 1950, he outlines Pick's argument.  He  refers to Pick's paper, but in "Ztschr. d. Vereins 'Lotos' in Prag".  Steinhaus also cites his own paper, above.

J. F. Reeve.  On the volume of lattice polyhedra.  Proc. London Math. Soc. 7 (1957) 378‑395.  Deals with the failure of the obvious form of Pick's theorem in 3‑D and finds a valid generalization.

Ivan Niven & H. S. Zuckerman.  Lattice points and polygonal area.  AMM 74 (1967) 1195‑1200.  Straightforward proof.  Mention failure for tetrahedra.

D. W. De Temple & J. M. Robertson.  The equivalence of Euler's and Pick's theorems.  MTr 67 (1974) 222‑226.  ??NYS.

W. W. Funkenbusch.  From Euler's formula to Pick's formula using an edge theorem.  AMM 81 (1974) 647‑648.  Easy proof though it could be easier.

R. W. Gaskell, M. S. Klamkin & P. Watson.  Triangulations and Pick's theorem.  MM 49 (1976) 35‑37.  A bit roundabout.

Richard A. Gibbs.  Pick iff Euler.  MM 49 (1976) 158.  Cites DeTemple & Robertson and observes that both Pick and Euler can be proven from a result on triangulations.

John Reay.  Areas of hex-lattice polygons, with short sides.  Abstracts Amer. Math. Soc. 8:2 (1987) 174, #832-51-55.  Gives a formula for the area in terms of the boundary and interior points and the characteristic of the boundary, but it is an open question to determine when this formula gives the actual area.

 

          6.I.     SYLVESTER'S PROBLEM OF COLLINEAR POINTS

 

          If a set of non‑collinear points in the plane is such that the line through any two points of the set contains a third point of the set, then the set is infinite.

 

J. J. Sylvester.  Question 11851.  The Educational Times 46 (NS, No. 383) (1 Mar 1893) 156.

H. J. Woodall & editorial comment.  Solution to Question 11851.  Ibid. (No. 385) (1 May 1893) 231.  A very spurious solution.

(The above two items appear together in Math. Quest. with their Sol. Educ. Times 59 (1893) 98‑99.)

E. Melchior.  Über Vielseite der projecktiven Ebene.  Deutsche Math. 5 (1940) 461‑475.  Solution, but in a dual form.

P. Erdös, proposer;  R. Steinberg, solver & editorial comment giving solution of T. Grünwald (later = T. Gallai).  Problem 4065.  AMM 50 (1943) 65  &  51 (1944) 169‑171.

L. M. Kelly.  (Solution.)  In:  H. S. M. Coxeter; A problem of collinear points; AMM 55 (1948) 26‑28.  Kelly's solution is on p. 28.

G. A. Dirac.  Note 2271:  On a property of circles.  MG 36 (No. 315) (Feb 1952) 53‑54.  Replace 'line' by 'circle' in the problem.  He shows this is true by inversion.  He asks for an independent proof of the result, even for the case when  two, three  are replaced by  three, four.

D. W. Lang.  Note 2577:  The dual of a well‑known theorem.  MG 39 (No. 330) (Dec 1955) 314.  Proves the dual easily.

H. S. M. Coxeter.  Introduction to Geometry.  Wiley, 1961.  Section 4.7: Sylvester's problem of collinear points, pp. 65-66.  Sketches history and gives Kelly'e proof.

W. O. J. Moser.  Sylvester's problem, generalizations and relatives.  In his:  Research Problems in Discrete Geometry 1981, McGill University, Montreal, 1981.  Section 27, pp. 27‑1 _ 27‑14.  Survey with 73 references.  (This problem is not in Part 1 of the 1984 ed. nor in the 1986 ed.)

 

          6.J.     FOUR BUGS AND OTHER PURSUIT PROBLEMS

 

          The general problem becomes too technical to remain recreational, so I will not try to be exhaustive here.

 

Arthur Bernhart. 

Curves of pursuit.  SM 20 (1954) 125‑141.

Curves of pursuit _ II.  SM 23 (1957) 49‑65.

Polygons of pursuit.  SM 24 (1959) 23‑50. 

Curves of general pursuit.  SM 24 (1959) 189‑206.

          Extensive history and analysis.  First article covers one dimensional pursuit, then two dimensional linear pursuit.  Second article deals with circular pursuit.  Third article is the 'four bugs' problem _ analysis of equilateral triangle, square, scalene triangle, general polygon, Brocard points, etc.  Last article includes such variants as variable speed, the tractrix, miscellaneous curves, etc.

 

Carlile.  Collection.  1793.  Prob. CV, p. 62.  A dog and a duck are in a circular pond of radius 40 and the swim at the same speed.  The duck is at the edge and swims around the circumference.  The dog starts at the centre and always toward the duck, so the dog and the duck are always on a radius.  How far does the dog swim in catching the duck.  He simply gives the result as  20π.  Letting  R  be the radius of the pond and  V  be the common speed, I find the radius of the dog,  r,  is given by  r = R sin Vt/R.  Since the angle,  θ,  of both the duck and the dog is given by  θ = Vt/R,  the polar equation of the dog's path is  r = R sin θ  and the path is a semicircle whose diameter is the appropriate radius perpendicular to the radius to the duck's initial position.

Cambridge Math. Tripos examination, 5 Jan 1871, 9 to 12.  Problem 16, set by R. K. Miller.  Three bugs in general position, but with velocities adjusted to make paths similar and keep the triangle similar to the original.

Lucas.  (Problem of three dogs.)  Nouvelle Correspondance Mathématique 3 (1877) 175‑176.  ??NYS _ English in Arc., AMM 28 (1921) 184‑185 & Bernhart.

H. Brocard.  (Solution of Lucas' problem.)  Nouv. Corr. Math. 3 (1877) 280.  ??NYS _ English in Bernhart.

Pearson.  1907.  Part II, no. 66: A duck hunt, pp. 66 & 172.  Duck swims around edge of pond;  spaniel starts for it from the centre at the same speed.

A. S. Hathaway, proposer and solver.  Problem 2801.  AMM 27 (1920) 31  &  28 (1921) 93‑97.  Pursuit of a prey moving on a circle.  Morley's and other solutions fail to deal with the case when the velocities are equal.  Hathaway resolves this and shows the prey is then not caught.

F. V. Morley.  A curve of pursuit.  AMM 28 (1921) 54-61.  Graphical solution of Hathaway's problem.

R. C. Archibald [Arc.] & H. P. Manning.  Remarks and historical notes on problems 19 [1894], 160 [1902], 273 [1909] & 2801 [1920].  AMM 28 (1921) 91-93.

Editor's note to Prob. 2 (proposed by T. A. Bickerstaff), National Mathematics Magazine (1937/38) 417 cites Morley and Archibald and adds that some authors credit the problem to Leonardo da Vinci _ e.g. MG (1930-31) 436 _ ??NYS

Nelson F. Beeler & Franklyn M. Branley.  Experiments in Optical Illusion.  Ill. by Fred H. Lyon.  Crowell, 1951, An illusion doodle, pp. 68-71, describes the pattern formed by four bugs starting at the corners of a square, drawing the lines of sight at (approximately) regular intervals.  Putting several of the squares together, usually with alternating directions of motion, gives a pleasant pattern which is now fairly common.  They call this 'Huddy's Doodle', but give no source.

J. E. Littlewood.  A Mathematician's Miscellany.  Op. cit. in 5.C.  1953.  'Lion and man', pp. 135‑136 (114‑117).  The 1986 ed. adds three diagrams and revises the text somewhat.  I quote from it.  "A lion and a man in a closed circular arena have equal maximum speeds.  What tactics should the lion employ to be sure of his meal?"  This was "invented by R. Rado in the late thirties" and "swept the country 25 years later".  [The 1953 ed., says Rado didn't publish it.]  The correct solution "was discovered by Professor A. S. Besicovitch in 1952".  [The 1953 ed. says "This has just been discovered ...; here is the first (and only) version in print."]

C. C. Puckette.  The curve of pursuit.  MG 37 (No. 322) (Dec 1953) 256‑260.  Gives the history from Bouguer in 1732.  Solves a variant of the problem.

R. H. Macmillan.  Curves of pursuit.  MG 40 (No. 331) (Feb 1956) 1‑4.  Fighter pursuing bomber flying in a straight line.  Discusses firing lead and acceleration problems.

Gamow & Stern.  1958.  Homing missiles.  Pp. 112‑114.

Howard D. Grossman, proposer;  unspecified solver.  Problem 66 _ The walk around.  In:  L. A. Graham; Ingenious Mathematical Problems and Methods; Dover, 1959, pp. 40 & 203‑205.  Four bugs _ asserts Grossman originated the problem.

I. J. Good.  Pursuit curves and mathematical art.  MG 43 (No. 343) (Feb 1959) 34‑35.  Draws tangent to the pursuit curves in an equilateral triangle and constructs various patterns with them.  Says a similar but much simpler pattern was given by G. B. Robison; Doodles; AMM 61 (1954) 381-386, but Robison's doodles are not related to pursuit curves, though they may have inspired Good to use the pursuit curves.

J. Charles Clapham.  Playful mice.  RMM 10 (Aug 1962) 6‑7.  Easy derivation of the distance traveled for  n  bugs at corners of a regular  n‑gon.  [I don't see this result in Bernhart.]

C. G. Paradine.  Note 3108:  Pursuit curves.  MG 48 (No. 366) (Dec 1964) 437‑439.  Says Good makes an error in Note 3079.  He shows the length of the pursuit curve in the equilateral triangle is  _  of the side and describes the curve as an equiangular spiral.  Gives a simple proof that the length of the pursuit curve in the regular  n‑gon is the side divided by  (1 ‑ cos 2π/n).

M. S. Klamkin & D. J. Newman.  Cyclic pursuit or "The three bugs problem".  AMM 78 (1971) 631‑639.  General treatment.  Cites Bernhart's four SM papers and some of the history therein.

P. K. Arvind.  A symmetrical pursuit problem on the sphere and the hyperbolic plane.  MG 78 (No. 481) (Mar 1994) 30-36.  Treats the  n  bugs problems on the surfaces named.

 

          6.K.    DUDENEY'S SQUARE TO TRIANGLE DISSECTION

 

Dudeney.  Weekly Dispatch (6 Apr, 20 Apr, 4 May, 1902) all p. 13.

Dudeney.  The haberdasher's puzzle.  London Mag. 11 (No. 64) (Nov 1903) 441 & 443.  (Issue with solution not found.)

Dudeney.  Daily Mail (1  &  8 Feb 1905) both p. 7.

Dudeney.  CP.  1907.  Prob. 25: The haberdasher's puzzle, pp. 49‑50 & 178‑180.

Western Puzzle Works, 1926 Catalogue.  No. 1712 _ unnamed, but shows both the square and the triangle.  Apparently a four piece puzzle.

Adams.  Puzzle Book.  1939.  Prob. C.153: Squaring a triangle, pp. 162 & 189.  Asserts that Dudeney's method works for any triangle, but his example is close to equilateral and I recall that this has been studied and only certain shapes will work??

Robert C. Yates.  Geometrical Tools.  (As:  Tools; Baton Rouge, 1941);  revised ed., Educational Publishers, St. Louis, 1949.  Pp. 40-41.  Extends to dissecting a quadrilateral to a specified triangle and gives a number of related problems.

 

          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 looks like a crossed ladders problem but is a simple right triangle problem.

Pacioli.  Summa.  1494.  Part 2, f. 56r, prob. 48.  (4, 6).

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.  This is actually a simple right triangle problem.

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 = Ö(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 Prolbems.  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. Jab_kowski.  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.  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.

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  Ö(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?

 

          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  Ö2/Ö3.  The largest square inscribable in this hexgon has edge  Ö6 - Ö2 = 1.03527618.  By passing the larger cube on a slant to the space diagonal, one can get the larger cube having edge  3Ö2/4 = 1.06066172.

 

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.)

Ozanam‑Montucla.  1778.  Percer un cube d'une ouverture, par laquelle peut paffer 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, Jena, 1834, p. 394, ??NYS.

P. Nieuwland.  (Finding of maximum cube which passes through another).  In:  van Swinden, op. cit., pp. 608‑610;  van Swinden‑Jacobi, op. cit., p. 542.  ??NYS

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 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.

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.

 

          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.  London, 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.

Mittenzwey.  1879?  Prob. 332, pp. 53 & 101.  Clear explanation.

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

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.

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.

L. Carroll.  Op. cit. in 5.B, 1899, pp. 316-317 (Collins: 231 and/or 232 (lacking in my copy)).  8 x 8  to  5 x 13.  Carroll may have stated this as early as 1888.

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.

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.

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.

 

          6.P.2. OTHER TYPES

 

Sebastiano Serlio.  Libro Primo d'Architettura.  1545.  This is the first part of his Architettura, 5 books, 1537-1547, first published together in 1584.  I have seen the following editions.

                    With French translation by Jehan Martin, no publisher shown, Paris, 1545, f. 22.r.  ??NX

                    1559.  F. 15.v.

                    Francesco Senese & Zuane(?) Krugher, Venice, 1566, f. 16.r.  ??NX

                    Jacomo de'Franceschi, Venice, 1619, f. 16.r.

                    Translated into Dutch by Pieter Coecke van Aelst as:  Den eerst_ vijfsten boeck van architectur_; Amsterdam, 1606.  This was translated into English as:  The Five Books of Architecture; Simon Stafford, London, 1611  = Dover, 1982.  The first Booke, f. 12v.

                    3 x 10  board is cut on a diagonal and slid to form a  4 x 7  table with  3 x 1  left over, but he doesn't actually put the two leftover pieces together nor notice the area change!

Pietro Cataneo.  L'Architettura di Pietro Cataneo Senese.  Aldus, Venice, 1567.  ??NX.  Libro Settimo.

P. 164, prop. XXVIIII: Come si possa accresciere una stravagante larghezza.  Gives a correct version of Serlio's process.

P. 165, prop. XXX: Falsa solutione del Serlio.  Cites p. xxii of Serlio.  Carefully explains the error in Serlio and says his method is "insolubile, & mal pensata".

Schwenter.  1636.  Discusses Serlio's dissection and observes the area change.  [??Schwenter has not yet been entered.]

I have a vague reference to the 1723 ed. of Ozanam, but I have not seen it in the 1725 ed. _ this may be an error for the 1778 ed. below.

Vyse.  Tutor's Guide.  1771?  Prob. 8, p. 317 & Key p. 358.  Lady has a table  27  square and a board  12 x 48.  She cuts the board into two  12 x 24  rectangles and cuts each rectangle along a diagonal.  By placing the diagonals of these pieces on the sides of her table, she makes a table  36  square.  Note that  362 = 1296  and  272 + 12 x 24  =  1305.  Vyse is clearly unaware that area has been created.  By dividing all lengths by  3,  one gets a version where one unit of area is lost.  Note that  4, 8, 9  is almost a Pythagorean triple.

William Hooper.  Rational Recreations.  1774.  Op. cit. in 4.A.1.  Vol. 4, pp. 286‑287: Recreation CVI _ The geometric money.  3 x 10  cut into four pieces which make a  2 x 6  and a  4 x 5.  (The diagram is shown in Gardner, MM&M, pp. 131‑132.)  (I recently saw that an edition erroneously has a  3 x 6  instead of a  2 x 6  rectangle.  This must be the 1st ed. of 1774, as it is correct in my 2nd ed. of 1782.)

Ozanam-Montucla.  1778.  Transposition de laquelle semble résulter que le tout peut être égal à la partie.  Prob. 21 & fig. 127, plate 16, 1778: 302-303 & 363;  1803: 298-299 & 361;  1814: 256 & 306;  1840: omitted.   3 x 11  to  2 x 7  and  4 x 5.  Remarks that M. Ligier probably made some such mistake in showing  172 = 2 x 122  and this is discussed further on the later page.

E. C. Guyot.  Nouvelles Récréations Physiques et Mathématiques.  Nouvelle éd.  La Librairie, Rue S. André‑des‑Arcs[sic], Paris, Year 7 [1799].  Vol. 2, Deuxième récréation: Or géométrique _ construction, pp. 41‑42 & plate 6, opp. p. 37.  Same as Hooper.

Minguét.  Engaños.  1822.  Pp. 145-146.  Same as Hooper.  Not seen in 1733 and 1755 eds.

Manuel des Sorciers.  1825.  Pp. 202-203, art. 19.  ??NX  Same as Hooper.

The Boy's Own Book.  The geometrical money.  1828: 413;  1828-2: 419;  1829 (US): 212;  1855: 566‑567;  1868: 669.  Same as Hooper.

Magician's Own Book.  1857.  Deceptive vision, pp. 258-259.  Same as Hooper.  = Book of 500 Puzzles, 1859, pp. 72-73.

Illustrated Boy's Own Treasury.  1860.  Optics: Deceptive vision, p. 445.  Same as Hooper.  Identical to Book of 500 Puzzles.

Wemple & Company (New York).  The Magic Egg Puzzle.  ©1880.  S&B, p. 144.  Advertising card, the size of a small postcard, but with ads for Rogers Peet on the back.  Cut into four rectangles and reassemble to make 6, 7, 8, 10, 11, 12  eggs.

R. March & Co. (St. James's Walk, Clerkenwell).   'The Magical Egg Puzzle', nd [c1890].  (I have a photocopy.)  Four rectangles which produce  6, 7, ..., 12  eggs.  As I recall, this is identical to the Wemple version.

Loyd.  US Patent 563,778 _ Transformation Picture.  Applied 11 Mar 1896;  patented 14 Jul 1896.  1p + 1p diagrams.  Simple rotating version using 8 to 7 objects.

Loyd.  Get Off the Earth.  Puzzle notices in the Brooklyn Daily Eagle (26 Apr ‑ 3 May 1896), printing individual Chinamen.  Presenting all of these at an office of the newspaper gets you an example of the puzzle.  Loyd ran discussions on it in his Sunday columns until 3 Jan 1897 and he also sold many versions as advertising promotions.  S&B, p. 144, shows several versions.

Loyd.  Problem 17: Ye castle donjon.  Tit‑Bits 31 (6  &  27 Feb  &  6  &  20 Mar 1897) 343,  401,  419  &  455.  = Cyclopedia, 1914, The architect's puzzle, pp. 241 & 372.  5 x 25  to area 124.

Dudeney.  Great puzzle crazes.  Op. cit. in 2.  1904.  Discusses and shows Get Off the Earth.

Ball.  MRE, 4th ed., 1905, pp. 50-51: Turton's seventy-seven puzzle.  Additional section describing Captain Turton's  7 x 11  to  7 x 11  with one projecting square, using bevelled cuts.  This is dropped from the 7th ed., 1917.

William F. White.  1907 & 1908.  See entries in 6.P.1.

Dudeney.  The world's best puzzles.  Op. cit. in 2.  1908.  Gives "Get Off the Earth" on p. 785.

Loyd.  Teddy and the Lions.  Gardner, MM&M, p. 123, says he has seen only one example, made as a promotional item for the Eden Musee in Manhattan.  This has a round disc, but two sets of figures _ 7 natives and 7 lions which become 6 natives and 8 lions.

Dudeney.  A chessboard fallacy.  The Paradox Party.  Strand Mag. 38 (No. 228) (Dec 1909) 676 (= AM, prob. 413, pp. 141 & 247).  (There is a solution in Strand Mag. 39 (No. 229) (Jan 1910) ??NYS.)  8 x 8  into 3 pieces which make a  9 x 7.

Loyd.  The gold brick puzzle.  Cyclopedia, 1914, pp. 32 & 342 (= MPSL1, prob. 24, pp. 22 & 129).  24 x 24  to  23 x 25.

Loyd.  Cyclopedia.  1914.  "Get off the earth", p. 323.  Says over 10 million were sold.  Offers prizes for best answers received in 1909.

Loyd Jr.  SLAHP.  1928.  "Get off the Earth" puzzle, pp. 5‑6.  Says 'My "Missing Chinaman Puzzle"' of 1896.  Gives a simple and clear explanation.

John Barnard.  The Handy Boy's Book.  Ward, Lock & Co., London, nd [c1930?].  Some interesting optical illusions, pp. 310-311.  Shows a card with 11 matches and a diagonal cut so that sliding it one place makes 10 matches.

W. A. Bagley.  Puzzle Pie.  Op. cit. in 5.D.5.  1944.  No. 24: A chessboard fallacy, pp. 28-29.  8 x 8  cut with a diagonal of a  8 x 7  region, then pieces slid and a triangle cut off and moved to the other end to make a  9 x 7.  Clear illustration.

Mel Stover.  From 1951, he devised a number of variations of both Get off the Earth (perhaps the best is his Vanishing Leprechaun) and of Teddy and the Lions (6 men and 4 glasses of beer become 5 men and 5 glasses).  I have examples of some of these from Stover and I have looked at his notebooks.  See  Gardner, MM&M, pp. 125-128.

John Fisher.  John Fisher's Magic Book.  Muller, London, 1968.

Financial Wizardry, pp. 18-19.  7 x 8  region with  £  signs marking the area.  A line cuts off a triangle of width 7 and height 2 at the top.  The rest of the area is divided by a vertical into strips of widths 4 and 3, with a small rectangle 3 by 1 cut from the bottom of the width 3 strip.  When the strips are exchanged, one unit of area is lost and one  £  sign has vanished.

Try-Angle, pp. 126-127.  This is one of Curry's triangles _ see Gardner, MM&M, p. 147.

Alco-Frolic!, pp. 148-149.  This is a form of Stover's  6 & 4  to  5 & 5  version.

D. E. Knuth.  Disappearances.  In:  The Mathematical Gardner; ed. by David Klarner; Prindle, Weber & Schmidt/Wadsworth, 1981.  P. 264.  An eight line poem which rearranges to a seven line poem.

Dean Clark.  A centennial tribute to Sam Loyd.  CMJ 23:5 (Nov 1992) 402‑404.  Gives an easy circular version with  11 & 12  astronauts around the earth and a  15 & 16  face version with three pieces, a bit like the Vanishing Leprechaun.

 

          6.Q.   KNOTTING A STRIP TO MAKE A REGULAR PENTAGON

 

Urbano d'Aviso.  Trattato della Sfera e Pratiche per Uso di Essa.  Col modeo di fare la figura celeste, opera cavata dalli manoscritti del. P. Bonaventura Cavalieri.  Rome, 1682.  ??NYS cited by Lucas (1895) and Fourrey.

Lucas.  RM2, 1883, pp. 202‑203.

Tom Tit.

Vol. 2, 1892.  L'Étoile à cinq branches, pp. 153-154.  = K, no. 5: The pentagon and the five pointed star, pp. 20‑21.  He adds that folding over the free end and holding the knot up to the light shows the pentagram.

Vol. 3, 1893.  Construire d'un coup de poing un hexagone régulier, pp. 159-161.  = K, no. 17: To construct a hexagon by finger pressure, pp. 49‑51.  Pressing an appropriate size Möbius strip flat gives a regular hexagon.

Vol. 3, 1893.  Les sept pentagones, pp. 165-166.  = K, no. 19: The seven pentagons, pp. 54‑55.  By tying five pentagons in a strip, one gets a larger pentagon with a pentagonal hole in the middle.

Somerville Gibney.  So simple!  The hexagon, the enlarged ring, and the handcuffs.  The Boy's Own Paper 20 (No. 1012) (4 Jun 1898) 573-574.  As in Tom Tit, vol. 3, pp. 159-161. 

Lucas.  L'Arithmétique Amusante.  1895.  Note IV:  Section II: Les Jeux de Ruban, Nos. 1 & 2:  Le nœud de cravate  &  Le nœud marin, pp. 220-222.  Cites d'Aviso and says he does both the pentagonal and hexagonal knots, but Lucas only shows the pentagonal one. 

E. Fourrey.  Procédés Originaux de Constructions Géométriques.  Vuibert, Paris, 1924.  Pp. 113 & 135‑139.  Cites Lucas and cites d'Aviso as Traitè de la Sphère and says he gives the pentagonal and hexagonal knots.  Fourrey shows and describes both, also giving the pictures on his title page.

F. V. Morley.  A note on knots.  AMM 31 (1924) 237-239.  Cites Knott's translation of Tom Tit.  Says the process generalizes to  (2n+3)‑gons by using  n  loops.  Gets even-gons by using two strips.  Discusses using twisted strips.

Robert C. Yates.  Geometrical Tools.  (As:  Tools; Baton Rouge, 1941);  revised ed., Educational Publishers, St. Louis, 1949.  Pp. 64-65 gives square (a bit trivial), pentagon, hexagon, heptagon and octagon.  Even case need two strips.

Donovan A. Johnson.  Paper Folding for the Mathematics Class.  NCTM, 1957, pp. 16-17: Polygons constructed by tying paper knots.  Shows how to tie square, pentagon, hexagon, heptagon and octagon.

James K. Brunton.  Polygonal knots.  MG 45 (No. 354) (Dec 1961) 299‑302.  All regular  n‑gons,  n > 4,  can be obtained, except  n = 6  which needs two strips.  Discusses which can be made without central holes.

Marius Cleyet-Michaud.  Le Nombre d'Or.  Presses Universitaires de France, Paris, 1973.  On pp. 47-48, he calls this the 'golden knot' (Le "nœud doré") and describes how to make it.

 

          6.R.    GEOMETRIC FALLACIES

 

          General surveys of such fallacies can be found in the following.  See also:  6.P, 10.A.1.

 

Ball.  MRE.  1st ed., 1892, pp. 31‑34, two examples, discussed below.  3rd ed., 1896, pp. 39‑46  =  4th ed., 1905, pp. 41-48, seven examples.  5th ed., 1911, pp. 44-52  =  11th ed., 1939, pp. 76-84, nine example.

Walther Lietzmann.  Wo steckt der Fehler?  Teubner, Stuttgart, (1950), 3rd ed., 1953.  (Strens/Guy has 3rd ed., 1963(?).)  (There are  2nd ed, 1952??;  5th ed, 1969;  6th ed, 1972.  MG 54 (1970) 182 says the 5th ed appears to be unchanged from the 3rd ed.)  Chap. B: V, pp. 87-99 has 18 examples. 

                    (An earlier version of the book, by Lietzmann & Trier, appeared in 1913, with 2nd ed. in 1917.  The 3rd ed. of 1923 was divided into two books:  Wo Steckt der Fehler?  and  Trugschlüsse.  There was a 4th ed. in 1937.  The relevant material would be in Trugschlüsse, but I have not seen any of the relevant books, though Northrop cites Lietzmann, 1923, three times _ ??NYS.)

Northrop.  Riddles in Mathematics.  1944.  Chap. 6,  1944: 97-116, 232-236 & 249-250;  1945: 91-109, 215-219 & 230-231;  1961: 98-115, 216-219 & 229.  Cites Ball, Lietzmann (1923), and some other individual items.

V. M. Bradis,  V. L. Minkovskii  &  A. K. Kharcheva.  Lapses in Mathematical Reasoning.  (As: [Oshibki v Matematicheskikh Rassuzhdeniyakh], 2nd ed, Uchpedgiz, Moscow, 1959.)  Translated by J. J. Schorr-Kon, ed. by E. A. Maxwell.  Pergamon & Macmillan, NY, 1963.  Chap. IV, pp. 123-176.  20 examples plus six discussions of other examples.

Edwin Arthur Maxwell.  Fallacies in Mathematics.  CUP, (1959), 3rd ptg., 1969.  Chaps. II-V, pp. 13-36, are on geometric fallacies.

Ya. S. Dubnov.  Mistakes in Geometric Proofs.  (2nd ed., Moscow?, 1955).  Translated by Alfred K. Henn & Olga A. Titelbaum.  Heath, 1963.  Chap 1-2, pp. 5-33.  10 examples.

_. _. __________ A. G. [Konforovich, A. G.]  (___________ _______ _ _________ [Matematichn_ Sof_zmi _ Paradoksi] (In Ukrainian).  _________ _____ [Radyans'ka Shkola], Kiev, 1983.)  Translated into German by Galina & Holger Stephan as: Andrej Grigorjewitsch Konforowitsch; Logischen Katastrophen auf der Spur _ Mathematische Sophismen und Paradoxa; Fachbuchverlag, Leipzig, 1990.  Chap. 4: Geometrie, pp. 102-189 has 68 exaples, ranging from the type considered here up through fractals and pathological curves.

S. L. Tabachnikov.  Errors in geometrical proofs.  Quantum 9:2 (Nov/Dec 1998) 37-39 & 49.  Shows:  every triangle is isosceles (6.R.1);  the sum of the angles of a triangle is 180o without use of the parallel postulate;  a rectangle inscribed in a square is a square;  certain approaching lines never meet (6.R.3);  all circles have the same circumference (cf Aristotle's Wheel Paradox in 10.A.1);  the circumference of a wheel is twice its radius;  the area of a sphere of radius  R  is  π2R2.

 

          6.R.1. EVERY TRIANGLE IS ISOSCELES

 

This is sometimes claimed to have been in Euclid's lost Pseudaria.

Ball.  MRE, 1st ed., 1892, pp. 33‑34.  On p. 32, Ball refers to Euclid's lost Fallacies and presents this fallacy and the one in 6.R.2:  "I do not know whether either of them has been published previously."  In the 3rd ed., 1896, pp. 42-43, he adds the heading:  To prove that every triangle is isosceles.  In the 5th ed., 1911, p. 45, he adds a note that he believes these two were first published in his 1st ed. and notes that Carroll was fascinated by them and they appear in The Lewis Carroll Picture Book _ see below.

Mathesis (1893).  ??NYS.  [Cited by Fourrey, Curiosities Geometriques, p. 145.]

L. Carroll.  Op. cit. in 5.B, 1899, pp. 264-265 (Collins: 190-191).  Every triangle is isosceles.  Carroll may have stated this as early as 1888.

Ahrens.  Mathematische Spiele.  Teubner.  Alle Dreiecke sind gleichschenklige.  2nd ed., 1911, chap. X, art. VI, pp. 108 & 119‑120.  3rd ed., 1916, chap. IX, art. IX, pp. 92-93 & 109-111.  4th ed., 1919  &  5th ed., 1927, chap IX, art. IX, pp. 99‑101 & 116‑118.

W. A. Bagley.  Puzzle Pie.  Op. cit. in 5.D.5.  1944.  Call Mr. Euclid _ No. 15: To prove all triangles are equilateral, pp. 16-17.  Clear exposition of the fallacy.

See Read in 6.R.4 for a different proof of this fallacy.

 

          6.R.2. A RIGHT ANGLE IS OBTUSE

 

Mittenzwey.  1879?  Prob. 231, pp. 53 & 101.

Ball.  MRE, 1st ed., 1892, pp. 32‑33.  See 6.R.1.  In the 3rd ed., 1896, pp. 40-41, he adds the heading:  To prove that a right angle is equal to an angle which is greater than a right angle.

L. Carroll.  Op. cit. in 5.B, 1899, pp. 266‑267 (Collins 191-192).  An obtuse angle is sometimes equal to a right angle. 

H. E. Licks.  1917.  Op. cit. in 5.A.  Art. 82, p. 56.

W. A. Bagley.  Puzzle Pie.  Op. cit. in 5.D.5.  1944.  Call Mr. Euclid _ No. 16: To prove one right angle greater than another right angle, pp. 18-19.  "Here again, if you take the trouble to draw an accurate diagram, you will find that the "construction" used for the alleged proof is impossible."

E. A. Maxwell.  Note 2121:  That every angle is a right angle.  MG 34 (No. 307) (Feb 1950) 56‑57.  Detailed demonstration of the error.

 

          6.R.3. LINES APPROACHING BUT NOT MEETING

 

Proclus.  5C.  A Commentary on the First Book of Euclid's Elements.  Translated by Glenn R. Morrow.  Princeton Univ. Press, 1970.  Pp. 289-291.  Gives the argument and tries to refute it.

van Etten/Henrion/Mydorge.  1630.  Part 2, prob. 7: Mener une ligne laquelle aura inclination à une autre ligne, & ne concurrera jamais contre l'Axiome des paralelles, pp. 13‑14.

Schwenter.  1636.  To be added.

Ozanam-Montucla.  1778.  Paradoxe géométrique des lignes ....  Prob. 70 & fig. 116-117, plate 13, 1778: 405-407;  1803: 411-413;  1814: 348-350.  Prob. 69, 1840: 180-181.  Notes that these arguments really produce a hyperbola and a conchoid.  Hutton adds that a great many other examples might be found.

Northrop.  Riddles in Mathematics.  1944.  1944: 209-211 & 239;  1945: 195‑197 & 222;  1961: 197‑198 & 222.  Gives the 'proof' and its fallacy, with a footnote on p. 253 (1945: 234; 1961: 233) saying the argument "has been attributed to Proclus."

Jeremy Gray.  Ideas of Space.  OUP, 1979.  Pp. 37-39 discusses Proclus' arguments in the context of attempts to prove the parallel postulate.

 

          6.R.4. OTHERS

 

Ball.  MRE, 3rd ed, 1896, pp. 44-45.  To prove that, if two opposite sides of a quadrilateral are equal, the other two sides must be parallel.  Cites Mathesis (2) 3 (Oct 1893) 224 _ ??NYS

Cecil B. Read.  Mathematical fallacies  &  More mathematical fallacies.  SSM 33 (1933) 575‑589  &  977-983.  There are two perpendiculars from a point to a line.  Part of a line is equal to the whole line.  Every triangle is isosceles (uses trigonometry).  Angle trisection (uses a marked straightedge).

P. Halsey.  Class Room Note 40:  The ambiguous case.  MG 43 (No. 345) (Oct 1959) 204‑205.  Quadrilateral  ABCD  with  angle A = angle C  and  AB = CD.  Is this a parallelogram?

 

          6.S.    TANGRAMS, ET AL.

 

          GENERAL HISTORIES.

 

Ronald C. Read.  Tangrams _ 330 Puzzles.  Dover, 1965.  The Introduction, pp. 1-6, is a sketch of the history.  Will Shortz says this is the first serious attempt to counteract the mythology created by Loyd and passed on by Dudeney.  Read cannot get back before the early 1800s and notes that most of the Loyd myth is historically unreasonable.  However, Read does not pursue the early 1800s history in depth and I consider van der Waals to be the first really serious attempt at a history of the subject.

Peter van Note.  Introduction.  IN: Sam Loyd; The Eighth Book of Tan; (Loyd & Co., 1903); Dover, 1968, pp. v-viii.  Brief debunking of the Loyd myth.

Jan van der Waals.  History  &  Bibliography.  In:  Joost Elffers; Tangram; (1973), Penguin, 1976.  Pp. 9‑27  &  29‑31.  Says the Chinese term "ch'i ch'ae" dates from the Chu era (‑740/‑330), but the earliest known Chinese book is 1813.  The History reproduces many pages from early works.  The Bibliography cites 8 versions of 4 Chinese books (with locations!) from 1813 to 1826 and 18 Western books from 1805 to c1850.

Hoffmann.  1893.  Chap III, pp. 74‑144.  Describes Tangrams and Richter puzzles at some length on pp. 74‑90.  Lots of photos in Hordern.

S&B.  1986.  Pp. 22‑33 discusses loculus of Archimedes, Chie no Ita, Tangrams and Richter puzzles.

 

Recent research by Jerry Slocum, backed up by The Admired Chinese Puzzle, indicates that the introduction of tangrams into Europe was done by a person or persons in Lord Amherst's 1815-1817 embassy to China, which visited Napoleon on St. Helena on its return voyage.  If so, then the conjectural dating of several items below needs to be amended.  I have amended my discussion accordingly and marked such dates with ??.  Although watermarking of paper with the correct date was a legal requirement at the time, paper might have been stored for some time before it was printed on, so watermark dates only give a lower bound for the date of printing.  I have seen several further items dated 1817, but it is conceivable that some material may have been sent back to Europe a few years earlier.

 

          SPECIFIC ITEMS

 

Kanchusen.  Wakoku Chiekurabe.  1727.  Pp. 9 & 28-29: a simple dissection puzzle with 8 pieces.  The problem appears to consist of a mitre comprising  ¾  of a unit square;  4 isoceles right triangles of hypotenuse 1  and  3 isoceles right triangles of side  ½,  but the solution shows that all the triangles are the same size, say having hypotenuse  1,  and the mitre shape is actually formed from a rectangle of size  1 x Ö2.

"Ganriken" [pseud., possibly of Fan Chu Sen].  Sei Sh_nagon Chie-no-Ita (The Ingenious Pieces by Sei Sh_nagon.) (In Japanese).  Kyoto Shobo, Aug 1742, 18pp, 42 problems and solutions.  Reproduced in a booklet, ed. by Kazuo Hanasaki, Tokyo, 1984, as pp. 19‑36.  Also reproduced in a booklet, transcribed into modern Japanese, with English pattern names and an English abstract, by Shigeo Takagi, 1989.  This uses a set of seven pieces different than the Tangram.  S&B, p. 22, shows these pieces.  Sei Sh_nagon (c965-c1010) was a famous courtier, author of The Pillow Book and renowned for her intelligence.  The Introduction is signed Ganriken.  S&B say this is probably Fan Chu Sen, but Takagi says the author's real name is unknown.

Utamaro.  Interior of an Edo house, from the picture‑book:  The Edo Sparrows (or Chattering Guide), 1786.  Reproduced in B&W in:  J. Hillier; Utamaro _ Colour Prints and Paintings; Phaidon Press, Oxford, (1961), 2nd ed., 1979, p. 27, fig. 15.  I found this while hunting for the next item.  This shows two women contemplating some pieces but it is hard to tell if it is a tangram‑type puzzle, or if perhaps they are cakes.  Hiroko and Mike Dean tell me that they are indeed cooking cakes.

Utamaro.  Woodcut.  1792.  Shows two courtesans working on a tangram puzzle.  van der Waals dated this as 1780, but Slocum has finally located it, though he has only been able to find two copies of it!  The courtesans are clearly doing a tangram-like puzzle with 12(?) pieces _ the pieces are a bit piled up and one must note that one of the courtesans is holding a piece.  They are looking at a sheet with 10 problem figures on it.

Early 19C books from China _ see Needham, p. 111.  ??NYS

Neues chinesisches Rätselspiel für Kinder, in 24 bildlichen und alphabetischen Darstellungen.  Pirna, Friese, c1805??.  ??NYS (van der Waals).

A New Invented Chinese Puzzle, Consisting of Seven Pieces of Ivory or Wood, Viz. 5 Triangles, 1 Rhomboid, & 1 Square, which may be so placed as to form the Figures represented in the plate.  Paine & Simpson, Boro'.  Undated, but the paper is watermarked 1806.  This consists of two 'volumes' of 8 pages each, comprising 159 problems with no solutions.  At the end are bound in a few more pages with additional problems drawn in _ these are direct copies of plates 21, 26, 22, 24, and 28 (with two repeats from plate 22) of The New and Fashionable Chinese Puzzle, 1817.  Bound in plain covers.  This is in Edward Hordern's collection and he has provided a photocopy.

H. F. Muller, ed.  Chinesisches Rätsel.  Enigmes chinoises.  Vienna, c1810??.  ??NYS (van der Waals).

Ch'i Ch'iao t'u ho‑pi (Harmoniously combined book of tangram problems).  1813.  (Bibliothek Leiden 6891, with an 1815 edition at British Library 15257 d 13.)  ??NYS (van der Waals).  323 examples.  The 1813 seems to be the earliest Chinese tangram book of problems, with the 1815 being the solutions.  Slocum says there was a solution book in 1815 and that the problem book had a preface by  Sang‑hsia K'o,  which was repeated in the solution book with the same date.  A version of this appears to have been the book given to Napoleon and to have started the tangram craze in Europe.  I have a version from c1820s which has 334 problems.

Shichi‑kou‑zu Gappeki [The Collection of Seven‑Piece Clever Figures].  Hobunkoku Publishing, Tokyo, 1881.  This is a Japanese translation of an 1813 Chinese book "recognized as the earliest of existing Tangram book", apparently the previous item.  [The book says 1803, but Jerry Slocum reports this is an error for 1813!]  Reprinted, with English annotations by Y. Katagiri, from N. Takashima's copy, 1989.  129 problems (but he counts 128 because he omits one after no. 124), all included in my version of the previous item, no solutions.

The Fashionable Chinese Puzzle.  Published by J. & E. Wallis, 42, Skinner Street and J. Wallis Junr, Marine Library, Sidmouth, nd [Mar 1817].  Photocopy from Jerry Slocum.  This has an illustrated cover, apparently a slip pasted onto the physical cover.  Slocum's copy has paper watermarked 1816.

PLUS 

A Key to the New and Fashionable Chinese Puzzle, Published by J. and E. Wallis, 42, Skinner Street, London, Wherein is explained the method of forming every Figure contained in That Pleasing Amusement.  Nd [Mar 1817].  PHOTOCOPY from the Bodleian Library, Oxford, catalogue number Jessel e.1176.  TP seems to made by pasting in the cover slip and has been bound in as a left hand page.  ALSO a PHOTOCOPY from Jerry Slocum.  In the latter copy, the apparent TP appears to be a paste down on the cover.  The latter copy does not have the Stanzas mentioned below.  Slocum's copy has paper watermarked 1815; I didn't check this at the Bodleian.

                    NOTE.  This is quite a different book than The New and Fashionable Chinese Puzzle published by Goodrich in New York, 1817.

                    Bound in at the beginning of the Fashionable Chinese Puzzle and the Bodleian copy of the Key is:  Stanzas, Addressed to Messrs. Wallis, on the Ingenious Chinese Puzzle, Sold by them at the Juvenile Repository, 42, Skinner Street.  In the Key, this is on different paper than the rest of the booklet.  The Stanzas has a footnote referring to the ex-Emperor Napoleon as being in a debilitated state.  (Napoleon died in 1821, which probably led to the Bodleian catalogue's date of c1820 for the entire booklet - but see below.  Then follow 28 plates with 323 numbered figures (but number 205 is skipped), solved in the Key.  In the Bodleian copy of the Key, these are printed on stiff paper, on one side of each sheet, but arranged as facing pairs, like Chinese booklets.

                    [Philip A. H. Brown; London Publishers and Printers  c. 1800-1870; British Library, 1982, p. 212] says the Wallis firm is only known to have published under the imprint J. & E. Wallis during 1813 and Ruth Wallis showed me another source giving 1813?-1814.  This led me to believe that the booklets originally appeared in 1813 or 1814, but that later issues or some owner inserted the c1820 sheet of Stanzas, which was later bound in and led the Bodleian to date the whole booklet as c1820.  Ruth Wallis showed me a source that states that John Wallis (Jun.) set up independently of his father at 186 Strand in 1806 and later moved to Sidmouth.  Finding when he moved to Sidmouth might help date this publication more precisely, but it may be a later reissue.  However, Slocum has now found the book advertised in the London Times in Mar 1817 and says this is the earliest Western publication of tangrams, based on the 1813/1815 Chinese work.  Wallis also produced a second book of problems of his own invention and some copies seem to be coloured.

                    In AM, p. 43, Dudeney says he acquired the copy of  The Fashionable Chinese Puzzle  which had belonged to Lewis Carroll.  He says it was "Published by J. and E. Wallis, 42 Skinner Street, and J. Wallis, Jun., Marine Library, Sidmouth" and quotes the Napoleon footnote, so this copy had the Stanzas included.  This copy is not in the Strens Collection at Calgary which has some of Dudeney's papers. 

                    Van der Waals cites two other works titled  The Fashionable Chinese Puzzle.  An 1818 edition from A. T. Goodridge [sic], NY, is in the American Antiquarian Society Library (see below) and another, with no details given, is in the New York Public Library.  Could the latter be the Carroll/Dudeney copy?

                    Toole Stott 823 is a copy with the same title and imprint as the Carroll/Dudeney copy, but he dates it c1840.  This version is in two parts.  Part I has 1 leaf text + 26 col. plates _ it seems clear that col. means coloured, a feature that is not mentioned in any other description of this book _ perhaps these were hand-coloured by an owner.  Unfortunately, he doesn't give the number of puzzles.  I wonder if the last two plates are missing from this??  Part II has 1 leaf text + 32 col. plates, giving 252 additional figures.  The only copy cited was in the library of J. B. Findlay _ I have recently bought a copy of the Findlay sale catalogue, ??NYR.

                    Toole Stott 1309 is listed with the title: Stanzas, ....  J. & F. [sic] Wallis ... and Marine Library, Sidmouth, nd [c1815].  This has 1 leaf text and 28 plates of puzzles, so it appears that the Stanzas have been bound in and the original cover title slip is lost or was not recognised by Toole Stott.  The date of c1815 is clearly derived from the Napoleon footnote but 1817 would have been more reasonable, though this may be a later reissue.  Again only one copy is cited, in the library of Leslie Robert Cole.

                    Plates 1-28 are identical to plates 1-28 of The Admired Chinese Puzzle, but in different order.  The presence of the Chinese text in The Admired Chinese Puzzle made me think the Wallis version was later than it.

                    Comparison of the Bodleian booklet with the first 27 plates of Giuoco Cinese, 1818?, reveals strong similarities.  5 plates are essentially identical, 17 plates are identical except for one, two or three changes and 3 plates are about 50% identical.  I find that 264 of the 322 figures in the Wallis booklet occur in  Giuoco Cinese,  which is about 82%.  However, even when the plates are essentially identical, there are often small changes in the drawings or the layout.

                    Some of the plates were copied by hand into Hordern's copy of  A New Invented Chinese Puzzle, c1806??.

The Admired Chinese Puzzle  A New & Correct Edition From the Genuine Chinese Copy.  C. Taylor, Chester, nd [1817].  Paper is clearly watermarked 1812, but the Prologue refers to the book being brought from China by someone in Lord Amherst's embassy to China, which took place in 1815-1817 and which visited Napoleon on St. Helena on its return.  Slocum dates this to after 17 Aug 1817, when Amherst's mission returned to England and this seems to be the second western book on tangrams.  Not in Christopher, Hall, Heyl or Toole Stott _ Slocum says there is only one copy known in England!  It originally had a cover with an illustration of two Chinese, titled  The Chinese Puzzle,  and one of the men holds a scroll saying  To amuse and instruct.  The Chinese text gives the title  Ch'i ch'iao t'u ho pi (Harmoniously combined book of tangram problems).  I have a photocopy of the cover from Slocum.  Prologue facing TP; TP; two pp in Chinese, printed upside down, showing the pieces;  32pp of plates numbered at the upper left (sometimes with reversed numbers), with problems labelled in Chinese, but most of the characters are upside down!  The plates are printed with two facing plates alternating with two facing blank pages.  Plate 1 has 12 problems, with solution lines lightly indicated.  Plates 2 - 28 contain 310 problems.  Plates 29-32 contain 18 additional "caricature Designs" probably intended to be artistic versions of some of the abstract tangram figures.  The Prologue shows faint guide lines for the lettering, but these appear to be printed, so perhaps it was a quickly done copperplate.  The text of the Prologue is as follows.

                                        This ingenious geometrical Puzzle was introduced into this Kingdom from China.

                    The following sheets are a correct Copy from the Chinese Publication, brought to England by a Gentleman of high Rank in the suit [sic] of Lord Amherst's late Embassy.  To which are added caricature Designs as an illustration, every figure being emblematical of some Being or Article known to the Chinese.

          The plates are identical to the plates in The Fashionable Chinese Puzzle above, but in different order and plate 4 is inverted and this version is clearly upside down.

Sy Hall.  A New Chinese Puzzle,  The Above Consists of Seven Pieces of Ivory or Wood, viz. 5 Triangles, 1 Rhomboid, and 1 Square, which will form the 292 Characters, contained in this Book; Observing the Seven pieces must be used to form each Character.  NB.  This Edition has been corrected in all its angles, with great care and attention.  Engraved by Sy Hall, 14 Bury Street, Bloomsbury.  Watermarked 1815.  31 plates with 292 problems.  Slocum and Hordern have copies.  Sy probably is an abbreviation of Sydney (or possibly Stanley?).  (I have seen a copy in the BL, bound with a large folding Plate 2 by Hall, which has 83 examples with solution lines drawn in (by hand??), possibly one of four sheets giving all the problems in the book.  However there is no relationship between the Plate and the book _ problems are randomly placed and often drawn in different orientation.  I have a photocopy of the plate on two A3 sheets.)

A New Chinese Puzzle.  Third Edition: Universally allowed to be the most correct that has been published.  1817.  Dalgety has a copy.

Miss Lowry.  A Key to the Only Correct Chinese Puzzle Which has Yet Been Published, with above a Hundred New Figures.  No. 1.  Drawn and engraved by Miss Lowry.  Printed by J. Barfield, London, 1817.  Hordern has a copy.

W. Williams.  New Mathematical Demonstrations of Euclid rendered clear and familiar to the minds of youth, with no other mathematical instruments than the triangular pieces, commonly called the Chinese Puzzle.  London, 1817.  ??NYS (van der Waals).

Anon.  Passe-temps Mathématique, ou Récréation à l'ile Sainte-Hélène.  Ce jeu qui occupé à qu'on prétend, les loisirs du fameux exilé à St.-Hélène.  Briquet, Geneva, 1817.  21pp.  [Copy advertised by Interlibrum, Vaduz, in 1990.]

Anonymous. 

The New and Fashionable Chinese Puzzle.  A. T. Goodrich & Co., New York, 1817.  TP, 1p of Stanzas (seems like there should be a second page??), 32pp with 346 problems.  Slocum has a copy.

[Key] to the Chinese Philosophical Amusements.  A. T. Goodrich & Co., New York, 1817.  TP, 2pp of stanzas (the second page has the Napoleon footnote and a comment which indicates it is identical to the material in the problem book), Index to the Key to the Chinese Puzzle, 80pp of solutions as black shapes with white spacing.  Slocum has a copy.

                    NOTE.  This is quite a different book than The Fashionable Chinese Puzzle published in London by Wallis in 1817.

                    Slocum writes: "Although the Goodrich problem book has the same title as the British book by Wallis and Goodrich has the "Stanzas" poem (except for the first 2 paragraphs which he deleted) the problem books have completely different layouts and Goodrich's solution book largely copies Chinese books."

Anon.  Buonapartes Geliefkoosste Vermaack op St. Helena, op Chineesch Raadsel.  1er Rotterdam by J. Harcke.  Prijs 1 - 4 ??.  2e Druck te(?) Rotterdam.  Ter Steendrukkery van F. Scheffers & Co.  Nanco Bordewijk has recently acquired this and Slocum has said it is a translation of one of the English items in c1818.  I have just a copy of the cover, and it uses many fancy letters which I don't guarantee to have read correctly.

Das grosse chinesische Rätselspiel für die elegante Welt.  Magazin für Industrie (Leipzig) (1818).  ??NYS (van der Waals).

Metamorfosi del Giuoco detto l'Enimma Chinese.  Gius. Landi, Florence, 1818.  100 shapes, some solved, then with elegant architectonic drawings in the same shapes.  See S&B, pp. 24‑25.

Al Gioco Cinese Chiamato Il Rompicapo Appendice di Figure Rappresentanti ... Preceduta da un Discorso sul Rompicapo e sulla Cina intitolato Passatempo Preliminare scritto dall'Autore Firenze All'Insegna dell'Ancora 1818.  Edward Hordern has a copy.  This has some similarities to Giuoco Cinese.

Gioco cinese chiamato il rompicapo.  Milan, 1818.  ??NYS (van der Waals).  Possibly the item above.

Al gioco cinese chiamato il rompicapo appendice.  Pietro & Giuseppe Vallardi, Milan, 1818. Possibly another printing of the item above with the same title, Firenze, 1818.  ??NYS.

Supplemento al nuovo giuoco cinese.  Fratelli Bettalli, Milan, 1818.  ??NYS.

Giuoco Cinese  Ossia  Raccolta di 364. Figure Geometrica [last letter is blurred] formate con un Quadrato diviso in 7. pezzi, colli quali si ponno formare infinite Figure diversi, come Vuomini[sic], Bestie, Ucelli[sic], Case, Cocchi, Barche, Urne, Vasi, ed altre suppelletili domestiche: Aggiuntovi l'Alfabeto, e li Numeri Arabi, ed altre nuove Figure.  Agapito Franzetti alle Convertite, Rome, nd [but 1818 is written in by hand].  Copy at the Warburg Institute, shelf mark FMH 4050.  TP & 30 plates.  It has alternate openings blank, apparently to allow you to draw in your solutions, as an owner has done in a few cases.  The first plate shows the solutions with dotted lines, otherwise there are no solutions.  There is no other text than on the TP, except for a florid heading  Alfabeto  on plate XXVIII.  The diagrams have no numbers or names.  The upper part of the TP is a plate of three men, intended to be Orientals, in a tent?  The one on the left is standing and cutting a card marked with the pieces.  The man on the right is sitting at a low table and playing with the pieces.  He is seated on a box labelled  ROMPI CAPO.  A third man is seated behind the table and watching the other seated man.  On the ground are a ruler, dividers and right angle.  The Warburg does not know who put the date 1818 in the book, but the book has a purchase note showing it was bought in 1913.  James Dalgety has the only other copy known.  Sotheby's told him that Franzetti was most active about 1790, but Slocum finds Sotheby's is no longer very definite about this.  I thought it possible that a page was missing at the beginning which gave a different form of the title, but Dalgety's copy is identical to this one.  The letters and numbers are quite different to those shown in Elffers and the other early works that I have seen, but there are great similarities to The New and Fashionable Chinese Puzzle, c1813, and some similarities to Al Gioco Cinese above.  I haven't counted the figures to verify the 364.

Ch'i Ch'iao pan.  c1820.  (Bibliothek Leiden 6891; Antiquariat Israel, Amsterdam.)  ??NYS (van der Waals).

Le Veritable Casse‑tete, ou Enigmes chinoises.  Canu Graveur, Paris, c1820.  BL.  ??NYS (van der Waals).

Ch'i Ch'iao ch'u pien ho‑pi.  After 1820.  (Bibliothek Leiden 6891.)  ??NYS (van der Waals).  476 examples.

Nouveau Casse‑Tête Français.  c1820 (according to van der Waals).  2pp with 58 figurative shapes.

Bestelmeier, 1823.  Item 1278: Chinese Squares.  It is not in the 1812 catalogue. 

Slocum.  Compendium.  Shows the above Bestelmeier entry.

Anonymous.  Ch'i ch'iao t'u ho pi (Harmoniously combined book of Tangram problems)  and  Ch'i ch'iao t'u chieh (Tangram solutions).  Two volumes of tangrams and solutions with no title page, Chinese labels of the puzzles, in Chinese format (i.e. printed as long sheets on thin paper, accordion folded and stitched with ribbon.  Nd [c1820s??], stiff card covers with flyleaves of a different paper, undoubtedly added later.  84 pages in each volume, containing 334 problems and solutions.  With ownership stamp of a cartouche enclosing EWSHING, probably a Mr. E. W. Shing.  Slocum says this is a c1820s reprint of the earliest Chinese tangram book which appeared in 1813 & 1815.  This version omits the TP and opening text.  I have a photocopy of the opening material from Slocum.  The original problem book had a preface by  Sang‑hsia K'o,  which was repeated in the solution book with the same date.  Includes all the problems of Shichi-kou-zu Gappeki, qv. 

New Series of Ch'i ch'iau puzzles.  Printed by Lou Chen‑wan, Ch'uen Liang, January 1826.  ??NYS.  (Copy at Dept. of Oriental Studies, Durham Univ., cited in R. C. Bell; Tangram Teasers.)

Child.  Girl's Own Book.  1833: 85;  1839: 72;  1842: 156.  "Chinese Puzzles _ These consist of pieces of wood in the form of squares, triangles, &c.  The object is to arrange them so as to form various mathematical figures."

Anon.  Edo Chiekata (How to Learn It??) (In Japanese).  Jan 1837, 19pp, 306 problems.  (Unclear if this uses the Tangram pieces.)  Reprinted in the same booklet as Sei Sh_nagon, on pp. 37‑55.

A Grand Eastern Puzzle.  C. Davenport & Co., London.  Nd.  ??NYS (van der Waals).

Augustus De Morgan.  On the foundations of algebra, No. 1.  Transactions of the Cambridge Philosophical Society 7 (1842) 287-300.  ??NX.  On pp. 289, he says "the well-known toy called the Chinese Puzzle, in which a prescribed number of forms are given, and a large number of different arrangements, of which the outlines only are drawn, are to be produced."

Crambrook.  1843.  P. 4, no. 4: Chinese Puzzle.  Chinese Books, thirteen numbers.  Though not illustrated, this seems likely to be the Tangrams _ ??

Boy's Treasury.  1844.  Puzzles and paradoxes, no. 16: The Chinese puzzle, pp. 426-427.  Instructions seem to intend the tangrams, but they give five shapes and say to make one copy of some and two copies of the others.  As I read it, this is incorrect, though it is intended to be the tangrams.  11  problem shapes given, no answers.  Most of the shapes occur in earlier tangram collections, particularly in A New Invented Chinese Puzzle.  This item is reproduced, complete with the error, as:  Magician's Own Book, 1857, prob. 49, pp. 289-290;  Landells, Boy's Own Toy-Maker, 1858, pp. 139-140;  Book of 500 Puzzles, 1859, pp. 103-104;  Boy's Own Conjuring Book, 1860, pp. 251-252;  Wehman, New Book of 200 Puzzles, 1908, pp. 34-35.

Leske.  Illustriertes Spielbuch für Mädchen.  1864? 

Prob. 584-11, pp. 288 & 405: Chinesisches Verwandlungsspiel.  Make a square with the tangram pieces.  Shows just five of the pieces, but correctly states which two to make two copies of.

Prob. 584-16, pp. 289 & 406.  Make an isosceles right triangle with the tangram pieces.

Prob. 584-18/25, pp. 289-291 & 407: Hieroglyphenspiele.  Form various figures from various sets of pieces, mostly tangrams, but the given shapes have bits of writing on them so the assembled figure gives a word.  Only one of the shapes is as in Boy's Treasury.

Prob. 588, pp. 298 & 410: Etliche Knackmandeln.  Another tangram problem like the preceeding, not equal to any in Boy's Treasury.

Adams & Co., Boston.  Advertisement in The Holiday Journal of Parlor Plays and Pastimes, Fall 1868.  Details?? _ xerox sent by Slocum.  P. 6: Chinese Puzzle.  The celebrated Puzzle with which a hundred or more symmetrical forms can be made, with book showing the designs.  Though not illustrated, this seems likely to be the Tangrams _ ??

J. Murray (editor of the OED).  Two letters to H. E. Dudeney (9 Jun 1910  &  1 Oct 1910).  The first inquires about the word 'tangram', following on Dudeney's mention of it in his "World's best puzzles" (op. cit. in 2).  The second says that 'tan' has no Chinese origin;  is apparently mid 19C, probably of American origin;  and the word 'tangram' first appears in Webster's Dictionary of 1864.  Dudeney, AM, 1917, p. 44, excerpts these letters.

F. T. Wang & C.‑S. Hsiung.  A theorem on the tangram.  AMM 49 (1942) 596‑599.  Determine the 20 convex regions which 16 isoceles right triangles can form and hence the 13 ones which the Tangram pieces can form.

Mitsumasa Anno.  Anno's Math Games.  (Translation of: Hajimete deau sugaku no ehon; Fufkuinkan Shoten, Tokyo, 1982.)  Philomel Books, NY, 1987.  Pp. 38-43 & 95-96 show a simplified  5-piece tangram-like puzzle which I have not seen before.  The pieces are: a square of side  1;  three isosceles right triangles of side  1;  a right trapezium with bases  1  and  2,  altitude  1  and slant side  Ö2.  The trapezium can be viewed as putting together the square with a triangle.  19 problems are set, with solutions at the back.

At the International Congress on Mathematical Education, Seville, 1996, the Mathematical Association gave out  The  3, 4, 5  Tangram,  a cut card tangram, but in a  6 x 8  rectangular shape, so that the medium sized triangle was a  3-4-5 triangle.  I modified this in Nov 1999, by stretching along a diagonal to form a rhombus with angles double the angles of a 3-4-5 triangle, so that four of the triangles are similar to 3-4-5 triangles.  Making the small triangles be actually 3-4-5, all edges are integral.  I made up 35 problems with these pieces.  I later saw that Hans Wierzorke has mentioned this dissection in CFF, but with no problems.  I distributed this as my present at the Fourth Gathering for Gardner, Feb 2000.

 

          6.S.1.  LOCULUS OF ARCHIMEDES

 

See S&B 22.  I recall there is some dispute as to whether the basic diagram should be a square or a double square.

E. J. Dijksterhuis.  Archimedes.  Munksgaard, Copenhagen, 1956;  reprinted by Princeton Univ. Press, 1987.  Pp. 408‑412 is the best discussion of this topic and supplies most of the classical references.

 

Archimedes.  Letter to Eratosthenes, c-250?.  Greek palimpsest, c975, on MS no. 355, from the Cloister of Saint Sabba (= Mar Saba), Jerusalem, then at Metochion of the Holy Sepulchre, Constantinopole.  [This MS disappeared in the confusion in Asia Minor in the 1920s but reappeared in 1998 when it was auctioned by Christie's in New York for c2M$.  Hopefully, modern technology will allow a facsimile and an improved transcription in the near future.]  Described by J. L. Heiberg (& H. G. Zeuthen); Eine neue Schrift des Archimedes; Bibliotheca Math. (3) 7 (1906‑1907) 321‑322.  Heiberg describes the MS, but only mentions the loculus.  The text is in Heiberg's edition of Archimedes; Opera; 2nd ed., Teubner, Leipzig, 1913, vol. II, pp. 415‑424, where it has been restored using the Suter MSS below.  Heath only mentions the problem in passing.  Heiberg quotes Marius Victorinus, Atilius Fortunatianus and cites Ausonius and Ennodius. 

H. Suter.  Der Loculus Archimedius oder das Syntemachion des Archimedes.  Zeitschr. für Math. u. Phys. 44 (1899) Supplement  = AGM 9 (1899) 491‑499.  This is a collation from two 17C Arabic MSS which describe the construction of the loculus.  They are different than the above MS.  The German is included in Archimedes Opera II, 2nd ed., 1913, pp. 420‑424.

Dijksterhuis discusses both of the above and says that they are insufficient to determine what was intended.  The Greek seems to indicate that Archimedes was studying the mathematics of a known puzzle.  The Arabic shows the construction by cutting a square, but the rest of the text doesn't say much.

 

Lucretius.  De Rerum Natura.  c‑70.  ii, 778‑783.  Quoted and discussed in H. J. Rose; Lucretius ii. 778‑83; Classical Review (NS) 6 (1956) 6‑7.  Brief reference to assembling pieces into a square or rectangle.

Decimus Magnus Ausonius.  c370.  Works.  Edited & translated by H. G. Evelyn White.  Loeb Classical Library, ??date.  Vol. I, Book XVII: Cento Nuptialis (A Nuptial Cento), pp. 370-393 (particularly the Preface, pp. 374-375) and Appendix, pp. 395-397.  Refers to 14 little pieces of bone which form a monstrous elephant, a brutal boar, etc.  The Appendix gives the construction from the Arabic version, via Heiberg, and forms the monstrous elephant.

Marius Victorinus.  4C.  VI, p. 100 in the edition of Keil, ??NYS.  Given in Archimedes Opera II, 2nd ed., 1913, p. 417.  Calls it 'loculus Archimedes' and says it had 14 pieces which make a ship, sword, etc.

Ennodius.  Carmina: De ostomachio eburneo.  c500.  In:  Magni Felicis Ennodii Opera; ed. by F. Vogel, p. 340.  In:  Monumenta Germaniae Historica, VII (1885) 249.  ??NYS.  Refers to ivory pieces to be assembled.

Attilius Fortunatianus.  6C.  ??NYS  Given in Archimedes Opera II, p. 417.  Same comment as for Marinus Victorinus.

E. Fourrey.  Curiositiés Géométriques.  (1st ed., Vuibert & Nony, Paris, 1907);  4th ed., Vuibert, Paris, 1938.  Pp. 106‑109.  Cites Suter, Ausonius, Marius Victorinus, Attilius Fortunatianus.

Collins.  Book of Puzzles.  1927.  The loculus of Archimedes, pp. 7-11.  Pieces made from a double square.

 

          6.S.2.  OTHER SETS OF PIECES

 

          See Hoffmann & S&B, cited at the beginning of 6.S, for general surveys.

          See Bailey in 6.AS.1 for an 1858 puzzle with 10 pieces and The Sociable and Book of 500 Puzzles, prob. 10, in 6.AS.1 for an 11 piece puzzle. 

          There are many versions of this idea available and some are occasionally given in JRM.

          The Richter Anchor Stone puzzles and building blocks were inspired by Friedrich FROEBEL (or Fröbel) (1782‑1852), the educational innovator.  He was the inventor of Kindergartens, advocated children's play blocks and inspired both the Richter Anchor Stone Puzzles and Milton Bradley.  The stone material was invented by Otto Lilienthal (1848‑1896) (possibly with his brother Gustav) better known as an aviation pioneer _ they sold the patent and their machines to F. Adolph RICHTER for 1000 marks.  The material might better be described as a kind of fine brick which could be precisely moulded.  Richter improved the stone and began production at Rudolstadt, Thüringen, in 1882; the plant closed in 1964.  Anchor was the company's trademark.  He made at least 36 puzzles and perhaps a dozen sets of building blocks which were popular with children, architects, engineers, etc.  The Deutsches Museum in Munich has a whole room devoted to various types of building blocks and materials, including the Anchor blocks.  There is an Anker Museum in the Netherlands (Stichting Ankerhaus (= Anker Museum); Opaalstraat 2‑4 (or Postf. 1061), NL-2400 BB Alphen aan den Rijn, NETHERLANDS; tel: 01720‑41188) which produces replacement parts for Anker stone puzzles.  Modern facsimiles of the building sets are also being produced.

 

In 1996 I noticed the ceiling of the room to the south of the Salon of the Ambassadors in the Alcazar of Seville.  This 15C? ceiling was built by workmen influenced by the Moorish tradition and has 112 square wooden panels in a wide variety of rectilineal patterns.  One panel has some diagonal lines and looks like it could be used as a 10 piece tangram-like puzzle.  Consider a  4 x 4  square.  Draw both diagonal lines, then at two adjacent corners, draw two lines making a unit square at these corners.  At the other two corners draw one of these two lines, namely the one perpendicular to their common side.  This gives six isosceles right triangles of edge  1;  two pentagons with three right angles and sides  1, 2, 1, Ö2, Ö2;  two quadrilaterals with two right angles and sides  2, 1, Ö2, 2Ö2.  Since geometric patterns and paneling are common features of Arabic art, I wonder if there are any instances of such patterns being used for a tangram-like puzzle?

Jackson.  Rational Amusement.  1821.  Geometrical Puzzles, nos. 20-27, pp. 27-29 & 88-89 & plate II, figs. 15-22.  This is a set of 20 pieces of 8 shapes used to make a square, a right triangle, three squares, etc.

Crambrook.  1843.  P. 4, no. 1: Pythagorean Puzzle, with Book.  Though not illustrated, this is probably(??) the puzzle described in Hoffmann, below, which was a Richter Anchor puzzle No. 12 of the same name and is still occasionally seen.  See S&B 28.

Edward Hordern has a Circassian Puzzle, c1870, with many pieces.

Hoffmann.  1893.  Chap. III, no. 3: The Pythagoras Puzzle, pp. 83-85 & 117-118.  This has 7 pieces and is quite like the Tangram _ see comment under Crambrook.

C. Dudley Langford.  Note 1538:  Tangrams and incommensurables.  MG 25 (No. 266) (Oct 1941) 233‑235.  Gives alternate dissections of the square and some hexagonal dissections.

C. Dudley Langford.  Note 2861:  A curious dissection of the square.  MG 43 (No. 345) (Oct 1959) 198.  There are 5 triangles whose angles are multiples of  π/8 = 22½o.  He uses these to make a square.

See items at the end of 6.S.

 

          6.T.    NO THREE IN A LINE PROBLEM

 

          See also section 6.AO.2.

 

Loyd.  Problem 14: A crow puzzle.  Tit‑Bits 31 (16 Jan  &  6 Feb 1897) 287  &  343.  = Cyclopedia, 1914, Crows in the corn, pp. 110 & 353.  = MPSL1, prob. 114, pp. 113 & 163‑164.  8 queens with no two attacking and no three in any line.

Dudeney.  The Tribune (7 Nov 1906) 1.  ??NX.  = AM, prob. 317, pp. 94 & 222.  Asks for a solution with two men in the centre 2 x 2 square.

Loyd.  Sam Loyd's Puzzle Magazine, January 1908.  ??NYS.  (Given in A. C. White; Sam Loyd and His Chess Problems; 1913, op. cit. in 1; p. 100, where it is described as the only solution with 2 pieces in the 4 central squares.)

Ahrens, MUS I 227, 1910, says he first had this in a letter from E. B. Escott dated  1 Apr 1909.  (Moser, below, refers this to the 1st ed., 1900, but this must be due to his not having seen it.)

C. H. Bullivant.  Home Fun, 1910, op. cit. in 5.S.  Part VI, Chap. IV: No. 2: Another draught puzzle, pp. 515 & 520.  The problem says "no three men shall be in a line, either horizontally or perpendicularly".  The solution says "no three are in a line in any direction" and the diagram shows this is indeed true.

Loyd.  Picket posts.  Cyclopedia, 1914, pp. 105 & 352.  = MPSL2, prob. 48, pp. 34 & 136.  2 pieces initially placed in the 4 central squares.

Blyth.  Match-Stick Magic.  1921.  Matchstick board game, p. 73.  6 x 6  version phrased as putting "only two in any one line: horizontal, perpendicular, or diagonal."  However, his symmetric solution has three in a row on lines of slope 2.

King.  Best 100.  1927.  No. 69, pp. 28 & 55.  Problem on the  6 x 6  board _ gives a symmetric solution.  Says "there are two coins on every row" including "diagonally across it", but he has three in a row on lines of slope 2. 

Loyd Jr.  SLAHP.  1928.  Checkers in rows, pp. 40 & 98.  Different solution than in Cyclopedia.

Adams.  Puzzle Book.  1939.  Prob. C.83: Stars in their courses, pp. 144 & 181.  Same solution as King, but he says "two stars in each vertical row, two in each horizontal row, and two in each of the the two diagonals ....  There must not be more than two stars in the same straight line", but he has three in a row on lines of slope 2.

W. O. J. Moser & J. Pach.  No‑three‑in‑line problem.  In:  100 Research Problems in Discrete Geometry 1986; McGill Univ., 1986.  Problem 23, pp. 23.1 _ 23.4.  Survey with 25 references.  Solutions are known on the  n x n  board for  n £ 16  and for even  n £ 26.  Solutions with the symmetries of the square are only known for  n = 2, 4, 10.

 

          6.U.   TILING

 

          6.U.1. PENROSE PIECES

 

R. Penrose.  The role of aesthetics in pure and applied mathematical research.  Bull. Inst. Math. Appl. 10 (1974) 266‑272.

M. Gardner.  SA (Jan 1977).  Extensively rewritten as Penrose Tiles, Chaps. 1 & 2.

R. Penrose.  Pentaplexity.  Eureka 39 (1978) 16‑22.  =  Math. Intell. 2 (1979) 32‑37.

D. Shechtman, I. Blech, D. Gratias & J. W. Cohn.  Metallic phase with long‑range orientational order and no translational symmetry.  Physical Rev. Letters 53:20 (12 Nov 1984) 1951‑1953.  Describes discovery of 'quasicrystals' having the symmetry of a Penrose‑like tiling with icosahedra.

David R. Nelson.  Quasicrystals.  SA 255:2 (Aug 1986) 32‑41 & 112.  Exposits the discovery of quasicrystals.  First form is now called 'Shechtmanite'.

Kimberly-Clark Corporation has taken out two patents on the use of the Penrose pattern for quilted toilet paper as the non-repetition prevents the tissue from 'nesting' on the roll.  In Apr 1997, Penrose issued a writ against Kimberly Clark Ltd. asserting his copyright on the pattern and demanding damages, etc.

John Kay.  Top prof goes potty at loo roll 'rip-off'.  The Sun (11 Apr 1997) 7.

Patrick McGowan.  It could end in tears as maths boffin sues Kleenex over design.  The Evening Standard (11 Apr 1997) 5.

Kleenex art that ended in tears.  The Independent (12 Apr 1997) 2.

For a knight on the tiles.  Independent on Sunday (13 Apr 1997) 24.  Says they exclusively reported Penrose's discovery of the toilet paper on sale in Dec 1996.

 

          6.U.2. PACKING BRICKS IN BOXES

 

          In two dimensions, it is not hard to show that  a x b  packs  A x B  if and only if  a  divides either  A or B;  b  divides either  A or B;  A and B  are both linear combinations of  a and b.  E.g.  2 x 3  bricks pack a  5 x 6  box.

          See also 6.G.1.

 

Manuel H. Greenblatt (  -1972, see JRM 6:1 (Winter 1973) 69).  Mathematical Entertainments.  Crowell, NY, 1965.  Construction of a cube, pp. 80‑81.  Can  1 x 2 x 4  fill  6 x 6 x 6?  He asserts this was invented by R. Milburn of Tufts Univ.

N. G. de Bruijn.  Filling boxes with bricks.  AMM 76 (1969) 37‑40.  If  a1 x ... x an  fills  A1 x ... x An  and  b  divides  k  of the  ai,  then  b  divides at least  k  of the  Ai.  He previously presented the results, in Hungarian, as problems in Mat. Lapok 12, pp. 110‑112, prob. 109 and 13, pp. 314‑317, prob. 119.  ??NYS.

D. A. Klarner.  Brick‑packing puzzles.  JRM 6 (1973) 112‑117.  General survey.  In this he mentions a result that I gave him _ that  2 x 3 x 7  fills a  8 x 11 x 21,  but that the box cannot be divided into two packable boxes.  However, I gave him the case  1 x 3 x 4  in  5 x 5 x 12  which is the smallest example of this type.

 

          6.V.    SILHOUETTE AND VIEWING PUZZLES

 

          Viewing problems must be common among draughtsmen and engineers, but I haven't seen many examples.  I'd be pleased to see further examples.

 

2 silhouettes.

Circle  &  triangle  _  van Etten,  Ozanam,  Guyot,  Magician's Own Book (UK version)

Circle  &  square  _  van Etten

Circle  &  rhombus  _  van Etten,  Ozanam

Rectangle with inner rectangle  &  rectangle with notch  _  Diagram Group.

3 silhouettes.

Circle,  circle,  circle  _  Madachy

Circle,  cross,  square  _  Wyatt,  Perelman

Circle,  oval,  rectangle  _  van Etten,  Ozanam,  Guyot,  Magician's Own Book (UK version)

Circle,  oval,  square  _  van Etten,  Ozanam,  Ozanam‑Montucla,  Badcock,  Jackson,  Endless  Amusement II,  Young Man's Book

Circle,  rhombus,  rectangle  _  Ozanam,  Alberti

Circle,  square,  triangle  _  Catel,  Bestelmeier,  Jackson,  Boy's Own Book,  Crambrook,  Family Friend,  Magician's Own Book,  Book of 500 Puzzles,  Boy's Own Conjuring Book,  Illustrated Boy's Own Treasury,  Riecke,  Elliott,  Tom Tit,  Handy Book,  Hoffmann,  Williams,  Wyatt,  Perelman,  Madachy

Square,  tee,  triangle  _  Perelman

4 silhouettes.

Circle,  square,  triangle,  rectangle with curved ends  _  Williams

3 views.

Madachy,  Ranucci

 

van Etten.  1624. 

Prob. 22 (misnumbered 15 in 1626) (Prob. 20), pp. 19‑20 & figs. opp. p. 16 (pp. 35‑36): 2 silhouettes _  one circular, the other triangular, rhomboidal or square.  (English ed. omits last case.)  The 1630 Examen says the author could have done better and suggests:  isosceles triangle, several scalene triangles, oval or circle, which he says can be done with an elliptically cut cone and a scalene cone.  I am not sure I believe these.  It seems that the authors are allowing the object to fill the hole and to pass through the hole moving at an angle to the board rather than perpendicularly as usually understood.  In the English edition the Examination is combined with that of the next problem. 

Prob. 23 (21), pp. 20‑21 & figs. opp. p. 16 (pp. 37‑38): 3 silhouettes _ circle, oval and square or rectangle.  The 1630 Examen suggests:  square, circle, several paralellograms and several ellipses, which he says can be done with an elliptic cylinder of height equal to the major diameter of the base.  The English Examination says "a solid colume ... cut Ecliptick-wise" _ ??

Dudeney.  Great puzzle crazes.  Op. cit. in 2.  1904.  He says  square, circle and triangle  is in a book in front of him dated 1674.  I suspect this must be the 1674 English edition of van Etten, but I don't find the problem in the English editions I have examined.  Perhaps Dudeney just meant that the idea was given in the 1674 book, though he is specifically referring to the square, circle, triangle version.

Ozanam.  1725.  Vol. II, prob. 58 & 59, pp. 455‑458 & plate 25* (53 (note there is a second plate with the same number)).  Circle and triangle;  circle and rhombus;  circle, oval, rectangle;  circle, oval, square.  Figures are very like van Etten.  See Ozanam-Montucla, 1778.

Ozanam.  1725.  Vol. IV.  No text, but shown as an unnumbered figure on plate 15 (17).  3 silhouettes:  circle, rhombus, rectangle.

Simpson.  Algebra.  1745.  Section XVIII, prob. XXIX, pp. 279-281.  (1790: prob. XXXVII, pp. 306-307.  Computes the volume of an ungula obtained by cutting a cone with a plane.  Cf. Riecke, 1867.

Alberti.  1747.  No text, but shown as an unnumbered figure on plate XIIII, opp. p. 218 (112), copied from Ozanam, 1725, vol IV.  3 silhouettes:  circle, rhombus, rectangle.

Ozanam-Montucla.  1778.  Faire passer un même corps par un trou quarré, rond & elliptique.  Prob. 46, 1778: 347-348;  1803: 345-346;  1814: 293.  Prob. 45, 1840: 149-150.  Circle, ellipse, square.

Catel.  Kunst-Cabinet.  1790.  Die mathematischen Löcher, p. 16 & fig. 42 on plate II.  Circle, square, triangle.

E. C. Guyot.  Nouvelles Récréations Physiques et Mathématiques.  Op. cit. in  6.P.2.  1799.  Vol. 2, Quatrième récréation, p. 45 & figs. 1‑4, plate 7, opp. p. 45.  2 silhouettes:  circle & triangle;  3 silhouettes:  circle, oval, rectangle.

Bestelmeier. 

1801.  Item 536: Die 3 mathematischen Löcher.  (See also the picture of Item 275, but that text is for another item.)  Square, triangle and circle.

1807.  Item 1126: Tricks includes the  square, triangle and circle.

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  P. 14, no. 23: How to make a Peg that will exactly fit three different kinds of Holes.  "Let one of the holes be circular, the other square, and the third an oval; ...."  Solution is a cylinder whose height equals its diameter.

Jackson.  Rational Amusement.  1821.  Geometrical Puzzles.

No. 16, pp. 26 & 86.  Circle, square, triangle,  with discussion of the dimensions:  "a wedge, except that its base must be a circle".

No. 29, pp. 30 & 89-90.  Circle, oval, square.

Endless Amusement II.  1826?  P. 62:  "To make a Peg that will exactly fit three different kinds of Holes."  Circle, oval, square.  c= Badcock.

The Boy's Own Book.  The triple accomodation.  1828: 419;  1828-2: 424;  1829 (US): 215;  1855: 570;  1868: 677.  Circle, square and triangle.

Young Man's Book.  1839.  Pp. 294-295.  Circle, oval, square.  Identical to Badcock.

Crambrook.  1843.  P. 5, no. 16: The Mathematical Paradox _ the Circle, Triangle, and Square.  Check??

Family Friend 3 (1850) 60 & 91.  Practical puzzle _ No. XII.  Circle, square, triangle.  This is repeated as Puzzle 16 _ Cylinder puzzle in (1855) 339 with solution in (1856) 28.

Magician's Own Book.  1857.  Prob. 21: The cylinder puzzle, pp. 273 & 296.  Circle, square, triangle.  = Book of 500 Puzzles, 1859, prob. 21, pp. 87 & 110.  = Boy's Own Conjuring Book, 1860, prob. 20, pp. 235 & 260.

Illustrated Boy's Own Treasury.  1860.  Practical Puzzles, No. 42, pp. 403 & 442.  Identical to Magician's Own Book, with diagram inverted.

F. J. P. Riecke.  Op. cit. in 4.A.1, vol. 1, 1867.  Art. 33: Die Ungula, pp. 58‑61.  Take a cylinder with equal height and diameter.  A cut from the diameter of one base which just touches the other base cuts off a piece called an ungula (Latin for claw).  He computes the volume as  4r3/3.  He then makes the symmetric cut to produce the circle, square, triangle shape, which thus has volume  (2π ‑ 8/3) r3.  Says he has seen such a shape and a board with the three holes as a child's toy.  Cf. Simpson, 1745.

Magician's Own Book (UK version).  1871.  The round peg in the square hole:  To pass a cylinder through three different holes, yet to fill them entirely, pp. 49-50.  Circle, oval, rectangle;  circle & (isosceles) triangle.

Alfred Elliott.  Within‑Doors.  A Book of Games and Pastimes for the Drawing Room.  Nelson, 1872.  [Toole Stott 251.  Toole Stott 1030 is a 1873 ed.]  No. 4: The cylinder puzzle, pp. 27‑28 & 30‑31.  Circle, square, triangle.

Tom Tit, vol. 2.  1892.  La cheville universelle, pp. 161-162.  = K, no. 28: The universal plug, pp. 72‑73.  = R&A, A versatile peg, p. 106.  Circle, square, triangle.

Handy Book for Boys and Girls.  Op. cit. in 6.F.3.  1892.  Pp. 238-242: Captain S's peg puzzle.  Circle, square, triangle.

Hoffmann.  1893.  Chap. X, no. 20: One peg to fit three holes, pp. 344 & 381‑382.  Circle, square, triangle.

Williams.  Home Entertainments.  1914.  The plug puzzle, pp. 103-104.  Circle, square, triangle and rectangle with curved ends.  This is the only example of this four-fold form that I have seen.  Nice drawing of a board with the plug shown in each hole, except the curve on the sloping faces is not always drawn down to the bottom.

E. M. Wyatt.  Puzzles in Wood, 1928, op. cit. in 5.H.1. 

The "cross" plug puzzle, p. 17.  Square, circle and cross.

The "wedge" plug puzzle, p. 18.  Square, circle and triangle.

Perelman.  FMP.  c1935?  One plug for three holes;  Further "plug" puzzles, pp. 339‑340 & 346.  6 simple versions;  3 harder versions:  square, triangle, circle;  circle, square, cross;  triangle, square, tee.  The three harder versions are also in FFF, 1957: probs. 69-71, pp. 112 & 118-119;  1979: probs. 73‑75, pp. 137 & 144  = MCBF: probs. 73-75, pp. 134-135 & 142-143.

Joseph S. Madachy.  3‑D in 2‑D.  RMM 2 (Apr 1961) 51‑53  &  3 (Jun 1961) 47.  Discusses 3 view and 3 silhouette problems.

                    3 circular silhouettes, but not a sphere.

                    Square, circle, triangle.

Ernest R. Ranucci.  Non‑unique orthographic projections.  RMM 14 (Jan‑Feb 1964) 50.  3 views such that there are 10 different objects with these views.

The Diagram Group.  The Family Book of Puzzles.  The Leisure Circle Ltd., Wembley, Middlesex, 1984.  Problem 114, with Solution at the back of the book.  Front view is a rectangle with an interior rectangle.  Side view is a rectangle with a rectangular notch on front side.  Solution is a short cylinder with a straight notch in it.  This is a fairly classic problem for engineers but I haven't seen it in print elsewhere.

Marek Penszko.  Polish your wits _ 3: Loop the loop.  Games 11:2 (Feb/Mar 1987) 28 & 58.  Draw lines on a glass cube to produce three given projections.  Problem asks for all three projections to be the same.

 

          6.W.  BURR PUZZLES

 

          When assembled, a burr looks like three sticks crossing orthogonally, forming a 'star' with six points at the vertices of an octahedron.  Slocum says Wyatt [Puzzles in Wood, 1928, op. cit. in 5.H.1] is the first to use the word 'burr'.  Collins, Book of Puzzles, 1927, p. 135, calls them "Cluster, Parisian or Gordian Knot Puzzles" and states: "it is believed that they were first made in Paris, if, indeed, they were not invented thre."

          See S&B, pp. 62‑85.

          See also 6.BJ.

 

          6.W.1.          THREE PIECE BURR

 

          Most of these have three pieces which are rectangular in cross-section with slots of the same size and some of the pieces have notches from the slot to the outside.  When one piece is pushed, it slides, revealing its notch.  When placed properly, this allows a second piece to slide off and out.  In the 1990s, a more elaborate type of three piece burr has appeared.  These have three  3 x 3 x 5  pieces which intersect in a central  3 x 3 x 3  region.  Within this region, some of the unit cubes are not present, which allows sliding of the pieces.  Some versions of the puzzle permit twisting of pieces though this usually requires a bit of rounding of edges and the actual examples tend to break, so these are not as acceptable.

 

Crambrook.  1843.  P. 5, no. 4:  Puzzling Cross  3 pieces.  This seems likely to be a three piece burr, but perhaps is in 6.W.3 _ ??  It is followed by  "Maltese Cross  6 pieces".

Edward Hordern has examples in ivory from 1850-1900.

Hoffmann.  1893.  Chap. III, no. 35: The cross‑keys or three‑piece puzzle, pp. 106 & 139.  (Hordern, p. 67, has a photo.)  One piece has an extra small notch which does not appear in other versions where the dimensions are better chosen.

Benson.  1904.  The cross keys puzzle, pp. 205‑206.

Pearson.  1907.  Part III, no. 56: The cross‑keys, pp. 56 & 127‑128.

Arthur Mee's Children's Encyclopedia 'Wonder Box'.  The Children's Encyclopedia appeared in 1908, so this is probably 1908 or soon thereafter.  3-Piece Mortise with thin pieces.

Anon.  Woodwork Joints.  Evans, London, (1918), 2nd ed., 1919.  [I have also seen a 4th ed., 1925, which is identical to the 2nd ed., except for advertising pages at the end.]  A mortising puzzle, pp. 197‑199.

Collins.  Book of Puzzles.  1927.  Pp. 136-137: The cross‑keys puzzle.

E. M. Wyatt.  Three piece cross.  Puzzles in Wood, 1928, op. cit. in 5.H.1, pp. 24‑25.

A. S. Filipiak.  Burr puzzle.  Mathematical Puzzles, 1942, op. cit. in 5.H.1, p. 101.

Dic Sonneveld seems to be the first to begin designing three piece burrs of the more elaborate style, perhaps about 1985.  Trevor Wood has made several examples for sale.

Bill Cutler.  Email announcement to NOBNET on 27 Jan 1999.  He has begun analysing the newer style of three piece burr, excluding twist moves.  His first stage has examined cases where the centre cube of the central region is occupied and the piece this central cube belongs to has no symmetry.  He finds  202 x 109  assemblies (I'm not sure if this is an exact figure) and there are  33  level-8 examples (i.e. where it takes  8  moves to remove the first piece);  6674  level-7 examples;  73362  level-6 examples.  He thinks this is about  70%  of the total and it is already about six times the number of cases considered for the six piece burr (see 6.W.2).

Bill Cutler.  Christmas letter of 4 Dec 1999.  Says he has completed the above analysis and found  25 x 1010  possibilities, which took 225 days on a workstation.  The most elaborate examples require 8 moves to get a piece out and there are 80 of these.  He used one for his IPP19 puzzle.

 

          6.W.2.          SIX PIECE BURR  =  CHINESE CROSS

 

          See also 6.W.7.

 

Minguét.  Engaños.  1733.  Pp. 103-105 (1755: 51-52; 1822: 122-124).  Pieces diagrammed.  One plain key piece.

Catel.  Kunst-Cabinet.  1790.  Die kleine Teufelsklaue, p. 10 & fig. 16 on plate I.  Figure shows it assembled and fails to draw one of the divisions between pieces.  Description says it is 6 pieces, 2 inches long, from plum wood and costs 3 groschen (worth about an English penny of the time).  (See also pp. 9-10, fig. 20 on plate I for Die grosse Teufelsklaue _ the 'squirrelcage'.)

Bestelmeier.  1801.  Item 147: Die kleine Teufelsklaue.  (Note _ there is another item 147 on the next plate.)  Only shows it assembled.  Brief text may be copying part of Catel.  See also the picture for item 1099 which looks like a six‑piece burr included in a set of puzzles.  (See also Item 142: Die grosse Teufelsklaue.)

Edward Hordern has examples, called The Oak of Old England, from c1840.

Crambrook.  1843.  P. 5, no. 5:  Maltese Cross   6 [pieces], three sorts.  Not clear if these might be here or in 6.W.4 or 6.W.5 _ ??

Magician's Own Book.  1857.  Prob. 1: The Chinese cross, pp. 266-267 & 291.  One plain key piece.  Not the same as in Minguét.

Landells.  Boy's Own Toy-Maker.  1858.  Pp. 137-139.  Identical to Magician's Own Book.

Book of 500 Puzzles.  1859.  1: The Chinese cross, pp. 80-81 & 105.  Identical to Magician's Own Book.

A. F. Bogesen (1792‑1876).  In the Danish Technical Museum, Helsingør (= Elsinore) are a number of wooden puzzles made by him, including a 6 piece burr, a 12 piece burr, an Imperial Scale? and a complex (trick??) joint.

Illustrated Boy's Own Treasury.  1860.  Practical Puzzles, No. 23: The Chinese Cross, pp. 399 & 439.  Identical to Magician's Own Book, except one diagram in the solution omits two labels.

Boy's Own Conjuring Book.  1860.  Prob. 1: The Chinese cross, pp. 228 & 254.  Identical to Magician's Own Book.

Hoffmann.  1893.  Chap. III, no. 36: The nut (or six‑piece) puzzle, pp. 106 & 139‑140.  Different pieces than in Minguét and Magician's Own Book.

Dudeney.  Prob. 473 _ Chinese cross.  Weekly Dispatch (23 Nov  &  7 Dec 1902), both p. 13.  Different than one known to his correspondents.

Dudeney.  Great puzzle crazes.  Op. cit. in 2.  1904.  "... the "Chinese Cross," a puzzle of undoubted Oriental origin that was formerly brought from China by travellers as a curiosity, but for a long time has had a steady sale in this country."

Wehman.  New Book of 200 Puzzles.  1908.  The Chinese cross, pp. 40-41.  = Magician's Own Book.

Dudeney.  The world's best puzzles.  1908.  Op. cit. in 2.  P. 779 shows a '"Chinese Cross" which ... is of great antiquity.'

Oscar W. Brown.  US Patent 1,225,760 _ Puzzle.  Applied 27 Jun 1916;  patented 15 May 1917.  3pp + 1p diagrams.  Coffin says this is the earliest US patent, with several others following soon after.

Anon.  Woodwork Joints, 1918, op. cit. in 6.W.1.  Eastern joint puzzle, pp. 196‑197: Two versions using different pieces.  Six‑piece joint puzzle, pp. 199‑200.  Another version.

Western Puzzle Works, 1926 Catalogue.  No. 86: 6 piece Wood Block.  Several other possible versions _ see 6.W.7.

E. M. Wyatt.  Six‑piece burr.  Puzzles in Wood, 1928, op. cit in 5.H.1, pp. 27‑28.  Describes 17 versions from 13 types of piece.

A. S. Filipiak.  Mathematical Puzzles, 1942, op. cit. in 5.H.1, pp. 79‑87.  73 versions from 38 types of piece.

William H. [Bill] Cutler.  The six‑piece burr.  JRM 10 (1977‑78) 241‑250.  Complete, computer assisted, analysis, with help from T. H. O'Beirne and A. C. Cross.  Pieces are considered as 'notchable' if they can be made by a sequence of notches, which are produced by two saw cuts and then chiseling out the space between them.  Otherwise viewed, notches are what could be produced by a wide cutter or router.  There are  25  of these which can occur in solutions.  (In 1994, he states that there are a total of  59  notchable pieces and diagrams all of them.)  One can also have more general pieces with 'right-angle notches' which would require four chisel cuts _ e.g. to cut a single  1 x 1 x 1  piece out of a  2 x 2 x 8  rod.  Alternatively, one can glue cubes into notches.  There are  369  which can occur in solutions.  (In 1994, he states that there are  837  pieces which produce  2225  different oriented pieces, and he lists them all.)  He only considers solid solutions _ i.e. ones where there are no internal holes.  He finds and lists the  314  'notchable' solutions.  There are  119,979  general solutions.

C. Arthur Cross.  The Chinese Cross.  Pentangle, Over Wallop, Hants., UK, 1979.  Brief description of the solutions in the general case, as found by Cutler and Cross.

S&B, p. 83, describes holey burrs.

W. H. [Bill] Cutler.  Christmas letter, 1987.  Sketches results of his (and other's) search for holey burrs with notchable pieces.

Bill Cutler.  Holey 6‑Piece Burr!  Published by the author, Palatine, Illinois.  (1986);  with addendum, 1988, 48pp.  He is now permitting internal holes.  Describes holey burrs with notchable pieces, particularly those with multiple moves to release the first piece. 

Bill Cutler.  A Computer Analysis of All 6-Piece Burrs.  Published by the author, ibid., 1994.  86pp.  Sketches complete history of the project.  (I have included a few details in the description of his 1977/78 article, above.)  In 1987, he computed all the notchable holey solutions, using about 2 months of PC AT time, finding  13,354,991  assemblies giving  7.4 million solutions.  Two of these were level 10 _ i.e. they require 10 moves to remove the first piece (or pieces), but the highest level occurring for a unique solution was 5.  After that he started on the general holey burrs and estimated it would take 400 years of PC AT time _ running at 8 MHz.  After some development, the actual time used was about 62.5 PC AT years, but a lot of this was done on by Harry L. Nelson during idle time on the Crays at Lawrence Livermore Laboratories, and faster PCs became available, so the whole project only took about 2½ years, being completed in Aug 1990 and finding  35,657,131,235  assemblies.  He hasn't checked if all assemblies come apart fully, but he estimates there are 5.75 billion solutions.  He estimates the project used 45 times the computing power used in the proof of the Four Color Theorem and that the project would only take two weeks on the eight RS6000 workstations he now supervises.  Some 70,000 high-level solutions were specifically saved and can be obtained on disc from him.  The highest level found was 12 and the highest level for a unique solution was 10.  See 6.W.1 for a continuation of this work.

 

          6.W.3.          THREE PIECE BURR WITH IDENTICAL PIECES

 

          See S&B, p. 66.

 

Crambrook.  1843.  P. 5, no. 4:  Puzzling Cross  3 pieces.  This seems likely to be a three piece burr, but perhaps is in 6.W.1 _ ??  It is followed by  "Maltese Cross  6 pieces".

Wilhelm Segerblom.  Trick wood joining.  SA (1 Apr 1899) 196.

 

          6.W.4.          DIAGONAL SIX PIECE BURR  =  TRICK STAR

 

          This version often looks like a stellated rhombic dodecahedron.  It has two basic forms, one with a key piece;  the other with all pieces identical, which assembles as two groups of three.

          See S&B, p. 78.

 

Crambrook.  1843.  P. 5, no. 5:  Maltese Cross   6 [pieces], three sorts.  Not clear if these belong here or in 6.W.2 or 6.W.5 _ ??

Slocum.  Compendium.  Shows Star Puzzle from The Youth's Companion, 1875.  The picture does not show which form it is.

S. P. Chandler.  US Patent 393,816 _ Puzzle.  Patented 23 Apr 1888.  1p + 1p diagrams, but the text page is missing from my copy _ get??.  Coffin says this is the earliest version, but it is more complex than usual, with  12  pieces, and has a key piece.

John S. Pinnell.  US Patent 774,197 _ Puzzle.  Applied 9 Oct 1902;  patented 8 Nov 1904.  2pp + 2pp diagrams.  Coffin notes that this extends the idea to  102  pieces!

William E. Hoy.  US Patent 766,444 _ Puzzle‑Ball.  Applied 16 Oct 1902;  patented 2 Aug 1904.  2pp + 2pp diagrams.  Spherical version with a key piece.

George R. Ford.  US Patent 779,121 _ Puzzle.  Applied 16 May 1904;  patented 3 Jan 1905.  1p + 1p diagrams.  With square rods, all identical.  He shows assembly by inserting a last piece rather than joining two groups of three.

E. M. Wyatt.  Woodwork puzzles.  Industrial Arts Magazine 12 (1923) 326‑327.  Version with a key piece and square rods.

Collins.  Book of Puzzles.  1927.  The bonbon or nut puzzle, pp. 137-139. 

Iffland Frères (Lausanne).  Swiss patent 245,402 _ Zusammensetzspiel.  Received 19 Nov 1945;  granted 15 Nov 1946;  published 1 Jul 1947.  2pp + 1p diagrams.  Stellated rhombic dodecahedral version with a key piece.  (Coffin says this is the first to use this shape, although Slocum has a version c1875.)

 

          6.W.5.          SIX PIECE BURR WITH IDENTICAL PIECES

 

          One form has six identical pieces and all move outward or inward together.  Another form with flat notched pieces has one piece with an extra notch or an extended notch which allows it to fit in last, either by sliding or twisting, but this is not initially obvious.  This form is sometimes made with equal pieces so that it can only be assembled by force, perhaps after steaming, and it then makes an unopenable money box.  This might be considered under 11.M.

 

Edward Hordern has a version with one piece a little smaller than the rest from c1800.

Crambrook.  1843.  P. 5, no. 5:  Maltese Cross   6 [pieces], three sorts.  Not clear if these belong here or in 6.W.2 or 6.W.4 _ ??

C. Baudenbecher catalogue, c1850s.  Op. cit. in 6.W.7.  This has an example of the six equal flat pieces making an unopenable(?) money box.

F. Chasemore.  Some mechanical puzzles.  In:  Hutchison; op. cit. in 5.A; 1891, chap. 70, part 1, pp. 571‑572.  Item 5: The puzzle box, p. 572.  Six U pieces make a uniformly expanding cubical box.

Hoffmann.  1893.  Chap. III, no.33: The bonbon nut puzzle, pp. 104 & 138.  One piece has an extra notch to simplify the assembly.  Photo in Hordern, p. 66.

Burnett Fallow.  How to make a puzzle money-box.  The Boy's Own Paper 15 (No. 755) (1 Jul 1893) 638.  Equal flat notched pieces forced together to make an unopenable box.

Burnett Fallow.  How to make a puzzle picture-frame.  The Boy's Own Paper 16 (No. 815) (25 Aug 1894) 749.  Each corner has the same basic forced construction as used in the puzzle money-box.

Benson.  1904.  The bonbon nut puzzle, p. 204.

Bartl.  c1920.  Several versions on p. 306.

Western Puzzle Works, 1926 Catalogue.  Last page shows 20 Chinese Wood Block Puzzles, High Grade.  Some of these are of the present type.

Collins.  Book of Puzzles.  1927.  The bonbon or nut puzzle, pp. 137-139.  As in Hoffmann.

Iona & Robert Opie and Brian Alderson.  Treasures of Childhood.  Pavilion (Michael Joseph), London, 1989.  P. 158 shows a "cluster puzzle which Professor Hoffman [sic] names the 'Nut (or Six‑piece) Puzzle', but which is usually called 'The Maltese Puzzle'." 

 

          6.W.6.          ALTEKRUSE PUZZLE

 

William Altekruse.  US Patent 430,502 _ Block-Puzzle.  Applied 3 Apr 1890;  patented 17 Jun 1890.  1p + 1p diagrams.  Described in S&B, p. 72.  The standard version has 12 pieces, but variations discovered by Coffin have  14, 36 & 38  pieces.

Western Puzzle Works, 1926 Catalogue.  No. 112: 12 piece Wood Block.  Possibly Altekruse.

 

          6.W.7.  OTHER BURRS

 

          See also 6.BJ for other 3D dissections.  I have avoided repeating items, so 6.BJ should also be consulted if you are reading this section.

 

Catel.  Kunst-Cabinet.  1790.  Die grosse Teufelsklaue, pp. 9-10 & fig. 20 on plate I.  24 piece 'squirrel cage'.  Cost 16 groschen.

Bestelmeier.  1801.  Item 142: Die grosse Teufelsklaue.  The 'squirrelcage', identical to Catel, with same drawing, but reversed.  Text may be copying some of Catel.

C. Baudenbecher, toy manufacturer in Nuremberg.  Sample book or catalogue from c1850s.  Baudenbecher was taken over by J. W. Spear & Sons in 1919 and the catalogue is now in the Spear's Game Archive, Ware, Hertfordshire.  It comprises folio and double folio sheets with finely painted illustrations of the firm's products.  One whole folio page shows about 20 types of wooden interlocking puzzles, including most of the types mentioned elsewhere in this section and in 6.W.5 and 6.BJ.  Until I get a picture, I can't be more specific.

Slocum.  Compendium.  Shows a 'woodchuck' type puzzle, called White Wood Block Puzzle, from The Youth's Companion, 1875.  I can't see how many pieces it has:  12 or 18??

Slocum.  Compendium.  Shows: "Mystery", Magic "Champion Puzzle" and "Puzzle of Puzzles" from Bland's Catalogue, c1890.

The first looks like a 6 piece burr with circular segments added to make it look like a ball.  So it may be a 6 piece burr in disguise.  See also Hoffmann, pp. 107‑108 & 141‑142  = Benson, p. 205.

The second is a six piece puzzle, but the pieces are flattish and it may be of the type described in 6.W.5. 

The third is complex, with perhaps 18 pieces.

Bartl.  c1920.  Several versions on pp. 306-307, including some that are in 6.W.5 and some 'Chinese block puzzles'.

Western Puzzle Works, 1926 Catalogue.  Shows a number of burrs and similar puzzles.

No. 86: 6 piece Wood Block.

No. 112: 12 piece Wood Block.  Possibly Altekruse.

No. 212: 11 piece Wood Block

The last page shows 20 Chinese Wood Block Puzzles, High Grade.  Some of these are burrs.

Collins.  Book of Puzzles.  1927.  Other cluster puzzles, pp. 139-142.  Describes and illustrates:  The cluster;  The cluster of clusters;  The gun cluster;  The point cluster;  The flat cluster;  The cluster (or secret) table;  The barrel;  The Ball;  The football.  All of these have a key piece.

Jan van de Craats.  Das unmögliche Escher-puzzle.  (Taken from:  De onmogelijke Escher-puzzle; Pythagoras (Amsterdam) (1988).)  Alpha 6 (or:  Mathematik Lehren / Heft 55 _??) (1992) 12-13.  Two Penrose tribars made into an impossible 5-piece burr.

 

          6.X.    ROTATING RINGS OF POLYHEDRA

 

          Generally, these have edge to edge joints.  'Jacob's ladder' joints are used by Engel _ see 11.L for other forms of this joint.

 

I am told these may appear in Fedorov (??NYS).

Max Brückner.  Vielecke und Vielfläche.  Teubner, Leipzig, 1900.  Section 162, pp. 215‑216 and Tafel VIII, fig. 4.  Describes rings of  2n  tetrahedra joined edge to edge, called stephanoids of the second order.  The figure shows the case  n = 5.

Paul Schatz.  UK Patent 406,680 _ Improvements in or relating to Boxes or Containers.  Convention Date (Germany): 10 Dec 1931;  application Date (in UK): 19 Jul 1932;  accepted: 19 Feb 1934.  6pp + 6pp diagrams.  Six and four piece rings of prisms which fold into a box.

Paul Schatz.  UK Patent 411,125 _ Improvements in Linkwork comprising Jointed Rods or the like.  Convention Date (Germany): 31 Aug 1931;  application Date (in UK): 31 Aug 1932;  accepted: 31 May 1934.  3p + 6pp diagrams.  Rotating rings of six tetrahedra and linkwork versions of the same idea, similar to Flowerday's Hexyflex.

Sidney Melmore.  A single‑sided doubly collapsible tessellation.  MG 31 (No. 294) (1947) 106.  Forms a Möbius strip of three triangles and three rhombi, which is basically a flexagon (cf 6.D).  He sees it has two distinct forms, but doesn't see the flexing property!!  He describes how to extend these hexagons into a tessellation which has some resemblance to other items in this section.

Wallace G. Walker invented his "IsoAxis" ® in 1958 while a student at Cranbrook Academy of Art, Michigan.  This is approximately a ring of ten tetrahedra.  He obtained a US Patent for it in 1967 _ see below.  In 1973(?) he sent an example to Doris Schattschneider who soon realised that the basic idea was a ring of tetrahedra and that Escher tessellations could be adapted to it.  They developed the idea into "M. C. Escher Kaleidocycles", published by Ballantine in 1977 and reprinted several times since.

Douglas Engel.  Flexahedrons.  RMM 11 (Oct 1962) 3‑5.  These have 'Jacob's ladder' hinges, not edge‑to‑edge hinges.  He says he invented these in Fall, 1961.  He formed rings of  4, 6, 7, 8  tetrahedra and used a diagonal joining to make rings of 4 and 6 cubes.

Wallace G. Walker.  US Patent 3,302,321 _ Foldable Structure.  Filed 16 Aug 1963;  issued 7 Feb 1967.  2pp. + 6pp. diagrams.

Joseph S. Madachy.  Op. cit. in 6.D, 1966.  Solid Flexagons, pp. 81‑84.  Based on Engel, but only gives the ring of 6 tetrahedra.

D. Engel.  Flexing rings of regular tetrahedra.  Pentagon 26 (Spring 1967) 106‑108.  ??NYS _ cited in Schaaf II 89 _ write Engel.

Paul Bethell.  More Mathematical Puzzles.  Encyclopædia Britannica International, London, 1967.  The magic ring, pp. 12-13.  Gives diagram for a ten-tetrahedra ring, all tetrahedra being regular.

J. Slothouber & W. Graatsma.  Cubics.  1970.  Op. cit. in 6.G.1.  ??NYS.  Presents versions of the flexing cubes and the 'Shinsei Mystery'.

J. Slothouber.  Flexicubes _ reversible cubic shapes.  JRM 6 (1973) 39‑46.  As above.

Frederick George Flowerday.  US Patent 3,916,559 _ Vortex Linkages.  Filed: 12 Aug 1974 (23 Aug 1973 in UK);  issued: 4 Nov 1975.  Abstract + 2pp + 3pp diagrams.  Mostly shows his Hexyflex, essentially a six piece ring of tetrahedra, but with just four edges of each tetrahedron present.  He also shows his Octyflex which has eight pieces.  Text refers to any even number  ³ 6.

Naoki Yoshimoto.  Two stars in a cube (= Shinsei Mystery).  Described in Japanese in:  Itsuo Sakane; A Museum of Fun; Asahi Shimbun, Tokyo, 1977, pp. 208‑210.  Shown and pictured as Exhibit V‑1 with date 1972 in:  The Expanding Visual World _ A Museum of Fun; Exhibition Catalogue, Asahi Shimbun, Tokyo, 1979, pp. 102 & 170‑171.  (In Japanese).  ??get translated??

Lorraine Mottershead.  Investigations in Mathematics.  Blackwell, Oxford, 1985.  Pp. 63-66.  Describes Walkers IsoAxis and rotating rings of six and eight tetrahedra.

 

          6.Y.    ROPE ROUND THE EARTH

 

          The first few examples illustrate what must be the origin of the idea in more straghtforward situations.

 

Lucca 1754.  c1390.  F. 8r, pp. 31‑32.  This mentions the fact that a circumference increases by  44/7  times the increase in the radius.

Muscarello.  1478. 

Ff. 932-93v, p. 220.  A circular garden has outer circumference  150  and the wall is    thick.  What is the inner circumference?  Takes  π  as  22/7.

F. 95r, p. 222.  The internal circumference of a tower is  20  and its wall is  3  thick.  What is the outer circumference?  Again takes  π  as  22/7.

Pacioli.  Summa.  1494.  Part 2, f. 55r, prob. 33.  Florence is 5 miles around the inside.  The wall is    braccia wide and the ditch is  14  braccia wide _ how far is it around the outside?  Several other similar problems.

William Whiston.  Edition of Euclid, 1702.  Book 3, Prop. 37, Schol. (3.).  ??NYS _ cited by "A Lover" and Jackson, below.

"A Lover of the Mathematics."  A Mathematical Miscellany in Four Parts.  2nd ed., S. Fuller, Dublin, 1735.  The First Part is:  An Essay towards the Probable Solution of the Forty five Surprising PARADOXES, in GORDON's Geography, so the following must have appeared in Gordon.  Part I, no. 73, p. 56.  "'Tis certainly Matter of Fact, that three certain Travellers went a Journey, in which, Tho' their Heads travelled full twelve Yards more than their Feet, yet they all return'd alive, with their Heads on."

Carlile.  Collection.  1793.  Prob. XXV, p. 17.  Two men travel, one upright, the other standing on his head.  Who "sails farthest"?  Basically he compares the distance travelled by the head and the feet of the first man.  He notes that this argument also applies to a horse working a mill by walking in a circle; the outside of the horse travels about six times the thickness of the horse further than the inside on each turn.

Jackson.  Rational Amusement.  1821.  Geographical Paradoxes, no. 54, pp. 46 & 115-116.  "It is a matter of fact, that three certain travellers went on a journey, in which their heads travelled full twelve yards more than their feet; and yet, they all returned alive with their heads on."  Solution says this is discussed in Whiston's Euclid, Book 3, Prop. 37, Schol. (3.).  [This first appeared in 1702.]

K. S. Viwanatha Sastri.  Reminiscences of my esteemed tutor.  In:  P. K. Srinivasan, ed.; Ramanujan Memorial Volumes:  1: Ramanujan _ Letters and Reminiscences;  2: Ramanujan _ An Inspiration;  Muthialpet High School, Number Friends Society, Old Boys' Committee, Madras, 1968.  Vol. 1, pp. 89-93.  On p. 93, he relates that this was a favourite problem of his tutor, Srinivasan Ramanujan.  Though not clearly dated, this seems likely to be c1908-1910, but may have been up to 1914.  "Suppose we prepare a belt round the equator of the earth, the belt being    feet longer, and if we put the belt round the earth, how high will it stand?  The belt will stand  1  foot high, a substantial height."

Dudeney.  The paradox party.  Strand Mag. 38 (No. 228) (Dec 1909) 673‑674 (= AM, p. 139).

Ludwig Wittgenstein was fascinated by the problem and used to pose it to students.  Most students felt that adding a yard to the rope would raise it from the earth by a negligible amount _ which it is, in relation to the size of the earth, but not in relation to the yard.  See:  John Lenihan;  Science in Focus;  Blackie, 1975, p. 39.

Ernest K. Chapin.  Loc. cit. in 5.D.1.  1927.  Prob. 5, p. 87 & Answers p. 7.  A yard is added to a band around the earth.  Can you raise it 5 inches?  Answer notes the size of the earth is immaterial.

Collins.  Book of Puzzles.  1927.  The globetrotter's puzzle, pp. 68‑69.  If you walk around the equator, how much farther does your head go?

Abraham.  1933.  Prob. 33 _ A ring round the earth, pp. 12 & 24 (9 & 112).

Perelman.  FMP.  c1935??  Along the equator, pp. 342 & 349.  Same as Collins.

Sullivan.  Unusual.  1943.

Prob. 20: A global readjustment.  Take a wire around the earth and insert an extra 40 ft into it _ how high up will it be?

Prob. 23: Getting ahead.  If you walk around the earth, how much further does your head go than your feet?

W. A. Bagley.  Puzzle Pie.  Op. cit. in 5.D.5.  1944.  Things are seldom what they seem _ No. 42a, 43, 44, pp. 50-51.  42a and 43 ask how much the radius increases for a yard gain of circumference.  No. 44 asks if we add a yard to a rope around the earth and then tauten it by pulling outward at one point, how far will that point be above the earth's surface?

 

          6.Z.    LANGLEY'S ADVENTITIOUS ANGLES

 

          Let ABC be an isosceles triangle with  Ð B = Ð C = 80o.  Draw  BD  and  CE,  making angles  50o  and  60o  with the base.  Then  Ð CED = 20o.

 

JRM 15 (1982‑83) 150 cites Math. Quest. Educ. Times 17 (1910) 75.  ??NYS

Peterhouse and Sidney Entrance Scholarship Examination.  Jan 1916.  ??NYS.

E. M. Langley.  Note 644:  A Problem.  MG 11 (No. 160) (Oct 1922) 173.

Thirteen solvers, including Langley.  Solutions to Note 644.  MG 11 (No. 164) (May 1923) 321‑323.

Gerrit Bol.  Beantwoording van prijsvraag No. 17.  Nieuw Archief voor Wiskunde (2) 18 (1936) 14‑66.  ??NYS.  Coxeter (CM 3 (1977) 40) and Rigby (below) describe this.  The prize question was to completely determine the concurrent diagonals of regular polygons.  The  18‑gon is the key to Langley's problem.  However Bol's work was not geometrical.

Birtwistle.  Math. Puzzles & Perplexities.  1971.  Find the angle, pp. 86-87.  Short solution using law of sines and other simple trigonometric relations.

Colin Tripp.  Adventitious angles.  MG 59 (No. 408) (Jun 1975) 98‑106.  Studies when РCED  can be determined and all angles are an integral number of degrees.  Computer search indicates that there are at most  53  cases.

CM 3 (1977) 12  gives 1939 & 1950 reappearances of the problem and a 1974 variation.

D. A. Q. [Douglas A. Quadling].  The adventitious angles problem: a progress report.  MG 61 (No. 415) (Mar 1977) 55-58.  Reports on a number of contributions resolving the cases which Tripp could not prove.  All the work is complicated trigonometry _ no further cases have been demonstrated geometrically.

CM 4 (1978) 52‑53 gives more references.

D. A. Q. [Douglas A. Quadling].  Last words on adventitious angles.  MG 62 (No. 421) (Oct 1978) 174-183.  Reviews the history, reports on geometric proofs for all cases and various gneralizations.

J[ohn]. F. Rigby.  Adventitious quadrangles: a geometrical approach.  MG 62 (No. 421) (Oct 1978) 183-191.  Gives geometrical proofs for almost all cases.  Cites Bol and a long paper of his own to appear in Geom. Dedicata (??NYS).  He drops the condition that  ABC  be isosceles.  His adventitious quadrangles correspond to Bol's triple intersections of diagonals of a regular  n-gon.

MS 27:3 (1994/5) 65  has two straightforward letters on the problem, which was mentioned in ibid. 27:1 (1994/5) 7.  One letter cites 1938 and 1955 appearances.  P. 66 gives another solution of the problem.  See next item.

Douglas Quadling.  Letter: Langley's adventitious angles.  MS 27:3 (1994/5) 65‑66.  He was editor of MG when Tripp's article appeared.  He gives some history of the problem and some life of Langley (d. 1933).  Edward Langley was a teacher at Bedford Modern School and the founding editor of the MG in 1894-1895.  E. T. Bell was a student of Langley's and contributed an obituary in the MG (Oct 1933) saying that Langley was the finest expositor he ever heard _ ??NYS.  Langley also had botanical interests and a blackberry variety is named for him.

 

          6.AA. NETS OF POLYHEDRA

 

Albrecht Dürer.  Underweysung der messung mit dem zirckel u_ richtscheyt, in Linien ebnen unnd gantzen corporen.  Nürnberg, 1525, revised 1538.  German facsimile/English translation:  The  Painter's Manual; trans. by Walter L. Strauss; Abaris Books, NY, 1977.  Figures 29‑41 (pp. 316-347, Dürer's 1525 ff. M-iii-v - N-v-r) show a net of each of the 5 regular polyhedra and seven Archimedean ones.  (See 6.AT.3 for details.)  (Panofsky's biography of Dürer asserts that Dürer invented the concept of a net _ this is excerpted in The World of Mathematics I 618‑619.)  In the revised version of 1538, the icosidodecahedron and great rhombi-cubo-octahedron are added (figures 43 & 43a, pp. 414-419).

Albrecht Dürer.  Elementorum Geometricorum (?) _ the copy of this that I saw at the Turner Collection, Keele, has the title page missing, but Elementorum Geometricorum is the heading of the first text page and appears to be the book's title.  This appears to be a Latin translation of Unterweysung der Messung ....  Christianus Wechelus, Paris, 1532.  Liber quartus, fig. 29-43, pp. 145-158 shows the same material as in the 1525 edition. 

Cardan.  De Rerum Varietate.  1557, ??NYS  = Opera Omnia, vol. III, pp. 246-247.  Liber XIII.  Corpora, qua regularia diei solent, quomodo in plano formentur.  Shows nets of the regular solids, except the two halves of the dodecahedron have been separated to fit into one column of the text.

E. Welper.  Elementa geometrica, in usum geometriae studiosorum ex variis Authoribus collecta.  J. Reppius, Strassburg, 1620.  ??NYS _ cited, with an illustration of the nets of the octahedron, icosahedron and dodecahedron, in Lange & Springer Katalog 163 _ Mathematik & Informatik, Oct 1994, item 1350 & illustration on back cover, but the entry gives Trassburg.

Athanasius Kircher.  Ars Magna, Lucis et Umbrae.  Rome, 1646.  ??NX.  Has net of a rhombicubocatahedron.

Pike.  Arithmetic.  1788.  Pp. 458-459.  "As the figures of some of these bodies would give but a confused idea of them, I have omitted them; but the following figures, cut out in pasteboard, and the lines cut half through, will fold up into the several bodies."  Gives the regular polyhedra.

Dudeney.  MP.  1926.  Prob. 146: The cardboard box, pp. 58 & 149 (= 536, prob. 316, pp. 109 & 310).  All  11  nets of a cube.

Perelman.  FMP.  c1935?  To develop a cube, pp. 179 & 182‑183.  Asserts there are  10  nets and draws them, but two "can be turned upside down and this will add two more ...."  One shape is missing.  Of the two marked as reversible, one is symmetric, hence equal to its reverse, but the other isn't.

C. Hope.  The nets of the regular star‑faced and star‑pointed polyhedra.  MG 35 (1951) 8‑11.  Rather technical.

H. Steinhaus.  One Hundred Problems in Elementary Mathematics.  (As:  Sto Zada_, PWN _ Polish Scientific Publishers, Warsaw, 1958.)  Pergamon Press, 1963.  With a Foreword by M. Gardner; Basic Books, NY, 1964.  Problem 34: Diagrams of the cube, pp. 20 & 95‑96.  (Gives all  11  nets.)  Gardner (pp. 5‑6) refers to Dudeney and suggests the four dimensional version of the problem should be easy.

M. Gardner.  SA (Nov 1966) c= Carnival, pp. 41‑54.  Discusses the nets of the cube and the Answers show all  11  of them.  He asks what shapes these  11  hexominoes will form _ they cannot form any rectangles.  He poses the four dimensional problem;  the Addendum says he got several answers, no two agreeing.

Charles J. Cooke.  Nets of the regular polyhedra.  MTg 40 (Aut 1967) 48‑52.  Erroneously finds  13  nets of the octahedron.

Joyce E. Harris.  Nets of the regular polyhedra.  MTg 41 (Winter 1967) 29.  Corrects Cooke's number to  11.

A. Sanders & D. V. Smith.  Nets of the octahedron and the cube. MTg 42 (Spring 1968) 60‑63.  Finds  11  nets for the octahedron and shows a duality with the cube.

Peter Turney.  Unfolding the tesseract.  JRM 17 (1984‑85) 1‑16.  Finds  261  nets of the  4‑cube.  (I don't believe this has ever been confirmed.)

P. Light & D. Singmaster.  The nets of the regular polyhedra.  Presented at New York Acad. Sci. Graph Theory Day X, 213 Nov 1985.  In Notes from New York Graph Theory Day X, 23 Nov 1985; ed. by J. W. Kennedy & L. V. Quintas; New York Acad. Sci., 1986, p. 26.  Based on Light's BSc project in 1984-1984 under my supervision.  Shows there are  43,380  nets for the dodecahedron and icosahedron.  I may organize this into a paper, but several others have since verified the result.

 

          6.AB. SELF‑RISING POLYHEDRA

 

H. Steinhaus.  Mathematical Snapshots.  Stechert, NY, 1938.  (= Kalejdoskop Matematyczny.  Ksi__nica‑Atlas, Lwów and Warsaw, 1938, ??NX.)  Pp. 74-75 describes the dodecahedron and says to see the model in the pocket at the end, but makes no special observation of the self-rising property.  Described in detail with photographs in OUP, NY, eds:  1950: pp. 161-164;  1960: pp. 209‑212;  1969 (1983): pp. 196-198.

Donovan A. Johnson.  Paper Folding for the Mathematics Class.  NCTM, 1957, p. 29: Pop-up dodecahedron.

M. Kac.  Hugo Steinhaus _ a reminiscence and a tribute.  AMM 81 (1974) 572‑581.  Material is on pp. 580‑581, with picture on p. 581.

A pop‑up octahedron was used by Waddington's as an advertising insert in a trade journal at the London Toy Fair about 1981.  Pop-up cubes have also been used.

 

          6.AC. CONWAY'S LIFE

 

          There is now a web page devoted to Life run by Bob Wainwright _ address is:

http://members.aol.com/life1ine/life/lifepage.htm [sic!].

 

M. Gardner.  Solitaire game of "Life".  SA (Oct 1970).  On cellular automata, self‑reproduction, the Garden‑of‑Eden and the game of "Life".  SA (Feb 1971).  c= Wheels, chap. 20-22.  In the Oct 1970 issue, Conway offered a $50 prize for a configuration which became infinitely large _ Bill Gosper found the glider gun a month later.  At the Second Gathering for Gardner, Atlanta, 1996, Bob Wainwright showed a picture of Gosper's telegram to Garnder on 4 Nov 1970 giving the coordinates of the glider gun.  I wasn't clear if Wainwright has this or Gardner still has it.

Robert T. Wainwright, ed. (12 Longue Vue Avenue, New Rochelle, NY, 10804, USA).  Lifeline (a newsletter on Life), 11 issues, Mar 1971 _ Sep 1973.  ??NYR.

John Barry.  The game of Life: is it just a game?  Sunday Times (London) (13 Jun 1971).  ??NYS _ cited by Gardner.

Anon.  The game of Life.  Time (21 Jan 1974).  ??NYS _ cited by Gardner.

Carter Bays.  The Game of Three‑dimensional Life.  Dept. of Computer Science, Univ. of South Carolina, Columbia, South Carolina, 29208, USA, 1986.  48pp.

A. K. Dewdney.  The game Life acquires some successors in three dimensions.  SA 256:2 (Feb 1987) 8‑13.  Describes Bays' work.

Bays has started a quarterly 3‑D Life Newsletter, but I have only seen one (or two?) issues.  ??get??

Alan Parr.  It's Life _ but not as we know it.  MiS 21:3 (May 1992) 12-15.  Life on a hexagonal lattice.

 

          6.AD. ISOPERIMETRIC PROBLEMS

 

          There is quite a bit of classical history which I have not yet entered.  Magician's Own Book notes there is a connection between the Dido version of the probelm and Cutting a card so one can pass through it, Section. 6.BA.

 

Virgil.  Aeneid.  ‑19.  Book 1, lines 360‑370.  (p. 38 of the Penguin edition, translated by W. F. Jackson Knight, 1956.)  Dido came to a spot in Tunisia and the local chiefs promised her as much land as she could enclose in the hide of a bull.  She cut it into a long strip and used it to cut off a peninsula and founded Carthage.  This story was later adapted to other city foundations.  John Timbs; Curiosities of History; With New Lights; David Bogue, London, 1857, devotes a section to Artifice of the thong in founding cities, pp. 49-50, relating that in 1100, Hengist, the first Saxon King of Kent, similarly purchased a site called Castle of the Thong and gives references to Indian, Persian and American versions of the story as well as several other English versions.

Pappus.  c290.  Synagoge [Collection].  Book V, Preface, para. 1‑3, on the sagacity of bees.  Greek and English in SIHGM, Vol. II, pp. 588‑593.  A different, abridged, English version is in HGM II 389‑390.

The 5C Saxon mercenary, Hengist or Hengest, is said to have requested from Vortigern: "as much land as can be encircled by a thong".  He "then took the hide of a bull and cut it into a single leather thong.  With this thong he marked out a certain precipitous site, which he had chosen with the greatest possible cunning."  This is reported by Geoffrey of Monmouth in the 12C and this is quoted by the editor in:  The Exeter Book Riddles;  8-10C (Bryant (op cit in 9.E) gives last quarter of the 10C) _ ??;  Translated and edited by Kevin Crossley-Holland;  (As: The Exeter Riddle Book, Folio Society, 1978, Penguin, 1979);  Revised ed., Penguin, 1993; pp. 101-102.

Lucca 1754.  c1390.  Ff. 8r‑8v, pp. 31‑33.  Several problems, e.g. a city  1 by 24  has perimeter  50  while a city  8 by 8  has perimeter  32  but is  8/3  as large;  stitching two sacks together gives a sack  4  times as big.

Calandri.  Arimethrica.  1491.  F. 97v.  Joining sacks which hold  9  and  16  yields a sack which holds  49!!

Pacioli.  Summa.  1494.  Part 2, ff. 55r-55v.  Several problems, e.g. a cord of length 4 encloses 100 ducats worth, how much does a cord of length 10 enclose?  Also stitching bags together.

Buteo.  Logistica.  1559.  Prob. 86, pp. 298-299.  If 9 pieces of wood are bundled up by    feet of cord, how much cord is needed to bundle up 4 pieces?  5 pieces?

Pitiscus.  Trigonometria.  Revised ed., 1600, p. 223.  ??NYS _ described in: Nobuo Miura; The applications of trigonometry in Pitiscus: a preliminary essay; Historia Scientarum 30 (1986) 63-78.  A square of side  4  and triangle of sides  5, 5, 3  have the same perimeter but different areas.  Presumably he was warning people not to be cheated in this way.

J. Kepler.  The Six‑Cornered Snowflake, op. cit. in 6.AT.3.  1611.  Pp. 6‑11 (8‑19).  Discusses hexagons and rhombic interfaces, but only says "the hexagon is the roomiest" (p. 11 (18‑19)).

van Etten.  1624.  Prob. 90 (87).  Pp. 136‑138 (214‑218).  Compares fields  6 x 6  and  9 x 3.  Compares 4 sacks of diameter 1 with 1 sack of diameter 4.  Compares 2 water pipes of diameter 1 with 1 water pipe of diameter 2.

Ozanam.  1725.

Question 1, 1725: 327.  Question 3, 1778: 328;  1803: 325;  1814: 276;  1840: 141.  String twice as long contains four times as much asparagus.

Question 2, 1725: 328.  If a cord of length 10 encloses 200, how much does a cord of length 8 enclose?

Question 3, 1725: 328.  Sack 5 high by 4 across versus 4 sacks 5 high by 1 across.  c=  Q. 2, 1778: 328;  1803: 324;  1814: 276;  1840: 140-141, which has sack 4 high by 6 around versus two sacks 4 high by 3 around.

Question 4, 1725: 328‑329.  How much water does a pipe of twice the diameter deliver?

Les Amusemens.  1749. 

Prob. 211, p. 376.  String twice as long contains four times as much asparagus.

Prob. 212, p. 377.  Determine length of string which contains twice as much asparagus.

Prob. 223-226, pp. 386-389.  Various problems involving changing shape with the same perimeter.  Notes the area can be infinitely small.

Ozanam‑Montucla.  1778.

Question 1, 1778: 327;  1803: 323-324;  1814: 275-276;  1840: 140.  Square versus oblong field of the same circumference.

Prob. 35, 1778: 329-333;  1803: 326-330;  1814: 277-280;  1840: 141-143.  Les alvéoles des abeilles (On the form in which bees construct their combs).

Jackson.  Rational Amusement.  1821.  Geometrical Puzzles.

No. 30, pp. 30 & 90.  Square field versus oblong (rectangular?) field of the same perimeter.

No. 31, pp. 30 & 90-91.  String twice as long contains four times as much asparagus.

Magician's Own Book (UK version).  1871.  To cut a card for one to jump through, p. 124, says:  "The adventurer of old, who, inducing the aborigines to give him as much land as a bull's hide would cover, and made it into one strip by which acres were enclosed, had probably played at this game in his youth."  See 6.BA.

 

          6.AD.1.        LARGEST PARCEL ONE CAN POST

 

          New section.  Are there older examples?

 

Richard A. Proctor.  Greatest content with parcels' post.  Knowledge 3 (3 Aug 1883) 76.  Height + girth  £  6 ft.  States that a cylinder is well known to be the best solution.  Either for a cylinder or a box, the optimum has  height = 2,  girth = 4,  with optimum volumes  2  and  8/π = 2.54... ft3.

R. F. Davis.  Letter:  Girth and the parcel post.  Knowledge 3 (17 Aug 1883) 109-110, item 897.  Independent discussion of the problem, noting that  length  £  3½ ft  is specified, though this doesn't affect the maximum volume problem.

H. F.  Letter:  Parcel post problem.  Knowledge 3 (24 Aug 1883) 126, item 905.  Suppose 'length' means "the maximum distance in a straight line between any two points on its surface".  By this he means the diameter of the solid.  Then the optimum shape is the intersection of a right circular cylinder with a sphere, the axis of the cylinder passing through the centre of the sphere, and this has the 'length' being the diameter of the sphere and the maximum volume is then  2_ ft3

Algernon Bray.  Letter:  Greatest content of a parcel which can be sent by post.  Knowledge 3 (7 Sep 1883) 159, item 923.  Says the problem is easily solved without calculus.  However, for the box, he says "it is plain that the bulk of half the parcel will be greatest when [its] dimensions are equal".

Pearson.  1907.  Part II, no. 20: Parcel post limitations, pp. 118 & 195.  Length  £  3½ ft;  length + girth  £  6 ft.  Solution is a cylinder.

Adams.  Puzzle Book.  1939.  Prob. B.86: Packing a parcel, pp. 79 & 107.  Same as Pearson, but first asks for the largest box, then the largest parcel.

T. J. Fletcher.  Doing without calculus.  MG 55 (No. 391) (Feb 1971) 4‑17.  Example 5, pp. 8‑9.  He says only that  length + girth  £  6 ft.  However, the optimal box has length 2, so the maximal length restriction is not critical.

I have looked at the current parcel post regulations and they say  length £ 1.5m  and  length + girth £ 3m,  for which the largest box is  1 x ½ x ½,  with volume  1/4.  The largest cylinder has length  1  and radius  1/π  with volume  1/π.

I have also considered the simple question of a person posting a fishing rod longer than the maximal length by putting it diagonally in a box.  The longest rod occurs at a boundary maximum, at  3/2 x 3/4 x 0  or  3/2 x 0 x 3/4,  so one can post a rod of length  3Ö5/4  =  1.677..., which is about  12%  longer than  1.5m.  In this problem, the use of a cylinder actually does worse!

 

          6.AE.  6"  HOLE THROUGH SPHERE LEAVES CONSTANT VOLUME

 

Hamnet Holditch.  Geometrical theorem.  Quarterly J. of Pure and Applied Math. 2 (1858) ??NYS, described by Broman.  If a chord of a closed curve, of constant length  a+b,  be divided into two parts of lengths  a, b  respectively, the difference between the areas of the closed curve, and of the locus of the dividing point as the chord moves around the curve, will be  πab.  When the closed curve is a circle and  a = b,  then this is the two dimensional version given by Jones, below.  A letter from Broman says he has found Holditch's theorem cited in 1888, 1906, 1975 and 1976.

Samuel I. Jones.  Mathematical Nuts.  1932.  P. 86.  ??NYS.  Cited by Gardner, (SA, Nov 1957) = 1st Book, chap. 12, prob. 7.  Gardner says Jones, p. 93, also gives the two dimensional version:  If the longest line that can be drawn in an annulus is  6"  long, what is the area of the annulus?

L. Lines.  Solid Geometry.  Macmillan, London, 1935;  Dover, 1965.  P. 101, Example 8W3:  "A napkin ring is in the form of a sphere pierced by a cylindrical hole.  Prove that its volume is the same as that of a sphere with diameter equal to the length of the hole."  Solution is given, but there is no indication that it is new or recent.

L. A. Graham.  Ingenious Mathematical Problems and Methods.  Dover, 1959.  Prob. 34: Hole in a sphere, pp. 23 & 145‑147.  [The material in this book appeared in Graham's company magazine from about 1940, but no dates are provided in the book.  (??can date be found out.)]

M. H. Greenblatt.  Mathematical Entertainments, op. cit. in 6.U.2, 1965.  Volume of a modified bowling ball, pp. 104‑105.

C. W. Trigg.  Op. cit. in 5.Q.  1967.  Quickie 217: Hole in sphere, pp. 59 & 178‑179.  Gives an argument based on surface tension to see that the ring surface remains spherical as the hole changes radius.  Problem has a 10" hole.

Andrew Jarvis.  Note 3235:  A boring problem.  MG 53 (No. 385) (Oct 1969) 298‑299.  He calls it "a standard problem" and says it is usually solved with a triple integral (??!!).  He gives the standard proof using Cavalieri's principle.

Birtwistle.  Math. Puzzles & Perplexities.  1971.

Tangential chord, pp. 71-73.  10"  chord in an annulus.  What is the area of the annulus?  Does traditionally and then by letting inner radius be zero.

The hole in the sphere, pp. 87-88 & 177-178.  Bore a hole through a sphere so the remaining piece has half the volume of the sphere.  The radius of the hole is approx.  .61  of the radius of the sphere.

Another hole, pp. 89, 178 & 192.  6"  hole cut out of sphere.  What is the volume of the remainder?  Refers to the tangential chord problem.

Arne Broman.  Holditch's theorem: An introductory problem.  Lecture at ICM, Helsinki, Aug 1978.  Broman then sent out copies of his lecture notes and a supplementary letter on 30 Aug 1978.  He discusses Holditch's proof (see above) and more careful modern versions of it.  His letter gives some other citations.

 

          6.AF.  WHAT COLOUR WAS THE BEAR?

 

          A hunter goes 100 mi south, 100 mi east and 100 mi north and finds himself where he started.  He then shoots a bear _ what colour was the bear?

          Square versions:  Perelman;  Klamkin, Breault & Schwarz;  Kakinuma, Barwell & Collins;  Singmaster.

          I include other polar problems here.  See also 10.K for related geographical problems.

 

"A Lover of the Mathematics."  A Mathematical Miscellany in Four Parts.  2nd ed., S. Fuller, Dublin, 1735.  The First Part is:  An Essay towards the Probable Solution of the Forty five Surprising PARADOXES, in GORDON's Geography, so the following must have appeared in Gordon.  Part I, no. 10, p. 9.  "There is a particular Place of the Earth where the Winds (tho' frequently veering round the Compas) do always blow from the North Point."

Philip Breslaw (attrib.).  Breslaw's Last Legacy; or the Magical Companion: containing all that is Curious, Pleasing, Entertaining and Comical; selected From the most celebrated Masters of Deception: As well with Slight of Hand, As with Mathematical Inventions.  Wherein is displayed The Mode and Manner of deceiving the Eye; as practised by those celebrated Masters of Mirthful Deceptions.  Including the various Exhibitions of those wonderful Artists, Breslaw, Sieur, Comus, Jonas, &c.  Also the Interpretation of Dreams, Signifcation of Moles, Palmestry, &c.  The whole forming A Book of real Knowledge in the Art of Conjuration.  (T. Moore, London, 1784, 120pp.)  With an accurate Description of the Method how to make The Air Balloon, and inject the Inflammable Air.  (2nd ed., T. Moore, London, 1784, 132pp;  5th ed., W. Lane, London, 1791, 132pp.)  A New Edition, with great Additions and Improvements.  (W. Lane, London, 1795, 144pp.)  Facsimile from the copy in the Byron Walker Collection, with added Introduction, etc., Stevens Magic Emporium, Wichita, Kansas, 1997, HB.  [This was first published in 1784, after Breslaw's death, so it is unlikely that he had anything to do with the book.  There were versions in  1784, 1791, 1792, 1793, 1794, 1795, 1800, 1806, c1809, c1810, 1811, 1824.  Hall, BCB 39-43, 46-51.  Toole Stott 120-131, 966‑967.  Heyl 35-41.  This book went through many variations of subtitle and contents _ the above is the largest version.].  I will cite the date as  1784?.

                    Geographical Paradoxes.

Paradox I, p. 35.  Where is it noon every half hour?  Answer: At the North Pole in Summer, when the sun is due south all day long, so it is noon every moment!

Paradox II, p. 36.  Where can the sun and the full moon rise at the same time in the same direction?  Answer: "Under the North Pole, the sun and the full moon, both decreasing in south declination, may rise in the equinoxial points at the same time; and under the North Pole, there is no other point of compass but south."  I think this means at the North Pole at the equinox.

Carlile.  Collection.  1793.  Prob. CXVI, p. 69.  Where does the wind always blow from the north?

Jackson.  Rational Amusement.  1821.  Geographical Paradoxes.

No. 7, pp. 36 & 103.  Where do all winds blow from the north?

No. 8, pp. 36 & 110.  Two places  100  miles apart, and the travelling directions are to go  50  miles north and  50  miles south.

Mr. X.  His Pages.  The Royal Magazine 10:3 (Jul 1903) 246-247.  A safe catch.  Airship starts at the North Pole, goes south for seven days, then west for seven days.  Which way must it go to get back to its starting point?  No solution given.

Pearson.  1907.

Part II, no. 21: By the compass, pp. 18 & 190.  Start at North Pole and go  20  miles southwest.  What direction gets back to the Pole the quickest?  Answer notes that it is hard to go southwest from the Pole!

Part II, no. 15: Ask "Where's the north?" _ Pope, pp. 117 & 194.  Start  1200  miles from the North Pole and go  20  mph due north by the compass.  How long will it take to get to the Pole?  Answer is that you never get there _ you get to the North Magnetic Pole.

Ackermann.  1925.  P. 116.  Man at North Pole goes  20  miles south and  30  miles west.  How far, and in what direction, is he from the Pole?

H. Phillips.  Week‑End.  1932.  Prob. 8, pp. 12 & 188.  = his Playtime Omnibus, 1933, prob. 10: Popoff, pp. 54 & 237.  House with four sides facing south.

H. Phillips.  The Playtime Omnibus.  Faber & Faber, London, 1933.  Section XVI, prob. 11: Polar conundrum, pp. 51 & 234.  Start at the North Pole, go  40  miles South, then  30  miles West.  How far are you from the Pole.  Answer:  "Forty miles.  (NOT thirty, as one is tempted to suggest.)"  Thirty appears to be a slip for fifty??

Perelman.  FFF.  1934.  1957: prob. 6, pp. 14-15 & 19-20: A dirigible's flight;  1979: prob. 7, pp. 18-19 & 25-27: A helicopter's flight.  MCBF: prob 7, pp. 18-19 & 25-26: A helicopter's flight.  Dirigible/helicopter starts at Leningrad and goes  500km  N,  500km  E,  500km  S,  500km  W.  Where does it land?  Cf Klamkin et seq, below.

Phillips.  Brush.  1936.  Prob. A.1: A stroll at the pole, pp. 1 & 73.  Eskimo living at North Pole goes 3 mi south and 4 mi east.  How far is he from home?

J. R. Evans.  The Junior Week‑End Book.  Gollancz, London, 1939.  Prob. 9, pp. 262 & 268.  House with four sides facing south.

Jules Leopold.  At Ease!  Op. cit. in 4.A.2.  1943.  A helluva question!, pp. 10 & 196.  Hunter goes  10  mi south,  10  mi west, shoots a bear and drags it  10  mi back to his starting point.  What colour was the bear?  Says the only geographic answer is the North Pole.

Northrop.  Riddles in Mathematics.  1944.  1944: 5-6;  1945: 5-6;  1961: 15‑16.  He starts with the house which faces south on all sides.  Then he has a hunter that sees a bear 100 yards east.  The hunter runs 100 yards north and shoots south at the bear _ what colour ....  He then gives the three‑sided walk version, but doesn't specify the solution.

E. J. Moulton.  A speed test question; a problem in geography.  AMM 51 (1944) 216 & 220.  Discusses all solutions of the three-sided walk problem.

W. A. Bagley.  Puzzle Pie.  Op. cit. in 5.D.5.  1944.  No. 50: A fine outlook, pp. 54-55.  House facing south on all sides used by an artist painting bears!

Leeming.  1946.  Chap. 3, prob. 32: What color was the bear?, pp. 33 & 160.  Man walks  10  miles south, then  10  miles west, where he shoots a bear.  He drags it  10  miles north to his base.  What color ....  He gives only one solution.

Darwin A. Hindman.  Handbook of Indoor Games & Contests.  (Prentice‑Hall, 1955);  Nicholas Kaye, London, 1957.  Chap. 16, prob. 4: The bear hunter, pp. 256 & 261.  Hunter surprises bear.  Hunter runs  200  yards north, bear runs  200  yards east, hunter fires south at bear.  What colour ...

Murray S. Klamkin, proposer;  D. A. Breault & Benjamin L. Schwarz, solvers.  Problem 369.  MM 32 (1958/59) 220  &  33 (1959/60) 110  &  226‑228.  Explorer goes  100  miles north, then east, then south, then west, and is back at his starting point.  Breault gives only the obvious solution.  Schwartz gives all solutions, but not explicitly.  Cf Perelman, 1934.

Benjamin L. Schwartz.  What color was the bear?.  MM 34 (1960) 1-4.  ??NYS _ described by Gardner, SA (May 1966) = Carnival, chap. 17.  Considers the problem where the hunter looks south and sees a bear  100  yards away.  The bear goes  100  yards east and the hunter shoots it by aiming due south.  This gives two extra types of solution.

Yasuo Kakinuma, proposer;  Brian Barwell and Craig H. Collins, solvers.  Problem 1212 _ Variation of the polar bear problem.  JRM 15:3 (1982‑83) 222  &  16:3 (1983-84) 226‑228.  Square problem going one mile south, east, north, west.  Barwell gets the explicit quadratic equation, but then approximates its solutions.  Collins assumes the earth is flat near the pole.

David Singmaster.  Bear hunting problems.  Submitted to MM, 1986.  Finds explicit solutions for the general version of Perelman/Klamkin's problem.  [In fact, I was ignorant of (or had long forgotten) the above when I remembered and solved the problem.  My thanks to an editor (Paul Bateman ??check) for referring me to Klamkin.  The Kakinuma et al then turned up also.]  Analysis of the solutions leads to some variations, including the following.

David Singmaster.  Home is the hunter.  Man heads north, goes ten miles, has lunch, heads north, goes ten miles and finds himself where he started.

                    Used as:  Explorer's problem by Keith Devlin in his Micromaths Column; The Guardian (18 Jun 1987) 16  &  (2 Jul 1987) 16. 

                    Used by me as one of:  Spring term puzzles; South Bank Polytechnic Computer Services Department Newsletter (Spring 1989) unpaged [p. 15].

                    Used by Will Shortz in his National Public Radio program 6? Jan 1991. 

                    Used as:  A walk on the wild side, Games 15:2 (No. 104) (Aug 1991) 57 & 40.

                    Used as:  The hunting game, Focus 3 (Feb 1993) 77 & 98.

                    Used in my Puzzle Box column, G&P, No. 11 (Feb 1995) 19  &  No. 12 (Mar 1995) 41.

Bob Stanton.  The explorers.  Games Magazine 17:1 (No. 113) (Feb 1993) 61 & 43.  Two explorers set out and go  500  miles in each direction.  Madge goes N, W, S, E, while Ellen goes E, S, W, N.  At the end, they meet at the same point.  However, this is not at their starting point.  How come?  and how far are they from their starting point, and in what direction?  They are not near either pole.

Yuri B. Chernyak & Robert S. Rose.  The Chicken from Minsk.  BasicBooks, NY, 1995.  Chap. 11, prob. 9: What color was that bear? (A lesson in non-Euclidean geometry), pp. 97 & 185-191.  Camper walks south 2 km, then west 5 km, then north 2 km; how far is he from his starting point?  Solution analyses this and related problems, finding that the distance  x  satisfies  0 £ x £ 7.183,  noting that there are many minimal cases near the south pole and if one is between them, one gets a local maximum, so one has to determine one's position very carefully. 

David Singmaster.  Symmetry saves the solution.  IN: Alfred S. Posamentier & Wolfgang Schulz, eds.; The Art of Problem Solving: A Resource for the Mathematics Teacher; Corwin Press, NY, 1996, pp. 273-286.  Sketches the explicit solution to Klamkin's problem as an example of the use of symmetric variables to obtain a solution.

Anonymous.  Brainteaser B163 _ Shady matters.  Quantum 6:3 (Jan/Feb 1996) 15 & 48.  Is there anywhere on earth where one's shadow has the same length all day long?

 

          6.AG. MOVING AROUND A CORNER

 

          There are several versions of this.  The simplest is moving a ladder or board around a corner _ here the problem is two-dimensional and the ladder is thin enough to be considered as a line.  There are slight variations _ the corner can be at a  T  or  +  junction;  the widths of the corridors may differ;  the angle may not be a right angle;  etc.  If the object being moved is thicker _ e.g. a table _ then the problem gets harder.  If one can use the third dimension, it gets even harder.

 

H. E. Licks.  Op. cit. in 5.A, 1917.  Art. 110, p. 89.  Stick going into a circular shaft in the ceiling.  Gets  [h2/3 + d2/3)]3/2  for maximum length, where  h  is the height of the room and  d  is the diameter of the shaft.  "A simple way to solve a problem which has proved a stumbling block to many."

Abraham.  1933.  Prob. 82 _ Another ladder, pp. 37 & 45 (23 & 117).  Ladder to go from one street to another, of different widths.

E. H. Johnson, proposer;  W. B. Carver, solver.  Problem E436.  AMM 47 (1940) 569  &  48 (1941) 271‑273.  Table going through a doorway.  Obtains  6th order equation.

J. S. Madachy.  Turning corners.  RMM 5 (Oct 1961) 37,  6 (Dec 1961) 61  &  8 (Apr 1962) 56.  In 5, he asks for the greatest length of board which can be moved around a corner, assuming both corridors have the same width, that the board is thick and that vertical movement is allowed.  In 6, he gives a numerical answer for his original values and asserts the maximal length for planar movement, with corridors of width  w  and plank of thickness  t,  is  2 (wÖ2 ‑ t).  In vol. 8, he says no two solutions have been the same.

L. Moser, proposer;  M. Goldberg and J. Sebastian, solvers.  Problem 66‑11 _ Moving furniture through a hallway.  SIAM Review 8 (1966) 381‑382  &  11 (1969) 75‑78  &  12 (1970) 582‑586.  "What is the largest area region which can be moved through a "hallway" of width one (see Fig. 1)?"  The figure shows that he wants to move around a rectangular corner joining two hallways of width one.  Sebastian (1970) studies the problem for moving an arc.

J. M. Hammersley.  On the enfeeblement of mathematical skills ....  Bull. Inst. Math. Appl. 4 (1968) 66‑85.  Appendix IV _ Problems, pp. 83‑85, prob. 8, p. 84.  Two corridors of width  1  at a corner.  Show the largest object one can move around it has area  < 2 Ö2 and that there is an object of area  ³  π/2 + 2/π  =  2.2074. 

                    Partial solution by T. A. Westwell, ibid. 5 (1969) 80, with editorial comment thereon on pp. 80‑81.

T. J. Fletcher.  Easy ways of going round the bend.  MG 57 (No. 399) (Feb 1973) 16‑22.  Gives five methods for the ladder problem with corridors of different widths.

Neal R. Wagner.  The sofa problem.  AMM 83 (1976) 188‑189.  "What is the region of largest area which can be moved around a right‑angled corner in a corridor of width one?"  Survey.

R. K. Guy.  Monthly research problems, 1969‑77.  AMM 84 (1977) 807‑815.  P. 811 reports improvements on the sofa problem.

J. S. Madachy & R. R. Rowe.  Problem 242 _ Turning table.  JRM 9 (1976‑77) 219‑221.

G. P. Henderson, proposer;  M. Goldberg, solver;  M. S. Klamkin, commentator.  Problem 427.  CM 5 (1979) 77  &  6 (1979) 31‑32 & 49‑50.  Easily finds maximal area of a rectangle going around a corner.

Research news:  Conway's sofa problem.  Mathematics Review 1:4 (Mar 1991) 5-8 & 32.  Reports on Joseph Gerver's almost complete resolution of the problem in 1990.  Says Conway asked the problem in the 1960s and that Moser is the first to publish it.  Says a grop at a convexity conference in Copenhagen improved Hammersley's results to  2.2164.  Gerver's analysis gives an object made up of 18 segments with area  2.2195.  The analysis depends on some unproven general assumptions which seem reasonable and is certainly the unique optimum solution given those assumptions.

A. A. Huntington.  More on ladders.  M500 145 (Jul 1995) 2-5.  Does usual problem, geting a quartic.  The 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 6.AG is related to 6.L.]

 

          6.AH. TETHERED GOAT

 

          A goat is grazing in a circular field and is tethered to a post on the edge.  He can reach half of the field.  How long is the rope?  There are numerous variations obtained by modifying the shape of the field or having buildings within it.  In recent years, there has been study of the form where the goat is tethered to a point on a circular silo in a large field _ how much area can he graze?

 

Ladies Diary, 1748.  P. 41.  ??NYS 

Dudeney.  Problem 67: Two rural puzzles _ No. 67: One acre and a cow.  Tit‑Bits 33 (5 Feb  &  5 Mar 1898) 355  &  432.  Circular field opening onto a small rectangular paddock with cow tethered to the gate post so that she can graze over one acre.  By skilful choice of sizes, he avoids the usual transcendental equation.

Arc. [R. A. Archibald].  Involutes of a circle and a pasturage problem.  AMM 28 (1921) 328‑329.  Cites Ladies Diary and it appears that it deals with a horse outside a circle.

J. Pedoe.  Note 1477:  An old problem.  MG 24 (No. 261) (Oct 1940) 286-287.  Finds the relevant area by integrating in polar coordinates centred on the post.

A. J. Booth.  Note 1561:  On Note 1477.  MG 25 (No. 267) (Dec 1941) 309‑310.  Goat tethered to a point on the perimeter of a circle which can graze over  ½, _, ¼  of the area.

Howard P. Dinesman.  Superior Mathematical Puzzles.  Op. cit. in 5.B.1.  1968.

No. 8: "Don't fence me in", pp. 87.  Equilateral triangular field of area 120.  Three goats tethered to the corners with ropes of length equal to the altitude.  Consider an area where  n  goats graze as contributing  1/n  to each goat.  What area does each goat graze over?

No. 53: Around the silo, pp. 71 & 112-113.  Goat tethered to the outside of a silo of diameter 20 by a rope of length  10π,  i.e. he can just get to the other side of the silo.  How big an area can he graze?  The curve is a semicircle together with two involutes of a circle, so the solution uses some calculus.

Marshall Fraser.  A tale of two goats.  MM 55 (1982) 221‑227.  Gives examples back to 1894.

Marshall Fraser.  Letter:  More, old goats.  MM 56 (1983) 123.  Cites Arc[hibald].

Bull, 1998, below, says this problem has been discussed by the Internet newsgroup  sci.math  some years previously.

Michael E. Hoffman.  The bull and the silo: An application of curvature.  AMM 105:1 (Jan 1998) ??NYS _ cited by Bull.  Bull is tethered by a rope of length  L  to a circular silo of radius  R.  If  L £ πR,  then the grazeable area is  L3/3R + πL2/2.  This paper considers the problem for general shapes. 

John Bull.  The bull and the silo.  M500 163 (Aug 1998) 1-3.  Improves Hoffman's solution for the circular silo by avoiding polar coordinates and using a more appropriate variable, namely the angle between the taut rope and the axis of symmetry. 

 

          6.AI.  TRICK JOINTS

 

          S&B, pp. 146‑147, show several types.

          These are often made in two contrasting woods and appear to be physically impossible.  They will come apart if one moves them in the right direction.  A few have extra complications.  The simplest version is a square cylinder with dovetail joints on each face _ called common square version below.  There are also cases where one thinks it should come apart, but the wood has been bent or forced and no longer comes apart _ see also 6.W.5.

 

See Bogesen in 6.W.2 for a possible early example.

Johannes Cornelus Wilhelmus Pauwels.  UK Patent 15,307 _ Improved Means of Joining or Fastening Pieces of Wood or other Material together, Applicable also as a Toy.  Applied 9 Nov 1887;  complete specification 9 Aug 1888;  accepted: 26 Oct 1888.  2pp + 1p diagrams.  It says Pauwels is a civil engineer of The Hague.  Common square version.

Tom Tit, vol. 2.  1892.  Assemblage paradoxal, pp. 231-232.  = K, no. 155: The paradoxical coupling, pp. 353‑354.  Common square version with instructions for making it by cutting the corners off a larger square.

Emery Leverett Williams.  The double dovetail and blind mortise.  SA (25 Apr 1896) 267.  The first is a trick T‑joint.

T. Moore.  A puzzle joint and how to make it.  The Woodworker 1:8 (May 1902) 172.  S&B, p. 147, say this is the earliest reference to the common square version.  "... the foregoing joint will doubtless be well-known to our professional readers.  There are probably many amateur woodworkers to whom it will be a novelty."

Dudeney.  The world's best puzzles.  Op. cit. in 2.  1908.  Shows the common square version "given to me some ten years ago, but I cannot say who first invented it."  He previously published it in a newspaper.  ??look in Weekly Dispatch.

Dudeney.  AM.  1917.  Prob. 424: The dovetailed block, pp. 145 & 249.  Shows the common square version _ "... given to me some years ago, but I cannot say who first invented it."  He previously published it in a newspaper.  ??as above

Anon.  Woodwork Joints, 1918, op. cit. in 6.W.1.  A curious dovetail joint, pp. 193, 195.  Common square version.  Dovetail puzzle joint, pp. 194‑195.  A singly mortised T‑joint, with an unmortised second piece.

E. M. Wyatt.  Woodwork puzzles.  Industrial‑Arts Magazine 12 (1923) 326‑327.  Doubly dovetailed tongue and mortise T‑joint called 'The double (?) dovetail'.

Sherman M. Turrill.  A double dovetail joint.  Industrial‑Arts Magazine 13 (1924) 282‑283.  A double dovetail right angle joint, but it leaves sloping gaps on the inside which are filled with blocks.

Collins.  Book of Puzzles.  1927.  Pp. 134‑135: The dovetail puzzle.  Common square version.

E. M. Wyatt.  Puzzles in Wood, 1928, op. cit. in 5.H.1.

The double (?) dovetail, pp. 44‑45.  Doubly dovetailed tongue and mortise T‑joint.

The "impossible" dovetail joint, p. 46.  Common square version.

Double‑lock dovetail joint, pp. 47‑49.  Less acceptable tricks for a corner joint.

Two‑way fanned half‑lap joint, pp. 49‑50.  Corner joint.

A. B. Cutler.  Industrial Arts and Vocational Education (Jan 1930).  ??NYS.  Wyatt, below, cites this for a triple dovetail, but I could not not find it in vols. 1‑40.

R. M. Abraham.  Prob. 225 _ Dovetail Puzzle.  Winter Nights Entertainments.  Constable, London, 1932, p. 131.  (= Easy‑to‑do Entertainments and Diversions with coins, cards, string, paper and matches; Dover, 1961, p. 225.)  Common square version.

Abraham.  1933.  Prob. 304 _ Hexagon dovetail;  Prob. 306 _ The triangular dovetail, pp. 142‑143 (100 & 102).

Bernard E. Jones, ed.  The Practical Woodworker.  Waverley Book Co., London, nd [1940s?].  Vol. 1: Lap and secret dovetail joints, pp. 281‑287.  This covers various secret joints _ i.e. ones with concealed laps or dovetails.  Pp. 286-287 has a subsection: Puzzle dovetail joints.  Common square version is shown as fig. 28.  A pentagonal analogue is shown as fig. 29, but it uses splitting and regluing to produce a result which cannot be taken apart.

E. M. Wyatt.  Wonders in Wood.  Bruce Publishing Co., Milwaukee, 1946.

Double‑double dovetail joint, pp. 26‑27.  Requires some bending.

Triple dovetail puzzle, pp. 28‑29.  Uses curved piece with gravity lock.

S&B, p. 146, reproduces the above Wyatt and shows a 1948 example.

W. A. Bagley.  Puzzle Pie.  Op. cit. in 5.D.5.  1944.  Dovetail deceptions, p. 64.  Common square version and a tapered T joint.

Allan Boardman.  Up and Down Double Dovetail.  Shown on p. 147 of S&B.  Square version with alternate dovetails in opposite directions.  This is impossible!

I have a set of examples which belonged to Tom O'Beirne.  There is a common square version and a similar hexagonal version.  There is an equilateral triangle version which requires a twist.  There is a right triangle version which has to be moved along a space diagonal!  [One can adapt the twisting method to  n-gons!]

Dick Schnacke (Mountain Craft Shop, American Ridge Road, Route 1, New Martinsville, West Virginia, 26155, USA) makes a variant of the common square version which has two dovetails on each face.  I bought one in 1994.

 

          6.AJ.  GEOMETRIC ILLUSIONS

 

Anonymous 15C French illustrator of Giovanni Boccaccio, De Claris Mulieribus, MS Royal 16 Gv in the British Library.  F. 54v: Collecting cocoons and weaving silk.  ??NYS _ reproduced in: The Medieval Woman  An Illuminated Book of Postcards, HarperCollins, 1991.  This shows a loom(?) frame with uprights at each corner and the crosspieces joining the tops of the end uprights as though front and rear are reversed compared to the ground.

L. A. Necker.  LXI. Observations on some remarkable optical phœnomena seen in Switzerland; and on an optical phœnomenon which occurs on viewing a figure of a crystal or geometrical solid.  Phil. Mag. (3) 1:5 (Nov 1832) 329-337.  This is a letter from Necker, written on 24 May 1832.  On pp. 336-337, Necker describes the visual reversing figure known as the Necker cube which he discovered in drawing rhomboid crystals.  This is also quoted in Ernst; The Eye Beguiled, pp. 23-24].  Richard L. Gregory [Mind in Science; Weidenfeld and Nicolson, London, 1981, pp. 385 & 594] and Ernst say that this was the first ambiguous figure to be described.

See Thompson, 1882, in 6.AJ.2, for illusions caused by rotations.

F. C. Müller-Lyer.  Optische Urtheilstusehungen.  Arch. Physiol. Suppl. 2 (1889) 263-270.  ??NYS _ cited by Gregory in The Intelligent Eye.  But cf below.

Wehman.  New Book of 200 Puzzles.  1908.  The cube puzzle, p. 37.  A 'baby blocks' pattern of cubes, which appears to show six cubes piled in a corner one way and seven cubes the other way.  I don't recall seeing this kind of puzzle in earlier sources??

Lietzmann, Walther  &  Trier, Viggo.  Wo steckt der Fehler?  3rd ed., Teubner, 1923.  [The Vorwort says that Trier was coauthor of the 1st ed, 1913, and contributed most of the Schülerfehler (students' mistakes).  He died in 1916 and Lietzmann extended the work in a 2nd ed of 1917 and split it into Trugschlüsse and this 3rd ed.  There was a 4th ed., 1937.  See Lietzmann for a later version combining both parts.]  II. Täuschungen der Anschauung, pp. 7-13.

Lietzmann, Walther.  Wo steckt der Fehler?  3rd ed., Teubner, Stuttgart, (1950), 1953.  (Strens/Guy has 3rd ed., 1963.)  (See: Lietzmann & Trier.  There are 2nd ed, 1952??;  5th ed, 1969;  6th ed, 1972.  Math. Gaz. 54 (1970) 182 says the 5th ed appears to be unchanged from the 3rd ed.)  II. Täuschungen der Anschauung, pp. 15-25.  A considerable extension of the 1923 ed.

Williams.  Home Entertainments.  1914.  Colour discs for the gramophone, pp. 207-212.  Discusses several effects produced by spirals and eccentric circles on discs when rotated.

Gerald H. Fisher.  The Frameworks for Perceptual Localization. Report of MOD Research Project70/GEN/9617, Department of Psychology, University of Newcastle upon Tyne, 1968.  Good collection of examples, with perhaps the best set of impossible figures.

                    Pp. 42‑47 _ reversible perspectives.

                    Pp. 56‑65 _ impossible and ambiguous figures.

                    Appendix 6, p.190 _ 18 reversible figures.

                    Appendix 7, pp. 191‑192 _ 12 reversible silhouettes.

                    Appendix 8, p. 193 _ 12 impossible figures.

                    Appendix 14, pp. 202‑203 _ 72 geometrical illusions.

Harvey Long.  "It's All In How You Look At It".  Harvey Long & Associates, Seattle, 1972.  48pp collection of examples with a few references.

Bruno Ernst [pseud. of J. A. F. Rijk].  (Avonturen met Onmogelijke Figuren; Aramith Uitgevers, Holland, 1985.)  Translated as:  Adventures with Impossible Figures.  Tarquin, Norfolk, 1986.  Describes tribar and many variations of it, impossible staircase, two‑pronged trident.  Pp. 76‑77 reproduces an Annunciation of 14C in the Grote Kerk, Breda, with an impossible perspective.  P. 78 reproduces Print XIV of Giovanni Battista Piranesi's "Carceri de Invenzione", 1745, with an impossible 4‑bar.

Diego Uribe.  Catalogo de impossibilidades.  Cacumen (Madrid) 4 (No. 37) (Feb 1986) 9‑13.  Good summary of impossible figures.  15 references to recent work.

Bruno Ernst.  Escher's impossible figure prints in a new context.  In:  H. S. M. Coxeter, et al., eds.; M. C. Escher _ Art and Science; North‑Holland (Elsevier), Amsterdam, 1986, pp. 124‑134.  Pp. 128‑129 discusses the Breda Annunciation, saying it is 15C and quoting a 1912 comment by an art historian on it.  There is a colour reproduction on p. 394.  P. 130 shows and discusses briefly Bruegel's "The Magpie on the Gallows", 1568.  Pp. 130‑131 discusses and illustrates the Piranesi.

Bruno Ernst.  (Het Begoochelde Oog, 1986?.)  Translated by Karen Williams as:  The Eye Beguiled.  Benedikt Taschen Verlag, Köln, 1992.  Much expanded version of his previous book, with numerous new pictures and models by new artists in the field.  Chapter 6: Origins and history, pp. 68-93, discusses and quotes almost everything known.  P. 68 shows a miniature of the Madonna and Child from the Pericope of Henry II, compiled by 1025, now in the Bayersche Staatsbibliothek, Munich, which is similar in form to the Breda Annunciation (stated to be 15C).  (However, Seckel, below, reproduces it as 2© and says it is c1250.)  P. 69 notes that Escher invented the impossible cube used in his Belvedere.  P. 82 is a colour reproduction of Duchamp's 1916-1917 'Apolinère enameled' which has an impossible bed frame containing an impossible triangle.  Pp. 83-84 shows and discusses Piranesi.  Pp. 84-85 show and discuss Hogarth's 'False Perspective' of 1754.  Reproduction and brief mention of Brueghel (= Bruegel) on p. 85.  Discussion of the Breda Annunciation on pp. 85-86.  Pp. 87-88 show and discuss a 14C Byzantine Annunciation in the National Museum, Ochrid.  Pp. 88-89 show and discuss Scott Kim's impossible four-dimensional tribar. 

J. R. Block & Harold E. Yuker.  Can You Believe Your Eyes?  Brunner/Mazel, NY, 1992.  Excellent survey of the field of illusions, classified into 17 major types _ e.g. ambiguous figures, unstable figures, ..., two eyes are better than one.  They give as much information as they can about the origins.  They give detailed sources for the following _ originals NYS??.  These are also available as two decks of playing cards.

W. E. Hill.  My wife and my mother-in-law.  Puck, (6 Nov 1915) 11.  [However, Julian Rothenstein & Mel Gooding; The Paradox Box; Redstone Press, London, 1993; include a reproduction of a German visiting card of 1888 with a version of this illusion.  The English caption by James Dalgety is:  My Wife and my Mother-in-law.  Cf Seckel, below.]  Ernst, just above, cites Hill and says he was a cartoonist, but gives no source.  Long, above, asserts it was designed by E. G. Boring, an American psychologist.

G. H. Fisher.  Mother, father and daughter.  Amer. J. Psychology 81 (1968) 274-277.

G. Kanisza.  Subjective contours.  SA 234:4 (Apr 1976) 48-52.  (Kanisza triangles.)

Al Seckel.  Illusions in Art.  Two decks of playing cards in case with notes.  Deck 1 _ Classics.  Works from Roman times to the middle of the 20th Century.  Deck 2 _ Contemporary.  Works from the second half of the 20th Century.  Y&B Associates, Hempstead, NY, 1997.  This gives further details on some of the classic illusions _ some of this is entered above and in 6.AU and some is given below.

10¨: Rabbit/Duck.  Devised by Joseph (but notes say Robert) Jastrow, c1900.

10§: My Wife and My Mother-in-Law, anonymous, 1888.

 

Here I make some notes about origins of other illusions, but I have fewer details on these.

 

The Müller-Lyer Illusion _  <->  vs  >---<  was proposed by Zollner in 1859 and described by Johannes Peter Müller (1801-1858) & Lyer in 1889.  But cf above.  Lietzmann & Trier, p. 7, date it as 1887.

The Bisection Illusion _ with a vertical segment bisecting a horizontal segment, but above it _ was described by Albert Oppel (1831-1865) and Wilhelm Wundt (1832-1920) in 1865.

Zollner's Illusion _ parallel lines crossed by short lines at  45o,  alternately in opposite directions _ was noticed by Johann K. F. Zollner (1834-1882) on a piece of fabric with a similar design.

Hering's Illusion _ with parallel lines crossed by numerous lines through a point between the lines _ was invented by Ewals Hering (1834-1918) in 1860.

In a lecture, Al Seckel said the spiral circles illusion is due to James Fraser, c1906.

 

          6.AJ.1          TWO PRONGED TRIDENT

 

Oscar Reutersvård.  Letters quoted in Ernst, 1992, pp. 69-70, says he developed an equivalent type of object, which he calls impossible meanders, in the 1930s.

R. L. Gregory says this is due to a MIT draftsman (= draughtsman) about 1950??

California Technical Industries.  Advertisement.  Aviation Week and Space Technology 80:12 (23 Mar 1964) 5.  Standard form.  (I wrote them but my letter was returned 'insufficient address'.)

Hole location gage.  Analog Science Fact • Science Fiction 73:4 (Jun 1964) 27.  Classic Two pronged trident, with some measurements given.  Editorial note says the item was 'sent anonymously for some reason' and offers the contributor $10 or a two year subscription if he identifies himself.  (Thanks to Peter McMullen for the Analog items, but he doesn't recall the contributor ever being named.)

Edward G. Robles, Jr.  Letter (Brass Tacks column).  Analog Science Fact • Science Fiction 74:4 (Dec 1964) 4.  Says the Jun 1964 object is a "three-hole two slot BLIVIT" and was developed at JPL (Jet Propulsion Laboratory, Pasadena) and published in their Goddard News.  He provides a six-hole five-slot BLIVIT, but as the Editor comments, it 'lacks the classic simple elegance of the Original.'

Sergio Aragones.  A Mad look at winter sports.  Mad Magazine (?? 1964);  reprinted in:  Mad Power;  Signet, NY, 1970, pp. 120‑129.  P. 124 shows a standard version.

Bob Clark, illustrator.  A Mad look at signs of the times.  Loc. cit. under Aragones, pp. 167‑188.  P. 186 shows standard version.

D. H. Schuster.  A new ambiguous figure: a three‑stick clevis.  Amer. J. Psychol. 77 (1964) 673.  Cites Calif. Tech. Ind. ad.  [Ernst, 1992, pp. 80-81 reproduces this article.]

Mad Magazine.  No. 93 (Mar 1965).  Cover.  Miniature reproduction in:  Maria Reidelbach; Completely Mad _ A History of the Comic Book and Magazine; Little, Brown & Co., Boston, 1991, p. 82.  Shows a standard version.  Al Seckel says they thought it was an original idea and they apologised in the next issue _ to whom??

Reveille (a UK weekly magazine) (10 Jun 1965).  ??NYS _ cited by Briggs, below _ standard version.

Don Mackey.  Optical illusion.  Skywriter (magazine of North American Aviation) (18 Feb 1966).  ??NYS _ cited by Conrad G. Mueller et al.; Light and Vision; Time-Life Books Pocket Edition, Time-Life International, Netherlands, 1969, pp. 171 & 190.  Standard version with nuts on the ends.

Heinz Von Foerster.  From stimulus to symbol: The economy of biological computation.  IN:  Sign Image Symbol;  ed. Gyorgy Kepes;  Studio Vista, London, 1966, pp. 42-60.  On p. 55, he shows the "Triple-pronged fork with only two branches" and on p. 54, he notes that although each portion is correct, it is impossible overall, but he gives no indication of its history or that it is at all new.

G. A. Briggs.  Puzzle and Humour Book.  Published by the author, Ilkley, 1966.  Pp. 17-18 shows the unnamed trident in a version from Adcock & Shipley (Sales) Ltd., machine tool makers in Leicester.  Cites Reveille, above.  Standard versions.

Harold Baldwin.  Building better blivets.  The Worm Runner's Digest 9:2 (1967) 104‑106.  Discusses relation between numbers of slots and of prongs.  Draws a three slot version and 2 and 4 way versions.

Charlie Rice.  Challenge!  Op. cit. in 5.C.  1968.  P. 10 shows a six prong, four slot version, called the "Old Roman Pitchfork".

Roger Hayward.  Blivets; research and development.  The Worm Runner's Digest 10 (Dec 1968) 89‑92.  Several fine developments, including two interlaced frames and his monumental version.  Cites Baldwin.

M. Gardner.  SA (May 1970) = Circus, pp. 3‑15.  Says this became known in 1964 and cites Mad & Hayward, but not Schuster.

D. Uribe, op. cit. above, gives several variations.

 

          6.AJ.2.         TRIBAR AND IMPOSSIBLE STAIRCASE

 

Silvanus P. Thompson.  Optical illusions of motion.  Brain 3 (1882) 289-298.  Hexagon of non‑overlapping circles. 

Thomas Foster; Illusions of motion and strobic circles; Knowledge 1 (17 Mar 1882) 421-423;  says Thompson exhibited these illusions at the British Association meeting in 1877.

Pearson.  1907.  Part II, no. 3: Whirling wheels, p. 3.  Gives Thompson's form, but the wheels are overlapping, which makes it look a bit like an ancestor of the tribar.

Oscar Reutersvård.  Omöjliga Figure [Impossible Figures _ In Swedish].  Edited by Paul Gabriel.  Doxa, Lund, (1982);  2nd ed., 1984.  This seems to be the first publication of his work, but he has been exhibiting since about 1960 and some of the exhibitions seem to have had catalogues.  P. 9 shows and discusses his Opus 1 from 1934, which is an impossible tribar made from cubes.  (Reproduced in Ernst, 1992, p. 69 as a drawing signed and dated 1934.  Ernst quotes Reutersvård's correspondence which describes his invention of the form while doodling in Latin class as a schoolboy.  A school friend who knew of his work showed him the Penroses' article in 1958 _ at that time he had drawn about 100 impossible objects _ by 1986, he had extended this to some 2500!)  He has numerous variations on the tribar and the two‑pronged trident.

Oscar Reutersvård.  Swedish postage stamps for 25, 50, 75 kr.  1982, based on his patterns from the 1930s.  The 25 kr. has the tribar pattern of cubes which he first drew in 1934.  (Also the 60 kr.??)

L. S. & R. Penrose.  Impossible objects: A special type of visual illusion.  British Journal of Psychology 49 (1958) 31‑33.  Presents tribar and staircase.  Photo of model staircase.  [Ernst, 1992, pp. 71-73, quotes conversation with Penrose about his invention of the Tribar and reproduces this article.  Penrose, like the rest of us, only learned about Reutersvård's work in the 1980s.]

Anon.(?)  Don't believe it.  Daily Telegraph (24 Mar 1958) ?? (clipping found in an old book).  "Three pages of the latest issue of the British Journal of Psychology are devoted to "Impossible Objects.""  Shows both the tribar and the staircase.

M. C. Escher.  Lithograph:  Belvedere.  1958.

L. S. & R. Penrose.  Christmas Puzzles.  New Scientist (25 Dec 1958) 1580‑1581 & 1597.  Prob. 2: Staircase for lazy people.

M. C. Escher.  Lithograph:  Ascending and Descending.  1960.

M. C. Escher.  Lithograph:  Waterfall.  1961.

Uribe, op. cit. above, gives several variations, including a perspective tribar.

Jan van de Craats.  Das unmögliche Escher-puzzle.  (Taken from:  De onmogelijke Escher-puzzle; Pythagoras (Amsterdam) (1988).)  Alpha 6 (or:  Mathematik Lehren / Heft 55 _ ??) (1992) 12-13.  Two Penrose tribars made into an impossible 5-piece burr.

 

          6.AJ.3.         CAFÉ WALL ILLUSION

 

          This is the illusion seen in alternatingly coloured brickwork where the lines of bricks distinctly seem tilted.  I suspect it must be apparent in brickwork going back to Roman times.

 

The illusion is apparent in the polychrome brick work on the side wall inside Keble College Chapel, Oxford, by William Butterfield, completed in 1876 [thanks to Deborah Singmaster for observing this].

Lietzmann & Trier, op. cit. at 6.AJ, 1923.  Pp. 12-13 has a striking version of this, described as a 'Flechtbogen der Kleinen'.  I can't quite translate this _ Flecht is something interwoven but Bogen could be a ribbon or an arch or a bower, etc.  They say it is reproduced from an original by Elsner.  See Lietzmann, 1953.

Ogden's Optical Illusions.  Cigarette card of 1927.  No. 5.  Original ??NYS _ reproduced in:  Julian Rothenstein & Mel Gooding; The Paradox Box; Redstone Press, London, 1993  AND in their:  The Playful Eye; Redstone Press, London, 1999, p. 56.  Vertical version of this illusion.

B. K. Gentil.  Die optische Täuschung von Fraser.  Zeitschr. f. math. u. naturw. Unterr. 66 (1935) 170 ff.  ??NYS _ cited by Lietzmann.

Nelson F. Beeler & Franklyn M. Branley.  Experiments in Optical Illusion.  Ill. by Fred H. Lyon.  Crowell, 1951, p. 42, fig. 39, is a good example of the illusion.

Lietzmann, op. cit. at 6.AJ, 1953.  P. 23 is the same as above, but adds a citation to Gentil, listed above.

Leonard de Vries.  The Third Book of Experiments.  © 1965, probably for a Dutch edition.  Translated by Joost van de Woestijne.  John Murray, 1965; Carousel, 1974.  Illusion 10, pp. 58-59, has a clear picture and a brief discussion.

Richard L. Gregory & Priscilla Heard.  Border locking and the café wall illusion.  Perception 8 (1979) 365‑380.  ??NYS _ described by Walker, below.  [I have photos of the actual café wall in Bristol.]

Jearl Walker.  The Amateur Scientist: The café‑wall illusion, in which rows of tiles tilt that should not tilt at all.  SA 259:5 (Nov 1988) 100‑103.  Good summary and illustrations.

 

          6.AK. POLYGONAL PATH COVERING  N x N  LATTICE OF POINTS,

                                        QUEEN'S TOURS, ETC.

 

          For magic circuits, see 7.N.4.

3x3 problem:  Loyd (1907),  Pearson,  Bullivant,  Goldston,  Loyd (1914),  Blyth,  Abraham,  Hedges,  Evans,  Piggins & Eley

4x4 problem:  King,  Abraham,  Adams,  Evans,  Depew,  Meyer

Queen's tours:  Loyd (1867, 1897, 1914),  Loyd Jr.

Bishop's tours:  Dudeney (1932),  Doubleday,  Obermair

Rook's tours:  Loyd (1878),  Proctor,  Loyd (1897),  Bullivant,  Loyd (1914),  Filipiak,  Hartswick,  Barwell,  Gardner,  Peters,  Obermair

Other versions:  Prout

 

Loyd.  ??Le Sphinx (Mar 1867 _ but the Supplement to Sam Loyd and His Chess Problems corrects this to 15 Nov 1866).  = Chess Strategy, Elizabeth, NJ, 1878, no. or p. 336(??).  = A. C. White; Sam Loyd and His Chess Problems; 1913, op. cit. in 1; no. 40, pp. 42‑43.  Queen's circuit on  8 x 8  in  14  segments.  (I.e. closed circuit, not leaving board, using queen's moves.)  No. 41 & 42 of White give other solutions.  White quotes Loyd from Chess Strategy, which indicates that Loyd invented this problem.  Tit‑Bits No. 31 & SLAHP: Touring the chessboard, pp. 19 & 89, give No. 41.

Loyd.  Chess Strategy, 1878, op. cit. above, no. or p. 337 (??)  (= White, 1913, op. cit. above, no. 43, pp. 42‑43.)  Rook's circuit on  8 x 8  in  16  segments.  (I.e. closed circuit, not leaving board, using rook's moves, and without crossings.)

Richard A. Proctor.  Gossip column.  Knowledge 10  (Dec 1886) 43  &  (Feb 1887) 92.  6 x 6  array of cells.  Prisoner in one corner can exit from the opposite corner if he passes "once, and once only, through all the  36  cells."  "... take the prisoner into either of the cells adjoining his own, and back into his own, ....  This puzzle is rather a sell, ...."  Letter and response [in Gossip column, Knowledge 10 (Mar 1887) 115-116] about the impossibility of any normal solution.

Loyd.  Problem 15: The gaoler's problem.  Tit‑Bits 31 (23 Jan  &  13 Feb 1897) 307  &  363.  Rook's circuit on  8 x 8  in  16  segments, but beginning and ending on a central square.  Cf. The postman's puzzle in the Cyclopedia, 1914.

Loyd.  Problem 16: The captive maiden.  Tit‑Bits 31 (30 Jan  &  20 Feb 1897) 325  &  381.  Rook's tour in minimal number of moves from a corner to the diagonally opposite corner, entering each cell once.  Because of parity, this is technically impossible, so the first two moves are into an adjacent cell and then back to the first cell, so that the first cell has now been entered.

Loyd.  Problem 20: Hearts and darts.  Tit‑Bits 31 (20 Feb,  13  &  20 Mar 1897) 381,  437,  455.  Queen's tour on  8 x 8,  starting in a corner, permitting crossings, but with no segment going through a square where the path turns.  Solution in  14  segments.  This is No. 41 in White _ see the first Loyd entry above.

Ball.  MRE, 4th ed., 1905, p. 197.  At the end of his section on knight's tours, he states that there are many similar problems for other kinds of pieces.

Loyd.  Sam Loyd's Puzzle Magazine (Apr 1908) _ ??NYS, reproduced in:  A. C. White; Sam Loyd and His Chess Problems; 1913, op. cit. in 1; no. 56, p. 52.  = Problem 26: A brace of puzzles _ No. 26: A study in naval warfare; Tit‑Bits 31 (27 Mar 1897) 475  &  32 (24 Apr 1897) 59.  = Cyclopedia, 1914, Going into action, pp. 189 & 364.  = MPSL1, prob. 46, pp. 44 & 138.  = SLAHP: Bombs to drop, pp. 86 & 119.  Circuit on  8 x 8  in  14  segments, but with two lines of slope  2.  In White, p. 43, Loyd says an ordinary queen's tour can be started "from any of the squares except the twenty which can be represented by  d1, d3 and d4."  This problem starts at  d1.  However I think White must have mistakenly put down twenty for twelve??

Loyd.  In G. G. Bain, op. cit. in 1, 1907.  He gives the  3 x 3  lattice in four lines as the Columbus Egg Puzzle.

Pearson.  1907.  Part I, no. 36: A charming puzzle, pp. 36 & 152‑153.  3 x 3  lattice in  4  lines.

C. H. Bullivant.  Home Fun, 1910, op. cit. in 5.S.  Part VI, Chap. IV.

No. 1: The travelling draught‑man, pp. 515 & 520.  Rook's circuit on  8 x 8  in 16 segments, different than Loyd's.

No. 3: Joining the rings.  3 x 3  in 4 segments.

Will Goldston.  More Tricks and Puzzles without Mechanical Apparatus.  The Magician Ltd., London, nd [1910?].  (BMC lists Routledge & Dutton eds. of 1910.)  (There is a 2nd ed., published by Will Goldston, nd [1919].)  The nine‑dot puzzle, pp. 127‑128 (pp. 90‑91 in 2nd ed.).

Loyd.  Cyclopedia, 1914, pp. 301 & 380.  = MPSL2, prob. 133 _ Solve Christopher's egg tricks, pp. 93 & 163 (with comment by Gardner).  c= SLAHP: Milkman's route, pp. 34 & 96.  3 x 3  case.

Loyd.  Cyclopedia, 1914, pp. 293 & 379.  Queen's circuit on  7 x 7  in  12  segments.

Loyd.  The postman's puzzle.  Cyclopedia, 1914, pp. 298 & 379.  Rook's circuit on  8 x 8  array of points, with one point a bit out of line, starting and ending at a central square, in  16  segments.  P. 379 also shows another  8 x 8  circuit, but with a slope  2  line.  See also pp. 21 & 341 and SLAHP, pp. 85 & 118, for two more examples.

Loyd.  Switchboard problem.  Cyclopedia, 1914, pp. 255 & 373.  (c= MPSL2, prob. 145, pp. 102 & 167.)  Rook's tour with minimum turning.

Blyth.  Match-Stick Magic.  1921.  Four-way game, pp. 77-78.  3 x 3  in  4  segments.

King.  Best 100.  1927.  No. 16, pp. 12 & 43.  4 x 4  in  6  segments, not closed, but easily can be closed.

Loyd Jr.  SLAHP. 1928.  Dropping the mail, pp. 67 & 111.  4 x 4  queen's tour in  6  segments.

Collins.  Book of Puzzles.  1927.  The star group puzzle, pp. 95-96.  3 x 3  in  4  segments.

Dudeney.  PCP.  1932.  Prob. 264: The fly's tour, pp. 82 & 169.  = 536, prob. 422, pp. 159 & 368.  Bishop's path, with repeated cells, going from corner to corner in  17  segments.

Sid G. Hedges.  More Indoor and Community Games.  Methuen, London, 1937.  Nine spot, p. 110.  3 x 3.  "Of course it can be done, but it is not easy."  No solution given.

Abraham.  1933.  Probs. 101, 102, 103, pp. 49 & 66 (30 & 118).  3 x 3,  4 x 4  and  6 x 6  cases.

Adams.  Puzzle Book.  1939.  Prob. C.64: Six strokes, pp. 140 & 178.  4 x 4  array in  6  segments which form a closed path, though the closure was not asked for.

J. R. Evans.  The Junior Week‑End Book.  Op. cit. in 6.AF.  1939.  Probs. 30 & 31, pp. 264 & 270.  3 x 3  &  4 x 4  cases in  4 & 6  segments, neither closed nor staying within the array.

Depew.  Cokesbury Game Book.  1939.  Drawing, p. 220.  4 x 4  in  6  segments, not closed, not staying within the array.

Meyer.  Big Fun Book.  1940.  Right on the dot, pp. 99 & 732.  4 x 4  in  6  segments.

A. S. Filipiak.  Mathematical Puzzles, 1942, op. cit. in 5.H.1, pp. 50‑51.  Same as Bullivant, but opens the circuit to make a 15 segment path.

M. S. Klamkin, proposer and solver;  John L. Selfridge, further solver.  Problem E1123 _ Polygonal path covering a square lattice.  AMM 61 (1954) 423  &  62 (1955) 124 & 443.  Shows  N x N  can be done in  2N‑2  segments.  Selfridge shows this is minimal.

W. Leslie Prout.  Think Again.  Frederick Warne & Co., London, 1958.  Joining the stars, pp. 41 & 129.  5 x 5  array of points.  Using a line of four segments, pass through  17  points.  This is a bit like the  3 x 3  problem in that one must go outside the array.

R. E. Miller & J. L. Selfridge.  Maximal paths on rectangular boards.  IBM J. Research and Development 4:5 (Nov 1960) 479-486.  They study rook's paths where a cell is deemed visited if the rook changes direction there.  They find maximal such paths in all cases.

F. Gregory Hartswick.  In:  H. A. Ripley & F. Gregory Hartswick, Detectograms and Other Puzzles, Scholastic Book Services, NY, 1969.  Prob. 4, pp. 42‑43 & 82.  Asks for  8 x 8  rook's circuit with minimal turning and having a turn at a central cell.  Solution gives two such with  16  segments and asserts there are no others.

Brian R. Barwell.  Arrows and circuits.  JRM 2 (1969) 196‑204.  Introduces idea of maximal length rook's tours.  Shows the maximal length on a  4 x 4  board is  38  and finds there are  3  solutions.  Considers also the  1 x n  board.

Solomon W. Golomb & John L. Selfridge.  Unicursal polygonal paths and other graphs on point lattices.  Pi Mu Epsilon J. 5 (1970) 107‑117.  Surveys problem.  Generalizes Selfridge's 1955 proof to  M x N  for which  MIN(2M, M+N‑2)  segments occur in a minimal circuit.

Doubleday - II.  1971.  Path finder, pp. 95-96.  Bishop's corner to corner path, same as Dudeney, 1932.

M. Gardner.  SA (May 1973) c= Knotted, chap. 6.  Prob. 1: Find rook's tours of maximum length on the  4 x 4  board.  Cites Barwell.  Knotted also cites Peters, below.

Edward N. Peters.  Rooks roaming round regular rectangles.  JRM 6 (1973) 169‑173.  Finds maximum length on  1 x N  board is  N2/2  for  N  even;  (N‑1)2/2 + N‑1  for  N  odd, and believes he has counted such tours.  He finds tours on the  N x N  board whose length is a formula that reduces to  4 BC(N+1, 3) ‑ 2[(N‑1)/2].  I am a bit unsure if he has shown that this is maximal.

David Piggins & Arthur D. Eley.  Minimal path length for covering polygonal lattices:  A review.  JRM 14:4 (1981‑82) 279‑283.  Mostly devoted to various trick solutions of the  3 x 3  case.

Obermair.  Op. cit. in 5.Z.1.  1984.

Prob. 19, pp. 23 & 50.  Bishop's path on  8 x 8  in  17  segments, as in Dudeney, PCP, 1932.

Prob. 41, p. 72.  Rook's path with maximal number of segments, which is  57.  [For the  2 x 2,  3 x 3,  4 x 4  boards, I get the maximum numbers are  3, 6, 13.]

 

          6.AL.  STEINER‑LEHMUS THEOREM

 

          This has such an extensive history that I will give only a few items.

 

C. L. Lehmus first posed the problem to Jacob Steiner in 1840.

Rougevin published the first proof in 1842.  ??NYS.

Jacob Steiner.  Elementare Lösung einer Aufgabe über das ebene und sphärische Dreieck.  J. reine angew. Math. 28 (1844) 375‑379 & Tafel III.  Says Lehmus sent it to him in 1840 asking for a purely geometric proof.  Here he gives proofs for the plane and the sphere and also considers external bisectors.

Theodor Lange.  Nachtrag zu dem Aufsatze in Thl. XIII, Nr. XXXIII.  Archiv der Math. und Physik 15 (1850) 221‑226.  Discusses the problem and gives a solution by Steiner and two by C. L. Lehmus.  Steiner also considers the external bisectors.

H. S. M. Coxeter.  Introduction to Geometry.  Wiley, 1961.  Section 1.5, ex. 4, p. 16.  An easy proof is posed as a problem with adequate hints in four lines.

M. Gardner.  SA (Apr 1961) = New MD, chap. 17.  Review of Coxeter's book, saying his brief proof came as a pleasant shock.

G. Gilbert & D. MacDonnell.  The Steiner‑Lehmus theorem.  AMM 70 (1963) 79‑80.  This is the best of the proofs sent to Gardner in response to his review of Coxeter.

Léo Sauvé.  The Steiner‑Lehmus theorem.  CM 2:2 (Feb 1976) 19‑24.  Discusses history and gives 22 references, some of which refer to 60 proofs.

Charles W. Trigg.  A bibliography of the Steiner‑Lehmus theorem.  CM 2:9 (Nov 1976) 191‑193.  36 references beyond Sauvé's.

David C. Kay.  Nearly the last comment on the Steiner‑Lehmus theorem.  CM 3:6 (1977) 148‑149.  Observes that a version of the proof works in all three classical geometries at once and gives its history.

 

          6.AM.          MORLEY'S THEOREM

 

          This also has an extensive history and I give only a few items.

 

T. Delahaye and H. Lez.  Problem no. 1655 (Morley's triangle).  Mathesis (3) 8 (1908) 138‑139.  ??NYS.

E. J. Ebden, proposer;  M. Satyanarayana, solver.  Problem no. 16381 (Morley's theorem).  The Educational Times (NS) 61 (1 Feb 1908) 81  &  (1 Jul 1908) 307‑308  = Math. Quest. and Solutions from "The Educational Times" (NS) 15 (1909) 23.  Asks for various related triangles formed using interior and exterior trisectors to be shown equilateral.  Solution is essentially trigonometric.  No mention of Morley.

Frank Morley.  On the intersections of the trisectors of the angles of a triangle.  (From a letter directed to Prof. T. Hayashi.)  J. Math. Assoc. of Japan for Secondary Education 6 (Dec 1924) 260‑262.  (= CM 3:10 (Dec 1977) 273‑275.

Frank Morley.  Letter to Gino Loria.  22 Aug 1934.  Reproduced in:  Gino Loria; Triangles équilatéraux dérivés d'un triangle quelconque.  MG 23 (No. 256) (Oct 1939) 364‑372.  Morley says he discovered the theorem in c1904 and cites the letter to Hayashi.  Loria mentions other early work and gives several generalizations.

H. F. Baker.  Note 1476:  A theorem due to Professor F. Morley.  MG 24 (No. 261) (Oct 1940) 284‑286.  Easy proof and reference to other proofs.  He cites a related result of Steiner.

Dan Pedoe.  Notes on Morley's proof of his theorem on angle trisectors.  CM 3:10 (Dec 1977) 276‑279.  "... very tentative ... first steps towards the elucidation of his work."

C. O. Oakley & Charles W. Trigg.  A list of references to the Morley theorem.  CM 3:10 (Dec 1977) 281‑290  &  4 (1978) 132.  169 items.

André Viricel (with Jacques Bouteloup).  Le Théorème de Morley.  L'Association pour le Développement de la Culture Scientifique, Amiens, 1993.  [This publisher or this book was apparently taken over by Blanchard as Blanchard was selling copies with his label pasted over the previous publisher's name in Dec 1994.]  A substantial book (180pp) on all aspects of the theorem.  The bibliography is extremely cryptic, but says it is abridged from Mathesis (1949) 175  ??NYS.  The most recent item cited is 1970.

 

          6.AN. VOLUME OF THE INTERSECTION OF TWO CYLINDERS

 

Archimedes.  The Method:  Preface, 2.  In:  T. L. Heath; The Works of Archimedes, with a supplement "The Method of Archimedes"; (originally two works, CUP, 1897 & 1912)  = Dover, 1953.  Supplement, p. 12, states the result.  The proof is lost, but pp. 48‑51 gives a reconstruction of the proof by Zeuthen.

Liu Hui.  Jiu Zhang Suan Chu Zhu (Commentary on the Nine Chapters of the Mathematical Art).  263.  ??NYS _ described in Li & Du, pp. 73‑74 & 85.  He shows that the ratio of the volume of the sphere to the volume of Archimedes' solid, called mou he fang gai (two square umbrellas), is  π/4,  but he cannot determine either volume.

Zu Geng.  c500.  Lost, but described in:  Li Chunfeng; annotation to Jiu Zhang (= Chiu Chang Suan Ching) made c656.  ??NYS.  Described on  pp. 86‑87 of:  Wu Wenchun; The out‑in complementary principle; IN:  Ancient China's Technology and Science; compiled by the Institute of the History of Natural Sciences, Chinese Academy of Sciences; Foreign Languages Press, Beijing, 1983, pp. 66‑89.  [This is a revision and translation of parts of:  Achievements in Science and Technology in Ancient China [in Chinese]; China Youth Publishing House, Beijing(?), 1978.]

                    He considers the shape, called fanggai, within the natural circumscribed cube and shows that, in each octant, the part of the cube outside the fanggai has cross section of area  h2  at distance  h  from the centre.  This is equivalent to a tetrahedron, whose volume had been detemined by Liu, so the excluded volume is  _  of the cube.

                    Li & Du, pp. 85‑87, and say the result may have been found c480 by Zu Geng's father, Zu Chongzhi.

Lam Lay-Yong & Shen Kangsheng.  The Chinese concept of Cavalieri's Principle and its applications.  HM 12 (1985) 219-228.  Discusses the work of Liu and Zu. 

Shiraishi Ch_ch_.  Shamei Sampu.  1826.  ??NYS _ described in Smith & Mikami, pp. 233-236.  "Find the volume cut from a cylinder by another cylinder that intersects is orthogonally and touches a point on the surface".  I'm not quite sure what the last phrase indicates.  The book gives a number of similar problems of finding volumes of intersections.

P. R. Rider, proposer;  N. B. Moore, solver.  Problem 3587.  AMM 40 (1933) 52 (??NX) &  612.  Gives the standard proof by cross sections, then considers the case of unequal cylinders where the solution involves complete elliptic integrals of the first and second kinds.  References to solution and similar problem in textbooks.

Leo Moser, solver;  J. M. Butchart, extender.  MM 25 (May 1952) 290  &  26 (Sep 1952) 54.  ??NX.  Reproduced in Trigg, op. cit. in 5.Q: Quickie 15, pp. 6 & 82‑83.  Moser gives the classic proof that  V = 16r3/3.  Butchart points out that this also shows that the shape has surface area  16r2.

 

          6.AO. CONFIGURATION PROBLEMS

 

          NOTATION:  (a, b, c)  denotes the configuration of  a  points in  b  rows of  c  each.  The index below covers articles other than the surveys of Burr et al. and Gardner.

 

(  5,   2, 3):   Sylvester

(  6,   3, 3):   Mittenzwey

(  7,   6, 3):   Criton

(  9,   8, 3):   Sylvester;  Criton

(  9,   9, 3):   Criton

(  9, 10, 3):   Jackson;  Family Friend;  Parlour Pastime;  Magician's Own Book;  The Sociable;  Book of 500 Puzzles;  Charades etc.;  Boy's Own Conjuring Book;  Hanky Panky;  Carroll;  Crompton;  Berkeley & Rowland;  Hoffmann;  Dudeney (1908);  Wehman;  Williams;  Loyd Jr;  Blyth;  Rudin;  Brooke;  Putnam;  Criton

(10,   5, 4):   The Sociable;  Book of 500 Puzzles;  Carroll;  Hoffmann;  Dudeney (1908);  Wehman;  Wiliams;  Dudeney (1917);  Blyth;  King;  Rudin;  Putnam

(10, 10, 3):   Sylvester

(11, 11, 3):   The Sociable;  Book of 500 Puzzles; Wehman

(11, 12, 3):   Hoffmann;  Williams

(11, 13, 3):   Prout

(11, 16, 3):   Wilkinson _ in Dudeney (1908 & 1917);  Macmillan

(12,   4, 5) _ Trick version of a hollow  3 x 3  square with doubled corners, as in 7.Q:  Family Friend (1858);  Illustrated Boy's Own Treasury;  Secret Out; 

(12,   6, 4):   Endless Amusement II;  The Sociable;  Book of 500 Puzzles;  Boy's Treasury;  Cassel's;  Hoffmann;  Wehman;  Rudin;  Criton

(12,   7, 4) _ Trick version of a  3 x 3  square with doubled diagonal:  Hoffmann (1876);  Mittenzwey;  Hoffmann (1893), no. 8

(12,   7, 4):   Dudeney (1917);  Putnam

(12, 19, 3):   Macmillan

(13,   9, 4):   Criton

(13, 12, 3):   Criton

(13, 18, 3):   Sylvester

(13, 22, 3):   Criton

(15, 15, 3):   Jackson

(15, 16, 3):   The Sociable;  Book of 500 Puzzles;  Wehman

(15, 23, 3):   Jackson

(15, 26, 3):   Woolhouse

(16, 10, 4):   The Sociable;  Book of 500 Puzzles;  Hoffmann;  Wehman

(16, 12, 4):   Criton

(16, 15, 4):   Dudeney (1899, 1902, 1908);  Brooke;  Putnam; Criton

(17, 24, 3):   Jackson

(17, 28, 3):   Endless Amusement II;  Pearson

(17, 32, 3):   Sylvester

(18, 18, 4):   Macmillan

(19, 19, 4):   Crition

(19,   9, 5):   Endless Amusement II;  The Sociable;  Book of 500 Puzzles;  Proctor;  Hoffmann;  Clark;  Wehman;  Ripley;  Rudin;  Putnam;  Criton

(19, 10, 5):   Proctor

(20, 18, 4):   Loyd Jr

(20, 21, 4):   Criton

(21,   9, 5):   Magician's Own Book;  Book of 500 Puzzles;  Boy's Own Conjuring Book;  Blyth;  Depew

(21, 10, 5):   Mittenzwey

(21, 11, 5):   Putnam

(21, 12, 5):   Dudeney (1917);  Criton

(21, 30, 3):   Hoffmann

(21, 50, 3):   Sylvester

(22, 15, 5):   Macmillan

(22, 20, 4):   Dudeney (1899)

(22, 21, 4):   Dudeney (1917);  Putnam

(24, 28, 3):   Jackson;  Parlour Pastime

(24, 28, 4):   Jackson;  Héraud;  Benson;  Macmillan

(24, 28, 5):   Jackson

(25, 12, 5):   Endless Amusement II;  Young Man's Book;  Proctor;  Criton

(25, 30, 4):   Macmillan

(25, 72, 3):   Sylvester

(26, 21, 5):   Macmillan

(27,   9, 6):   The Sociable;  Book of 500 Puzzles;  Hoffmann;  Wehman

(27, 10, 6):   The Sociable;  Book of 500 Puzzles;  Wehman

(27, 15, 5):   Jackson

(29, 98, 3):   Sylvester

(30, 12, 7):   Criton

(30, 22, 5):   Criton

(30, 26, 5):   Macmillan

(31,   6, 6) _ with 7 circles of 6:  The Sociable;  Book of 500 Puzzles;  Magician's Own Book (UK version);  Wehman

(31, 15, 5):   Proctor

(36, 55, 4):   Macmillan

(37, 18, 5):   Proctor

(37, 20, 5):   The Sociable;  Book of 500 Puzzles;  Illustrated Boy's Own Treasury;  Hanky Panky; Wehman

(49, 16, 7):   Criton

 

Trick versions _ with doubled counters:  Family Friend (1858),  Illustrated Boy's Own Treasury,  Secret Out,  Hoffmann (1876),  Mittenzwey,  Hoffmann (1893), nos. 8 & 9,  Pearson,  Home Book ....  These could also be considered as in 7.Q.2 or 7.Q.

 

Jackson.  Rational Amusement.  1821.  Trees Planted in Rows, nos. 1-10, pp. 33-34 & 99-100 and plate IV, figs. 1-9.  [Brooke and others say this is the earliest statement of such problems.]

1.  (9, 10, 3).  Quoted in Burr, below.

                    "Your aid I want, nine trees to plant

                              In rows just half a score;

                    And let there be in each row three.

                              Solve this: I ask no more."

2.  (n, n, 3),  He does the case  n = 15.

3.  (15, 23, 3).

4.  (17, 24, 3).

5.  (24, 24, 3)  with a pond in the middle.

6.  (24, 28, 4).

7.  (27, 15, 5)

8.  (25, 28, c)  with  c = 3, 4, 5.

9.  (90, 10, 10)  with equal spacing _ decagon with 10 trees on each side.

10. Leads to drawing square lattice in perspective with two vanishing points, so the diagonals of the resulting parallelograms are perpendicular.

Endless Amusement II.  1826? 

Prob. 13, p. 197.  (19, 9, 5).

Prob. 14, p. 197.  (12, 6, 4).

Prob. 26, p. 202.  (25, 12, 5).  Answer is a  5 x 5  square array.

          Ingenious artists, how may I dispose

          Of five-and-twenty trees, in just twelve rows;

          That every row five lofty trees may grace,

          Explain the scheme _ the trees completely place.

Prob. 35, p. 212.  (17, 28, 3).  [This is the problem that is replaced in the 1837 ed.]

Young Man's Book.  1839.  P. 239.  Identical to Endless Amusement II.

Crambrook.  1843.  P. 5, no. 15: The Puzzle of the Steward and his Trees.  This may be a configuration problem _ ??

Boy's Treasury.  1844.  Puzzles and paradoxes, no. 13, pp. 426 & 429.  (12, 6, 4).

Family Friend 1 (1849) 148 & 177.  Family Pastime _ Practical Puzzles _ 1. The puzzle of the stars.  (9, 10, 3). 

                    Friends of the Family Friend, pray show

                    How you nine stars would so bestow

                    Ten rows to form _ in each row three _

                    Tell me, ye wits, how this can be?

                                                                Robina.

          Answer has

                    Good-tempered Friends! here nine stars see:

                    Ten rows there are, in each row three!

W. S. B. Woolhouse.  Problem 39.  The Mathematician 1 (1855) 272.  Solution:  ibid. 2 (1856) 278‑280.  ??NYS _ cited in Burr, et al., below, who say he does  (15, 26, 3).

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Mechanical puzzles.

No. 1, p. 176 (1868: 187).  (9, 10, 3). 

          Ingenious artist pray disclose,

          How I nine trees can so dispose,

          That these ten rows shall formed be,

          And every row consist of three?

No. 12, p. 182 (1868: 192-193).  (24, 28, 3),  but with a central pond breaking 4 rows of 6 into 8 rows of 3.

Magician's Own Book.  1857.

Prob. 33: The puzzle of the stars, pp. 277 & 300.  (9, 10, 3),

          Friends one and all, I pray you show

          How you nine stars would so bestow,

          Ten rows to form _ in each row three _

          Tell me, ye wits, how this can be?

Prob. 41: The tree puzzle, pp. 279 & 301.  (21, 9, 5),  unequally spaced on each row.  Identical to Book of 500 Puzzles, prob 41.

The Sociable.  1858.  = Book of 500 Puzzles, 1859, with same problem numbers, but page numbers decreased by 282.

Prob. 3: The practicable orchard, pp. 286 & 302.  (16, 10, 4).

Prob. 8: The florist's puzzle, pp. 289 & 303-304.  (31, 6, 6)  with 7 circles of 6.

Prob. 9: The farmer's puzzle, pp. 289 & 304.  (11, 11, 3).

Prob. 12: The geometrical orchard,  p. 291 & 306.  (27, 9, 6).

Prob. 17: The apple-tree puzzle, pp. 292 & 308.  (10, 5, 4).

Prob. 22: The peach orchard puzzle, pp. 294 & 309.  (27, 10, 6).

Prob. 26: The gardener's puzzle, pp. 295 & 311.  (12, 6, 4)  two ways.

Prob. 27: The circle puzzle, pp. 295 & 311.  (37, 20, 5)  equally spaced along each row.

Prob. 29: The tree puzzle, pp. 296 & 312.  (15, 16, 3)  with some bigger rows.

Prob. 32: The tulip puzzle, pp. 296 & 314.  (19, 9, 5).

Prob. 36: The plum tree puzzle, pp. 297 & 315.  (9, 10, 3).

Family Friend (Dec 1858) 359.  Practical puzzles _ 2.  "Make a square with twelve counters, having five on each side."  (12, 4, 5).  I haven't got the answer, but presumably it is the trick version of a hollow square with doubled corners, as in 7.Q.  See Illustrated Boy's Own Treasury, 1860.

Book of 500 Puzzles.  1859.  Prob.  3, 9, 12, 17, 22, 26, 27, 29, 32, 36  are identical to those in The Sociable, with page numbers decreased by 282.

Prob. 33: The puzzle of the stars, pp. 91 & 114.  (9, 10, 3),  identical to Magician's Own Book, prob. 33.

Prob. 41: The tree puzzle, pp. 93 & 115.  (21, 9, 5),  identical to Magician's Own Book, prob. 41.  See Illustrated Boy's Own Treasury.

Charades, Enigmas, and Riddles.  1859?: prob. 13, pp. 58 & 61;  1862?: prob. 557, pp. 105 & 152.  (9, 10, 3).  (The 1865 has slightly different typography.)

                    Sir Isaac Newton's Puzzle (versified).

                              Ingenious Artist, pray disclose

                              How I, nine Trees may so dispose,

                              That just Ten Rows shall planted be,

                              And every Row contain just Three.

Boy's Own Conjuring Book.  1860.

Prob. 40: The tree puzzle, pp. 242 & 266.  (21, 9, 5),  identical to Magician's Own Book, prob 41.

Prob. 42: The puzzle of the stars, pp. 243 & 267.  (9, 10, 3),  identical to Magician's Own Book, prob. 33, with commas omitted.

Illustrated Boy's Own Treasury.  1860.

Prob. 2, pp. 395 & 436.  (37, 20, 5),  equally spaced on each row, identical to The Sociable, prob. 27.

Prob. 13, pp. 397 & 438.  "Make a square with twelve counters, having five on each side."  (12, 4, 5).  Trick version of a hollow square with doubled corners.  Presumably identical to Family Friend, 1858.  Same as Secret Out.

J. J. Sylvester.  Problem 2473.  Math. Quest. from the Educ. Times 8 (1867) 106‑107.  ??NYS _ Burr, et al. say he gives  (10, 10, 3),  (81, 800, 3)  and  (a, (a‑1)2/8, 3).

The Secret Out.  Op. cit. in 4.A.1.  1871?  The square of counters, p. 9.  (12, 4, 5) _ trick version.  Same as Illustrated Boy's Own Treasury, prob. 13.

Magician's Own Book (UK version).  1871.  The solution to The florist's puzzle (The Sociable, prob. 8) is given at the bottom of p. 284, apparently to fill out the page as there is no relevant text anywhere. 

Hanky Panky.  1872. 

To place nine cards in ten rows of three each, p. 291.  I.e.  (9, 10, 3).

Diagram with no text, p. 128.  (37, 20, 5),  equally spaced on each line as in The Sociable, prob. 27.

Hoffmann.  Modern Magic.  (George Routledge, London, 1876);  reprinted by Dover, 1978.  To place twelve cards in rows, in such a manner that they will count four in every direction, p. 58.  Trick version of a  3 x 3  square with extras on a diagonal, giving a form of  (12, 7, 4).

Lewis Carroll.  MS of 1876.  ??NYS _ described in:  David Shulman; The Lewis Carroll problem; SM 6 (1939) 238-240. 

Given two rows of five dots, move four to make 5 rows of 4.  Shulman describes this case, following Dudeney, AM, 1917, then observes that since Dudeney is using coins, there are further solutions by putting a coin on top of another.  He refers to Hoffmann and Loyd.

(9, 10, 3).  Shulman quotes from Robert T. Philip; Family Pastime; London, 1852, p. 30, ??NYS, but this must refer to the item in Family Friend, which was edited by Robert Kemp Philp.  BMC indicates Family Pastime may be another periodical.  Shulman then cites Jackson and Dudeney.

Mittenzwey.  1879? 

Prob. 174, pp. 33 & 82.  (6, 3, 3).  (6, 4, 3)  by a trick.

Prob. 175, pp. 33 & 82.  Arrange 16 pennies as a  3 x 3  square so each row and column has four in it.  Solution shows a  3 x 3  square with extras on the diagonal _ but this only uses 12 pennies!  So this the trick version of  (12, 7, 4)  as in Hoffmann (1876).

Prob. 176, pp. 33 & 82.  (21, 10, 5).

Cassell's.  1881.  P. 92: The six rows puzzle.  = Manson, 1911, p. 146.

J. J. Sylvester.  Problem 2572.  Math. Quest. from the Educ. Times 45 (1886) 127‑128.  ??NYS _ cited in Burr, below.  Obtains good  examples of  (a, b, 3)  for each  a.  In most cases, this is still the best known.

Richard A. Proctor.  Some puzzles;  Knowledge 9 (Aug 1886) 305-306  &  Three puzzles;  Knowledge 9 (Sep 1886) 336-337.  (19, 9, 5).  Generalises to  (6n+1, 3n, 5).

Richard A. Proctor.  Our puzzles.  Knowledge 10  (Nov 1886) 9  &  (Dec 1886) 39-40.  Gives several solutions of  (19, 9, 5)  and asks for  (19, 10, 5).  Gossip column, (Feb 1887) 92, gives another solution

William Crompton.  The odd half-hour.  The Boy's Own Paper 13 (No. 657) (15 Aug 1891) 731-732.  Sir Isaac Newton's puzzle (versified).  (9, 10, 3).

                              Ingenious artist pray disclose

                              How I nine trees may so dispose

                              That just ten rows shall planted be

                              And every row contain just three.

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Card Puzzles No. IV, p. 3.  (9, 10, 3).

Hoffmann.  1893.  Chap. VI, pp. 265‑268 & 275‑281.

No. 1: (11, 12, 3).

No. 2: (9, 10, 3).

No. 3: (27, 9, 6).

No. 4: (10, 5, 4).

No. 5: (12, 6, 4).

No. 6: (19, 9, 5).

No. 7: (16, 10, 4).

No. 8: (12, 7, 4)  _  Trick version of a  3 x 3  square with extras on a diagonal.

No. 9: 9 red + 9 white,  form  10 + 8  lines of 3 each.  Puts a red and a white point at the same place, so this is a trick version.

No. 11: (10, 8, 4)  _  counts in 8 'directions', so he counts each line twice!

No. 12: (13, 12, 5)  _  with double counting as in no. 11.

No. 15: (21, 30, 3)  _  but points must lie on a given figure.

Dudeney.  A batch of puzzles.  Royal Magazine 1:3 (Jan 1899)  &  1:4 (Feb 1899) 368-372.  (22, 20, 4)  with trees at lattice points of a  7 x 10  lattice.  Compare with AM, prob. 212.

Anon. & Dudeney.  A chat with the Puzzle King.  The Captain 2 (Dec? 1899) 314-320  &  2:6 (Mar 1900) 598-599  &  3:1 (Apr 1900) 89.  (16, 15, 4).  Cf. 1902.

Dudeney.  "The Captain" puzzle corner.  The Captain 3:2 (May 1900) 179.  This gives a solution of a problem called Joubert's guns, but I haven't seen the proposal.  (10, 5, 4)  but wants the maximum number of castles to be inside the walls joining the castles.  Manages to get two inside.  = Dudeney; The puzzle realm; Cassell's Magazine ?? (May 1908) 713-716; no. 6: The king and the castles.  = AM, 1917, prob. 206: The king and the castles, pp. 56 & 189. 

Dudeney.  The ploughman's puzzle.  In:  The Canterbury Puzzles, London Magazine 9 (No. 49) (Aug 1902) 88‑92  &  (No. 50) (Sep 1902) 219.  = CP; 1907; no. 21, pp. 43‑44 & 175‑176.  (16, 15, 4).  Cf. 1899.

A. Héraud.  Jeux et Récréations Scientifiques _ Chimie, Histoire Naturelle, Mathématiques.  Baillière et Fils, Paris, 1903.  P. 307: Un paradoxe mathématique.  (24, 28, 4).  I haven't checked for this problem in the 1884 ed.

Clark.  Mental Nuts.  1904: no. 91: The lovers' grove.  (19, 9, 5).

          I am required to plant a grove

          To please the lady whom I love.

          This simple grove to be composed

          Of nineteen trees in nine straight rows;

          Five trees in each row I must place,

          Or I shall never see her face.

                    Cf Ripley, below.

Pearson.  1907. 

Part I, no. 77: Lines on an old sampler, pp. 77 & 167.  (17, 28, 3).

Part II, no. 83: For the children, pp. 83 & 177.  Trick version of  (12, 4, 5),  as in Family Friend (1858).

Dudeney.  The world's best puzzles.  Op. cit. in 2.  1908.  He says  (9, 10, 3)  "is attributed to Sir Isaac Newton, but the earliest collection of such puzzles is, I believe, in a rare little book that I possess _ published in 1821."  [This must refer to Jackson.]  Says Rev. Mr. Wilkinson gave  (11, 16, 3)  "some quarter of a century ago" and that he, Dudeney, published  (16, 15, 4)  in 1897 (cf under 1902 above).  He leaves these as problems but doesn't give their solutions in the next issue.

Wehman.  New Book of 200 Puzzles.  1908.

P. 4: The practicable orchard.  (16, 10, 4).  = The Sociable, prob. 3.

P. 7: The puzzle of the stars.  (9, 10, 3).  = Magician's Own Book, prob. 33.

P. 8: The apple-tree puzzle.  (10, 5, 4).  = The Sociable, prob. 17.

P. 8: The peach orchard puzzle.  (27, 10, 6).  = The Sociable, prob. 22.

P. 8: The plum tree puzzle.  (9, 10, 3).  = The Sociable, prob. 36.

P. 12: The farmer's puzzle.  (11, 11, 3).  = The Sociable, prob. 9.

P. 19: The gardener's puzzle.  (12, 6, 4)  two ways.  = The Sociable, prob. 26.

P. 26: The circle puzzle.  (37, 20, 5)  equally spaced along each row.  = The Sociable, prob. 27.

P. 30: The tree puzzle.  (15, 16, 3)  with some bigger rows.  = The Sociable, prob. 29.

P. 31: The geometrical orchard.  (27, 9, 6).  = The Sociable, prob. 12.

P. 31: The tulip puzzle.  (19, 9, 5).  = The Sociable, prob. 32.

P. 41: The florist's puzzle.  (31, 6, 6)  with seven circles of six.  = The Sociable, prob. 8.

J. K. Benson, ed.  The Pearson Puzzle Book.  C. Arthur Pearson, London, nd [c1910, not in BMC or NUC].  [This is almost identical with the puzzle section of Benson, but has 13 pages of different material.]  A symmetrical plantation, p. 99.  (24, 28, 4).

Williams.  Home Entertainments.  1914.  Competitions with counters, p. 115.  (11, 12, 3);  (9, 10, 3);  (10, 5, 4).

Dudeney.  AM.  1917.  Points and lines problems, pp. 56-58 & 189-193.

Prob. 206: The king and the castles.  See The Captain, 1900.

Prob. 207: Cherries and plums.  Two  (10, 5, 4)  patterns among 55 of the points of an  8 x 8  array.

Prob. 208: A plantation puzzle.  (10, 5, 4)  among 45 of the points of a  7 x 7  array.

Prob. 209: The twenty-one trees.  (21, 12, 5).

Prob. 210: The ten coins.  Two rows of five.  Move four to make  (10, 5, 4).  Cf. Carroll, 1876.  Shows there are 2400 ways to do this.  He shows that there are six basic solutions of the  (10, 5, 4)  which he calls:  star, dart, compasses, funnel, scissors, nail and he describes the smallest arrays on which they can fit.

Prob. 211: The twelve mince-pies.  12 points at the vertices and intersections of a Star of David.  Move four to make  (12, 7, 4).

Prob. 212: The Burmese plantation.  (22, x, 4)  among the points of a  7 x 7  array.  Finds  x = 21.  Cf. 1899.

Prob. 213: Turks and Russians, pp. 58 & 191‑193.  Complicated problem leading to  (11, 16, 3)  _ cites his Afridi problem in Tit-Bits and attributes the pattern to Wilkinson 'some twenty years ago', cf 1908.

Blyth.  Match-Stick Magic.  1921.

Four in line, p. 48.  (10, 5, 4).

Three in line, p. 77.  (9, 10, 3).

Five-line game, pp. 78-79.  (21, 9, 5).

King.  Best 100.  1927.  No. 62, pp. 26 & 54.  = Foulsham's no. 21, pp. 9 & 13.  (10, 5, 4).

Loyd Jr.  SLAHP.  1928.  Points and lines puzzle, pp. 20 & 90.  Says Newton proposed  (9, 10, 3).  Asks for  (20, 18, 4)  on a  7 x 7  array.

R. Ripley.  Believe It or Not!  Book 2.  Op. cit. in 5.E, 1931.  The planter's puzzle, p. 197, asks for  (19, 9, 5)  but no solution is given.  See Clark, above, for a better version of the verse.

          "I am constrained to plant a grove

          For a lady that I love.

          This ample grove is too composed;

          Nineteen trees in nine straight rows.

          Five trees in each row I must place,

          Or I shall never see her face."

Rudin.  1936.  Nos. 105-108, pp. 39 & 99-100.

No. 105: (9, 10, 3).

No. 106: (10, 5, 4)  _ two solutions.

No. 107: (12, 6, 4)  _ two solutions.

No. 108: (19, 9, 5).

Depew.  Cokesbury Game Book.  1939.  The orange grower, p. 221.  (21, 9, 5).

The Home Book of Quizzes, Games and Jokes.  Op. cit. in 4.B.1, 1941.  P. 147, prob. 1 & 2.  Place six coins in an  L  or a cross and make two rows of four, i.e.  (6, 2, 4),  which is done by the simple trick of putting a coin on the intersection.

R. H. Macmillan.  Letter:  An old problem.  MG 30 (No. 289) (May 1946) 109.  Says he believes Newton and Sylvester studied this.  Says he has examples of  (11, 16, 3), (12, 19, 3),  (18, 18, 4),  (24, 28, 4),  (25, 30, 4),  (36, 55, 4),  (22, 15, 5),  (26, 21, 5),  (30, 26, 5).

W. Leslie Prout.  Think Again.  Frederick Warne & Co., London, 1958.  Thirteen rows of three, pp. 45 & 132.  (11, 13, 3).

Maxey Brooke.  Dots and lines.  RMM 6 (Dec 1961) 51‑55.  Cites Jackson and Dudeney.  Says Sylvester showed that  n  points can be arranged in at least  (n‑1)(n‑2)/6  rows of three.  Shows  (9, 10, 3)  and  (16, 15, 4).

S. A. Burr, B. Grünbaum & N. J. A. Sloane.  The orchard problem.  Geometria Dedicata 2 (1974) 397‑424.  Establishes good examples of  (a, b, 3)  slightly improving on Sylvester, and establishes some special better examples.  Gives upper bounds for  b  in (a, b, 3). Sketches history and tabulates best values and upper bounds for  b  in (a, b, 3), for  a = 1 (1) 32. 

                    The following have the maximal possible value of  b  for given  a  and  c.

          (3, 1, 3);  (4, 1, 3);  (5, 2, 3);  (6, 4, 3);  (7, 6, 3);  (8, 7, 3);  (9, 10, 3);  (10, 12, 3);  (11, 16, 3);  (12, 19, 3);  (16, 37, 3).

                    The following have the largest known value of  b  for the given  a  and  c. 

          (13, 22, 3);  (14, 26, 3);  (15, 31, 3);  (17, 40, 3);  (18, 46, 3);  (19, 52, 3);  (20, 57, 3);  (21, 64, 3);  (22, 70, 3);  (23, 77, 3);  (24, 85, 3);  (25, 92, 3);  (26, 100, 3);  (27, 109, 3);  (28, 117, 3);  (29, 126, 3);  (30, 136, 3);  (31, 145, 3);  (32, 155, 3).

M. Gardner.  SA (Aug 1976).  Surveys these problems, based on Burr, Grünbaum & Sloane.  He gives results for  c = 4. 

                    The following have the maximal possible value of  b  for the given  a  and  c.

          (4, 1, 4);  (5, 1, 4);  (6, 1, 4);  (7, 2, 4);  (8, 2, 4);  (9, 3, 4);  (10, 5, 4);  (11, 6, 4);  (12, 7, 4).

                    The following have the largest known value of  b  for the given  a  and  c.

          (13, 9, 4);  (14, 10, 4);  (15, 12, 4);  (16, 15, 4);  (17, 15, 4);  (18, 18, 4);  (19, 19, 4);  (20, 20, 4).

Putnam.  Puzzle Fun.  1978.  Nos. 17-23: Bingo arrangements, pp. 6 & 29-30.  (21, 11, 5),  (16, 15, 4),  (19, 9, 5),  (9, 10, 3),  (12, 7, 4),  (22, 21, 4),  (10, 5, 4).

S. A. Burr.  Planting trees.  In:  The Mathematical Gardner; ed. by David Klarner; Prindle, Weber & Schmidt/Wadsworth, 1981.  Pp. 90‑99.  Pleasant survey of the 1974 paper by Burr, et al.

Michel Criton.  Des points et des Lignes.  Jouer Jeux Mathématiques 3 (Jul/Sep 1991) 6-9.  Survey, with a graph showing  c  at  (a, b).  Observes that some solutions have points which are not at intersections of lines and proposes a more restirctive kind of arrangement of  b  lines whose interesections give  a  points with  c points on each line.  He denotes these with square brackets which I write as  [a, b, c].  Pictures of  (7, 6, 3),  [9, 8, 3],  (9, 9, 3),  (12, 6, 4),  [13, 9, 4],  (13, 12, 3),  (13, 22, 3), (16, 12, 4),  (19, 19, 4),  (19, 19, 5),  (20, 21, 4),  [21, 12, 5],  (25, 12, 5),  (30, 12, 7),  (30, 22, 5),  (49, 16, 7)  and mentions of  (9, 10, 3),  (16, 15, 4),

 

          6.AO.1.        PLACE FOUR POINTS EQUIDISTANTLY  =  MAKE FOUR

                                                  TRIANGLES WITH SIX MATCHSTICKS

 

Endless Amusement II.  1826?  Prob. 21, p. 200.  "To place 4 poles in the ground, precisely at an equal distance from each other."  Uses a pyramidal mound of earth.

Young Man's Book.  1839.  P. 235.  Identical to Endless Amusement II.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Mechanical puzzles, no. 6, p. 178 (1868: 189).  Plant four trees at equal distances from each other.

Frank Bellew.  The Art of Amusing.  1866.  Op. cit. in 5.E.  1866: pp. 97-98 & 105-106;  1870: pp. 93‑94 & 101‑102.

Mittenzwey.  1879?  Prob. 184, pp. 34 & 83.  Problem only says six sticks _ solution is a rectangle with its diagonals.  Prob. 186, pp. 34 & 83.  Problem says six equally long sticks _ solution is a tetrahedron.

F. Chasemore.  Loc. cit. in 6.W.5.  1891.  Item 3: The triangle puzzle, p. 572.

Hoffmann.  1893.  Chap. VII, no. 15, pp. 290 & 298.  Chap. X, no. 19: The four wine glasses, pp. 344 & 381.

Loyd.  Problem 34: War‑ships at anchor.  Tit‑Bits 32 (22 May  &  12 Jun 1897) 135  &  193.  Place four warships equidistantly so that if one is attacked, the others can come to assist it.  Solution is a tetrahedron of points on the earth's oceans.

Pearson.  1907.  Part III, no. 77: Three squares, p. 77.  Make three squares with nine matches.  Solution is a triangular prism!

Williams.  Home Entertainments.  1914.  Tricks with matches: To form four triangles with six matches, p. 106.

Blyth.  Match-Stick Magic.  1921.  Four triangle puzzle, p. 23.  Make four triangles with six matchsticks.

King.  Best 100.  1927.  No. 59, pp. 24 & 53.  = Foulsham's no. 20, pp. 8 & 12.  Use six matches to make four triangles.

 

          6.AO.2.        PLACE AN EVEN NUMBER ON EACH LINE

 

          See also section 6.T.

          Sometimes the diagonals are considered.

 

Leske.  Illustriertes Spielbuch für Mädchen.  1864? 

Prob. 564-31, pp. 254 & 396.  From a  6 x 6  array, remove  6  to leave an even number in each row.  (The German 'Reihe' can be interpreted as row or column or both.)  If we consider this in the first quadrant with coordinates going from 1 to 6, the removed points are:  (1,2), (1,3), (2,1), (2,2), (6,1), (6,3).  The use of the sixth column is peculiar and has the effect of making both diagonals odd, while the more usual use of the third column would make both diagonals even.

Prob. 583-5, pp. 285 & 403: Von folgenden 36 Punkten sechs zu streichen.  As above, but each file ('Zeile') in 'all four directions' has four or six points.  Deletes:  (1,1), (1,2), (2,2), (2,3), (6,1), (6,3)  which makes one diagonal even and one odd.

Mittenzwey.  1879?  Prob. 177, pp. 33 & 82.  Given a  4 x 4  array, remove 6 to leave an even number in each row and column.  Solution removes a  2 x 3  rectangle from a corner.

Hoffmann.  1893. 

Chap. VI, no. 22: The thirty‑six puzzle, pp. 271 & 285.  Place 30 counters on a  6 x 6  board so each horizontal and each vertical line has an even number.  Solution places the six blanks in a  3 x 3  corner in the obvious way.  This also makes the diagonals have even numbers. 

Chap. VI, no. 23: The "Five to Four" puzzle, pp. 272 & 285.  Place 20 counters on a  5 x 5  board subject to the above conditions.  Solution puts blanks on the diagonal.  This also makes the diagonals have even number.

Dudeney.  The puzzle realm.  Cassell's Magazine ?? (May 1908) 713-716.  The crack shots.  10 pieces in a  4 x 4  array making the maximal number of even lines _ counting diagonals and short diagonals _ with an additional complication that pieces are hangine in in vertical strings.  The picture is used in AM, prob. 270.

Loyd.  Cyclopedia.  1914.  The jolly friar's puzzle, pp. 307 & 380.  (= MPSL2, no. 155, pp. 109 & 172.  = SLAHP: A shifty little problem, pp. 64 & 110.)  10 men on a  4 x 4  board _ make a maximal number of even rows, including diagonals and short diagonals.  This is a simplifications of Dudeney, 1908.

King.  Best 100.  1927.  No. 72, pp. 29 & 56.  As in Hoffmann's No. 22, but specifically asks for even diagonals as well.

Rudin.  1936.  No. 151, pp. 53-54 & 111.  Place 12 counters on a  6 x 6  board with two in each row, column and main diagonal.

Adams.  Puzzle Book.  1939.  Prob. C.179: Even stars, pp. 169 & 193.  Same as Loyd.

Obermair.  Op. cit. in 5.Z.1.  1984.  Prob. 37, pp. 38 & 68.  52 men on an  8 x 8  board with all rows, columns and diagonals (both long and short) having an even number.

 

          6.AP. DISSECTIONS OF A TETRAHEDRON

 

          6.AP.1.         TWO PIECES

 

Richard A. Proctor.  Our puzzles;  Knowledge 10 (Feb 1887) 83  &  Solutions of puzzles;  Knowledge 10 (Mar 1887) 108-109.  "Puzzle XIX.  Show how to cut a regular tetrahedron (equilateral triangular pyramid) so that the face cut shall be a square: also show how to plug a square hole with a tetrahedron."  Solution shows the cut clearly.

Edward T. Johnson.  US Patent 2,216,915 _ Puzzle.  Applied 26 Apr 1939;  patented 8 Oct 1940.  2pp + 1p diagrams.  Described in S&B, p. 46.

E. M. Wyatt.  Wonders in Wood.  Op. cit. in 6.AI.  1946.  Pp. 9 & 11: the tetrahedron or triangular pyramid.  P. 9 is reproduced in S&B, p. 46.

Donovan A. Johnson.  Paper Folding for the Mathematics Class.  NCTM, 1957, p. 26: Pyramid puzzle.  Gives instructions for making the pieces from paper.

Claude Birtwistle.  Editor's footnote.  MTg 21 (Winter 1962) 32.  "The following interesting puzzle was given to us recently."

Birtwistle.  Math. Puzzles & Perplexities.  1971.  Bisected tetrahedron, pp. 157-158.  Gives the net so one can make a drawing, cut it out and fold it up to make one piece. 

 

          6.AP.2.         FOUR PIECES

 

          These dissections usually also work with a tetrahedron of spheres and hence these are related to ball pyramid puzzles, 6.AZ.

          The first version I had in mind dissects each of the two pieces of 6.AP.1 giving four congruent rhombic pyramids.  Alternatively, imagine a tetrahedron bisected by two of its midplanes, where a midplane goes halfway between a pair of opposite edges.  This puzzle has been available in various versions since at least the 1970s, including one from Stokes Publishing Co., 1292 Reamwood Avenue, Sunnyvale, California, 94089, USA., but I have no idea of the original source.  The same pieces are part of a more complex dissection of a cube, PolyPackPuzzle, which was produced by Stokes in 1996.  (I bought mine from Key Curriculum Press.)

          In 1997, Bill Ritchie, of Binary Arts, sent a quadrisection of the tetrahedron that they are producing.  Each piece is a hexahedron.  The easiest way to describe it is to consider the tetrahedron as a pile of spheres with four on an edge and hence  20  altogether.  Consider a planar triangle of six of these spheres with three on an edge and remove one vertex sphere to produce a trapezium (or trapezoid) shape.  Four of these assemble to make the tetrahedron.  Writing this has made me realise that Ray Bathke has made and sold these  5-sphere  pieces as Pyramid 4 for a few years.  However, the solid pieces used by Binary Arts are distinctly more deceptive.

          Len Gordon produced another quadrisection of the  20  sphere tetrahedron                       0      0

using the planar shape at the right.  This was c1980??                                                                  0  0  0 

 

David Singmaster.  Sums of squares and pyramidal numbers.  MG 66 (No. 436) (Jun 1982) 100-104.  Consider a tetrahedron of spheres with  2n  on an edge.  The quadrisection described above gives four pyramids whose layers are the squares  1, 4, ..., n2.  Hence  four times the sum of the first  n  squares is the tetrahedral number for  2n,  i.e. 4 [1 + 4 + ... + n2]  =  BC(n+2, 3). 

 

          6.AQ. DISSECTIONS OF A CROSS,  T  OR  H

          The usual dissection of a cross has two diagonal cuts at  45o  to the sides and passing through two of the reflex corners of the cross and yielding five pieces.  The central piece is six-sided, looking like a rectangle with its ends pushed in.  Depending on the relative lengths of the arms, head and upright of the cross, the other pieces may be isosceles right triangles or right trapeziums.  Removing the head of the cross gives the usual dissection of the  T  into four pieces _ then the central piece is five-sided.  Sometimes the central piece is split in halves.  Occasionaly the angle of the cuts is different than  45o.  Dissections of an  H  have the same basic idea of using cuts at  45o  _ the result can be a bit like two  Ts  with overlapping stems and the number of pieces depends on the relative size and positioning of the crossbar of the  H  _ see: Rohrbough.

 

S&B, pp. 20‑21, show several versions.  They say that crosses date from early 19C.  They show a 6‑piece Druid's Cross, by Edwards & Sons, London, c1855.  They show several  T‑puzzles _ they say the first is an 1903 advertisement for White Rose Ceylon Tea, NY _ but see 1898 below.  They also show some  H‑puzzles.

 

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. 4 is "Analysis of the Essay of Games".  F. 4.v has a cross cut into 5 pieces in the usual way.

Endless Amusement II.  1826?  Prob. 30, p. 207.  Usual five piece cross.  The three small pieces are equal.

Crambrook.  1843.  P. 4.

No. 10: Five pieces to form a Cross.

No. 11: The new dissected Cross.

          Without pictures, I cannot tell what dissections are used??

Boy's Treasury.  1844.  Puzzles and paradoxes, no. 2, pp. 424 & 428.  Usual five piece cross, very similar to Endless Amusement.  Three pieces of fig. 2.

Family Friend 2 (1850) 58 & 89.  Practical Puzzle _ No. II.  = Illustrated Boy's Own Treasury, 1860, No. 32, pp. 401 & 440.  Usual five piece cross to "form that which, viewed mentally, comforts the afflicted."  Three pieces of fig. 1.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Mechanical puzzles, no. 7, p. 178-179 (1868: 189).  Five piece dissection of a cross, but the statement of the problem doesn't say which piece to make multiple copies of.

Magician's Own Book.  1857.  Prob. 17: The cross puzzle, pp. 272 & 295.  Usual 5 piece cross, essentially identical to Family Friend, except this says to "form a cross."  = Book of 500 Puzzles, 1859, prob. 17, pp. 86 & 109.  = Boy's Own Conjuring Book, 1860, prob. 16, pp. 234 & 258. 

Charades, Enigmas, and Riddles.  1859?: prob. 33, pp. 60 & 66;  1862?: prob. 577, pp. 108 & 156.  Usual five piece cross, showing all five pieces.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 584-12, pp. 288 & 406: Ein Kreuz.  Begins as the usual five piece cross, but the central piece is then bisected into two mitres and the base has two bits cut off to give an eight piece puzzle.

Frank Bellew.  The Art of Amusing.  1866.  Op. cit. in 5.E.  1866: pp. 239-240;  1870: pp. 236‑238.  Usual five piece cross.

Elliott.  Within‑Doors.  Op. cit. in 6.V.  1872.  Chap. 1, no. 1: The cross puzzle, pp. 27 & 30.  Usual five piece cross, but instructions say to cut three copies of the wrong piece.

Mittenzwey.  1879?  Prob. 213, pp. 37 & 87.  10 piece dissection of a cross obtained by further dissecting the usual five pieces.

Lemon.  1890.  A card board puzzle, no. 33, pp. 8 & 98.  Usual five piece cross.

Hoffmann.  1893.  Chap. III, no. 12: The Latin cross puzzle, pp. 93 & 126.  As in Indoor & Outdoor.  Photo in Hordern, p. 59, showing Druid's Cross Puzzle.

Lash, Inc. _ Clifton, N.J. _ Chicago, Ill. _ Anaheim, Calif.  T  Puzzle.  Copyright Sept. 1898.  4‑piece  T  puzzle to be cut out from a paper card, but the angle of the cuts is about  35o  instead of  45o  which makes it less symmetric and less confusing than the more common version.  The resulting  T  is somewhat wider than usual, being about  16%  wider than it is tall.  It advertises:  Lash's Bitters  The Original Tonic Laxative.  Photocopy sent by Slocum.

Benson.  1904.  The cross puzzle, pp. 191‑192.  Usual 5 piece version.

Arthur Mee's Children's Encyclopedia 'Wonder Box'.  The Children's Encyclopedia appeared in 1908, so this is probably 1908 or soon thereafter.  Usual 5 piece cross.

Wehman.  New Book of 200 Puzzles.  1908.  The cross puzzle, p. 17.  Usual 5 piece version.

A. Neely Hall.  Op. cit. in 6.F.5.  1918.  The T‑puzzle, pp. 19‑20.  "A famous old puzzle ...."  Usual 4‑piece version, but with long arms.

Western Puzzle Works, 1926 Catalogue.  No. 1394: Four pieces to form Letter  T.  The notched piece is less symmetric than usual.

Collins.  Book of Puzzles.  1927.  The crusader's cross puzzle, pp. 1-2.  The three small pieces are equal.

A. F. Starkey.  The  T  puzzle.  Industrial Arts and Vocational Education 37 (1938) 442.  "An interesting novelty ...."

Rohrbough.  Puzzle Craft.  1932.  The "H" Puzzle, p. 23.  Very square  H  _ consider a  3 x 3  board with the top and bottom middle cells removed.  Make a cut along the main diagonal and two shorter cuts parallel to this to produce four congruent isosceles right triangles and two odd pentagons. 

See Rohrbough in 6.AS.1 for a very different  T  puzzle.

 

          6.AR. QUADRISECTED SQUARE PUZZLE

 

          This is usually done by two perpendicular cuts through the centre.  A dissection proof of the Theorem of Pythagoras described by Henry Perigal (Messenger of Mathematics 2 (1873) 104) uses the same shapes.  For sides  a < b,  the perpendicular cuts are done in the square of side  b  so they meet the sides at distance  (b-a)/2  from a corner.  These pieces then fit around the square of side  a  to make a square of side  c.  Perigal is ??NYS, but described in Elisha Scott Loomis; The Pythagorean Proposition; 2nd ed., NCTM, 1940, pp. 104-105 & 214, where some earlier possible occurrences are mentioned.

          The pieces make a number of other different shapes.

 

Crambrook.  1843.  P. 4, no. 17: Four pieces to form a Square.  This might be the dissection being considered here??

A. Héraud.  Jeux et Récréations Scientifiques _ Chimie, Histoire Naturelle, Mathématiques.  (1884);  Baillière, Paris, 1903.  Pp. 303‑304: Casse‑tête.  Uses two cuts which are perpendicular but are not through the centre.  He claims there are  120  ways to try to assemble it, but his mathematics is shaky _ he adds the numbers of ways at each stage rather than multiplying!  Also, as Strens notes in the margin of his copy (now at Calgary), if the crossing is off-centre, then many of the edges have different lengths and the number of ways to try is really only one.  Actually, I'm not at all sure what the number of ways to try is _ Héraud seems to assume one tries each orientation of each piece, but some intelligence sees that a piece can only fit one way beside another.

Handy Book for Boys and Girls.  Op. cit. in 6.F.3.  1892.  P. 14: The divided square puzzle.  Crossing is off-centre.

Tom Tit, vol 3.  1893.  Carré casse-tête, pp. 179-180.  = K, no. 26: Puzzle squares, pp. 68‑69.  = R&A, Puzzling squares, p. 99.  Not illustrated, but described:  cut a square into four parts by two perpendicular cuts, not necessarily through the centre.

A. B. Nordmann.  One Hundred More Parlour Tricks and Problems.  Wells, Gardner, Darton & Co., London, nd [1927 _ BMC].  No. 77: Pattern making, pp. 69-70 & 109.  Make five other shapes.

Adams.  Puzzle Book.  1939.  Prob. C.12: The broken square, pp. 125 & 173.  As above, but notes that the pieces also make a square with a square hole.

 

          6.AS.  DISSECTION OF SQUARES INTO A SQUARE

 

Lorraine Mottershead.  Investigations in Mathematics.  Blackwell, Oxford, 1985.  P. 102 asserts that dissections of squares to various hexagons and heptagons were known c1800 while square to rectangle dissections were known to Montucla _ though she illustrates the latter with examples like 6.Y, she must mean 6.AS.5.

 

          6.AS.1.         TWENTY 1, 2, Ö5 TRIANGLES MAKE A SQUARE

                                                  OR FIVE EQUAL SQUARES TO A SQUARE

 

          The basic puzzle has been varied in many ways by joining up the 20 triangles into various shapes, but I haven't attempted to consider all the modern variants.  A common form is a square with a skew  #  in it, with each line joining a corner to the midpoint of an opposite side, giving the 9 piece version.  This has four of the squares having a triangle cut off.  For symmetry, it is common to cut off a triangle from the fifth square, giving 10 pieces, though the assembly into one square doesn't need this.  See Les Amusemens for details.

          If the dividing lines are moved a bit toward the middle and the central square is bisected, we get a 10 piece puzzle, having two groups of four equal pieces and a group of two equal pieces, called the Japan square puzzle.  I have recently noted the connection of this puzzle with this section, so there may be other examples which I have not previously paid attention to _ see:  Magician's Own Book,  Book of 500 Puzzles,  Boy's Own Conjuring Book,  Illustrated Boy's Own Treasury,  Landells,  Hanky Panky,  Wehman.

 

Les Amusemens.  1749.  P. xxxii.  Consider five  2 x 2  squares.  Make a cut from a corner to the midpoint of an opposite side on each square.  This yields five  1, 2, Ö5  triangles and five pieces comprising three such triangles.  The problem says to make a square from five equal squares.  So this is the 10 piece version.

Vyse.  Tutor's Guide.  1771?  Prob. 6, p. 317 & Key p. 357.  2 x 10  board to be cut into five pieces to make into a square.  Cut into a  2 x 2  square and four  2, 4, 2Ö5  triangles.

Ozanam‑Montucla.  1778.  Avec cinq quarrés égaux, en former un seul.  Prob. 18 & fig. 123, plate 15, 1778: 297;  1803: 292-293;  1814: 249-250;  1840: 127.  9 piece version.  Remarks that any number of squares can be made into a square _ see 6.AS.5.

Catel.  Kunst-Cabinet.  1790.

Das mathematische Viereck, pp. 10-11 & fig. 15 on plate I.  10 piece version with solution shown.  Notes these make five squares.

Das grosse mathematische Viereck, p. 11 & fig. 14 on plate I.  Cut the larger pieces to give five more  1, 2, Ö5  triangles and five  Ö5, Ö5, 2  triangles.  Again notes these make five squares.

Guyot.  Op. cit. in 6.P.2.  1799.  Vol. 2: première récréation: Cinq quarrés éqaux étant sonnés, en former un seul quarré, pp. 40‑41 & plate 6, opp. p. 37.  10 piece version.  Suggests cutting another triangle off each square to give 10 triangles and 5 parallelograms.

Bestelmeier.  1801.  Item 629: Die 5 geometrisch zerschnittenen Quadrate, um aus 5 ein einziges Quadrat zu machen.  As in Les Amusemens.  S&B say this is the first appearance of the puzzle.  Only shown in a box with one small square visible.

Jackson.  Rational Amusement.  1821.  Geometrical Puzzles.

No. 8, pp. 25 & 84 & plate I, fig. 5, no. 1.  = Vyse.

No. 10, pp. 25 & 84-85 & plate I, fig. 7, no. 1.  Five squares to one.  Nine piece version.

Minguét.  Engaños.  1822.  Pp. 145-146.  Not in 1733 or 1755 eds.  9  piece version.  Also a  15  piece version where triangles are cut off diagonally opposite corners of each small square leaving parallelogram pieces as in Guyot.

Manuel des Sorciers.  1825.  Pp. 201-202, art. 18.  ??NX  Five squares to one _ usual 10 piece form and 15 piece form as in Guyot.

Endless Amusement II.  1826? 

[1837 only]  Prob. 35, p. 212.  20 triangles to form a square.

Prob. 37, p. 215.  10 piece version.

Boy's Own Book.  The square of triangles.  1828: 426;  1828-2: 430;  1829 (US): 222;  1855: 576;  1868: 676.  Uses 20 triangles cut from a square of wood.

Nuts to Crack IV (1835), no. 195.  20 triangles _ part of a long section: Tricks upon Travellers.  The problem is used as a wager and the smart-alec gets it wrong.

The Riddler.  1835.  The square of triangles, p. 8.  Identical to Boy's Own Book, but without illustration, some consequent changing of the text, and omitting the last comment.

Crambrook.  1843.  P. 4.

No. 7: Egyptian Puzzle.  Probably the 10 piece version as in Les Amusemens.  See S&B below, late 19C.  Check??

No. 23: Twenty Triangles to form a Square.  Check??

Boy's Treasury.  1844.  Puzzles and paradoxes, no. 5, pp. 425 & 429.  "Cut twenty triangles out of ten square pieces of wood;" and make a square.  The solution shows that he means 'out of five square pieces'.  The phrasing is very similar to Boy's Own Book.

Magician's Own Book.  1857. 

How to make five squares into a large one without any waste of stuff, p. 258.  9 piece version.

Prob. 29: The triangle puzzle, pp. 276 & 298.  Identical to Boy's Treasury.

Prob. 35: The Japan square puzzle, pp. 277 & 300.  Make two parallel cuts and then two perpendicular to the first two so that a square is formed in the centre.  This gives a 9 piece puzzle, but here the central square is cut by a vertical through its centre to give a 10 piece puzzle.  = Landells, Boy's Own Toy-Maker, 1858, pp. 145-146. 

Charles Bailey (manufacturer in Manchester, Massachusetts).  1858.  An Ingenious Puzzle for the Amusement of Children ....  The 10 pieces of Les Amusemens, with 19 shapes to make, a la tangrams.  Sent by Jerry Slocum _ it is not clear if there were actual pieces with the printed material.

The Sociable.  1858. 

Prob. 10: The protean puzzle, pp. 289 & 305-306.  Cut a  5 x 1  into 11 pieces to form eight shapes, e.g. a Greek cross.  It is easier to describe the pieces if we start with a  10 x 2.  Then three squares are cut off.  One is halved into two  1 x 2  rectangles.  Two squares have two  1, 2, Ö5  triangles cut off leaving triangles of sides  2, Ö5, Ö5.  The remaining double square is almost divided into halves each with a  1, 2, Ö5  triangle cut off, but these two triangles remain connected along their sides of size  1,  thus giving a  4, Ö5, Ö5  triangle and two trapeziums of sides  2, 2, 1, Ö5.  = Book of 500 Puzzles, 1859, prob. 10, pp. 7 & 23-24.

Prob. 42: The mechanic's puzzle, pp. 298 & 317.  Cut a  10 x 2  in five pieces to make a square, as in Vyse.  = Book of 500 Puzzles, 1859, prob. 16, pp. 16 & 35.

Book of 500 Puzzles.  1859.

Prob. 10: The protean puzzle, pp. 7 & 23-24.  As in The Sociable.

Prob. 42: The mechanic's puzzle, pp. 16 & 35.  As in The Sociable.

How to make five squares into a large one without any waste of stuff, p. 72.  Identical to Magician's Own Book.

Prob. 29: The triangle puzzle, pp. 90 & 113.  Identical to Boy's Treasury.

Prob. 35: The Japan square puzzle, pp. 91 & 114. 

Indoor & Outdoor.  c1859.  Part II, prob. 11: The mechanic's puzzle, pp. 130-131.  Identical to The Sociable.

Boy's Own Conjuring Book.  1860.

Prob. 28: The triangle puzzle, pp. 238 & 262.  Identical to Boy's Treasury and Magician's Own Book.

Prob. 34: The Japan square puzzle, pp. 240 & 264.  Identical to Magician's Own Book.

Illustrated Boy's Own Treasury.  1860.

Prob. 9, pp. 396 & 437.  [The Japan square puzzle.]  Almost identical to Magician's Own Book.

Optics: How to make five squares into a large one without any waste of stuff, p. 445.  Identical to Book of 500 Puzzles, p. 72.

Leske.  Illustriertes Spielbuch für Mädchen.  1864? 

Prob. 174, pp. 87-88.  Nine piece version.

Prob. 584-6, pp. 287 & 405.  Ten piece version of five squares to one.

Hanky Panky.  1872.

The puzzle of five pieces, p. 118.  9 piece version.

Another [square] of four triangles and a square, p. 120.  10 x 2  into five pieces to make a square.

[Another square] of ten pieces, pp. 121-122.  Same as the Japan square puzzles in Magician's Own Book.

[Another square] of twenty triangles, p. 122.  Similar to Boy's Treasury, but with no diagram and less text, making it quite cryptic.

Mittenzwey.  1879?  Prob. 200‑204, pp. 35 & 84‑85.  10 pieces as in Les Amusemens used to make a square and 9 other shapes, e.g. a  4 x 5  rectangle.

S&B, pp. 11 &