Sources page biographical material

6.AJ.5. IMPOSSIBLE CRATE.
This is like a Necker Cube where all the edges are drawn as wooden slats in an impossible configuration.
Escher. Man with Cuboid, which is essentially a detail from Belvedere, both 1958, are apparently the first examples of this impossible object.

Chuck Mathias. Letter Mad Magazine 95 (Jun 1965) 2. Gives an impossible crate.

Jerry Andrus developed his actual model in 1981 and it appeared on the cover of Omni in 1981. But Al Seckel's exhibition says the first physical example was The Feemish Crate, due to C. F. Cochran.

Seckel, 2002a, figs. 27 A&B, pp. 36-37 (= 2002b, figs. 169 A&B, pp. 186-187), shows and discusses Andrus' crate from two viewpoints.

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, Anon., Bullivant, Goldston, Loyd (1914), Blyth, Abraham, Hedges, Evans, Doubleday - 1, Piggins & Eley

4x4 problem: King, Abraham, M. Adams, Evans, Depew, Meyer, Ripley's,

Queen's tours: Loyd (1867, 1897, 1914), Loyd Jr.

Bishop's tours: Dudeney (1932), Doubleday - 2, Obermair

Rook's tours: Loyd (1878), Proctor, Loyd (1897), Bullivant, Loyd (1914), Filipiak, Hartswick, Barwell, Gardner, Peters, Obermair

Other versions: Prout, Doubleday - 1

Trick solutions: Fixx, Adams, Piggins, Piggins & Eley

Thanks to Heinrich Hemme for pointing out Fixx, which led to adding most of the material on trick solutions.
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. 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.

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??

Anon. Prob. 67. Hobbies 31 (No. 782) (8 Oct 1910) 39 & (No. 785) (29 Oct 1910) 94. 3 x 3 lattice in 4 lines "brought under my notice some time back".

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.

Abraham. 1933. Probs. 101, 102, 103, pp. 49 & 66 (30 & 118). 3 x 3, 4 x 4 and 6 x 6 cases.

The Bile Beans Puzzle Book. 1933. No. 4: The puzzled milkman. 3 x 3 array in four lines.

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.

M. 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.

Ripley's Puzzles and Games. 1966. Pp. 72-73, item 2. 4 x 4 cases with closed solution symmetric both horizontally and vertically.

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.

Doubleday - 1. 1969. Prob. 60: Test case, pp. 75 & 167. = Doubleday - 4, pp. 83-84. Two 3 x 3 arrays joined at a corner, looking like the Fore and Aft board (cf 5.R.3), to be covered in a minimum number of segments. He does it in seven segments by joining two 3 x 3 solutions.

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 - 2. 1971. Path finder, pp. 95-96. Bishop's corner to corner path, same as Dudeney, 1932.

James F. Fixx. More Games for the Superintelligent. (Doubleday, 1972); Muller, (1977), 1981. 6. Variation on a variation, pp. 31 & 87. Trick solution in three lines, assuming points of finite size.

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.

James L. Adams. Conceptual Blockbusting. Freeman, 1974, pp. 16-22. 3rd ed., (A-W, 1986), Penguin, 1987, pp. 24-33. Trick solution of 3 x 3 case in three lines, assuming points of finite size, which he says was submitted anonymously when he and Bob McKim used the puzzle on an ad for a talk on problem-solving at Stanford. Also describes a version using paperfolding to get all nine points into a line. The material is considerably expanded in the 3rd ed. and adds several new versions. From the references in Piggins and Eley, it seems that these all appeared in the 2nd ed of 1979 -- ??NYS.

Cut out the 3 x 1 parts and tape them into a straight line.

Take the paper and roll it to a cylinder and then draw a slanting line on the cylinder which goes through all nine, largish, points.

Cut out bits with each point on and skewer the lot with a pencil.

Place the paper on the earth and draw a line around the earth to go through all nine points. One has to assume the points have some size.

Wodge the paper, with large dots, into a ball and stick a pencil through it. Open up to see if you have won -- if not, try again!

Use a very fat line, i.e. as thick as the spacing between the edges of the array.

David J. Piggins. Pathological solutions to a popular puzzle. JRM 8:2 (1975-76) 128-129. Gives two trick solutions.

Three parallel lines, since they meet at infinity.

Put the figure on the earth and use a slanting line around the earth. This works in the limit, but otherwise requires points of finite size, a detail that he doesn't mention.

No references for these versions.

David J. 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. They cite Piggins' solution with three parallel lines. They say that Gardner sent them the trick solution in 1973 and then cite Adams, 1979. They give solutions using points of different sizes, getting both three and two segment solutions and mention a two segment version that depends on the direction of view. They then give the solution on a sphere, citing Adams, 1979, and Piggins. They give several further versions using paper folding, including putting the surface onto a twisted triangular prism joined at the ends to make the surfaces into a Möbius strip -- Zeeman calls this a umbilical bracelet or a Möbius bar.

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.]

Nob Yoshigahara. Puzzlart. Tokyo, 1992. Section: The wisdom of Solomon, pp. 40-47, abridged from an article by Solomon W. Golomb in Johns Hopkins Magazine (Oct 1984). Classic 3 x 3 problem. For the 4 x 4 case: 1) find four closed paths; 2), says there are about 30 solutions and gives 19 beyond the previous 4. Find the unique 5 segment closed path on the 3 x 4. Gives 3 solutions on 5 x 5. 10-segment solution on 6 x 6 which stays on the board. Loyd's 1867? Queen's circuit. Queen's circuit on 7 x 7, attributed to Dudeney, though my earliest entry is Loyd, 1914 -- ??CHECK.

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.

N. J. Chignell. Note 1031: A difficult converse. MG 16 (No. 219) (Jul 1932) 200-202. [The author's name is omitted in the article but appears on the cover.] 'Three fairly simple proofs', due to: M. J. Newell; J. Travers, improving J. H. Doughty, based on material in Lady's and Gentleman's Diary (1859) 87-88 & (1860) 84-86; Wm. Mason, found by Doughty, in Lady's and Gentleman's Diary (1860) 86.

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. A later source says this turned out to be identical to Lehmus' original proof!

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.

Anonymous [R. P.] Morley's trisector theorem. Eureka 16 (Oct 1953) 6-7. Short proof, working backward from the equilateral triangle.

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 determined by Liu, so the excluded volume is ⅓ 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 Chōchū. 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; Carroll; Criton

( 9, 9, 3): Carroll; Bridges; 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; Young World; Brooke; Putnam; Criton

(10, 5, 4): The Sociable; Book of 500 Puzzles; Carroll; Hoffmann; Dudeney (1908); Wehman; Williams; Dudeney (1917); Blyth; King; Rudin; Young World; Hutchings  & Blake; Putnam

(10, 10, 3): Sylvester

(11, 11, 3): The Sociable; Book of 500 Puzzles; Wehman

(11, 12, 3): Hoffmann; Williams; Young World

(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); Secret Out; Illustrated Boy's Own Treasury;

(12, 6, 4): Endless Amusement II; The Sociable; Book of 500 Puzzles; Boy's Own Book; Cassell's; Hoffmann; Wehman; Rudin; Criton

(12, 7, 4) -- Trick version of a 3 x 3 square with doubled diagonal: Secret Out; 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; H. D. Northrop; 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

(17, 7, 5): Ripley's

(18, 18, 4): Macmillan

(19, 19, 4): Criton

(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, 12, 5): trick method: Doubleday - 3

(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): Secret Out; 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, 18, 5): Bridges

(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); Secret Out; Illustrated Boy's Own Treasury; Hoffmann (1876); Mittenzwey; Hoffmann (1893), nos. 8 & 9; Pearson; Home Book ....; Doubleday - 3. These could also be considered as in 7.Q.2 or 7.Q.

A different type of configuration problem is considered by Shepherd, 1947.

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). = New Sphinx, c1840, p. 135.

Prob. 14, p. 197. (12, 6, 4). = New Sphinx, c1840, p. 135.

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 Own Book. 1843 (Paris): 438 & 442, no. 15: "Is it possible to place twelve pieces of money in six rows, so as to have four in each row?" I. e. (12, 6, 5). = Boy's Treasury, 1844, pp. 426 & 429, no. 13. = de Savigny, 1846, pp. 355 & 358, no. 11.

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. Solution is a 3 x 4 array with three extra trees halfway between the points of the middle line of four.

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 Secret Out, 1859 & 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.

The Secret Out. 1859.

To place twelve Cards in such a manner that you can count Four in every direction, p. 90. (12, 7, 4) trick of a 3 x 3 array with doubling along a diagonal. 'Every direction' must refer to just the rows and columns, but one diagonal also works.

The magical arrangement, pp. 381-382 = The square of counters, (UK) p. 9. (12, 4, 5) -- trick version. Same as Family Friend & Illustrated Boy's Own Treasury, prob. 13.

The Sphynx, pp. 385-386. (21, 30, 3). = Hoffmann, no. 15.

Charades, Enigmas, and Riddles. 1860: prob. 13, pp. 58 & 61; 1862: prob. 13, pp. 133 & 139; 1865: prob. 557, pp. 105 & 152. (9, 10, 3). (The 1862 and 1865 have 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).

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). Same as Secret Out.

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. The same problem is in Carroll-Wakeling, prob. 1: Cakes in a row, pp. 1-2 & 63, but undated and the answer mentions the possibility of stacking the counters.

(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 which may be another periodical. Shulman then cites Jackson and Dudeney. Carroll-Wakeling, prob. 2: More cakes in a row, pp. 3 & 63, gives the problems (9, 8, 3), (9, 9, 3), (9, 10, 3), undated.

Mittenzwey. 1880.

Prob. 151, pp. 31 & 83; 1895?: 174, pp. 36 & 85; 1917: 174, pp. 33 & 82. (6, 3, 3) in three ways.

Prob. 152, pp. 31 & 83; 1895?: 175, pp. 36 & 85; 1917: 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 Secret Out & Hoffmann (1876).

Prob. 153, pp. 31 & 83; 1895?: 176, pp. 36 & 85; 1917: 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 = Hoffmann-Hordern, pp. 174-182, with photo.

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). Photo on p. 177 shows L'Embarras du Brigadier, by Mauclair-Dacier, 1891 1900, which has a board with a 7 x 6 array of holes and 12 pegs. The horizontal spacing seems closer than the vertical spacing.

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 as in Secret Out, Hoffmann (1876) & Mittenzwey.

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, which is the same as in The Secret Out.

Clark. Mental Nuts. 1897, no. 19: The apple orchard; 1904, no. 91: The lovers' grove. (19, 9, 5). 1897 just has "Place an orchard of nineteen trees so as to have nine rows of five trees each." 1904 gives a poem.

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.

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.

H. D. Northrop. Popular Pastimes. 1901. No. 11: The tree puzzle, pp. 68 & 73. = The Sociable, no. 29.

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.

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).

G. C. Shephard. A problem in orchards. Eureka 9 (Apr 1947) 11-14. Given k points in n dimensions, the general problem is to draw N(k, n) hyperplanes to produce k regions, each containing one point. The most common example is k = 7, n = 2, N = 3. [See Section 5.Q for determining k as a function of n and N.] The author investigates the question of determining the possible locations of the seventh point given six points. He gives a construction of a set T such that being in T is necessary and sufficient for three such lines to exist.

J. Bridges. Potter's orchard. Eureka 11 (Jan 1949) 30 & 12 (Oct 1949) 17. Start with an orchard (9, 9, 3). Add 16 trees to make (25, 18, 5). The nine trees are three points in a triangle, with the three midpoints of the sides and the three points halfway between these. Six of the new trees are one third of the way along the sides of the original triangle; another six are one third of the way along the lines joining the midpoints of the original triangle; one point is the centre of the original triangle and the last three are easily seen.

W. Leslie Prout. Think Again. Frederick Warne & Co., London, 1958. Thirteen rows of three, pp. 45 & 132. (11, 13, 3).

Young World. c1960. Pp. 10-11.

Three coin lines. (9, 10, 3).

Five coin lines. (10, 5, 4).

Eleven coin trick. (11, 12, 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).

R. L. Hutchings & J. D. Blake. Problems drive 1962. Eureka 25 (Oct 1962) 20-21 & 34-35. Prob. F. (10, 5, 4) with points in the centres of cells of a chess board. Actually only needs a 7 x 7 board.

Ripley's Puzzles and Games. 1966. Pp. 18-19, item 4. (17, 7, 5).

Doubleday - 3. 1972. Count down, pp. 125-126. Start with a 4 x 4 array of coins. Add four coins so that each row, column and diagonal has the same number. Solution doubles the coins in the 1, 3, 4, 2 positions in the rows.

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 restrictive kind of arrangement of b lines whose intersections 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
I am adding the problem of making three squares with nine matchsticks here a it uses the same thought process -- see Mittenzwey and see the extended discussion at Anon., 1910.
Pacioli. De Viribus. c1500. Ff. 191r - 192r. LXXX. Do(cumento). commo non e possibile piu ch' tre ponti o ver tondi spere tocarse in un piano tutti (how it is not possible for more than three points or discs or spheres to all touch in a plane). = Peirani 252-253. Says you can only get three discs touching in the plane, but you can get a fourth so they are all touching by making a pyramid.

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. 1880.

Prob. 161, pp. 32 & 84; 1895?: 184, pp. 37 & 86; 1917: 184, pp. 34 & 83. Use six sticks to make four congruent triangles. Solution is a rectangle (should be a square) with its diagonals, but then two of the sticks have to be longer than the others.

Prob. 163, pp. 32 & 84; 1895?: 186 & 194, pp. 37 & 86-87; 1917: 186 & 194, pp. 34 & 83-84. Use six equally long sticks to make four congruent triangles -- solution is a tetrahedron. The two problems in the 1895? are differently phrased, but identical in content, while the first solution is a picture and the second is a description.

Prob. 171, pp. 33 & 85; 1895?: 195, pp. 38 & 87; 1917: 195, pp. 34 & 84. Use nine equal sticks to make three squares. Solution is three faces of a cube.

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 = Hoffmann-Hordern, pp. 195. Four matches.

Chap. X, no. 19: The four wine glasses, pp. 344 & 381 = Hoffmann-Hordern, pp. 238 239, with photo on p. 239 of a version by Jaques & Son, 1870-1900. I usually solve the second version by setting one glass on top of the other three, but here he wants the centre of the feet of the glasses to be equally spaced and he turns one glass over and places it in the centre of the other three, appropriately spaced.

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.

Parlour Games for Everybody. John Leng, Dundee & London, nd [1903 -- BLC], p. 30. "With 6 matches form 4 triangles of equal size."

Pearson. 1907. Part III, no. 77: Three squares, p. 77. Make three squares with nine matches. Solution is a triangular prism!

Anon. Prob. 66. Hobbies 31 (No. 781) (1 Oct 1910) 2 & (No. 784) (22 Oct 1910) 68. Use nine matches to make three squares. "... the only possible solution" is to make two adjacent squares with seven matches, then bisect each square to produce a third square which overlaps the other two.

I re-invented this problem in Apr 1999 and posted it on NOBNET on 19 Apr 1999. Solution (1) is the idea I had when I made up the puzzle, but various friends gave more examples and then I found solution (3).

(1). Arrange the nine matches to form the following.

_

|_| |_|

| |
Then 4 is a square, 9 is a square and 49 is a square.

(2). Use the matches to form a triangular prism. One may object that this also makes two triangles.

(3). Make three squares forming three faces of a cube, all meeting at one corner. Cf Mittenzwey 171.

(4). Make two adjacent squares with seven of the matches. Now bisect each of the squares with a match parallel to the common edge of the squares. This produces a row of four adjacent half-squares as below. The middle two form a new square. Here one may object that the squares are overlapping.

─── ───

│ │ │ │ │

─── ───

(5). Use the matches to make the figures 0, 1 and 4.

One can use the matches to make squares whose edge is half the match length, but one only needs eight matches to make three squares.

There are other solutions which use the fact that matches have squared off ends and have square cross-section, but these properties do not hold for paper matches torn from a matchbook or for other equivalent objects like toothpicks and hence I don't consider them quite reasonable.

Anon. Prob. 76. Hobbies 31 (No. 791) (10 Dec 1910) 256 & (No. 794) (31 Dec 1910) 318. Make as many triangles as possible with six matches. From the solution, it seems that the tetrahedron was expected with four triangles, but many submitted the figure of a triangle with its altitudes drawn, but only one solver noted that this figure contains 16 triangles! However, if the altitudes are displaced to give an interior triangle, I find 17 triangles!!

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, but it is not always clear what is intended.