Sources page biographical material

6.AY.1. O'BEIRNE'S STEPS
This is a cube dissected into 6 pieces which form 6 cuboids, each of which can be 'staircased' in two ways. There is a 6-cycle through the cuboids, with relative sizes: 12 x 12 x 12, 8 x 12 x 18, 8 x 9 x 24, 12 x 6 x 24, 16 x 6 x 18, 16 x 9 x 12. I have a fine example using six different woods, that had been made for Tom O'Beirne, from Mrs. O'Beirne.
Richard K. Guy. Op. cit. in 5.H.2. 1960. Pp. 151-152 describes O'Beirne's invention.

T. H. O'Beirne. Puzzles and paradoxes -- 9: A six-block cycle for six step-cut pieces. New Scientist 9 (No. 224) (2 Mar 1961) 560-561.

6.AY.2. SWISS FLAG PUZZLE
This appears to be a 7 x 5 flag with a Greek cross X X X X X O O X X X X X

of 5 cells removed from the middle as in the first figure X X X O O O X X X O O

at the right. One has to cut it into two pieces to make a X X O O X X O O O

perfect square. This is done by cutting along a 'staircase' X X X O O O X X X O O

as shown. However, this seems to produce a 5 x 6 flag, X X O O O O O X X O O O

not a square. But there is usually a swindle -- the diagram O O O O O

is not drawn with the unit cells square, but instead the unit

cells are 6/5 as wide as they are tall. Normally the reader would not recognize this and the diagrams are often rather imprecise.

Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co., NY), nd [1895]. P. 5: The flag puzzle. Starts with a square and asks to make a Swiss flag. The square is actually 31.5 mm = 1¼ in on an edge and the flag is 26 x 37 mm = 1 x 1½ in with the Greek cross formed from squares of edge 5 mm = ¼ in, so the areas do not add up! No solution.

Loyd. Problem 4: The Swiss flag puzzle. Tit-Bits 31 (31 Oct 1896) 75 & (21 Nov 1896) 131. c= Cyclopedia, pp. 250 & 373. Swiss flag puzzle with the flag at an angle and a slight wiggle in the edges, so the solution requires an extra cut to make it square.

Loyd. Cyclopedia, 1914, pp. 14 & 341: A Swiss puzzle -- part 2. = MPSL2, no. 144, pp. 101 & 166. = SLAHP, pp. 48 & 102: How was this flag made? Starts with the Swiss flag which is 47 x 30 mm and the Greek cross has cells 6.5 x 5.5 mm, so this is approximately the correct shape to make a square, but the resulting square is not drawn.
6.AZ. BALL PYRAMID PUZZLES
This section is largely based on Gordon's Notes, cf below. See also 6.AP.2 for dissections of a tetrahedron in general. This section is now expanding to consider all polysphere puzzles.

See also S&B, p. 42, which mentions Hein and some other versions.

Piet Hein. Pyramystery. Made by Skjøde of Skjern, Denmark, 1970. With leaflet saying it was "recently invented by Piet Hein.... Responding to numerous requests, the inventor has therefore obliged the many admirers of the puzzle by also inventing its history". He then gives a story about Cheops. Peter Hajek and Jerry Slocum have different examples!!

Hajek's example has four planar rectangular pieces: 1 x 4, 2 x 3, 3 x 2, 4 x 1 rectangles. It is the same as Tut's Tomb -- see below. It has a 4pp English leaflet marked © Copyright Piet Hein 1970.

Slocum's example has 6 planar pieces: 4 3-spheres and 2 4-spheres. The leaflet is 34pp (?? -- Slocum only sent me part of it) with 3pp of instructions in each of 9 languages and then 6pp of diagrams of planar and 3-D problems. It is marked © 1970 Aspila, so perhaps this is a later development from the above?? The story part of the text is very similar to the above, but slightly longer. The pieces make an order 4 tetrahedron or two order 3 tetrahedra or two order 4 triangles and one can also divide them into two groups of three pieces such that one group makes an order 3 tetrahedron, but the other does not.

Advertising leaflet for Pyramystery from Piet Hein International Information Center, ©1976, describes the puzzle as having six pieces.

Mag Nif Inc. Tut's Tomb. c1972. Same as the first Pyramystery.

Akira Kuwagaki & Sadao Takenaka. US Patent 3,837,652 -- Solid Puzzle. Filed: 1 May 1973; patented: 24 Sep 1974. 2pp + 4pp diagrams. Four planar 3-spheres and a 2-sphere to make a square pyramid of edge 3. 11 planar 4-spheres to make an octahedron shape of edge 4. Cites a 1936 Danish patent -- Hein ??NYS

Len Gordon. Perplexing Pyramid. 1974. Makes a edge 4 tetrahedron with 6 planar right-angled pieces: domino; straight and L trominoes; I, L, Y tetrominoes.

Patrick A. Roberts. US Patent 3,945,645 -- Tangential Spheres Geometric Puzzle. Filed: 28 Jun 1976; patented: 29 Nov 1977. 3pp + 3pp diagrams. 8 4-spheres and a 3-sphere to make a tetrahedron of side 5. 5 of the 4-spheres are non-planar.

Robert E. Kobres. US Patent 4,060,247 -- Geometric Puzzle. Filed: 28 Jun 1976; patented: 29 Nov 1977. 1p + 2pp diagrams. 5 pieces which make a 4 x 5 rhomboid or a tetrahedron. Two pieces have the form of a 2 x 3 rhombus; two pieces are 2-spheres and the last piece is the linear 4-sphere.

Len Gordon. Some Notes of Ball Pyramid and Related Puzzles. Revised version, 10 Jul 1986, 14pp. Available from the author, 2737 N. Nordic Lane, Tucson, Arizona, 85716, USA.

Ming S. Cheng. US Patent 4,988,103 -- Geometric Puzzle of Spheres. Filed: 2 Oct 1989; patented: 29 Jan 1991. Front page, 5pp diagrams, 4pp text. A short version is given in Wiezorke, 1996, p. 64. 7 planar 5 spheres to make a tetrahedron; a hexagon with sides 3, 4, 4, 3, 4, 4; an equilateral triangle lacking one vertex.

Bernhard Wiezorke. Puzzling with Polyspheres. Published by the author (Lantzallee 18, D 4000 Düsseldorf 30, Germany), Mar 1990, 10pp.

Bernhard Wiezorke. Compendium of Polysphere Puzzles. (1995); Second Preliminary Edition, as above, Aug, 1996. 64pp, reproducing the short versions of the above patents. Despite Wiezorke's searches, nothing earlier than Hein's 1970 puzzles has come to light.

Torsten Sillke & Bernhard Wiezorke. Stacking identical polyspheres. Part 1: Tetrahedra. CFF 35 (Dec 1994) 11-17. Studies packing of tetrahedra with identical polysphere pieces, with complete results for tetrahedra of edges 4 - 8 and polyspheres of 3, 4, 5 spheres. Some of the impossibility results have only been done by computer, but others have been verified by a proof.

6.BA. CUTTING A CARD SO ONE CAN PASS THROUGH IT
Ozanam. 1725. 1725: vol. IV, prob. 34, pp. 436 437 & fig. 40, plate 12 (14).

Minguet. 1733. Pp. 115-117. (1755: 83-84; 1864: 112-113; not noted in 1822, but it's likely to be at p. 138.) Similar to Ozanam, 1725.

Alberti. 1747. Art. 34, p. 208-209 (110) & fig. 42, plate XI, opp. p. 210 (109). Copied from Ozanam, 1725.

Family Friend 3 (1850) 210 & 241. Practical puzzle -- No. XVII. Shows a 3 inch by 5 inch card. Repeated as Puzzle 15 -- The wonder puzzle in (1855) 339 with solution in (1856) 28.

Magician's Own Book. 1857. Prob. 10: The cardboard puzzle, pp. 269 & 294. Problem shows 3 inch by 5 inch card. Answer calls it "the cut card puzzle". c= Landells, Boy's Own Toy-Maker, 1858, p. 142. = Book of 500 Puzzles, 1859, pp. 83 & 108. = Boy's Own Conjuring Book, 1860, prob. 9, pp. 230 & 256.

Indoor & Outdoor. c1859. Part II, prob. 7: The cardboard puzzle, p. 129. No diagram, so the solution is a bit cryptic.

The Secret Out. 1859. How to Cut a Visiting Card for a Cat to Jump through it, p. 382.

Illustrated Boy's Own Treasury. 1860. No. 27, pp. 400 & 440. Identical to Magician's Own Book, but solution omits the sentence: "A laurel leaf may be treated in the same manner."

Magician's Own Book (UK version). 1871. To cut a card for one to jump through, p. 124. He adds: "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.AD.

Elliott. Within-Doors. Op. cit. in 6.V. 1872. Chap. 1, no. 2: The cardboard puzzle, pp. 27 & 30 31. No diagram, so the solution is a bit cryptic.

Lemon. 1890. Cardboard puzzle, no. 140, pp. 23 & 102. = Sphinx, no. 467, pp. 65 & 113.

J. B. Bartlett. How to walk through a laurel leaf. The Boy's Own Paper 12 (No. 587) (12 Apr 1890) 440.

Hoffmann. 1893. Chap. X, no. 28: The cut playing card, pp. 346 & 385 386 = Hoffmann Hordern, p. 243.

Benson. 1904. The elastic cardboard puzzle, pp. 200 201.

Dudeney. Cutting-out paper puzzles. Cassell's Magazine ?? (Dec 1909) 187-191 & 233-235. With photo of Dudeney going through the card.

Collins. Book of Puzzles. 1927. Through a playing card, pp. 16-17.

6.BB. DOUBLING AN AREA WITHOUT CHANGING ITS HEIGHT OR

WIDTH
The area is usually a square, but other shapes are possible. If one views it as a reduction, one can reduce the area to any fraction of the original!
The Sociable. 1858. Prob. 41: The carpenter puzzled, pp. 298 & 316. 3 x 3 square of wood with holes in it forming a 4 x 4 array with the corner holes at the corners of the board. Claims one can cut 1/4 of the board out of the centre without including any holes. But this only gets 2/9 of the area -- double the central square. = Book of 500 Puzzles, 1859, prob. 41, pp. 16 & 34. = Secret Out, 1859, pp. 386-387.

Indoor & Outdoor. c1859. Part II: prob. 14: The carpenter puzzled, pp. 133 134. Almost identical with The Sociable.

Hanky Panky. 1872. P. 226 shows the same diagram as the solution in The Sociable, but there is no problem or text.

Lewis Carroll. Letter of 15 Mar 1873 to Helen Feilden. = Carroll-Collingwood, pp. 212-215 (Collins 154-155), without solution. Cf Carroll-Wakeling, prob. 28: The square window, pp. 36-37 & 72. Halve the area of a square window. Wakeling and Carroll-Gardner, p. 52, give the surname as Fielden, but it is Feilden in Carroll-Collingwood and in Cohen, who sketches her life. Wakeling writes that Feilden is correct.

Mittenzwey. 1880. Prob. 216-217, pp. 38-39 & 90; 1895?: 241-242, pp. 43 & 92; 1917: 241 242, pp. 39 & 88. Divide a rectangle or square into two pieces with the same height and width as the square. Solution is to draw a diagonal.

Lemon. 1890. A unique window, no. 444, pp. 58 & 114. The philosopher's puzzle, no. 660, pp. 82 & 121.

Don Lemon. Everybody's Scrap Book of Curious Facts. Saxon, London, 1890. P. 82 quotes an article from The New York World describing this as 'an excellent, if an old, puzzle'.

Hoffmann. 1893. Chap. IX, no. 28: A curious window, pp. 319 & 327 = Hoffmann-Hordern, pp. 211-212. Notes that either a diamond or a triangle in appropriate position can be so doubled.

Clark. Mental Nuts. 1897, no. 40. The building lot. "Have a lot 50 x 100. Want to build a house 50 x 100 and have the yard same size. How?" Solution shows 50 x 100 with a diagonal drawn.

Pearson. 1907. Part II, no. 79: At a duck pond, pp. 79 & 176. A square pond is to be doubled without disturbing the duckhouses at its corners.

Wehman. New Book of 200 Puzzles. 1908. The carpenter puzzled, p. 39. = The Sociable.

Will Blyth. Handkerchief Magic. C. Arthur Pearson, London, 1922. Doubling the allotment, pp. 23-24.

Hummerston. Fun, Mirth & Mystery. 1924. Some queer puzzles, Puzzle no. 76, part 6, pp. 164 & 183. Solution notes that a window in the shape of a diamond or a right triangle or an isosceles triangle can be doubled in area without changing its width or height.

King. Best 100. 1927.

No. 2, pp. 7 8 & 38. Same as Indoor & Outdoor, with the same error.

No. 4, pp. 8 & 39. Halve a square window. See Foulsham's.

Foulsham's Games and Puzzles Book. W. Foulsham, London, nd [c1930]. No. 2, pp. 5 & 10. Double a window without changing its height or width. (This is one of the few cases where the problem is not quite identical to King.)

M. Adams. Puzzle Book. 1939. Prob. B.117: Enlarging the allotment, pp. 86 & 110. Double a square allotment without disturbing the trees at the corners.

6.BC. HOFFMAN'S CUBE
This consists of 27 blocks, a x b x c, to make into a cube a+b+c on a side. It was first proposed by Dean Hoffman at a conference at Miami Univ. in 1978. See S&B, p. 43. The planar version, to use 4 rectangles a x b to make a square of side a + b is easy. These constructions are proofs of the inequality of the arithmetic and geometric means. Sometime in the early 1980s, I visited David Klarner in Binghamton and Dean Hoffman was present. David kindly made me a set of the blocks and a three-sided corner to hold them.
D. G. Hoffman. Packing problems and inequalities. In: The Mathematical Gardner, op. cit. in 6.AO, 1981. Pp. 212 225. Includes photos of Carl Klarner assembling the first set of the blocks. Asks if there are analogous packings in n dimensions.

Berlekamp, Conway & Guy. Winning Ways. 1982. Vol. 2, pp. 739 740 & 804 806. Shows all 21 inequivalent solutions.

6.BD. BRIDGE A MOAT WITH PLANKS
In the simplest case, one has a 3 x 3 moat with a 1 x 1 island in the centre. One wants to get to the island using two planks of length 1 or a bit less than 1. One plank is laid diagonally across the corner of the moat and the second plank is laid from the centre of the first plank to the corner of the island. If the width of the moat is D and the planks have length L, then the method works if 3L/2 > D2, i.e. L > 22 D/3 = .94281.. D. One really should account for the width of the planks, but it is not clear just how much overlap is required for stability. Depew is the only example I have seen to use boards of different lengths. With more planks, one can reach across an arbitrarily large moat, but the number of planks needed gets very large. In this situation, the case of a circular moat and island is a bit easier to solve.

The Magician's Own Book (UK version) version is quite different and quite erroneous.

Magician's Own Book (UK version). 1871. The puzzle bridge, p. 123. Stream 15 or 16 feet across, but none of the available planks is more than 6 feet long. He claims that one can use a four plank version of the three knives make a support problem (section 11.N) to make a bridge. However the diagram of the solution clearly has the planks nearly as long as the width of the stream. In theory, one could build such a bridge with planks slightly longer than half the width of the stream, but to get good angles (e.g. everything crossing at right angles or nearly so), one needs planks somewhat longer than 2/2 of the width. E.g. for a width of 16 ft, 12 ft planks would be adequate.

Mittenzwey. 1880. Prob. 298, pp. 54 & 105; 1895?: 330, pp. 58 & 106; 1917: 330, pp. 52 53 & 101. 4 m gap bridged with two boards of length 3¾ m. He only gives a diagram. In fact this doesn't work because the ratio of lengths is 15/16 = .9375

Lucas. RM2. 1883. Le fossé du champ carré. Bridge the gap with two planks whose length is exactly 1. Notes this works because 3/2 > 2.

Hoffmann. 1893. Chap. VII, no. 9, pp. 289 & 295. Matchstick version = Hoffmann-Hordern, p. 193.

Benson. 1904. The moat puzzle, p. 246. Same as Hoffmann, but the second plank is shown under the first!!

Dudeney. CP. 1907. No. 54: Bridging the ditch, pp. 83-85 & 204. Eight 9' planks to cross a 10' ditch where it makes a right angle.

Pearson. 1907. Part I, no. 34: Across the moat, pp. 122 & 186.

Blyth. Match-Stick Magic. 1921. Boy Scouts' bridge, p. 21. Ordinary version done with matchsticks.

J. C. Cannell. Modern Conjuring for Amateurs. C. Arthur Pearson, London, nd [1930s?]. Boy Scouts' bridge, pp. 68-69. As in Blyth.

Depew. Cokesbury Game Book. 1939. Crossing the moat, pp. 225-226. Square moat 20 feet wide to be crossed with boards of width 18 and 15 ft. In fact this doesn't work -- one needs L₁ + L₂/2 > D2.

"Zodiastar". Fun with Matches and Matchboxes. Op. cit. in 4.B.3. Late 1940s? The bridge, pp. 66-67 & 83. Matchstick version of the square moat & square island problem.

F. D. Burgoyne. Note 3106: An n plank problem. MG 48 (No. 366) (Dec 1964) 434 435. The island is a point in the centre of a 2 x 2 lake. Given n planks of length s, can you get to the island? He denotes the minimal length as s(n) and computes s(1) = 1, s(2) = 2 2/3, s(3) = .882858... and says s() = 2/2, [but I believe it is 0, i.e. one can get across an arbitrarily large moat with a fixed length of plank].

Jonathan Always. Puzzling You Again. Tandem, London, 1969.

Prob. 10: A damsel in distress, pp. 15 & 70 71. Use two planks of length L to reach a point in the centre of a circular moat of radius R. He finds one needs L²  4R²/5.

Prob. 11: Perseus to the rescue again, pp. 15 16 & 71 72. Same with five planks. The solution uses only four and needs L²  2R²/3.

C. V. G.[?] Howe? Mathematical Pie 75 (Summer 1975) 590 & 76 (Autumn 1975) 603. How big a square hole can be covered with planks of unit length? Answer says there is no limit, but the height of the pile increases with the side of the square.

Highlights for Children (Columbus, Ohio). Hidden Picture Favorites and Other Fun. 1981. Brain Buster 4, pp. 12 & 32: Plink, plank, kerplunk? Two children arrive at a straight(!) stream 4m wide with two planks 3m long. Solution: extend one plank about 1¼ m over the stream and one child stands on the land end. The second child carries the other plank over the stream and extends it to the other side and crosses. He then pulls the plank so it extends about 1¼ m over the stream. The first child now extends her plank out to rest on the second plank and crosses, pulling up her plank and taking it with her. I theory this technique will work if L > ⅔D, but one needs some overlap space, and the children may not have the same weight.

Richard I. Hess. Puzzles from Around the World. The author, 1997. (This is a collection of 117 puzzles which he published in Logigram, the newsletter of Logicon, in 1984-1994, drawn from many sources. With solutions.) Prob. 64. Usual problem with D = 10, but he says the board have width W = 1 and so one use the diagonal of the board in place of L. In my introduction, we saw that the standard version leads to L²  8D²/9, so Hess's version leads to L²  8D²/9 - W² and we can get across with slightly shorter planks, but we have to tread very carefully!

6.BE. REVERSE A TRIANGULAR ARRAY OF TEN CIRCLES
One has a triangle of ten coins with four on an edge. Reverse its direction by moving only three coins. New section -- I'm surprised not to have seen older examples.
Sid G. Hedges. More Indoor and Community Games. Methuen, London, 1937. The triangle trick, p. 54. Uses peas, buttons or nuts.

M. Adams. Puzzle Book. 1939. Prob. B.34: Pitching camp, pp. 66 & 103. Array of tents.

Evelyn August. The Black-Out Book. Op. cit. in 5.X.1. 1939. The General inspects the balloons, pp. 106 & 214. Array of 10 barrage balloons.

Leopold. At Ease! 1943. 1, 2, 3 -- shift!, pp. 20 & 198. Thanks to Heinrich Hemme for the lead to this.

Joseph Leeming. Games with Playing Cards Plus Tricks and Stunts. Franklin Watts, 1949. ??NYS -- but two abridged versions have appeared.

Games and Fun with Playing Cards. Dover, NY, 1980. This contains everything except the section on bridge.

Tricks and Stunts with Playing Cards Plus Games of Solitaire. Gramercy Publishing, NY, nd [1960s?]. This includes all the tricks, stunts, puzzles and solitaire games.

25 Puzzles with Cards, 8th puzzle. Tricky triangle. Dover: pp. 154-155 & 172. Gramercy: pp. 45-46 & 65. Both have fig. 25 & 42.

Young World. c1960. P. 7: fifteen coin problem. Reverse a triangle with five on a side by moving five coins.

Robert Harbin. Party Lines. Op. cit. in 5.B.1. 1963. Birds in flight, p. 34. Says this problem is described by Gardner, but gives no specific source.

Maxey Brooke. (Fun for the Money, Scribner's, 1963); reprinted as: Coin Games and Puzzles, Dover, 1973. Prob. 4: Bottoms up, pp. 15 & 75. On p. 6, he acknowledges Leopold as his source. Thanks to Heinrich Hemme for this reference.

D. B. Eperson. Triangular (old) pennies. MG 54 (No. 387) (Feb 1970) 48 49. The number of pennies which must be moved to reverse a triangle with n on a side is [T(n)/3], where T(n) is the n th triangular number, which is the number in the array.

James Bidwell. The ten coin triangle. MTg 54 (1971) 21 22. How many coins must be moved to reverse the triangle with n on an edge? His students find the same value as Eperson, but they weren't sure they had proved it.

Putnam. Puzzle Fun. 1978. No. 25: Triangular reverse, pp. 6 & 31. Usual 10 coin triangle.

6.BF. PYTHAGOREAN RECREATIONS
6.L might be considered as part of this section. There are some examples of problems with ladders which look like crossed ladders, but are simple Pythagorean problems.

See also 6.AS.2 for dissection proofs of the theorem of Pythagoras. I will include here only some interesting ancient examples. See Elisha Scott Loomis; The Pythagorean Proposition; 2nd ed., NCTM, 1940, for many proofs.

Aryabhata I, v. 17, states the Theorem of Pythagoras and the related theorem that if ABC is a diameter of a circle and LBM is a chord perpendicular to it, then LB² = AB x BC; Bhaskara I's commentary applies the latter in several forms where modern algebra would make it more natural to use the former. Brahmagupta, v. 41, states LM² = AB x BC.

I had overlooked the examples in Mahavira -- thanks to Yvonne Dold for pointing them out.

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

Vyse. Tutor's Guide. 1771?

Prob. 9, 1793: p. 178, 1799: p. 189 & Key p. 224. 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. There is a misprint of 9 for 6 in the answer.

Prob. 17 (in verse), 1793: 179, 1799: p. 190 & Key p. 228. A variation of the Broken Bamboo problem, cf below, with D = 30, H - X = 63, which is a simple right triangle problem.

Hutton. A Course of Mathematics. 1798?

Prob. VIII, 1833: 430; 1857: 508. = Vyse, prob. 19.

Prob. IX, 1833: 430; 1857: 508. = Vyse, prob. 17 with D = 15, H - X = 39.
6.BF.1 THE BROKEN BAMBOO
A bamboo (or tree) of height H breaks at height X from the ground so that the broken part reaches from the break to the ground at distance D from the foot of the bamboo. In fact the quadratic terms drop out of the solution, leaving a linear problem. This may be of Babylonian origin?? The hawk and rat problems of 6.BF.3 are geometrically the same problem viewed sideways.

In all cases below, H and D are given and X is sought, so I will denote the problem by (H, D).

See Tropfke, p. 620.
Chiu Chang Suan Ching (Jiu Zhang Suan Shu). c 150? Chap. IX, prob. 13, p. 96. [English in Mikami, p. 23 and in Swetz & Kao, pp. 44 45, and in HM 5 (1978) 260.] (10, 3).

Bhaskara I. 629. Commentary to Aryabhata, chap. II, v. 17, part 2. Sanskrit is on pp. 97 103; English version of the examples is on pp. 296-300. The material of interest is examples 4 and 5. In the set-up described under 6.BF.3, the bamboo is BOC which breaks at O and the point C reaches the ground at L.

Ex. 4: (18, 6). Shukla notes this is used by Chaturveda.

Ex. 5: (16, 8).

Chaturveda. 860. Commentary to the Brahma sphuta siddhanta, chap. XII, section IV, v. 41, example 2. In Colebrooke, p. 309. Bamboo: (18, 6).

Mahavira. 850. Chap. VII, v. 190-197, pp. 246-248.

v. 191. (25, 5), but the answer has H - X rather than X.

v. 192. (49, 21), but the answer has H - X rather than X.

v. 193. (50, 20), but with the problem reflected so the known leg is vertical rather than horizontal.

v. 196. This modifies the problem by imagining two trees of heights H and h, separated by D. The first, taller, tree breaks at height X from the ground and leans over so its top reaches the top of the other tree. If we subtract h from X and H, then │X - h│ is the solution of the problem (H - h, D). Because the terms are squared, it doesn't matter whether X is bigger or smaller than h. He does the case H, h, D = 23, 5, 12.

Bhaskara II. Lilavati. 1150. Chap. VI, v. 147 148. In Colebrooke, pp. 64 65. (32, 16).

Bhaskara II. Bijaganita. 1150. Chap. IV, v. 124. In Colebrooke, pp. 203 204. Same as Lilavati.

Needham, p. 28, is a nice Chinese illustration from 1261.

Gherardi. Libro di ragioni. 1328. Pp. 75 76: Regolla di mesura. (40, 14).

Pseudo-dell'Abbaco. c1440. No. 166, p. 138 with B&W reproduction on p. 139. Tree by stream. (60, 30). I have a colour slide of this.

Muscarello. 1478. F. 96v, pp. 224-225. Tree by a stream. (40, 30).

Calandri. Aritmetica. c1485. Ff. 87v-88r, pp. 175-176. Tree by a stream. (60, 30). = Pseudo-dell'Abbaco.

Calandri. Arimethrica. 1491. F. 98r. Tree by a river. (50, 30). Nice woodcut picture. Reproduced in Rara, 48.

Pacioli. Summa. 1494. Part II, f. 55r, prob. 31. (30, 10). Seems to say this very beautiful and subtle invention is due to Maestro Gratia.

Clark. Mental Nuts. 1897, no. 78. The tree and the storm. (100, 30). [I have included this as this problem is not so common in the 19C and 20C as in earlier times.]

N. L. Maiti. Notes on the broken bamboo problem. Gaņita-Bhāratī [NOTE: ņ denotes an n with an underdot] (Bull. Ind. Soc. Hist. Math.) 16 (1994) 25-36 -- ??NYS -- abstracted in BSHM Newsletter 29 (Summer 1995) 41, o/o. Says the problem is not in Brahmagupta, though this has been regularly asserted since Biot made an error in 1839 (probably a confusion with Chaturveda -- see above). He finds eight appearances in Indian works, from Bhaskara I (629) to Raghunath-raja (1597).