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6.BF.2. SLIDING SPEAR = LEANING REED
A spear (or ladder) of length H stands against a wall. Its base moves out B from the wall, causing the top to slide down D. Hence B² + (H - D)² = H².

The leaning reed has height H. It reaches D out of the water when it is straight up. When it leans over, it is just submerged when it is B away from its upright position. This is identical to the sliding spear turned upside down.

In these problems, two of H, B, D are given and one wants the remaining value. I will denote them, by e.g. H, D = 30, 6.

See Tropfke, p. 619 & 621.

BM 85196. Late Old Babylonian tablet in the British Museum, c 1800. Transcribed, translated and commented on by O. Neugebauer; Mathematische Keilschrift-texte II; Springer, Berlin, 1935, pp. 43++. Prob. 9 -- translation on pp. 47-48, commentary on p. 53. Quoted in B. L. van der Waerden; Science Awakening; OUP, 1961, p. 76. See also: J. Friberg; HM 8 (1981) 307-308. Sliding beam(?) with H, D = 30, 6 and with H, B = 30, 18.

BM 34568. Seleucid period tablet in the British Museum, c 300. Transcribed, translated and commented on by O. Neugebauer; Mathematische Keilschrift-texte III, Springer, Berlin, 1937, pp. 14-22 & plate 1. Prob. 12 -- translation on p. 18, commentary on p. 22. Quoted in van der Waerden, pp. 76 77. See also: J. Friberg, HM 8 (1981) 307-308. Sliding reed or cane, B, D = 9, 3.

Papyri Cairo J. E. 89127 30 & 89137 43. c 260. Shown and translated in: Richard A. Parker; Demotic Mathematical Papyri; Brown Univ. Press, Providence, 1972; pp. 1, 3 4, 35 40 & Plates 9 10.

Prob. 24 26: H, B = 10, 6; 14½, 10; 10, 8.

Prob. 27 29: H, D = 10, 2; 14½, 4; 10, 4.

Prob. 30 31: B, D = 6, 2; 10, 4.

Chiu Chang Suan Ching (Jiu Zhang Suan Shu). c 150? Chap. IX.

Prob. 6, p. 92. [English in Mikami, p. 22.] Leaning reed, B, D = 5, 1.

Prob. 7, p. 93. Version with a rope hanging and then stretched giving B, D = 8, 3.

Prob. 8, p. 93. [English in Mikami, p. 22 and in Swetz & Kao, pp. 30 32 and in HM 4 (1977) 274.] Ladder, but with vertical and horizontal reversed. B, D = 10, 1. Mikami misprints the answer as 55 rather than 50.5.

Bhaskara I. 629. Commentary to Aryabhata, chap. II, v. 17, part 2. Sanskrit is on pp. 97 103, with reproductions of original diagrams on pp. 101-102; English version of the examples is on pp. 296-300. The material of interest is examples 6 and 7. In the setup of 6.BF.3, the lotus is OBA and LBM is the water level.

Ex. 6: B, D = 24, 8. Shukla notes this is used by Chaturveda.

Ex. 7: B, D = 48, 6.

Chaturveda. 860. Commentary to the Brahma sphuta siddhanta, chap. XII, section IV, v. 41, example 3. In Colebrooke, pp. 309 310. Leaning lotus: B, D = 24, 8.

Tabari. Miftāh al-mu‘āmalāt. c1075. P. 124, no. 42. ??NYS - cited by Tropfke 621.

Bhaskara II. Lilavati. 1150. Chap. VI, v. 152 153. In Colebrooke, pp. 66. Leaning lotus: B, D = 2, ½.

Bhaskara II. Bijaganita. 1150. Chap. IV, v. 125. In Colebrooke, p. 204. Same as Lilavati.

Fibonacci. 1202. P. 397 (S: 543). Sliding ladder: H, B = 20, 12.

Leonardo Fibonacci. La Practica di Geometria. Volgarizzata da Cristofano di Gherardo di Dino, cittadino pisano. Dal Codice 2186 della Biblioteca Riccardiana di Firenze, 1448. Ed. by Gino Arrighi, Domus Galilaeana, Pisa, 1966. P. 37 and fig. 18. Same as Fibonacci 1202.

Zhu Shijie. Siyuan Yujian (Precious Mirror of the Four Elements). 1303. ??NYS -- English given in Li & Du, p. 179. Questions in verse, no. 1. Two reeds, 14 apart, which reach 2½ and 1 out of the water. When they lean together, they just touch at the water surface. The water is assumed to have the same depth at both reeds, which reduces the problem from two variables to one variable.

Gherardi?. Liber habaci. c1310. Pp. 139 140. H, B = 20, 12

Gherardi. Libro di ragioni. 1328. Pp. 77 78. Ship's mast with H, D = 131, 4. B&W picture on p. 77, from f. 46v.

Columbia Algorism. c1350.

No. 135, pp. 138 139. Sliding ladder: H, B = 10, 6.

No. 140, pp. 149 150. Leaning tree: H, B = 20, 10.

(I have colour slides of the illustrations to these problems.)

Pseudo-dell'Abbaco. c1440. No. 167, pp. 138-140, with B&W picture on p. 139. Sliding spear: H, D = 30, 4. I have a colour slide of this.

della Francesca. Trattato. c1480. F. 44r (108). B, D = 6, 2. English in Jayawardene.

Pacioli. Summa. 1494. Part II, ff. 54v-55v.

Prob. 25 (misprinted 52). B, D = 6, 2. = della Francesca.

Prob. 26. H, H - D = 10, 8.

Prob. 27. H, D = 10, 4.

Prob. 28. H, D = 10, B/3.

Prob. 29. D + H = 12, DH = 12.

Prob. 30. H = D + 4, DH = 12.

Prob. 41. Tree of height 40 with a rope of length 50 tied to the top which reaches to the ground at a point 30 away. Length 10 of the rope is pulled, causing the tree to lean. How high is the top of the tree now? We now have a triangle of sides 40, 40, 30 and want the altitude to the side of length 30.

PART II.

F. 68r, prob. 98. Tree of height 30 has a rope of length 50 tied to the top, so it reaches the ground 40 away. How much rope has to be pulled to move the top of the tree to being 8 away from the vertical? He neglects to give the value 50 in the problem statement and the ground distances of 8 and 32 are interchanged in the diagram.

van Etten. 1624. Prob. 89 (86), part V (4), p. 135 (214). Sliding ladder: H, B = 10, 6.

Ozanam. 1694. Prob. 42 & fig. 48, plate 10, 1696: 123-124; 1708: 129; 1725: 320-321 & plate 10 (11). Ladder 25 long with foot 7 from wall. Foot is pulled out 8 more -- how much does the top come down?

Vyse. Tutor's Guide. 1771? Prob. 16 (in verse), 1793: p. 179, 1799: p. 190 & Key p. 228. H, B = 100, 10.

Mittenzwey. 1880. Prob. 294, pp. 53-54 & 104; 1895?: 324, pp. 57 & 106; 1917: 324, pp. 52 & 100. Leaning reed, B, D = 5, 1. [I have included this and the next entry as this problem is not so common in the 19C and 20C as in earlier times.]

Clark. Mental Nuts. 1904, no. 69; 1916, no. 95. The boatman's puzzle. Leaning pole, B, D = 12, 6. Find H - D.