Sources page biographical material

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 (S: 543-544) looks like a crossed ladders problem but is a simple right triangle problem.

Pacioli. Summa. 1494. Part II.

F. 56r, prob. 48. (4, 6).

F. 60r, prob. 64. (10, 15).

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. How wide is the street? This is actually a simple right triangle problem.

Victor Katz reports that Hutton's problem, with values 60; 37, 23 appears in a notebook of Benjamin Banneker (1731-1806).

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 = (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 Problems. 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. Jabłkowski. 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.