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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.