Miscellaneous Problem 1 The crossed ladder problem. In an alley a ladder length 3 is propped up against one wall with its foot at the bottom of the other wall. It crosses another ladder length 2 leaning the other way a distance 1 from the ground. How wide is the alley? [In other words we are given a quadrilateral ABCD. Angles ADC and DCB are right angles. AC = 2, BD = 3 and AC, BD intersect at X a distance 1 from CD. Find CD.]
1.2311857... (positive real root to x 8 - 22x 6 + 163x 4 - 454x 2 + 385 = 0)  It is not hard to derive the answer to this problem. The only curious thing is that it is such an awkward answer! Let Y be the foot of the perpendicular from X to CD. CYX, CBA are similar, so 1/AB = CY/CB. Similarly 1/CD = BY/BC. Hence 1/AB + 1/CD = 1. But AB 2 = 4 - x 2 , CD 2 = 9 - x 2 , where x is the width of the alley. So we immediately get a quartic in x 2 . But there is no obvious simplification. One can solve the quartic numerically, or write down the root explicitly (either way, it is obvious which root you need). But the explicit expression is horrendous! The solutions given on this site are not always complete, they are designed to be sufficient for anyone who has thought hard about the problem. Miscellaneous © John Scholes jscholes@kalva.demon.co.uk 24 Dec 1998