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Re: Ladder Problem
Posted:
May 11, 2000 1:55 PM


In article <8fem3m$9b1$1@nnrp1.deja.com>, achava@hotmail.com writes: > In article <8fd3d0$hsp$1@nnrp1.deja.com>, > rayfil@mydeja.com wrote: > > An old man's memories! > > > > Problem: > > > > A narrow street is lined with tall buildings. A > > 30ft ladder is rested at the base of the > > building on the right side of the street and > > leans on the building on the left side. A 40ft > > ladder is rested at the base of the building on > > the left side of the street and leans on the > > building on the right side. The point where the > > two ladders cross is exactly 10 feet from the > > ground. > > > > What if the width of the street? > > An interesting problem indeed! Implicit in the solutions that you > received is a geometrical fact that to me is far more interesting than > simply getting the answer. > > I can't draw in a posting very well, so I will attempt to state it in > words. If we let u be the vertical elevation of the high point of the > 40 foot ladder, and let v be the vertical elevation of the high point of > the 30 foot ladder, then > > 1/u + 1/v = 1/10 > > which pops out of a simple calculation using similar triangles. From > here the Pythagorean theorem gives us u and v in terms of x, but the > nice and (to me) surprising symmetry of the relations between the > vertical elevations of the ladders and the height of the intersection > point is the "real" geometrical result here.
This suggests a nomogram for performing calculations to locate the images of objects produced by lenses and curved mirrors.
Has somebody invented this before?
 Robert Hill



