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Topic: Ladder Problem
Replies: 16   Last Post: May 13, 2000 5:03 PM

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Robert Hill

Posts: 529
Registered: 12/8/04
Re: Ladder Problem
Posted: May 11, 2000 1:55 PM
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In article <8fem3m$9b1$1@nnrp1.deja.com>, achava@hotmail.com writes:
> In article <8fd3d0$hsp$1@nnrp1.deja.com>,
> rayfil@my-deja.com wrote:

> > An old man's memories!
> >
> > Problem:
> >
> > A narrow street is lined with tall buildings. A
> > 30-ft 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 40-ft
> > 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







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