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Topic: Crossed Ladders Problem (again)
Replies: 6   Last Post: Dec 29, 2012 2:37 AM

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John Conway

Posts: 2,238
Registered: 12/3/04
Re: Crossed Ladders Problem (again)
Posted: Aug 23, 2001 12:46 PM
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On 23 Aug 2001, Randall L. Rathbun wrote:

> My GP-Pari program will solve for the roots, using some type of numerical
> algorithm to hundreds of thousands of digits. But what I want to know, is
> the very nature of this problem intrinsically involve a quartic so that no
> matter how hard we try, it will always pop up?

I think the answer is "yes", even though I'm not quite sure exactly
what your question means.

Let me say what my "yes" means. If you're given the lengths a and
b of the two ladders, and the height h of their intersection,
then the distance x between the two vertical walls is given by a quartic
equation that is in general irreducible, although for particular values of
a,b,h it might just happen to have a rational root.

It's easy to find such values: we can suppose the bottom of the
picture looks like this:

\ /
15/ | \13
/ 12 \
9 5
for instance. Then obviously we have x = 14 and since the top half of
the figure will look like:

| \13u /|
12u| \ 5v/ |4v
| \ / |
| 5u X 3v|

we must have

a = 13u + 13, where 5u = 9 and b = 15 + 5v, where 3v = 5,

giving a = 182/5, b = 70/3, h = 12, (for x = 14).

I rescale this to get a simpler case: multiplying by 15/14

a = 39, b = 25, h = 90/7

for a problem whose answer is x = 15.

I wonder what the simplest rational solution is?

John Conway

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