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A Ladder Puzzle


Date: 10/20/2000 at 14:40:10
From: Ralf Dueck
Subject: Geometry

A cube is placed flat against a wall. A ladder of length 10m is 
leaning against the wall, just touching an edge of the cube. At what 
height does the end of the ladder touch the wall?

I tried with Pythagoras, but I didn't find the answer. I know that 
there must be two answers.


Date: 10/21/2000 at 05:17:36
From: Doctor Floor
Subject: Re: Geometry

Dear Ralph,

Thanks for your question. It seems that you forgot to give dimensions 
(measures) of the cube. Let me take 3m for the edge lengths of the 
cube.

This is a very curious and nice problem. I saw it for the first time 
in a booklet with a collection of problems from a Dutch newspaper (NRC 
Handelsblad) 15-20 years ago. I can't find it now, but I know that I 
didn't really appreciate the solution given there, because it didn't 
benefit from the symmetry of the problem. The symmetry here is that if 
the ladder fits to a certain height, then it also fits when it is at 
that distance on the floor.

Let me sketch the solution:

The cube divides the ladder in two parts, which I will take to have 
lengths 5+x and 5-x, and I will try to solve for x (computation of the 
height you are looking for is straightforward then.) I chose this use 
of x because of the symmetry. If x is a solution, then of course -x is 
also a solution. We have the following picture:

      |
      |\
      | \
      |  \5+x
     t|   \
      |    \
      |__3__\
      |     |\
     3|    3| \5-x
      |_____|__\___
         3    u

We know with help of the Pythagorean theorem:

     u^2 = (5-x)^2 - 9 = x^2 - 10x + 16
     t^2 = (5+x)^2 - 9 = x^2 + 10x + 16

The slopes of the two parts of the ladder must of course be equal, so 
the following must be true:

     t/3 = 3/u
     tu = 9
     t^2*u^2 = 81
     (x^2-10x+16)*(x^2+10x+16) = 81
     x^4 - 68x^2 + 175 = 0

This equation is quadratic in x^2, and you can find four solutions for 
x. But not all solutions are valid. Of course 5-x as well as 5+x must 
be greater than 3, so you must have -2 < x < 2. In fact there are two 
solutions left, as you already said.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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