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Geometry problem

  1. Nov 10, 2004 #1

    Galileo

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    Consider an alley with two ladders placed crosswise.
    Kinda like this sucky diagram below:

    [tex]
    \begin{array}{cccccccc}
    |&* & & & & & &| \\
    |& &* & & & & &| \\
    |& & &* & & &&/| \\
    |& & & &* &/& &| \\
    |& & &/& &* & &| \\
    |&/& & & & &* &|
    \end{array}
    [/tex]
    ( :yuck: Ugh, I hope you get the idea)

    One ladder has a length of 2 m. The other is 3 m long.
    They cross each other at a height of 1 m.
    What is the width of the alley?

    It's not hard to see there's a unique solution to this question.
    Good luck :wink:
     
    Last edited: Nov 10, 2004
  2. jcsd
  3. Nov 10, 2004 #2

    Gokul43201

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    [tex]\sqrt{9-w^2} + \sqrt{4-w^2} = \sqrt{(9-w^2)(4-w^2)} [/tex]

    This might look pretty, but it's both messy and inelegant...there's got to be a nicer way.

    PS : That gives me w = about 1.231m
     
    Last edited: Nov 10, 2004
  4. Nov 10, 2004 #3
    The solution to these 'crossed ladder' problems, except in trivial cases, is a quartic. They are almost always messy:

    spoiler - highlight or Ctrl-A to view

    w^8 - 22w^6 + 163w^4 - 454w^2 + 385 = 0

    w is about 1.2312 m.


    Best found (IMO) by successive approximation, ie. guessing.
     
    Last edited: Nov 10, 2004
  5. Nov 10, 2004 #4

    Galileo

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    You're right. (so is Ceptimus), but how did you figure that?
    On the left side I see the height of the 3m ladder plus the height of the 2m ladder and on the right side the product. Why are these equal?
     
  6. Nov 10, 2004 #5

    NateTG

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    I don't know how they did things, but there are lots of similar triangles to pick from. It seems like that should be enough to work things out.
     
  7. Nov 23, 2004 #6
    could someone explain this one for me? this intrigues me
     
  8. Nov 24, 2004 #7

    Galileo

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    Me too. C'mon Gokul43201! :smile:
     
  9. Nov 24, 2004 #8
    Galileo,
    this is sort of a hint and i am not sure whether Gokul did it this way...

    Let the ladders be L1 and L2.
    Let H_L1 be the height at which ladder L1 meets the wall
    Let H_L2 be the height at which ladder L2 meets the wall

    can u show that,
    1/H_L1 + 1/H_L2 = 1

    -- AI
     
  10. Nov 24, 2004 #9

    GCT

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    Here's a hint: highlight to see

    Use trigonometry (cos, sin, etc.......). It's not a difficult problem as you will see, the solution is quite "simple."
     
    Last edited: Nov 24, 2004
  11. Nov 24, 2004 #10

    Gokul43201

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    That's what I did. But if you want a complete solution, I'll post it a little later...no time now.
     
  12. Dec 18, 2004 #11

    Eureka !! :!!) I have discovered a truly unremarkable Al Gore Rythym to solve this teaser. :smile: *.102m Unfortunatley, :frown: it will not fit in this small margin. :devil: Get the idea? Thanx
    *encrypted so not to spoil it for others>12 clock arithmetic.
     
    Last edited by a moderator: Dec 18, 2004
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