Solve Geometry Problem: Alley w/2 Ladders of 2m & 3m, Cross at 1m

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The forum discussion centers on solving a geometry problem involving two ladders of lengths 2 meters and 3 meters that cross at a height of 1 meter in an alley. The unique solution for the width of the alley, denoted as 'w', is approximately 1.2312 meters, derived from the quartic equation w^8 - 22w^6 + 163w^4 - 454w^2 + 385 = 0. The problem can be approached using trigonometric relationships, specifically the equation 1/H_L1 + 1/H_L2 = 1, where H_L1 and H_L2 are the heights at which the ladders meet the wall. The discussion emphasizes the use of successive approximation for finding the solution.

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

<br /> \begin{array}{cccccccc}<br /> |&amp;* &amp; &amp; &amp; &amp; &amp; &amp;| \\<br /> |&amp; &amp;* &amp; &amp; &amp; &amp; &amp;| \\<br /> |&amp; &amp; &amp;* &amp; &amp; &amp;&amp;/| \\<br /> |&amp; &amp; &amp; &amp;* &amp;/&amp; &amp;| \\<br /> |&amp; &amp; &amp;/&amp; &amp;* &amp; &amp;| \\<br /> |&amp;/&amp; &amp; &amp; &amp; &amp;* &amp;|<br /> \end{array}<br />
( 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:
 
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\sqrt{9-w^2} + \sqrt{4-w^2} = \sqrt{(9-w^2)(4-w^2)}

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[/color]
 
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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.
 
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Gokul43201 said:
\sqrt{9-w^2} + \sqrt{4-w^2} = \sqrt{(9-w^2)(4-w^2)}
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?
 
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.
 
could someone explain this one for me? this intrigues me
 
Me too. C'mon Gokul43201! :smile:
 
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
 
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."
 
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  • #10
GeneralChemTutor said:
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."

That's what I did. But if you want a complete solution, I'll post it a little later...no time now.
 
  • #11
Galileo said:
Consider an alley with two ladders placed crosswise.
Kinda like this sucky diagram below:

( 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:


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.
 
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