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

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The discussion centers around a mathematical problem involving two ladders of different lengths (2 m and 3 m) crossing each other at a height of 1 m in an alley. Participants explore the unique solution for the width of the alley, with one contributor providing a complex quartic equation that leads to an approximate width of 1.231 m. The conversation touches on the use of trigonometry to derive the relationship between the heights at which the ladders meet the walls and hints at simpler methods for solving the problem. There is a sense of intrigue and encouragement among participants to delve deeper into the solution, with some expressing a desire for clarification on the underlying principles. Overall, the discussion highlights the complexity and elegance of solving crossed ladder problems.
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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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