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

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Discussion Overview

The discussion revolves around a geometry problem involving two ladders of different lengths (2 m and 3 m) that cross each other at a height of 1 m in an alley. Participants are exploring methods to determine the width of the alley, engaging in both mathematical reasoning and conceptual clarification.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a mathematical equation involving square roots to relate the lengths of the ladders and the width of the alley, suggesting a solution of approximately 1.231 m.
  • Another participant notes that the solution to crossed ladder problems typically results in a quartic equation, indicating the complexity of the problem.
  • There is a discussion about the equality of two expressions involving the heights of the ladders and their relationship to the width, with one participant questioning the reasoning behind this equality.
  • Some participants suggest using similar triangles as a potential method for solving the problem.
  • Hints involving trigonometric functions are provided, with participants expressing intrigue and requesting further clarification on the problem-solving approach.
  • A participant mentions discovering a method to solve the problem, but does not disclose the details, indicating a playful approach to the solution.

Areas of Agreement / Disagreement

Participants express varying methods and approaches to the problem, with no consensus on a single solution or method. Some participants agree on the complexity of the problem, while others propose different strategies without resolving the overall question.

Contextual Notes

The discussion contains unresolved mathematical steps and assumptions regarding the relationships between the heights and widths involved in the problem. The exact method for arriving at a solution remains unclear, with multiple approaches suggested.

Who May Find This Useful

Readers interested in geometry, mathematical problem-solving, and the exploration of complex relationships in physical scenarios may find this discussion engaging.

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

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