Geometry problem

[Galileo]

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


\begin{array}{cccccccc}
|&* & & & & & &| \\
|& &* & & & & &| \\
|& & &* & & &&/| \\
|& & & &* &/& &| \\
|& & &/& &* & &| \\
|&/& & & & &* &|
\end{array}

( :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:

[Gokul43201]

\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

[ceptimus]

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.

[Galileo]

\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?

[NateTG]

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.

[relativelyslow]

could someone explain this one for me? this intrigues me

[Galileo]

Me too. C'mon Gokul43201! :smile:

[TenaliRaman]

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

[GCT]

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."

[Gokul43201]

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.

[alleycat2]

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

( :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:


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.