# Geometry problem

1. Nov 10, 2004

### 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

Last edited: Nov 10, 2004
2. Nov 10, 2004

### Gokul43201

Staff Emeritus
$$\sqrt{9-w^2} + \sqrt{4-w^2} = \sqrt{(9-w^2)(4-w^2)}$$

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

Last edited: Nov 10, 2004
3. Nov 10, 2004

### 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.

Last edited: Nov 10, 2004
4. Nov 10, 2004

### Galileo

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?

5. Nov 10, 2004

### 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.

6. Nov 23, 2004

### relativelyslow

could someone explain this one for me? this intrigues me

7. Nov 24, 2004

### Galileo

Me too. C'mon Gokul43201!

8. Nov 24, 2004

### 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

9. Nov 24, 2004

### 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."

Last edited: Nov 24, 2004
10. Nov 24, 2004

### Gokul43201

Staff Emeritus
That's what I did. But if you want a complete solution, I'll post it a little later...no time now.

11. Dec 18, 2004

### alleycat2

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

Last edited by a moderator: Dec 18, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?