Two vertical walls, AB and CD are distance 2 metres apart.

A       C
|      /|
|    / |
|  /   |
|/     |
B       D  One ladder (1)rests at the foot of wall AB and leans against wall CD

Another ladder (2)rests at the foot of wall CD and leans against wall AB

If top of ladder (1) is 5.2 metres up wall CD, and top of ladder (2) is 3.8 metres up wall AB 

At what height do they cross?

Simplest solution required

2.2m or have i missed the point.

Not that simple I suspect, I'll get a five year old to whom these things are so much easier to dp the calculus. And report back.

2.35m. Break it all down into triangles and then work out amgles and lenghts using tan, sine and cosine equations.

QED - wrong again
How about - (AB x CD)/(AB + CD) = 5.2*3.8/9 = 2.19555

think I would take the lift!

As the simplest solution is required, draw it on graph paper. I drew it on plain paper and got 2.2, given ladders are at least 0.05 thick, I think this is as accurate as you need. An took about 15 seconds to figure out.

'ere!!!!

Can I have a job like you lot have got, where no work needs to be done?  Is the graph paper part of your package, Alan, or is it a taxable benefit in kind?

GGGGGGGG

Come on GGGGG how many hours do you truly work? We all need diversion from defending the relm, or running multinationals.

Correct answer 2.195

Tis equivalent to two resistances in parallel


1/5.2  + 1/3.8 = 1/answer

0.193 +  0.263= 0.456

1/0.456 = 2.195

Seems like 2.195 has it.

Ok, some us us are still working.

Quickest one to answer if the walls are not 2 meters apart but 2 miles apart (note change to imperial). If you got that in, then what is the answer if the walls are infinity apart?

same answer

And thre proof in the infinite case is ......