Workshop garrotte

[Raggle]

Marv, I first came across that problem with the exact figures when I first got on the 'net in '98 (20ft ladder, 6ft cube) and have been entertained by it ever since, though I gained a bad reputation for asking all and sundry for an answer.

My bro-in-law, freshly in receipt of his pure maths degree, came back with a 6 line method in 2 days. But wildly wrong answer. He hadn't done the easy bit, checking with Pythagoras. He swore his maths was right.

I posed the question on a site which was set up to organise our centenary school reunion and a young whippersnapper from 2 years below me came back with a superb method but got his answer one foot out. Fell on the simple arithmetic.

No, I'm still not competent enough to do it, though I can follow this chap's method and am still in awe. I have seen a shorter method since.

I will post the whippersnapper's method, on request, but not until this thread runs its course.

Ray

[Tennessee Whiskey]

Zing, good one Jerry. Nope really just 18

[Raggle]

Oops! Your ladder is 25, mine is 20

Still, the problem is no easier in either case

Ray

[mklotz]

Ray,

Yeah, it's been around for quite a while.  An oldie but goodie, so to speak.

I'm hoping our resident genii here on the forum will come up with a spectacularly concise new solution.

When I first attempted it, I came up with a quartic equation.  It was the first time in my whole career that I had legitimately arrived at a quartic.  With some clever substitutions, though, one can work around the quartic and only have to solve quadratics.

[Captain Jerry]

Please clarify what is meant by 6 unit cubical box. 6 units per edge, side?
Mosey

Each edge is 6 units long.

The way that I read the problem was that the 6 unit cubical box was a box of 6 cubical unit volume so  sides (and edges) would be equal to the cube root of 6. Aren't sides and edges the same? And  it seems that there should be two solutions but I don't have it yet.

Jerry

[Pete49]

the answer is......42......
Pete

[Raggle]

Here's where I found it all those years ago

http://www.stirlingsouth.com/richard/math.htm

Though the question in question is not trig

Ray

[Nickle]

This is a tough one. I got to the point of arriving at the quartic and am looking for another approach. Its been a while since I've had a shot at a set of simultaneous equations and I'm not sure those skills are still present. Lucky you didn't ask us to account for sag in the ladder due to gravity.

Regards

Nick

[Raggle]

That's why the crate is there, Nick

Ray

[mklotz]

And  it seems that there should be two solutions but I don't have it yet.

Very insightful, Jerry, and quite correct.  Think about it.  There are two ways to place the ladder so it's simultaneously touching the box, the wall and the ground.  In one position the ladder is close to vertical, reaching as far up the wall as possible.  In the other it's more nearly horizontal, with the ground end far from the wall.  Actually, these two cases are really the same case with the roles of the wall and ground interchanged.

Either solution is "correct" but by convention it's usually the first that is quoted.

[mklotz]

Aren't sides and edges the same?

No, they aren't.  A cube has 6 sides (Faces) and 12 Edges.  A point where Edges meet is called a Vertex and the cube has 8 of them.

Euler gave us a relation among these values for regular solids...

V - E + F = 2

8 - 12 + 6 = 2

As an example, a tetrahedron (4 equilateral triangle Faces) has 6 Edges and 4 Vertices.

4 - 6 + 4 = 2

Try it for an icosahedron (I don't have the patience.)

[Mosey]

This is a tough but interesting problem, but I admit that I am more interested in how light falls on the cube and shadows are cast across it by the ladder. I'm still working on the geometry.
Mosey

[Maryak]

I gave up because I couldn't remember how to manipulate quadratic equations and too lazy to look it up.

Best Regards
Bob

[Captain Jerry]

If the distance from the floor to the top of the ladder is 'A' then
 
(A-6)² + (6 + 36/(a-6) )² - 625 = 0

or A is a little more than 23 ¹/₂ but that is as far as I can go.  My eyes just slammed shut.

Jerry

[Alan Haisley]

Marv,
This one had me going for a while. After a long time demonstrating that a+b=b+a and similar silliness,    I came up with an answer that I liked. 
By the way, two other answers come to mind: 31 and 30.27