The Ladder & the Box

  • Thread starter eldrick
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  • #1
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A classic problem - one of the highest ever on

diffuculty solving / describing simple problem

scale


lets see how good you are :

A ladder 4m long , is leaning against a wall in such a way that it just touches a box, 1m by 1m.

How high is the top of the ladder above the floor ?
 

Answers and Replies

  • #2
Curious3141
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eldrick said:
A classic problem - one of the highest ever on

diffuculty solving / describing simple problem

scale


lets see how good you are :

A ladder 4m long , is leaning against a wall in such a way that it just touches a box, 1m by 1m.

How high is the top of the ladder above the floor ?

It's not difficult at all, just tedious. You get a quartic that can be solved using one of the methods detailed in the tutorial I linked : https://www.physicsforums.com/showthread.php?t=119284
 
  • #3
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Curious3141 said:
It's not difficult at all, just tedious. You get a quartic that can be solved using one of the methods detailed in the tutorial I linked : https://www.physicsforums.com/showthread.php?t=119284

Maybe tedious, but it's still intriguing !

Let's see the detailed solution...
 
  • #4
Curious3141
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eldrick said:
Maybe tedious, but it's still intriguing !

Let's see the detailed solution...

I'm sorry, but I'm not a masochist. :biggrin: I know it can be done, I know how to do it algebraically, and I know how to estimate it numerically. That's good enough for me.

If you'd like to edify others, you could work the thing out in full. :smile:
 
  • #5
DaveC426913
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There is not enough information given to complete the solution.

At least, there isn't unless we make some huge assumptions - ones that you really need to state in the problem.

See attached diagram for arrangements that meet the criteria as specified in the problem.
 
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  • #6
DaveC426913
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On the other hand, the problem also didn't state that it had a unique solution.

Thus, my answer:

The top of the ladder above the floor can range between 0m and 5m.
QED.

Where's my prize!
 
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  • #7
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Curious3141 said:
I'm sorry, but I'm not a masochist. :biggrin: I know it can be done, I know how to do it algebraically, and I know how to estimate it numerically. That's good enough for me.

If you'd like to edify others, you could work the thing out in full. :smile:

i'd prefer to see the "brute-force" solution

i've only got the lateral-thinking "similar triangles" solution in my book :

i want to see "brute-force" !
 
  • #8
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listen boyz

it has a unique solution :

~ 3.76m

now show us this algebraically !!!
 
  • #9
DaveC426913
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eldrick said:
it has a unique solution :

~ 3.76m
Not unless you restate the problem...
 
  • #10
Curious3141
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eldrick said:
i'd prefer to see the "brute-force" solution

i've only got the lateral-thinking "similar triangles" solution in my book :

i want to see "brute-force" !

Similar triangles gives a quartic in one of the dimensions. The form of the quartic depends on which length you take.

The quartic is reducible to a quadratic with some manipulation.

There is no unique answer because it's not stipulated whether the ladder is higher up the wall or farther up the wall (mutually exclusive conditions). Therefore the answer can be 1.362 meters OR 3.761 meters. The exact answer is [tex]h = \frac{4}{2 \pm \sqrt{5 - \sqrt{17}}}[/tex].

(1.362^2 + 3.761^2 = 4^2, this is the obvious symmetry).
 
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  • #11
matt grime
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Please can I ask that you stop posting these 'classical' puzzles. I don't think that this is the correct arena for doing that. At least use the math Q and A game thread that was set up for this very purpose.
 
  • #12
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matt grime said:
Please can I ask that you stop posting these 'classical' puzzles. I don't think that this is the correct arena for doing that. At least use the math Q and A game thread that was set up for this very purpose.

if you are the moderator of this forum, i humbly apologise & will do so

if you are not the moderator, then i suggest you mind your own business
 
  • #13
matt grime
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No, I am not a moderator, it was just a polite request. (Prefaced by the word 'please'.) And I did point you in the direction of a stickied thread that would almost serve you perfectly before you start a new post every well known puzzle from a Martin Gardner book, or whereever these are from. You're more than welcome to ask a moderator to berate me for overstepping the mark (they're the ones indicated as moderators on the forum list page next to the forum name). There really ought to be a puzzles section or something for this kind of thing.
 
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  • #14
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then i suggest you do mind your own business

when you become a moderator, then tell me what to do

this is not from a gardner book ( he is NOT the the only puzzle setter in the world )

the only puzzles i post are the ones that i found extremely stimulating & intellectualy challenging
 
  • #15
Curious3141
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eldrick said:
then i suggest you do mind your own business

when you become a moderator, then tell me what to do

this is not from a gardner book ( he is NOT the the only puzzle setter in the world )

the only puzzles i post are the ones that i found extremely stimulating & intellectualy challenging

Dude, why all the hostility ? It's not good for the heart, and you should know, being a specialist in that organ. :smile:

Matt isn't saying don't post puzzles, he's just saying this is not the best place for it. And he's right - there's a complete section devoted to Brain Teasers and another for specific Math challenges (which is the thread he referred you to). This forum is more for "genuine" problems that are unrelated to homework - in the sense that the poster does not know the answer or needs some help or someone wants to discuss an important open question. Not for well-worn puzzles.
 
  • #16
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Curious3141 said:
Dude, why all the hostility ? It's not good for the heart, and you should know, being a specialist in that organ. :smile:

Matt isn't saying don't post puzzles, he's just saying this is not the best place for it. And he's right - there's a complete section devoted to Brain Teasers and another for specific Math challenges (which is the thread he referred you to). This forum is more for "genuine" problems that are unrelated to homework - in the sense that the poster does not know the answer or needs some help or someone wants to discuss an important open question. Not for well-worn puzzles.

point taken - & i will do so in future

back to the puzzle - i'm not sure if you can have 2 solutions - the answer in the book is 2.76m : the alternative answer of 1.36m looks intuitively difficult as the box is already 1m high & for the "small" angle involved between ladder & box, it woud seem that the ladder woud have to be a lot longer than 4m to satisfy the criteria ( i.e. touch the ground ) ?
 
  • #17
Curious3141
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eldrick said:
point taken - & i will do so in future

back to the puzzle - i'm not sure if you can have 2 solutions - the answer in the book is 2.76m :

3.76


the alternative answer of 1.36m looks intuitively difficult as the box is already 1m high & for the "small" angle involved between ladder & box, it woud seem that the ladder woud have to be a lot longer than 4m to satisfy the criteria ( i.e. touch the ground ) ?

Sure it's possible, just rotate the picture 90 degrees. Either answer is possible as long as there's enough friction between the ladder and the floor to hold that position.
 
  • #18
shmoe
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eldrick said:
back to the puzzle - i'm not sure if you can have 2 solutions - the answer in the book is 2.76m : the alternative answer of 1.36m looks intuitively difficult as the box is already 1m high & for the "small" angle involved between ladder & box, it woud seem that the ladder woud have to be a lot longer than 4m to satisfy the criteria ( i.e. touch the ground ) ?

When the top of the ladder is 3.76... high on the wall, how far from the wall is the base of the ladder?
 
  • #19
MathematicalPhysicist
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eldrick said:
listen boyz

it has a unique solution :

~ 3.76m

now show us this algebraically !!!
i got this system of equations:
(1+x)^2+h^2=4^2
1^2+x^2=y^2
(h-1)^2+1^2=(4-y)^2
where h is the height, x is the horizontal distance from the wall, and y is part of length of the ladder.
 
  • #20
3.76090563295442m (1.36219999266324m from base of the wall) or 1.36219999266324m (3.76090563295442m from base of the wall) :)

I wrote some software to calculate it, all variables and angles (Attached), the software errors if you give it impossible numbers (Only very basic error handlers).

If anyone is interested in a mind-numbingly simple explaination, I'm willing to elaborate :)
 

Attachments

  • LadderProblem.zip
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  • #21
https://www.physicsforums.com/attachment.php?attachmentid=17013&stc=1&d=1230570459

Known:
(A+B) = 4
H = 1

Workings:
(P+Q) = -H + Sqr(H ^ 2 + (A+B) ^ 2)

P = (P+Q) / 2 + Sqr((P+Q) ^ 2 / 4 - H ^ 2)
Q = (P+Q) / 2 - Sqr((P+Q) ^ 2 / 4 - H ^ 2)

A = (A+B) / 2 + Sqr((A+B) ^ 2 / 4 - H * (P+Q))
B = (A+B) / 2 - Sqr((A+B) ^ 2 / 4 - H * (P+Q))

Height (Given H (Box Side Length) and (A + B) Length of ladder):
Height = H + (-H + Sqr(H ^ 2 + (A+B) ^ 2)) / 2 + Sqr((-H + Sqr(H ^ 2 + (A+B) ^ 2)) ^ 2 / 4 - H ^ 2)
or
Height = H + (-H + Sqr(H ^ 2 + (A+B) ^ 2)) / 2 - Sqr((-H + Sqr(H ^ 2 + (A+B) ^ 2)) ^ 2 / 4 - H ^ 2)

And in numerical form:
Height = 1 + (-1 + Sqr(1 ^ 2 + 4 ^ 2)) / 2 + Sqr((-1 + Sqr(1 ^ 2 + 4 ^ 2)) ^ 2 / 4 - 1 ^ 2)
or
Height = 1 + (-1 + Sqr(1 ^ 2 + 4 ^ 2)) / 2 - Sqr((-1 + Sqr(1 ^ 2 + 4 ^ 2)) ^ 2 / 4 - 1 ^ 2)​

Angle Calculation (DEG):
Angle1 = ArcSin((P+H) / (A+B)) * 57.2957795130823
Angle2 = 180 - (Angle1 + 90)
Angle3 = 90
 

Attachments

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