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23 replies to this topic

#11 Jkyle1980

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Posted 29 February 2008 - 10:30 PM

I am new to the site, so I hope I am posting correctly. Let me say that I don't have the solution to what I am posting. It is something my dad presented to me as a child and I have never been able to find anyone able to give me a mathematical solution to the problem. I am hoping to get it here. It goes like this....

Imagine a building that is 10 foot x 10 foot x 10 foot on level ground. Running along side of the building is a 100 foot flag pole. The flag pole breaks at such a point that it just touches the opposite corner and the top of the flag pole lands touching the ground. At what point did the flag pole break?


I feel like there is a variable missing, possibly the distance from the corner of the building to the tip of the flag pole. After looking at it, I have a right triangle of unknown angles and sides. I have a point on the hypotenuse at (200^(.5), 10). The flagpole obviously broke less than halfway up.

Divide the tangent by the cosine of negative infinity times .2 and I think that the flagpole broke 48.5 feet above the ground. This would give the upper segment of the pole a length of 51.5 and the top of the pole would be on the ground 17.6 feet from the bottom of the flagpole. I sit back and idly wait for someone to disprove me; I am looking forward to it, in fact.
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#12 emeraldcity

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Posted 01 March 2008 - 05:26 AM

I feel like there is a variable missing, possibly the distance from the corner of the building to the tip of the flag pole. After looking at it, I have a right triangle of unknown angles and sides. I have a point on the hypotenuse at (200^(.5), 10). The flagpole obviously broke less than halfway up.

Divide the tangent by the cosine of negative infinity times .2 and I think that the flagpole broke 48.5 feet above the ground. This would give the upper segment of the pole a length of 51.5 and the top of the pole would be on the ground 17.6 feet from the bottom of the flagpole. I sit back and idly wait for someone to disprove me; I am looking forward to it, in fact.

Can you give me the equation and a drawing? I remember that I "trialed and errored" it, so to speak, when I was a kid and in order for the flag pole to break at such a point and have it touch the edge of the cube and then the top of the flag pole to just touch the ground it broke at about 12 feet.
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#13 Jkyle1980

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Posted 01 March 2008 - 06:31 AM

Can you give me the equation and a drawing? I remember that I "trialed and errored" it, so to speak, when I was a kid and in order for the flag pole to break at such a point and have it touch the edge of the cube and then the top of the flag pole to just touch the ground it broke at about 12 feet.


I was waiting for someone much better at this than me to post an equation to test my numbers. I just used trial and error to the nearest tenth of a foot. I have no equation, but the graphics match up. I would not be surprised if there were two answers: mine with the y-axis being longer than the x-axis and another with the x-axis being longer than the y-axis. That is why I thought there should be another given dimension. But with the tools I have over here, Microsoft Office and the basic Windows calculator, I can't be sure if I'm right.
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#14 bonanova

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Posted 03 March 2008 - 10:00 AM

Spoiler for Three solutions ...

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#15 Duh Puck

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Posted 03 March 2008 - 10:54 AM

[6] -200B3 + 13800B2 - 220000B + 1000000 = 0

Nice work. That very equation is written on a piece of paper at home, and I didn't know how to solve it. Didn't think of finding a website that would do it for me. <_<

You said that your first approach was iterative. I suspected that there would be a way to use calculus limits (I never took calculus, btw, but I have studied a bit on my own) to solve this, but I didn't know the math for it. For my edification, can you show a bit more how you arrived at a solution in this manner? Or did you just progressively plug numbers in until they got closer? That's how I would have done it programmatically, but that feels like cheating.

Incidentally, while I understand that there are three solutions to the equation, wouldn't it be more accurate to say that there are only two solutions to the problem, B1 and B2?
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#16 Lost in space

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Posted 03 March 2008 - 11:00 AM

At what point did the flag pole break?

Structural integrity failed at a point:

1a - time of failure (compromised structual integrity)
1b - time untill faliure occured - in situ.
2 - How did it become weak? (least strength - if wooden then, at lowest density, natural defects, manufacturing defect, decay, infestation, felling defects etc. - if metal then, rust, poor manufacture, weak weld etc)
3 - where along the section/axis did failure occur

Most people will answer for No. 3 probably due to the information available.
The others are not excluded other than by lack of information.
Is this a deliberate attempt to get people to respond, request more information?
Is it a flawed puzzle?

Spoiler for Sorry for waffle!

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#17 heatherlovesjade

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Posted 03 March 2008 - 12:09 PM

where exactly is the flag pole located on the side of the building- an angle is important.
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#18 bonanova

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Posted 03 March 2008 - 01:40 PM

where exactly is the flag pole located on the side of the building- an angle is important.

Judging from the discussion in the first few posts, the flagpole is on one corner of the building.
It then falls diagonally toward the building so that the broken part passed directly above the center of the roof.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#19 bonanova

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Posted 03 March 2008 - 01:49 PM

You said that your first approach was iterative. I suspected that there would be a way to use calculus limits (I never took calculus, btw, but I have studied a bit on my own) to solve this, but I didn't know the math for it. For my edification, can you show a bit more how you arrived at a solution in this manner? Or did you just progressively plug numbers in until they got closer? That's how I would have done it programmatically, but that feels like cheating.

Incidentally, while I understand that there are three solutions to the equation, wouldn't it be more accurate to say that there are only two solutions to the problem, B1 and B2?

I did the iteration by hand using an interpretive program - took only about 10 iterations.
You can also use Newton-Raphson method to get convergence in 3-4 iterations given a good 1st guess.
Didn't feel like being that elegant.

This is not cheating by the way -- don't feel guilty about going this route.
It's very legitimate numerical analysis technique.
If the equations were transcendental, you couldn't arrive at a closed form solution.
Numerical methods would be all that you had.

Sure. Only two of the three are physically realizable.
Whether "solution" refers to the puzzle or to the equations is up for grabs, but if it's the former there are only two.
If I had drawn the third line, it would have started on the upper right corner then gone to the ground
to the left of the picture, passing through the pole at 8.65... feet.
That would have satisfied the equations I set up, but not the actual conditions of the puzzle.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#20 Lost in space

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Posted 03 March 2008 - 02:18 PM

Judging from the discussion in the first few posts, the flagpole is on one corner of the building.
It then falls diagonally toward the building so that the broken part passed directly above the center of the roof.


It says running along the side - not clear to me, the puzzle would be simple otherwise I think


Spoiler for for - I think

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