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   Stabilise the square!
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   Author  Topic: Stabilise the square!  (Read 3274 times)
Wonderer
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Stabilise the square!  
« on: Dec 17th, 2005, 12:47am »
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Stabilise the square!
 
You are given 4 identical chopsticks; each has a small connecter at both ends.  With the connectors, you can use the 4 chopsticks to form a perfect square.  However, this square is not stable.  Applying pressure will force the square to change shape.  
 
Now, you are given an infinite amount of such chopstick.  Can you find a way so that this square can be stabilised?
 
Please note, you can only connect chopsticks with their ends.  
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JocK
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Re: Stabilise the square!  
« Reply #1 on: Dec 17th, 2005, 1:18am »
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With four more chopsticks I'd build a pyramid with a square base.
 
 
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xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Re: Stabilise the square!  
« Reply #2 on: Dec 17th, 2005, 1:36am »
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on Dec 17th, 2005, 1:18am, JocK wrote:
With four more chopsticks I'd build a pyramid with a square base.
 
 

 
No, I don't think so.  This was also the first answer I came across,  bust later realized that the square can still change shape into a 3D object.  Even building a Octhedron structure could not stabilize the square.
 
The square has to be stabilized in 2D.
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Re: Stabilise the square!  
« Reply #3 on: Dec 17th, 2005, 2:27am »
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on Dec 17th, 2005, 1:36am, Wonderer wrote:

 
No, I don't think so.  This was also the first answer I came across,  bust later realized that the square can still change shape into a 3D object.  Even building a Octhedron structure could not stabilize the square.
 
The square has to be stabilized in 2D.

 
Are you saying you can build an octahedron with equal edge lengths that is not regular?  Huh
 
 
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.

xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Re: Stabilise the square!  
« Reply #4 on: Dec 17th, 2005, 3:11am »
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on Dec 17th, 2005, 2:27am, JocK wrote:

 
Are you saying you can build an octahedron with equal edge lengths that is not regular?  Huh
 
 

 
Oooops, sorry. My mistake.  What was I thinking .  Octahedron does work.
 
Now, what if you are not allowed to build 3D structures?  You can only connect chopsticks in 2D.  Can you still stabilize the angles in the square?
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Re: Stabilise the square!  
« Reply #5 on: Dec 17th, 2005, 7:16am »
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The only rigid structure that you can make is a triangle. Further, since any side longer than 1 that you build will have a hinge in the middle, the only rigid triangle that you can build is an equilateral triangle of sidelength one.
 
Can you build a square from equilateral triangles? No. All angles will be multiples of 60o. So right angles cannot be constructed.
 
Now, I have assumed that your chopsticks can only meet at their ends, since crossing elsewhere would require 3-D to allow one to go over the other. If you accept this minor departure from 2-D, however, you get more freedom, and can actually build 90o angles. I do not believe you can make them rigid, but this is harder to see.
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Re: Stabilise the square!   stablesquare.zip
« Reply #6 on: Dec 17th, 2005, 8:24am »
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The only rigid structure you can make without having surrounding structures is a triangle, as Icarus said.
 
But, you can expand on this.
 
You can also have two linked diamonds stable.
 
How to build two stable diamonds?
 
Start with a regular hexagon with radius one chopstick.
 
remove one chopstick, notice it's still stable? [in 2d].
 
Now, if you put two diamonds in this place, with angle 30*... you notice these diamonds aren't stable, but they will be when you put a triangle on the  open part of the diamonds.
 
Now...
 
place another diamond in the other direction, meeting back at the hexagon. Notice a 90* angle?
 
Now, we don't have a stable square yet, but we're getting there.
 
Just add two more triangles to your structure, and a square.
 
At least, I think it is stable... can anyone doublecheck?
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Joe Fendel
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Re: Stabilise the square!  
« Reply #7 on: Dec 17th, 2005, 11:15am »
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I like your creativity, Sjoerd, but it doesn't look stable to me.  I could be wrong, though.
 
It looks to me like the lower-left two-triangle diamond can, for example, be "pushed" up against the almost-hexagon.  This has the effect of pushing the upper-left triangle up and to the right and also deforming the square into a rhombus.
 
But hey, I don't have any better ideas!
« Last Edit: Dec 17th, 2005, 11:16am by Joe Fendel » IP Logged
towr
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Re: Stabilise the square!  
« Reply #8 on: Dec 17th, 2005, 3:11pm »
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What if you surround a hexagon with 6 squares (connected to the hexagon) and 6 triangles (connecting the squares)?
« Last Edit: Dec 17th, 2005, 3:13pm by towr » IP Logged

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Re: Stabilise the square!  
« Reply #9 on: Dec 17th, 2005, 3:12pm »
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What about the following?

1) Create a chain of equilateral triangles to get a stable straight line of 3 chopsticks
2) do the same for 4 chopsticks
3) and once again for 5 chopsticks
4) form a triangle with sides 3,4 and 5, making sure the construction chopsticks are all outside the triangle; if memory serves well, this should form a right angled triangle;  
5) add two more chopsticks in the right corner of the triangle to form the square
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Re: Stabilise the square!  
« Reply #10 on: Dec 17th, 2005, 3:14pm »
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on Dec 17th, 2005, 3:12pm, JohanC wrote:
What about the following?

1) Create a chain of equilateral triangles to get a stable straight line of 3 chopsticks
2) do the same for 4 chopsticks
3) and once again for 5 chopsticks
4) form a triangle with sides 3,4 and 5, making sure the construction chopsticks are all outside the triangle; if memory serves well, this should form a right angled triangle;  
5) add two more chopsticks in the right corner of the triangle to form the square

 
Bingo!!  Correct answer!!! Cheesy
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Re: Stabilise the square!  
« Reply #11 on: Dec 17th, 2005, 3:16pm »
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That obviously works, very good
So the next question is, do you really need to use that many chopsticks?
« Last Edit: Dec 17th, 2005, 3:17pm by towr » IP Logged

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Re: Stabilise the square!   StableSquare.png
« Reply #12 on: Dec 17th, 2005, 3:42pm »
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on Dec 17th, 2005, 3:16pm, towr wrote:
That obviously works, very good

Thanks.
I'm attaching a drawing.
 
on Dec 17th, 2005, 3:16pm, towr wrote:
So the next question is, do you really need to use that many chopsticks?

Probably not.
Still some food for thought, while waiting for the chopstick food to be served ....
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Re: Stabilise the square!  
« Reply #13 on: Dec 17th, 2005, 11:32pm »
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Nicely done, Johan!
 
BTW, are the chopsticks allowed to cross?
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Re: Stabilise the square!   rigid_squares.png
« Reply #14 on: Dec 18th, 2005, 7:37am »
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Here's a drawing for the solution I offered a minute before JohanC gave his (inspired by Sjoerd's solution).
It needs fewer chopsticks, and also has more squares.  
And it has a certain elegance imo.
 
Of course you can still remove at least one more chopstick, from the hex.
« Last Edit: Dec 18th, 2005, 7:50am by towr » IP Logged


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Re: Stabilise the square!  
« Reply #15 on: Dec 18th, 2005, 8:49am »
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on Dec 18th, 2005, 7:37am, towr wrote:
Of course you can still remove at least one more chopstick, from the hex.

In fact, can't you remove all six inner chopsticks from the hex?  displacing any of the corners of the hex requires distorting the equilateral triangle outsie that corner, yes?
 
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Re: Stabilise the square!  
« Reply #16 on: Dec 18th, 2005, 8:57am »
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on Dec 18th, 2005, 8:49am, SMQ wrote:
In fact, can't you remove all six inner chopsticks from the hex?  displacing any of the corners of the hex requires distorting the equilateral triangle outsie that corner, yes?

No, if you remove the six inner sticks, you get 4 parallel chopsticks in a row, with nothign to stop them tilting. And that three times in different directions.
 
Maybe two, on the outside the hex, one on one side and the other opposite.
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Re: Stabilise the square!  
« Reply #17 on: Dec 18th, 2005, 9:26am »
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on Dec 18th, 2005, 7:37am, towr wrote:
Here's a drawing for the solution I offered a minute before JohanC gave his (inspired by Sjoerd's solution).
It needs fewer chopsticks, and also has more squares.  
And it has a certain elegance imo.
 
Of course you can still remove at least one more chopstick, from the hex.

 
This also looks unstable to me.  It seems like you could rotate all 6 outer triangles simultaneously so that the shape is a star-of-david, with 12 chopsticks doubled, for example.
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Re: Stabilise the square!  
« Reply #18 on: Dec 18th, 2005, 9:38am »
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Damn.. you're right.. Embarassed
 
Could you stabilize it by overlapping two or three of these? (Where the overlap between two is a square with a triangle on both sides)
« Last Edit: Dec 18th, 2005, 9:42am by towr » IP Logged

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Re: Stabilise the square!   rigid_square_3.png
« Reply #19 on: Dec 18th, 2005, 9:49am »
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Two are probably enough, but certainly three should be stable, right?
 
I should get out my Lego, I think Tongue
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Re: Stabilise the square!  
« Reply #20 on: Dec 18th, 2005, 10:29am »
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meh.. doesn't work either Sad
I suppose I should just give up this approach..
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Re: Stabilise the square!  
« Reply #21 on: Dec 18th, 2005, 11:06am »
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on Dec 18th, 2005, 10:29am, towr wrote:
meh.. doesn't work either Sad
I suppose I should just give up this approach..

It might not work, but it certainly looks cool...
 
Seems like the 3-4-5 is the only certainty now...
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Re: Stabilise the square!  
« Reply #22 on: Dec 18th, 2005, 11:13am »
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on Dec 18th, 2005, 11:06am, Sjoerd Job Postmus wrote:
Seems like the 3-4-5 is the only certainty now...
And any other Pythagorean triple (e.g. 5 12 13), but those are all much larger.
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Re: Stabilise the square!  
« Reply #23 on: Dec 18th, 2005, 11:31am »
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There exist configurations with fewer chopsticks. How fewer - depends on whether it is allowed or not to cross them.
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Re: Stabilise the square!   33sticks.jpg
« Reply #24 on: Dec 19th, 2005, 12:21pm »
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Well, if they can cross, I think I can do it with 33 sticks.
« Last Edit: Dec 19th, 2005, 2:40pm by towr » IP Logged

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