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Topic: Stabilise the square! (Read 3369 times) |
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Wonderer
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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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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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Wonderer
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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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JocK
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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?
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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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Wonderer
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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? |
| 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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Icarus
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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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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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Sjoerd Job Postmus
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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!
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« Last Edit: Dec 17th, 2005, 11:16am by Joe Fendel » |
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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)?
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« Last Edit: Dec 17th, 2005, 3:13pm by towr » |
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JohanC
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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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Wonderer
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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!!!
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towr
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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?
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« Last Edit: Dec 17th, 2005, 3:17pm by towr » |
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JohanC
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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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Barukh
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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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towr
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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.
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« Last Edit: Dec 18th, 2005, 7:50am by towr » |
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SMQ
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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? --SMQ
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--SMQ
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towr
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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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« Last Edit: Dec 18th, 2005, 9:01am by towr » |
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Joe Fendel
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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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towr
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Re: Stabilise the square!
« Reply #18 on: Dec 18th, 2005, 9:38am » |
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Damn.. you're right.. 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)
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« Last Edit: Dec 18th, 2005, 9:42am by towr » |
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towr
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Two are probably enough, but certainly three should be stable, right? I should get out my Lego, I think
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« Last Edit: Dec 18th, 2005, 9:50am by towr » |
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towr
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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 I suppose I should just give up this approach..
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Sjoerd Job Postmus
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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 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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towr
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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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Barukh
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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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Joe Fendel
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Well, if they can cross, I think I can do it with 33 sticks.
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« Last Edit: Dec 19th, 2005, 2:40pm by towr » |
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