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Thread: Ouch! Paper cut.

  1. #1 Ouch! Paper cut. 
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    Let's play a few paper games. I will ask you to find a way to achieve something with paper using basic geometric knowledge (like a compass & ruler step by step process). The person to figure it out can propose his / her own problem. Then the person to solve that can do his / her own. (if it is unsolvable using only the given resources, then impossible is an acceptable answer) All the problems involve paper and its dimensions. Only the stated objects can be used. Got it? Let's begin.

    --------------------------------------------------------------------------------------------------------------------------------

    A simple one:

    Materials: a letter size sheet of paper, and a ruler and protractor.

    Conditions: The ruler only measures discretely, as in only 1 unit, 2 units, 3 units, etc. are labelled. You cannot measure a length other than the non-positive integers on the ruler. Folding is allowed.

    Task: Create a length that you know is irrational.

    Hint: It's a little tricky since you can't measure an irrational length with your ruler. So you'll have to make something you know is irrational based off the properties of what you can make. Pretty easy.


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  3. #2  
    Forum Senior TheObserver's Avatar
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    You could just fold the paper until you get a square with side lengths of 1 unit measured with the ruler. Then fold the paper from corner to opposite corner to get a root 2 hypotenuse.


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  4. #3  
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    Yeah, pretty much what I was thinking. In general, you can define a rectangle with lengths and so that is known to be irrational. Basically, any possible answer involves an irreducible square root or some multiple of it. I'm actually interested if there's an alternate algorithm instead of using Pythagorean identities.

    Do you have a task in mind that might prove puzzling?
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  5. #4  
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    I won't participate in this but I must say this is really fun
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    Quote Originally Posted by halorealm View Post
    Do you have a task in mind that might prove puzzling?
    Thats an even more interesting problem than solving yours. Let me think about it and try to come up with something cool.
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    New challenge. ( not ignoring you Observer, whenever you're ready )

    Materials: a rectangular 2 by 4 sheet of paper

    Task: With the basic arithmetical operations ( + - x / ), using only edge length measurements of the paper, find some relationship between them that equals the Golden ratio.

    Conditions: You can only make a relationship between the lengths of actual edges of the paper. Folding is allowed. Any transformations of the original sheet by folding is allowed, and thus you can use the length of the edges of that transformation.

    Again, in summary: You have a 2 by 4 rectangle. Try to find a relationship between the edge lengths that uses only the 4 basic operations so that it equals the Golden Ratio. A little trickier than before. Good luck!
    Last edited by halorealm; July 26th, 2012 at 04:58 PM. Reason: typo
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  8. #7  
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    No takers? Okay, here's a hint for doing it one way.

    You have to fold the paper three times before you can get the constants you must relate arithmetically so that you end up with the Golden ratio. I won't specify how you have to fold the paper though. Also, think about what algebraic representations of the Golden ratio exist. This one happens to use the most common. Now it should be quite easy Good luck.
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  9. #8  
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    Okay. This hasn't gotten much attention, so I'll just throw out the solution.

    2 x 4 rectangle. Fold in half so now you have a 1 x 4 rectangle. Fold in half again the other way so now you have a 1 x 2 rectangle. Fold the paper so that a diagonal is made corner to opposite corner. Now you have a 1, 2, right triangle. Now we have the edge lengths needed for our Golden ratio. Add the diagonal and the side length 1, and divide the sum by the side length 2. .

    If anyone has something they want to share, have a go.
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  10. #9  
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    If anyone can prove that the three angular bisectors of a triangle intersect at one point; the three medians of a triangle intersect at one point; the three heights of a triangle intersect at one point, WITHOUT GOOGLING FOR ANSWERS, I will be impressed.

    P.S. I myself have proved the above statements before, but it was pretty hard, especially the last one.
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  11. #10  
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    Quote Originally Posted by Wise Man View Post
    If anyone can prove that the three angular bisectors of a triangle intersect at one point; the three medians of a triangle intersect at one point; the three heights of a triangle intersect at one point, WITHOUT GOOGLING FOR ANSWERS, I will be impressed.

    P.S. I myself have proved the above statements before, but it was pretty hard, especially the last one.
    Sounds like 9th Grade geometry all over again. I'll see what I can do.
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  12. #11  
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    ya right. Believe me, it's not that easy. You might have to use vectors if you are a dunce at geometry.
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