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Math Help - dividing an n-gon into rhombuses

  1. #1
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    dividing an n-gon into rhombuses

    I'm trying to prove that any regular and convex 2n-polygon can be divided into rhombuses.

    I proceed by induction.
    BASE CASE: For n=2 this is obviously true because a square is a rhombus.
    INDUCTIVE HYPOTHESIS: suppose the statement is true for a polygon with 2n sides.

    Now to make a polygon with 2(n+1) sides, we take the 2n-polygon which has already been divided into rhombuses. We take any n sides and replace them with n+2 sides, thus 'elongating' the 2n-polygon (so to speak) in such a way that we draw n lines parallel to those n sides, at some distance. And then join them with the 2n-polygon with a line at each end. Hence n more rhombuses are added to the 2n-polygon and this new shape is a 2n+2-polygon.

    This completes my proof.
    I have a feeling there's some loophole in this argument :s Can someone kindly help?
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  2. #2
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    Sorry, I can't visualize your procedure because there are too many possible variants.

    Now to make a polygon with 2(n+1) sides, we take the 2n-polygon which has already been divided into rhombuses. We take any n sides
    Absolutely any n sides? Not necessarily connected?
    we draw n lines parallel to those n sides, at some distance
    This looks suspicious because if those lines are not drawn at a precise distance, then the resulting figure will not all have sides equal, i.e., will not be a regular polygon.
    And then join them with the 2n-polygon with a line at each end.
    There are so many ways to "join them with the 2n-polygon". It is not cleat what you mean.

    Could you draw a picture how you transform a hexagon into an octagon?
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  3. #3
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    I just tried doing it on paper and I realized that this way the polygon i come up with is extremely lopsided its not regular at all.

    To answer your questions first, I take any n consecutive sides and draw lines parallel to them at a distance that is equal to any one side of the 2n-gon.

    The main thing that I missed before is, that when we go from a 2n-gon to 2(n+1)-gon.... EACH interior angle must change too. So we can't replace just half the sides with new ones like I was doing.

    So how then should I proceed? Any hints?
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  4. #4
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    From google search, the answer is here

    http://mathcircle.berkeley.edu/newsi...002examsol.pdf

    I haven't looked at it though.
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  5. #5
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    Thanks. I think I understood solution #2 on p. 4.

    There is also a , but I have not figured out yet how to turn into a precise proof.
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  6. #6
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    @undefined: thank you for that link! I'm glad to see that my initial attempt at the proof was pretty close!

    @emakarov: So the outline of the proof is that at nth step we add n new rhombuses starting from the top, in clockwise direction. But to complete the proof we need to:
    1) prove that at every step the polygon is regular.
    2) We need to give the angle at which we construct the new rhombus at every step.
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  7. #7
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    So the outline of the proof is that at nth step we add n new rhombuses starting from the top, in clockwise direction. But to complete the proof we need to:
    1) prove that at every step the polygon is regular.
    2) We need to give the angle at which we construct the new rhombus at every step.
    Hmm, I am not sure what solution you are talking about. Both solutions on p. 4 of the PDF document the link to which appears above start with the bottom side of the polygon. Further, solution #1 does not use induction (or, if it does, its statement is different from the problem). Solution #2, which I found clearer, does not mention steps. Finally, solution #2 works with more general EP-gons, which are not necessarily regular. It often happens in proofs by induction that one has to generalize the induction statement.
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