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Polygon with 2007 angles

  1. Jan 31, 2007 #1
    I have to find a solution to this problem until next week, so perhaps someone can help:

    There is a constant polygon with 2007 angles. Put the natural numbers 1,2,.. 4014 on each angle and the center of each side of the polygon, so that the amount of the 3 numbers (angle + center + corner) is equal on each side of the polygon.
  2. jcsd
  3. Jan 31, 2007 #2
    I don't quite understand what you're asking. What do you mean by "angle + center + corner"? By corner, do you mean the other angle?

    Try thinking about doing this with a smaller polygon with a manageable number of sides. What relationships do you notice that the numbers would need to have?
  4. Jan 31, 2007 #3

    Tom Mattson

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    Yes, and what's a "constant polygon"? I assume that it's a "regular polygon", but I'd like for you to confirm that.
  5. Feb 1, 2007 #4


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    This is an interesting question. I haven't figured it out yet, though I can do small polygons with a computer program. I can't quite see the pattern. But the geometry of the polygon really doesn't matter. It's really just an exercise in ordering numbers 1 to N so selected groups of three add up to a constant.
  6. Feb 2, 2007 #5
    Dick the original description by Cavemiss is at best ambiguous but you have obviously made a well defined problem out of it. I would be interested in the statement of the problem you are attempting to solve.
    Last edited: Feb 2, 2007
  7. Feb 2, 2007 #6


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    I interpreted it to be this. Take an N sided polygon. Label each edge and each vertex with one of the numbers 1 to 2N with no repeated numbers. The challenge is to do this in such a way that for each edge, the sum of the number on the edge and the numbers on the two adjacent vertices is the same for each edge.
  8. Feb 2, 2007 #7
    Does look interesting. Perhaps it needs a wider audience than will be found in the homework forum. Perhaps you could restate in in General Maths
  9. Feb 6, 2007 #8
  10. Feb 6, 2007 #9
  11. Feb 6, 2007 #10
    ha... nice question. there is actually a simple way to construct something like this (there are probably many other ways)..

    In order to not give it away while providing some help, I'll demonstrate what happen if you have a 7-gon, and you need to label them 1 to 14 so that each vertex-edge-vertex pair have the same sum.

    the idea is simple, what you want to do is first put 1-7 on the corners so that the sum of each corner-corner pair is one less than the previous corner-corner pair.

    to explain this idea... take for instance, in7-gon. label the corners (_ indicate blank):
    1 _ 2 _ 3 _ 4

    continuing on,
    1 5 2 6 3 7 4

    you see, the sum of the two pairs are:
    and since the polygon goes in cycle, 4+1=5, we have:

    now we put the number on the edges. I don't want to give it all away... so as an exercise... how can you put the number 8-14 on the edges so that the sums are the same?
    applying the same idea, what about 2007-gon?

    Last edited: Feb 6, 2007
  12. Feb 7, 2007 #11


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    So THAT'S the clever way of organizing this that I have been so completely missing. Thanks for the tip!
  13. Feb 7, 2007 #12
    But if I put the numbers 1-7 on each center of the polygon and the numbers 8-14 on each corner I got this:


    The same happens with a 3-gon and 5-gon
    For example 3-gon:(1-3 on each center,4-6 on the corners)

    I've tried a lot and I got for one side of the 2007-gon the amount: 7028
    3013+1+4014= 7028

    Is it right or is there a mistake?Does the solution change if I put the smaller number on the corners as you did at the 7-gon?
  14. Feb 7, 2007 #13
    of correct you are correct. As i said... the solutions are not unique. this is just one nice way of constructing the solution. you can put 1-7 or 8-14 on the corners... either way is fine. As long as you have corner pairs that have sums one less than the previous corners pair, then you can put any arithmetic sequence with difference 1 in the middle and make it work.
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