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No. of diagonals of a polygon

  1. Jan 5, 2016 #1
    1. The problem statement, all variables and given/known data
    A polygon with n sides has a total of 1/p . n . (n-q) diagonals, where p and q are integers.

    (i) Find the values of p and q.

    2. Relevant equations


    3. The attempt at a solution
    Can someone just help me to start on this? I know that q = 3 and p = 2 but how?
     
  2. jcsd
  3. Jan 5, 2016 #2

    Samy_A

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    Take a polygon with n=10 (for example). How many diagonals can you draw from one of the corners?
     
  4. Jan 5, 2016 #3
    9
     
  5. Jan 5, 2016 #4

    Samy_A

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    No. A diagonal links the corner to one of the other non-adjacent corners. How many of these are there (still with n=10)?
     
  6. Jan 5, 2016 #5
    8
     
  7. Jan 5, 2016 #6

    Samy_A

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    No. A corner has 2 adjacent corners, and there is no diagonal linking a corner with itself. So, still with n=10, how many diagonals are there starting in one corner?
     
  8. Jan 5, 2016 #7
    If you take a triangle with 3 sides (obviously), there are no diagonals, right?
     
  9. Jan 5, 2016 #8

    Samy_A

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    Yes, that is correct. A corner of a triangle has two adjacent corners, and that's it. No possibility to draw a diagonal there.
     
  10. Jan 5, 2016 #9
    So a 4 sided polygon as only 1 diagonal, am I correct?
     
  11. Jan 5, 2016 #10

    Mark44

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    In future posts, you need to show more of an effort than this.
     
  12. Jan 5, 2016 #11

    Samy_A

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    No. A 4 sided polygon has 2 diagonals.
     
  13. Jan 5, 2016 #12
    But apart from trial and error is there any other way I can do this?

    4 sides polygon
    1/2 . 4 . (4-3) = 2 diagonals
    5 sides polygon
    1/2 . 5 . (5-3) = 5 diagonals
    6 sides polygon
    1/2 . 6 . (6-3) = 9 diagonals

    And so on...
     
  14. Jan 5, 2016 #13
    Apologies
     
  15. Jan 5, 2016 #14

    Samy_A

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    Yes, there is. I was trying to get you there by reasoning.

    So, still with n=10, how many diagonals are there starting in one corner?
     
  16. Jan 5, 2016 #15
    4 sided polygon
    1/2 . 4 . (4-3) = 2 diagonals
    5 sided polygon
    1/2 . 5 . (5-3) = 5 diagonals
    6 sided polygon
    1/2 . 6 . (6-3) = 9 diagonals
    ...
    10 sided polygon
    1/2 . 10 . (10-3) = 35 diagonals
     
  17. Jan 5, 2016 #16
    Is there anyway I can do this problem without doing it like i have, guessing what p and q are...

    How can one prove it?
     
  18. Jan 5, 2016 #17

    Samy_A

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    By reasoning how many diagonals there are. It is possible, it is easy, but you have to try it.

    Outline:
    Take one corner, and compute the number of diagonals it lies on. We were almost there: there is a diagonal linking that corner to any other corner, except itself and its two adjacent corners. That makes n - 1 - 2 = n -3 diagonals from that one corner.
    Now, this is the case for each one of the n corners. So now you can compute the total number of diagonals, but don't forget that each diagonal connects two corners.

    Good night.
     
  19. Jan 5, 2016 #18
    Got it thanks!
     
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