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Thread: Polygon of n Sides

  1. #1
    Senior Member harpazo's Avatar
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    Polygon of n Sides

    If a polygon of n sides has (n/2)(n - 3) diagonals, how many sides will a polygon with 65 diagonals have? Is there a polygon with 80 diagonals?

    Must I equate (n/2)(n - 3) to 65 and then again to 80 and solve for n individually?
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    Re: Polygon of n Sides

    Yes that's correct.

    When you've done the solving, check your answers back in the context of the question.
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  3. #3
    Senior Member harpazo's Avatar
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    Re: Polygon of n Sides

    Quote Originally Posted by Debsta View Post
    Yes that's correct.

    When you've done the solving, check your answers back in the context of the question.
    Ok. Thank you.
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  4. #4
    Senior Member harpazo's Avatar
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    Re: Polygon of n Sides

    (n/2)(n - 3) = 65

    2[(n/2)(n - 3)] = 65(2)

    n(n - 3) = 130

    n^2 - 3n = 130

    n^2 - 3n - 130 = 0

    (n - 13)(n + 10) = 0

    n - 13 = 13

    n + 10 = -10 is rejected.

    A polygon with 65 diagonals has 13 sides.

    What about a polygon with 80 diagonals?

    (n/2)(n - 3) = 80

    n(n - 3) = 160

    n^2 - 3n - 160 = 0

    After using the quadratic formula, I got the following values for n:

    n = (3/2) - [(sqrt{649})/2]

    n = (3/2) + [(sqrt{649})/2]

    The question now is: which one is correct?
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    Re: Polygon of n Sides

    Quote Originally Posted by harpazo View Post
    ((n/2)(n - 3) = 80

    n(n - 3) = 160

    n^2 - 3n - 160 = 0

    After using the quadratic formula, I got the following values for n:

    n = (3/2) - [(sqrt{649})/2]

    n = (3/2) + [(sqrt{649})/2]

    The question now is: which one is correct?
    Irrelevant. The discriminant must equal a positive integer
    in order to get a valid solution; sqrt(649) <> integer.
    So answer here is: no polygon has 80 diagonals.

    Let d = number of diagonals.
    d = n(n-3) / 2
    d = (n^2 - 3n) / 2
    So (n^2 - 3n) / 2 must equal an integer.

    Since minimum n = 3 (triangle, 3 sides), you get this series:
    n,d
    3,0 (triangle)
    4,2 (rectangle)
    5,5 (5 sided)
    6,9 (6 sided
    7,14 (7 sided)
    ....and so on.
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    Re: Polygon of n Sides

    Quote Originally Posted by harpazo View Post
    (n/2)(n - 3) = 65

    2[(n/2)(n - 3)] = 65(2)

    n(n - 3) = 130

    n^2 - 3n = 130

    n^2 - 3n - 130 = 0

    (n - 13)(n + 10) = 0

    n - 13 = 13

    n + 10 = -10 is rejected.

    A polygon with 65 diagonals has 13 sides.

    What about a polygon with 80 diagonals?

    (n/2)(n - 3) = 80

    n(n - 3) = 160

    n^2 - 3n - 160 = 0

    After using the quadratic formula, I got the following values for n:

    n = (3/2) - [(sqrt{649})/2]

    n = (3/2) + [(sqrt{649})/2]

    The question now is: which one is correct?
    That's what I meant by:
    When you've done the solving, check your answers back in the context of the question.
    If a polygon has n sides, then n must be an positive integer (greater than or equal to 3 because the smallest polygon has three sides).

    Both values you have found are NOT integers (the first one BTW is negative so that's out for two reasons).

    So even though you have found the solutions to the equation, it does not answer the question. Because of this, you can conclude that there is no polygon with 80 diagonals.
    Again, in word problems especially, always take your answer back to the context of the question.
    Thanks from harpazo
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    Re: Polygon of n Sides

    Quote Originally Posted by DenisB View Post
    Irrelevant. The discriminant must equal a positive integer … you mean a perfect square
    in order to get a valid solution; sqrt(649) <> integer.
    So answer here is: no polygon has 80 diagonals.

    Let d = number of diagonals.
    d = n(n-3) / 2
    d = (n^2 - 3n) / 2
    So (n^2 - 3n) / 2 must equal an integer.

    Since minimum n = 3 (triangle, 3 sides), you get this series:
    n,d
    3,0 (triangle)
    4,2 (rectangle)
    5,5 (5 sided)
    6,9 (6 sided
    7,14 (7 sided)
    ....and so on.
    corner?
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  8. #8
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    Re: Polygon of n Sides

    Yes Deb, I goofed...joining you in the corner:
    bringing pencils and paper so we can play X&O; I start!!
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  9. #9
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    Re: Polygon of n Sides

    Quote Originally Posted by harpazo View Post
    (n/2)(n - 3) = 65

    2[(n/2)(n - 3)] = 65(2)

    n(n - 3) = 130

    n^2 - 3n = 130

    n^2 - 3n - 130 = 0

    (n - 13)(n + 10) = 0

    n - 13 = 13

    n + 10 = -10 is rejected.


    A polygon with 65 diagonals has 13 sides.

    What about a polygon with 80 diagonals?

    (n/2)(n - 3) = 80

    n(n - 3) = 160

    n^2 - 3n - 160 = 0

    After using the quadratic formula, I got the following values for n:

    n = (3/2) - [(sqrt{649})/2]

    n = (3/2) + [(sqrt{649})/2]

    The question now is: which one is correct?
    No! No! No! I would, and have, taken points off for a statement like this.

    If n - 13 = 13 then n = 26. Write it out. n - 13 = 0 means (implies, gives us, whatever) n = 13.

    -Dan
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  10. #10
    Forum Admin topsquark's Avatar
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    Re: Polygon of n Sides

    Quote Originally Posted by DenisB View Post
    Yes Deb, I goofed...joining you in the corner:
    bringing pencils and paper so we can play X&O; I start!!
    I would pay to see this. She's gonna kick your butt.

    -Dan
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  11. #11
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    Re: Polygon of n Sides

    Quote Originally Posted by topsquark View Post
    I would pay to see this. She's gonna kick your butt.

    -Dan
    Why am I in the corner anyway?
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  12. #12
    Senior Member harpazo's Avatar
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    Re: Polygon of n Sides

    Thank you everyone.
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