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Product of Diagonals of Regular Polygon?

  1. Mar 31, 2012 #1
    So any help would be really appreciated! I really have no idea where to start, and I can use any help.

    So essentially the problem is we have a regular polygon P inscribed in a unit circle. This regular polygon has n vertices. Fix one vertex and take the product of the lengths of diagonals drawn from the one vertex to each of the other vertices.

    Show that the product of these diagonals is equal to n.

    ...so I guess what I have so far (which is quite minimal) is this:

    So I suppose I figure as much that the vertices of the regular polygon are nth roots of unity. And so I also suppose that one vertex is always 1. (1 to the nth power is always 1)

    So let's label the vertices of the polygon v0, v1, v2, ... v(n-1), where v0=1.

    So essentially we want to find the product |v1-1||v2-1|...|v(n-1)-1| -- and we want to show that equals n.

    I guess this is what I have so far, but I'm not really sure where to go from there.

    Again, any help will be greatly appreciated!
     
  2. jcsd
  3. Apr 1, 2012 #2

    AlephZero

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    It might be easier to show that the square of the product of the diagonals = ##n^2##. That gets rid of all the modulus functions.

    Think about the properties of the n'th roots of unity. They are the roots of a polynomial equation, so it is easy to show that the sum of the n'th roots = 0. There are similar results for the sums of products and powers of the n'th roots.
     
  4. Apr 1, 2012 #3
    hmmm so I'm a little conrfused...so I guess okay, the nth roots of unity are the roots of a polynomial equation..(first of all I don't think I really get that)

    And, I don't seem to understand how the sum of them equal 0...
     
  5. Apr 1, 2012 #4

    AlephZero

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    The nth roots of unity, ##\cos 2 \pi k/n + i \sin 2 \pi k/n## for k = 0, 1, ... n-1 are the roots of the equation ##z^n - 1 = 0##.

    If ##z## is a root , then ##z^2##, ##z^3##, ... are also roots.

    ##z^n -1 = (z-1)(z^{n-1} + z^{n-2} \cdots + 1)##

    So if ##z \ne 1## is a root of ##z^n - 1 = 0##, ##z^{n-1} + z^{n-2} \cdots + 1 = 0##.

    See http://en.wikipedia.org/wiki/Root_of_unity (or Google for "roots of unity").
     
  6. Nov 18, 2012 #5
    I dont really get that...can you restate?
     
  7. Nov 18, 2012 #6

    AlephZero

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    ##z^n - 1 = (z-1)(z^{n-1} + z^{n-2} \cdots + 1)##. (To show that, just multiply the two factors).

    If ##z## is a root of ##z^n - 1##, then ##(z^n - 1) = 0##.

    So either ##(z-1) = 0## or ##(z^{n-1} + z^{n-2} \cdots + 1) = 0##.

    If ##z \ne 1##, then ##(z-1) \ne 0##.

    So ##(z^{n-1} + z^{n-2} \cdots + 1) = 0##.
     
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