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Homework Help: Symmetry with respect to a circle

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data

    Could someone please just explain a step in a proof for me?

    Suppose x and y are symmetric with respect to a circle, and a,b and c are three points on the circle. This means that:

    (x,a,b,c) = ([tex]\bar{y}[/tex],[tex]\bar{a}[/tex],[tex]\bar{b}[/tex],[tex]\bar{c}[/tex])

    The writer then says:

    "symmetry only depends on the circle, and not on the points, so if x and y are symmetric then:

    [tex]\frac{x-b}{a-b}[/tex] = [tex]\frac{\bar{y}-\bar{b}}{\bar{a}-\bar{b}}[/tex]"

    I understand why this would be if the points were on a straight line, as then you could choose c to be the point at infinity, and the cross ratio would reduce to this. But why is it also true for a circle?
  2. jcsd
  3. Oct 25, 2008 #2


    Staff: Mentor

    You need to give us some more information. Are we talking about a circle in the real plane? What's the significance of [tex]\bar{y}[/tex]? That suggests conjugates, so maybe we're not talking about circles in the real plane. And what does (x, a, b, c) denote?
  4. Oct 26, 2008 #3
    Fair enough!

    This is a complex analysis problem - the circle is in the complex plane.

    If x, a, b, c are complex numbers, then (x,a,b,c) denotes the cross ratio:


    Two points x and y being symmetric with respect to a circle means that if a, b, c are any three points on that circle, then:

    (x,a,b,c) = [tex]\overline{(y,a,b,c)}[/tex]

    (ie. the cross ratio of x, a, b, c is equal to the complex conjugate of the cross ratio of y, a, b, c ).

    So: [tex]\frac{(x-b)(a-c)}{(x-c)(a-b)}[/tex] = [tex]\frac{\overline{(y-b)}\overline{(a-c)}}{\overline{(y-c)}\overline{(a-b)}}[/tex]

    But the text says that we can choose the points a, b, c such that some of these terms cancel and we get:

    [tex]\frac{(x-b)}{(a-b)}[/tex] = [tex]\frac{\overline{(y-b)}}{\overline{(a-b)}}[/tex]

    I don't understand this - as far as I can see this would only be possible by choosing c to be infinity, but the circle does not contain the poiint at infinity.
    Last edited: Oct 26, 2008
  5. Oct 27, 2008 #4
    Is it considered bad form to post a reply to your own question to get it back to the top of the board?
  6. Oct 28, 2008 #5


    Staff: Mentor

    I had one class on complex analysis in grad school, but I have to confess we didn't cover any of the ideas you've presented, such as symmetry wrt a circle or the cross-ratio business.

    So looking at what you have and where you want to go, it seems that you want the a -c and x -c factors to cancel in the ratio on the left, and the a -c and y -c (conjugate) factors to cancel in the ratio on the right. For the one on the right, if you choose a = y those factors cancel, but on the left what seems to work is choosing a = x. You said that one could choose a, b, and c, but I don't think you can choose a to be simultaneously equal to both x and y, so that doesn't seem to help.

    What bothers me is that the word "symmetry" is used in this problem, but I don't have a clear picture in mind of where x and y are located on the circle. I would think that x and y being symmetric with respect to the circle means one of the following:
    1. Re(x) = Re(y) (the points are directly across the real axis from one another so that x = conj(y))
    2. Im(x) = Im(y) (the points are directly across the imaginary axis from one another)
    3. Arg(x) = Arg(y) + k*pi, where k is an integer (symmetry about the origin)
    Is symmetry with respect to a circle what I have described here?

    Sorry I'm not able to provide more help.
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