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Homework Help: P is a point in E

  1. Dec 15, 2009 #1
    1. The problem statement, all variables and given/known data

    If P is a point in E (euclidean Plane) Then the "Inversion through P" is
    Vp = {(x,y)| x,y in E and either,
    1. x=y=p, or
    2. p is the midpoint of segment xy}

    What's Fix(Vp)

    Show that Vp composed of Vp = The identity

    2. Relevant equations

    3. The attempt at a solution
    Vp composed of Vp = The identity because Vp is a rigid motion and two rigid motions are equal to the identity?? (not so sure on this one)

    I believe is that Fix(Vp) is every point because given point P in E and it's inversion through point P gives us that the inversion is = x=y=p so wouldn't that mean (p,p)?

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Dec 15, 2009
  2. jcsd
  3. Dec 16, 2009 #2


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    Science Advisor

    so Vp is the set of all pairs of points satisfying those conditions?

    What does "Fix" mean? My first thought was "fixed point" but you find fixed points for a function from a set to itself and here V takes points to sets of pairs of points- it can't have any fixed points. And, further, it appears you are applying "Fix" to Vp, not V. Again, what is "Fix"?

    The way you have defined it, Vp is a set of pairs of points, not a function- you compose functions, not sets.

  4. Dec 16, 2009 #3
    The only way I can help you with this question is to give you an example of another composition

    given a point P not lying on a line L, Reflect point P over line L to give you point P' which is the Rl(p) (the reflection of P.)
    Then Rl(p) composed of Rl(P) will bring you right back to the beginning point P and therefore
    Rl composed of Rl is the identity
  5. Dec 16, 2009 #4
    OK, you have the formalities here, but you don't seem to be thinking of [tex]V_p[/tex] as a function or mapping, which is what you really need. If [tex]x = p[/tex], what is [tex]V_p(x)[/tex]? What about if [tex]x \neq p[/tex]; what is [tex]V_p(x)[/tex], in a rough geometrical description?

    It may help you to impose coordinates; in that case it is simpler to start by supposing [tex]p = (0, 0)[/tex], and then figure out how to generalize.

    It is true that [tex]V_p[/tex] is a rigid motion of the plane, but it is not true that a rigid motion composed with itself always equals the identity (think of translation by any nonzero distance in any direction, or rotation by say [tex]\pi/4[/tex]).

    What is [tex]\mathop{\mathrm{Fix}}(V_p)[/tex]? It is the set of fixed points of [tex]V_p[/tex], that is, [tex]\mathop{\mathrm{Fix}}(V_p) = \{ x \in E : V_p(x) = x \}[/tex]. You are right to consider whether [tex]p \in \mathop{\mathrm{Fix}}(V_p)[/tex] separately from whether [tex]x \in \mathop{\mathrm{Fix}}(V_p)[/tex] for [tex]x \notin p[/tex], but to build a correct argument, you need to think more carefully of [tex]V_p[/tex] as a function, and what it does to each point.
  6. Dec 16, 2009 #5
    Ok I'm going to write what I came up with, I'm not the best with notation so see if this makes sense to you...
    Describe Fix(Vp) and proof - I now say that there's only one point fixed in Vp and that's when X=Y=P.
    Part 1. Given an element X and and a point P which is X, The inversion of element X through P will result in X therefore Vp(x)=x

    Part 2. Given a element X and a point P that's not X in E. The inversion through P will produce a point X' where P is the midpoint of XX'. I say that it's impossible to have the inversion through P = P because if X isn't = P then the inversion will result in another element that's equal distance from P that X is.

    Vp composed of Vp = Identity
    Here's what I came up with, again sorry for the notation.
    In the definition of Vp if x=y=p then we know that if x is equal to p then the inversion will give us p', then once again we take the inversion of p' it'll take us back to P. Where P=P'.

    Now if X=Y=P, then you have your element x and a point p. the inversion through P will result in a point X' where XP congruent to X'P. because p is the midpoint. So by taking the inversion of X' will result back to point X where p is once again the midpoint.
  7. Dec 17, 2009 #6
    Most of this is right in essence; I'll show you how to phrase some of it more carefully.
    You could just say "by the definition of [tex]V_p[/tex], [tex]V_p(p) = p[/tex]".
    This is correct. You might end by observing "and this distance is nonzero, because [tex]x \neq p[/tex]".
    This is correct, but it is simpler to say that since [tex]V_p(p) = p[/tex], also [tex]V_p(V_p(p)) = V_p(p) = p[/tex].
    I think you mean to begin with "if [tex]x \neq p[/tex]". The essence of your argument is correct. [tex]V_p(x)[/tex] is the other endpoint of the segment having [tex]x[/tex] as one endpoint and [tex]p[/tex] as midpoint. [tex]V_p(V_p(x))[/tex] is the other endpoint of the segment having [tex]V_p(x)[/tex] as one endpoint and [tex]p[/tex] as midpoint. But these are one and the same segment; so [tex]V_p(V_p(x)) = x[/tex].

    Therefore [tex]V_p(V_p(x)) = x[/tex] for all [tex]x\in E[/tex], whether [tex]x = p[/tex] or [tex]x \neq p[/tex]; that is, [tex]V_p \circ V_p[/tex] is the identity map.
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