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Rigid Motion Woes (Geometry)

  1. Apr 7, 2009 #1
    Working in a Hilbert plane, show that any rigid motion that fixes at least three noncollinear points must be the identity.

    I am certain that I can claim that:
    (i) any translation of the plane will fix none of the points
    (ii) any rotation will fix a single point
    (iii) any reflection will fix only the points on the line about which the plane is reflected

    The trouble is I don't know how to prove that no composition of these could fix only three points in the plane...

    Where do I go next?
     
  2. jcsd
  3. Apr 7, 2009 #2
    I'm unfamiliar with the axioms of a Hilbert plane. If this question were posed in the context of an ordinary Euclidean plane, however, this is how I would approach it. Rigid motions preserve distances (i.e., the distance between points [tex] f(a) [/tex] and [tex] f(b) [/tex] is the same as that between [tex] a [/tex] and [tex] b [/tex], where [tex] f [/tex] is a rigid motion). Given a two points [tex] A [/tex] and [tex] B [/tex] and a third point [tex] X [/tex] not on [tex] \overline{AB} [/tex], there exists exactly one other point [tex] Y \neq X [/tex] such that [tex] d(A,Y) = d(A,X) [/tex] and [tex] d(B,Y) = d(B,X) [/tex] (where [tex] d(\; , \;) [/tex] denotes the distance function). Furthermore, [tex] Y = r_{AB}(X) [/tex], the reflection of [tex] X [/tex] about [tex] \overline{AB} [/tex]. If [tex] X [/tex] is on [tex] \overline{AB} [/tex], then [tex] X [/tex] is the unique point in the plane satisfying these equations. (Neither of these assertions is hard to prove.)

    Let [tex] f [/tex] be a rigid motion fixing the three noncollinear points [tex] A,B,C [/tex]. From the last observation, we know that [tex] f [/tex] fixes all of [tex] \overline{AB}, \overline{AC} [/tex], and [tex] \overline{BC} [/tex]. Let [tex] X [/tex] be a point not on any of these lines. Suppose [tex] f(X) \neq X [/tex]; then, by distance conservation, we must have [tex] f(X) = r_{AB}(X) = r_{AC}(X) = r_{BC}(X) [/tex] simultaneously, a contradiction since (by assumption) [tex] \overline{AB} \neq \overline{AC} \neq \overline{BC} [/tex].
     
  4. Apr 8, 2009 #3
    A Hilbert Plane is just a Euclidean Plane, but without the Parallel Axiom and the Circle–Circle Intersection Property.

    So, yeah, your proof works.
     
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