1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Similar Threads for Rigid Motion Woes
Motion of a non-linear pendulum with air resistance