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Homework Help: Existence of some special kind of isometry in R^n-1

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Is there a theorem that states the following?

    Let [itex]P= \{ P_{1}, . . . , P_{n} \}[/itex] be the set of n distinct points in [itex]\mathbb{R}^{n-1}[/itex] and [itex]P'= \{ P'_{1}, . . . , P'_{n} \}[/itex] also a set of points in [itex]\mathbb{R}^{n-1}[/itex]. If for all [itex]i,j[/itex] [itex]|P_{i} - P_{j}|=|P'_{i} - P'_{j}|[/itex] then there is an isometry [itex]f: \mathbb{R}^{n-1} \to \mathbb{R}^{n-1}[/itex] such that for all [itex]i[/itex] [itex]f(P_{i})=P'_{i}[/itex].

    Or at least is it ture? If true how to prove it?

    2. Relevant equations

    3. The attempt at a solution
    Geometrically, it seems plausible especially in some cases where such an isometry is a rotation, a translation, a reflection, or a combination of them.
  2. jcsd
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