# Existence of some special kind of isometry in R^n-1

1. Apr 24, 2012

### julypraise

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

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

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.