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Show that for an isometry $T:{\mathbb{R}}^n \rightarrow {\mathbb{R}}^n$, if

$ X \mapsto \mathrm{dist}(X,T(X)) \quad (X \in {\mathbb{R}}^n)$

is a constant map, $T$ is a parallel translation.

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    @joriki: (Sorry to be so slow: I got side-tracked.) I finally realized that after thinking about it for a while. The other problem was that my visualization was a little different: in $\mathbb{R}^3$ I was looking at a cylinder.2011-08-25

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Isometries of $R^n$ are known to be linear functions of the form $T(x)=A(x)+b$ where $A$ is a linear isometry fixing 0 and $b$ is a vector. For an isometry $T$ the displacement $f(x) = T(x) - x$ is a linear vector-valued function of $x$. If $f$ is of constant length then $f$ must be constant, otherwise there is a pair of points $p,q$ with $f(p) \neq f(q)$ and then $f$ would be unbounded on the line through $p$ and $q$.