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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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    The condition says that $T$ moves every point the same distance. You also know that it preserves distances between points: for all $X,Y\in\mathbb{R}^n$, $\text{dist}(T(X),T(Y))=\text{dist}(X,Y)$. If two points are moved the same distance and *not* in parallel directions, can they stay the same distance apart?2011-08-23
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    @Brian: Imagine the spheres around $X$ and $Y$ on which the mapped points must lie, pick a point on the sphere around $X$ and form the sphere with radius given by the constant map around it; that intersects the sphere around $Y$ on an $(n-2)$-sphere.2011-08-23
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    @joriki: I’m sorry, but I’m missing something in that description: I don’t see why the sphere about the new point need intersect the one about $Y$ at all.2011-08-23
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    @zyx: It’s not possible in the plane for the question that I intended to ask. I see now that I phrased the question very badly, however. Despite what I wrote in the question proper, I was assuming that all points were moved the same distance, and in the plane it’s not hard to see that that does force parallel translation of $X$ and $Y$: just look at their midpoint.2011-08-23
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    @Brian: Sorry, I mixed up the radii, I meant "form the sphere with radius $\text{dist}(X,Y)$". If I understand your last comment correctly, you're saying that looking only at two points isn't enough but it's enough to also consider their midpoint? I think that's true, but it's not entirely obvious to me. Is there an easy argument without calculation why the midpoint couldn't also work out for any of the other solutions on the $(n-2)$-sphere?2011-08-23
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    In any dimension, a line intersects a sphere in no more than two points, so any third point on line XY would do [considering the sphere of possible displacements and the line joining displacement(X) to displacement(Y), which intersects this sphere twice].2011-08-23
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    @zyx: I see, thanks.2011-08-23
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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 $q$.