Let $\phi: \mathcal{E}\to \mathcal{E}$ be a mapping defined on a Euclidean affine space. Assume that $\phi$ preserves the distances: $$d(\phi(A),\phi(B)) = d(A,B).$$ Prove that $\phi$ is affine, and hence is an isometry.
Usually when defining isometry, we require the mapping to be affine or bijective. But does being distance-preserving alone force the mapping to be affine?
On
For the case when $\mathcal{E}$ is the plane, this is true. I believe it is true for higher dims, too. Let $p$ be a point in the plane, and let $q$ be a point distance $1$ away from $p$. Let $r$ lie on the ray starting at $p$ directed towards $q$. Say it is a distance $d$ away from $p$. Since $\phi$ preserves distances, $d(\phi(p),\phi(q))=1$ and $d(\phi(p),\phi(r))=d$ and $d(\phi(q),\phi(r))=d-1$. Then $\phi(r)$ must lie on both the circle of radius $d$ about $p$ and the circle of radius $d-1$ about $q$. By the triangle inequality, these should only intersect in one point - the point on the ray starting at $\phi(p)$ to $\phi(q)$ that is a distance $d$ away. Therefore $phi$ maps lines through $p$ to lines through $\psi(p)$, and because it is an isometry, when restricted to any line, it is an affine map.
Now, let $q$ be as above, and let $r$ be a point a distance $1$ away from $p$ so that the line from $p$ to $r$ is orthogonal to the line from $p$ to $q$. We need only show that $\phi(r+q) = \phi(r) + \phi(q)$. (with addition relative to $p$ and $\phi(p)$ as origins, respectively). But $\phi(r+q)$ must be mapped to a point distance $1$ away from each of $p, r,$ and $q$. So it must lie on the intersection of three distinct circles. These circles intersect at the single point $\phi(r) + \phi(q)$, and we're done.
For higher dims, we can show an isometry maps lines to lines, just as we did here. To show the linearity property, I think we just note that $\phi(r+q)$ must lie in the intersection of many different spheres. Now, just simply pick enough spheres so that this point is unique.
The answer is yes - this ought to be proved in any linear algebra book.
After applying a translation, we can assume that $\phi$ has a fixed point $O$. Taking $O$ as the origin, we can identify $\phi$ with a mapping of the underlying vector space $E$, which fixes the origin and preserves distances.
Now we prove $\phi$ preserves scalar products. We have $$\phi(u) \cdot \phi(v) = \frac{1}{2}\left[|\phi(u)|^2 + |\phi(v)|^2 - |\phi(u) - \phi(v)|^2 \right] = \frac{1}{2} \left[|u|^2 + |v|^2 - |u - v|^2 \right] = u \cdot v.$$
Let $(e_i)$ be an orthonormal basis of $E$. Let $f_i = \phi(e_i)$. Since $\phi$ preserves scalar products, $(f_i)$ is also an orthonormal basis of $E$.
Now if $u$ is any vector, let its coordinates in the basis $(e_i)$ be $(x_i)$. Since $x_i = u \cdot e_i$ and $\phi$ preserves scalar products, it follows that the coordinates of $\phi(u)$ in the basis $(f_i)$ are also $(x_i)$. This proves that $\phi$ is linear.