[proofplan] We first remove the translation part by setting $a=F(0)$ and studying the translated map $G(x)=F(x)-a$. The distance-preserving hypothesis then implies that $G$ preserves norms, and the real polarization identity converts norm preservation of differences into preservation of the Euclidean [inner product](/page/Inner%20Product). From inner product preservation we prove additivity and homogeneity directly by showing the relevant error vectors have squared norm zero. This makes $G$ linear, its standard matrix is orthogonal, and the converse follows by a direct norm computation for affine maps with orthogonal linear part. [/proofplan] [step:Translate the map so that the origin is fixed] Assume first that $F$ preserves Euclidean distances. Define the vector $a \in \mathbb{R}^n$ by \begin{align*} a := F(0) \end{align*} and define the map \begin{align*} G: \mathbb{R}^n \to \mathbb{R}^n, \qquad x \mapsto F(x)-a. \end{align*} Then $G(0)=0$. For every $x,y \in \mathbb{R}^n$, [translation invariance](/theorems/4911) of the Euclidean norm gives \begin{align*} |G(x)-G(y)| = |F(x)-F(y)| = |x-y|. \end{align*} Taking $y=0$ and using $G(0)=0$, we obtain \begin{align*} |G(x)| = |x| \end{align*} for every $x \in \mathbb{R}^n$. [/step] [step:Use polarization to show that the translated map preserves inner products] Let $\langle \cdot,\cdot\rangle$ denote the Euclidean inner product on $\mathbb{R}^n$. For every $u,v \in \mathbb{R}^n$, the real polarization identity is \begin{align*} \langle u,v\rangle = \frac{1}{2}\left(|u|^2+|v|^2-|u-v|^2\right). \end{align*} Applying this identity with $u=G(x)$ and $v=G(y)$, and using the norm and distance preservation already proved for $G$, gives \begin{align*} \langle G(x),G(y)\rangle = \frac{1}{2}\left(|G(x)|^2+|G(y)|^2-|G(x)-G(y)|^2\right). \end{align*} Substituting $|G(x)|=|x|$, $|G(y)|=|y|$, and $|G(x)-G(y)|=|x-y|$, we get \begin{align*} \langle G(x),G(y)\rangle = \frac{1}{2}\left(|x|^2+|y|^2-|x-y|^2\right)=\langle x,y\rangle. \end{align*} Thus $G$ preserves the Euclidean inner product. [guided] The distance condition gives information about lengths of differences. To prove that $G$ is linear, it is more useful to know that $G$ preserves inner products. The bridge between these two pieces of information is the real polarization identity: \begin{align*} \langle u,v\rangle = \frac{1}{2}\left(|u|^2+|v|^2-|u-v|^2\right) \end{align*} for all $u,v \in \mathbb{R}^n$. We apply this identity to the two vectors $G(x)$ and $G(y)$, where $x,y \in \mathbb{R}^n$ are arbitrary. Since the previous step proved $|G(x)|=|x|$, $|G(y)|=|y|$, and $|G(x)-G(y)|=|x-y|$, we obtain \begin{align*} \langle G(x),G(y)\rangle = \frac{1}{2}\left(|G(x)|^2+|G(y)|^2-|G(x)-G(y)|^2\right). \end{align*} Replacing each norm involving $G$ by the corresponding norm in the domain gives \begin{align*} \langle G(x),G(y)\rangle = \frac{1}{2}\left(|x|^2+|y|^2-|x-y|^2\right). \end{align*} Applying the same polarization identity to $x$ and $y$ identifies the right-hand side as $\langle x,y\rangle$. Therefore \begin{align*} \langle G(x),G(y)\rangle = \langle x,y\rangle \end{align*} for every $x,y \in \mathbb{R}^n$. This is the key structural fact: the translated distance-preserving map preserves angles and lengths, not only distances. [/guided] [/step] [step:Derive additivity from preservation of the inner product] Let $x,y \in \mathbb{R}^n$ be arbitrary, and define the error vector $r \in \mathbb{R}^n$ by \begin{align*} r := G(x+y)-G(x)-G(y). \end{align*} Using the Euclidean identity $|r|^2=\langle r,r\rangle$ and expanding by bilinearity of the inner product, we obtain \begin{align*} |r|^2 = |G(x+y)|^2+|G(x)|^2+|G(y)|^2-2\langle G(x+y),G(x)\rangle-2\langle G(x+y),G(y)\rangle+2\langle G(x),G(y)\rangle. \end{align*} By inner product preservation, this becomes \begin{align*} |r|^2 = |x+y|^2+|x|^2+|y|^2-2\langle x+y,x\rangle-2\langle x+y,y\rangle+2\langle x,y\rangle. \end{align*} Expanding the remaining inner products in $\mathbb{R}^n$ gives \begin{align*} |r|^2 = 0. \end{align*} Since the Euclidean norm is positive definite, $r=0$. Therefore \begin{align*} G(x+y)=G(x)+G(y) \end{align*} for every $x,y \in \mathbb{R}^n$. [/step] [step:Derive homogeneity from preservation of the inner product] Let $t \in \mathbb{R}$ and $x \in \mathbb{R}^n$ be arbitrary, and define the error vector $s \in \mathbb{R}^n$ by \begin{align*} s := G(tx)-tG(x). \end{align*} Expanding the squared norm and using inner product preservation, we get \begin{align*} |s|^2 = |G(tx)|^2-2t\langle G(tx),G(x)\rangle+t^2|G(x)|^2. \end{align*} Thus \begin{align*} |s|^2 = |tx|^2-2t\langle tx,x\rangle+t^2|x|^2. \end{align*} Since $|tx|^2=t^2|x|^2$ and $\langle tx,x\rangle=t|x|^2$, the right-hand side is $0$. Hence $s=0$, and therefore \begin{align*} G(tx)=tG(x) \end{align*} for every $t \in \mathbb{R}$ and every $x \in \mathbb{R}^n$. Together with additivity, this proves that $G$ is a [linear map](/page/Linear%20Map). [/step] [step:Represent the translated map by an orthogonal matrix] Let $e_1,\dots,e_n \in \mathbb{R}^n$ denote the standard ordered basis, and let $Q \in \mathbb{R}^{n \times n}$ be the matrix whose $j$-th column is $G(e_j)$. Since $G$ is linear, its standard matrix is $Q$, so \begin{align*} G(x)=Qx \end{align*} for every $x \in \mathbb{R}^n$. For $1 \le i,j \le n$, the $(i,j)$-entry of $Q^\top Q$ is \begin{align*} (Q^\top Q)_{ij}=\langle G(e_i),G(e_j)\rangle. \end{align*} By inner product preservation, \begin{align*} (Q^\top Q)_{ij}=\langle e_i,e_j\rangle. \end{align*} The standard basis is orthonormal, so $\langle e_i,e_j\rangle$ is the $(i,j)$-entry of the identity matrix $I_n$. Hence $Q^\top Q=I_n$, and therefore $Q \in O(n)$. Since $G(x)=F(x)-a$, we conclude \begin{align*} F(x)=Qx+a \end{align*} for every $x \in \mathbb{R}^n$. [/step] [step:Verify directly that every affine orthogonal map preserves distances] Conversely, suppose that there exist $Q \in O(n)$ and $a \in \mathbb{R}^n$ such that $F(x)=Qx+a$ for every $x \in \mathbb{R}^n$. Since $Q \in O(n)$, we have $Q^\top Q=I_n$. For arbitrary $x,y \in \mathbb{R}^n$, \begin{align*} |F(x)-F(y)|^2 = |Qx+a-Qy-a|^2. \end{align*} Thus \begin{align*} |F(x)-F(y)|^2 = |Q(x-y)|^2=\langle Q(x-y),Q(x-y)\rangle. \end{align*} Using $Q^\top Q=I_n$, this becomes \begin{align*} |F(x)-F(y)|^2=\langle x-y,Q^\top Q(x-y)\rangle=\langle x-y,x-y\rangle=|x-y|^2. \end{align*} Both sides are non-negative, so taking square roots gives \begin{align*} |F(x)-F(y)|=|x-y|. \end{align*} Therefore $F$ preserves Euclidean distances. This proves both directions of the equivalence. [/step]