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  "body": "\n## Proof\n\nWe first write $f$ as $x \\mapsto Mx + b$, and then sort the cases by $M$.\n\n1. Put $g(x) = f(x) - f(0)$. Since $f$ is a [[Euclidean isometry|0J1N3WTKM4]], $\\norm{g(x) - g(y)} = \\norm{f(x) - f(y)} = \\norm{x - y}$ for all $x, y$. With $y = 0$, $\\norm{g(x)} = \\norm{x}$.\n2. By the [[polarisation identity|05FKHRT0B4]], $g(x) \\cdot g(y) = \\frac{1}{2}(\\norm{g(x)}^2 + \\norm{g(y)}^2 - \\norm{g(x) - g(y)}^2)$. By step 1 this equals $x \\cdot y$, the [[dot product|0S52TNNTH9]] of $x$ and $y$.\n3. Let $e_1$, $e_2$ be the standard basis and $c_j = g(e_j)$. By step 2, $c_1$, $c_2$ are orthonormal. An [[orthonormal family is linearly independent|0JWM6GJNQW]], so $c_1$, $c_2$ form a [[basis|096V7H13H9]] of $\\mathbb{R}^2$.\n4. By the [[coordinates in an orthonormal basis|09DZ8CD1AK]], $g(x) = (g(x) \\cdot c_1)c_1 + (g(x) \\cdot c_2)c_2$. By step 2, this is $x_1 c_1 + x_2 c_2$ for $x = (x_1, x_2)$.\n5. So $f(x) = Mx + b$, where $b = f(0)$ and $M$ is the matrix with the orthonormal columns $c_1$, $c_2$. Thus $M$ is an [[orthogonal matrix|0W02QESPJQ]].\n6. Since $\\norm{c_1} = 1$, the [[polar form|0BRXFNQ5JV]] of the complex number with the coordinates of $c_1$ and [[Euler's formula|04R3CY41VJ]] give a real $\\theta$ with $c_1 = (\\cos\\theta, \\sin\\theta)$.\n7. Put $m = (-\\sin\\theta, \\cos\\theta)$. By the [[Pythagorean identity|03H7WZ4T3K]], $c_1$, $m$ are orthonormal. Expanding $c_2$ in this basis gives $c_2 = (c_2 \\cdot c_1)c_1 + (c_2 \\cdot m)m = \\lambda m$, with $\\lambda = c_2 \\cdot m$.\n8. From $\\norm{c_2} = 1$ we get $\\abs{\\lambda} = 1$, so $\\lambda = 1$ or $\\lambda = -1$.\n9. Case $\\lambda = 1$ and $\\theta \\in 2\\pi\\mathbb{Z}$. Then $M$ is the identity matrix and $f(x) = x + b$. If $b = 0$, $f$ is the identity. Otherwise, $f$ is the [[translation|087VJM5Q77]] by $b \\neq 0$.\n10. Case $\\lambda = 1$ and $\\theta \\notin 2\\pi\\mathbb{Z}$. Then $M = \\begin{pmatrix} \\cos\\theta \u0026 -\\sin\\theta \\\\ \\sin\\theta \u0026 \\cos\\theta \\end{pmatrix}$, and $\\cos\\theta \\neq 1$.\n11. The [[determinant|00KDZRW4C0]] of $M - I$ is $(\\cos\\theta - 1)^2 + \\sin^2\\theta = 2 - 2\\cos\\theta \u003e 0$. So $I - M$ is invertible.\n12. Put $c = (I - M)^{-1} b$. Then $c + M(p - c) = Mp + (I - M)c = Mp + b = f(p)$. So $f$ is the [[rotation|0HFREYRSZP]] about $c$ through the angle $\\theta \\notin 2\\pi\\mathbb{Z}$.\n13. Case $\\lambda = -1$. Then $M = \\begin{pmatrix} \\cos\\theta \u0026 \\sin\\theta \\\\ \\sin\\theta \u0026 -\\cos\\theta \\end{pmatrix}$. Put $s = \\sin\\frac{\\theta}{2}$, $k = \\cos\\frac{\\theta}{2}$ and the unit vector $n = (s, -k)$.\n14. By the [[angle addition formulas|028XEQCE88]] and the Pythagorean identity, $\\cos\\theta = 1 - 2s^2 = 2k^2 - 1$ and $\\sin\\theta = 2sk$. So $M = I - 2nn^T$, that is, $Mx = x - 2(x \\cdot n)n$.\n15. Put $\\beta = b \\cdot n$ and $w = b - \\beta n$, so $w \\cdot n = 0$. Then $f(x) = x - 2(x \\cdot n - \\tfrac{\\beta}{2})n + w$.\n16. Let $H = \\set{x : x \\cdot n = \\beta/2}$. Since $n \\cdot n = 1$, step 15 says $f = T_w \\circ \\sigma_H$, with the [[reflection|06BRDF3F0E]] $\\sigma_H$ and the translation $T_w$.\n17. If $w = 0$, then $f = \\sigma_H$ is a reflection in a line. If $w \\neq 0$, then $f$ is a [[glide reflection|0167F9B344]] with the nonzero glide vector $w$, since $w \\cdot n = 0$.\n18. So $f$ has at least one of the five types. It remains to see that no map has two types. We count fixed points.\n19. The identity fixes every point. A translation by $v \\neq 0$ fixes none, since $p + v \\neq p$.\n20. A rotation $p \\mapsto c + M(p - c)$ with $\\theta \\notin 2\\pi\\mathbb{Z}$ fixes $p$ if and only if $(M - I)(p - c) = 0$. By step 11, this means $p = c$. So it has exactly one fixed point.\n21. A reflection $\\sigma_H(x) = x - 2\\frac{x \\cdot a - \\kappa}{a \\cdot a} a$, with $a \\neq 0$ and $H = \\set{x : x \\cdot a = \\kappa}$, fixes $x$ if and only if $x \\cdot a = \\kappa$. So its fixed set is the line $H$, which has more than one point but is not the whole plane.\n22. Let $T_w \\circ \\sigma_H$ be a glide reflection, with $w \\neq 0$ and $w \\cdot a = 0$. If $x$ were fixed, then $w = 2\\frac{x \\cdot a - \\kappa}{a \\cdot a} a$. Taking the dot product with $w$ would give $\\norm{w}^2 = 0$, a contradiction. So it has no fixed point.\n23. By steps 19 to 22, only a translation and a glide reflection could coincide. For a translation $f$, $f(x) - f(0) = x$ for all $x$.\n24. For a glide reflection, $f(x) - f(0) = \\sigma_H(x) - \\sigma_H(0) = x - 2\\frac{x \\cdot a}{a \\cdot a} a$. At $x = a$ this is $-a \\neq a$. So a glide reflection is not a translation, and $f$ has exactly one of the five types.\n\n## Idea\n\nAn isometry that fixes the origin keeps dot products, so it is an orthogonal matrix. A $2 \\times 2$ orthogonal matrix is a rotation matrix or the matrix of a reflection. Then the translation part either moves the centre of the rotation, or splits into a shift of the mirror and a glide along it. Fixed points tell the five types apart.\n",
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