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I'm having some difficulty wrapping my head around rigid motions in a plane. In particular, I'm trying to solve this following problem:

In a Euclidean plane, show that the product of two rotations around different points is equal to either a rotation around a third point or a translation. Hint: Show that it has at most one fixed point.

I working in a plane $\Pi$ over an Euclidean ordered field, so rotations are transformations defined by \begin{cases} x'=cx-sy \\ y'=sx+cy \end{cases} where $c^2+s^2=1$. I simply take two rotations $\psi$ and $\phi$ such that \phi=\begin{cases} x'=cx-dy\\ y'=dx+cy \end{cases} and \psi=\begin{cases} x'=ex-fy\\ y'=fx+ey \end{cases} where $c^2+d^2=1$ and $e^2+f^2=1$. Composing them, I see $ \begin{align*} \psi\phi(x,y) &= (ecx-edy-fdx-fcy,fcx-fdy+edx+ecy) \\ &= ((ec-fd)x-(ed+fc)y,(fc+ed)x+(ec-fd)y) \end{align*} $ but $ \begin{align*} (ec-fd)^2+(ed+fc)^2 &= e^2c^2-2ecfd+f^2d^2+e^2d^2+2edfc+f^2c^2 \\ &= (e^2+f^2)c^2+(e^2+f^2)d^2 \\ &= c^2+d^2=1 \end{align*} $ so $\psi\phi$ is a rotation. I feel I have done something wrong, since the problem seems to be prodding me in a different direction. What is the correct way to show this? Thanks.

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    @Alexander, thanks, I'll try to redo my argument.2011-03-21

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With no coordinate system, or at least, as long as possible without one.

An affine transformation of $\mathbb{R}^n$ consists of a linear transformation followed by a translation $x \mapsto A x+ b$, which we denote by $(A,b)$. The transformation $(A,b)$ is a rotation if and only if $A$ is an element of the special orthogonal group $SO(n)$ and $A$ is not identity. It is a translation (or the identity) if $A$ is identity.

The composition of two rotations $(A,b)$ and (A',b') is x\mapsto (A'A)x+(A'b+b'). One knows that A'A is an element of $SO(n)$. Hence, if A'A is not the identity, the composition is a rotation and if A'A is the identity, this is a translation.

In dimension $2$, any element $A$ of the group $SO(2)$ can be written as $ A=\begin{pmatrix} \cos(u) & \sin(u)\\ -\sin(u) & \cos(u)\end{pmatrix} $ with $u$ real, hence $A$ is identity if and only if $u$ is a multiple of $2\pi$. The product A'A, which you need to compute the composition of $(A,b)$ and (A',b'), is A'A=\begin{pmatrix} \cos(u+u') & \sin(u+u')\\ -\sin(u+u') & \cos(u+u')\end{pmatrix} (this is the addition formula for sines and cosines). The result is a proper rotation when u+u' is not a multiple of $2\pi$, and a translation when it is.

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    @yunone Hence the links to wikipedia pages in my post. If the post was useful in the end, everything is fine.2011-03-21
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All isometries (rigid transformations) of a plane can be expressed as a composition of 3 or fewer reflections.

  • A composite of two reflections over intersecting lines is a rotation about the point of intersection of the lines of reflection.
  • A composite of two reflections over parallel lines is a translation perpendicular to the lines of reflection.
  • A composite of three reflections is a "glide reflection" (which can be expressed as a reflection followed by a translation).

Isometries that can be expressed as a composite of an even number of reflections preserve orientation; those that can be expressed as a composite of an odd number of reflections reverse orientation. Since rotations preserve orientation, composing rotations also preserves orientation, which means that the result must be a rotation or a translation (the result must be an isometry and cannot be an orientation-reversing isometry, so cannot be a reflection or glide-reflection).

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    Thank you for your answer Isaac, I was not as aware of how prevalent reflections are. The last paragraph of your post seems like it's hinting at the next exercise I want to try, so much thanks for that too!2011-03-21