Question: Identify the combination formed by first applying the glide reflection $γ_{A,B}$ and then applying $γ_{C,D}$ where $A = (0, 0), B = (2, 0), C = (1, −2), D = (1, 0)$.

Before, I jump in to show my work, I just want to say that what is stopping me from solving this problem is there is going to be a rotation involved which is not centered at the origin and I'm not sure how to set it up. In other words, I am having trouble finding the center of rotation of : $$R_{C,180} \circ \tau_{AB}$$

which shall be seen below.

In addition, $ γ$ =glide reflections, $\tau$=translations, and $\rho$=reflections

We have:

$$ γ_{C,D} \circ γ_{A,B}$$

which is equivalent to:

$$ (\rho_{C,D} \circ \tau_{C,D})\circ (\rho_{A,B} \circ \tau_{A,B})$$

or we can reverse the translations and reflections since glide reflections are commutative.

Here is the picture that shall be used to understand below:

enter image description here

Now, I worked each part. I started off with $\rho_{A,B} \circ \tau_{A,B}$. Observe there is a translation involved and I know a translation can be represented as two reflections. Same with $ \rho_{C,D} \circ \tau_{C,D}$.

I let every translation be rewritten as two reflections which are parallel to each other

$$ (\rho_{C,D} \circ \tau_{C,D})\circ (\rho_{A,B} \circ \tau_{A,B})$$ $$ (\rho_{m} \circ \rho_{v} \circ \rho_{u})\circ (\rho_{n} \circ \rho_{s} \circ \rho_{r})$$.

Note line $u=n$, so their reflections will cancel out and we are left with:

$$ (\rho_{m} \circ \rho_{v} \circ \rho_{s} \circ \rho_{r})$$.

Note that m and v intersect, so this will be a rotation with the center $(1,-2)=C$ and angle of rotation is 180 degrees. Now note that lines s and r are parallel, so this is a translation $\tau_{AB}$

So now it reduces to:

$$R_{C,180} \circ \tau_{AB}$$

where $c=(1,-2)$

Now I attempt to solve: $$R_{C,180} \circ \tau_{AB}$$

where $c=(1,-2)$

Let the following be represented as:


Now note that the rotation is not at the origin, therefore for some point (x,y) we must shift it to the origin, rotate and then shift back.


With our rotation and translation being identified:

$$R_{C,180} \circ \tau_{AB}(x,y)=R_{C,180}(x+2,y)=(-(x+2)+2,-y-4)=(-x,-y-4)$$

To find the center:

$$(x,y)=(-x,-y-4)$$ then: $$x=-x$$ $$y=-y-4$$

but then, this doesn't work. My professor said it is a rotation centered at $(0,1)$ and I can't seem to find where I went wrong. If anyone can help?

If anyone can help get pass this point, it would be really helpful!


Your computation of the center of rotation of $R_{C,180} \circ \tau_{AB}$ is correct, but $\gamma_{C,D} \circ \gamma_{A,B}$ is not equal to $R_{C,180} \circ \tau_{AB}$. You made a mistake when you wrote $\tau_{CD}$ and $\tau_{AB}$ as compositions of reflections : it is not true that $\tau_{CD} = \rho_v \circ \rho_u$. You should get the correct answer if you use the line $y = 1$ instead of $v$.

Another way to solve the problem would be to write $\rho_{CD}$, $\rho_{AB}$, $\tau_{CD}$ and $\tau_{AB}$ in coordinates \begin{align} \rho_{CD}(x,y) &= (2-x,y) \\ \rho_{AB}(x,y) &= (x,-y) \\ \tau_{CD}(x,y) &= (x,y+2) \\ \tau_{AB}(x,y) &= (x+2,y) \end{align} and compute the composition \begin{align} \rho_{CD} \circ \tau_{CD} \circ \rho_{AB} \circ \tau_{AB}(x,y) &= \rho_{CD} \circ \tau_{CD} \circ \rho_{AB} (x+2,y) \\ &= \rho_{CD} \circ \tau_{CD} (x+2,-y) \\ &= \rho_{CD} (x+2,2-y) \\ &= (-x,2-y) \end{align} By solving $x = -x$ and $y = 2 - y$, we find that the center of rotation is $(0,1)$, as expected.

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  • $\begingroup$ this way was super simple! thanks a lot! $\endgroup$ – Mark May 6 '15 at 2:10

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