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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:

$$\tau_{AB}=(x+2,y)$$

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.

$$R_{C,180}=(x,y)=(x-1,y+2)=(-x+1,-y-2)=((-x+1)+1,(-y-2)-2)=(-x+2,-y-4)$$

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!

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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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