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Let $ABCDEF$ be a regular hexagon that is oriented clockwise (so that a rotation from $A$ to $B$ to $C$ to $D$ to $E$ to $F$ is clockwise).

i) Identify $R_{D,120} \circ R_{A,60}$

which are two rotations meaning the end result will also be a rotation of $180$ but I can't seem to find the center.

ii) Identify $R_{F,180} \circ \rho_{ED} \circ R_{D,120}$

This one is confusing since it is a rotation composed with a reflection composed with a different rotation. I overheard two people mentioning how the composition is a glide reflection but I thought glide reflections where compositions of reflection and translation.

If anyone can help me with these two problems

UPDATE: Here is my work for ii)

enter image description here

Identify $R_{F,180} \circ \rho_{ED} \circ R_{D,120}$

I started out with:

$$\rho_{ED} \circ R_{D,120}$$

where I let:

$$\rho_{ED}=\rho_{m}$$ $$ R_{D,120}=\rho_{m} \circ \rho_{n}$$

$$ \text{ m: be represented by the line ED }$$ $$ \text{ n: be represented by the line that goes through D and A}$$

Then we have:

$$\rho_{ED} \circ R_{D,120}=\rho_{m} \circ (\rho_{m} \circ \rho_{n})=(\rho_{m} \circ \rho_{m} ) \circ \rho_{n}=\rho_{n} $$

So now we have:

$$R_{F,180} \circ \rho_{n}$$

I let $$R_{F,180}=\rho_{k} \circ \rho_{r}$$

where:

$$ \text{ k: be represented by the line BF}$$ $$ \text{ r: be represented by the line EF }$$

$$R_{F,180} \circ \rho_{n}=(\rho_{k} \circ \rho_{r}) \circ \rho_{n}=\rho_{k} \circ (\rho_{r} \circ \rho_{n})=\rho_{BF} \circ \tau_{BF}=\gamma_{BF}$$

which is a glide reflection

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    $\begingroup$ You are worried that a glide reflection can be expressed as a composition of group elements in more than one way. But if you think of it from a group theory point of view, you'll realize that something similar true in any group you can imagine, such as the group of integers under addition: $7 + 3 + 5 = 6 + 9$. $\endgroup$ – Lee Mosher Apr 12 '15 at 14:14
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Denote by $O$ the middle of the segment $[A,D]$. You can check that all the triangles $OAF, OBA, OCB, ODC, OED, OFE$ are equilateral triangles. I shall denote by $(AD)$ the straight line through $A$ and $D$, etc. It is obvious that $(AD)=(DA)$. The notation $\mathrm{Id}$ stands for the identity map.

i) In order to identify $R_{D,120}\circ R_{A,60}$ we can make the following decompositions: $$ R_{D,120}=\rho_{\Delta_1}\circ \rho_{DA},\quad R_{A,60}=\rho_{AD}\circ\rho_{\Delta_2}, $$ with $$ \Delta_1\cap(DA)=\{D\},\, \angle(DA,\Delta_1)=60,\, \Delta_2\cap(AD)=\{A\},\, \angle(AD,\Delta_2)=30. $$ Since $\angle(\overrightarrow{DA},\overrightarrow{DE})=60$, we deduce that $\Delta_1=(DE)$. Similarly we have $\Delta_2=(AE)$ because $\angle(\overrightarrow{AE},\overrightarrow{AD})=30$.

Hence $$ R_{D,120}\circ R_{A,60}=\left(\rho_{DE}\circ \rho_{DA}\right)\circ\left(\rho_{AD}\circ\rho_{AE}\right)=\rho_{DE}\circ\underbrace{\left(\rho_{DA}\circ\rho_{AD}\right)}_{\mathrm{Id}}\circ\rho_{AE}=\rho_{DE}\circ\rho_{AE}=\rho_{DE}\circ\rho_{AE}. $$ Now notice that $(DE)\cap(AE)=\{\color{red}{E}\}$ and $$ \angle(\overrightarrow{EA},\overrightarrow{ED})=\angle(\overrightarrow{EA},\overrightarrow{EO})+\angle(\overrightarrow{EO},\overrightarrow{ED})=30+60=90=\frac{\color{red}{180}}{2}, $$ and therefore $$ R_{D,120}\circ R_{A,60}=R_{\color{red}{E},\color{red}{180}}. $$ ii) In order to identify $R_{F,180}\circ\rho_{\color{blue}{ED}}\circ R_{\color{blue}{D},120}$, we first make the following decomposition: $$ R_{D,120}=\rho_{\color{blue}{DE}}\circ\rho_{\Delta_3}, $$ with $$ \Delta_3\cap(DE)=\{\color{blue}{D}\},\, \angle(\Delta_3,DE)=60. $$ Therefore $\Delta_3=(DA)$, because $OED$ is an equilateral triangle. It follows that

\begin{eqnarray} R_{F,180}\circ\rho_{ED}\circ R_{D,120}&=&R_{F,180}\circ\left(\rho_{\color{blue}{ED}}\circ R_{\color{blue}{D},120}\right)=R_{F,180}\circ\left[\rho_{\color{blue}{ED}}\circ\left(\rho_{\color{blue}{DE}}\circ\rho_{\color{green}{DA}}\right) \right]\\ &=&R_{F,180}\circ\left[\underbrace{\left(\rho_{\color{blue}{ED}}\circ\rho_{\color{blue}{DE}}\right)}_{\mathrm{Id}}\circ\rho_{\color{green}{DA}}\right]=R_{F,180}\circ\rho_{\color{green}{DA}}. \end{eqnarray} Since $\overrightarrow{FB}\perp\overrightarrow{FE}$, i.e. $\angle(\overrightarrow{FB},\overrightarrow{FE})=90$ we can write: $$ R_{F,180}=\rho_{FB}\circ\rho_{FE}. $$ Thus $$ R_{F,180}\circ\rho_{ED}\circ R_{D,120}=\left(\rho_{FB}\circ\rho_{FE}\right)\circ\rho_{DA}=\rho_{FB}\circ\underbrace{\left(\rho_{FE}\circ\rho_{DA}\right)}_{T_{\overrightarrow{BF}}}=\rho_{FB}\circ T_{\overrightarrow{BF}}, $$ i.e. $$ R_{F,180}\circ\rho_{ED}\circ R_{D,120}=\rho_{BF}\circ T_{\overrightarrow{BF}}=T_{\overrightarrow{BF}}\circ\rho_{BF} $$ is a glide reflection.

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  • $\begingroup$ I worked out this problem a while back and it's good to know we can agree with part i. I got the same thing but for ii) I got different lines. For the part where you said: $R_{F,180}∘ρ_{CD}$, I got $R_{F,180}∘ρ_{AD}$. I'll update with a picture of what I have and I am going to work out yours as well. $\endgroup$ – Mark Apr 22 '15 at 19:55
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    $\begingroup$ @Mark You were right, I just fixed it. $\endgroup$ – Mercy King Apr 26 '15 at 20:07
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To do this problem geometrically, starting with the ABCDEF hexagon, reflect it repeatedly through its sides to produce a "tiling" by hexagons. Now do some calculations: what vertex of the tiling is the image of, say, $B$ under the composition $R_{D,120} \circ R_{A,60}$? What vertex is the image of $C$? of $D$? Once you have done a few of these calculations, you should be able to infer what map this composition is.

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