Translation or rotation? Identify $R_{C,-120} \circ R_{B,-60} \circ R_{A,-180}$

Question

Let $ABC$ be a right triangle that is oriented clockwise and has angles of $90, 30, 60$ at vertices $A,B,C$.

Identify $R_{C,-120} \circ R_{B,-60} \circ R_{A,-180}$

I started out with:

$R_{B,-60} \circ R_{A,-180}=R_{C,-240}$

$R_{C,-120} \circ R_{C,-240}$

However, now I have a problem because I have two theorem that fit this but both say different thngs.

One of my theorem says, when two rotations have the same center, then the end result will be the addition of their angles. So from this theorem, I get :

$$R_{C,-120} \circ R_{C,-240}=R_{C,-360}$$

However, another theorem says if the angles of two rotations is 360 or a multiple of 360 then it is a translation.

I don't think it is a translation because translation involves vectors and my end result is a point.

Any insights?

here is my triangle:

Answer 1

Recall that any rotation $R_{\Omega,\theta}$ of $\theta$ about a point $\Omega\in \mathbb{R}^2$ can be decomposed as the product $S_{\Delta_2}\circ S_{\Delta_1}$ of two reflections provided that the two straight lines $\Delta_1$ and $\Delta_2$ intersect at $\Omega$ and the (oriented) angle between them is $\theta/2$: $$\tag{1} R_{\Omega,\theta}=S_{\Delta_2}\circ S_{\Delta_1},\quad \Delta_1\cap\Delta_2=\{\Omega\},\quad \angle(\Delta_1,\Delta_2)=\frac{\theta}{2}. $$

We can write $$ R_{C,-120}=S_{\Delta_2}\circ S_{\Delta_1},\quad R_{B,-60}=S_{\Delta_1}\circ S_{\Delta_3}, $$ with:

• $\Delta_1$ the straight line trough the points $B$ and $C$;
• $\Delta_2$ the straight line trough the point $C$, and such that $\angle(\Delta_1,\Delta_2)=-60$;
• $\Delta_3$ the straight line through the point $B$, and such that $\angle(\Delta_3,\Delta_1)=-30$.

One can check that $\Delta_2\cap\Delta_3=\{S_{\Delta_1}(A)\}$ and $\angle(\Delta_2,\Delta_3)=90$. Therefore, setting $A_1=S_{\Delta_1}(A)$, we have: \begin{eqnarray} R_{C,-120}\circ R_{B,-60}&=&\left(S_{\Delta_2}\circ S_{\Delta_1}\right)\circ\left(S_{\Delta_1}\circ S_{\Delta_3}\right)=S_{\Delta_2}\circ\underbrace{\left(S_{\Delta_1}\circ S_{\Delta_1}\right)}_{\mbox{ identity } }\circ S_{\Delta_3}=S_{\Delta_2}\circ S_{\Delta_3}\\ &=&R_{A_1,180}. \end{eqnarray} Now we have $$ R_{C,-120}\circ R_{B,-60}\circ R_{A,-180}=R_{A_1,180}\circ R_{A,-180}. $$ Also, we can write $$ R_{A_1,180}=S_{\Delta_5}\circ S_{\Delta_4},\quad R_{A,-180}=S_{\Delta_4}\circ S_{\Delta_6} $$ with

• $\Delta_4$ the straight line through the points $A$ and $A_1$;
• $\Delta_5$ the straight line through the point $A_1$, and perpendicular to $\Delta_4$;
• $\Delta_6$ the straight line through the point $A$, and perpendicular to $\Delta_4$.

We have $$ R_{A_1,180}\circ R_{A,-180}=\left(S_{\Delta_5}\circ S_{\Delta_4}\right)\circ\left(S_{\Delta_4}\circ S_{\Delta_6}\right)=S_{\Delta_5}\circ \underbrace{\left(S_{\Delta_4}\circ S_{\Delta_4}\right)}_{\mbox{ identity }}\circ S_{\Delta_6}=S_{\Delta_5}\circ S_{\Delta_6}. $$ Since both lines $\Delta_5$ and $\Delta_6$ are perpendicular to $\Delta_4$, they are therefore parallel. Thus we have $$ R_{A_1,180}\circ R_{A,-180}=S_{\Delta_5}\circ S_{\Delta_6}=T_{\vec{v}}, \quad, \vec{v}=2\overrightarrow{AA_1}=4\overrightarrow{AH}, $$ with $H$ the orthogonal projection of $A$ onto the segment $[B,C]$.