# Fixed point in rotations

• Oct 18th 2010, 07:07 AM
Showcase_22
Fixed point in rotations
Quote:

Let $P,Q \in \mathbb{E}^2$ be distinct points, and $\alpha, \beta$ angles such that $\alpha+\beta \neq 0, 2 \pi$.

Let OPQO' be a quadrilateral such that $OPO'= \alpha$, $O'Qo= \beta$ are directed angles (both clockwise as interior angles), and PQ halves both. Find the unique point $S \in \mathbb{E}^2$ for which $\text{Rot}(P, \alpha)(S)= \text{Rot}(q, -\beta)(S)$.
I've drawn a picture and I think I know what the method is:

$\text{Rot}(P, \alpha)(S)=\begin{pmatrix}{\cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{pmatrix} \begin{pmatrix}{s_1-p_1 \\ s_2-p_2} \end{pmatrix}+ \begin{pmatrix}{p_1 \\ p_2} \end{pmatrix}$

$\text{Rot}(Q, -\beta)(S)=\begin{pmatrix}{\cos \beta & -\sin \beta \\ -\sin \beta & -\cos \beta \end{pmatrix} \begin{pmatrix}{s_1-q_1 \\ s_2-q_2} \end{pmatrix}+ \begin{pmatrix}{q_1 \\ q_2} \end{pmatrix}$

So the next step was to equate the two and solve it for values of $s_1$ and $s_2$.

However, this is really long!!

Is there a shorter way to do this?
• Oct 18th 2010, 07:59 AM
Failure
Quote:

Originally Posted by Showcase_22
I've drawn a picture and I think I know what the method is:

$\text{Rot}(P, \alpha)(S)=\begin{pmatrix}{\cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{pmatrix} \begin{pmatrix}{s_1-p_1 \\ s_2-p_2} \end{pmatrix}+ \begin{pmatrix}{p_1 \\ p_2} \end{pmatrix}$

$\text{Rot}(Q, -\beta)(S)=\begin{pmatrix}{\cos \beta & -\sin \beta \\ -\sin \beta & -\cos \beta \end{pmatrix} \begin{pmatrix}{s_1-q_1 \\ s_2-q_2} \end{pmatrix}+ \begin{pmatrix}{q_1 \\ q_2} \end{pmatrix}$

So the next step was to equate the two and solve it for values of $s_1$ and $s_2$.

However, this is really long!!

Is there a shorter way to do this?

I sure would not want to write everything down in coordinates. Instead I would introduce the notation $\mathrm{Rot}(\varphi)$ for a rotation by $\varphi$ around the origin and then write $\mathrm{Rot}(P,\alpha)\vec{s}=\mathrm{Rot}(\alpha) (\vec{s}-\vec{p})+\vec{p}$, similarly for $\mathrm{Rot}(Q,-\beta)\vec{s}$.
Then I could proceed purely algebraically (without coordinates) until I have isolated $\vec{s}$ on one side of the equation, provided $\mathrm{Rot}(\alpha)-\mathrm{Rot}(-\beta)$ is invertible.
• Oct 18th 2010, 10:07 AM
Showcase_22
(It was funny when I got your post back: "Failure has replied to your post" (Crying) )

I followed it through and it was much easier than using coordinates.

$\vec{s}=(\text{Rot}(\alpha)-\text{Rot}(- \beta))^{-1}(\vec{q}-\vec{p}-\text{Rot}(-\beta) \vec{q}+\text{Rot}(\alpha) \vec{p})$

I could write it out in terms of coordinates, but I don't think it would get any simpler.

I'm also sorry for springing this on you, but I hoped this would help me solve part ii) of the question. Unfortunately, i'm still lost:

Quote:

Hence calculate the composite $\text{Rot}(Q, \beta) \circ \text{Rot}(P, \alpha).$
So I know that:

$\mathrm{Rot}(P,\alpha)\vec{s}=\mathrm{Rot}(\alpha) (\vec{s}-\vec{p})+\vec{p}$

and:

$\mathrm{Rot}(Q,\beta)\vec{s}=\mathrm{Rot}(\beta)(\ vec{s}-\vec{q})+\vec{q}$

Therefore:

$\text{Rot}(Q, \beta) \circ \text{Rot}(P, \alpha) \vec{x}=\mathrm{Rot}(\beta)(\mathrm{Rot}(\alpha)(\ vec{x}-\vec{p})+\vec{p}-\vec{q})+\vec{q}$

(putting the first one into the second one, for some vector $\vec{x}$).

Expanding this out gives:

$\text{Rot}(Q, \beta) \circ \text{Rot}(P, \alpha) \vec{x}=\text{Rot}(\beta). \text{Rot}(\alpha)(\vec{x}-\vec{p})+\text{Rot}(\beta) \vec{p}-\text{Rot}(\beta) \vec{q}+\vec{q}$

If it didn't say "hence calculate...", I probably would have left it there.

At this point I thought I could group the terms together to get $\vec{s}$ appearing somewhere (or at least $\text{Rot}(\alpha)-\text{Rot}(-\beta)) \vec{s}$). A big obstacle is that $\vec{s}$ doesn't really pop up. I tried seeing if there was a link between $\text{Rot}(Q, \beta)$ and $\text{Rot}(Q, -\beta)$, but writing them out doesn't produce anything.
• Oct 18th 2010, 11:01 AM
Failure
Quote:

Originally Posted by Showcase_22
(It was funny when I got your post back: "Failure has replied to your post" (Crying) )

I followed it through and it was much easier than using coordinates.

$\vec{s}=(\text{Rot}(\alpha)-\text{Rot}(- \beta))^{-1}(\vec{q}-\vec{p}-\text{Rot}(-\beta) \vec{q}+\text{Rot}(\alpha) \vec{p})$

I could write it out in terms of coordinates, but I don't think it would get any simpler.

I'm also sorry for springing this on you, but I hoped this would help me solve part ii) of the question. Unfortunately, i'm still lost:

So I know that:

$\mathrm{Rot}(P,\alpha)\vec{s}=\mathrm{Rot}(\alpha) (\vec{s}-\vec{p})+\vec{p}$

and:

$\mathrm{Rot}(Q,\beta)\vec{s}=\mathrm{Rot}(\beta)(\ vec{s}-\vec{q})+\vec{q}$

Therefore:

$\text{Rot}(Q, \beta) \circ \text{Rot}(P, \alpha) \vec{x}=\mathrm{Rot}(\beta)(\mathrm{Rot}(\alpha)(\ vec{x}-\vec{p})+\vec{p}-\vec{q})+\vec{q}$

(putting the first one into the second one, for some vector $\vec{x}$).

Expanding this out gives:

$\text{Rot}(Q, \beta) \circ \text{Rot}(P, \alpha) \vec{x}=\text{Rot}(\beta). \text{Rot}(\alpha)(\vec{x}-\vec{p})+\text{Rot}(\beta) \vec{p}-\text{Rot}(\beta) \vec{q}+\vec{q}$

If it didn't say "hence calculate...", I probably would have left it there.

At this point I thought I could group the terms together to get $\vec{s}$ appearing somewhere (or at least $\text{Rot}(\alpha)-\text{Rot}(-\beta)) \vec{s}$). A big obstacle is that $\vec{s}$ doesn't really pop up. I tried seeing if there was a link between $\text{Rot}(Q, \beta)$ and $\text{Rot}(Q, -\beta)$, but writing them out doesn't produce anything.

The condition that S must satisfy makes it a fixed point of the concatenation of the two rotations. Hence one might guess that the concatenation is the rotation $\mathr{Rot}(S,\alpha + \beta)$. Given this idea you might try to transform what you have into $\mathr{Rot}(S,\alpha + \beta)$.

Note that it would suffice to show that it is a rotation by $\alpha+\beta$ around some point, since the center of rotation is the only fixed point it has (except if one happens to be rotating by 0 of course) that center has to be S.