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Math Help - Fixed point in rotations

  1. #1
    Super Member Showcase_22's Avatar
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    Fixed point in rotations

    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?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    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.
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  3. #3
    Super Member Showcase_22's Avatar
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    (It was funny when I got your post back: "Failure has replied to your post" )

    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:

    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.
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    (It was funny when I got your post back: "Failure has replied to your post" )

    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.
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