1
$\begingroup$

Let $n$ be a positive integer number and $P$ be a point in a plane. Let $A_1$, $A_2$, $\cdots$, $A_m$ be $m$ points in the plane, we take modulo $m$ for $A_j$ (it is mean $A_{m+i}=A_{i}$ for $i=1, 2, \cdots$).

Now we rotate $P$ clockwise above $A_1$ with angle $\frac{\pi}{n}$ we have point $P_1$, rotate $P_1$ clockwise above $A_2$ with angle $\frac{\pi}{n}$ we have point $P_2$ $,\cdots,$ rotate $P_k$ clockwise above $A_{k+1}$ with angle $\frac{\pi}{n}$ we have point $P_{k+1}$....

By my computation with $(m,n)=(3,1), (3,2), (3,3), (4,3), (3,4), (2,3), (2,4), (2,5)$....I see that exist $k$ such that $P_{k+1}\equiv P$

My question: What condition of $(m,n)$ such that exist $k$ so that $P_{k+1}\equiv P$? Version of this problem for Euclidean space (or non-Euclidean geometries)?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Here is a counterexample. Let $(m,n)=(2,1)$ and choose $A_1=(0,0)$ and $A_2\neq A_1$ arbitrary. Then the rotation around $A_1$ is given by $P\mapsto -P$ and the rotation around $A_2$ is given by $P\mapsto 2A_2-P$. In particular, the composition of both rotations is the translation $P\mapsto 2A_2+P$. Hence, whenever we start with a point $P$ that does not lie on the line defined by $A_1$ and $A_2$, then $P$ is not equal to $\pm (P+2\lambda A_2)$ with $\lambda\in\mathbb{Z}\backslash\{0\}$. $\endgroup$
    – Philipp Lampe
    Commented Jul 31, 2018 at 16:42
  • $\begingroup$ Thank You very much, but $(m,n)$=$(3,1)$, $(3, 2)$, $(3,3)$, $(4, 3)$, $(3,4)$... are true @PhilippLampe $\endgroup$
    – Đào Thanh Oai
    Commented Jul 31, 2018 at 17:17
  • $\begingroup$ $(m, n)$=$(2,3)$, $(2, 4)$ are also true......@PhilippLampe $\endgroup$
    – Đào Thanh Oai
    Commented Jul 31, 2018 at 17:25
  • $\begingroup$ Thank You very much for your comment. I have just improved my question from a conjecture to an open problem. @PhilippLampe $\endgroup$
    – Đào Thanh Oai
    Commented Jul 31, 2018 at 17:42
  • 2
    $\begingroup$ Using complex numbers, the rotation around $A_i$ is given by $P\mapsto (P-A_i)\omega+A_i=P\omega+(1-\omega)A_i$ where $\omega$ in a primitive $(2n)$-th root of unity. Hence, a sufficient condition for periodicity after $2n$ full cycles is $1+\omega^m+\ldots+\omega^{(2n-1)m}=0$. This fails when $m$ is a multiple of $2n$. $\endgroup$
    – Philipp Lampe
    Commented Jul 31, 2018 at 17:53

1 Answer 1

Reset to default
2
$\begingroup$

А composition of rotations is ether rotation or translation (a special case of Chasles' theorem) depending on the total rotation angle which is $k\pi /n$, so $P_k=R_0^{k\pi /n}(O_k)$ (rotation with some center $O_k$) for $k\not\equiv 0\pmod {2n}$ and $P_k=T_{v_k}(P)$ (translation) for $k\equiv 0\pmod{ 2n}$ (This rotation angle can be controled by complex numbers as in Philipp Lampe's comment). So fixed point exists only if $k\not\equiv 0\pmod {2n}$ (this point is $O_k$) or $k\equiv 0\pmod{ 2n}$ and $v_k=0$ (all points are fixed). The last case can be checked starting from arbitrary $P$ and checking whether $P_k=P.$

Improve this answer
$\endgroup$
4
  • $\begingroup$ Do You mean always exist $k$ if $m\not\equiv 0\pmod {2n}$ ? What is formular of $k=f(m,n) ?$ $\endgroup$
    – Đào Thanh Oai
    Commented Aug 1, 2018 at 2:22
  • $\begingroup$ @ĐàoThanhOai Any $k$ such that $k\not\equiv 0\pmod {2n}$ and $k\equiv 0\pmod {2n}$ depending on positions of $A_i$'s. $\endgroup$
    – Alexey Ustinov
    Commented Aug 1, 2018 at 3:10
  • $\begingroup$ I don't think so, $k$ is not depending on positions of $A_i$. My comment is true with $(m,n)=(3,1),(3,2),(3,3),(4,3),(3,4),(2,3),(2,4),(2,5)$ @AlexeyUstinov $\endgroup$
    – Đào Thanh Oai
    Commented Aug 1, 2018 at 3:15
  • 1
    $\begingroup$ @ĐàoThanhOai The answer depends on positions of $A_i$ only if $k\equiv 0\pmod {2n}$. $\endgroup$
    – Alexey Ustinov
    Commented Aug 1, 2018 at 3:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .