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Let $P$ be a closed polygon defined by the sequence
$p_0,\,\dots,\,p_{n-1},p_0$ of points.

Question:

how can one construct, with straightedge and compass alone, another sequence of points $q_0,\,\dots,q_{n-1}$ such that:

  • $q_i$ lies on the bisector of $p_i$ and $p_{i+1}$
  • $q_i,p_{i+1}$ and $q_{i+1}$ are collinear
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  • $\begingroup$ Do I understand it correctly that $q_i$ is just the midpoint between $p_i$ and $p_{i+1}$? If so, what is the difficulty? Or, perhaps, I don't understand what you mean by the bisector of points. $\endgroup$
    – Iosif Pinelis
    Commented Jan 27, 2023 at 15:37
  • $\begingroup$ @IosifPinelis my apologies; I mixed up the $p$ and $q$; it should now be correct. $\endgroup$
    – Manfred Weis
    Commented Jan 27, 2023 at 16:04
  • $\begingroup$ I see. Do you know anything about the existence and uniqueness of the $q_i$'s? About their expressions in terms of the $p_i$'s? $\endgroup$
    – Iosif Pinelis
    Commented Jan 27, 2023 at 19:36
  • $\begingroup$ What is the bisector of two points? $\endgroup$
    – Christophe Leuridan
    Commented Jan 27, 2023 at 19:40
  • $\begingroup$ @IosifPinelis in the case of regular $2n$-gons the solution is not unique; would be another interesting problem to characterise the polygons with ambiguous solution - I suspect that periodicity of angles and sidelengths plays a role. The existence seems however to be granted - if one starts with $q_0$ arbitrarily chosen and proceed on the repeated sequence of polygon points to get $q_{i+1}$ from $q_i$ and $p_{i+1}$ one will observe a kind of spiraling behavior where $q_0'\ne q_0$ in general. Using $\left(q_0'+q_0\right)/2$ as the starting point for the next iteration will solve the problem. $\endgroup$
    – Manfred Weis
    Commented Jan 28, 2023 at 10:09

1 Answer 1

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Let $L_i$ be the bisector of $p_i$ and $p_{i+1}$, and let $f_i \colon L_i \to L_{i+1}$ be the central projection through $p_{i+1}$. This is a projective transformation with constructible coefficients. Your question is about the fixed points of the projective transformation $f_n \circ \cdots \circ f_1$. The coefficients of this projective transformation are also constructible. The fixed points are found by solving a quadratic equation, which can be done by compass and ruler.

It follows also that the number of solutions is among $0, 1, 2$ or $\infty$, although it is not clear whether there are examples with no solutions.

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