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I am trying to prove that all the isometries of a regular polygon that map the polygon back onto itself must map vertices to vertices. I nearly have the proof, but I need to prove one more statement:

Take a regular $n$ sided polygon, with $n>3$ and sides of length $l$. Pick any vertex $A$. Then any point on the polygon $x$ such that $|A-x| < l$ (i.e. the distance from $A$ to $x$ is less than $l$) must lie on an edge incident to $A$ or is equal to $A$.

This statements seems to be true to me, just through visualising polygons, but I can't think of an elegant proof.

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  • $\begingroup$ What about the case where $A = x$? $\endgroup$ – MathApprentice Jun 5 '16 at 11:50
  • $\begingroup$ @MathApprentice I've rephrased to question to cover that case. I was assuming that edges include their incident vertices, but now I've removed the ambiguity $\endgroup$ – texasflood Jun 5 '16 at 12:06
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    $\begingroup$ suggestion: take an outer node with its 2 edges. fix any 2 points on these edges. together with your node you get a triangle. now on this triangle you apply triangle inequality on the side connecting the points you chose. for nodes this inequality is always strict. by isometry you can map these strict inequalities from pre-image to image and the other way around. therefore nodes get mapped to nodes. $\endgroup$ – Max Jun 5 '16 at 12:07
  • $\begingroup$ @Max I can follow your reasoning, except for the last statement. I can't see how this shows that nodes get mapped to nodes. Say $A$ is the node and we choose points $B$ and $C$, and $\phi$ is the isometry. Do you assume that $\phi(A)$ must be a node in order for $\phi(A)$, $\phi(B)$ and $\phi(C)$ to always satisfy this strict inequality? $\endgroup$ – texasflood Jun 5 '16 at 12:46
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    $\begingroup$ $\phi\left(A\right)$ is the node, yes. if a node gets mapped to a non-node, then for the non-node you can chose the points such that the inequality is an equality (a degenerate triangle) - transforming it back it with the isometry gives an equality in triangle equaltion in a node; wich would be a contradiction, since for such a node you cannot choose a degenerate triangle. $\endgroup$ – Max Jun 5 '16 at 12:50
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I haven't spent any time on the subsidiary Q, but as to the problem of an isometry $f$ of the polygon $P$ to itself, observe that the isometric image $f(S)$ of a closed bounded line segment $S$ of positive length $l$ must be a closed bounded line segment of equal length, because

(1)..... For any $\{x_1,x_2,x_3\}\subset S$ and for some $\{i,j,k\}=\{1,2,3\}$ we have $d(x_i,x_j)+d(x_j,x_k)=d(x_i,x_k)$ and hence $d(f(x_i),f(x_j))+d(f(x_j),f(x_k))=d(f(x_i),f(x_k)).$..... Therefore

($\bullet$) Any 3 points in $f( S)$ are collinear.

(2)..... Let $e_1,e_2$ be the endpoints of $S$. Let $T$ be the closed line segment from $f(e_1)$ to $f(e_2).$ Let $x$ be any point of $S$. Then $l= d(e_1,e_2)=d(f(e_1),f(e_2)),$ while $d(f(x),f(e_1))+ d(f(x),f(e_2))=d(x,e_1)+d(x,e_2)=l$ ...... So by ($\bullet$), we have

($\bullet \bullet)$ $f(S) \subset T.$ And $f(S)$ includes the endpoints of $T,$ and the length of $T$ is $l$.

(3)..... For any $y\in T ,$ there exists $x\in S$ with $d(x,e_i)=d(y,f(e_i))$ for $i\in \{1,2\},$ and by ($\bullet \bullet$) we have $f(x)\in T.$ So we must have $f(x)=y.$..... Hence $f(S)\supset T.$ Together with $f(S)\subset T$ from (2), we have $$ f(S)=T.$$

For the polygon $P,$ and an isometry $f:P\to P$, let $S$ be any of its sides. Then $f(S)$ is a closed straight-line segment of equal length and $f(S)\subset P,$ so $f(S)$ is a side of $P.$ And from ($\bullet \bullet$), $f$ maps the endpoints of $S,$ which are vertices of $P,$ to the endpoints of $f(S),$ which are also vertices of $P$, as they are the endpoints of the side $f(S). $

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In a regular $n$-gon (considered in your question as a "one-dimensional complex") the inner angle between subsequent edges is $\bigl(1-{2\over n}\bigr)\pi$. If $n=4$ this angle is ${\pi\over2}$, so that the two edges not adjacent to $A$ are just tangents to the circle with center $A$ and radius $\ell$. If $n\geq5$ this angle is $>{\pi\over2}$, so that the next two edges are "running away" from the mentioned circle. The edges "further down" should cause no problems.

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