Cutting a symmetric polygon into two congruent parts

Cutting a symmetric polygon into two congruent parts
This question is a continuation of the MSE question: Union of two disjoint congruent polygons is centrally symmetric. Must the polygons differ by a 180 degree rotation? Consider a property of plane polygons $A,B,P$ defined as follows. $\textbf{Con}(A,B,P)$: “$A,B$ are congruent polygons that tile a polygon $P$, such that $P$ is centrally symmetric across point $O$, but $A,B$ are not centrally symmetric images of each other across point $O$.” Note: "Tiling" is treated in the sense that we don't care about what happens on the boundary. There are various cases of $\textbf{Con}(A,B,P)$ that is known to be possible or impossible. I've made a table below. $\textbf{Con}(A,B,P)$ is... $P$ may have holes $P$ is simple $A,B$ may be disconnected Possible${}^1$ Possible${}^1$ $A,B$ may have holes Unresolved Impossible${}^2$ $A,B$ are simple Unresolved Impossible${}^2$ The following example is given by M. Bidva and Yu. Markelov. Reference: A. Yu. Sadovnichii, Problem of cutting a polygon into two congruent parts, Matematicheskie Zametki 117 (2025), no. 3, 443-452 When $A,B,P$ are simple polygons, $\textbf{Con}(A,B,P)$ is proven impossible by Omri Zemer in the linked MSE post. Let's resolve the remaining case(s)! Question: Do there exist interior-connected but not necessarily simply connected (i.e. with holes) polygons $A,B,P$ such that $\textbf{Con}(A,B,P)$ holds? Some personal notes: By Omri Zemer's proof in the linked post, it's necessary for $A,B$ to be congruent by a nontrivial rotation, i.e. $A,B$ are a rotation apart by an angle $0^\circ<\theta<180^\circ$ and around a center that's different from point $O$. Omri Zemer's impossibility proof does not cover this remaining case. The reason is that the proof invokes the ping-pong lemma on the two relevant transformations of the Euclidean plane. This requires the boundary of $P$ to be partitioned by $A,B$ into two pieces that are centrally symmetric images of each other across point $O$. I do not see a proof of this criterion in the unresolved case. Due to the above reason, I suspect an example similar to the Russian example exists. That is, there is some example $A,B,P$ that are in fact orthogonally connected polyominos where $A,B$ differ by exactly a $90^\circ$ rotation. I hope such an example can be found by a deep computer search or a really insightful human ;) Feel free to prove my hope wrong by proving this is also impossible.

Take Your Experience to the Next Level

New

Download our mobile app for a faster and better experience.

Comments

0
U

Join the discussion

Sign in to leave a comment

0:000:00