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I asked a related question here on MO without any answers yet.

The question is in the title - give an example of a convex $n$-gon that cannot be subdivided into $k>1$ congruent convex polygons. Even better, give a family that solves this for all combinations of $(n,k)$.

Intuitively, any generic $n$-gon should work, but the crux is in the details - I am very curious about what methods one can use to rigorously prove that a subdivision is impossible.

There are of course many variations, e.g., drop the convex restriction, and remove the restriction that the pieces are polygons.

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    $\begingroup$ Many triangles into two pieces should work. One of the original vertices will be unaffected by a cut, so the other piece must have a congruent angle, which means the first piece with the original vertex must have the supplementary angle in addition if one cuts a side, or a small angle if a cut is made through a vertex. Gerhard "That's How I'd Start It" Paseman, 2018.06.20. $\endgroup$
    – Gerhard Paseman
    Commented Jun 20, 2018 at 18:53
  • $\begingroup$ Also, you should make it clear what subdivided means. There are results that allow for equidecomposition, such as Dudeney's dissection puzzles. If it is your intent, you should explicitly disallow rearranging or reassembling of pieces. Gerhard "Otherwise There Are Many Solutions" Paseman, 2018.06.20. $\endgroup$
    – Gerhard Paseman
    Commented Jun 20, 2018 at 18:59
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    $\begingroup$ Finally, there are such decompositions for regular polygons, in particular (n,n) and (n,n2^k) and (3,4). It might be nice to show for a given n-gon P that the allowable values of k are multiplicatively closed, or mostly so. Gerhard "Find The Edges Of Proof" Paseman, 2018.06.20. $\endgroup$
    – Gerhard Paseman
    Commented Jun 20, 2018 at 19:08
  • $\begingroup$ Right the intention is to make precisely k pieces, no dissection and reassembly allowed. $\endgroup$
    – Per Alexandersson
    Commented Jun 20, 2018 at 19:22
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    $\begingroup$ What do you know about a scalene triangle? Certainly $(3,t^2)$ is possible with triangles congruent to the large one. I can't think of anything else at the moment. $\endgroup$
    – Aaron Meyerowitz
    Commented Jun 20, 2018 at 22:20

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Here is a cheap argument for a generic $n$-gon with $n>6$.

Assume that it is decomposed into $k$ congruent convex $p$-gons. Then the dimension of the subspace of $\mathbb R$ generated by the angles of the $n$-gon over $\mathbb Q$ is generically $n$ while all of them are in the linear span of angles of the $p$-gon whence $p\ge n$. That part does not require convexity or connectedness but the next ones do.

Let $s$ be the number of inside points where several vertices meet and $t$ be the number of inside points where several vertices meet on the side of another polygon (including the side of the big one). The angle count yields $$ 2\pi s+\pi t+ \pi(n-2)=\pi k(p-2) $$ i.e., $$ 2s+t+n-2=k(p-2)\,. $$ On the other hand at least $3$ polygons meet by a vertex at those $s$ points and at least $2$ at those $t$ points, so the vertex count gives $$ 3s+2t+n\le kp $$ so $s+t\le 2k-2$ and, since $p\ge n$, we get $4(k-1)\ge (k-1)(n-2)$, so when $n\ge 7$, we have no chance.

You can improve this argument a bit but I will also be very interested in an argument that would show that in the generic case $n=3$, $k=m^2$ is the only real option for convex partitions (and $n=3$ is the only real option with the connectedness and convexity restrictions dropped). Note that some decompositions are highly non-trivial. For instance, you can split the equilateral triangle into $5$ congruent polygonal pieces (not connected though)

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  • $\begingroup$ I was trying to come up with simple examples that would require a large number of congruent polygons for a proper decomposition. I decided to try a near square, or a square with a small right triangle glued to one edge so that it was half of a 1 by (2+epsilon) rectangle. Imagine my surprise when I found a dissection into 3 pieces. It makes me wonder if such polygons (with k=3) can be classified. Gerhard "Also Reminds Me Of Rep-tiles" Paseman, 2018.06.21. $\endgroup$
    – Gerhard Paseman
    Commented Jun 22, 2018 at 5:10

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