TITLE: Example of convex n-gon that cannot be decomposed into k congruent convex polygons
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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)