Can a dodecagon be cut into $n$ congruent pieces for any $n$ not of the form $1,2,3,4,6,8,12k^2,24k^2$?

Question

Suppose I want to cut a regular dodecagon into $n$ congruent simply-connected pieces. For which $n$ is this possible?

I can cut it into 24 right triangles, by cutting from the center to each vertex and the midpoint of each edge; by gluing consecutive such triangles together, I can also get any factor of $24$.

If I cut into either 12 or 24 triangular sectors, I can subdivide each of those triangles into $k^2$ congruent pieces for any positive integer $k$, giving $n=12k^2$ or $n=24k^2$.

Can anything else be done? Obviously if the answer is "no" this is likely to be near-impossible to prove, but I would accept any evidence of someone investigating this problem in the literature and failing to exhibit additional possibilities. (I'd also be interested in partial impossibility proofs - for instance, the case of triangular tiles seems likely to be tractable using the sorts of methodologies Michael Beeson has applied to the problem of tiling a larger triangle.)

I ask about a dodecagon in particular because I recently found a nontrivial tiling of it, shown below:

but this tiling uses $48$ pieces and so doesn't add anything new to the possible values of $n$.

Edit: This dissection appears in C. Dudley Langford's Tiling Patterns for Regular Polygons, from 1960.

Answer 1

I. In the figure below, $AOB$, $BOC$ are two adjacent isosceles triangles of the twelve that can tile the dodecagon. Join $AC$, and on $AO$, $CO$ construct isosceles triangles $AEO$, $CDO$ with base angle $=\angle BAC=15^o$, and join $AD$. This produces two quadrilaterals$$AEOD\cong ABCD$$with angles of $45^o$, $150^o$, $60^o$, $105^o$.

Since in area the two quadrilaterals together exceed $\triangle OAB+\triangle OBC$ by $\triangle AEO$, and fall short by $\triangle CDO\cong \triangle AEO$, then$$AEOD=AOB=ABCD=BOC$$and the dodecagon can be tiled with twelve of these congruent quadrilaterals, just as it can be tiled by the twelve congruent isosceles triangles based on the sides of the dodecagon with common vertex at the center.

This transformation of $n$ congruent isosceles triangles into $n$ congruent quadrilaterals which can tile a regular n-gon is evidently possible only for the dodecagon. For congruent triangles $AEO$, $CDO$ must also be congruent with similar triangle $ABC$, but $AC=AO$ only for the dodecagon, since only the regular hexagon has its side equal to the radius of the circumcircle.

II. As OP illustrates, the dodecagon can also be tiled by $48$ tiles similar to the above but of one-fourth the area. In the figure below, taking midpoints $Q,F,M,J$ of quadrilateral $ABCD$ and constructing the four congruent quadrilaterals $BQGF$, $JQGH$, $MKGF$, $HGKL$, evidently $ABCD$ exceeds the four quadrilaterals by $\triangle JAQ+\triangle MFC$, but falls short by figure $JHLKMD$.

But joining $HK$ and $DG$, evidently $\triangle LHK\cong \triangle MFC$, and since it can be shown that $\triangle JNH\cong \triangle GPD$ and $\triangle MKP\cong\triangle GDN$, then remaining area $JHKMD=\triangle KHG\cong \triangle AQJ$. Thus we go from a 12-tiling to a 48-tiling. And dividing $AB$, $BC$ into $3$, $4$, or other number of equal parts will yield a tiling by $108$, $192$, and generally $12k^2$ congruent quadrilaterals, which are thus among those already noted by OP.

III. If a dodecagon is the only regular polygon that can be non-trivially tiled by congruent quadrilaterals, then it seems the answer is “No, nothing else can be done.”

Addendum: In response to OP's comment: Part II. of my answer above might be improved by carrying the construction in the second figure all the way to the center, so as to better compare it with the first figure.

Further, it may help to note what may be obvious, that the quadrilateral in question is the fusion of an equilateral triangle with the side of a triangular half-square. It is well known that the equilateral triangle, square, and regular hexagon can tile the plane. Further however, squares and equilateral triangles in combination can tile the regular dodecagon. The triangles are double the squares in number, so pairing two triangles with each square yields quadrilaterals equal in number to their component triangles, as seen in the next figure.

At the risk of confusing the eye, part II. above may be more persuasive if we keep the square/triangle fusion and carry the construction up to the center $O$, i.e. for a full one-sixth of the dodecagon, as in the figure below.
Superimpose on the two congruent quadrilaterals and equal isosceles triangles of my first figure the eight congruent quadrilaterals $BQGF$, $JQGH$, $MKGF$, $HGKL$, $LHSO$, $TSHJ$, $JQRT$, $AQRV$. Together they exceed the two isosceles triangles $AOB+BOC$ by triangles $TSO+ART+RAV$, but fall short of them by the respectively equal triangles $KLO+CMK+MCF$.

Since these eight tile one-sixth of the dodecagon, then forty-eight tile the whole.