We all know the "hat" monotile by now, right? It can obviously be dissected into 8 congruent kites, or 4 congruent pentagons. Can it be dissected at all into 2 congruent shapes? What about 3?

I wish this was smaller

In general, characterize the values of n for which the hat can be dissected into n congruent pieces.

(I don't know the answer.)

  • $\begingroup$ Incidentally, is there a googleable name for this specific kind of "dissect into congruent figures" puzzle (e.g. here and here and here)? The search term "dissection" more often turns up situations (equidissection, congruent-by-dissection) where the pieces themselves do not have to be congruent, which makes it hard to find info on this kind of puzzle. $\endgroup$ Jun 7, 2023 at 21:08
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    $\begingroup$ Tilers generally just say something like 'can you tile this shape with copies of a single tile'. If the shape is similar to the tile, 'is this tile a reptile'. I don't know of any specific shortcut terminology beyond that except for specific instances like 'squaring the square'. $\endgroup$ Jun 8, 2023 at 1:12
  • $\begingroup$ Answering my own comment above: On MathOverflow, @NandakumarR uses the term "congruent partition" for this kind of puzzle; and that makes etymological sense to me. $\endgroup$ Jun 23, 2023 at 16:38

3 Answers 3


Building on @theonetruepath's answer, a drafter can be divided in three pieces like so:

enter image description here

so in addition to $32k^2$, we have $2^{4+m}3^nk^2$.


Obviously you can do $n=16$ pieces just by splitting the kites into drafters, assuming flipping is allowed which it generally is with 'hat' constructions. A drafter is the 30-60-90 triangle you get when you dissect a kite into two congruent pieces.

Furthermore, you can use four drafters to tile a 'doubled' drafter. So that gives $n=64$ as well, in fact n can be $16*4^m (m=1,2,3,...)$ by dividing drafters into smaller and smaller pieces.

And as Jaap points out, you can use a great many triangle divisions, not just 2^2. If we write down just the new ones, ie avoiding overlaps, we get $4, 8, 16(p*p)^n$ where p is primes starting at 2 and n is all positive integers. And with OP's division into four 30-30-120 triangles, add $2*16(p*p)^n$ and since those are symmetric triangles, add $4*16(p*p)^n$ as well. Some overlaps there but I think both terms add new numbers. Possibly there are further divisions of kites and kite-pairs into triangles or other simple shapes as well.

I see no dissections into 2,3,5,etc shapes (yet). But I can see plenty of searching to fill in some of the larger missing numbers.

Actually I realise I've missed out a bunch of possibilities. We should write $(1,2,4)16(p*p)^a(q*q)^b(r*r)^c,...$ where $p,q,r,...$ are primes starting at $2$ and $a,b,c,...$ are non-negative integers.

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    $\begingroup$ Actually, $16 n^2$ for any $n$, cause a triangle can be cut into $n^2$ congruent pieces. $\endgroup$ Jun 7, 2023 at 9:17
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    $\begingroup$ You can get $16m^2$ by dividing the kite into two drafters; and you can get $32m^2$ by dividing the kite into four 30-30-120 triangles. So we've now got all the powers of 2, starting with 4 (pentagons), 8 (kites), and then 16,32,64,128... (triangles). All that's left are the interesting cases. ;) $\endgroup$ Jun 7, 2023 at 21:02

Just to summarize all the previous answers in one place: The hat can be dissected into

  • 4 pentagons
    • Bisect those pentagons into kites (8 kites)
      • Bisect those kites into drafters (16 drafters)
        • Repeatedly trisect and/or $k^2$-sect those drafters into smaller drafters ($16\cdot 3^m k^2$ drafters)
      • Quadrisect those kites into 30-30-120 triangles (32 triangles)
        • $k^2$-sect those triangles ($32 k^2$ triangles)
          • Bisect those triangles into drafters; repeatedly trisect and/or $k^2$-sect those drafters ($64\cdot 3^m k^2$ drafters again)

I'll add to this answer as more dissections/triangulations are discovered.

The fact that the hat can be broken down into 48 drafters suggests that maybe you could build those 48 drafters back up into 12 congruent tetradrafters, or 6 congruent octodrafters. Can anyone do it, or alternatively prove that it's hopeless? In general, can we atomize the hat into smaller and smaller triangles, but never put any of those triangles back together into larger groupings?

Alternatively, can a drafter, 30-30-120, or kite be broken down into a clever number of non-triangular pieces? "Triangles that can be cut into mutually congruent and non-convex polygons" is just barely relevant here. Also this answer which links to "Triangle Tiling I" (Michael Beeson, 2012); "II", "III"(!), "IV", "V", but I think those all deal with triangles with integer sides, so, no drafters or 30-30-120s.

  • $\begingroup$ (Consider making this answer a community wiki...) $\endgroup$
    – Rubio
    Aug 12, 2023 at 0:25

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