# Dissections of the "hat" aperiodic monotile

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

• 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. Jun 7 at 21:08
• 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'. Jun 8 at 1:12
• 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. Jun 23 at 16:38

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

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.

• Actually, $16 n^2$ for any $n$, cause a triangle can be cut into $n^2$ congruent pieces. Jun 7 at 9:17
• 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. ;) Jun 7 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) 