Cutting a 30-60-90 triangle

[Ed Pegg]

How many ways can a 30-60-90 triangle be split into n self-similar triangles?  

1, 1, 3, 17

This gets very complicated quickly, so I thought I'd ask if I'm missing anything.  I would not be surprised if I did.

I hope I can get up to 8, so I can include a variation of the 1 size 1, 2 size 2, 3 size 3, 4 size 4 30-60-90 partridge solution that both John Conway and R K Guy managed to solve.  
https://erich-friedman.github.io/mathmagic/0802.html  

For a different problem, I built similar triangle solutions using sporadic generators ... very strange cores around a vertex. The 30-60-90 can get extremely strange.

[Ed Pegg]

There are at least 113 ways to divide a 30-60-90 drafter triangle into five self-similar parts. This is 112 of them. A tweak of the last triangle is missing, but I might have missed other dissections.

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[Ed Pegg]

The number of distinct areas so far: 2, 6, 16, 43. 

2-part:  {1/4, 3/4};

3-part: {1/16, 3/16, 1/4, 1/3, 9/16, 3/4};

4-part: {1/64, 3/64, 1/16, 1/12, 1/9, 9/64, 9/49, 3/16,
  12/49, 1/4, 16/49, 1/3, 27/64, 4/9, 9/16, 3/4};  

5-part: {1/256, 1/100, 3/256, 1/81, 1/64, 1/48, 1/36, 3/100,
9/256, 1/27, 1/25, 9/196, 3/64, 4/81, 3/49, 1/16,
4/49, 1/12, 27/256, 1/9, 3/25, 27/196, 9/64, 4/27,
4/25, 9/49, 3/16, 16/81, 12/49, 1/4, 81/256, 16/49,
1/3, 25/64, 27/64, 4/9, 12/25, 9/16, 16/27, 16/25,
36/49, 3/4, 49/64}  

A curious predominance of powers in the denominators.

[M F Hasler]

On Friday, September 25, 2026 at 8:30:44 AM UTC-4 edp...@gmail.com wrote:

There are at least 113 ways to divide a 30-60-90 drafter triangle into five self-similar parts. This is 112 of them. A tweak of the last triangle is missing, but I might have missed other dissections.

On Fri, Sep 25, 2026 at 12:51 AM Ed Pegg <edp...@gmail.com> wrote:

How many ways can a 30-60-90 triangle be split into n self-similar triangles?  

1, 1, 3, 17

Nice! 

I wondered whether there are associated OEIS sequences for this and the next.

There are 120 matches for "1,1,3,17", but none if "dissection" is added.

Oh, but 2 matches if the initial 1 is removed:

A295634

Triangle read by rows: T(n,k) = number of nonequivalent dissections of an n-gon into k polygons by nonintersecting diagonals up to rotation and reflection.

1, 1, 1, 1, 1, 1, 1, 2, 3, 3, 1, 2, 6, 7, 4, 1, 3, 11, 24, 24, 12, 1, 3, 17, 51, 89, 74, 27, 1, 4, 26, 109, 265, 371, 259, 82, 1, 4, 36, 194, 660, 1291, 1478, 891, 228, 1, 5, 50, 345, 1477, 3891, 6249, 6044, 3176, 733, 1, 5, 65, 550, 3000, 10061, 21524, 29133, 24302, 11326, 2282

A179300

a(n) is the number of corner-rooted hexangulations of girth 6 with n inner faces.

Probably not relevant, though.

I hope I can get up to 8, so I can include a variation of the 1 size 1, 2 size 2, 3 size 3, 4 size 4 30-60-90 partridge solution that both John Conway and R K Guy managed to solve.  
https://erich-friedman.github.io/mathmagic/0802.html  

That's fascinating!

I wondered whether there are there solutions for all sizes beyond the partridge number,

and whether it could be feasible to count the *number* of ("inequivalent") solutions.

I guessed it's "yes" to both of these, and I think that's confirmed by this sequence I found :

A381976

a(n) is the number of distinct solutions to the Partridge Puzzle of size n.

1, 0, 0, 0, 0, 0, 0, 2332, 216285, 36349315

Interesting links there!

(In the Matt Parker video there's also the link to an interactive version, https://www.mscroggs.co.uk/squares/.)

There's also an "uncovered" / "at most" variant :

A369891

Minimum possible uncovered area when at most k squares of side k, k = 1..n, are packed into a square of side n*(n+1)/2 = A000217(n).

+10
3

0, 0, 4, 4, 16, 13, 8, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

- Maximilian

[Allan Wechsler]

The 30-60-90 triangle is not the only one that has "extra" tilings by similar copies, where by "extra" I mean "besides linear transformations of tilings that any triangle would have". Here, by "any triangle", we can use a triangle with incommensurate angles (like the 3-4-5 triangle); any such triangle is as good as any other.

Another "special" triangle is the equilateral triangle; then there's the 45-45-90 right triangle. These are the only examples of "order-3" and "order-4" special triangles, respectively, where by "order" I mean "the minimum k such that all the angles of the triangle are multiples of pi/k". 

Then there are two order-5 triangles, with angles 36-36-108 and 36-72-72. (These correspond to the three-part partitions of 5, 3+1+1 and 2+2+1.)

The 30-60-90 triangle is order-6 (3+2+1), but there is another order-6 example, with angles 30-30-120 (4+1+1). Note that we don't count the equilateral triangle (2+2+2) as being of order 6, because it has a simpler representation of order 3.

Not every partition of k into 3 parts yields a special triangle. For example, the triangle whose angles are 3,2, and 2 times pi/7, is almost certainly not special; my intuition is that it doesn't allow any dissections that a completely generic triangle wouldn't also allow. It feels like the key is whether any of the angles is a nontrivial sum of copies of the other two angles.

I have almost certainly fouled up this reasoning in various places. For example, I'm pretty sure the equilateral triangle is special; I have a sketch of a dissection of the equilateral triangle into 9 smaller equilateral triangles (sides 4/7, 2x3/7, 3x2/7, 3x1/7 the side length of the containing triangle) which I am almost sure isn't "generic". But I am not certain of this, and it might turn out that the equilateral triangle allows no "special" dissections after all.

There are lots of sequences hiding here; the most obvious is the number of special triangles of order n. Then, each special triangle gives rise to a separate sequence of the kind Ed has been quarrying for the order-6 3+2+1 case.

-- Allan

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[Ed Pegg]

Allan, 
One triangle I'm curious about comes from vertices 1,2,4 of a regular heptagon.  The angles are 1/7, 2/7, 4/7 Pi.  It seems like something could be done that's not possible with a general non-right triangle, but so far I haven't found anything special. I vaguely remember Andrzej ̇Zak proved something about dissections for triangles with rational fractions of Pi. 

Here are some of the sporadic triangles I know that have weird dissections. But I also need to enumerate general isosceles triangles and the 45-45-90. The 30-60-90 seems to be the weirdest, though, which surprises me.

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[D. S. McNeil]

I seem to find 18.  Are you missing this one, or do I have an angle wrong?

Doug

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[Allan Wechsler]

Doug,

I'm puzzled by your diagram. I don't see any 30-60-90 triangles in it. Is this the correct image? If so, how should it be interpreted?

-- Allan

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[D. S. McNeil]

Ah, sorry.  To make everything rational I rescaled y by sqrt(3), and didn't undo that for the picture.  The dissection positions are:

  (0,√3/5), (4/5,√3/5), (0,√3)   
  (1/5,0), (4/5,√3/5), (1,0)          
  (0,√3/5), (1/5,0), (4/5,√3/5)      

  (0,0), (1/5,0), (0,√3/5) 

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[Ed Pegg]

Yes, I missed that one!  It's even a Symmedian dissection.  It's the 4th triangle in this area.

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[Ed Pegg]

With help from Andrew Hudson and D. S. McNeil, the new count is 118 for five pieces. 
Sequence so far:   1, 1, 3, 18, 118. 

[Ed Pegg]

We've converged on a count of 121 dissections of a 30-60-90 triangle into five 30-60-90 triangles. 
D.S. McNeil sent me 3 more, at about the same time I was finding them. Andrew Hudson also joined in.   
Of these, six dissections are independent and not derivable from solution of the same size or smaller.
Sequence so far:   1, 1, 3, 18, 121.  

A good excuse to make some lovely graphics. The triangle connection graph isn't as pretty.

[Ed Pegg]

Number of ways to dissect a 30-60-90 triangle into n drafter triangles.
Sequence so far:   1, 1, 3, 18, 121, 844  
There are 23 independent dissection. These are my favorites.

[Ed Pegg]

A compressed form of the 844 six-piece solutions.  Assume the original 30-60-90 triangle has area 1.

The first triangle is a-9-23.  The 9 is the common denominator of the area. There are 23 different ways to arrange these six triangles into a 30-60-90.  The triangle areas are 1/9 x 5, 4/9.
The second triangle is a-12-12.  The 12 is the common denominator of the area. There are 12 different ways to arrange these six triangles into a 30-60-90.  
The third triangle is b-12-16.  The 12 is the common denominator of the area. There are 16 different ways to arrange these six triangles into a 30-60-90. 
a-256-1 through x-256-1 gives the largest group of shared area denominators.
a-1444-1 and b-1444-1  have the largest area denominator. 
b-400-1 and a-1024-1 each have largest triangles with the same area, which get the same color.   

[Ed Pegg]

A similar compressed form for 5-piece.