Cutting a triangle into similar parts

[Ed Pegg]

I'm working on sequences A399765  and  A399770 

A399765   Ways to cut a non-right triangle into n similar parts 
1, 0, 0, 4, 0, 36, 64  

A399770   Ways to cut a right triangle in n similar parts  
1, 1, 2, 9, 48, 240    

... but I've also gotten this wrong numerous times.  

[Ed Pegg]

A399765   Ways to cut a non-right triangle into n similar parts 

1, 0, 0, 4, 0, 36, 64, 468, 1672

A399770   Ways to cut a right triangle in n similar parts  

1, 1, 2, 9, 48, 260, 1366, 9601, 69273  

Took me a while to convince myself I was collecting the numbers correctly. Would anyone like to look over my mathematica code?

260 right triangles 468 nonright triangles 

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[Ed Pegg]

I'm an idiot.  Brian Trease pointed out a right triangle has the 8th dissection here... a method missed by my code.  So there are 10 ways to divide a right triangle into 10 triangles similar to the original triangle.  I knew this would be tricky.

[Fred Lunnon]

EP:  << Brian Trease pointed out a right triangle has the 8th dissection here... a method missed by my code.   >> 

  Yet your code found both #4 & # 6 ... ?! 

WFL

_

On Sat, Sep 12, 2026 at 9:38 PM Ed Pegg <edp...@gmail.com> wrote:

I'm an idiot.  Brian Trease pointed out a right triangle has the 8th dissection here... a method missed by my code.  So there are 10 ways to divide a right triangle into 10 triangles similar to the original triangle.  I knew this would be tricky.

... 

[L. Edson Jeffery]

Ed, you are not an idiot. Your work is very beautiful and interesting; don't give up. Dissection problems like this one can be brutally difficult to enumerate. Maybe it is in some way equivalent to enumerating a certain class of integer partitions for which one might be able find a nice generating function. As it is now, if you otherwise happen to find a recurrence relation that matches the first n terms, then its correctness could also be very difficult to prove.

Ed Jeffery 

[Ed Pegg]

On Sat, Sep 12, 2026 at 6:13 PM L. Edson Jeffery <lejef...@gmail.com> wrote:

Ed, you are not an idiot. Your work is very beautiful and interesting; don't give up. Dissection problems like this one can be brutally difficult to enumerate.   

Thanks!  Yeah, I know ... but I didn't expect *this* problem to clobber me.  

Brian Trease has right triangles at 1, 1, 2, 10, 60, 363, 2346, 16581 
I've only gotten to 58 triangles for 5 parts.  Embarrassing.  But also very 
good that someone else took a look.

[Ed Pegg]

 There are sixty ways to cut a right triangle into five self-similar parts. With help from Brian Trease.

I think I could have solved this faster back when I was ten, just using pencil and paper.

[Elijah Beregovsky]

Dear Ed!
Is there a reason the variations of triangle cuts are sorted in the way they are on your picture? I’m asking, because there are patterns in the cuts that could be useful for calculations, and would be more apparent with a reorder of the results. I suppose you’re using some of them in your code, and maybe something else?

If I had to calculate it by hand I’d’ve used a lot of recursion, beacuse when cutting a triangle off a similar triangle there are only three options:

1. The cut passes through the right-angle vertex, so the triangle is split into two similar ones
2. The cut is parallel to one of the sides of the big triangle, so what’s left is a trapezium
3. The cut is not parallel to any of the sides, in which case what’s left is a cyclic quadrilateral. 

The first option is the easiest to tackle: when considering cuts with n triangles you instantly get twice the number of the cuts with n-1 triangles. Moreover, I’m pretty sure they’re definitely not isomorphic to each other, so you don’t even have to compare them. Then I’d probably look at all ways of constructing a trapezium out of n-1 similar right angle triangles, and remove any ones that have a cut through the right-angle vertex that splits off a single triangle. Might need to check for isomorphisms within the group, but there shouldn’t be too many. Then I think I’d look at the ways to construct a pentagon with three right angles (two next to each other and one across from them) out of n-2 triangles and remove the ones where the lone right angle is cut off with a line parallel to the opposite side, as they’d already be counted at the previous step. I thiiiink step three only becomes relevant at n>=6, cuz I don’t see a single cut like that in your set for n=5. Steps 2 and 3 also constrain the angles of trapezia and pentagons under consideration because the original triangle is not automatically similar to the small ones you used for construction, but I dunno how hard/annoying it would be to account for that. 

Best wishes,

Elijah

[Allan Wechsler]

Are we sure there aren't any "wild" dissections where none of the cuts go all the way through the triangle? I fear that Elijah's taxonomy might throw out some babies with the bathwater.

-- Allan

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