# Triangle dissection, no shared edges

It's possible to divide a triangle into smaller triangles such that no edge lengths are shared. Alternately, no two internal triangles share two vertices. The top three are the known simplest solutions. The bottom three were found via doodling.

The triangles in the first solution could be subdivided the same way, so another criterion is added: no subdivided internal triangles are allowed.

What are some other solutions with 7 to 18 internal triangles?

EDIT. Of the prime 7 dissections, and the prime-8, no new solutions are found.

New solutions might be found by analyzing Schlegel diagrams of planar graphs, and picking three vertices of each face to be a triangle. If no two triangles share two faces, and if excised facial vertices can be moved between triangular vertices, then it's a potential solution. For example, this graph leads to the upper right solution, with triangles 234, 478, 456, 138, 125, 167.

One more solution, with nine of the triangles having area 1.

• What does it mean, "no edge lengths are shared"? If the outer triangle is equilateral with side $s$ (do we know it isn't?) then each one of the dissections in the question has at least two subtriangles that each have one side equal to $s$. Do we actually mean that no two triangles share two vertices? (That is, not just the length is the same but the edges are exactly the same collection of points.) Jun 9, 2016 at 17:06
• As for "the next few", is there some sequence in which we can enumerate these? Or are we just looking for a few more dissections? Jun 9, 2016 at 17:10
• I looked at the prime-5 dissections on vicher.cz/puzzle/triangles/triangles.htm and there was no dissection into 5 triangles that meets the no-shared-edge criterion. Jun 9, 2016 at 17:34
• I fixed the errors and added a new solution. Jun 10, 2016 at 17:55
• The singular of "criteria" is "criterion". Feb 9 at 15:29

## 1 Answer

Here is the $$10$$-triangle version of a construction which produces solutions with $$4,6,8,\ldots$$ triangles (including the first two shown in the original post):

We can apply a similar inflation to any nonconvex quadrilateral region in one of these dissections in which two nonadjacent sides of the quadrilateral are not a side of any exterior triangle, and fill it in with such "ladders" of any even size. As an example, here's the same construction applied to a quadrilateral inside the final 10-triangle example of the OP:

A different way to extend quadrilaterals by two triangles is exhibited by the top right example; we can iterate this procedure to produce constructions like the following ones:

Additionally, here's a 14-triangle construction not given by either of the above methods:

(This one is also produced by a generalizable "trick", though it's a little harder to describe - I noticed that the bottom-left construction in the OP contained at its base almost all of the structure of the top left construction, but had stuck some extra things on top and "pulled back" the original construction's left-hand triangle, and applied the same method again to the bottom left construction. This maneuver adds 6 triangles.)

• This is one of the best answers I've seen to a difficult math question. It's instantly visually obvious. I have to wonder .. did you solve this with fancy math or doodling? Feb 9 at 15:23
• @EdPegg I started off thinking about ways to encode incidence structures in a SAT solver, but pretty quickly gave up and went back to doodling - I was inspired by an observation in the comments from a similar question of mine that let you extend lots of tilings with a few more pieces, so I was primed to look for strategies like that. Once I saw a thing that I expected to work I'd sketch out one or two iterations of it until the pattern was obvious. Feb 9 at 17:01