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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.

no shared edges

The triangles in the first solution could be subdivided the same way, so another criteria 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.

Schlegel

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

enter image description here

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  • $\begingroup$ 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.) $\endgroup$ – David K Jun 9 '16 at 17:06
  • $\begingroup$ 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? $\endgroup$ – David K Jun 9 '16 at 17:10
  • $\begingroup$ You can assume the initial triangle is scalene. "No two triangles share two vertices" is another way of saying it. $\endgroup$ – Ed Pegg Jun 9 '16 at 17:11
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    $\begingroup$ 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. $\endgroup$ – Joffan Jun 9 '16 at 17:34
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    $\begingroup$ I fixed the errors and added a new solution. $\endgroup$ – Ed Pegg Jun 10 '16 at 17:55
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If you have an equilateral triangle in the solution, you may just copy into it any of the other solutions thus having number of solutions as you desire.

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  • $\begingroup$ "another criteria is added: no subdivided internal triangles are allowed." $\endgroup$ – Ed Pegg Jul 23 '16 at 18:03

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