You have a machine that produces random triangles of perimeter $1$ in the following way. On a stick of length $1$, the machine chooses two independent uniformly random points. If breaking the stick at those points would produce three fragments that could form a triangle (which has probability $\frac14$), then the machine breaks the stick at those points and forms the triangle. Otherwise, the machine erases the points and starts over.
You have a very large circular table (much larger than any single triangle). The machine produces a triangle, and you place it on the table. The machine places another triangle, and you place it on the table. And so on. The triangles may not overlap, and they must lie completely on the table.
Every time you place a new triangle, you may rearrange the triangles. You may not discard any triangle.
Your goal is fit as many triangles on the table as possible.
How should you arrange the triangles, to optimize the number of triangles that can fit on the table?
It seems that there should be some optimal general strategy, but it seems quite dificult to work out and prove. (The only source I found is about packing equilateral triangles in a circle.)
We could replace the circular table with, say, a square table, and the optimal strategy would be the same, because the table is so large that gaps near the edge of the table don't really matter.
One strategy would be to sort the triangles into groups, for example:
- Those that are almost equilateral: arrange them as close as possible to a triangular grid.
- Those that are almost right triangles: arrange them in pairs to form quasi-rectangles, and then arrange the quasi-rectangles of similar height in rows.
- Those that are long and narrow: arrange them in parallel, like pencils in a pencil case.
Then there will be three regions on the table, one with each group.
Context: This question was inspired by another question, which shows that the expected area of the triangles is $\frac{\pi}{105}$.
Edit: @Marcus Mitchell commented that this problem is NP-hard and is likely not to have a general solution. If we cannot get the best solution, I would still be interested in good solutions.
