Recently I was given the following problem at work: Given a 5 pointed star, draw two straight lines through it so that there are 10 minimal triangles within the drawing. It took some work but I eventually came up with this answer seen here:
Then I started thinking about a simple case of a single equilateral triangle and adding two lines to it and seeing how many minimal triangles I could get from it and came up with 3. I then started wondering about what happens when I add more lines to it, say 3 straight lines and got 6 minimal triangles. I did some searching online and was wondering if there is a pattern/strategy to use for this or a proof stating what the maximum number of minimal triangles formed from partitioning an equilateral triangle (or more advanced shape like the star above) with n straight lines?
I just thought this problem was interesting and am looking on additional resources for understanding similar triangle partitioning cases. Sorry if this isn't the correct exchange site as this question might be more combinatorics related but I thought I'd try my luck here.
It seems the following.
We can obtain upper bounds for the maximum number $Shape(n)$ of minimal triangles as follows. Assume that the shape has $n_0$ straight sides. Extend all of them to straight lines. Remark, that this partition does not decrease a number of minimal triangles, because a triangle partitioned by a straight line always has one of the partition parts again a triangle. So, now we have a partition of the plane by $N=n+n_0$ straight lines. In is well known and easy to prove by induction that this partition has at most $(N^2+N+2)/2$ parts, among them are $2N$ unbounded. And I highly suspect that this bound can be improved if we count not all parts of the partition, but only triangles. Nevertheless, even this weak bound yields $$Shape(n)\le \frac {(n_0+n)^2-3(n_0+n)+2}2.$$
In particular, for triangle
$$\Delta(n)\le \frac {n^2+3n+2}2,$$
and for the pentagram $$\star(n)\le \frac {n^2+7n+12}2.$$
To be continued...