Consider two flat triangles, choose a vertex in each of them and glue them together by identifying these two distinguished vertices. Equip the resulting space $P$ with the quotient metric. You get a metric bouquet. As such and since flat triangles are non-negatively and non-positively curved the metric bouquet $P$ is non-positively curved (see e.g. Proposition 4.2.9 in [BBI]=[Burago/Burago/Ivanov: "A Course in Metric Geometry"], http://www.math.psu.edu/petrunin/papers/alexandrov/bbi.pdf. On the other hand, $P$ is a 2-dimensional polyhedral space and the sum of the angles around the connecting vertex is clearly not greater than $2\pi$. Hence Theorem 4.2.14 of [BBI] asserts that $P$ is of non-negative curvature.
So, is there any mistake in this argumentation or is $P$ both non-negatively and non-positively curved? If $P$ was both non-negatively and non-positively curved, however, triangles would have to have the same angles as their Euclidean comparison triangles. But this I cannot see. Instead I can see a lot of triangles which are thinner then their comparison traingles ...
Now, if I glue together in the same way as above 90 flat isoscele triangles (at one common vertex), then the angles at the gluing point add up to $30\pi$. Hence, according to the cited theorem the resulting polyhedral space is not non-negatively curved but, as a metric bouquet, it is non-positively curved. Right?
So does curvature here really depend on the number of triangles I glue together? Intuitively that doesn't make much sense to me ...
What is a nice visual example of a 2-dimensional polyhedral space which is non-negatively curved but NOT non-positively curved? Any link to nice illustrations known?
NB: I am aware that these questions are rather trivial (possibly even silly) and not on research level. Simply, I don't know a more suitable place to ask them. So I would be very grateful if someone could shed some light on this!
