TITLE: Is there a triangle which makes dense set of angles by drawing medians?
QUESTION [10 upvotes]: This problem is a restatement of this question, first announced in MathStackExchange.
We start with a triangle $T$ in the Euclidean plane and we define $A_n$ as the set of angles of the $6^n$ triangles obtained from the $n$-th barycentric subdivision of $T$. (Angles are geometric angles, i.e., the angles of $T$ and its children lie in $[0, \pi]$).
What we can say about set $A=\bigcup_{n=0}^{\infty}A_n$? 
Is there a specific predefined triangle for which $A$ is dense on $(0,\pi)$?
If no, does there exist another simple alternative to this iterative procedure, i.e., barycentric subdivision, using other concurrent lines in triangles, which achieves our goal?

REPLY [15 votes]: The answer to the second question is yes, for any non-flat triangle $T$, the set of angles $A$ is dense in $(0, \pi)$. This follows from a stronger result of Barany et al. [Theorem 1, 1]:


Theorem. Successive barycentric subdivisions of a non-flat triangle contain
    triangles which, to within a similarity, approximate arbitrarily closely any
    given triangle.


Here is a nice divulgative article  on the dynamics of iterated barycentric subdivisions. It also has a list of further readings on the topic. 
(The joy of barycentric subdivision, by Bill Casselman)- 

[1] I. Barany, A. Beardon and T. Carne, "Barycentric subdivision of triangles and semigroups of Möbius maps", 1996. MR1401715.