Is there a triangle which makes dense set of angles? We start from a predefined triangle in a plane,
at first we have 3 angles of the triangle in set $A$.
in each level we add angles of triangles created by drawing medians of each triangle created from the latter level to $A$. 
What can we say about $A$? 
Is there a specific triangle which lead us to a dense set of angles between 0 and $\pi$ through this procedure?
If no ,does there exist another simple alternative to this procedure(medians) like another defined concurance lines in triangles, which achieves our goal?
Note: 
There are at least two possible interpretations of the allowed operations.


*

*3 x 2 x 1/2 - In each step, pick a triangle $T$ from previous step. Split $T$ into $2$ pieces by cutting $T$ along one of the $3$ medians. You are leave with $6$ smaller triangles whose area is one-half of that of $T$. 

*6 x 1/6 - In each step, pick a triangle $T$ from previous step. Split $T$
in $6$ pieces by cutting $T$ along all $3$ medians. You are leave with $6$ smaller triangles whose area is one-sixth of that of $T$.
Update:
Problem is solved under interpretation 3 x 2 x 1/2 by achille hui below.
For  6 x 1/6 interpretation the problem is reanounced in Mathoverflow here and answered there through a nice article.
 A: There are at least two possible interpretations of the allowed operations.


*

*3 x 2 x 1/2 - In each step, pick a triangle $T$ from previous step. Split $T$ into $2$ pieces by cutting $T$ along one of the $3$ medians. You are leave with $6$ smaller triangles whose area is one-half of that of $T$. 

*6 x 1/6 - In each step, pick a triangle $T$ from previous step. Split $T$
in $6$ pieces by cutting $T$ along all $3$ medians. You are leave with $6$ smaller triangles whose area is one-sixth of that of $T$.
I'm going to show under interpretation (3 x 2 x 1/2),  $A$ is dense in $(0,\pi)$.   

For any two points $P,Q$, let


*

*$\verb/mid/(P,Q)$ be the mid-point of $P$ and $Q$.

*$\verb/seg/(P,Q)$ be the line segment joining $P$ (inclusive) and $Q$ (exclusive)


Start from any predefined triangle $\triangle BCD$ and angle $\alpha \in (0,\pi)$,
Construct a sequence of triangles $\triangle B_pCD$ by setting 
$$B_p = \begin{cases} B, & p = 0\\
\verb/mid/(B_{p-1},D) & p > 0\end{cases}$$ 
Since $\lim\limits_{p\to\infty} \angle B_pCD \to 0$, we can pick a $p$ with $\angle B_pCD < \min(\alpha,\pi-\alpha)$. Call this $B_p$ as $E$.
Construct another sequence of triangles $\triangle E_qCD$ by setting
$$E_q = \begin{cases} E, & q = 0\\
\verb/mid/(E_{q-1},C), & q > 0\end{cases}$$
Since $\lim\limits_{q\to\infty} \angle CDE_q = 0$, we can pick a $q$ with $\angle CDE_q < \min(\alpha,\pi-\alpha)$. Call this $E_q$ as $F$.
In $\triangle FCD$, since both angles $\angle FCD, \angle CDF < \min(\alpha,\pi-\alpha)$, there is a point $G \in \verb/seg/(C,D)$ with $\angle FGD = \alpha$. Construct
yet another sequence of triangles $\triangle FC_rD_r$ by setting
$$(C_r, D_r) = \begin{cases}
(C_0,D_0), & r = 0\\
(C_{r-1},M_{r-1}), & r > 0, G \in \verb/seg/(C_{r-1},M_{r-1})\\
(M_{r-1},D_{r-1}), & r > 0, G \in \verb/seg/(M_{r-1},D_{r-1})
\end{cases}
\quad\text{ where }\quad 
M_{r-1} = \verb/mid/(C_{r-1},D_{r-1})
$$
In each step, $\verb/seg/(C_r,D_r)$ is a segment containing $G$. The whole
process is essentially a procedure which locates $G$ by bisection of line segment
containing it. 
All the triangles constructed so far are created by drawing medians of some existing triangles.
This implies $\angle FC_rD_r \in A$. As $r$ increases, both $C_r$ and $D_r$ converges to $G$. Furthermore,
$\angle FC_rD_r$ converges to $\alpha$ from below. 
This means $\alpha$ belongs to closure of $A$.
Since $\alpha$ can be any number from $(0,\pi)$, we can conclude
under interpretation (3 x 2 x 1/2), $A$ is dense over $(0,\pi)$.
I have no idea what happens under interpretation (6 x 1/6).
