Find probability that random triangle covers centre of circumscribed circle We are given the equilateral triangle A.
 On each edge of the triangle we pick a point: 


randomly (probability distribution is uniform)  
independently of others  


We construct new triangle B from randomized points.
Task is to find the chance of B containing the centre of circle circumscribed around triangle A.
  

I would appreciate hints and pointers. Thanks!

Edit
Suppose $P_0$ is chosen on the bottom line of triangle A. $P_O = (x_0,0), x_0 \in [0,1]$
Let's find geometrical place of points $P_1, P_2$.
If $P_1$ is on the upper left edge, that $P_1 = (x_1,\sqrt 3 x_1), x_1 \in [0, \frac 1 2]$.
Then $P_2$ is upper right edge, and $P_2 = (x_2, \sqrt 3 (1 - x_2)), x_2 \in [\frac 1 2, 1]$.  

Centre point is $C = (\frac 1 2, \frac 1 {2 \sqrt 3 })$.
Now, we shall find the constraints for points.
For fixed $x_0$, $P_1$ and $P_2$ must lay below the line between $P_0$ and $C$.
But I wonder, if there is a way to describe constraint better?
Since this most trivial way suffers from situation when line $CP_0$ and one of the edges intersects outside of range $x_m$.
Edit 2
I have listened to the useful hints about advantages of angle-based view on the problem, and believe that now I have the geometry part figured:
As earlier, We fixed $P_0$ on the bottom edge and on the step 1 we pick $P_1$ on the left edge.
Consider $\angle \alpha$ line between bottom edge and line $CP_0$. Then $\alpha \in [\frac \pi 6, \frac \pi 2]$.
Then prohibited sector for left edge has angle $\frac \pi 2 - \alpha$.
Using same logic on the step 2 we obtain prohibited sector for point $P_2$.
  

And now it looks like I have to get down to all that integration, so your suggestions are welcome!
Let $a_1, a_2, a_3$ be the vertices of $A$.
Choose a coordinate system such that
$$a_1 = (0,0),\quad a_2 = (1,0),\quad a_3 = \left(\frac12,\frac{\sqrt{3}}{2}\right)$$
The centroid of triangle $A$ will be located at 
$$c = \frac{a_1 + a_2 + a_3}{3} = \left(\frac12, \frac{1}{2\sqrt{3}}\right)$$

Let $b_1, b_2, b_3$ be the vertices of $B$. By definition of triangle $B$, there exists
three uniform random numbers $u_1, u_2, u_3$ taking values over $[0,1]$ such that
$$b_1 = (1-u_1) a_1 + u_1 a_2,\quad b_2 = (1-u_2) a_2 + u_2 a_3,\quad b_3 = (1-u_3) a_3 + u_3 a_1$$

Consider following 4 events concerning the relative position of $c$ with respect to these 6 vertices.
$$\begin{cases}
\mathcal{E}_0 &amp;\stackrel{def}{=} \text{Event}( c \in B = \triangle b_1 b_2 b_3 )\\
\mathcal{E}_1 &amp;\stackrel{def}{=} \text{Event}( c \in \triangle a_1 b_1 b_3 )\\
\mathcal{E}_2 &amp;\stackrel{def}{=} \text{Event}( c \in \triangle a_2 b_2 b_1 )\\
\mathcal{E}_3 &amp;\stackrel{def}{=} \text{Event}( c \in \triangle a_3 b_3 b_2 )\\
\end{cases}$$
With probability one, these $4$ events are mutually disjoint. More precisely,
$$\text{Prob}( \mathcal{E}_i \,\cap\, \mathcal{E}_j ) = 0\quad\text{ for } 0 \le i &lt; j \le 3$$
Notice $\text{Prob}( \bigcup_{i=0}^3 \mathcal{E}_i ) = 1$ and by symmetry, 
$$\text{Prob}( \mathcal{E}_1 ) = \text{Prob}( \mathcal{E}_2 ) = \text{Prob}( \mathcal{E}_3 )$$
The probability we want is given by
$$\text{Prob}( c \in B ) = \text{Prob}(\mathcal{E}_0) = 1 - 3\text{Prob}(\mathcal{E}_1)$$

Let $u = u_1$ and $v = 1-u_3$, we have 
$b_1 = (u,0), b_3 = \left(\frac{v}{2},\frac{v\sqrt{3}}{2}\right)$.
The line passing through $b_1, b_3$ is given by
$$\left|\begin{matrix}
x &amp; y &amp; 1\\
u &amp; 0 &amp; 1\\
\frac{v}{2} &amp; \frac{v\sqrt{3}}{2} &amp; 1\\
\end{matrix}\right| = 0$$
Plugging the coordinates of $c$ into LHS and treat the resulting expression as a function
of $u,v$. It is not hard to see the condition for $c \in \triangle a_1 b_1 b_3$ is simply

$$\left|\begin{matrix}
\frac12 &amp; \frac{1}{2\sqrt{3}} &amp; 1\\
u &amp; 0 &amp; 1\\
\frac{v}{2} &amp; \frac{v\sqrt{3}}{2} &amp; 1\\
\end{matrix}\right| \ge 0
\quad\iff\quad
3uv - u - v \ge 0
\quad\iff\quad
u \ge \frac12 \,\land\, v \ge \frac{u}{3u-1}
$$
This leads to
$$\text{Prob}(\mathcal{E}_1) = \int_{\frac12}^1 \left(1 - \frac{u}{3u-1}\right) du
= \frac{3 - \log 4}{9}$$
and hence
$$\text{Prob}( c \in B ) = 1 - 3\left(\frac{3 - \log 4}{9}\right) = \frac{\log 4}{3}
\approx 0.4620981203733$$

As a double check, I have performed a numerical simulation of this problem.
Out of $10^9$ copies of triangle $B$ generated, $462114483$ of them contains $c$.
This leads to an estimate of the probability $c \in B$ at 
$0.462114 \pm 0.000016\;(\leftarrow 1 \sigma)$.
This is consistent with what we have just derived.
I may be mistaken but I dont believe there is a solution to this problem. It depends entirely on how you construct the triangle. Possibly. Have you heard of Bertrand's Paradox? Its a paradox in probability theory that asks a similar question: Take an equilateral triangle inscribed in a circle, what is the probability that a random chord of the circle is longer than the side of the triangle?  Its simple enough to understand, and easy enough to answer, but it can be shown that you get drastically distinct answers depending on how you construct the chord.

Maybe Im wrong though. Maybe this is nothing like Bertrand's Paradox.