Maximizing the area of a region involving triangles I thought of a question while making up an exercise sheet for high school students, and posted it on MathStackExchange but did not receive an answer (the original post is here), so I thought perhaps MO would be a more suitable place to attract answers.
Let $T$ be an equilateral triangle of unit area, with vertices $A_1, A_2, A_3$. Place triangles $T_1, T_2, T_3$ each of unit area such that the centroid $G_i$ of $T_i$ is equal to $A_i$ for $i = 1,2,3$. What is the maximum possible value of the area of the region $T_1 \cap T_2 \cap T_3$, and what configuration and shape of $T_1, T_2, T_3$ achieves this maximum?
 A: I find an area of $.245463$ for the intersection when each $T_i$ has a vertex at distance $3.15835$ from the center of $T$. This seems to be maximal among the symmetric options.
The gray triangle in the diagram is $T$, the black triangles are the $T_i$ (with their specified far vertices outside the area of the diagram), and the blue nonagon is the desired intersection.

If the distance is $(1+2b)R$, where $R$ is the circumradius of $T$, we can write the area exactly as
$$4\left(2b - b^2 - \frac{3}{4b^2-1} - \frac{8b^2-4b-2}{12b^2 -1}\right)$$
which is easy to maximize numerically.
For a quicker approximate answer, we can assume that the area of intersection is a circle, tangent to the two long sides of $T_1$ at $(r,\theta)$ in polar coordinates. Then we can solve for $r,R,\theta$ satisfying
\begin{align}
3R^2 \sqrt{3} / 4 &= \,1, \text{ for circumradius of }T;\\
r(\sec \theta - 2)/3 &= R, \text{ for centroid of }T_1;\\
r^2(1+\sec \theta)^2 \cot \theta &= \,1, \text{ for area of }T_1
\end{align}
and the area and distance quoted above are $\pi r^2\simeq .225$ and $r \sec \theta \simeq 3.17$.
