Smallest Circumcircle of Three Triangles What is the minimum diameter of the circumcircle about the triangle formed by the center points of three congruent equilateral triangles that do not overlap?
The diagram is the best solution I've found so far.  If the triangles have a side of length 1, the circle has diameter 0.853.

 A: This is minimal. The tangent line is parallel to the edge of the triangle.
Diameter of circle is $\frac{10\sqrt{3}-6\sqrt{6}}{3}\approx0.8745232$

A: I took the right triangle (green) and rotated it around its center. I then moved it around so that it would not overlap with any of the other two triangles and found a position where its center is within the circumcircle of your solution. I think this means this layout of triangles has a smaller circumcircle solution. I'm not sure if this is due to the imperfections of my drawing or if it is indeed a smaller circumcircle.

Of course, this is not a conclusive answer because it's just another guess. But maybe this could help inspire to think in a different direction. Maybe there's something special about it when the edges of the triangles are touching.
A: Edit: This is not minimal.
Diameter of circle is $\frac{\sqrt{7}}{3}\simeq0.8819171$

A: As others have suggested, you slide the rightmost triangle down while maintaining contact between the triangle edges.
If you slide it until the three centres form a right angled triangle, the required diameter $D$ is the hypotenuse of this triangle. 
The distance from each triangle centre to its base is $\frac 13\times\frac{\sqrt{3}}{2}$ and the distance between the two lower triangle centres is, by application of similar triangles, $\frac 23$.
Then due to Pythagoras, 
$$D^2=(2\times\frac{\sqrt{3}}{6})^2+(\frac 23)^2=\frac 79$$
So in this case $D=\frac{\sqrt{7}}{3}$
While this does not prove that this value is minimum, it is less than the value which is found from the OP's original diagram, namely $\frac{9+\sqrt{3}}{12}$
