Smallest Circumcircle of Three Triangles with Equilateral Constraint

What is the minimum diameter of the circumcircle about an equilateral 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. Let the triangles have side 1. Applying the law of sines to ABC gives angle θ ≈ 14.5°. Triangle EDB now yields the target side x ≈ 0.809 and the circumcircle ≈ 0.934. Is EFG equilateral?

Here's another configuration: Applying the law of sines to ADC again gives θ ≈ 14.5° and x ≈ 0.809. Is there something better?

This problem came up in wood turning when arranging painter's pyramids to support small bowls. Painter's pyramids are tetrahedrons used to support objects while finishing. An equilateral triangle configuration of the supporting tips has the greatest stability. Isosceles and scalene configurations sacrifice stability for smaller bowl support.

• Yes, EFG is equilateral. Applying law of cosines to EBF and EAG gives $\overline{EF} = \overline{EG} \approx 0.809$ and by symmetry $\overline{FG} = \overline{EG}$ – Logophobic Jan 21 '16 at 17:41

By the sine law applied to triangle $ABC$ one gets $\sin\theta=1/4$, so that $\cos\theta=\sqrt{15}/4$. From triangle $BED$ one thus obtains: $$EF=2{1\over\sqrt3}\sin(30°+\theta)= {2\over\sqrt3}\left({1\over2}{\sqrt{15}\over4}+{\sqrt{3}\over2}{1\over4}\right) ={1+\sqrt{5}\over4}.$$ On the other hand one has $\angle BAC=30°-\theta$, so that $\angle EAG=150°-\theta$. By the cosine rule applied to $AEG$ one then gets: $$EG^2=\left({1\over\sqrt{3}}\right)^2+\left({1\over2\sqrt{3}}\right)^2 -2{1\over\sqrt{3}}{1\over2\sqrt{3}}\cos(150°-\theta) ={3+\sqrt{5}\over8}=\left({1+\sqrt{5}\over4}\right)^2.$$ So $EG=EF$ and triangle $EFG$ is equilateral.