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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.

Best So Far

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    $\begingroup$ I think you can get smaller by rotating the third triangle about the point at which it touches the other two, which brings the centre of the third triangle inside the circle you have drawn. $\endgroup$ – Mark Bennet Jan 19 '16 at 19:53
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    $\begingroup$ And then try sliding the third triangle up or down the side of the lower triangle until the radius is minimized. $\endgroup$ – David K Jan 19 '16 at 20:14
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    $\begingroup$ How did you get $0.853$? I may be mistaken, but I get $$\frac {9+\sqrt{3}}{12}\simeq0.894337567...$$ $\endgroup$ – David Quinn Jan 19 '16 at 20:38
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This is minimal. The tangent line is parallel to the edge of the triangle.

Diameter of circle is $\frac{10\sqrt{3}-6\sqrt{6}}{3}\approx0.8745232$

Minimal solution

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    $\begingroup$ Interesting! Can you show that it's minimal? $\endgroup$ – Brian Tung Jan 20 '16 at 19:43
  • $\begingroup$ @BrianTung Sliding the third triangle along the edge of the other moves its center along the tangent. Thus the center of the third triangle cannot be moved inside this circle, which would be necessary to define a smaller circle with the three centers. $\endgroup$ – Logophobic Jan 20 '16 at 19:48
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    $\begingroup$ This seems to be the "local" minimum, given this particular configuration of triangles. How do we know that we can't do better "globally" by not requiring two triangles to share an edge? $\endgroup$ – Blue Jan 20 '16 at 20:04
  • $\begingroup$ The offset up of the right triangle appears to be sqrt(2)-1 = 0.41421.... $\endgroup$ – RogerTaft Jan 22 '16 at 16:48
  • $\begingroup$ @RogerTaft That wasn't part of my original calculations, but I confirm that it is correct. $\endgroup$ – Logophobic Jan 23 '16 at 2:15
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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.

better (?) solution

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.

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  • $\begingroup$ You can improve this slightly by sliding the triangle down. If you look at the chord on the circle through the new and old centers of the triangle that was rotated, then the smallest circumcircle (obtained through this procedure) will be obtained by sliding the triangle to have it's center at the center of the this chord. $\endgroup$ – David Kleiman Jan 19 '16 at 20:35
  • $\begingroup$ Mea culpa. DK's value of my original configuration is the correct one. $\endgroup$ – RogerTaft Jan 20 '16 at 17:04
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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 $\frac{9+\sqrt{3}}{12}$

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Edit: This is not minimal.

Diameter of circle is $\frac{\sqrt{7}}{3}\simeq0.8819171$

Triangles

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