Seven years ago, one of my many contributions to the March 2010 edition of Erich Friedman's Math Magic was a packing of eight circles of unit diameter and one equilateral triangle of unit side length into as small a circle as I could manage.
To minimise the bounding circle's radius $r$, I had to determine the largest equilateral triangle externally tangent to the three circles adjacent to it, then numerically adjust $r$ until said triangle had side lengths $1$. This is not a trivial problem, and back then I did trial and error in GeoGebra. ($2r=3.4133707107\dots$)
It is easy to see that the circles can be shrunk and the equilateral triangle enlarged about its centre while preserving tangencies, so the problem is equivalent to finding the largest equilateral triangle $\Delta^*$ where each edge is incident to one of $A,B,C$ where $A,B,C$ are the circle centres. The following construction doesn't count since one edge does not meet $\triangle ABC$:
Now, having reinvigorated my interest in triangles, I believe I have found a construction for $\Delta^*$ for arbitrary $A,B,C$, and am asking here for a more traditional proof of its optimality. My construction proceeds as follows:
- Construct the first isogonic centre/Fermat point of the triangle, $X_{13}$ in the Encyclopedia of Triangle Centers (ETC). That is, construct outward equilateral triangles $A'BC, AB'C, ABC'$ on the sides of $\triangle ABC$, then $X_{13}$ is the concurrence of $AA',BB',CC'$.
- $\Delta^*$ is (conjecturally) the antipedal triangle of $X_{13}$.$^\dagger$ That is, $\Delta^*$ is the triangle formed by perpendiculars at $A,B,C$ of $AX_{13},BX_{13},CX_{13}$ respectively; it is guaranteed to be equilateral because the lines meeting at $X_{13}$ are evenly spaced around it.
The centroid of the triangle thus constructed is $X_{5463}$ in the ETC, the reflection of $X_{13}$ in the centroid of $\triangle ABC$, and its area is $\frac{a^2+b^2+c^2}{2\sqrt3}+2\operatorname{area}(\triangle ABC)$ where $a,b,c$ are the side lengths. (Alternatively, the area is $2\left(1+\frac{\cot\omega}{\sqrt3}\right)\operatorname{area}(\triangle ABC)$ where $\omega$ is the Brocard angle.)
How can I prove that my construction actually produces $\Delta^*$, the largest circumscribing equilateral triangle?
$^\dagger$Without loss of generality, assume $AB$ is the largest side, and let $P,Q$ be the midpoints of $AC',BC'$ respectively. If $\overline{CP}$ or $\overline{CQ}$ is less than $\frac12\overline{AB}$, my construction will not produce an equilateral triangle actually incident to all of $A,B,C$. In that case, $\Delta^*$ has a side collinear with the triangle's shortest side $s$ and one vertex coincident with the original triangle's smaller angle incident to $s$. If $A,B,C$ are circle centres in a packing problem, this means that a larger equilateral triangle can be drawn by introducing a point contact.