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Given a scalene triangle $ABC$ and an inscribed equilateral triangle whose vertices lie on different sides of $\triangle ABC$, what is the maximal ratio of the area of the equilateral triangle to that of the original triangle?

I would expect an answer as a ratio of polynomials in sines and cosines of the angles of $\triangle ABC$. I got such an expression via a clunky unsymmetrical method, but it but it was so messy that I gave up trying to simplify it. However, a more intelligent method might well yield a formula of reasonable length.

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    Are you aware of (http://math.stackexchange.com/q/186432) ? – 2017-01-18
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    @JeanMarie: Yes, but that is a different question. – 2017-01-19

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There isn't a single inscribed equilateral triangle: by choosing a point on a side, rotating another side by $60^\circ$ around that point and marking the intersection point with the third side, we always get an inscribed equilateral triangle. enter image description here

If you mean what is the maximum ratio $\frac{[ABC]}{[DEF]}$, the answer is given by a rescaled Napoleon triangle. We may notice that such ratio is also the ratio between the areas of the largest circumscribed equilateral triangle and the original triangle. On the ther hand, it is quite simple to describe the set of the circumscribed equilateral triangles: let we consider the Napoleon/Torricelli/Fermat configuration:

enter image description here

We have $O_B O_C\perp AV_A$ and so on. If we take a point $P$ on $\Gamma_A$ (the circumcircle of $BCV_A$), draw a line through $B$ till meeting $\Gamma_C$ at a new point $Q$, then draw a line through $A$ till meeting $\Gamma_B$ at a third point, the (grey) triangle defined this way is equilateral. It follows that we simply need the greatest $PQ$ segment among the segments built through the previous procedure. This segment has to be perpendicular to the radical axis of $\Gamma_A$ and $\Gamma_C$, so the Napoleon triangle of $ABC$ is the smallest inscribed equilateral triangle in $A'B'C'$. By rescaling, we get the solution. It is interesting to point out that the side length of $O_A O_B O_C$ is $\frac{1}{\sqrt{3}}AV_A$, and $AV_A=BV_B=CV_C$ is the length of the Steiner net of $ABC$, i.e. $FA+FB+FC$, with $F\in\Gamma_A\cap\Gamma_B\cap\Gamma_C$ being the Fermat point of $ABC$.

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    Sorry, I meant the former. I hope that the revised wording of the question is definitive now. – 2017-01-19
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    The counterexample you have given looks to be in the case of an isosceles triangle, the equilateral inscribed triangles being symmetrical with respect to the axis of symmetry of the triangle, thus with the same area. Can such a phenomena arise in the case of non isosceles triangle ? – 2017-01-19
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    The counterexample is general, the depicted triangle is not isosceles. And: are you sure you want the largest inscribed equilateral triangle? To find the smallest is way more interesting. – 2017-01-19
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    Have you had a look at my "solution" in terms of fixed point(s) of a certain geometrical function (that should be expressed, I think, as a certain homography, which could explain that there are always two solutions) ? – 2017-01-19
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    @JeanMarie: there are always an infinite number of solutions. Please see my updated answer. – 2017-01-19
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    @JohnBentin: with your actual wording, the interesting question is about the **minimal** ratio between the areas of the inscribed equilaterl triangle and the original triangle. The smallest inscribed equilateral triangle is given by a re-scaled Napoleon triangle. – 2017-01-19
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    Thank you for a great answer. – 2017-01-19
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    I have fully understood now where was my misunderstanding. A very small remark in your very neat presentation. It is about your first paragraph, more precisely when you say "by choosing a point on a side," it should be added that the chosen point must be on a side such that the opposite angle is $\leq \pi/3$, otherwise it is not always possible. – 2017-01-20
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This is intended as a different view, but doesn't constitute a complete answer.

We are going to work on the following figure, which has a classical aspect (first part of Miquel's theorem, see (https://en.wikipedia.org/wiki/Miquel%27s_theorem))

enter image description here

Instead of having a fixed triangle with given angles $\alpha, \beta, \gamma$, and equilateral triangles inscribed in it, let us completely reverse the problem into: being given a fixed equilateral triangle $UVW$ with unit side, consider a circumscribed triangle $ABC$ with these angles.

This makes sense as the issue is size and rotation invariant.

Being given these angles $\alpha, \beta, \gamma$, where can $A,B,C$ be situated?

Evidently, $A,B,C$ are restricted to be on circles (or more precisely their external arcs) from which one can see resp. line segments $VW$,$WU$ and $UV$ under angles $\alpha$, $\beta$ and $\gamma$ resp..

Taking an arbitrary point $A$ on the $\alpha$ circle; line AW intersects circle $\beta$ in $B$; then, line $BU$ intersects circle $\gamma$ in $C$; at last, points $C,V,A$ are aligned, as can be established by angle chasing in triangles $AWV$, $BUW$ and $CVU$. This is our manner to prove the existence of an infinite number of ways to inscribe an equilateral triangle in a given triangle.

Remark: We have also a global view of the areas ratio $R=[UVW]/[ABC]$ whose maximum is looked for (this ratio has not been modified by the transformation); it is equivalent to look for the

$$\text{Minimization of } \ \ 1/R=[ABC]/[UVW]=1+\frac12(h_A+h_B+h_C)$$

where $h_A,h_B,h_C$ are the altitudes issued from $A,B,C$ onto $VW$, $WU$, $UV$, resp.

As the figure of this text is very similar to the figure given by @Jack d'Aurizio (though obtained in a different spirit), I do not complete (at least for the moment) my proof because it would paraphrase his proof. I am indebted to Jack d'Aurizio because a first draft of this text was on bad tracks : I hadn't perceived that there is a - continuous - infinity of inscribed equilateral triangles in a given triangle.)