Three in four

The area of the equilateral triangle is directly proportional to the square of its side length $a$ . The triangle has its longest side length when one of its vertex coincident with a corner of the square. Let the corner be the top-left corner as shown (black triangle). Let the bottom triangle vertex be $x$ from the bottom corner of the square and the triangle left side forms with the base of the square be $\theta$ .

Then $a_{\text{max}} = \csc \theta$ and $x= \cot \theta$ , and

$\begin{aligned} a_{\text{max}} \cos (120^\circ - \theta) & = 1 - x \\ \csc \theta \left(-\frac 12 \cos \theta + \frac {\sqrt 3}2\sin \theta \right) & = 1 - \cot \theta \\ - \cot \theta + \sqrt 3 & = 2 - 2 \cot \theta \\ \cot \theta & = 2 - \sqrt 3 \end{aligned}$

$\begin{aligned} \implies a_{\text{max}} & = \csc \theta = \sqrt{1^2 + (2-\sqrt 3)^2} = \sqrt{8-4\sqrt 3} \end{aligned}$

We note that $a_{\text{min}} = 1$ , because any smaller, the two sides of the square would not hold the triangle (see the blue triangle in the figure). Since $A \propto a$ , we have:

$\begin{aligned} \frac {A_{\text{max}}}{A_{\text{min}}} & = \frac {a^2_{\text{max}}}{a^2_{\text{min}}} = 8-4\sqrt 3 \end{aligned}$ .

Therefore, $a+b+c = 8+4+3 = \boxed {15}$ .

The area of the triangle reaches the maximum when one of it's vertices is at one corner of the square. The maximum area is $\dfrac{\sqrt 3}{4}\times (8-4\sqrt 3)$ . The area attains it's minimum when one of it's vertices lie at the mid point of a side of the square. The minimum area is $\dfrac{\sqrt 3}{4}$ . Hence the required ratio is $8-4\sqrt 3$ , so that $a=8,b=4,c=3$ and $a+b+c=\boxed {15}$