Solution:

Notice that the given expression is defined and continuous not only on $0 < x < 0.5$, but also on $0 \leq x \leq 0.5$. Let $f(x)$ be the function representing the area of the (possibly degenerate) hexagon for $x \in [0, 0.5]$. Since $f(x)$ is equal to the given expression over $(0, 0.5)$, we can conclude that $f(0)$ and $f(0.5)$ will also be equal to the expression when $x = 0$ and $x = 0.5$ respectively. (In other words, $f(x)$ is equal to the expression over $[0, 0.5]$.)

In each of the cases, we can compute easily that $f(0) = \frac{\sqrt{3}}{4}$ and $f(0.5) = 0$, so by plugging them in, we get $\frac{c}{2 \cdot 1} = 1$ and $\frac{2 - b/2 + c}{(3/2) \cdot (3/2)} = 0$, which gives $b = 8$ and $c = 2$.

Let $P = AA_1 \cap CC_2$, $Q = AA_2 \cap BB_1$, $R = BB_1 \cap CC_2$. These three points are the points on the hexagon farthest away from $A$. For reasons of symmetry, the area of the hexagon (call it $H$ for convenience) is:

$$[H] = [ABC] - 3[BPRQC]$$

Also, $[BPC] = [BQC]$ by symmetry, so:

$$\begin{gathered} {[BPRQC] = [BPC] + [BQC] - [BRC]} \\ {[BPRQC] = 2[BPC] - [BRC].} \end{gathered}$$

$$[H] = \frac{8x^2 - 8x + 2}{(2-x)(x+1)} \cdot \frac{\sqrt{3}}{4}$$

thus giving us the answer $(b, c) = (8, 2)$.