6. Let $A B C$ be an equilateral triangle of side length 1 . For a real number $0<x<0.5$, let $A_{1}$ and $A_{2}$ be the points on side $B C$ such that $A_{1} B=A_{2} C=x$, and let $T_{A}=\triangle A A_{1} A_{2}$. Construct triangles $T_{B}=\triangle B B_{1} B_{2}$ and $T_{C}=\triangle C C_{1} C_{2}$ similarly.
There exist positive rational numbers $b, c$ such that the region of points inside all three triangles $T_{A}, T_{B}, T_{C}$ is a hexagon with area
$$
\frac{8 x^{2}-b x+c}{(2-x)(x+1)} \cdot \frac{\sqrt{3}}{4} \text {. }
$$

Find $(b, c)$.
Answer:
$(8,2)$
Solution 1:
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 $c=2$.
Solution 2:
Let $P=A A_{1} \cap C C_{2}, Q=A A_{2} \cap B B_{1}, R=B B_{1} \cap C C_{2}$. 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:
$$
[H]=[A B C]-3[B P R Q C] .
$$

Also, $[B P C]=[B Q C]$ by symmetry, so:
$$
\begin{array}{c}
{[B P R Q C]=[B P C]+[B Q C]-[B R C]} \\
{[B P R Q C]=2[B P C]-[B R C] .}
\end{array}
$$

From this, one can see that all we need to do is calculate the $A$-level of the points $P$ and $R$ in barycentric coordinates. Ultimately, the $A$-level of $P$ is $\frac{x}{x+1}$, and the $A$-level of $R$ is $\frac{x}{2-x}$. From this, straightforward calculation shows that:
$$
[H]=\frac{8 x^{2}-8 x+2}{(2-x)(x+1)} \cdot \frac{\sqrt{3}}{4},
$$
thus giving us the answer $(b, c)=(8,2)$.