8. Equilateral triangle $A B C$ of side length 2 is drawn. Three squares containing the triangle, $A B D E, B C F G$, and $C A H I$, are drawn. What is the area of the smallest triangle that contains these squares?
Solution: The equilateral triangle with sides lying on lines $D E, F G$, and $H I$ has minimal area. Let $J, K$, and $L$ be the vertices of this triangle closest to $D, H$, and $F$, respectively. Clearly, $D E=2$. Denote by $M$ the intersection of $C I$ and $B D$. Using the $30^{\circ}-30^{\circ}-120^{\circ}$ triangle $B C M$, we get $B M=2 / \sqrt{3}$, and thus $M D=2-2 / \sqrt{3}$. By symmetry, $M J$ bisects angle $D M I$, from which we see that $\triangle J M D$ is a $30^{\circ}-60^{\circ}-90^{\circ}$. We then get $J D=2 \sqrt{3}-2$, making the side length of $J K L 4 \sqrt{3}-2$, and its area $13 \sqrt{3}-12$.