12. A sapper uses a signal detector to detect landmines in an equilateral triangular area with a side length of 2 kilometers. It is known that the effective working distance of the detector is $\frac{\sqrt{3}}{2}$ kilometers. Starting from one vertex of the triangle, what is the minimum distance the sapper needs to walk to complete the detection task? (Reasoning is required)
As shown in the figure, with $A, B, C$ as centers and $\frac{\sqrt{3}}{2}$ as the radius, we get three arcs $\overparen{M N}, \overparen{P Q}, \overparen{R S}$. Clearly, the soldier must reach any point $X$ on $\overparen{M N}$ from point $B$ to detect point $A$. Then, turning towards point $C$ and reaching point $Y$ on $\overparen{P Q}$, the soldier can detect point $C$. When $X, Y, C$ are collinear, the detection path $B X Y$ is the shortest.

Construct the altitude $A D$ of $\triangle A B C$, intersecting $\overparen{M N}$ at point $E$. When the ellipse with foci at $B, C$ is tangent to $\overparen{M N}$ at point $E$, the rest of the points on $\overparen{M N}$ are outside the ellipse. Thus, when $X$ coincides with $E$, the path $B E F$ is the shortest, with the shortest distance being $2|B E|-|C F|=\sqrt{7}-\frac{\sqrt{3}}{2}$.

Assume that segment $B X$ intersects $\overparen{R S}$ at point $Z$. We need to prove that the three circles with centers at $X, Y, Z$ and radius $\frac{\sqrt{3}}{2}$ can cover $\triangle A B C$, i.e., $\odot X$ and $\odot Z$ cover side $A B$, $\odot X$ and $\odot Y$ cover side $A C$, and $\odot Z$ and $\odot Y$ cover side $B C$.

When $X$ moves on $\overparen{M N}$, at least one of $Y, Z$ is inside or on the boundary of $\triangle B E C$. Assume $Y$ is inside or on the boundary. Let $C M$ intersect $\overparen{P Q}$ at point $K$, then $Y \in \overparen{K F}$. Therefore, we only need to prove that $\odot M$ and $\odot K$ cover side $A C$, and $\odot S$ and $\odot F$ cover side $B C$.

Clearly, $\odot M$ covers segment $A N$. When $\odot K$ covers segment $N C$, we need $|K N| \leqslant \frac{\sqrt{3}}{2}$. Similarly, $\odot S$ covers segment $B R$, and when $\odot F$ covers segment $R C$, we need $|F R| \leqslant \frac{\sqrt{3}}{2}$.

Since $|K N|<|F R|$, and $\tan \angle E C D=\frac{\sqrt{3}}{2} \Rightarrow \cos \angle E C D=\frac{2}{\sqrt{7}}$, we have
$$
\begin{array}{l}
|F R|^{2}=\frac{3}{4}+\left(2-\frac{\sqrt{3}}{2}\right)^{2}-2 \cdot \frac{\sqrt{3}}{2} \cdot\left(2-\frac{\sqrt{3}}{2}\right) \cdot \frac{2}{\sqrt{7}} \\
=\frac{11}{2}-2 \sqrt{3}-\frac{4 \sqrt{3}-3}{\sqrt{7}} \leqslant \frac{3}{4} \Leftrightarrow \frac{19}{4} \leqslant 2 \sqrt{3}+\frac{4 \sqrt{3}-3}{\sqrt{7}} .
\end{array}
$$

Notice that $\frac{19}{4} \leqslant 2 \sqrt{3}+\frac{4 \sqrt{3}-3}{3}=\frac{10 \sqrt{3}-3}{3}=\frac{10-\sqrt{3}}{\sqrt{3}} \Rightarrow 19 \sqrt{3} \leqslant 40-4 \sqrt{3}$
$$
\Rightarrow 23 \sqrt{3} \leqslant 40 \Rightarrow 529 \cdot 3=1587<1600 \text {, }
$$

Thus, $\frac{19}{4}<2 \sqrt{3}+\frac{4 \sqrt{3}-3}{3}<2 \sqrt{3}+\frac{4 \sqrt{3}-3}{\sqrt{7}} \Rightarrow|F R|<\frac{\sqrt{3}}{2}$, and the proposition is proved.

In summary, the soldier needs to travel at least $\sqrt{7}-\frac{\sqrt{3}}{2}$ kilometers to complete the detection task.