G35 (15-4, Yugoslavia) A soldier wants to check for mines throughout a regular triangular area and its boundary. His detector has an effective radius equal to half the height of the equilateral triangle. The soldier starts his detection from one of the vertices of the triangle. What detection route should he follow to ensure the shortest path to cover the entire area?
Let the soldier start from vertex $A$ to explore the equilateral triangular area. The height of $\triangle A B C$ is $2 d$. With $B$ as the center and $d$ as the radius, draw a circle intersecting $A B, B C$ at $M, N$ respectively; with $C$ as the center and $d$ as the radius, draw a circle intersecting $A C, B C$ at $P, Q$ respectively. As shown in Figure 4-33.

By the problem's setup, the soldier must reach at least once on $\overparen{M N}$ and $\overparen{P Q}$. Without loss of generality, assume he first reaches point $D^{\prime}$ on $\overparen{M N}$ and then reaches point $E^{\prime}$ on $\overparen{P Q}$. Let $D$ be the midpoint of $\overparen{M N}$. Connect $A D, C D, C D^{\prime}$, intersecting $\overparen{P Q}$ at $E, E^{\prime \prime}$ respectively.

It is not difficult to see that the distance the soldier has traveled $\geqslant A D^{\prime}+D^{\prime} E^{\prime} \geqslant A D^{\prime}+D^{\prime} E^{\prime \prime} \geqslant A D + D E$.
The last inequality is due to $A D + D C \leqslant A D^{\prime} + D^{\prime} C$. This can be briefly explained as follows:
Draw a tangent line $l$ parallel to $A C$ through $D$, intersecting $A B, B C$ at $S, T$ respectively, then $\angle A D S = \angle C D T$. By symmetry, $D$ is the point on $l$ with the smallest sum of distances to $A$ and $C$. It is evident that $D^{\prime}$ and $C$ are on opposite sides of $l$. Let $C D^{\prime}$ intersect $l$ at $R$.
Then
\[
\begin{aligned}
A D + D C & \leqslant A R + R C \\
& \leqslant A D^{\prime} + D^{\prime} R + R C \\
& = A D^{\prime} + D^{\prime} C .
\end{aligned}
\]

Since the distance from $D$ to $A C$ is $d$, and the distance from any point on $A D$ to $A C$ is $\leqslant d$, it is easy to see that the vertical line from any point on the broken line $A D E$ to $A C$ ( $\leqslant d$ ) sweeps over the quadrilateral $A D E F$ (where $E F \perp A C$, and $F$ is the foot of the perpendicular, inside the sector $C$). As for the region within $\angle B A D$, outside $\overparen{M N}$, and the region below $D E$, outside the two sectors, it is easy to verify that for any point $G$ or $G^{\prime}$ in these regions, there must be a point $H$ or $H^{\prime}$ on $A D$ or $D E$ such that $G H, G^{\prime} H^{\prime} < d$. Therefore, $A D E$ is the shortest path, and there is another shortest path which is to first reach $\overparen{P Q}$ and then $\overparen{M N}$.
Noting that $\angle A B D = \angle C B D = 30^{\circ}$, by the cosine rule, the shortest distance is
\[
\begin{aligned}
& 2 \sqrt{\left(\frac{4}{\sqrt{3}} d\right)^{2} + d^{2} - \frac{8}{\sqrt{3}} d^{2} \cos 30^{\circ}} - d \\
= & \left(2 \sqrt{\frac{7}{3}} - 1\right) d \\
= & \frac{\sqrt{3}}{4} \left(2 \sqrt{\frac{7}{3}} - 1\right) A B \\
= & \left(\frac{\sqrt{7}}{2} - \frac{\sqrt{3}}{4}\right) A B .
\end{aligned}
\]