14. (YUG 5) ${ }^{\mathrm{IMO} 4} \mathrm{~A}$ soldier has to investigate whether there are mines in an area that has the form of an equilateral triangle. The radius of his detector is equal to one-half of an altitude of the triangle. The soldier starts from one vertex of the triangle. Determine the shortest path that the soldier has to traverse in order to check the whole region.
14. Suppose that the soldier starts at the vertex $A$ of the equilateral triangle $A B C$ of side length $a$. Let $\varphi, \psi$ be the arcs of circles with centers $B$ and $C$ and radii $a \sqrt{3} / 4$ respectively, that lie inside the triangle. In order to check the vertices $B, C$, he must visit some points $D \in \varphi$ and $E \in \psi$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-401.jpg?height=324&width=394&top_left_y=1496&top_left_x=887)  Thus his path cannot be shorter than the path $A D E$ (or $A E D$ ) itself. The length of the path $A D E$ is $A D + D E \geq A D + D C - a \sqrt{3} / 4$. Let $F$ be the reflection of $C$ across the line $M N$, where $M, N$ are the midpoints of $A B$ and $B C$. Then $D C \geq D F$ and hence $A D + D C \geq A D + D F \geq A F$. Consequently $A D + D E \geq A F - a \frac{\sqrt{3}}{4} = a\left(\frac{\sqrt{7}}{2} - \frac{\sqrt{3}}{4}\right)$, with equality if and only if $D$ is the midpoint of $\operatorname{arc} \varphi$ and $E = (C D) \cap \psi$.  Moreover, it is easy to verify that, in following the path $A D E$, the soldier will check the whole region. Therefore this path (as well as the one symmetric to it) is the shortest possible path that the soldier can take in order to check the entire field.