Let $ABC$ the triangle, with side $a$. Let $h=a\frac{\sqrt{3}}{4}$ the half altitude The soldier starts moving from $A$. Let $D,E,F$ the midpoints of $AB,BC,CA$ respectively. If he moves on the segment $DE$ (after $A\to D$), then he can reach the most points of the area but he will lose the vertice $C$. The problem is of course at the vertices. We construct the triangle $DEF$ and let $K,L$ the midpoints of $DE,EF$. From $K$, he can reach the point $B$ since $BK=h$. Also, from $L$ he can reach the point $C$. If he could start from $K$, then the better way would be $K\to L\to M$ where $M$ is the midpoint of $DF$. Actually, each of these points covers a large percentage of the triangle, for example the circle $(K,h)$ includes the triangle $BDE,DEF$ and much more area. By the way, if could "jump" from $K$ to $L$ and then to $M$, the area would be covered perfectly. I think that, since he must start from $A$, the most convinience way is to get first to the point $K$ and then go to the point $L$ The vertical distance from $A$ to $K$ is $2h-\frac{h}{2}=\frac{3h}{2}$ The horizontal distance is $\frac{a}{8}$ $AK^2 = (\frac{3h}{2})^2+(\frac{a}{8})^2 =$ $(\frac{3a\sqrt{3}}{8})^2 + (\frac{a}{8})^2 =$ $= a^2 \frac{9 \cdot 3 +1}{8^2}=$ $= a^2 \frac{7}{16}\Rightarrow$ $AK = a\frac{\sqrt{7}}{4}$ and $KL = \frac{a}{4}$ $AK+KL = a\frac{\sqrt{7}+1}{4}$ I don't know if this is the minimum or if this is the better way to face the problem. At least I tried Maybe it should be easier using calculus?

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This solution is wrong! Check http://www.mat.itu.edu.tr/gungor/IMO/http://www.kalva.demon.co.uk/imo/isoln/isoln734.html The answer is $ a\frac {2\sqrt {7} - \sqrt {3}}{4}$ which is less then $ a\frac {\sqrt {7} + 1}{4}$

Geometric Problems on MAXIMA and MINIMA (by Author "Titu Andreescu") "Let h be the length of the altitude of the given equilateral triangle ABC. Assume that the soldier's path starts at the point A. Consider the circles k1 and k2 with centers B and C, respectively, both with radius h/2. In order to check the points B and C, the soldier's path must have common points with both k1 and k2. Assume that the total length of the path is t and it has a common point M with k2 first and then a common point N with k1. Denote by D the common point of k2 and the altitude through C in triangle ABC and by L the line through D parallel to AB. Adding the constant h/2 to t and using the triangle inequality, one gets t+(h/2)≥AM+MN+NB=AM+MP+PN+NB≥AP+PB, where P is the intersection point MN and L. On the other hand, Heron's problem shows that AP+PB≥AD+DB, where equality occurs precisely when P=D. This implies t+(h/2)≥AD+DB, i.e. , t≥AD+DE, where E is the point of intersection of DB and k1." ⇒shortest path length=a×{2×(√7/4) - (√3/4)}, where a is the length of equilateral triangle ABC.

Let d be the distance from any of the vetices to the mid point of the triangle's altitude. Let h be the altitude of the equilateral triangle. In order to do his task, the soldier must travel the distance d plus the diference (d - h/2). Hence: displacement = d + d - h/2 ~ 0.8899 units of length, as 77ant suggested

Let h be the altitude of triangle ABC and t be the radius of the detector, $r=\frac{h}{2}$ Let the soldier start from A, so the critical points to detect are B and C --> the paths to be taken are $A\to D$ and $D\to E$, where D is a point with the distance $r=\frac{h}{2}$ from B and E is a point with the distance $r=\frac{h}{2}$ from C. The length of the paths is $s=\|A\to D\|+\|D\to E\|$ Let's imagine the soldier is already on D (although we don't know where). To take the shortest path from D to E, the soldier must walk in the direction to C --> E is on the line DC. So $\|D\to E\|=\|D\to C\| - \|E\to C\|=\|D\to C\| - \frac{h}{2}$ --> $s=\|A\to D\|+\|D\to C\|-r$ s is minimal, if and only if $\|A\to D\|+\|D\to C\|$ is minimal. $\|A\to D\|+\|D\to C\|$ is minimal, if and only if $\|A\to D\|=\|D\to C\|$. --> D must be on the line from B to F, where F is the midpoint of line AC. Because BF is an altitude of triangle ABC, so D must be the midpoint of line BF. We can calculate s as a function of h, but it's not the answer from the question. The correct answer is: The soldier must walk from A to D and then to E. D is the midpoint of line BF (the altitude of the triangle ABC from B). E is on line DC with $EC=\frac{h}{2}$