A soldier has to check for mines a region that has the form of an equilateral triangle. Let h be the length of an altitude of the triangle and h/2 the radius of activity of his mine detector. If the soldier starts at one of the vertices of the triangle, find the length (in terms of h) of the shortest path he could use to carry out his task.
here is how to construct my path of distance 1.5h, not sure if its the shortest but I figure its a good start. To be honest, right now I'm not even completely certain it covers the area, but I'm about 99% sure, just don't have time right now to complete my proof :-), so here it is.
Call the equilateral triangle ABC with being the point the soldier starts at. Follow this construction
1) draw altitude from A to BC
2) draw altitude from B to AC (could use C to AB if you prefer)
3) call the intersection of these two altitudes D
4) draw DC
5) move along DC a distance of (1/6)h and call this point E
6) draw EB
7) move along EB a distance of (2/3)h call this point F
now my path consists of moving along segments
AD,DE,EF for a total distance of (3/2)h or 1.5h
The reason I believe this covers the triangle is because after I move to D there only remains an area between D and B, and an area between D and C. I then move along DC a distance until I am 0.5h away from C thus the meter covers all the way to C and I need not travel any further. After that I do the same thing moving along a straight line to B until the edge of the area covered by the meter reaches B.
To be honest I doubt this is minimal but its a good start. I have a few other ideas that I will be posting when I have more time :-)
Posted by Daniel
on 2006-07-29 10:56:56