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 yet a shorter one,  once again triangle is ABC with A the starting point.

First follow this simple contruction

1) move along AC until you are a distance of 0.5h away from C and call this point D
2) move in a line perpindicular to AC from D in "upwards" direction (in other words moving closer to B) until you are a distance of 0.5h from B and call this point E

now my new path is AD then DE
let S3=Sqrt[3] then the distance is given by
(Sqrt[(S3-h)h]/S3)+(h/S3)-(1/2)

with h=1 this gives aproximately 0.571331 which is a great improvement upon my last path.  Still not sure if this is optimal but it sure seems like it :-D

Posted by Daniel on 2006-07-29 11:14:08