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.

my second submission may turn out to be correct.  Using mathematica I tried finding an aproximate solution for my minimal path proof when h=1.  While doing my calculations I found that if P or Q is on one of the sides then it causes divison by zero and thus I treated them as border values for the minimization.  The zero aproximater in Mathematica spat out that the x cooridnate of P would be about 0.72 and the x cooridnate for Q would be about 0.67.  Using these two values I get an aproximate total distance of 1.03 wich is about double the length of my second submission.

Now on further analysis my second submission is merely one of the border values where P is on AC and Q is on BC.

Now when considering the other border case of P being on BC and Q being on BC we get a distance of aproximately
1.16 with h=1

thus it would seem that if my proof is correct then my second submission would indeed be the minimal path.   Guess that will teach me to second guess myself :-)

Edited on July 30, 2006, 1:00 pm

Posted by Daniel on 2006-07-29 18:15:43