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Mine Detection (Posted on 2006-07-29) Difficulty: 3 of 5
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.

See The Solution Submitted by Bractals    
Rating: 3.0000 (3 votes)

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possible proof | Comment 6 of 7 |
I'm working on a proof for minimal path, but I've run into some road blocks so I'm going to post what I have thus far and maybe somebody can help me finish it.

I start once again with the triangle being ABC with starting point at A.

I've noted a couple of things that must be true about the minimal path.

1) the path must at some point come within 0.5h of point B, if it did not then point B would never be convered

2) the same is true for point C

using these 2 properties I've come up with this partial proof.

start with the triangle ABC.  Construct an arc of radius 0.5h centered at B going from side AB to side BC.  This arc sets the points that the path must reach in order to cover point B.  Now construct a similar arc centered at point C.

Now the path starting at A must come in contact with both arcs at some point.  Since ABC is equilateral then the order that the arcs are visited is not important so for simplicity I will say he visits the arc at C first then the arc at B.  Now the shortest distance between two points is a straight line so the minimal path would be a straight line from A to a point on arc C, call this point P.  Then from P to a point on arc B, call this point Q.

In order to calculate the length of the shortest path we must find P,Q in terms of h that minimizes the total length of AP and PQ.  I have been working on finding this minimum in terms of h, but that is where I hit the road block.  I have been trying to calculate this using Mathematica, the fact that Mathematica is having troubles worries me :-).  Unfortunately my knowledge of geometry is somewhat limited and so if there exists a geometrical shortcut to calculating this minimum it is beyond me.  Any assistance that can be offered would be apreciated :-D

  Posted by Daniel on 2006-07-29 17:37:56
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