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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.

  Submitted by Bractals    
Rating: 3.0000 (3 votes)
Solution: (Hide)

Let ABC be the triangle and A the starting vertex. Let B* and C* be circles of radius h/2 and centers B and C respectively. Clearly, to detect mines at B and C, his path must contain points on circles B* and C*. WOLOG let his path be the line segments AM and MN (where M and N are points on circles C* and B* respectively). Let A', B', and C' be the midpoints of sides BC, CA, and AB respectively. Let D be the intersection of CC' and A'B', E the intersection of DB with circle B*, and P the intersection of MN with A'B'.
  AD + DE + EB = AD + DB <= AP + PB <= AM + MP + PN + NB = AM + MN + NB


  AD + DE <= AM + MN.
Since M and N are arbitrary points on C* and B* respectively and D and E are points on C* and B* respectively, path AD + DE is our shortest path.

Disks of radius h/2 centered at points A, D, and E cover triangle ABC.
  |AD| + |DE| = |AD| + |DB| - |EB| = 2 |AD| - |EB|

              = 2 sqrt(|AC'|2 + |C'D|2) - |EB|

              = 2 sqrt(h2/3 + h2/4) - h/2

              = h (2 sqrt(21) - 3)/6

             ~= 1.02752523 h

Comments: ( You must be logged in to post comments.)
  Subject Author Date
interesting discoveryDaniel2006-07-29 18:15:43
possible proofDaniel2006-07-29 17:37:56
SolutionA possible solutionDej Mar2006-07-29 15:10:25
re(2): a shorter one :-)Daniel2006-07-29 12:44:29
Some Thoughtsre: a shorter one :-)Ady TZIDON2006-07-29 11:39:34
a shorter one :-)Daniel2006-07-29 11:14:08
get things started :-)Daniel2006-07-29 10:56:56
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