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 Mine Detection (Posted on 2006-07-29)
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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 A possible solution | Comment 5 of 7 |

The length I calculated is approximately 1.02752*h.

I labeled each vertex of the triangle A, B and C. I then disected angle B with a line segment of length h/2 and labeled the new end point D.

The length of side AB is equal to 2/3*SQRT(3)*h. Using the law of cosines, AD is approximately 0.76376*h. As ABC is an equilateral triangle, CD is equal to AD.

I then label point E as the point h/2 distant from C on the line segment CD.  The length of DE is then approximately (0.76376 - 0.5)*h, i.e., 0.26376*h.

The path of the mine detector would then be from A to D to E, for a total distance of approximately 1.02752*h.

By using the law of cosines and iteration, calculating the length of a line segment from B to any point h/2 distant from A and adding then the length of the line segment from that point to the nearest point to a point h/2 distant from C, it can be seen that the minimum length is derived from where the angle of the line segment of BD is 30 degrees (1/6 radians) from the line segment AB and BC.  Thus, the value given above is the minimal length path.

Edited on July 30, 2006, 12:18 am
 Posted by Dej Mar on 2006-07-29 15:10:25

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