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Walking the Edge Part 2 (Posted on 2006-04-20) Difficulty: 3 of 5
Four farm-hands need to carefully walk the entire perimeter of a large square field to check for signs of infestation. They can each walk separately and any section of edge need only be checked by one person. The field is 200m on an edge and they all start at the same corner. Each person can either walk normally at 2m/sec or walk while checking at only 1m/sec. Any person may cut through part of the field at a normal walking pace. They must all finish at the opposite corner of the field.
What is the shortest time in which they can check the entire edge?

Consider the same problem with a circular field of radius 100m.
How long would this take?

No Solution Yet Submitted by Jer    
Rating: 5.0000 (1 votes)

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Solution this is probably the best (fastest) strategy | Comment 1 of 8

In the square case, two of the hands can start examining the nearer two edges of the field, while the other two walk to the end of the same edges at a normal pace.  When these latter two reach the respective corners, they can start to inspect the farther two edges. A little while later, when the first two hands finish their edges, they'll also be at the corner, and head toward the far corner; they'll finish in that far corner at the same time as the other two, as each has traveled one side at 2 m/sec and another at 1 m/sec for a total of 300 seconds, or 5 minutes.

The circular situation should also be symmetrical, to allow all four hands to finish at the same time (no wasted time).  Two will start in opposite directions, checking 1/4 of the fence each, then cutting across a chord of the circle that subtends 90 degrees.  The other two will start with the chord, aiming for the point where the first two will stop their portion of the inspection. With the radius being 100 m, the chord will be 100*sqrt(2) m and will take 50*sqrt(2) sec at a normal pace. The quarter circumference that each will inspect will have length 50*pi and will take 50*pi sec to check. So each hand (and thus the entire process) will take 50*(sqrt(2)+pi) sec, or about 227.79 sec (or 3 min, 47.79 sec).


  Posted by Charlie on 2006-04-20 10:42:40
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