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 Walking the Edge (Posted on 2006-04-18)
Four farm-hands need to walk the entire perimeter of a very large square field to check the fence posts. They can each walk separately but every section of edge needs to be walked by at least one person. The field is 200m on an edge and they all start at the same corner and can walk at 2m/sec. They do not need to finish at the same place and they may cut through part of the field if necessary (and they walk at the same pace if they do).

What is the shortest time in which they can check the entire edge?

How long would it take if there were only 3 farm-hands?

Consider the same problem with a circular field of radius 100m.
How long would it take to check the entire perimeter with 4 farm-hands?
With only 3?

 No Solution Yet Submitted by Jer No Rating

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 solution | Comment 2 of 4 |

At first I thought that it would make sense for two of the hands to walk the diagonal of the square and upon reaching the opposite corner head toward the other two, who would start by walking the immediately adjacent fence and turn the first corner arrived at, until they meet the diagonal walkers.  While that's the easiest to describe, it makes more sense for the diagonal walkers to head, not for the opposite corner, but rather where they calculate the orthogonal walkers will need to finish, and then head toward the opposite corner along the fence.

Of course you want all four walkers to travel the same distance, so no one is idle, wasting time. For simplicity consider the 200-meter side length to be one unit.  Let x be the distance along the non-adjacent fence that the orthogonal walker will eventually travel. Then the orthogonal walker will walk 1+x units, while the diagonal walker will walk sqrt(x^2+1)+1-x units.  Set these equal:

1+x = sqrt(x^2+1)+1-x
2x = sqrt(x^2+1)
4x^2 = x^2+1
3x^2 = 1
x^2 = 1/3
x = 1/sqrt(3)

Translating back to meters, the point along the non-adjacent fence that the diagonal walkers should aim for, and the orthogonal walkers will reach at the same time that the diagonal walkers will reach the opposite corner, is about 115.47 meters past the corner turned by the orthogonal walkers. That means that each walker travels 315.47 meters, taking 157.735 seconds, or 2 min, 37.735 sec.

With only 3 farm hands, one will have to cover both sides of the opposite corner.  Since all the hands will again need to walk the same distance, the two orthogonal walkers will walk the same distance, making the diagram again symmetrical, and we can again use x to represent how far each goes past the corner.

This time, the diagonal walker will walk sqrt(x^2+1)+2(1-x) units, so

1+x = sqrt(x^2+1)+2(1-x)
3x - 1 = sqrt(x^2+1)
9x^2 - 6x + 1 = x^2 + 1
8x^2 - 6x = 0

Since x>0,

8x - 6 = 0

x = 3/4

That's 150 meters, so each hand walks 350 meters, taking 175 seconds, or 2 min., 55 sec.

In the circular case, let the radius be 1 unit, and the immediate edge walkers walk x units along the circumference, to the place where the chord walkers had started walking the circumference.

Those immediate edge walkers walk x radians.  The chord walkers walk 2 sin(x/2) along the chord and pi - x along the circumference.  So we want

x = 2 sin(x/2) + pi - x
2x = 2 sin(x/2) + pi
x = sin(x/2) + pi/2

Using

DEFDBL A-Z
pi = ATN(1) * 4
x = 1
FOR i = 1 TO 50
x = SIN(x / 2) + pi / 2
PRINT x
NEXT

x converges rapidly:

2.0502218653991
2.425568096050638
2.507391524103875
2.520939818998286
2.523030269532486
2.523348929213154
2.523397412985049
2.523404787625958
2.523405909299214
2.523406079903118
2.52340610585155
2.523406109798242
2.523406110398524
2.523406110489825
2.523406110503712
2.523406110505825
2.523406110506146
2.523406110506194
2.523406110506202
2.523406110506203
2.523406110506203
2.523406110506203
2.523406110506203
2.523406110506203
2.523406110506203

As the radius is really 100 meters, the final figure converts to 252.3406110506203 meters away from the starting point, traversed in about 126.17 seconds or 2 min, 6.17 sec.

With only 3 farm hands, the chord walker will have to cover both sides of the gap on the opposite side, so

x = 2 sin(x/2) + 2(pi - x)
3x = 2 sin(x/2) + 2 pi
x = 2 (sin(x/2) + pi) / 3

Then

DEFDBL A-Z
pi = ATN(1) * 4
x = 1
FOR i = 1 TO 50
x = 2 * (SIN(x / 2) + pi) / 3
PRINT x
NEXT

has x converge:

2.414012128129331
2.717431697323867
2.746125170476726
2.74807130021315
2.748198428319792
2.74820671124024
2.748207250815004
2.748207285964165
2.748207288253862
2.748207288403018
2.748207288412734
2.748207288413367
2.748207288413408
2.748207288413411
2.748207288413411
2.748207288413411
2.748207288413411
2.748207288413411
2.748207288413411
2.748207288413411

or 274.8207288413411 meters, traversed in about 137.41 seconds or 2 min, 17.41 sec.

 Posted by Charlie on 2006-04-18 10:16:08

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