- 27 Jun '06 22:04A farmer has a field 100 (metres, feet, yards, it does''t matter) in diameter. He also owns a goat and a rope. He ties the rope to a fence post at the edge of the field, and ties the goat to the other end of the rope, such that the goat can eat exactly half of the grass.

How long is the rope? - 27 Jun '06 22:27 / 1 edit

pi*R^2 = 2*pi*r^2*Originally posted by CauselessOne***A farmer has a field 100 (metres, feet, yards, it does''t matter) in diameter. He also owns a goat and a rope. He ties the rope to a fence post at the edge of the field, and ties the goat to the other end of the rope, such that the goat can eat exactly half of the grass.**

How long is the rope?

R = 100

solve for r.

Oops. I assumed the goat's post was in the center. Never mind.

Here's a start then:

http://mathworld.wolfram.com/Circle-CircleIntersection.html - 28 Jun '06 06:04Over 50, but less than 100. After lots of writing, figuring, and scribbling, the only way I can see to solve this is to focus on the fact that units don't matter. Therefore my other answer is that the rope should be 10 smomofos long, and it is up to someone else to determine the conversion factor from whatever units the diameter of the circle is measured in and the length of 1 smomofo.
- 28 Jun '06 06:20

Which corner of the circular field will you use?*Originally posted by ark13***The field is 100*100=10,000. The goat must eat 5,000 of it. Tied to a post at a corner, the goat's possible roaming area inside the field will be a quarter circle. Therefore,**

5000=.25*pi*r^2

20,000=pi*r^2

6366.2=r^2

79.79=r - 29 Jun '06 05:51 / 2 edits

35.35 units*Originally posted by CauselessOne***A farmer has a field 100 (metres, feet, yards, it does''t matter) in diameter. He also owns a goat and a rope. He ties the rope to a fence post at the edge of the field, and ties the goat to the other end of the rope, such that the goat can eat exactly half of the grass.**

How long is the rope?

nevermind never read the question right

GV - 29 Jun '06 06:37i have 3 answers to this problem....

1) the ropp is as long as it has to be and is not to long and not to short but lets the goat eat half of the filed

2) 50 meters

30 in order to work out the answer you need to draw the feild and then slect a piont on on of the edges where the rop has been tyed. then positsion a half squrle that has the area of 50% of the feild. then mesure the distance from the goat to the piont where the rope is tyed to the fence - 29 Jun '06 09:23

Using the formula from the page AThousandYoung posted. Substitute d=50, R=50, A=1250*pi and solve for r. Of course actually solving for r is pretty much impossible (Maple won't do it even) so the easiest way is a Newton's Method.*Originally posted by CauselessOne***Somewhere around there IIRC, but how did yoy calculate it?**

BTW, the only way I know is to draw the damned thing! - 29 Jun '06 11:34 / 1 edit

there is no solution thats why it won't solve it, the determinant < 0*Originally posted by XanthosNZ***Using the formula from the page AThousandYoung posted. Substitute d=50, R=50, A=1250*pi and solve for r. Of course actually solving for r is pretty much impossible (Maple won't do it even) so the easiest way is a Newton's Method.**

.5 sqrt (r^2(100-r) (100+r) = 1250pi

r^2 (100^2-r^2) = (2500pi)^2 ~ 625*10^5

0 = r^4 - 10^4 * r^2 + 625 * 10^5

r^2 = (+100^4 +/- SQRT(10^8 - 4 * 625* 10^5)) /2

10^8 - 4* 625* 10^5 =10^8 - 2.5 * 10^8 < 0

i think we violated some limiting condition of the problem at mathworld.