Goat

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?

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?

pi*R^2 = 2*pi*r^2

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

Over 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.

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

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

Which corner of the circular field will you use?

Originally posted by CauselessOne
Which corner of the circular field will you use?

Oh, whoops. I read it as rectangular. Serves me right for being up at 2:30 AM. Off to bed.

r=57.93642363

Originally posted by XanthosNZ
r=57.93642363

Somewhere around there IIRC, but how did yoy calculate it?

BTW, the only way I know is to draw the damned thing!

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?

35.35 units

nevermind never read the question right

GV

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!

http://mathworld.wolfram.com/Circle-CircleIntersection.html

i 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

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!

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 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.

there is no solution thats why it won't solve it, the determinant < 0
.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.