 #### Posers and Puzzles

1. 27 Oct '16 14:03
A farmer owns a circular field which has a radius of 20m.

He wishes to tether a goat to a post on the edge of the field (ie, the circumference of the circle) with a rope so that the goat will have access to exactly half of the field.

How long should he make the rope?
2. 28 Oct '16 11:49
I smell calculus.
3. 28 Oct '16 12:08
Originally posted by Blood On The Tracks
A farmer owns a circular field which has a radius of 20m.

He wishes to tether a goat to a post on the edge of the field (ie, the circumference of the circle) with a rope so that the goat will have access to exactly half of the field.

How long should he make the rope?
I've seen this before I think.
The divisions are going to be "half moon" shaped but my geometry knowledge is not good enough to apply to the problem
4. 28 Oct '16 12:38
Im probably not right, bu in my minds eye it looks correct.

If the radius is A and the circumference is B, then wouldnt you need to do A+.5A=Half the circle?
5. 28 Oct '16 12:54
Originally posted by Andrew Kern
Im probably not right, bu in my minds eye it looks correct.

If the radius is A and the circumference is B, then wouldnt you need to do A+.5A=Half the circle?
Hi

I think you are saying, for my 20m radius field, that it should be 30m rope?

Afraid that isnt correct ~ would be too long, letting goat eat much more than half
6. 28 Oct '16 17:25
14.14m
7. 28 Oct '16 17:45
Originally posted by yelrambob
14.14m
14.14m would be the answer for a full circle inside the circle. (So put the post 14.14 m from the border to get the goat to eat up to the border.
8. 28 Oct '16 22:54
14.14 isn't correct, as ponderable says, that is the radius of a circle with half the area of the field

But because the goat is tethered on the circumference, it doesn't eat in a full circle, it is basically 2 segments stuck together back to back.
9. 29 Oct '16 00:37
I know! Building a fence.

😉 😉 😉
10. 29 Oct '16 08:28
Goats will eat anything,so in the real world it'd probably chew thro' the rope!!
11. 31 Oct '16 00:384 edits
Originally posted by Blood On The Tracks
A farmer owns a circular field which has a radius of 20m.

He wishes to tether a goat to a post on the edge of the field (ie, the circumference of the circle) with a rope so that the goat will have access to exactly half of the field.

How long should he make the rope?
There probably is a cleaner way. ( to be honest I'm not going to find a numerical solution...because I'm lazy and I don't suppose you are after a numerical approximation, However, this should yield the answer if I followed through)

Let the radius of the field = R
Length of rope = L

The equation that needs to be solved:

1/2* Field Area = Area of Circular sector of radius "L" subtended by angle φ(L,R) + 2* Area Circular segment of Chord Length "L" subtended by angle ß(L,R)

1/2*π*R² = 1/2*arccos [L/( 2*R )]*L² + R²*( arcsin[ L/R*√(1 - ( L/( 2*R))² ) ] - L/R*√( 1 - ( L/( 2*R))² ))

Solve for L numerically ( maybe there is a compact analytic solution I'm missing?)
12. 01 Nov '16 12:141 edit
Originally posted by joe shmo
There probably is a cleaner way. ( to be honest I'm not going to find a numerical solution...because I'm lazy and I don't suppose you are after a numerical approximation, However, this should yield the answer if I followed through)

Let the radius of the field = R
Length of rope = L

The equation that needs to be solved:

1/2* Field Area = Area of Ci ...[text shortened]... 2*R))² ))

Solve for L numerically ( maybe there is a compact analytic solution I'm missing?)
Well, I went through with it. I found a small error in my equation when I tried to solve it.

Corrected version:

1/2*π*R² = arccos [L/( 2*R )]*L² + R²*( arcsin[ L/R*√(1 - ( L/( 2*R))² ) ] - L/R*√( 1 - ( L/( 2*R))² ))

13. 01 Nov '16 16:07
Nice work, Joe

23.17 it is

when I first met this, many years ago, I used Newton Raphson to hone in on the best angle

full solutions are available on the 'net'!
14. 23 Dec '16 00:092 edits
Never mind need to figure out curve.
15. 23 Dec '16 03:33
Originally posted by joe shmo
Well, I went through with it. I found a small error in my equation when I tried to solve it.

Corrected version:

1/2*π*R² = arccos [L/( 2*R )]*L² + R²*( arcsin[ L/R*√(1 - ( L/( 2*R))² ) ] - L/R*√( 1 - ( L/( 2*R))² ))