Posers and Puzzles

Posers and Puzzles

  1. Joined
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    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. Standard memberapathist
    looking for loot
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    28 Oct '16 11:49
    I smell calculus.
  3. Subscribervenda
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    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. Standard memberBongalloJoe
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    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. Joined
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    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. Brooklyn
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    28 Oct '16 17:25
    14.14m
  7. SubscriberPonderable
    chemist
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    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. Joined
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    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. Standard memberBongalloJoe
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    29 Oct '16 00:37
    I know! Building a fence.

    😉 😉 😉
  10. Subscribervenda
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    29 Oct '16 08:28
    Goats will eat anything,so in the real world it'd probably chew thro' the rope!!
  11. Subscriberjoe shmo
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    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. Subscriberjoe shmo
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    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 ))

    The answer is about 23.175 m.
  13. Joined
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    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. Joined
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    23 Dec '16 00:092 edits
    Never mind need to figure out curve.
  15. Standard memberBongalloJoe
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    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 ))

    The answer is about 23.175 m.
    dang! thats a lot. Neveer couldve figured that out nice work.
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