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Posers and Puzzles

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. Standard member apathist
    looking for loot
    28 Oct '16 11:49
    I smell calculus.
  3. Subscriber venda
    Dave
    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 member BongalloJoe
    Not Gone Yet
    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. Subscriber Ponderable
    chemist
    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. Standard member BongalloJoe
    Not Gone Yet
    29 Oct '16 00:37
    I know! Building a fence.

  10. Subscriber venda
    Dave
    29 Oct '16 08:28
    Goats will eat anything,so in the real world it'd probably chew thro' the rope!!
  11. Subscriber joe shmo
    Strange Egg
    31 Oct '16 00:38 / 4 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. Subscriber joe shmo
    Strange Egg
    01 Nov '16 12:14 / 1 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. 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 00:09 / 2 edits
    Never mind need to figure out curve.
  15. Standard member BongalloJoe
    Not Gone Yet
    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.