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Last geometry challenge (very difficult)

  1. Jul 15, 2004 #1

    GCT

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    This problem should be very difficult for those of you inexperienced with geometry, yet anyone who has taken geometry and trigonometry is capable of solving this problem. Feel free to post the answer/ discuss, hopefully more of the latter.

    There is a circle of grass with the radius R. We want to let a sheep eat the grass from that circle by attaching the sheep's leash on the edge of the circle. What must be the length of the leash for the sheep to eat exactly half of the grass?


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  2. jcsd
  3. Jul 15, 2004 #2
    I'm probably wrong, but here's my answer:

    (2R) / (sqrt(2))
     
  4. Jul 16, 2004 #3
    nah

    Nah..

    I think the answer is R/sqrt2...because you have x^2 = (R/2)^2 +(R/2)^2 and that is simply x^2 =(2R^2/4) and then x^2=R^2/2....so x=R/sqrt2

    :)
     
  5. Jul 16, 2004 #4

    Gokul43201

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    nah nah ... clearly the length needs to be greater than R
     
  6. Jul 16, 2004 #5

    Njorl

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    Assume leash is length P.

    Logically, R<P<2R.

    P is the radius of a large circle centered at the edge of the circle of grass at a point callled Q.

    We want to find the intersection of the areas of the 2 circles.

    (Note, I may have my nomenclature wrong. It has been a while. I am assuming a sector of a circle is a pie shaped wedge, and a segment is that wedge minus the triagular part. I am assuming a chord is the line from one point on a circle's perimeter to another.)

    At 2 points, A and B, the perimeters of the circles intersect.

    Define the chord L as the line segment that simultaneously cuts a segment from both circles - it goes from A to B.

    The area that the sheep consumes is equal to the sum of these two segments.

    You can express the areas of these segments in terms of R,P and L and set it equal to half the area of the grass circle. This is not enough info yet. You still have 2 unknowns.

    However, you also can express the segment of the grass circle as the sum of the isoceles triangle defined by ABQ and 2 segments of the grass circle, AQ and BQ. This will give you 2 equations and 2 unknowns, so you can solve for L and P.

    Still, it looks like a lot of work, and I ain't doin' it.

    Njorl
     
  7. Jul 16, 2004 #6

    NateTG

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    ...Making things a little bit more fun...:
    Assume that the Earth is a sphere of radius [tex]R[/tex] with [tex]R \geq \frac{2r}{\pi}[/tex]
     
  8. Jul 16, 2004 #7

    NateTG

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    Hmm.

    Let's say that the length of the rope is [tex]l[/tex]. Let
    [tex]\theta=\cos^{-1}\sqrt{\frac{l^2-r^2}{r^2}}[/tex]
    then the area covered by the sheep will be:
    [tex]\frac{r^2}{2} (\theta - \frac{1}{2} \sin {\theta}) + 4r^2(1-\cos \theta) (\pi - \theta)[/tex]
    so to find the length of the rope, solve:
    [tex]\frac{1}{2}=\frac{1}{2}(\theta - \frac{1}{2} \sin {\theta})+4(1-\cos \theta) (\pi - \theta)[/tex]
    ...
    Let's see if [tex]\phi=\pi-\theta[/tex] simplifies things...
    [tex]\sin(\theta)=\sin(\pi - \phi)= \sin \phi[/tex]
    and
    [tex]\cos(\theta}=\cos((\pi-\phi)=-cos \phi[/tex]
    so
    [tex]1=(\pi - \phi - \frac{1}{2} \sin \phi) + 4 \phi (1 + cos \phi)[/tex]

    This is fairily easy to solve numerically, but I'm not sure I see a algebraeic solution.
     
  9. Jul 16, 2004 #8
    I think numerical methods are needed to solve this. I get 1.1587285 R
     
  10. Jul 16, 2004 #9

    GCT

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  11. Jul 17, 2004 #10

    Galileo

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    Yuck, I got a solution after an hour of drawing and thinking. :yuck:
    It's awful and I think there's a more elegant solution. Anyway, here what I got:

    The length of the leash l should be:

    [tex]l=R\frac{\sin(\theta)}{\sin(\theta/2)} [/tex]

    Where [tex] \theta[/tex] is between zero and pi and is the solution to the following equation:

    [tex]\theta\left(\frac{\sin(\theta)}{\sin(\theta/2)}\right)^2 +\pi-\theta-\sin(\theta)=0[/tex]

    I can't solve it numerically, because I don't have a program for it...
    Can anyone help me with that numerical part and tell me my solution??
     
  12. Jul 17, 2004 #11

    jcsd

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    It's impossible to solve exactly.
     
  13. Jul 17, 2004 #12

    GCT

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  14. Jul 18, 2004 #13

    jcsd

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    I don't get you, the goat problem can't be solved exactly. Ceptimus has already given as good of an answer as you're going to get.
     
  15. Jul 18, 2004 #14

    GCT

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  16. Jul 18, 2004 #15

    GCT

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    Here's the solution for those who are interested:

    Let L denote the length of the leash. Let O be the center of the grass circle, and Q the location where the leash is fastened. Let P and P' be the two points on the circumference of the grass circle at distance L from Q. Let B denote the measure of angle PQO in radians, and (C = π - 2B) the measure of POQ. Because PQO is isosceles, we have L = 2 R cos B. The pie-shaped region emanating from O and reaching from P to P' has area (1/2) R2 (2C) = R2 C. The pie-shaped region emanating from Q and reaching from P to P' has area L2 B. Together these regions cover the sheep's eating area, but they both cover the quadrangle OPQP', so we must subtract its area, 2 ( (1/2) R L sin B) = R L sin B. We obtain ( R^2 C ) + ( L^2 B ) - R L sin B = (1/2) π R^2, from which ( R^2 ( π- 2B))+( 4 R^2 B cos2 B )-( 2 R^2 sin B cos B )=(1/2)π R^2, or π - 2B + 4 B cos2 B - 2 sin B cos B = π /2. We solve this numerically for B, and obtain B = 0.952848, C = 1.235897, L=1.158728R.


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    Last edited: Jul 18, 2004
  17. Jun 11, 2010 #16
    sine of 45 degrees TIMES radius = Maximum leash for the sheep.
    Sine45 degrees*r= x
     
  18. Dec 7, 2010 #17
    Wickedprodigy - you probably have misunderstood the problem. The grass covers the entire circle; not just the boundary.
     
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