The "half a field" problem?

Question

Some years back a friend of mine dug out this problem - I've a first-class honours degree in maths so I smiled smugly to myself, preparing to solve the problem in only a few minutes... and got horribly stumped (serves me right!).

Here's the problem: a farmer has a square field (let's say 100m on each side). He tethers a cow to the mid-point of one side. How long should the rope be so that the cow ends up eating exactly half the grass in the field?

I remember creating awful-looking trigonometric expressions constaining the answer, but never being able to reduce it to a

length of rope = (nasty expression)

kind of formula.

(BTW I'm not interested in approximations or iterative methods.)

I got it into my head that it was impossible to reduce to such a formula - and if so, that would be an interesting proof in itself.

Anyone got a solution or a way of proving that there's no simple expression?

Everyone - just to point out, because the cow is tethered to the mid-point of the edge of the field, it is NOT the case that the cow's circle lies wholly within the square field! It's only part of this circle, and it's not a straightforward arc because it's bounded by the square sides of the field...

(If you draw a quick picture it'll make sense)

Answer 1

There is a very similar question posted here, except the field was circular, not square. I was intrugued, and while I was playing around with the solution I had seen a response to the question with the answer provided in the link below. I decided not to bother posting another answer, and for that same reason I probably won't bother answering this one either, but I did think the link to the solution is worth a look for you. It's messy I guess, but it's not really all that difficult. It's more tedious than anything else. I think a good application of areal integration really. It's calc II stuff.

So, it's quite solveable, but still a good question. I may work on this on my own and see what the difference is with a square field as opposed to a circular one.

Here's another good one for you to ponder. I imagine you've seen this.

Take a sphere and drill a hole along a diameter. I don't remember the details exactly (sorry), but you are asked to find the volume of the resulting solid (sphere munus the hole material). Turns out it is a constant....like π/360, or π/386....something like that. So, regardless of how large the original sphere is, you always have this same resulting volume. Isn't that funky!?

If you want specifics on this, send a message and I will message you back. I know where to find it.

Answer 2

Put the base of the square field on the x-axis, and let (0,0) be the cow's tether point. The lower part of the square field is delimited by the lines y=0, x=-50 and x=+50. The cow has access to the area under the curve y=sqrt(r²-x²) within these boundaries. So

int(x=-50 to +50) sqrt(r²-x²)dx = 5000

is your equation. Solve for r.

You can also use trig: the area is the sum of a circular sector + two right triangles. Add up the three areas to express the total as a function of r. Solving this could be a ?%$*.

Answer 3

I don't think you can solve this problem without using iterative methods. Using calculus, I came up with the equation that has to be solved, but the radius (R) is stuck away in several different niches that I think is impossible to extract by any other means.
Letting a = side of field, the equation is :

a * sqrt(4R^2 - a^2) + 4R^2 * sin^-1[a /(2R)] = 2a^2

Edit : With a = 100 m, R = 58.28221624459555 m.

Answer 4

If I understand this problem you just need to calculate the length of the radius of a circle so half the area of the circle is also have the size of the field.

z^2 / 2 = pi * r^2 / 2

r^2 = z^2 / pi

In this case r = 56 m.

Ah now I see the problem :)
r is bigger than z / 2.
Not as easy as it looks.
I'll have to think harder.

Update:

The best I can come up with is:

x² = 2 ( x*sqr(1-x²) + asin(x)/pi )

with x = z / r ( z side of field, r is length of rope)

Answer 5

piR^2=1/2(a^2)

R=((a^2)/(2pi))^1/2
R=a/(2pi))^1/2
R=100m/(2 pi)^1/2=39.9m

Hmmm.. where is the problem?

Am I missing something?

Now in response to additional details.

Let
1.A be the angle in radians trough which the cow can move tethered to the middle of the side
2.R be the tether radius
3.x be maximum of the left ad the right side of the fence the cow can move along.
4.a be the side of the square field.

Then

AR^2+2(.5 a x)/2=.5a^2
R^2=.25a^2 + x^2
A=Pi(180-2arctan(2x/a))/180

Now we have three equations and three unknowns.

Reducing we have

[Pi(180-2arctan(2x/a))/180][ .25a^2 + x^2] + .5 a x =.5 a^2

And an analitical monster it is.

PS
SORRY
Your email is not 'confirmed' as far as Yahoo is concerned. We can't comunicate until it is.

Answer 6

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Answer 7

I think it should actually be z²/2 = pi * r² where z is the length of the side of the field and r is the length of the rope.
so z²/(2pi) = r²
and r = z/√(2pi)

btw, ² is - for me at least - alt+0178