Trickier problem than it seems. (goat in a circular paddock)
Moderators: gmalivuk, Moderators General, Prelates
Trickier problem than it seems. (goat in a circular paddock)
You have a goat, tied to the fence of a circular paddock. The paddock has a radius of 100M. You want the goat to eat HALF of the grass.
How long does the rope need to be?
Assume an even covering of grass, and don't worry about the goat's neck length etc.
How long does the rope need to be?
Assume an even covering of grass, and don't worry about the goat's neck length etc.
Wanna stalk me? Begin Here: http://protectorfire.com.au
Re: Trickier problem than it seems.
Simple reasoning:
But given it's "trickier than it seems": I have no idea if the goat can/wants to eat the grass fast enough to actually eat "half the grass". In any case it needs extra nutrition beside grass.
Spoiler:
But given it's "trickier than it seems": I have no idea if the goat can/wants to eat the grass fast enough to actually eat "half the grass". In any case it needs extra nutrition beside grass.
 Xenomortis
 Not actually a special flower.
 Posts: 1443
 Joined: Thu Oct 11, 2012 8:47 am UTC
Re: Trickier problem than it seems.
The rope is tied to the *edge* of the fence, so the region the goat can eat is actually the intersection of two discs.
You're looking for a radius r, of a disc centered on a point on another disc's boundary, such that the area of the intersection of the two discs is half the area of the second disc.
You're looking for a radius r, of a disc centered on a point on another disc's boundary, such that the area of the intersection of the two discs is half the area of the second disc.
Re: Trickier problem than it seems.
Goats never eat all the grass, leaving the roots intacts, so you can just let the goat roam free in the enclosure. It'll eat roughly the upper half of each blade of grass.
Ok, trick answers aside, it still seems straightforward enough.
Let r1 = 100m, r2 = length of the rope.
Our points are
C: center of the paddock
T: tying point on the fence
I1, I2: the two intersections between the circles (C, r1) and (T, r2)
alpha: angle I1, C, I2
beta: angle I1, T, I2
If you want to graze exactly half the grass, then C, I1, T, I2 is convex, so alpha < 180°, beta < 180°. That means that the grazed part is the sum of two circular segments, with a known formula for area:
A = r1^2/2 (alpha  sin(alpha)) + r2^2/2 (beta  sin(beta))
with A being the area of the grazed part, which is half of the paddock's area:
A = 1/2 * pi * r1^2
We have an equation with three variables r2, alpha, beta. The fourth, r1, is a constant.
To get rid of two of the variables, look at the triangle C, I1, T. It has two sides with length r1 and one with r2, one angle (alpha/2) and two angles (beta/2). It's an isosceles triangle, so you can find formulas for the relations between alpha, beta and r2. Substitute in the equation above, solve for alpha, beta or r2, whatever seems easiest. To solve for alpha, you could use:
beta/2 = (180°  alpha/2) / 2
r2 = 2*r1* sin(alpha/4)
Solving for anything is tricky, because the formulas are a tangled mess of triginometric functions that are difficult to untangle. At this point, I'd just solve numerically.
Ok, trick answers aside, it still seems straightforward enough.
Let r1 = 100m, r2 = length of the rope.
Our points are
C: center of the paddock
T: tying point on the fence
I1, I2: the two intersections between the circles (C, r1) and (T, r2)
alpha: angle I1, C, I2
beta: angle I1, T, I2
If you want to graze exactly half the grass, then C, I1, T, I2 is convex, so alpha < 180°, beta < 180°. That means that the grazed part is the sum of two circular segments, with a known formula for area:
A = r1^2/2 (alpha  sin(alpha)) + r2^2/2 (beta  sin(beta))
with A being the area of the grazed part, which is half of the paddock's area:
A = 1/2 * pi * r1^2
We have an equation with three variables r2, alpha, beta. The fourth, r1, is a constant.
To get rid of two of the variables, look at the triangle C, I1, T. It has two sides with length r1 and one with r2, one angle (alpha/2) and two angles (beta/2). It's an isosceles triangle, so you can find formulas for the relations between alpha, beta and r2. Substitute in the equation above, solve for alpha, beta or r2, whatever seems easiest. To solve for alpha, you could use:
beta/2 = (180°  alpha/2) / 2
r2 = 2*r1* sin(alpha/4)
Solving for anything is tricky, because the formulas are a tangled mess of triginometric functions that are difficult to untangle. At this point, I'd just solve numerically.
 Carlington
 Posts: 1588
 Joined: Sun Mar 22, 2009 8:46 am UTC
 Location: Sydney, Australia.
Re: Trickier problem than it seems.
On which note:
For the sake of dodging irrational numbers, I'm treating half the area of the paddock as being 15707.5m^{2}. Then, with help from research done on Wolfram MathWorld, I put together the equation:
I didn't want to solve that by hand, and Wolfram Alpha couldn't even interpret it to begin with. If anyone wants to solve that...or check it for mistakes...feel free.
EDIT: Ninja'd by Tub. I don't think that alpha and beta can both be convex. The radius of the paddock must less than the length of the rope, which means the centre of the paddock falls between the centre of the arc swept by the rope and the chord joining the points of intersection of the circles. We can't just be straightforwardly adding the area of two sectors, in any case.
For the sake of dodging irrational numbers, I'm treating half the area of the paddock as being 15707.5m^{2}. Then, with help from research done on Wolfram MathWorld, I put together the equation:
I didn't want to solve that by hand, and Wolfram Alpha couldn't even interpret it to begin with. If anyone wants to solve that...or check it for mistakes...feel free.
EDIT: Ninja'd by Tub. I don't think that alpha and beta can both be convex. The radius of the paddock must less than the length of the rope, which means the centre of the paddock falls between the centre of the arc swept by the rope and the chord joining the points of intersection of the circles. We can't just be straightforwardly adding the area of two sectors, in any case.
Kewangji: Posdy zwei tosdy osdy oady. Bork bork bork, hoppity syphilis bork.
Eebster the Great: What specifically is moving faster than light in these examples?
doogly: Hands waving furiously.
Eebster the Great: What specifically is moving faster than light in these examples?
doogly: Hands waving furiously.
Please use he/him/his pronouns when referring to me.
Re: Trickier problem than it seems.
Carlington wrote:The radius of the paddock must less than the length of the rope, which means the centre of the paddock falls between the centre of the arc swept by the rope and the chord joining the points of intersection of the circles.
That is true. And as the chord from I1 to I2 is between C and T, that means that the inner angles at both C and T must be smaller than 180°.
Carlington wrote:We can't just be straightforwardly adding the area of two sectors, in any case.
Not sectors, segments. Follow the link I gave. On each side of the chord is a circular segment.
Re: Trickier problem than it seems.
Xenomortis wrote:The rope is tied to the *edge* of the fence
Ah, that does make it trickier than it seems, when you misinterpret the question.
Alternatively to Tub, integrate over the radius with a lot of arcs (1:100 scale model):
(definite integration by parts, simplifying integrand)
Though something went wrong here, because the left hand never reaches an area of 1 and is negative by the time you fill the whole paddock, and consequently wolfram says its solutions are complex.

 Posts: 35
 Joined: Wed Apr 09, 2014 1:33 am UTC
Re: Trickier problem than it seems. (goat in a circular paddock)
The first approximate solution I have managed to come up with thus far is:
The length of the rope is approximately 2/√3 times the radius of the circular paddock fence.
The length of the rope is approximately 2/√3 times the radius of the circular paddock fence.
GENERATION 705  992 i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.
 Eebster the Great
 Posts: 3302
 Joined: Mon Nov 10, 2008 12:58 am UTC
 Location: Cleveland, Ohio
Re: Trickier problem than it seems. (goat in a circular paddock)
If I'm understanding the Mathworld formula correctly and haven't made any input errors, then setting d = r for their equation we get the following formula for the area, A, in terms of the radius of the paddock (the smaller circle), r, and the length of the rope (the radius of the larger circle) R: .
WolframAlpha easily simplifies this, under the assumption that r and R are positive real numbers, and then if I simpify it further by setting r = 1 (and therefore A = π) I get: .
Now, WolframAlpha really tries hard to solve this ugly but seemingly innocuous equation. I mean it really spends a tremendous amount of time computing considering I haven't paid for extra computing time. It doesn't seem to give up. But I can't seem to coax it into giving a solution.
Whatever the solution to this is, simply multiply by r = 100 m and that is the answer.
WolframAlpha easily simplifies this, under the assumption that r and R are positive real numbers, and then if I simpify it further by setting r = 1 (and therefore A = π) I get: .
Now, WolframAlpha really tries hard to solve this ugly but seemingly innocuous equation. I mean it really spends a tremendous amount of time computing considering I haven't paid for extra computing time. It doesn't seem to give up. But I can't seem to coax it into giving a solution.
Whatever the solution to this is, simply multiply by r = 100 m and that is the answer.
Last edited by Eebster the Great on Sun Jul 10, 2016 5:23 am UTC, edited 2 times in total.
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
 Posts: 5967
 Joined: Tue Nov 28, 2006 9:50 pm UTC
 Location: Sydney
Re: Trickier problem than it seems. (goat in a circular paddock)
It won't have a closed form. Its a polynomial intersecting a trig function. No closed form for that unless you're super lucky.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
 Eebster the Great
 Posts: 3302
 Joined: Mon Nov 10, 2008 12:58 am UTC
 Location: Cleveland, Ohio
Re: Trickier problem than it seems. (goat in a circular paddock)
I was hoping for a numerical approximation.

 Posts: 35
 Joined: Wed Apr 09, 2014 1:33 am UTC
Re: Trickier problem than it seems. (goat in a circular paddock)
Eebster the Great wrote:If I'm understanding the Mathworld formula correctly and haven't made any input errors, then setting d = r for their equation we get the following formula for the area, A, in terms of the radius of the paddock (the smaller circle), r, and the length of the rope (the radius of the larger circle) R: .
WolframAlpha easily simplifies this, under the assumption that r and R are positive real numbers, and then if I simpify it further by setting r = 1 (and therefore A = π) I get: .
Now, WolframAlpha really tries hard to solve this ugly but seemingly innocuous equation. I mean it really spends a tremendous amount of time computing considering I haven't paid for extra computing time. It doesn't seem to give up. But I can't seem to coax it into giving a solution.
Whatever the solution to this is, simply multiply by r = 100 m and that is the answer.
I don't think your equation is correct. By inspection, R=2 satisfies the equation and we are all fairly certain at this point that it is not an integer.
The negative solution is about 0.85, which I don't believe has a geometric interpretation in this scenario.
GENERATION 705  992 i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.
Re: Trickier problem than it seems. (goat in a circular paddock)
Eh, the solution won't be quite correct anyway since it probably is bound around the throat and can reach a bit further than the rope length. So, 100 is obviously to little, at 141 it includes the half you get from drawing a straight line through the center+quite a bit more. The missing area under the line must be as big as the extra area over the line. I think 120 would be a bit much since the not filled areas will be long but narrow I think. So I guess 115 or so, close enough for goat feeding.
I'm just curious how near guessing will be to the solution. (Oh and here my being bored and matching numbers solution:115.8728473018121 derived from matching the area of the small lune to the area of the intersection https://rechneronline.de/pi/lune.php To make the equations in this thread even longer I suggest equating the lune formula to the intersection formula for double the fun.)
I'm just curious how near guessing will be to the solution. (Oh and here my being bored and matching numbers solution:115.8728473018121 derived from matching the area of the small lune to the area of the intersection https://rechneronline.de/pi/lune.php To make the equations in this thread even longer I suggest equating the lune formula to the intersection formula for double the fun.)
 Soupspoon
 You have done something you shouldn't. Or are about to.
 Posts: 4043
 Joined: Thu Jan 28, 2016 7:00 pm UTC
 Location: 531
Re: Trickier problem than it seems. (goat in a circular paddock)
When we were given this in a secondary school maths class, in the '80s, the teacher ummed and ahhed a bit over this detail then proclaimed that the length of rope needed to tie the knot around the goat's neck/onto the goat's collar was (by complete coincidence) the same as the extra reach the goat could gain advantage of.PeteP wrote:Eh, the solution won't be quite correct anyway since it probably is bound around the throat and can reach a bit further than the rope length.
No, I can't remember our answer, but it was roughly calculated as above.
Re: Trickier problem than it seems. (goat in a circular paddock)
Btw wolfram alpha also has the goat problem itself if someone wants to compare their answer. http://mathworld.wolfram.com/GoatProblem.html
 Eebster the Great
 Posts: 3302
 Joined: Mon Nov 10, 2008 12:58 am UTC
 Location: Cleveland, Ohio
Re: Trickier problem than it seems. (goat in a circular paddock)
Paradoxica wrote:Eebster the Great wrote:If I'm understanding the Mathworld formula correctly and haven't made any input errors, then setting d = r for their equation we get the following formula for the area, A, in terms of the radius of the paddock (the smaller circle), r, and the length of the rope (the radius of the larger circle) R: .
WolframAlpha easily simplifies this, under the assumption that r and R are positive real numbers, and then if I simpify it further by setting r = 1 (and therefore A = π) I get: .
Now, WolframAlpha really tries hard to solve this ugly but seemingly innocuous equation. I mean it really spends a tremendous amount of time computing considering I haven't paid for extra computing time. It doesn't seem to give up. But I can't seem to coax it into giving a solution.
Whatever the solution to this is, simply multiply by r = 100 m and that is the answer.
I don't think your equation is correct. By inspection, R=2 satisfies the equation and we are all fairly certain at this point that it is not an integer.
Whoops, well no because I set A = the area of the enclosure instead of half that area. Obviously in that circumstance R = 2 ought to be a solution. But if we replace π with π/2, it should be right.
Re: Trickier problem than it seems. (goat in a circular paddock)
Second go at the integrals, this time not messing up straight and curved lines (the arctan would've made sense if I were integrating vertical segments, not curved segments). Once again geogebra comes to the rescue (and wolfram, but that's a given):
Spot the error in the integral calculation:
(luckily wolfram also wanted to calculate the integral of 2x*arccos(x/2) in the end, so the last expression is exactly that output and not what came out of integration by parts. But please, if you see the error I presumably made, tell me where it is.)
And huzzah, it's the same equation mathworld has! (given that arccos(1r^2/2)=2arcsin(r/2) for nonnegative reals)
In this form wolfram even allows you to get an approximation of 1.15872847301812....
But yeah, A133731 (linked on the mathworld page) has more digits.
Spot the error in the integral calculation:
(luckily wolfram also wanted to calculate the integral of 2x*arccos(x/2) in the end, so the last expression is exactly that output and not what came out of integration by parts. But please, if you see the error I presumably made, tell me where it is.)
And huzzah, it's the same equation mathworld has! (given that arccos(1r^2/2)=2arcsin(r/2) for nonnegative reals)
In this form wolfram even allows you to get an approximation of 1.15872847301812....
But yeah, A133731 (linked on the mathworld page) has more digits.
 Eebster the Great
 Posts: 3302
 Joined: Mon Nov 10, 2008 12:58 am UTC
 Location: Cleveland, Ohio
Re: Trickier problem than it seems. (goat in a circular paddock)
hehA133731  OEIS wrote:Decimal expansion of goat tether length to graze half a unit field.
Re: Trickier problem than it seems. (goat in a circular paddock)
We can simplify the equation for r a little. There's a simple relation between the two angles because they are in an isoceles triangle, so we can eliminate one of the cos^{1} calls.
Specifically,
cos^{1}(1  r^{2}/2) + 2cos^{1}(r/2) = π
Using 1 for the radius of the field, the equation for r, the tether length, can be written as
π / 2 + (r²  2) cos^{1}(r / 2)  r .sqrt(1  r² / 4) = 0
Here's some Python code that solves that equation using the 3rdparty arbitrary precision library mpmath
Output
The OEIS link given above doesn't list a formula for this constant, but one of the linked pages, http://oeis.org/A173201 does. The "goat tether number" r is closely related to psi, "the unique solution on (0,Pi) of sin(psi)psi*cos(psi) = Pi/2". The relationship is r = 2.cos(psi / 2)
The function for psi can easily be solved using NewtonRaphson. Here's some Python code that uses that method to calculate r to the precision supported by normal Python floats (IEEE 754 doubleprecision).
Output
Finally, here are 1000 decimal places, calculated using the earlier code, plus a little bit of extra code to do the formatting.
Specifically,
cos^{1}(1  r^{2}/2) + 2cos^{1}(r/2) = π
Using 1 for the radius of the field, the equation for r, the tether length, can be written as
π / 2 + (r²  2) cos^{1}(r / 2)  r .sqrt(1  r² / 4) = 0
Here's some Python code that solves that equation using the 3rdparty arbitrary precision library mpmath
Code: Select all
from mpmath import mp
pi = mp.pi
def func(r):
r2 = r * r
y = r * mp.sqrt(1  r2 / 4)
return pi / 2 + (r2  2) * mp.acos(r / 2)  y
mp.dps = 100
r = mp.findroot(func, 1)
print(r)
Output
Code: Select all
1.158728473018121517828233509933509149688292266492096511820695884820669802559196093199321610730860438
The OEIS link given above doesn't list a formula for this constant, but one of the linked pages, http://oeis.org/A173201 does. The "goat tether number" r is closely related to psi, "the unique solution on (0,Pi) of sin(psi)psi*cos(psi) = Pi/2". The relationship is r = 2.cos(psi / 2)
The function for psi can easily be solved using NewtonRaphson. Here's some Python code that uses that method to calculate r to the precision supported by normal Python floats (IEEE 754 doubleprecision).
Code: Select all
from math import sin, cos, pi
x = 2
for i in range(4):
sx = sin(x)
x = x  (sx  x * cos(x)  pi / 2) / (x * sx)
print(i, x)
r = 2 * cos(x / 2)
print(r)
Output
Code: Select all
0 1.906084209585137
1 1.9056957426183945
2 1.905695729309884
3 1.905695729309884
1.1587284730181215
Finally, here are 1000 decimal places, calculated using the earlier code, plus a little bit of extra code to do the formatting.
Code: Select all
1.
15872 84730 18121 51782 82335 09933 50914 96882 92266 49209
65118 20695 88482 06698 02559 19609 31993 21610 73086 04381
75967 49518 03405 18290 13925 08985 19099 26483 15448 46906
47994 14791 31748 52921 23564 02155 50163 23125 55506 86141
07819 72335 23550 17317 30653 43998 83035 05798 22481 55293
32171 41835 93309 12956 38485 59445 48041 27554 70688 16521
41840 20376 39610 49267 76710 56442 75329 13272 31265 32403
81347 70707 99486 14142 63256 46378 66611 22832 29089 75517
41115 73624 52541 70228 20719 55925 03623 94916 48880 32069
37051 06905 28405 19907 64163 92339 78104 68333 34144 96228
36912 51972 07948 89224 49372 78035 71735 33743 86181 60160
19796 46465 81839 20200 29405 17264 92245 39295 88320 29979
07620 31259 86973 61437 38544 37450 00060 31757 67277 53349
39964 80440 83983 48421 42312 19487 92589 57215 46425 81160
58695 65776 92540 47710 15738 25828 42835 43829 60688 13280
31687 48990 75429 37542 61432 91028 88349 02554 80282 11005
81989 45895 45781 48215 44625 12803 60480 55047 26965 79583
70011 43121 89242 63795 05106 75034 00669 45418 41932 37899
74734 05934 80088 31827 28764 03156 60630 74016 42923 26283
80151 50178 20537 15568 98600 74230 99366 62491 89439 58602
Re: Trickier problem than it seems. (goat in a circular paddock)
I've created a diagram in SVG for this puzzle.
Here's the SVG code. Just save it to a file with the .svg extension and you can view it in your browser.
Here's a PNG version.
And here's a simple interactive JavaScript / HTML5 diagram (so save it with the .html or .htm extension) that shows the grazing area for a range of tether lengths. See how well you can estimate the correct length by eye.
Here's the SVG code. Just save it to a file with the .svg extension and you can view it in your browser.
Code: Select all
<?xml version="1.0" encoding="UTF8" ?>
<svg xmlns="http://www.w3.org/2000/svg"
xmlns:xlink="http://www.w3.org/1999/xlink"
width="400" height="400" viewBox="0 0 240 240">
<! The goat tether problem
See http://forums.xkcd.com/viewtopic.php?f=17&t=117520
Created by PM 2Ring 2016.07.25 >
<defs>
<circle id="dot" r="0.75" fill="#000"/>
</defs>
<g transform="translate(120,120)">
<g style="fill: none; stroke: #000; strokewidth: 0.5">
<circle r="100"/>
<line x1="100" x2="100"/>
<line x1="32.86741629" y1="94.4443378" x2="32.86741629" y2="94.4443378"/>
<path d="M100, 0
L32.8674163, 94.4443378
A115.8728473,115.8728473 0 0,1 32.8674163, 94.4443378
Z"
style="stroke:#f00; fill:#f00; fillopacity:0.2;"/>
<path d="M0, 0
L32.8674163, 94.4443378
A100, 100 1 0,0 32.8674163, 94.4443378
Z"
style="stroke:#00f; fill:#00f; fillopacity:0.2;"/>
<! Dots >
<use xlink:href="#dot"/>
<use xlink:href="#dot" x="100" y="0"/>
<use xlink:href="#dot" x="32.8674163" y="94.4443378"/>
<use xlink:href="#dot" x="32.8674163" y="94.4443378"/>
<! <use xlink:href="#dot" x="15.8728473" y="0"/> >
</g>
<! Annotations >
<g style="stroke: none; fill: #000; fontsize: 10px">
<text x="5" y="0">O</text>
<text x="110" y="0">T</text>
<text x="40" y="98">A</text>
<text x="40" y="103">B</text>
</g>
</g>
</svg>
Here's a PNG version.
Spoiler:
And here's a simple interactive JavaScript / HTML5 diagram (so save it with the .html or .htm extension) that shows the grazing area for a range of tether lengths. See how well you can estimate the correct length by eye.
Code: Select all
<!DOCTYPE html>
<html>
<head>
<title>Goat Tether Problem</title>
<meta charset="utf8" />
<style>
#slider_div { marginleft: 25%; }
#slider {
width: 50%; border: 1px solid white; borderradius: 3px;
outline: 1px solid white; outlineoffset: 1px;
}
h3, #canvas_div { textalign: center; }
</style>
<script>
var Pi2 = 2 * Math.PI, ox, oy, bigrad, canvas,
outer_colour = "hsl(120, 80%, 60%)",
inner_colour = "hsl(90, 80%, 60%)";
function outputUpdate(rad)
{
document.querySelector('#radius').value = rad;
draw(bigrad * rad);
}
function setup()
{
var canvas_size;
canvas = document.getElementById('canvas');
if (canvas.getContext)
{
ctx = canvas.getContext('2d');
//Get window height & calculate canvas dimensions
canvas_size = 2 * Math.floor(0.8 * window.innerHeight / 2);
canvas.height = canvas.width = canvas_size;
}
else
{
alert("Sorry, I can't set up the canvas!");
return;
}
ox = oy = canvas_size / 2; bigrad = 0.95 * oy;
//Draw initial state
outputUpdate(0.50);
}
function circle(x, y, rad, fill, stroke)
{
ctx.beginPath();
ctx.arc(x, y, rad, 0, Pi2);
if(fill)
ctx.fill();
if(stroke)
ctx.stroke();
}
function draw(radius)
{
ctx.globalCompositeOperation = "sourceover";
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.fillStyle = outer_colour;
circle(ox, oy, bigrad, true, false);
ctx.globalCompositeOperation = "sourceatop";
ctx.fillStyle = inner_colour;
circle(ox  bigrad, oy, radius, true, true);
ctx.globalCompositeOperation = "sourceover";
circle(ox, oy, bigrad, false, true);
}
</script>
</head>
<body onload="setup();">
<h3>The Goat Tether Problem</h3>
<div id=canvas_div>
<canvas id="canvas">
If you can read this, your browser does not support the HTML5 Canvas.
</canvas>
</div>
<div id="slider_div">
<label for="slider">Length</label>
<input id="slider" type="range" min="0.50" max="1.50" value="0.50" step="0.001"
oninput="outputUpdate(value)">
<output for="slider" id="radius">50%</output>
</div>
</body>
</html>
Who is online
Users browsing this forum: No registered users and 3 guests