## The encircled goat puzzle

Yes, this one is going to be wickedly hard!

A field has the shape of a circle of radius 100m and is enclosed by a circular fence. A goat is attached by a rope to a hook, at a fixed point on the fence. To stop the goat from getting too fat, the farmer wants to make sure that it can only reach half of the grass in the field. How long should the rope be?

Big Thanks to Bilbao for submitting!

You can find Michaelc’s original Goat Problem right here

*Answers can be submitted below in the comment section. Will keep a list, though, of those who get it right:*

- Hex
- CJ
- Bizarette18
- Aaronlau-
- Michaelc
- Shawn
- APEX.JP
- AlexC- close
- Someone-close
- mileyfanchelsea- close

microu| Profile May 13th, 2009 - 1:16 amI am no good in maths, what I can guess is about 120-125 meters long would cover almost half area

alexc| Profile May 13th, 2009 - 1:56 amThe over lapping area of the two circles can be broken up into 2 areas by drawing a line from where the chain is attached to the fence to where the chain meets the fence again which forms a chord. The area between the chords can be area 1, and the angle the two chords make can be theta, then that leaves two equal areas of the chords to add to area 1 (chord area can be area 2)

x = chain length

So we have

Area 1 = Pi * x * theta / 360

Area 2 = 2 * chord area = 2 * 100^2 / 2 * (180 – theta – Sin(180 – theta))

So equation 1 = total area = A1 + A2

A = Pi * x * theta / 360 + 2 * 100^2 / 2 * (180 – theta – Sin(180 – theta))

Equation 2:

The chord length equals the chain length (cord length eqn: C = 2 sqrt(r^2 – t^2)

hence x = 2 sqrt (100^2 – t^2)

where t = 100 * Sin (theta/2)

Subbing t into x and x into A gives a long equation.

As stated in the problem A = half field area

therefore A = 100^2 * Pi /2

Solve for theta

we get theta = 178.4 degrees

Sub theta and A into eqn 1 and solve for x

x = 75.008m

hex| PUZZLE MASTER | Profile May 13th, 2009 - 5:16 amThis one is way easier than the goat problem if we use Calculus in order to calculate areas by integration.

In short, provided I have not fallen to silly mistakes, then a rope length of 115.8728401 m is to be used.

ruddwd| Profile May 13th, 2009 - 6:07 amWell the entire area is 31410 m^2.

1/2 that is 15705 m^2.

The Radius of that Circle is 70.71 m.

But that’s for the complete circle, if it were to be staked at the fixed point in the middle, this would be correct.

But as it is to be fixed to a point on the fence, it’s only a portion of that.

Not being able think past this point this early in the morning I am going to say that the rope has to be longer than 100m but at this point, I’m not sure by how long. Maybe after my coffee.

CJ| Profile May 13th, 2009 - 11:23 amThe rope should be 115.87m long.

aaronlau| Profile May 13th, 2009 - 12:17 pmGot the equation, but not good at algebra enough to simply and solve, so the crude method solution gives

109.766m

Shawn| PUZZLE GRANDMASTER | Profile May 13th, 2009 - 1:53 pmI get ~105.5 meters

bizarette18| PUZZLE MASTER | Profile May 13th, 2009 - 3:14 pmI think it should be around 115.87m

alexc| Profile May 14th, 2009 - 12:45 amAfter realizing how silly my first answer was i figured out my mistake, its not 180 – theta, but 90 – theta

Anyway, i get theta as 90 degrees

and

x = 125.33m

aaronlau| Profile May 14th, 2009 - 9:01 amnoticed an error in my equation.

will give it another shot

115.8729m

Hendy| Profile May 14th, 2009 - 9:34 amVery simple. The rope should be 2 times the diameter. Tie the rope from one side to the other creating a barrier at half the field. The other half of the rope should be tied to the goat and any point along the fence. One half of the field will be accessible and the other half will be cut off by the rope.

Caveat: I wouldn’t trust that a rope will stop a goat (as a barrier or as a restraint, it would probably be jumped over, crawled under or chewed to pieces), but it is a simple solution.

michaelc| Profile May 14th, 2009 - 6:05 pmI don’t think there’s a cool geometric solution to this one! I’m glad the old goat problem was inspiring however.

I’m getting ~115.8m as the length, that is if I didn’t make an error in my 3 pages of calculation. It sounds reasonable anyway.

The method I used was transforming the problem into polar coordinates with the origin at the tying off point. Integrating, and then creating an algorithim to find the right set of numbers that worked. That’s about as far as I will go into my 3 pages!

Great twist on the problem Bilbao. It was a good refresher for me for calculus that I hadn’t used in quite awhile…

Shawn| PUZZLE GRANDMASTER | Profile May 15th, 2009 - 1:41 pmHow about 115.87

Someone| Profile May 15th, 2009 - 4:09 pmYou are dividing a circle in half with an arc. If you divide the circle in half with a straight line, you can reasonably assume that this arc will not be completely past or completely before this line. Therefor, the length of the rope will be b/w 100 and 100 rad 2, so

rope length is an element of (100, 141.421356)

I found the area of reachable area(A) inside a unit circle as a function of the radius(r), and solved for A = pi/2

Area = arcsin(r/2) + r^2 * (pi/2 – arcsin(r/2)) – sin(4 * arcsin(r/2))

I got a radius of 123.1569539m

APEX.JP| Profile May 16th, 2009 - 1:52 pmWell… After detail evaluation of all important influencing factors… such as; height of the fixing point, fence post spacings & wire tension, breed of goat & distance from collar to teeth, tensile strength of the rope, type of knots used & extra rope length needed, percentage inedible due to trample & goat waste byproduct… It was probabilistically determined that the goat would chew through a normal rope, rendering the length irrelevant when computed in this model.. Therefore length = 0.

However as this only presents a theoretical result, the logical conclusion would be that the rope should be exactly 1000m long… or… in other words, remain un-used still packaged at its original length, still in the farmers shed.

Now… realising that the previously offered solution had inadvertently, yet convieniently, dismissed the farmers’ requirement for imposing of dietary restrictions upon the goat, a secondary more holistic evaluation of the problem was undertaken.

By using this approach it was determined that in order to establish boundary limitations to the goats’ grazing allocation then the rope could first be used to create a physical barrier by being stretched back & forward between the corresponding opposite fenceposts several times, creating the effect of a ‘three wire’ rope fence through the middle of the field. And then continuing on for an additional 200m to terminate at connection to the goat.

In detailed evaluation of this secondary holistic method the reductive probability calculated through causal determinism was subject to too-many variables which proved insufficient to provide a comprehensive length value.

So… After trying everything else, we finally arrive at this position; in which we have already used-up most of the farmers’ rope, the goat is still hungry, half the field is still unprotected… and we have still not determined an accurate value for our length of rope.

Therefore in conclusion what we are able to calculate; is 1000m that the farmer started with, the 850m we used-up trying to create a rope-fence, the 30m we used tying various knots… and not forgetting the bit the goat ate…

Leaves us with our final answer of: Rope length = 115.87285m…

If the farmer wants to tie-up a goat.. Everyone needs a hobby..

mileyfanchelsea| Profile May 17th, 2009 - 6:45 amthe rope should be 100m long

bilbao| Profile May 18th, 2009 - 5:41 amOriginal source:

“Elementary Calculus and Coordinate Geometryart II”, p.364 (1949) by C.G.Nobbs.

Hendy| Profile May 18th, 2009 - 12:56 pm70.7106781186548

within the precision of Excel

RK| Founder | Profile May 20th, 2009 - 1:49 pmThis one was very hard, will post Bilbao’s solution shortly. (although APEX.JP’s answer may enlighten you for the time being

RK| Founder | Profile May 20th, 2009 - 9:52 pmOk, here is Bilbao’s solution:

http://www.smart-kit.com/wp-co.....lution.jpg

MFox| Profile May 20th, 2009 - 11:35 pmThis puzzle got me all worked up, which is why I signed up for this site and am posting. I worked on this in consultation with my sister, and we managed to come up with an (admittedly ugly) algebraic solution.

It’s based on summing up the sector of the large circle contained within the field, with the two equal segments cut off by the sector. I have it all drawn up, and I’ll try to find a way to post it so you all can doublecheck me. The short version is that it is possible to put the equations for those areas all in terms of theta, where theta is the angle subtending the arc formed by the extended leash. Then setting the combined area equal to .5 x pi x 100m^2, it is possible to solve for theta, and back-calculate the length of the leash.

My sis is still chugging away at the actual numerical solution, but plugging 116m into these equations came pretty danged close, so I’m very optimistic we can get to the same place as all these calculus-heads.

MFox| Profile May 20th, 2009 - 11:47 pmA diagram which hopefully will explain some of my rambling in the previous comment can be found at:

http://www.tulane.edu/~venom/images/encircled goat problem.bmp

bilbao| Profile May 22nd, 2009 - 2:38 amHi Mfox, congrats to both you and your sister. I like it very much your approach, which may be followed easier than mine.

I have taken your equations to Maple and let it calculate the numerical solution.

I send to RK the results ;-)

RK| Founder | Profile May 22nd, 2009 - 6:48 amThanks Bilbao!, here we go:

http://www.smart-kit.com/wp-co.....ution2.jpg

michaelc| Profile June 2nd, 2009 - 8:12 pmHello MFox,

Thanks for sharing your solution to the puzzle! I like it!

Who needs calculus anyway?

I haven’t posted in awhile, as my workplace now blocks the website! I agree with Bilbao about it being easier to follow. I’m still wondering if I can follow my own solution to this thing!

MFox| Profile June 4th, 2009 - 9:17 amThanks for the feedback, and for carrying the approach out to a numeric solution. I’m very pleased to see that it worked.

I have to admit that my calculus is extremely weak, which is probably why I stuck to geometry/algebra. My sister is a super genius (I call her the queen of the nerds), and I was lucky to be able to work with her on it. She’s won her college’s annual math prize for two years in a row now, which involves developing the most elegant solution (or proof) for a difficult analytical or geometry problem very much along the lines of this one.

rmsilber| Profile July 1st, 2009 - 10:20 amWow!

I tried to write an equation for Q, the maximal angle the rope could create, using the fact that we had to create a region of area 5000*pi. (I actually decided to solve for the *uncovered* area — seemed a little easier.)

I got Q = 109.19 degrees.

Here was the equation I (well, really, my graphing calculator) solved to get Q:

5000*pi = 500*pi*Q/9 – 10000*sinQ*cosQ + 20000*(cos(Q/2))^2*sinQ – 1000*pi*Q*(cos(Q/2))^2/9

(To use my calc, I moved the constant to the right side, graphed, and found the “zero” of the graph.)

It should also make sense that r = 200cos(Q/2)

Using this, and my angle,

r = 115.87

Who knows if I did everything right?? I think that it all works, with the possible exception of careless mistakes. It was lots of fun, either way!

PabloRamirez| Profile November 8th, 2009 - 7:18 amalpha: angle betwen two reacheable points of fence and the circle center. [radians]

radius^2*( sin(alpha/2)*cos(alpha/2) + pi – alpha/2 ) = pi*radius^2/4

2*radius*( 1 – cos(alpha/2) ) = rope lenght

then numericaly obtained:

alpha = 2,30988146 radians

rope lenght 119,2054493 meters

judycorstjens| Profile April 15th, 2013 - 7:56 amThe goat can eat an area created by the intersection of a circle radius R (the length of the rope) and the circle of the field (radius r). The eating area (E) consists of a segment of the new circle, say angle 2a (a in rads, also call pi p, and let b be the complimentary angle to a). E = R^2.a plus two areas behind the chords in the ‘field’. The area of the chords is r^2.b minus the two triangles height r.cos(b) and base r.sin(b), or r^2.sinb.cosb. Also R=2.r.sin(b) so putting E equal to half of the field we get: (2rsinb)^2.(p/2-b) + 2r^2(b-sinb.cosb) = p.r^2/2

Divide through by r^2 to get 4(sinb)^2.(p/2-b) +2(b – sinb.cos.b) -p/2 = 0 Not elegant, but solvable on a TI-83, and you get b is about 0.61767 rads, so 2sinb is 1.15827 which is the ratio of R to r, i.e the Rope must be about 116m long, as far any farmer cares.

Guggs| Profile July 17th, 2013 - 2:09 pmThe answer to this goat in a firld problem is 115.87 m and doesn’t require the radius of the feild to be used in the equation, since once the radio of the field radius and goat tethher is computes it’s a simple mater of multiplying by 100 – see proof at link to http://www.guggs.co.uk.