## The Involuting Goat

A farmer owns a round barn of radius 12m. A goat is attached by a rope at a fixed point on the wall outside the barn (see image).

The farmer wants to make sure that it can only graze an area equal to that of the barn. How long should the rope be?

*For this (very) hard math puzzle, you may enter your answers into the comment section below. Depending on how many solvers we get, will aim to unmask next week!*

RK| Founder | Profile January 6th, 2010 - 11:34 pmThe goats are back!! Thank you Bilbao for submitting

The Original Goat Problem: http://www.smart-kit.com/s2010/the-goat-problem/

The Encircled Goat: http://www.smart-kit.com/s2231.....at-puzzle/

hex| PUZZLE MASTER | Profile January 7th, 2010 - 6:07 pmA very hard math puzzle indeed!

Having worked at this for a few minutes, it is not very difficult to transform it into a calculus problem. We’ll get 3 areas, one of which is a part of a circle, and the 2 others which are identical…well.. are hard to describe without equations. As their boundaries involve theta and trigonometrical functions of theta (polar coordinates), I doubt that this could be solved without resorting to numerical analysis software which I don’t have.

I will try to work it out this weekend. In the meantime, I can only say that the rope length has to be between 12 and 24m.

Thanks Bilbao

aaronlau| Profile January 8th, 2010 - 6:41 amif my geometry does not fail me.. a bit rusty now..

got the two equations.

Solving it gave length of rope = 10.96264m

Gotta be one twisted farmer to do so much math just for one goat.

Shawn| PUZZLE GRANDMASTER | Profile January 8th, 2010 - 11:09 amSeveral arc cosines and square roots later, I get a value of 15 feet for the rope.

Actually, I get 15.0145 feet, but I attribute the extra bit to Excel and rounding issues.

bizarette18| PUZZLE MASTER | Profile January 8th, 2010 - 12:56 pmJust under 15m?

bizarette18| PUZZLE MASTER | Profile January 8th, 2010 - 5:09 pmI think I should have put just over 15m – around 15.2m – but that’s just approximate.

james turner| Profile January 11th, 2010 - 12:04 pmInteresting Problem:

I assume that the barn radius is denoted rb. Further assume that the rope, when in contact with the barn area makes an unknown angle, theta (i.e., the unknown), with respect to the line connecting the center of the barn circle to the attach point of the rope. The length of the perimeter rope is rb*theta. Assume that along the tangent to the barn circle one stretches the rope so that the maximum length is rg = rb*theta. Assume the goat sweeps out half a circle of radius rg, yielding an area of pi*rg*rg/2. Next assume that one starts along the tangent line and begins wrapping the rope along the edge of the barn area, for a given angle phi, the length of the remaining rope is r = rb*(theta-phi) (which is the maximum length the goat can achieve). There are two identical areas that must be approximated (i.e. areas between the barn perimeter and the tangent line).

A differential area needs a height x base. The height is: h = rb*(theta-phi). The base is found by (1) fixing phi and computing the x-axis project (i.e., axis perpendicular to the tangent line at circle), yielding rb*cos(phi), and (2) decreasing phi by dphi and projecting on the x-axis (yielding a +differential increment) rb*cos(phi-dphi). The x-axis differential increment then follows as: dx ~ rb( cos(phi-dphi) – cos(phi) ) ~ rb*sin(phi)*dphi +O(dphi*dhpi). Now we can integrate both side areas as:

Aside =

2*integrate( h*dx, phi, 0, theta )

2*integrate( rb*(theta-phi)*rb*sin(phi), phi, 0, theta)

2*rb*rb(theta-sin(theta)

The necessary condition defining the solution for theta, so that the barn and goat areas are equal is given by

N(theta) = pi*rb*rb – pi*(rb*theta)*(rb*theta)/2 – 2*rb*rb(theta-sin(theta))

= rb*rb( pi(1-theta*theta/2) – 2(theta-sin(theta))

No closed form solution is possible because the equation is transcendental. As a result, a successive approximation strategy is employed. A starting guess for theta is obtained by plotting N as a function of theta, which indicates that the desired theta is (i.e., n~0): theta~1.26. Polishing this estimate using newton’s method yields:

theta ~ 1.2659638

This implies that the length of the rope is:

rg = rb*theta ~ 12*1.2659638 ~ 15.1916 m

Hope I have not missed something obvious.

Jim Turner

RK| Founder | Profile January 11th, 2010 - 10:58 pmAfter seeing Bilbao’s solution, I honestly didn’t think many would get this.

Well, so far- the mighty Shawn, Bizarette18, and JamesTurner are within decimal points to what Bilbao holds the correct answer to be, and for all intents and purposes, probably have it correct.

Since this is so hard, will wait until Friday to announce all solvers in a special post.

bilbao| Profile January 12th, 2010 - 2:02 amTo encourage everybody to give it a try, I will say that I have made up this puzzle from scratch, inspired by other similar puzzles. So there isn’t a published solution anywhere.

I mean, my solution may be subject to analysis and debate, and once I post it you will have the chance to turn it down…if you can…hahaha!!!!

hex| PUZZLE MASTER | Profile January 12th, 2010 - 8:12 amThe tricky part in this puzzle is the fact that the goat will not describe a circular path when parts of the rope are in touch with the barn. I have setup the equations describing the area, but I am almost certain that they cannot be solved except numerically.

I would not want to consider approximating it to a circular path (which will make the puzzle way easier) as it would defeat the puzzle difficulty. Nevertheless, a quick calculation, without double checking, yields a rope length of 14.487 m

RK, do the submitted solutions make this approximation?

Bilbao, I am intrigued by your comment that your solution is debatable. Is it an exact solution?

bilbao| Profile January 12th, 2010 - 10:16 amHi Hex, my solution is exact and I bet it is correct.

But since I haven’t found it anywhere else, I guess the approach to the puzzle may be subject to approval…you know…kind of the debate with ‘the complex clock puzzle’ ;-)

hex| PUZZLE MASTER | Profile January 12th, 2010 - 4:48 pmHi Bilbao, The complex clock puzzle was one of the coolest math problems I have ever encountered. It was the concept of the question which was debated (How is the minimum defined for 3 variables).

Does your solution of this puzzle involve the use of maple or other equivalent software?

I have never explored such software to start with. Can it solve (numerically of course) an equation involving definite integrals? If yes (I think it does), then this problem would be a piece of cake as I have defined the equations of the goat’s motion. If not, well…rush and email me the solution

hex| PUZZLE MASTER | Profile January 12th, 2010 - 6:51 pmNew development:

I have acquired a mathematical analysis software that I tinkered with to get the following equation:

72*asin(cos(L/12))-(2592*sin(L/6)-216*L*cos(L/6)+432*L*cos(L/12)^2-L^3-648*L)/72+6*cos(L/12)*

sqrt(144-144*cos(L/12)^2)+(pi*L^2)/4-108*pi = 0

length of rope=15.079m

As I am new to this software, and it is very late, I have not double checked my results, nor was I able to get a better precision for the length of rope.

I can only say that with such formidable software, calculus became much easier!

RK| Founder | Profile January 12th, 2010 - 9:58 pmUpdate: Hex has submitted an answer that is

exactlythe same as Bilbao’s.bilbao| Profile January 13th, 2010 - 6:00 amHi Hex, yes, I am using maple v12. I learnt about it when facing ‘the complex clock puzzle’ and, as you, started fidling. I have never seen such a powerful math software. You can work out almost anything.

Luckily, the ball is still in your court when it comes to planning the correct approach to the puzzle…and maple comes later.

Shawn| PUZZLE GRANDMASTER | Profile January 13th, 2010 - 9:21 amI tried a second go at this, because “close” is never good enough!!

New result is 14.71115m

Shawn| PUZZLE GRANDMASTER | Profile January 13th, 2010 - 9:53 amOops, forgot to carry the one!

15.1069m

suineg| PUZZLE MASTER | Profile January 13th, 2010 - 1:38 pmOk Bilbao and hard equals trigonometry.

The problem itself is simple: You have the area of the barn: 144pi

then you have two circles:

Circle 1 that is the barn and does not change

Circle 2 that is the one formed with the rope attach to the goat, this one is variable for this problem

So now what you have to do is find a radius for Circle 2 so that the Area of Circle 1 is equal to the Area of Circle2 minus the Area of the intersection of this two Circles (sound easy)

the equation for this: AreaC1 = AreaC2 -AreaC1InterC2 this is:

(1)144 * pi = R2* pi – AreaC1InterC2- (this one is so hard to find)

In fact what you have to do next is put this intersection Area in function to R2 so that you can solved for this value:

For this you have to use a lot of trigonometry (man why man why!!)

For this Circle 1 has r0,the center is labeled A

Circle 2 has r1,the center is labeled B

C and D are the intersection points and c is the distance between A and B in our case is always r0

First you use the cosine formula ( my gosh):

r0^2 = r1^2 + c^2 – 2*r1*c*cos(CBA)

cos(CBA) = (r1^2 + c^2 – r0^2)/(2*r1*c)

Having found CBA, then (2) CBD = 2(CBA).

Similarly,

cos(CAB) = (r0^2 + c^2 – r1^2)/(2*r0*c)

and then (3) CAD = 2(CAB)

Express CBD and CAD in radian measure by multiply bi pi/180. Then we find the segment

of each of the circles cut off by the chord CD, by taking the area of

the sector of the circle BCD and subtracting the area of triangle BCD.

Similarly we find the area of the sector ACD and subtract the area of

triangle ACD.

AreaC1InterC2 = (1/2)(CBD)r1^2 – (1/2)r1^2*sin(CBD)

+ (1/2)(CAD)r0^2 – (1/2)r0^2*sin(CAD)

that was the area of intersection (WOW as hard as it can get)

Now put that in equation (1):

144pi = r1^2*pi – (1/2)(CBD)r1^2 – (1/2)r1^2*sin(CBD)

+ (1/2)(CAD)r0^2 – (1/2)r0^2*sin(CAD)

Now from (3) and (4) and knowing that c equals r0=12 you get:

144pi = r1^2*pi – (1/2)(2*acos(r1/24))r1^2 – (1/2)r1^2*sin(2*acos(r1/24)))

+ (1/2)(2*acos(288-r1^2)/288)*144 – (1/2)144*sin(2*acos(288-r1^2)/288)

This could be simplified more: sin(acosine(x)=(1- x^2)^(-1/2)

but I left it like that because acosine(x) is unsolvable manually

All is in function to r1 now you found the value, this is impossible algebraically at least for me and I read the comment of using maple so this is as far as I will go.

For me the difficulty in this problem relies to heavily in the knowlegde of trigonometry and the use of mathematical software and poorly on problem solving abilities, but cool man cool. Anyway there is an algebraic way of getting the value of arcosine(x) I mean manually of course, cool

suineg| PUZZLE MASTER | Profile January 13th, 2010 - 3:25 pmCBD and CAD should be multiplied by pi/180, missed that step at the end sorry

Bobo The Bear| PUZZLE MASTER | Profile January 13th, 2010 - 3:54 pmI found a description of this problem (as well as the previous gaot problem) at the following site:

http://mathworld.wolfram.com/GoatProblem.html

Using the information found there, I came up with a value of approximately 15.07882679m for the lenght of the rope. I hope this agrees with your results. Solving a cubic equation for L (the length of the rope) would yield an exact answer, but I have not done this yet.

Thanks for keeping our math skills sharp, bilbao!

hex| PUZZLE MASTER | Profile January 15th, 2010 - 6:41 amBobo The Bear’s link shows 2 things regarding my approach to solving this puzzle:

1- I wonder how I missed the method of calculating the area using polar coordinates (1/2 integral(r^2 dt)) which leads to a very simple solution without the need for math analysis software

2- When there is a will, there is a way!

#2 is way more important than #1

Bilbao, I researched the word involute and found out that it has several uses in your field. And I was wondering from where you coined the term hahaha

RK| Founder | Profile January 15th, 2010 - 9:18 amHere is Bilbao’s solution:

http://www.smart-kit.com/wp-co.....ution1.jpg

suineg| PUZZLE MASTER | Profile January 15th, 2010 - 11:00 amCool answers man, my approach was different from the one taken by Bilbao, Hex and Bobo the Bear that seems pretty similar, Polar cordinates gives back time, college, no responsabilities and only oportunities, good memories, anyway, I think that my approach was similar to the one took by Shawn for his comment about many arc cosine and square roots, So for the sake of my mathematical background I have a question for Bilbao is my approach inexact because I dont use integrals if so why, thanks in advance man if you can clarify me that doubt, cool.

hex| PUZZLE MASTER | Profile January 15th, 2010 - 4:47 pmSuineg, you have to take into consideration that the rope curves along with the barn. In other words, the goat curve is not a circular arc.

hex| PUZZLE MASTER | Profile January 15th, 2010 - 5:19 pmThe difference between the 3 solutions pertains to the involute part and is as follows:

Bilbao has divided the area into triangles and has summed them. By making these triangles infinitesimal, the sum turned into an integral. Equating the areas yields a 3rd degree polynomial/equation which could have been solved algebraically (too difficult but feasible), or by using math software as Bilbao did. I like this solution because it uses the fundamentals of calculus, uses a single integration to obtain the involute area, and yet reaches a 3rd degree equation.

Hex has derived the equations of the boundaries of the area and integrated using cartesian coordinates. Equating the areas yields a crazy equation with sines/cosines, practically unsolvable except by using math software. Using polar coordinates instead would have yielded eventually the same equation as Bilbao’s.

BoboTheBear used the results of the same puzzle in Wolfram, with a few modifications.

suineg| PUZZLE MASTER | Profile January 15th, 2010 - 7:36 pmThanks Hex, I missed that important detail, when reality meet maths, cool, the answer were pretty stunning man, However the intersection of the circles exercise geometrically was really hard to solve even without that detail, for me at the end was unsolvale algebraically.

Bilbao: You know what was really cool about this puzzle, that you give a clue in the title man and I did not even realized that, after Hex clarification I realized that this problem was an all around hard problem both mathematically and generally speaking so I take back my comment about difficulty been too focused in the mathematical aspect, but cool this help me read 2 books of trigonometry to learn some of what I missed back then.

hex| PUZZLE MASTER | Profile January 16th, 2010 - 3:39 pmSuineg, I too had not realized that the term involute describes the basic problem of this puzzle. It was only after I had a look at Bobo The Bear’s link and researched further that I found out it was a common issue in mechanical engineering, especially in involute gears where the gear tooth shape is derived similarly.

Check out http://en.wikipedia.org/wiki/Involute_gear and http://science.howstuffworks.com/gear8.htm for more info.

Another application is involute turbines.

As for intersection area of 2 circles, check out the encircled goat puzzle http://www.smart-kit.com/s2231.....at-puzzle/

bilbao| Profile January 18th, 2010 - 3:41 amA few little comments:

1. First of all, thanks hex for your detailed explanations

2. Congrats to everybody who tried the puzzle…

3. I remember learning at university how to design gears, and the very basis had to do with drawing involute and evolute curves to shape the gear teeth; the very same curve as in this puzzle, thus the title

4. At least with difficult puzzles I try to give a little clue in the title: ‘involuting goat’, ‘complex clock’,…these titles were not by chance…

5. ‘When there is a will, there is a way’. Wow, what a beautiful way to summarize how I face any puzzle or real-life problems.

Example: let’s suppose you face this problem with no knowledge about math software and you get to a very difficult equation…never give up…first feel satisfied to have reached this point…then, since there is always a way, look for alternatives…either ask yourself if a software may exist…or use excel!!…by using as many columns as you need you can build and solve almost any equation, piece by piece, and leave just one first cell to enter a value for the variable. You can get very exact solutions this way…

I always get deeper satisfaction whenever I manage to overcome an obstacle towards the solution than getting the exact solution itself…

james turner| Profile January 23rd, 2010 - 2:51 pmSeveral answers are very close, which is surprising given that the integration approaches yield exact solutions. This leads to a simple question: why are there differences? I believe that the differences can be traced to the curve sweep out from the tangent line to the circle. For example, Bilbao assumes that his polar differential area uses a triangle height that is measured tangent to the circle, whereas Turner uses a differential area where the height is parallel to the tangent line. As a result, the coordinates of the desired curve are given by:

Bilbao: r{cos(phi)-(theta-phi)*sin(phi), sin(phi)+(theta-phi)*cos(phi) }

Turner: r{cos(phi), sin(phi)+theta-phi}

where theta is the unknown max. angle and phi is variable.

The start and end points are the same. The y-axis components of the curve are different. The difference is small and accounts for the variation in answers. All this means is that the problem definition is vague, in the sense that it is unclear whether the “side” areas are to be minimized or maximized?

newguy4321| Profile August 4th, 2010 - 10:56 amFirst off I want to say great site and great question! I was looking over Bilbao’s solution and I had just one question. I am having trouble seeing how the angle dt in the blue circular region is directly equal to the dt in the green circular region. In the green region the angle dt is the angle formed between two tangent lines of the circle. Am I missing some geometrical relationship here? Thanks for your time!

Hex| PUZZLE MASTER | Profile August 6th, 2010 - 9:24 am@newguy4321, This issue bothered me at first when I saw Bilbao’s solution. But since the green lines and the blue lines are always perpendicular (radius vs tangent), it is ok.

tomlaidlaw| Profile May 28th, 2013 - 11:20 pmL=length of rope

R=radius

As other answers have shown, this is a calculus problem. After reviewing many sites with this problem I have found an easy equation for deriving the area of the involute of the circle. Using that makes this puzzle easy. The formula is:

L^3/3R+1/2L^2*pi=R^2*pi

This cubic equation gives 3 roots, two of which are negative so unusable. The positive root is 15.0788

Please see http://tomlaidlaw.com/goatandsilo for more discussion.