Brain Teasers ? or 50 Years On........... ?
Posted by: Don Atkinson on 02 June 2015
50 Years on…….
50 years ago, I was doing what many 18 year olds are doing this week and over the next few weeks……………….their A-Levels.
Mine were Pure Maths; Applied Maths; Physics and Chemistry. We also had a new subject called The Use of English.
About 10 years ago I started a few “Brain Teaser” threads on this forum. One or two people complained that many of the so-called Brain Teasers were no more than A-Level maths dressed up. That was true of a few teasers, but most were real teasers, especially the ones like “The Ladder” posted by Bam and also the one about the maximum number of 1cm diameter spheres that can be packed into a 10x5x5 cm box.
Any way, never mind Brains or Teasers, I guess one or two other Forumites are also looking back 50 years and would be delighted to tease their brains with calculus, probability, spherical geometry, geometric progressions, Newton’s Laws of Motion ……………………….no ? Then probably best if you drink your weekly 21 units tonight and wake up in the Music Room tomorrow to recover from the nightmare !
First one to follow shortly, and please, please add your own favourites !!
Hi Simon,
I posted a different one to the one I had in mind, simply because it was easy to produce and doesn't need a diagram.
It looks simple enough........................
TETHERED GOAT
Near a small town in Wiltshire with a big church, the Ex MD of a premium Hifi manufacturer named Paul supplements his meagre retirement pension with part-time farming. Paul owns a circular field of grass of 100m radius and he also owns a very hungry goat from which he derives milk for cheese. Paul wishes to tie the goat to a post on the circumference of the field and wants to manage the growth of the grass so the goat doesn't decimate it too quickly. He therefore decides to use just enough rope so that the goat can only eat 50% of the grass.
What must the length of the rope be? Assume the distance between the jaws of the goat and the point of attachment of the rope is insignificant.
The whole thing fails as goats are browsers not grazers!
Ahh clearer and more taxing.. Let me ponder..
TETHERED GOAT
Near a small town in Wiltshire with a big church, the Ex MD of a premium Hifi manufacturer named Paul supplements his meagre retirement pension with part-time farming. Paul owns a circular field of grass of 100m radius and he also owns a very hungry goat from which he derives milk for cheese. Paul wishes to tie the goat to a post on the circumference of the field and wants to manage the growth of the grass so the goat doesn't decimate it too quickly. He therefore decides to use just enough rope so that the goat can only eat 50% of the grass.
What must the length of the rope be? Assume the distance between the jaws of the goat and the point of attachment of the rope is insignificant.
The whole thing fails as goats are browsers not grazers!
........ and if you've ever reared goats you would know that rope is on the menu.
TETHERED GOAT
Near a small town in Wiltshire with a big church, the Ex MD of a premium Hifi manufacturer named Paul supplements his meagre retirement pension with part-time farming. Paul owns a circular field of grass of 100m radius and he also owns a very hungry goat from which he derives milk for cheese. Paul wishes to tie the goat to a post on the circumference of the field and wants to manage the growth of the grass so the goat doesn't decimate it too quickly. He therefore decides to use just enough rope so that the goat can only eat 50% of the grass.
What must the length of the rope be? Assume the distance between the jaws of the goat and the point of attachment of the rope is insignificant.
The whole thing fails as goats are browsers not grazers!
........ and if you've ever reared goats you would know that rope is on the menu.
made me smile Mike !
Before I forget...........the circumference of an ellipse "c"................to save you looking it up in wikipedia
C = 2Пa [1 – (1/2)^2(e^2) – (1.3/2.4)^2((e^4)/3) – (1.3.5/2.4.6)^2((e^6)/5) …..]
a = semi-mjor axis
e = eccentricity
..........so can we get a bigger lorry park ?
Are you expecting a proof that zero eccentricity is a local minimum (an ellipse with small eccentricity has a larger length to volume ratio than a circle) or a global minimum (an ellipse of any non zero eccentricity has a larger length to volume ratio)?
Simon, Peter, anybody else who is trying,
Apologies for not posting earlier, but i've been working late and Mrs D is away in Canada so I't always a bit late at night that I get to look in here and only for a short while !!
Yes, this goat thing looks simple, doesn't it !
I blame Simon He asked for "today's" puzzle and this was the easiest one that I had to hand (from a ready-to-post point of view !)
I still keep running winky's "more than two squares" across my mind...........
Don, put us out of our misery.. how do you solve the tethered goat puzzle?
Use an electric fence across the diameter rather than using a rope!
Actually that probably won't work, knowing goats they'd just jump over it.
Don, put us out of our misery.. how do you solve the tethered goat puzzle?
Use an electric fence across the diameter rather than using a rope!
Actually that probably won't work, knowing goats they'd just jump over it.
Sell the goat to the Ja Curry Goat Cook-House & buy a grass cutter
Well Don
Congratulations for keeping us all on tenterhooks !!
I have been thinking about the problem a little more - but sadly am not really any nearer the right answer. I am 100% confident that the rope length lies between 100 metres and 100 * SQRT 2 metres. Just need to tie it down a little better.
As a first approximation I get SQRT (100^2 + 50^2) which is 111 metres.
I am pretty sure my first approach would work (ie determining the area of a sector + 2 segments) - but I clearly made an error somewhere in my calculations. And in any case having to use an iterative method is pretty close to cheating ! A desperate man clutching at straws.
I look forward to finding out what I have overlooked.
Regards,
Peter
Peter, Simon,
There is no unique solution, ie no directly solvable formula such as y = ax + b
You look at the geometry of intersecting circles and derive an equality, make a stab at the solution, then a series of iterations to get a progressively more accurate answer.
I'll dig out my solution, for tomorrow, if possible...................unless you get there first !
On re-reading this post, I feel I need to clarify things.
There is an equation, (derived through geometry)
It can be expressed in a number of different ways.
It contains only one unknown.
That unknown is "L" ie the length of the rope.
The equation can't be manipulated to make "L" the identity.
The only way to solve the equation is by iteration
ie you "guess" a value of L
put it into the equation and evaluate the result
If the result is Zero, you have found the correct value of L
If not, then try a better guess at L depending on whether the first output was > 0 or <0
Keep on trying better guesses until you have enough decimals to satisfy your mind
[its an "L" of a process ]
The answer to the equation is 1.15 and a lot more decimals
Multiply !by 100 and you have 115m (plus as many decimals as you are prepared to iterate - I think I stopped at eight decimals)
I used Excel to help with the iterations, but its a manual iteration process.
Once we have sorted out the equation, i'll provide a copy of my spreadsheet to anyone who requests it.
Hopefully, somebody will then produce an automatic (goal-seeker ?) version of the spreadsheet.
Hope this helps and inspires a solution.
Cheers
Don
Don
Thanks for partially putting me out of my misery !
I had another look at my equation set and realised I had made a silly mistake (confusing my thetas and 2 thetas !). Bottom line is that when I re-run my numbers I get a rope length of 115.9 metres. So I probably still haven't got it right.
Congratulations on a great question that really had me racking my brain to remember: Area of sector; area of segment; relationship for Sin 2x.
That said, it would have been a hell of a lot easier if you had tethered the goat in the centre of the field !
Regards,
Peter
Hi Peter, & Simon
Peter, your 115.9m is probably correct. I'm at work at present so don't have my answer to hand. My recollection is that the next decimals gave 1.1587 or 1.1578 so your "rounding off" to 1.159 looks good to me.
The equation, in whichever form you see it, doesn't look pretty. Its only the initial concept that looks pretty, such is life !
Perhaps you would like to share your equation with Simon and help put him out of his misery
Well Don
Congratulations for keeping us all on tenterhooks !!
I have been thinking about the problem a little more - but sadly am not really any nearer the right answer. I am 100% confident that the rope length lies between 100 metres and 100 * SQRT 2 metres. Just need to tie it down a little better.
As a first approximation I get SQRT (100^2 + 50^2) which is 111 metres.
I am pretty sure my first approach would work (ie determining the area of a sector + 2 segments) - but I clearly made an error somewhere in my calculations. And in any case having to use an iterative method is pretty close to cheating ! A desperate man clutching at straws.
I look forward to finding out what I have overlooked.
Regards,
Peter
Peter, Simon,
There is no unique solution, ie no directly solvable formula such as y = ax + b
You look at the geometry of intersecting circles and derive an equality, make a stab at the solution, then a series of iterations to get a progressively more accurate answer.
I'll dig out my solution, for tomorrow, if possible...................unless you get there first !
On re-reading this post, I feel I need to clarify things.
There is an equation, (derived through geometry)
It can be expressed in a number of different ways.
It contains only one unknown.
That unknown is "L" ie the length of the rope.
The equation can't be manipulated to make "L" the identity.
The only way to solve the equation is by iteration
ie you "guess" a value of L
put it into the equation and evaluate the result
If the result is Zero, you have found the correct value of L
If not, then try a better guess at L depending on whether the first output was > 0 or <0
Keep on trying better guesses until you have enough decimals to satisfy your mind
[its an "L" of a process ]
The answer to the equation is 1.15 and a lot more decimals
Multiply !by 100 and you have 115m (plus as many decimals as you are prepared to iterate - I think I stopped at eight decimals)
I used Excel to help with the iterations, but its a manual iteration process.
Once we have sorted out the equation, i'll provide a copy of my spreadsheet to anyone who requests it.
Hopefully, somebody will then produce an automatic (goal-seeker ?) version of the spreadsheet.
Hope this helps and inspires a solution.
Cheers
Don
Don, one answer to this is a process called simulated annealing. I don't have time at present to properly describe it, but it's one of the techniques in 'Press et al, "Numerical Recipes" '.
There are other solutions available, such as forming a quadratic or polynomial* and solving for the roots - there'll be at least a real and an imaginary root.
* I think it'll be quadratic, but it could end up being polynomial, I haven't worked it out due to time constraint. If it's quadratic I'd recommend Newton's method, if it's a polynomial I'd recommend a downhill simplex to obtain the solution.