Maths problem just for fun

[JAParticipant @ja]

Goat in field – my method:

O is centre of paddock

AP is the fence with P as the tether point.

I halfed the "model".

The area of a ribbon of grass is R x T x OPA (angle in radians). The first ribbon starts when R = T/2 (very close to P) and its area is the start of the total area. The next ribbon has R = R (previous) + T and the resulting area is added to the previous total area. And so on.

Obviously there is an error because the end of the ribbon does not match the fence. However if T is small this is small. Some will recognise this as integral calculus.

This only calculates a general area, it does not know when to stop. However if a spreadsheet is used (the obvious way of doing the above) you just run the sheet until the required area is reached.

The angle OPA can be easily obtained from trigonometry, the triangle OPA is isosceles and can be split in two to give right angled triangles. Therefore Cos(OPA) = AP / (2 x OP) = length of tether / diameter of paddock. This is very easily incorporated in to the spreadsheet.

Quite enough of this.

I wish everyone a Happy Christmas and just hope next year is better than the last two.

JA

[SillyOldDufferModerator @sillyoldduffer]

Posted by Martin Connelly on 18/12/2021 13:10:33:

Rob, farmer Giles needs a tether at least 25 yards long just to get the goat to the centre of the field and that is clearly not going to let it graze half the field. The actual required length is a smidgen under 29 yards. You have to get a formula for the sector that the tether gives access to then a formula for the two edge segments that need adding to this sector to give the total grazed area. It's just a bit messy needing pi for areas, inverse cosine (or inverse sine depending on which angle you chose to calculate) for the sector angle and Pythagoras as well. Then solve for this formula equalling half the total area of the field.

Martin C

Far too busy to be messing with maths problems, but here's a way of solving it without complicated maths. Needs a computer though!

I used a Monte Carlo method. The idea is to generate a random list of points inside a rectangle that contains a circle representing the paddock. The count of points falling inside the paddock circle is proportional to its area.

Then a second circle is drawn with its centre somewhere on the paddock circle's boundary. The point count inside it is proportional to the circular area a tethered goat can roam.

Here's the fiendish bit: the number of points inside both circles is proportional to area the tethered goat can reach inside the paddock, i.e. the intersection.

My simple program generated a thousand random points inside the rectangle, of which 324 fell inside the paddock. The length of the goat's tether (r) was increased a yard at a time producing the following table of intersection counts:

The result is In broad agreement with Martin's calculation. r=30 produces an intersection count of about half the paddock count.

The accuracy of a Monte Carlo method depends mainly on the number of random points generated. I used a thousand, which could be done manually by dropping Smarties inside a big frame, but millions would be better. Computers are good at counting big numbers quickly.

My implementation is flawed. To keep it simple, the program steps in whole yards across the target, which is crude. To do better the algorithm should home in on a decimal value by progressively reducing the increment as it approaches the correct number.

The data collected by Monte Carlo for 'r' could be used to establish the underlying goat formula by applying an automated regression analysis. So much is easy when you know how, which I usually don't!!!

Dave

[Grindstone CowboyParticipant @grindstonecowboy]

All of the above just reminds why I failed mathematics…

Merry Christmas to all.

Rob

[lee websterParticipant @leewebster72680]

I don't know if this conumdrum is of interest, but I have nothing else to do but annoy you with it!

Three students sharing a flat decide to club together and buy a television. They got one for £30 at the local cheap tv shop, that's £10 each. Later that day the manager realised the tele was actually in a sale for £25. He gave the £5 to one of the shop workers with instructions to give it to the students. On the way to the flat the worker decided to keep £2 and give the students £1 each, which he did. This means that the students each paid £9 towards the tv and the shop worker kept £2. Add the students total of 3 times £9 = £27 and the shop workers £2 and you get £29. Where is the other pound?

[Nicholas FarrParticipant @nicholasfarr14254]

Hi Lee, the students paid £30.00 – £5.00 returned = £25.00 / 3 = £8.33333> + £1.00 = £9.3333> x 3 = £28.00 + £2.00 = £30.00 original payment.

Regards Nick.

[pgk pgkParticipant @pgkpgk17461]

The real question here is what sort of TV was it? If the students were economy minded then extra for a battery run TV and they could have watched it on parent's licence..otherwise they're stuck with needing to front another £53 each. Of course if they were smart they'd have hard-wired it to a nicked battery backup system from the uni computer lab even though it had originally been a mains unit…

[GreensandsParticipant @greensands]

.I have been playing around with the Ladder against the Wall problem and by following the link given for the solution to the problem using the Equation to the a Straight Line approach and solving the resulting quartic equation using an Excel utility you arrive at an answer of X=9.677. When this value is put back into an ACAD construction the resulting length of the ladder comes out at 10.0000 which is good enough for me. However, I would like to be able to generate the quartic equation from the three quadratic equations established by JA in the previous posting. Has anyone been able to do this?

[VersabossParticipant @versaboss]

Maybe you find this one interesting too, I found it in a magazine some days ago, and to be honest I could not find the answer, except perhaps if I would unearth some long gone geometry formulas. But from the type of mag I suspect one should find the answer 'just by thinking hard' – if you know what I mean.

The question is: how much larger is the area of the outer triangle than the area of the small one.
The triangles are equilateral, should that not be clear

Ignore the yellow line, that's an artefact of my CAD drawing.

[RobCoxParticipant @robcox]

The small triangle is 1/4 the area of the large one I believe.

The circle contacts the large triangle at the midpoints of the lines, so the triangle formed by the contact points has sides half as long as the large triangle. The small triangle must be identical to the contact point triangle being formed in the same circle. Half the linear dimension gives 1/4 the area.

[Nicholas FarrParticipant @nicholasfarr14254]

Hi, I could see how four of the small triangles would fit inside the large triangle, the give away is the four small circles and the line from the centre to the right hand side of the large triangle, therefore if you move the small one up to the top of the large one, you could flip it vertically and the top would touch the bottom of the large one, you can then get a small triangle in each of the bottom two corners of the large one, thus 1 / 4.

Regards Nick.

[duncan webster 1Participant @duncanwebster1]

four

edit Beaten to it I see

Edited By duncan webster on 30/12/2021 18:48:45

[Nigel Graham 2Participant @nigelgraham2]

I take my grease-top off to all you mathematicians for being able to solve those problems!

The secret appears to be knowing how to derive the methods – the rest is standard (though often very hard) algebra and arithmetic.

A couple of days ago Gary Wooding challenged me to solve by CAD a purely-geometrical version of the Ladder, giving me the length of the "ladder" and the size of the obstruction it has to just touch.

After some experimenting it proved quite easy, in TurboCAD, by incremental adjustments and re-drawing, to close the problem to 4 decimal places. His point really was that other CAD packages can solve it almost directly by a tool I don't think is in TurboCAD.

I had though realised I could not have solved it mathematically – and proved that by looking at it here.

'

I worked for a company whose research and design was intensely mathematical. One day I found a report on an experiment, one of whose eye-watering equations had no fewer than 5 topped-and-tailed integral signs at its head. The whole page looked like a stylised swannery.

They made me wonder, not how to solved such hard sums (analysing readings from an experiment), but how in general one assesses which techniques are needed to solve them. After all, when we learnt or were taught, Arithmetic and later Mathematics at school we were given set ways to answer set questions. Not to derive the ways, ways to find the right maths to make sense of a crate of very non-linear numbers. A computer can make the algebraic swarf but someone still needs know which algebra to translate into computerese first.

Ironically, the test-rig in that experiment, about shock-vibrations in certain materials, was merely a wooden metre rule with a carpenter's hammer fastened to one end, penduluming about a screw at the other.

It came back to me in a discussion with one of my nephews. He said airily you don't need learn maths because it's all in a computer or calculator. I think I convinced him that the electronics only does the arithmetic: you need tell it what arithmetic to do, so need understand the maths!

Similarly with the puzzles above: it's not enough to know how to "do" advanced geometry, trigonometry, algebra and aritthmetic. You need know how to select and combine the right tools from those areas; just as need know which cutting-tool at what speed for what metal.

'

This delightful equation is printed round a tea-mug I found, but assume came from the gift-shop of a museum dedicated to the mathematician George Green. it is evidently quoted from his works. Read the question-marks as Integral signs: it's a .jpg image but something somewhere turned them into " ? " signs.

I have no idea what Green was showing by that Eqn.3' , though I can hazard a vague guess at the topic and no doubt someone here would understand the whole treatise right through …..

….. But has anyone identified yet that strange "circular slide rule " of peculiar numbers and fractions, that had us all foxed nearly a year ago?

[Martin ConnellyParticipant @martinconnelly55370]

Greensands, this is from my post on 17 Dec. The quartic equation I came up with and which I used my old calculator to solve for the ladder against the wall.

y^4+4y^3-92y^2+16y+16=0

Martin C

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