Maths problem just for fun

[Martin ConnellyParticipant @martinconnelly55370]

There is one other method of solving the resistor cube. It involves splitting two of the resistors into two in parallel 2ohm resistors. Then the cube can be split into two mirror images with simple series and parallel paths to calculate. Calculate one side then half the value (parallel sum of the two sides) to get the final answer.

Martin C

[Nicholas FarrParticipant @nicholasfarr14254]

Hi, while having a cuppa, I resolved the ladder problem by a different solution, and you could say being Model Engineering, I did to scale. The ratio of 10' and 2' is five, therefore 5" is equal to 10' and 1" is equal to 2'. Put that together and I came up with this.

So being this is a scale of 1 / 24th, 24 x 4.852 = 116.448" then divide that by 12 and I get 9.704 Feet, now being that the wall is probably not square to the floor, I reckon I'm not Farr out.

Regards Nick.

 

Edited By Nicholas Farr on 17/12/2021 14:18:04

[DMBParticipant @dmb]

Nick,

I don't think so unless I'm missing something.

The rule appears to touch the corner of the cube at the 1" mark, and the 5" mark touches the caliper point. 5 – 1 = 4. ??????

John

[JAParticipant @ja]

I had another attempt, this time using coordinate geometry. The manipulation of the numbers is easier but you will alway end up with a quartic equation. These are very difficult to solve except by numerical methods.

I really can't resist setting a simple problem: Farmer Giles has a fenced, grass circular paddock, 50 yards in diameter. He wants to tether his goat to the fence so that it eats half the grass. How long should the tether be? He does not really care about the RSPCA so the tether is attached to the goat's mouth.

[Nicholas FarrParticipant @nicholasfarr14254]

Posted by DMB on 17/12/2021 15:07:42:

Nick,

I don't think so unless I'm missing something.

The rule appears to touch the corner of the cube at the 1" mark, and the 5" mark touches the caliper point. 5 – 1 = 4. ??????

John

Hi John, the corner of the cube was at 1 – 3/128" the best as I could tell, when I made the measurement, however I had to rotate the mat a quarter of a turn because the display was in the shadow and didn't show on the first photo I took, so the cube or the rule may have moved slightly. My attempt was a light hearted example, as the thread had "just for fun" in the title and my answer wasn't meant to be taken as the correct answer, but I don't think it is very far off.

Regards Nick.

Edited By Nicholas Farr on 17/12/2021 18:26:01

[Martin ConnellyParticipant @martinconnelly55370]

Duncan and Rob, you are correct it does come down to a quartic equation. Using similar triangles and Pythagoras' equation if you take the base of the ladder to the near corner of the box as y you can show that the top of the ladder to the near corner of the box is 4/y. The quartic equation is y^4+4y^3-92y^2+16y+16=0. Luckily I still have my trusty Texas Instruments TI-68 which solves quartic equations and gives the following values for y: 0.521036706, 7.677002321, -0.33724556, -11.86079346

The two negative numbers do not work as the top of the ladder will be 2+11.86 above the ground and requires a ladder 14.06 feet in length. The other two give the answers that work. x equals 2+7.677002321=9.677002321 or x equals 2+0.521036706=2.521036706, the latter being a bit silly.

Somewhere I have got a spreadsheet I did that solves polynomials using the matrix method that I put together just to see how well it worked.

Correction, solves simultaneous equations, not polynomials,

Martin C

Edited By Martin Connelly on 17/12/2021 19:10:10

Edited By Martin Connelly on 17/12/2021 19:46:52

[Grindstone CowboyParticipant @grindstonecowboy]

Posted by JA on 17/12/2021 15:12:00:

….How long should the tether be? He does not really care about the RSPCA so the tether is attached to the goat's mouth.

 

About eleven foot nine and a half inches, I think?

Rob

Edit – of course it isn't, that would be too simple. I'll go and think again

Edited By Grindstone Cowboy on 17/12/2021 19:48:35

[John HaineParticipant @johnhaine32865]

Oh well then. Iron has a density of about 8 gm/cm^3, so assume a 10 cm cube will weigh 8 Kg. If three holes are drilled right through, one along each axis of the cube perpendicular to the faces and meeting at the exact centre, what weight of iron remains?

[Pete RimmerParticipant @peterimmer30576]

Posted by John Haine on 17/12/2021 22:51:22:

Oh well then. Iron has a density of about 8 gm/cm^3, so assume a 10 cm cube will weigh 8 Kg. If three holes are drilled right through, one along each axis of the cube perpendicular to the faces and meeting at the exact centre, what weight of iron remains?

Surely that depends on the size of the holes?

[John HaineParticipant @johnhaine32865]

Oops! Posting too late at night. 5cm diameter in a cube 10 x 10 x 10 cm.

[pgk pgkParticipant @pgkpgk17461]

Posted by John Haine on 18/12/2021 08:30:47:

Oops! Posting too late at night. 5cm diameter in a cube 10 x 10 x 10 cm.

Weight of iron is the same. Weight of cube is another matter which I may ponder later…gonna be a pig to figure out those intersecting curved bits in the middle..
(still smarting over my earlier stupidity of squaring x+2 to x^2+4. Doh!)

pgk

[Martin ConnellyParticipant @martinconnelly55370]

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

[JAParticipant @ja]

Martin

I remember trying to solve the problem about 45 years. It was a Christmas problem set at work. Using nasty integral calculus I did get an answer. It took hours and many sheets of paper

With a spread sheet it is not difficult, no Pythagoras and just a right angled triangle (Cosine). I made it 28 yards and 35 inches but that is approximate because of the method. On the computer it takes less than 10 minutes of setting up the spread sheet.

JA

[duncan webster 1Participant @duncanwebster1]

goat problem: I nearly got the answer but then the goat bit through the rope and ate all the grass

[duncan webster 1Participant @duncanwebster1]

But seriously, can I bring the attention of sums fans to Smath Studio, which is a free version of MathCad. Using this solving intractable equations becomes easy, example. The only bit to note is using the Boolean = sign in the line beginning ans:= ……..

Edited By duncan webster on 18/12/2021 21:38:06

[Gary WoodingParticipant @garywooding25363]

Posted by pgk pgk on 18/12/2021 08:56:21:

Posted by John Haine on 18/12/2021 08:30:47:

Oops! Posting too late at night. 5cm diameter in a cube 10 x 10 x 10 cm.

Weight of iron is the same. Weight of cube is another matter which I may ponder later…gonna be a pig to figure out those intersecting curved bits in the middle..
(still smarting over my earlier stupidity of squaring x+2 to x^2+4. Doh!)

pgk

4.6721Kg

[John HaineParticipant @johnhaine32865]

Oh gosh! Now I'll have to work out the answer again!

[John HaineParticipant @johnhaine32865]

I did the goat problem again – when I first did it years back I used an integral but this time the method outlined above to get a complicated equation involving an arcsine then got Excel to solve it The answer is 1.158728517 times the radius of the field (I think). When I first saw this it was in an "elementary" puzzle book but I don't think it has a quick and neat solution.

Unlike this one – a hole 1 inch long is drilled right through the centre of a sphere. What volume of sphere remains?

Edited By John Haine on 19/12/2021 13:57:45

[Martin ConnellyParticipant @martinconnelly55370]

I got 4.701kg for the three holes in the block problem. The hardest part was visualising what was left in the middle. What is taken out of the centre is a 50mm cube less eight pointy corners with the three curves on them and six Ø50 cylinders 25mm long. The pointy corners have a 7.322mm cube in them. The three times eight points were the problem that then needs calculating. Once I figured that these points had a square cross section it is relatively easy to do an approximate volume by slicing them at 1mm spacing from the point (0.1 to 17.1mm since 17.1mm is 0.5mm less than the point where the internal cube is in these corners) and calculating the size of the square (Pythagoras again) and summing the areas of these squares.

The numbers (in cubic centimetres) come out as follows:

Original volume 1000

Six cylinders to remove Ø5 x 2.5 = 294.524

Central 50mm cube to remove = 125

Eight 7.322mm cubes to put back in 3.14

24 points to put back in 3.998

Total = 1000-294.524-125+3.14+3.998=587.614

The weight is then 8*587.614=4700.912g, 4.7kg

Martin C

[John HaineParticipant @johnhaine32865]

Very good Martin – the exact answer to 6 decimal places is 4.701825 kg. I first saw this problem before the web was common and spent a couple of happy hours at work puzzling it out with a colleague – much better than working! The exact formula and its derivation is here.

[JasonBModerator @jasonb]

Sorry, I've given up on the maths questions in favour of actually removing iron from a block Anyone want to calculate what's left here?

[Nicholas FarrParticipant @nicholasfarr14254]

Hi JasonB, a lighter piece of iron.

Regards Nick.

[TSHParticipant @tsh73987]

Posted by John Haine on 19/12/2021 13:57:14:

I did the goat problem again – when I first did it years back I used an integral but this time the method outlined above to get a complicated equation involving an arcsine then got Excel to solve it The answer is 1.158728517 times the radius of the field (I think). When I first saw this it was in an "elementary" puzzle book but I don't think it has a quick and neat solution.

Unlike this one – a hole 1 inch long is drilled right through the centre of a sphere. What volume of sphere remains?

Edited By John Haine on 19/12/2021 13:57:45

Nice one! Taking the question at face value, the implication is that the answer depends only on the length of the hole. Then the answer is clearly the volume of a 1 inch diameter sphere (the hole would have a vanishingly small diameter). So pi/6 cu inches. A direct calculation confirms that the answer is indeed independent of everything except the length of the hole and is pi.D^3 /6 for a sphere of diameter D.

Trevor

[Martin ConnellyParticipant @martinconnelly55370]

The hole through a sphere was one I expected to appear here. I remember the solution as stated above but at the time I first came across it the solution was a bit above my maths education level.

Martin C

[TSHParticipant @tsh73987]

Posted by TSH on 20/12/2021 09:46:25:

Posted by John Haine on 19/12/2021 13:57:14:

I did the goat problem again – when I first did it years back I used an integral but this time the method outlined above to get a complicated equation involving an arcsine then got Excel to solve it The answer is 1.158728517 times the radius of the field (I think). When I first saw this it was in an "elementary" puzzle book but I don't think it has a quick and neat solution.

Unlike this one – a hole 1 inch long is drilled right through the centre of a sphere. What volume of sphere remains?

Edited By John Haine on 19/12/2021 13:57:45

Nice one! Taking the question at face value, the implication is that the answer depends only on the length of the hole. Then the answer is clearly the volume of a 1 inch diameter sphere (the hole would have a vanishingly small diameter). So pi/6 cu inches. A direct calculation confirms that the answer is indeed independent of everything except the length of the hole and is pi.D^3 /6 for a sphere of diameter D.

Trevor

Oops. What I meant to say in my last sentence is that the volume remaining is pi.L^3 /6 for a hole of length L.

Best wishes for a safe Christmas and New Year,

Trevor

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