Brain Teasers at Xmas

Posted by: Don Atkinson on 23 December 2010

Brain Teasers at Xmas

Xmas time - and many of us will be off on holiday or meeting up with friends and family. For those with a bit of time on their hands, I thought I would post a dozen (12 days of Xmas ?) brain teasers to help keep your minds in gear.

All are taken from the Brain Teaser thread from a few years back and credit is given to the original poster where this is other than me. Ken c still posts here, but BAM and Matthew T….??

None of these teasers are “trick” questions, but you will need to read them carefully in order to understand some of them. Discuss at leisure......

I know it’s not as exciting as today’s hi-tech electronic games, but I hope it helps to keep some of you amused over Xmas, and as Ken c would say – “enjoy”

Cheers

Don

Posted on: 01 February 2011 by Don Atkinson

"So y=-1 + sqrt(1+L^2) in your solution."

Yep, same as me.

BAM, who set the original puzzle, gave me 5/10 for elegance !! but it gives answers easily and without the need for log tables, a calculator or a computer. We never did find out what BAM's 10/10 looked like, he left the forum before revealing!

Cheers

Don

Posted on: 01 February 2011 by Don Atkinson

"Have you figured out the value of i^i as I posed earlier?"

ah!, forgot about that one....

possibly....e^(-pi/2) but I can't check it out in Excel !!

Cheers

Don

Posted on: 01 February 2011 by Don Atkinson

ok, I put e^(pi/2) into Excel and it gave 0.20788.

ISTR this figure for i^i but am far from certain

Cheers

Don

Posted on: 01 February 2011 by Ian G.

e^(-pi/2) is right. It entertains me that i^i is a simple real number !

busy with the goat and the crop circles but must stop as it is a school night...

Posted on: 02 February 2011 by Don Atkinson

" It entertains me that i^i is a simple real number !"

Ian, that had never ocurred to me. Intersting !

But I suppose that since i*i is a simple, real number, I don't feel too bad about not previously appreciating that i^i is a real number. However, we passed the limit of my understanding of maths ages ago. I can remember bits and pieces. I can feed info in and get practical reults out, but understand it? ...........whoahhh!!

Cheers

Don

Posted on: 08 February 2011 by Don Atkinson

"busy with the goat and the crop circles"

...you will ned your trancendental techniques for this one.

The "Ladder" has algebraic solutions that provide precise dimensions.

The crop circle, AFAIK, doesn't.........

....so an approximation (let's say 6 decimal places?) will do!

Cheers

Don

Posted on: 09 February 2011 by Ian G.

ok - didn't realise I could search for the answer

How's about a radius for the goat tether of 1.1587285.

Ian

Posted on: 09 February 2011 by Don Atkinson

"ok - didn't realise I could search for the answer "

aahhh! ....."search".....?

Well, I think just looking up the naked answer on the internet is sort of cheating (well, IMHO anyway) and I doubt if that's all that you did!

However, how do we learn? by looking up "how to do things" often using a library or bookshop, or university lecture/notes, or even the internet, then by practising...

So "searching" for an answer is ok, and I have no idea where to draw "the line" between what is morally acceptable in a quiz and what isn't. That's up to each individual !

I have seen several different forms of the equation that can be used to find the answer using numerical analysis.

Your answer is spot on so you must have found a good form of equation - well done !!

Cheers

Don

Posted on: 09 February 2011 by Ian G.

no no I meant I developed an expression for the area and then used a spreadsheet to evaluate it as a function of radius until I found the right value-i.e. I searched my expression for the value not searched the internet for the solution!

Working is all at work now but basically by finding that the intersection of the two circles
occurs at a height of r^2 /2 then integrating the two pieces of area separately one gets a bit of an ugly expression involving two arcsin functions and some surds

I had this expression ages ago but couldn't get an explicit solution so kept searching for a cleverer way of solving it. I still have a feeling a geometrical construction mught be possible but haven't had any luck in finding it

cheers

Ian

Posted on: 09 February 2011 by Don Atkinson

"I still have a feeling a geometrical construction mught be possible"

hmmm. I keep returning to this problem from time to time and have a similar feeling. But I've never managed to derive such a solution and i've never come across one in any text book.

Anyway, i'm glad I mentioned trancendental techniques - you might still have been looking for a unique solution for years............

Cheers

Don

Posted on: 12 February 2011 by Don Atkinson

To limit the goat to eating one third of the grass, the rope would be 0.911598r where r is the radius of the field.

And for two-thirds 1.393007r

I will try to write up my formula for the goat/grass and the Ladder so that we can find them again in the future (assuming the "search" function is restored)

Cheers

Don

Posted on: 12 February 2011 by alainbil

Originally Posted by Ian G.:

Don,

A bit of both, used the lack of a given radius to hint it might not be necessary, but then worked out the volume of the drilled out section (cylindrical core + two identical end cap sections) from the 6inch length constraint. This showed the leftover part is indeed independent of the radius which is neat.

Not sure how one could do it without knowing some calculus though, so is there a cute geometry based shortcut I'm missing?

cheers

Ian

You can compute directly the volume left,

It is given by a simple integral and the result is 4/3 3^3 as it should.

Alain

Posted on: 12 February 2011 by Don Atkinson

"so is there a cute geometry based shortcut I'm missing?"

There isn't any cute geometry based shortcut of which i'm aware, but as Alain says, the integration needed is straight forward.

I'll try to post it, along with my goat/grass and ladder solutions once I get the hang of this new forum format.

Cheers

Don

Posted on: 12 February 2011 by Don Atkinson

Solution to The Ladder

Well, I hope. There should be two attachments - a diagram and the working out.

Bam insisted on a single equation. I prefer using two equations - Number 6 with number 4 providing the value of "y"

Fingers crossed

Don

Posted on: 13 February 2011 by Adam Meredith

The solution I posted to the 12 Coins problem was my father's - with some grunt work done by the Childe Adame.

My father suggested using base3 with 0, 1 and 2 representing the three possible results of a balance test.

Results 000, 111 and 222 are uninformative - leaving 24 useful sequences. It remained only to devise a weighing strategy.

I think I was sent off to work this out and came up with the table through slog rather any underlying logic. It works but I always wondered if it could have been given a slight, spurious elegance through a different choice of weighings.

Here's a, possibly very slight, problem he set for me years earlier. I had the mathematics for the solution then but am now mainly looking for confirmation of my moment of inspiration (cheat). The maths has long atrophied through lack of beatings.

A hemisphere, 1 metre in diameter, is filled with water and rests with its 'equator' on a flat surface. A hole is made through the 'pole' (do these 'terms' apply to hemispheres?).

What force must one exert on the top, avoiding the hole, to prevent water seeping out?

I hope I have remembered this correctly.

I was going to share an elegant solution to Nim - based on binary - but discover (damn the internet) that it is well known and available to fools and donkeys.

What value education and the spongy mind?

Posted on: 13 February 2011 by winkyincanada

Assuming a massless hemisphere, the needed force applied equals the area of the circle defined by the equator times the water pressure at the table level. So: pi x (0.5)^2 x 1(density) x 9.81("g") x 0.5(height). (I'd have to think about the units here, but the form is right, I think)

After thinking about complex integrals of force vectors and pressure applying to the surface of the hemisphere, I thought the other round and considered "how hard must you push to hold the table onto the bottom of the hemisphere" ? The force at the table interface actually also includes the force needed to hold everything up, but that is taken care of by the table pushing against gravity - you don't need to apply that force again to the hemisphere. The hole in the pole eliminates atmospheric pressure from net consideration, as it means that the atmospheric pressure on the hemisphere is balanced by the same increase in absolute pressure in the water.

Posted on: 13 February 2011 by Don Atkinson

Will give this one some thought whilst out at pictures this evening (Mrs D wants to see something, don't know what, but it helps to keep things sweet).

My initial thoughts are that you don't need any force (lots of assumptions) other than "just enough" to ensure a perfect seal between the bowl and the table.

As I say, i'll give it a bit more thought.

Nice teaser Adam.

Cheers

Don

Posted on: 14 February 2011 by Don Atkinson

Well, the picture we went to see was a light-hearted romantic comedy called "just go with it"

So I am going to "just go with it" ie my initial thoughts.....ie, no force (other than that required to ensure an effective seal between bowl (hemisphere) and table (equator))

Different point of view to winky.

Anybody agree? or have I completely lost it!!!!

Cheers

Don

Posted on: 14 February 2011 by Don Atkinson

Originally Posted by winkyincanada:

The hemispherical bowl would contain approximately quarter ton (250kg) water. A force of quater ton is about the same as 2,500 Newtons.

I have assumed that pi = 3 and g =10N/s/s which is good enough to get an order of magnitude.

So you would need a fairly strong table on which to do the experiment !!

Cheers

Don

Posted on: 14 February 2011 by Adam Meredith

Originally Posted by Don Atkinson:

So you would need a fairly strong table on which to do the experiment !!

It's all in the mind - there is no table, there is no water.

I'm feeling doubtful I could answer this now but I think my initial insight still stands - I just can't remember what I then did with it. Hρg dynes rings some bells.

Posted on: 16 February 2011 by Don Atkinson

Originally Posted by Adam Meredith:

It's all in the mind - there is no table, there is no water.

Ah, not in my shed. In there, is real water. A real bowl. And a large bucket and syphon.

.......And I am trying to amas at least 131Kg of load (half of what winky was thinking about).....

on the other hand, perhaps i'll just give it a bit more thought.

Cheers

Don

Posted on: 19 February 2011 by Don Atkinson

Originally Posted by Don Atkinson:

.......And I am trying to amas at least 131Kg of load (half of what winky was thinking about).....

on the other hand, perhaps i'll just give it a bit more thought.

hhhhmmmm.....I've given it a bit more thought and.......

I still get 131kg

ok, 125xPi/3kg........but..

rather than set up an integration, I have created a little spread sheet to estimate the lifting pressure on thin slices of the hemisphere - well, that's all integration is after all.

I have assumed that water weighs 1000kg/cu.m
I have also assumed that horizontal slices through the hemisphere have equal surface area - this took a bit of figuring, but i'm pretty sure its right. (*)
I then estimated the pressure at the depth of each slice (1kg/sqm per mm depth). this gave the pressure on each slice, normal to the surface of the hemisphere (pressure equal in all directions)
Pressure on slice x area of slice gives load normal to surface of hemisphere
Estimated vertical component of load based on vectors/geometry
Summed vertical loads on aech slice to give total uplift
total uplift = load required to prevent seepage

Based on 100 slices, spread sheet gave just under 131kg

A bit of lateral thinking re geometry of sphere and cylinder suggested 125xPi/3

Anybody else got any thoughts?

Cheers

Don

(*) The hemisphere is hollow, so the surface area i'm referring to is the inner surface of the hemisphere associated with the each slice.

Posted on: 06 March 2011 by Don Atkinson

Well, a couple of weeks have passed and I haven't had any new ideas regarding Adam's inverted bowl.

I'm not even sure about my basic concept. The arithmetic for my concept is right AFICT - I have done it three different ways. But that would be rather pointless if the basic concept is wrong !

So, at present I have a mental block. I can't convince myself that my concept is definitely right, but I can't come up with any other concept either.

Any thoughts ?

Cheers

Don

Posted on: 06 March 2011 by Adam Meredith

I am having serious doubts but I imagined a cylinder of same height and diameter as the hemisphere sitting around it. I filled that with imaginary water to the top - pressure equalised.

That seemed to suggest the answer.

Posted on: 07 March 2011 by Ian G.

I think Adam's last post nails it.

One can imagine adding water to his imaginary cylinder as the means of exerting the force on the hemisphere. To get the pressure balanced you need to fill the cylinder up to the level of the top of the hemisphere. You can calculate the mass of the water you need to do this from the volume difference

Vol of hemisphere 2*pi*(0.5)^3 / 3 = pi/12
Vol of (imaginary) cylinder 0.5*pi*(0.5)^2 = pi/8

Difference in volume = pi/24.

The mass of that pi/24 water = density*volume = 1000*pi/24 = 130.9 kg (seem familiar to anyone :-) )

So the downward force exerted by it on the hemisphere is mg = 1292N.

Ian