Brain Teaser No 1
Posted by: Don Atkinson on 16 November 2001
THE EXPLORER
An explorer set off on a journey. He walked a mile south, a mile east and a mile north. At this point he was back at his start. Where on earth was his starting point? OK, other than the North Pole, which is pretty obvious, where else could he have started this journey?
Cheers
Don
Posted on: 06 February 2003 by steved
BALLS!
272
STEVE D
272
STEVE D
Posted on: 06 February 2003 by steved
BALLS! (again)
For some reason I cannot edit my answer above. It assumes that the top row of balls cannot exceed the height of the box, ie there is a notional lid!
STEVE D
For some reason I cannot edit my answer above. It assumes that the top row of balls cannot exceed the height of the box, ie there is a notional lid!
STEVE D
Posted on: 06 February 2003 by Matthew T
Yes there is a top.
Posted on: 06 February 2003 by Matthew T
Dobbin,
I reckon you have some sneaky small balls!
Matthew
I reckon you have some sneaky small balls!
Matthew
Posted on: 06 February 2003 by steved
BALLS
Thanks Matthew. Am I right then?
BOATS
I reckon 5 trips if the dog can tow the empty boat, or 9 trips if there has to be a human rowing it.
STEVE D
Thanks Matthew. Am I right then?
BOATS
I reckon 5 trips if the dog can tow the empty boat, or 9 trips if there has to be a human rowing it.
STEVE D
Posted on: 06 February 2003 by Dan M
quote:
Originally posted by hlhopffgarten:
actually, i think the answer is one (1) and meant my jest as a hint. hello? hello? matthew, dan, don, dobbin, et al ... are u out there?
No one need get wet, and the answer is in the single digits but not one! Floating cars are not necessary.
cheers
Dan
Posted on: 06 February 2003 by Dan M
quote:
Originally posted by steved:
I reckon 5 trips if the dog can tow the empty boat, or 9 trips if there has to be a human rowing it.
Oops, didn't notice your answer! Yes, the answer is 9 - well done. Care to share your solution?
cheers,
Dan
Posted on: 06 February 2003 by Dobbin
OK Mine's a fag packet solution where the error grows as the box's volume decreases.
I'm assuming that my available volume is 10r*10r*20r=2000r^3 where r=0.5 (You didn't say you couldn't deform the box!)
By Kepler close packed spheres occupy 74% of the space of their less closed packed cousins.
Sphere's vol is (4/3)*Pi*r^3
Therefore (how do you do dot dot dot here?)
max no of balls which can fit into the available 'volume' is:
.74*2000*(0.5)^3/((4/3)*Pi*r^3)=350 ish
If the box were of infinite dimensions we could of course accommodate an infinite number of infinitely large balls - or 552/500 combinations!
I'm assuming that my available volume is 10r*10r*20r=2000r^3 where r=0.5 (You didn't say you couldn't deform the box!)
By Kepler close packed spheres occupy 74% of the space of their less closed packed cousins.
Sphere's vol is (4/3)*Pi*r^3
Therefore (how do you do dot dot dot here?)
max no of balls which can fit into the available 'volume' is:
.74*2000*(0.5)^3/((4/3)*Pi*r^3)=350 ish
If the box were of infinite dimensions we could of course accommodate an infinite number of infinitely large balls - or 552/500 combinations!
Posted on: 06 February 2003 by Matthew T
And if we turn the box inside out then we can put more balls then I would care to count on the inside, actually it would be really hard to keep them out of the box.
I managed 288.
Lynn, I think you may ahve error in you workings.
Matthew
I managed 288.
Lynn, I think you may ahve error in you workings.
Matthew
Posted on: 06 February 2003 by steved
BALLS
First layer 10x5 = 50 (1cm)
Second layer 9*4 = 36 (extra 0.707cm)
Third layer 10*5 = 50 (extra 0.707cm)
Fourth layer 9*4 = 36 (extra 0.707cm)
Fifth layer 10*5 = 50 (extra 0.707cm)
Sixth layer 10*5 = 50 (extra 1 cm)
TOTAL = 272 (4.828cm) - first answer
I now think we can replace one of the 9*4 layer with 9x5. This layer is 0.866cm.
Therefore the total becomes 281 balls (4.987cm)
STEVE D
First layer 10x5 = 50 (1cm)
Second layer 9*4 = 36 (extra 0.707cm)
Third layer 10*5 = 50 (extra 0.707cm)
Fourth layer 9*4 = 36 (extra 0.707cm)
Fifth layer 10*5 = 50 (extra 0.707cm)
Sixth layer 10*5 = 50 (extra 1 cm)
TOTAL = 272 (4.828cm) - first answer
I now think we can replace one of the 9*4 layer with 9x5. This layer is 0.866cm.
Therefore the total becomes 281 balls (4.987cm)
STEVE D
Posted on: 06 February 2003 by steved
BOATS
CLEVER DOG!
Cross1 Mother in boat, with dog.
Cross2 Dog swims back pulling boat (Family shout to it).
Cross3 Father in boat, dog swims.
Cross4 Same as cross2
Cross5 Son and Daughter in boat, dog swims.
USELESS DOG!
Cross1 Son&Daughter in boat, Daughter gets off
Cross2 Son rows back, and gets off.
Cross3 Father in boat, and gets off.
Cross4 Daughter gets back in boat and rows back.
Cross5 Son& Daughter in boat, Daughter gets off.
Cross6 Son rows back and gets off.
Cross7 Mother in boat and gets off.
Cross8 Daughter rows back, but stays in.
Cross9 Son and Daughter in boat, both get off.
Dog is no use whatsoever. Swims across, chases seagulls, barks loudly, shakes itself and wets entire family.
Now if it had been a Yorkshire Ferrier .......!
STEVE D
CLEVER DOG!
Cross1 Mother in boat, with dog.
Cross2 Dog swims back pulling boat (Family shout to it).
Cross3 Father in boat, dog swims.
Cross4 Same as cross2
Cross5 Son and Daughter in boat, dog swims.
USELESS DOG!
Cross1 Son&Daughter in boat, Daughter gets off
Cross2 Son rows back, and gets off.
Cross3 Father in boat, and gets off.
Cross4 Daughter gets back in boat and rows back.
Cross5 Son& Daughter in boat, Daughter gets off.
Cross6 Son rows back and gets off.
Cross7 Mother in boat and gets off.
Cross8 Daughter rows back, but stays in.
Cross9 Son and Daughter in boat, both get off.
Dog is no use whatsoever. Swims across, chases seagulls, barks loudly, shakes itself and wets entire family.
Now if it had been a Yorkshire Ferrier .......!
STEVE D
Posted on: 06 February 2003 by Dan M
Steve D,
Yes, in fact the swimming dog was a red herring
-Dan
Yes, in fact the swimming dog was a red herring
-Dan
Posted on: 07 February 2003 by Matthew T
It's much worse then that!
Posted on: 07 February 2003 by Don Atkinson
BALLS (& TETRAHEDRONS)
It's much worse then that!
I think we're in danger of loosing our marbles on this one....
Cheers
Don
It's much worse then that!
I think we're in danger of loosing our marbles on this one....
Cheers
Don
Posted on: 09 February 2003 by Don Atkinson
Balls
Matthew recons on 288.
As Lynn says, standing the box on end and using alternating layers of 5x5 and 4x4 we get 271 balls and a height of 9.484.
By substituting one of the 4x4 layers for a 5x5 layer we can increase the number of balls to 280 and the height only goes up to 9.777.
Steved suggested an intermediate layer with a thickness 0.866. I haven't fully visualised this yet, but if it works we could substitute another 4x4 layer, this time with an intermediate 5x4 layer. The balls go up to 284 and the height to 9.936.
Now somewhere I had a box containing at least 290 marbles.........
Cheers
Don
Matthew recons on 288.
As Lynn says, standing the box on end and using alternating layers of 5x5 and 4x4 we get 271 balls and a height of 9.484.
By substituting one of the 4x4 layers for a 5x5 layer we can increase the number of balls to 280 and the height only goes up to 9.777.
Steved suggested an intermediate layer with a thickness 0.866. I haven't fully visualised this yet, but if it works we could substitute another 4x4 layer, this time with an intermediate 5x4 layer. The balls go up to 284 and the height to 9.936.
Now somewhere I had a box containing at least 290 marbles.........
Cheers
Don
Posted on: 09 February 2003 by Don Atkinson
By putting FEWER balls in each layer, the effective layer becomes THINNER.
Perhaps there is an OPTIMUM spacing of balls in each layer such that the aggregated thickness of the layers matches the dimensions of the box.
We have tried aggregating 4x4; 5x5 and 5x4 layers with thicknessess of 0.707; 1.000 and 0.866. Perhaps there are other density/thicknessess that might help??
Cheers
Don
Perhaps there is an OPTIMUM spacing of balls in each layer such that the aggregated thickness of the layers matches the dimensions of the box.
We have tried aggregating 4x4; 5x5 and 5x4 layers with thicknessess of 0.707; 1.000 and 0.866. Perhaps there are other density/thicknessess that might help??
Cheers
Don
Posted on: 09 February 2003 by Don Atkinson
On reflection, I think my last post qualifies me for my MBA.....Master of the Bleeding Apparant.........
Cheers
Don
Cheers
Don
Posted on: 10 February 2003 by Matthew T
A clue
Every wall was touched to get 288 balls in the box and there was no scope for moving them around.
Another clue
FCC but the last C is a little misleading.
Matthew
Every wall was touched to get 288 balls in the box and there was no scope for moving them around.
Another clue
FCC but the last C is a little misleading.
Matthew
Posted on: 10 February 2003 by Don Atkinson
TETRAHEDRONS
would certainly appreciate someone verifying the height of the tetrahedron for me/us.
Lynn, your figure of 0.8165 is absolutely correct (to 4 dp) which is plenty good enough for this forum!!!!
A little bit of geometry on a tetrahedron of side 1.0 generates a figure of h = sqrt(2/3) = sqrt(0.6666667) = 0.816497 etc
Cheers
Don
[This message was edited by Don Atkinson on MONDAY 10 February 2003 at 23:13.]
would certainly appreciate someone verifying the height of the tetrahedron for me/us.
Lynn, your figure of 0.8165 is absolutely correct (to 4 dp) which is plenty good enough for this forum!!!!
A little bit of geometry on a tetrahedron of side 1.0 generates a figure of h = sqrt(2/3) = sqrt(0.6666667) = 0.816497 etc
Cheers
Don
[This message was edited by Don Atkinson on MONDAY 10 February 2003 at 23:13.]
Posted on: 10 February 2003 by Don Atkinson
Another fact, or two...
The most dense arrangement of solid spheres, in an infinite space, is formed by the tetrahedron stack, ie one ball on top of three etc.
As Dobbin rightly pointed out above, in such a stack the ratio of the volume of solids, to the combined volume of solids and entrapped air, is 74%
BUT, there always is a BUT!!!! we aren't dealing with an infinite space, Matthew, bless him, gave us a box!!!!
When filling a given volume (eg a box) it is sometimes more effective to use a COMBINATION of stacking patterns, such as tetrahedrons and pyramids, to make better use of the enclosed space available.
I think, Lynn, this is what you have done.
Cheers
Don
The most dense arrangement of solid spheres, in an infinite space, is formed by the tetrahedron stack, ie one ball on top of three etc.
As Dobbin rightly pointed out above, in such a stack the ratio of the volume of solids, to the combined volume of solids and entrapped air, is 74%
BUT, there always is a BUT!!!! we aren't dealing with an infinite space, Matthew, bless him, gave us a box!!!!
When filling a given volume (eg a box) it is sometimes more effective to use a COMBINATION of stacking patterns, such as tetrahedrons and pyramids, to make better use of the enclosed space available.
I think, Lynn, this is what you have done.
Cheers
Don
Posted on: 10 February 2003 by Don Atkinson
Even more (useless) facts
OK, I'm having to recall distant memory cells here, so these might not be facts......Kepler, as Dobbin says above, sussed that the tetrahedron stack was the most dense way of stacking balls and did the sums to show the 'density' was 74%, BUT, here we go again... BUT, it wasn't until about 1985 (yes, only a few years back) that some genius of a mathematician, PROVED that this was the most dense stack.......
Talk about get a life.....
Cheers
Don
PS Kepler was more famous for doing something about the motion of the planets, IIRC
OK, I'm having to recall distant memory cells here, so these might not be facts......Kepler, as Dobbin says above, sussed that the tetrahedron stack was the most dense way of stacking balls and did the sums to show the 'density' was 74%, BUT, here we go again... BUT, it wasn't until about 1985 (yes, only a few years back) that some genius of a mathematician, PROVED that this was the most dense stack.......
Talk about get a life.....
Cheers
Don
PS Kepler was more famous for doing something about the motion of the planets, IIRC
Posted on: 10 February 2003 by Don Atkinson
Don't fill each layer
I can get (say) 55 balls in a 10x5 layer. If they were tightly packed they would fit into a rectangle 9.928x5.
If I don't pack them tightly, as would be the case in a 10x5 rectangle, then the next 'layer' could sit slightly 'lower' in the box.....
Just a thought....
Cheers
Don
I can get (say) 55 balls in a 10x5 layer. If they were tightly packed they would fit into a rectangle 9.928x5.
If I don't pack them tightly, as would be the case in a 10x5 rectangle, then the next 'layer' could sit slightly 'lower' in the box.....
Just a thought....
Cheers
Don
Posted on: 10 February 2003 by Dan M
quote:
Originally posted by Matthew T:
Another clue
FCC but the last C is a little misleading.
Matthew, this one is lost on me - FCC?
Families with Children from China
Farm Credit Canada
Feline Conservation Center
Florida Christian College
and of course
Federal Communications Commission
-Dan
Posted on: 11 February 2003 by Matthew T
FCC is a acronym used in the area of mateiral science to describe a method of packing (in the case of materials science this is for atoms).
FCC stands for face centred cubic which is one of two ways to get the maximum packing density, the otehr being hexagonal close packing. The obvious advantage of FCC is that you have a very high degree of symmetry.
More clues will follow later.
Matthew
FCC stands for face centred cubic which is one of two ways to get the maximum packing density, the otehr being hexagonal close packing. The obvious advantage of FCC is that you have a very high degree of symmetry.
More clues will follow later.
Matthew
Posted on: 14 February 2003 by Matthew T
Another clue for the weekend.
There are an equal number of balls in each layer, and each layer is identical, alternate layers line up with one another.
More clues next week if this remains unresolved.
Matthew
There are an equal number of balls in each layer, and each layer is identical, alternate layers line up with one another.
More clues next week if this remains unresolved.
Matthew