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Posted by snow (Member # 5804) on :
 
I'm quite bored, so im asking for all the clever folk out there to contribute some brain teasers/random questions about math, physics, logic puzzels, anything.

First I'll contribute my own (no cheating by looking up the answers)

1. If you have a really big box of spherical marbles, whats the largest possible percentage of the space the marbles could take up?

2. Consider a soccer ball shape (truncated icosahedron), what are the angles between the pentagonal-hexagonal intersections and between the hexagonal-hexagonal intersectional, and what is the volume in terms of a unit side length.

3. Imagine there is a prize behind one of three doors, and you choose one door. Afterwards, the game show host reveals one of the doors you did not choose, it is empty. He gives you the option to change doors. Should you switch or stay with the door you orginally choose, or does it even matter?

Enjoy, Id like to see some good ones myself.
 

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Posted by Hobbes (Member # 433) on :
 
3) Switch

Hobbes
 

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Posted by snow (Member # 5804) on :
 
yea, probably everyone here has heard of the third one, if they've read the thread that had it in the past.
 

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Posted by Teshi (Member # 5024) on :
 
I don't know number three...
 

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Posted by xnera (Member # 187) on :
 
You asked for it. Solve this:

quote:

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1. There are five houses.
2. Each house has its own unique color.
3. All house owners are of different nationalities.
4. They all have different pets.
5. They all drink different drinks.
6. They all smoke different cigarettes.
7. The Englishman lives in the red house.
8. The Swede has a dog.
9. The Dane drinks tea.
10. The green house is on the left side of the white house.
11. In the green house they drink coffee.
12. The man who smokes Pall Mall has birds.
13. In the yellow house they smoke Dunhill.
14. In the middle house they drink milk.
15. The Norwegian lives in the first house.
16. The man who smokes Blend lives in the house next to the house with cats.
17. In the house next to the house where they have a horse, they smoke Dunhill.
18. The man who smokes Blue Master drinks beer.
19. The German smokes Prince.
20. The Norwegian lives next to the blue house.
21. They drink water in the house that lays next to the house where they smoke Blend.

So, who owns the zebra?

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Posted by rivka (Member # 4859) on :
 
Actually, you should only switch if the host KNEW which ones were empty. If he chose at random, and happened to open an empty one, then it doesn't matter whether you switch or not.
 

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Posted by Dragon (Member # 3670) on :
 
We've been having fun with these lately...

Fill in the rest of the sequence:

1) O T T F F

2) J F M

3) S M H
 

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Posted by kaydubb (Member # 5953) on :
 
My guess for the houses/pets/cigarettes etc puzzle is that the German has the zebra. However, I think my logic/matrix skills are lacking because I ended up with some people in houses with incorrect drinks/pets..
 

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Posted by xnera (Member # 187) on :
 
Dragon:

1) S S E N T
2) A M J J A S O N D
3) D W M Y
 

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Posted by butterfly (Member # 5898) on :
 
The German has the zebra?
 

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Posted by xnera (Member # 187) on :
 
Yup, the German has the zebra.

I got this logic puzzle five years ago on a particular slow day at work, so out of boredom I wrote a two-and-a-half page proof. Yeah, I was really, really bored.
 

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Posted by saxon75 (Member # 4589) on :
 

quote:

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If you have a really big box of spherical marbles, whats the largest possible percentage of the space the marbles could take up?

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Er... I believe this question needs a little clarification. There are a couple of possible scenarios. Two equivalent cases would be a box of finite size with infinitely small marbles or marbles of finite size with an infinitely large box. In either case, of course, the number of marbles is infinite. But neither of these correspond to a real-world model (unless we can consider the universe to be an infinitely large box). Are the marbles and/or the box of finite size? And, if so, what model of the universe are you using? For example, string theory rules out point particles; everything must be bigger than the Planck Length. And are we assuming that all of the marbles are the same size? And are they constructed of a rigid material?
 

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Posted by fugu13 (Member # 2859) on :
 
saxon -- a hint: the question isn't what's the highest number, its what's the highest percentage.

Approached from there, its not too hard to solve (I'm pretty sure I can prove it over the general category of finite boxes, but not necessarily for a single arbitrary box, though that wouldn't take much more reasoning, I just haven't gone through that reasoning yet).

Also, I think its reasonable to assume that, given we are presented with physical objects to work with (marbles and boxes), we are working finitely, and that spheres and points are different constructs in the geometry, a sphere necessarily having 3 dimensions to a point's none, the point being merely the limit of a sphere as it approaches radius zero, yet not a member of the set of spheres.
 

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Posted by saxon75 (Member # 4589) on :
 
Point taken on spheres vs. points. And maybe I'm just approaching this too much like an engineer (on an engineering exam if you see "really big box" you are usually meant to read it as "infinitely big box"). I'm pretty sure that if you can make the marbles inconsequentially small compared to the size of the box, epsilon approaches zero (where epsilon is the amount of leftover space in the box). Additionally, if you can make the marbles of different sizes, or if the marbles are squishy, that changes things as well, wouldn't you say?
 

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Posted by fugu13 (Member # 2859) on :
 
Okay, no squishy marbles, and marbles are all the same size. Otherwise, yes, one can make the volume filled arbitrarily approach the volume of the box. However, think again on the idea of just making them smaller to fill up space. It may appear that they fill up space if you look at a mass of them, but consider how much empty space is necessitated by inability of two marbles to occupy the same space.
 

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Posted by fugu13 (Member # 2859) on :
 
As a simple exercise, consider a (cubic) box with one marble in it, marble diameter equal to box side-length. Then consider the same box with 8 marbles in it (perfectly packed and as large as possible, so marble diameter equal to one half box side-length), and notice what happens to the space filled by the marbles in the second situation compared to the first. Then notice that you can generalize very easily to all numbers of marbles in powers of four (consider each marble to be in an imaginary box and divide it as before), and also quite easily to any set of marbles that perfectly fills the box.

Then just prove that any set of marbles, of a number which can be packed perfectly into a cube, which can be packed in a box can be packed perfectly into a cube of equal or lesser volume to the (not necessarily cubic) box, and you're nearly done.

I leave the final steps as an exercise for the reader .

[ December 22, 2003, 01:48 PM: Message edited by: fugu13 ]
 

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Posted by fugu13 (Member # 2859) on :
 
And if you haven't guessed, there's a joke hidden in the above which you will only get if you do at least part of the derivation .
 

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Posted by snow (Member # 5804) on :
 
Actually, fugu has the right idea, sort of, but the equivelent unit-cell for the best packing ratio isn't cubic. The answer you get with a cubic packing is far from optimal.
 

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Posted by Nato (Member # 1448) on :
 
Would the most efficient use of box space be to stack marbles perfectly on top of one another?

I'm not sure how to go about doing it, but it seems like it would be better to stack them so that each marble would lay with its center above the spot exactly between marbles on the layer below.

code:

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 ()
()()

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quote:

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How many non-overlapping circles of diameter 1 can you fit in the box?

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1 what?
 

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Posted by Dragon (Member # 3670) on :
 
good job xenra!
 

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Posted by Papa Moose (Member # 1992) on :
 
One inch, you dweeb. You're just pointing out my faults -- like that's a feat.... *smile*
 

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Posted by saxon75 (Member # 4589) on :
 
Well, in that case you can fit 2000, unless this is some sort of trick question. You can't fit more than two rows, even if you stagger them. If you do stagger them, you actually can only fit 1999. So the best fit is just lining them up in squares.
 

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Posted by Papa Moose (Member # 1992) on :
 
Nope. You can fit more than 2,000. You just need to figure out how.
 

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Posted by fugu13 (Member # 2859) on :
 
Heh, here's a "hint": think of dice, and the number five.

Note, I haven't even bothered to calculate the answer for my hint, but it should at least start someone thinking on the right track even if its wrong .
 

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Posted by snow (Member # 5804) on :
 
I agree with Saxon, i dont see any way to fit more then 2000. If the answer is something like, chop them all up into nice little bits and fit them in that way, ill be pissed.
And fugu, I believe you are mistaken, that configuration doesnt seem to work as well as just a plain square packing. only about 1730.
 

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Posted by Juliette (Member # 5484) on :
 
(saxon here)

The densest possible configuration is a hexagonal configuration. However, if we are limited to whole circles that must all be contained within the rectangle, a hexagonal configuration allows at most two rows. A square configuration clearly allows two rows of 1000 circles. The total height of the two rows will be 2d = 4r, where r is the radius of the circle and d is the diameter, or 2 inches. In the hexagonal configuration we get one row of 1000 and one row of 999. The total height of the two rows is r * sqrt(3), or approximately 1.73 inches. Clearly, there is no more room for any more circles above that row (adding another row in hexagonal configuration requires a total height of about 3.46 inches).

The only way you can fit in more circles is if you allow fractional circles, or circles of different sizes.

[Edit: Excuse me, the actual total height of two staggered rows is (r * sqrt(3)) + 2r = 1.87", and the total height of three rows is (2r * sqrt(3)) + 2r = 2.73".]

[Edit 2: Including a top row consisting of fractional circles gives about another 79.8 circles. Just need the two fractional circles from the middle row...]

[Edit 3: Symmetry is nice. You get one half circle on either end of the middle row. So, using a hexagonal configuration gives you a total of 2079.81914 circles. But I think including fractional circles is cheating.]

[ December 23, 2003, 01:57 AM: Message edited by: Juliette ]
 

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Posted by Mike (Member # 55) on :
 
I can fit 2007. It's pretty cool that it works out. I can't prove that it's optimal, though. Yet.

Saxon, you're right that the densest configuration is hexagonal, but it doesn't end up being very efficient over a large scale because of the shape of the container. So I figured I'd try little chunks of hexagonally packed circles.

If you do chunks of four circles like so:

code:

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 ()()
()()

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but tilt the whole chunk up so that it touches the top and bottom of the containing box (about 5.26 degrees), you shave a little bit of space (about 0.0856 inches per chunk). Not only that, you can do even better than the 2.414 (actually, exactly 1 + sqrt(2)) inches, since the chunks can overlap slightly (in the x direction -- the circles themselves don't overlap). This lets you fit four circles every 1.99156... inches, after paying the initial 0.414-inch cost of not using a square lattice.

You can squeeze 501 of these 4-circle chunks into the box, with enough room for another 3 circles at the end. That's 2007. It'll be a good problem for the ARML competition in three and a half years.


 

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Posted by Papa Moose (Member # 1992) on :
 
Well done, Mike. 2,007 is the best I know of (done slightly differently), but I've never seen it proven that it's the highest. Way to think outside the box inside the box. *smile*
 

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Posted by Mike (Member # 55) on :
 


Well, don't keep me waiting... how do you fit 2007?

*goes off to figure out question 2 in the original post*
 

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Posted by Papa Moose (Member # 1992) on :
 
Triangles rather than diamonds -- 1,2,3 at far left, pull circles 4,5,6 up and slightly left, then 7,8,9 move left, then 10,11,12 up and left, etc. Still fits 2,007.

--Pop
 

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Posted by Mike (Member # 55) on :
 
Interesting. Looks like you have to fudge the last set of three to fit them. Actually, I'm not quite convinced you can fit more than two in the space left over, but I could be wrong.
 

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Posted by Mike (Member # 55) on :
 
Original post question (2):

First part -- angles. The angle between adjacent hexagonal faces is the same as in an icosahedron, or about 138.19 degrees. The angle between hexagonal and pentagonal faces is about 114.80 degrees. I hope these are the angles you're looking for. (If you want to know how I figured it out, I'd be glad to explain.)

Second part -- volume. Um, lemme work on this for a sec.
 

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Posted by Mike (Member # 55) on :
 
Ooh, made a mistake. Disregard those angles. Will try again later. D'oh.
 

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Posted by Mike (Member # 55) on :
 
Hmmm, not sure if anyone else is really interested in this. But hey, I am and that's all that matters, right?

Pop, it actually looks like your solution fits 2011 circles, so you beat me by 4 (my dad figured this one out, though he needed a couple of hints before he believed anything over 2000 was possible).

OK, I was right about the angle between the hexagonal faces (~138.19 degrees). The angle between hexagonal and pentagonal faces is 142.62 degrees and change.

The volume of an icosahedron with side length 1 is about 2.18169. A truncated icosahedron has 1/3 the side length of the original, so we have to multiply by 27, and then subtract the 12 truncated pentagonal pyramids. I might have made a mistake, but the final result I get is 58.77176.

Now, can anyone back me up here?
 

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Posted by Papa Moose (Member # 1992) on :
 
That's what I get for trusting someone else's math (I figured out the method but he did the calculations). I'm gonna have to calculate it myself now to be confident of the answer. As to the others, no, I can't bring myself to care much about question #2. It's not a brain teaser -- it's a math problem. Pure calculation with no logical puzzles. So no, can't back you up.

--Pop
 

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