agaricus5
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Hmm...

You can go smaller than that. Remember that all cubes are cuboids, but not all cuboids are cubes.

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Oneiromancer
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I can get 749 cm^3, by assuming a close-packed Face-centered-cubic (FCC) structure (Hexagonal-close-packed, HCP, should give the same or similar answer but the math is a little harder and I didn't want to derive it).

The final arrangement of the spheres is to first put 30 down in a 5x6 arrangement. Then put 20 down on top of those 30 in the crevices created by the spheres being so closely packed. This base is 12 cm x 10 cm, or 120 cm^2. The height is another matter, but you can calculate it using just the Pythagorean theorem a couple of times (c^2 = a^2 + b^2). First imagine 4 spheres close together, making a small 2x2 square. Place 1 sphere in the center of those 4. The center of that sphere is exactly in the middle of those 4 spheres, and it touches all 4 spheres at some point. So we can connect a line between the centers of all the spheres that are touching and we know that the length of all of those lines are equal to the diameters of the spheres. Now, I set the origin of my "graph" to be equal to the center of one of my 4 base spheres. The center of the center sphere will be at R in both the x and y directions, so I can draw a small right triangle which has a hypotenuse of sqrt(2)R. Now I can draw another right triangle, the base of which is sqrt(2)R, the hypotenuse of which is 2R, and the height of which is the unknown height of the center of the center sphere. Using Pythagorean's theorem again, I get that the height is equal to sqrt(2)R also (perhaps this should have been obvious). And since there are 4 layers of spheres, I need to take three of these heights and also add another full diameter (2R) to account for the fact that I was doing my height calculation from the center of one of the base spheres. So, since R = 1 cm (diameter = 2 cm), the total height of my box is [3*sqrt(2)+2]*R = 6.24 cm. Now multiply that by the 120 cm^2 base and you get 749 cm^3.

So, in conclusion, yes, that was a math problem, and as such wasn't really appropriate. Hopefully I actually got it right...I always screw up little things somewhere.

In case someone wants to work out the HCP volume, instead of using a sample base of 4 spheres and putting 1 sphere in the middle, you have instead 3 spheres and put 1 sphere in the middle. Over the course of an infinite atom, the FCC close-packed and the HCP end up having the same density, or "packing fraction", but I'm not sure about a finite number like 100. So that's another way in which my answer could be wrong.

Oh, and in case you're wondering, I didn't have to look anything up to solve that (I looked up the packing fraction value afterwards), and I didn't have the answer memorized, so I did have to solve it by myself so I'm not breaking any of my own rules about not answering if you know the answer already.

Game on,

[Edited by Oneiromancer on 01-21-2004 at 02:38 AM GMT: Stupid math mistakes]

[Edited by Oneiromancer on 01-21-2004 at 09:51 PM GMT: Forgot to update first sentence]

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DiMono
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793.5 cm^3, eh? You realize that all that math saves you 6.5 cm^3 over sticking them in a cuboid tube 2*2*200?

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Scott
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Its a maths problem all right because I didn't even realise that the term you used didn't mean a cube.

Oneiromancer
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DiMono wrote:
793.5 cm^3, eh? You realize that all that math saves you 6.5 cm^3 over sticking them in a cuboid tube 2*2*200?

Thanks for pointing that out, I noticed that I made two bad math mistakes. The first one I noticed was only adding 1R instead of 2R to account for the extra space above and below the centers. (I was thinking in diameters instead of radii.) But that would have given my more volume than what you had, and I knew that couldn't be right. So then I looked over it again and saw my second mistake, which was that I again was thinking in terms of diameters, and I set the center of the center sphere at (x,y) = (R/2,R/2) instead of (x,y) = (R, R), which makes a pretty big difference. It means that the center of the center sphere is a height sqrt(2)R above the centers of the spheres below it, which is much more reasonable and symmetric. And my final answer is even more less than your trivial answer, although I wonder if I could improve it even more. I think if I were to extend this more and more balls I would get closer and closer to 0.74 of the evenly spaced value, which is the idealized "packing fraction" which I forgot to give in the above answer (right now I am at 0.75).

Whew! Sorry about that. I am pretty confident that I am correct now, since it is so close to the theoretical value, but I'll wait for confirmation from Wesley.

Game on,

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agaricus5
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Apologies for the problem being too mathematical. These are about the only types of logical puzzle I know.

Oneiromancer wrote:
I can get 793.5 cm^3, by assuming a close-packed Face-centered-cubic (FCC) structure (Hexagonal-close-packed, HCP, should give the same or similar answer but the math is a little harder and I didn't want to derive it)....

Nope. You can still go smaller than that.

So, in conclusion, yes, that was a math problem, and as such wasn't really appropriate. Hopefully I actually got it right...I always screw up little things somewhere.

Agreedly it is hard to a point, but the only things you need for this puzzle are:

1. Knowing how to find the volume of a cuboid.
2. Knowing how to use Pythagoras' Theorem.
3. Having a visual imagination.

For example, when I made this, I wasn't actually thinking about metallic bonding and packing of atoms in solid states of matter and was totally unaware about any "packing equation".

It's not really breaking the rules too much, is it?


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Oneiromancer
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Ack, agaricus, read my post again, I edited it to fix an error and forgot to fix the first sentence.

Game on,

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agaricus5
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Oneiromancer wrote:
Ack, agaricus, read my post again, I edited it to fix an error and forgot to fix the first sentence.

Game on,

I really don't like to say this, but mine is still a bit smaller (Not much, but still smaller).

The hundreds value is correct, though.

[Edited by agaricus5 on 01-21-2004 at 10:14 PM GMT]

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Oneiromancer
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I can get 688.4 cm^3 if I assume triangular packing (hexagonal). That is, I have 4 layers of 25 spheres, and each layer is staggered from the one below it. The first and third layers, and the second and fourth layers, are directly above each other. Each layer is also staggered within itself, so that each center sphere is touching six other spheres (so basically a honeycome shape, which is why this is hexagonal packing). Since the spheres are staggered, the "width" of the base, which is the direction where the spheres are all in a line, is 11 cm (5 spheres in a row, and the next row sticks out an additional half-sphere). The "length" of the base is similar to the height of the previous way I did it, and although there are 5 staggered rows, on the next layer up I will have additional "length" because of the way the spheres fit in the bottom layer. So I have to add an additional row to the base. Thus, because the rows are staggered like the height of my previous solution, the "length" is now [5*sqrt(2)+2]*R, or 9.07 cm. Finally, the "height" does not follow the same rules as before, because the much tighter packing of the base layer means that the successive layers must sit higher than before. My math leads me to conclude that the center of each layer is sqrt(8/3)*R above the previous layer. Thus, with 4 layers, and accounting for the extra height from the centers of the top and bottom layers, the total height is [3*sqrt(8/3)+2]*R, or 6.90 cm. Thus, the total volume is 11*9.07*6.90 = 688.4 cm^3.

I'm not completely confident about the "width" being exactly 11 cm, although it certainly seems right to me. If I err on the side of caution and say that the width is instead 10+sqrt(2), which might also be reasonable (but I think is false when I envision the setup), then I get a total volume of 714 cm^3, which at least has the same hundreds digit as agaricus's answer.

I'm gonna check on my math soon, but I'll still post this first, since I don't want to have to copy and paste it out to correct it later. My apologies for being lazy.

Game on,

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agaricus5
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I believe your first answer stands. Your answer is far better than mine - 717.4153...cm^3 or 9*(((3^(1/2))*4)+ 2)^2). I think I was going for an almost cubic answer that involved space at the top and neglected to spot that 100 = (2^2) * (5^2)

Well done! I think it's your turn now.

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Oneiromancer
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Well, if you have time, please feel free to check my answer. I have to go to a seminar on muon spin rotation right now and I won't be able to post a puzzle for a few hours. I'll be checking my math as well during the seminar, since physicists are well known for forgetting to carry the 2, metaphorically speaking.

Game on,

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"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

Oneiromancer
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Ack, sorry I didn't do it for so long, I got distracted. Here's the next puzzle, another series, hopefully not as easy as the last one I did.

Give the next three letters in this series:

J
W
Y
M
?
?
?

Game on,

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Scott
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O A C



[Edited by Scott on 01-22-2004 at 11:26 PM GMT]

Oneiromancer
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Incorrect, although if you tell me your reasoning behind your answer I can offer hints, if I think you are way too far off.

Game on,

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eytanz
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P
H
C



What, no?

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Oneiromancer
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Here's a hint for everyone: Eytan is not even close. If this pea were the Earth, and this basketball were the Sun, then Eytan would still be wrong.



Um, yeah, you know what I mean.

Game on,

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Schik
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T I I?


That's a pretty horrible guess, but I can't come up with anything better yet.


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Oneiromancer
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Nope, sorry. Here's a hint: these aren't the first letters of a phrase. The letters are significant in and of themselves. There is a pattern, something that makes their sequence make sense. But some manipulation might be necessary first.

Game on,

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DiMono
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How about N C C?


Am I close?

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Oneiromancer
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Nope, not correct...and not really that close, either.

I can't think of any hints that are actually useful, without giving it away, so I'll just add the next letter in the series and ask people to find the last two:

J
W
Y
M
D
?
?

Game on,

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"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

bibelot
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All right, I got it.
The next two letters are E, N.

Oneiromancer
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bibelot wrote:
All right, I got it.
The next two letters are E, N.

Shift each one back one letter in the alphabet.

That's correct (which is why I removed the secret tags), but when you post your next puzzle why don't you explain why this is important, in case other people don't see the significance of this?

Game on,

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"He who is certain he knows the ending of things when he is only beginning them is either extremely wise or extremely foolish; no matter which is true, he is certainly an unhappy man, for he has put a knife in the heart of wonder." -- Tad Williams

bibelot
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Shifting by one letter gives I, V, X, L, C, D, M, the Roman numerals for 1, 5, 10, 50, 100, 500, 1000.

Here's a cute one:

Your doctor has prescribed for you two medicines, X and Y. You must take one pill of each together each day. One day, you pour out a pill from jar X into your hand. Then you get the other jar, but when you try to pour out a Y pill, you accidentally pour two. In a poor design move, the company that makes the pills has made them all look identical, and you can't remember which is the X pill. Moreover, they are quite expensive, so you don't want to throw these three pills out, and of course you don't want to overdose. What's your best course of action?

Schik
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Take another X out, crush the four pills, mix it all together, and take half one day and the other half the next day?


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bibelot
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That's almost right, but what happens if your mixing is quite poor and you don't realize it?

[Edited by bibelot on 01-23-2004 at 07:34 PM GMT]

Schik
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mix it in water? Put the pill dust and water into a blender for a minute or two, that should mix it up really good.

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bibelot
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There's still a tiny chance that you could be taking much more of X than Y, but since you've basically got the right idea and it works in almost all circumstances, you may as well take it, unless you want to keep guessing. (What I was looking for is below.)

Oneiromancer
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I think Schik's method has a much less chance of taking much more X than Y, because it's more reasonable to assume that the pills will dissolve and mix uniformly in the blender than to assume that you'll be able to accurately cut the pills in half, not accidentally mix them up again after cutting them up, etc.

But in terms of "logic puzzle-ness", your intended solution is more appropriate.

Game on,

[Edited by Oneiromancer on 01-23-2004 at 08:14 PM GMT]

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eytanz
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Actually, mixing the half-pills shouldn't be much of a problem if you're careful.

But accurately cutting a pill exactly in half? Four times in a row? That's far less likely than a blender achieving practically uniform distribution.

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Schik
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I was about to say the same thing as Onei, but he beat me to it while I was thinking of my next puzzle. Also, not all pills are uniform throughout - cutting in half might leave all the medication part in one half and all the filler in the other. But I digress. I'll just take it and run to the next puzzle:

You have an infinite number of brand new, unsharpened pencils at your disposal. Ordinary #2 pencils. For arguments sake let's say they're 20cm long and about .5 cm thick.

How many of them can you arrange such that each pencil is touching every other pencil? For this problem we'll assume that pencils are perfectly rigid - they won't bend at all.

Edit: Only two pencils may be touching at any given point. It's easier to just make this a rule than to explain why.



[Edited by Schik on 01-23-2004 at 09:58 PM GMT]

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The greatness of a nation and its moral progress can be judged by the way it treats its animals.
--Mahatma Gandhi