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Main - Spatula - Best math problem ever New thread | New reply


boingboingsplat
Posted on 12-14-08 05:36 PM Link | Quote | ID: 96759


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http://nrich.maths.org/public/viewer.php?obj_id=5749

The name of the problem says it all.

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James Freeman
Posted on 12-16-08 10:47 PM Link | Quote | ID: 96976


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Last view: 4296 days 		Cheese cutting.

...

That's a horrible joke.

But it's still valid mathematics, so what the hell.

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jargon
Posted on 12-17-08 04:57 AM (rev. 2 of 12-17-08 04:59 AM) Link | Quote | ID: 97012


Ninji
Banned until 2010-10-15 for an utterly psychedelic posting style
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Last view: 4608 days 		i just used 4 cuts and made 15 pieces. didn't want to mess with it more.

obviously, you cant make 2x2x2x2 pieces in 3d space with 4 cuts.

(what seems like what would produce 16 pieces only makes 10.)

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blackhole89
Posted on 12-20-08 02:28 AM Link | Quote | ID: 97191


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15 is the maximum, yeah, but I can't think of an overly elegant way (apart from a long series of proofs for all possible cases based on points) of proving it now*, nor see any immediate algorithm to maximize the piece number for larger numbers of cuts yet.

*I just realized it's fairly easy. Consider a trivial lemma that you won't get even 15 if any of the first three cuts are parallel, then consider the point where the three first cuts intersect accordingly. You'll see that, no matter how you rotate or translate the fourth cutting plane, owing to the perpendicular of the intersection point onto the plane, at least one of the eight cubes that have said point as a vertex isn't going to be intersected by it.

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NightKev
Posted on 12-20-08 09:01 AM (rev. 2 of 12-22-08 03:28 AM) Link | Quote | ID: 97207


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Posted by blackhole89
15 is the maximum, yeah, but I can't think of an overly elegant way (apart from a long series of proofs for all possible cases based on points) of proving it now*, nor see any immediate algorithm to maximize the piece number for larger numbers of cuts yet.

*I just realized it's fairly easy. Consider a trivial lemma that you won't get even 15 if any of the first three cuts are parallel, then consider the point where the three first cuts intersect accordingly. You'll see that, no matter how you rotate or translate the fourth cutting plane, owing to the perpendicular of the intersection point onto the plane, at least one of the eight cubes that have said point as a vertex isn't going to be intersected by it.
This is not English, what is wrong with you?!() (also, "lemma"? is that actually a word? (not silly German nonsense )).

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blackhole89
Posted on 12-20-08 09:10 AM Link | Quote | ID: 97208


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Did you read the problem?

Do you have a basic concept of maths?

Could you at least be bothered to use Wikipedia before making bad replies?

(though I concede my sentence structure was somewhat confusing in places - for instance, the "immediate" in the first sentence should by relocated before the "see" and augmented with a -ly)

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NightKev
Posted on 12-20-08 09:24 AM Link | Quote | ID: 97210


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I'm posting for the people who don't know much about math, because you know they will post eventually.

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Komaru
Posted on 12-20-08 05:23 PM Link | Quote | ID: 97223


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Posted by NightKev
This is not English, what is wrong with you?! (also, "lemma"? is that actually a word? (not silly German nonsense )).


Yeah, lemma is a word. It's sort of a "mini-theorem" to making a theorem.

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Skeet Urchin
(post deleted) ID: 99028

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