You should be able to solve this.
>>8078451
A³
>>8078451
>filename
are you serious?
>>8078454
What, can't solve it?
>>8078451
(4/3)*pi*(A/2)^3
>>8078451
The answer is right there on the pic
>>8078451
This is a really stupid question that wastes too many words explaining shit that's irrelevant to it and effectively noise.
There's basically no information on the actual task except that there's a cube and you're supposed to figure out its volume. The picture attached to it says the length of the cube is a.
>>8078458
>D[0] is defined as the set of all points closer to A[0] than any other point
That's defining a sphere since we're in 3D space. In the context of the diagram, it's a sphere with diameter A.
>>8078459
No, it's not a sphere because the corners of the cube are still closer to the center of that cube than they are to the center of any other cube.
>>8078459
This is wrong.
>>8078452
FINDS A WAY
>>8078460
I can't tell which was stupider, the arena fights or this scene.
>>8078452Stop posting this because it already found a way.https://www.youtube.com/watch?v=JcVGDV67L-ghttps://www.youtube.com/watch?v=o_rz1bluG_k
>>8078463
ignore this, misread the question
>>8078452
>>8078463
I sure hope you are not really this stupid.
Its to the center, not to an arbitrarily chosen point.
Red and black are the same length, by the way.
>>8078465
>In the question, A and a are different things.
No, they aren't.
>>8078470
Yeah I missed the center points.
>>8078451
You should be able to solve THIS.P.S. It has a single, definite solution.
>>8078469
I wasted 5 minutes trying to solve this.
Fuck you
The answer is simple once you consider the following:
>all points in the lattice must be either closer to one atom or fall on an edge/vertex between multiple atoms
>the lattice structure is homogenous
That means the shape whose volume we're looking for tesselates; i.e. space can be filled by stacking them on top of one another.
We also know that the atom is surrounded by 14 other atoms (eight vertices of the cuba surrounding it plus the atoms up, down, north, south, east, and west, if you follow)
There's only one 14gon that tesselates by itself.
The answer is the volume of a truncated octahedron of edge length A/sqrt(2).
Oh wait, that's not simple at all.
The easy way to do it is to ask yourself- what fraction of a cube's volume is closer to the inside than the outside? What fraction of the volume is closer than the halfway point to the surface?half of the cube, (a^3)/2
>>8078475
Oops, *14-hedron.
>>8078467
Ah, I misread the question. It's still not a sphere though as that would still lead to non-overlapping area that isn't bounded by any sphere. It should be a diamond like structure.
>>8078473
sauce
>>8078452Threadly reminder that you should stop posting that because it's a thing.https://www.fictionpress.com/s/3206139/1/
>>8078469
Finds a way
>>8078452
>>8078469
ALWAYS
>>8078451
This is fucking stupid it flat out tells you that D is 0.
>>8078469
What are the hints?
>>8078471
A represents the entire lattice and a represents a single cell.
>>8078465
The answer is not a^3 because there are atoms at the corners of each cube.
The question is asking how much volume of the cube is closer to the center than it is to the vertices. Obviously a point is closer to the center when it is on the center-side of the half way point. This half way point applies to each of the vertices, so the answer is (a^3)/2, that is, half of the volume is closer to the center than the outside. Really pretty simple when you ignore all the extraneous bullshit the question gives you. Of course that's an important part of problem solving, too.
>>8078469
ALWAYS
>>8078469
I spend all my hint coins on the super hint
>>8078451
How did the author even make this up?
inb4 it's an actual question he encountered
>>8078469
>>8078452
>Keit愛 OST
OP: https://www.youtube.com/watch?v=ycfdfinG_P8
ED1: https://www.youtube.com/watch?v=hXDNGS9V4Us
ED2: https://www.youtube.com/watch?v=m4DyTjrruVo
OP Full: https://www.youtube.com/watch?v=4NShEKYoPXw
ED 1 Full: https://www.youtube.com/watch?v=IIX9aoN7g4o
ED 2 Full: https://www.youtube.com/watch?v=woiMYzVTl9c
Who /chemistry/ here?
>>8078485
Will it not, however, be spherical, then?
>>8078490
>Will it not, however, be spherical, then?
Why would it be?
>>8078485
You would be right if there were only the centre and the 8 vertices. However, in the diagram we are given 27 points in total, including the centre, and what we want to calculate is the area of the domain consisting of all the points closer to the centre point than any of the other 26 points.
>>8078490
It's not spherical because you have to have something that tessellates completely.
>>8078491
Because in a crystal lattice, you'd be measuring the radius that is closest to a single atom. Because you get a halfway point between two atoms, that is going to be the radius used for the sphere around an atom.
4/3 pi r^3 is the volume of the sphere, and the radius is half of edge A.
So 4/3 pi (A)^3 is the answer.
>>8078495
You're not measuring atomic radius; you're measuring the domain, which is defined to include points outside of the atomic radii that are more proximal to one of the centers.
>>8078492
I fucked up the edge length of the polyhedron, btw. Should be a/2sqrt(2)
>>8078497
Oh, I see. So then the domain would be in an octahedron shape. My lazy ass just went with a sphere instead. Now that I look back in the thread, an anon already got that.
>>8078473
Solved.
Each atom is the vertex of eight cubes, so for one vertex, its domain in one cube is 1/8a. The central atom's domain is thus a³ - a.
Is that about right?
>>8078497
Also fuck this question, it's pretty much pointless. It's useless to know the domain it's asking about because it's ignoring the dimensions of the atoms in the lattice.
Most questions of this type would ask for atomic radius, or even the determination of packing efficiency, as opposed to the arbitrary determination of the domain of an atom.
Unless, of course, this domain can be used to determine the ideal behaviors of ions that could form a body centered cubic lattice. In that case, I guess it's useful, but usually atomic radii ratio can indicate that.
>>8078500
Oops*
>>8078504
You beat me to it
>>8078502
It purely serves as a trick question to remind you that, given a density of n atoms per X volume, and asked how much volume there is per atom, you can stop reading there and just do the arithmetic.
>>8078501
Disregard this, I'm stupid. I was just trying to remember how they worked it out in the show. Half the cube is right.
>>8078505
So there's only one solution ?
>>8078508
There must be only one solution if you do it without guessing.
>>8078458
>This is a really stupid question that wastes too many words explaining shit that's irrelevant to it and effectively noise.
Maybe that's the point. Read the filename
Puzzles are not a measure of intelligence but autism.
>>8078488
It's definitely a question I've seen before, though I don't recall the particulars.
>>8078509
What do you mean?
There's no guessing, just backtracking.
>>8078506
Well yeah, but even that's not perfect. That's usually how it's worded, and what I feel like the question wants to ask, but the fact that it words it in such a way where the answer would be that of a trunkated octahedron is bullshit. Atoms themselves are spherical, or are considered as such, so this just is irritating.
>>8078514
The way you worded "if you do it without guessing" implies there are other possible solutions.
>>8078512
>>8078488
I took a materials science class last semester; basically we learned about what happens to metals at the atomic level when they're heated, stretched, strained, etc. We had similar questions, although the more interesting thing to solve would be the packing fraction.
http://www.science.uwaterloo.ca/~cchieh/cact/c123/bcc.html
>>8078516
Intended as "If you do it without guessing, you show yourself that there must be only one solution."
The mathematical proof is beyond my ability though.
>>8078518
The mathematical proof would be interesting.
I'm not certain, because even if you follow the rules, maybe this precise problem gives only one solution, but maybe on a larger board, or another board, applying the same rules may result in multiple possible solutions.
Why don't you guys have these threads during the weekend instead of the night before the workweek starts?
>>8078519
You could probably create a puzzle with these rules that have multiple solutions. I don't really have any math to back this up, but this problem itself seems inherently similar to the 8 queens problem mixed with Sudoku, both of which can have multiple solutions.
![tumblr_o6crjzE5hk1v0pigno1_500[1].jpg](https://i.imgur.com/6SBls9v.jpg "tumblr_o6crjzE5hk1v0pigno1_500[1].jpg")
>>8078521
>Implying
>>8078522
No, a sudoku by definition only has a single solution.
>>8078524
That's not true at all, and there is literally no math to back up your statement. Just because people don't publish puzzles with multiple solutions doesn't mean they aren't possible. I could create a Sudoku puzzle which has literally nothing filled in from the start and it would still be Sudoku.
>>8078523
>tumblr
embarassing
>>8078525
https://en.wikipedia.org/wiki/Sudoku
>which for a well-posed puzzle has a unique solution.
I'm talking about the definition.
You can make a grid like you want, but unless it only has one solution, it's not a sudoku.
It's spelled seppuku, retards
>>8078527
That Wikipedia article is literally disagreeing with the point you are trying to make. It's saying that a /good/ puzzle has one solution, not that all of them need.
>>8078515
This is just the 1BZ of each atom in reciprocal space though, so the question makes perfect sense if you've ever done any solid state physics/lattice mechanics.
>>8078526
>[1].jpg
I don't actually have a folder on my computer dedicated to smug amphibians. I just googled for it and took the first thing that popped up.
>>8078454
It's the final question on a test from Assassination Classroom.
>>8078529
>a sudoku is not a sudoku
>a not-a-sudoku is a sudoku
>>8078527
>In June 2008 an Australian drugs-related jury trial costing over A$1 million was aborted when it was discovered that five of the twelve jurors had been playing Sudoku instead of listening to evidence.[37]
Heh.
To make things simpler, let's solve the 2D version of this question first (because 2D > 3D).
So we have a square of side A (see left) instead of a cube, and they're arranged as shown. So if we have 4 squares stacked in 2x2, we get the following 2 figures.
Most people I see here are calculating as if the 4 points in the middle of the sides of the combined square are removed (see middle). If you calculate it as such, then the answer is (2a)²/2 = 2a². However, this is incorrect because it fails to take into account said 4 points.
So when we actually do take those points into account, we get a much smaller area (see right). Now we extend to 3D and it becomes (2a)³/2 = 4a³ and a³ respectively.
Those spherefags do the same and try to explain why the 3D domain should be a sphere and not a cube. I'm guessing you'll fail.
>>8078535
>side A
side a*
>>8078498
Try (sqrt(3(a^2)/4))^3
Think three dimensionally.I came up with your answer before, but the side length is different and I realize it was wrong
>>8078535
Then there are also those that get the answer 0.5a² for 2D (0.5a³ for 3D) because they take a square and calculate how much of the area is closer to the point the centre than to any of the corners. The problem with this working is that the point in the centre does not exist in the question. Here's why it doesn't work out.
>>8078458
>This is a really stupid question that wastes too many words explaining shit that's irrelevant to it and effectively noise
That's the point. They put these questions on tests specifically to weed out memorization plebs from people who can actually analyze.
>>8078458
>The picture attached to it says the length of the cube is a.
But that's different from the cube who's volume you're asked to actually work out.
>>8078451
(A/2)units cubed
I honestly found that question to be hilarious, because it's such a good question. The answer is blatantly obvious and simple, but people still get tricked into thinking that it's complicated. One of my favourite moments of Ass Class.
>>8078541
I found the question to be poorly posed. What it was apparently asking did not give the answer it wanted.
>>8078535
>Most people I see here are calculating as if the 4 points in the middle of the sides of the combined square are removed (see middle).
Re-read the question. They are, and a is twice the length you are implying it it. You're not doing it right.
>>8078542
>I found the question to be poorly posed. What it was apparently asking did not give the answer it wanted.
It's a trick question, but that's not a valid criticism of it. The answer is obvious as long as you take the question literally.
>(1/8)(a^3)(sqrt(27))
That's not even a length.
>>8078541
That's wrong though. It's (a^3)/2. Every unit cell of a^3 has two atoms associated with it.
>>8078451
>>8078549
I was talking about the edge length of the truncated octahedron D[0].
A cube of equivalent volume would have sides of length a times the third root of 1/2.
[a(1/2)^(1/3)]
>>8078469
Fuck this question.
>>8078451
Fuck that. It's a waste of time.
This information is completely useless.
Especially since pure alkaline metals do not occur naturally due to their radical-like status.
>>8078451
I knew I saved this for a reason.
>>8078554
>>8078553
>Especially since pure alkaline metals do not occur naturally due to their radical-like status.
And that's disregarding that they're practically fluid from Rubidium onwards.
>>8078478Phi Brain: Kami no Puzzle
Think Yu-Gi-Oh except with PUZZLES
>>8078451
aw man that takes me back
did that stuff in 1st year material science
fun times
>>8078558
Rip sysadmin ;_;
>>8078469
What's the answer?
>>8078554
I miss the fun threads
I must be dumb.
>>8078559
What?
>>8078451
It's the set of points closest to each atom is half the volume of each cube. The layout is a two sphere tessellation and through basic logic the points closest to each atom must be evenly distributed and have a total volume equal to the volume of a cube, thus D[0]=a^3 / 2.
I don't see how the answer is not a sphere according to you guys.
The closest atom to any atom is (0.5*sqrt(3))*a away because that is the distance from the center of the cube to a vertex.
The set of points closer than the closest atom is a sphere with radius (0.5*sqrt(3))*a and the volume turns out to be (0.5*sqrt(3))*pi*a^3.
No fun allowed
The volume is a^3. There is one atom per cube, and the answer has to be the same for every atom. Therefore, the volume must be divided evenly between the atoms.
>>8078643
Right idea, but at the corner there are 8 atoms, each at the interesection of 8 cubes. So for one cube, you have 1+8*1/8 = 2 cubes, giving a^3/2
>>8078488
He was likely to get it from ASSASSINATION CLASSROOM 2ND SEASON EPISODE 12
>>8078778
he said author, so he's talking about the manga instead of the anime obviously
>>8078643
>There is one atom per cube
Wrong. This body centered cubic, not primitive cubic.
>>8078451
[math]\frac{A^{3}}{2}[/math]
>>8078570
a sphere is an approximation of the answer but if you consider some corner cases you'll realize there will always be points under-/over-estimated by the boundaries of the sphere
>>8078475
i intuitively saw this but with the vague, engrish, pedantic wording, wasn't sure if that was the problem
>given atom A[0]
wow so intellectual, don't refer to them as 'points', that would be too clear
>>8078469
what game is this ?
>>8079604
I fucked up it's actually [math]a^3/2[/math]
watch assassination calssroom. This is the last problem on the Final
>>8079607
I might be retarded, but wouldn't the answer be [math] \frac{a^3}{3}[/math]? Since it seems that the volume of points that are closer to the arbitrary point form what looks to be 2 pyramids. The base of the pyramid is [math]a^2[/math], and the height would be [math]\frac{a}{2}[/math], since that is the furthest from the point that would be closest to the center. And the formula for a pyramid is area of base x height / 3. Which means:
[math]\frac{a^2\times\frac{a}{2}}{3}\times{2}[/math](because 2 pyramids)
So the end result would be: [math]\frac{a^3}{3}[/math]
>>8079606
Professor Layton. I think it might be from the first one, Curious Village. That image is edited, however. This is the original puzzle.
>>8078452
cancer finds a way (finds a way!) onto other boards
>>8078451
here is the procedure for solving the problem.
at every point in the lattice, there are only 14 points in the immediate vicinity of any center point O, up, down, front, back, left, right, up-front-right, up-front-left, up-back-right, up-back-left, down-front-right,down-front-left, down-back-right, and down-back-left. These 14 points are the only significant points to consider, as the cube of distance control about O (constructed from the first 6 points) is always inscribed inside the planes of equal distance between O and any other point EXCEPT the 8 diagonal points. The solid of distance control for the 8 diagonal points is an octahedron, and the volume bounded by the cube and the octahedron centered at O contains every point such that the point's distance from O is less than the distance between said point and any arbitrary point in the lattice (except O). Just find the volume of this intersection of 2 solids.
>>8078451
You should be able to solve this
>>8078451
You thought you had a solution, but it was [spoiler]I, D[0][/spoiler]
I am in agony
>>8078452
keit-ai finds a way
man that's melodic
>>8078451
You should be able to solve this.
>>8078451
>>8079695
>arbitrary point
It's in the center of the blue box.
The space you're solving volume for comes out to be a cube with corners at the midpoints of the blue box's faces.
The blue cube has a volume of [math]a^3[/math], the volume you're looking for will be half of that. So [math]\frac{a^3}{2}[/math]
If you want to look at it differently, it's the volume of the blue cube minus 8 pyramids of volume [math]{(\frac{a}{2})^{3}}/2[/math]. So again, [math]\frac{a^3}{2}[/math]
>>8078466
What is the real animu this footage is from though?
>>8080189
>"martensic" instead of "martensitic"
dropped
>>8080998
Well, whattissit?!
