You should be able to solve this.

# You should be able to solve this

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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.

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.

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.

~~Stop posting this because it already found a way.~~

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It's describing a body centered cubic until cell, meaning that there is an atom at the center and at every vertex of each cube depicted in the picture.

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.

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~~

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.

~~Threadly reminder that you should stop posting that because it's a thing.~~

A represents the entire lattice and a represents a single cell.

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.

How did the author even make this up?

inb4 it's an actual question he encountered

Keit愛 OST

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As pointed out in

Although the simplification isn't actually that hard, once you figure out the polyhedron

It's not a terribly hard simplification, as you pointed out, the tricky part is determining the polyhedron.

V=8*sqrt(2)*(a/sqrt(2))^3 = (a^3)/2

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.

It's not spherical because you have to have something that tessellates completely.

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.

Fuck, I meant 4/3 pi (A/2)^3, I'm tired.

Yeah, but it wouldn't be cubical either, and I'm tired, so I figured it'd be more accurate to represent it with a sphere, even though there's only 8 nearest neighbor atoms.

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.

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 user already got that.

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?

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.

Truncated octahedron, with 14 sides:

A sphere is actually a better approximation than a regular octahedron. The domain can't extend into its neighbor's box, visualising the box as the bonds drawn in OP's image.

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.

Disregard this, I'm stupid. I was just trying to remember how they worked it out in the show. Half the cube is right.

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

It's definitely a question I've seen before, though I don't recall the particulars.

You start out with all the possibilities and then rule them out by using the rules of the puzzle and logic.

This is true, logic isn't intelligence.

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.

The way you worded "if you do it without guessing" implies there are other possible solutions.

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.

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.

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?

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.

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.

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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.

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.

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.

[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.

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.

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~~

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.

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.

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.

(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.

I found the question to be poorly posed. What it was apparently asking did not give the answer it wanted.

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.

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.

That's wrong though. It's (a^3)/2. Every unit cell of a^3 has two atoms associated with it.

How is that a criticism of the answer? If I've made a mistake please correct me, but the question is effectively asking for the volume of a lattice of cube with side length:

(sqrt(3)/2)*a

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)]

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.

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.

aw man that takes me back

did that stuff in 1st year material science

fun times

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.