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41 votes
Accepted

How to calculate the surface area of a cube given a diagonal?

It is just the pythagorean theorem:
Seyed's user avatar
  • 8,943
40 votes

How to calculate the area of a 3D triangle?

Probably one of the best ways (and possibly least computationally intensive ways) to approach this problem is with vectors. In this case we have three points, I will keep them as arbitrary variables ...
Greg's user avatar
  • 561
37 votes

What is the equation for a 3D line?

When I originally asked this question, I was not expecting these seemingly indirect ways of describing a line, such as an intersection of two planes, or vector equations. Just like how $y=mx+b$ is the ...
Ovi's user avatar
  • 23.8k
37 votes
Accepted

Hexagons are best for tiling 2D space in terms of perimeter vs area. What's best for 3D space?

This is known as the Kelvin problem; the best known (and conjectured optimal) solution is the Weaire–Phelan structure, but proving this is likely very very hard. I don't know what the best results in $...
RavenclawPrefect's user avatar
30 votes

What's wrong with this derivation of the volume of a hemisphere?

You're computing the volume of a cone, not a hemisphere! More precisely, your integral gives the volume obtained by rotating the triangle $0 \le y \le x \le r$ about the $x$-axis. The hemisphere, on ...
Hans Lundmark's user avatar
29 votes

Hole inside cube with tetrahedrons at corners?

Here's a nearly-trivial proof that there is in fact a 'hole', or at least a piece left over when all the tetrahedra are removed: consider the tetrahedron with right vertex at the origin and its other ...
Steven Stadnicki's user avatar
28 votes
Accepted

What is the name of this 3D shape with 12 outer vertices?

The object is the first stellation of the rhombic dodecahedron. This is sometimes known as Escher's solid, for its use in a study for his Stars, though the final picture uses a compound of three ...
Parcly Taxel's user avatar
20 votes
Accepted

What's wrong with this derivation of the volume of a hemisphere?

The thing to keep in mind with disk integration is the fact that the variable of integration represents the thickness of each disk. You are trying to integrate the volume of a half-sphere, using: $$ \...
DotCounter's user avatar
  • 1,170
16 votes
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What does a flattened Teserract look like?

There are, in fact, 261 different ways to unfold the tesseract into 3D space, as proved in this reprint of a paper from Journal of Recreational Mathematics, Vol. 17(1), 1984-85. As far as I know, I'm ...
Mark McClure's user avatar
  • 30.8k
15 votes
Accepted

How would I go about calculating the center of a sphere based on points from it's surface?

The equation of a sphere with center $(x_0, y_0, z_0)$ and radius $r$ is $ (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2 $ Expanding, we get, $ x^2 + y^2 + z^2 + A x + B y + C z + D = 0 $ where $ A = -...
Quadrics's user avatar
  • 23.9k
13 votes

What's wrong with this derivation of the volume of a hemisphere?

Your problem seems to be that you don't have a clear concept of how to integrate the volume of an object by stacking disks on each other. In order to compute a volume by stacking disks, you need the ...
David K's user avatar
  • 99.6k
12 votes

Hexagons are best for tiling 2D space in terms of perimeter vs area. What's best for 3D space?

I'd guess that the answer is not known above 2 dimensions, but it is very likely that if the answer is known then it is known in dimensions 1, 2, 8 and 24 (1 is trivial), and possibly exactly those ...
Adam Chalcraft's user avatar
11 votes

How to calculate the surface area of a cube given a diagonal?

This problem has a simple solution: $s=2d^2$ where $s$ is the surface area and $d$ is the spacial diagonal. Explanation: I discovered through Google (while writing the question) that the side length ...
Travis's user avatar
  • 3,396
9 votes
Accepted

What is the name of the polyhedral shape of the Humanity Star?

Here I've put red dots on corners where six triangles meet and green dots on corners where five triangles meet: This has too many red dots together to be the pentakis icosidodecahedron suggested by ...
hmakholm left over Monica's user avatar
9 votes
Accepted

Minimal surfaces for planar octagons and nonagons

I believe I have an answer to one of your many sub-questions, and here it goes: No, the minimum for octagons is neither 15 nor 24. This guy has 12. Each face is one of the following shapes: The ...
Ivan Neretin's user avatar
  • 12.9k
9 votes
Accepted

Can a chain of solid round links be twisted without getting shorter?

Let me set $r=r_\text{major}$ and $\delta=r_\text{minor}$ and let $2a$ (with $a<r$) be the distance between the centres of the tori. If the tori are "orthogonal" to each other, then they are ...
Intelligenti pauca's user avatar
8 votes
Accepted

Tetrahedron Centers

For the comparatively-easy ones ... As in my other answer, given tetrahedron $PABC$, I use barycentric coordinates $\rho$, $\alpha$, $\beta$, $\gamma$ to parameterize the point of interest as $$\frac{\...
Blue's user avatar
  • 77.1k
8 votes
Accepted

Construct any solid with identical plane shapes and the resulting solid must be a fair die?

This statement is not true. At least, there are shapes for which it no longer makes "intuitive sense" that they would be fair dice, though rigorously proving unfairness would require a more ...
RavenclawPrefect's user avatar
7 votes

Can the inscribed angle theorem be generalized to solid angles in 3D? And beyond to n-dimensional space?

The answer to your question is no. The inscribed angle theorem does not work for inscribed solid angles in a sphere. To see why, I will start by stating an equivalent version of the inscribed angle ...
Jim Belk's user avatar
  • 49.6k
7 votes

Why is the volume of a cone one third of the volume of a cylinder?

Here is a derivation of the volume of a cone which does not use calculus, Cavalieri's principle, the method of exhaustion, or any other infinitesimal arguments. [Edit There is a flaw in this ...
Will's user avatar
  • 171
7 votes
Accepted

What shape is a Calippo?

It's hard to tell from the picture, but from the description "The shape itself consists of lines connecting the circle to line segment", I might represent the shape thusly: Suppose that the circle ...
Blue's user avatar
  • 77.1k
7 votes

Two individuals are walking around a cylindrical tower. What is the probability that they can see each other?

I will answer the version of the problem where each person will be on the outside edge of the lane. You can view the problem as two concentric circles of radii $A$ and $A+B$, with two points on the ...
angryavian's user avatar
  • 90.9k
7 votes

Cutting seven cuboids and forming a cube

Yes, it is possible, since the two sets of polyhedra both have Dehn invariant $0$; a result of Sydler (1965) shows that any two collections of polyhedra with the same Dehn invariant can be cut into ...
RavenclawPrefect's user avatar
7 votes
Accepted

Making a regular tetrahedron out of concrete

It is convenient to place the image of the tetrahedron with vertices at every other vertex of a cube. I am imagining the cube of side 10 units, whatever they might be. The seven sides of one of the ...
Will Jagy's user avatar
  • 141k
7 votes

Hole inside cube with tetrahedrons at corners?

The octahedron hole inside the cube: $\hspace{3cm}$
farruhota's user avatar
  • 31.6k
7 votes
Accepted

Viewing a circle from different angles - is the result always an ellipse?

This is a fairly standard result in computer vision and projective geometry: the image of any conic under a perspective transformation is another conic. If the image of a circle is a closed curve, ...
amd's user avatar
  • 53.8k
7 votes

4 Spheres all touching each other??

Not any size. If the fourth sphere is sufficiently small, it can fit in the hole in the middle, also resting on the table, without touching any of the other three spheres. I’m guessing this problem ...
MPW's user avatar
  • 44.1k
7 votes
Accepted

Volume of regular octahedron

The volume of a pyramid is $\frac 13 hA$ where $h$ is the height and $A$ is the area of the base. Using this for two pyramids suck together gives me $\frac{\sqrt 2}3\approx 0.47$ for a regular octagon ...
Mark Bennet's user avatar
7 votes
Accepted

Tetrahedron circumradius in high dimensions

The standard $n$ simplex is the convex hull in $\mathbb{R}^{n+1}$ of the points $$(1,0,0,\dots,0),\\ (0,1,0,\dots,0)\\ \vdots\\ (0,0,\dots,0,1)$$ That is, the set of all points in $\mathbb{R}^{n+1}$ ...
saulspatz's user avatar
  • 53.2k

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