Talk:Tetrahedron

Mathematics B‑class Mid‑priority

	Mathematics portal This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics
B	This article has been rated as B-class on Wikipedia's content assessment scale.
Mid	This article has been rated as Mid-priority on the project's priority scale.

( Written before TOC )

Maybe I'm just spacially challenged, but I can't agree with the following sentence from the article:

A tetrahedron can be embedded inside a cube so that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces.

If you obey "each vertex is a vertex of the cube", then the statement that "each edge is a diagonal of one of the cube's faces" is wrong, maximally 3 edges of the tetrahedron can be diagonals of the cube, the others must lie on the edges of the cube. WDYT? --snoyes 00:17 Feb 26, 2003 (UTC)

Consider vertices at (0, 0, 1), (0, 1, 0), (1, 0, 0) and (1, 1, 1) on the unit cube.

The edges of the tetrahedron made by these points are:

(0, 0, 1), (0, 1, 0)
(0, 1, 0), (1, 0, 0)
(0, 0, 1), (1, 0, 0)
(0, 0, 1), (1, 1, 1)
(0, 1, 0), (1, 1, 1)
(1, 0, 0), (1, 1, 1)

All face diagonals (I think: it's late, I'm tired). The Anome 00:35 Feb 26, 2003 (UTC)

Yip, I'm just spaTially (and spelling) impaired. Thanks for your time. --snoyes 01:24 Feb 26, 2003 (UTC)

The buckminster-fuller web site has a great animation that shows this :-) -- Tarquin 11:11 Feb 26, 2003 (UTC) ... and I can't find it on google. I have it on my HD though, email me Snoyes & I'll send it back to you :-)

A cube can be divided into 5 tetrahedra. The central one is a regular tetrahedron whose edges are the diagonals of the cube's faces. The other 4 tetrahedra are not regular with 3 edges diagonals of the cube and the other 3 edges edges of the cube. Thus everything said above is correct. You're just talking about different tetrahedra.

changed CTB to CTA in the volume formula

I have verified the changed formula is correct with a computer program, but not by mathematical proof. It would be good if the person who originally entered the formula could verify as well. Royappa 06:49, 1 January 2006 (UTC)[reply]

New stat table

I replace stat table with template version, which uses tricky nested templates as a "database" which allows the same data to be reformatted into multiple locations and formats. See here for more details: User:Tomruen/polyhedron_db_testing

Tom Ruen 00:52, 4 March 2006 (UTC)[reply]

edit of irregular tetrahedron symmetries

What has been here the last 6 months or more is the following:

An irregular tetrahedron (3-sided Pyramid (geometry)) with equilateral base and the top vertex above the center has 6 isometries, like an equilateral triangle.

A tetrahedron composed of two pairs of identical isosceles triangles is such that the two edges that adjoin identical triangles are opposite and perpendicular, and thus such a tetrahedron (or digonal disphenoid) has one twofold rotational axis passing through the centers of the two edges that adjoin identical triangles (in the case where all four triangles are identical and the figure is a tetragonal disphenoid this is also a fourfold improper rotational axis); there are also two mirror planes, each passing through one of these two edges and extending through the center of the opposite edge.

In other cases there is no rotational symmetry and at most one mirror plane.

This is inaccurate (according to this there are 3 or 4 cases, depending how you read it). I have put a more careful enumeration of the 7 possibilities. --Andrew Kepert 23:14, 10 August 2006 (UTC)[reply]

Proof for volume of tetrahedron

As stated above, someone requested the proof for the volume formula. I did not originally post the formula, but a proof can be found here:

http://www.mathpages.com/home/kmath424.htm.

I also think we should include this formula for the volume of a tetrahedron (found on http://mathworld.wolfram.com/Tetrahedron.html):

If a,b,c, are the three polyhedron edge vectors from a given polyhedron vertex, the volume is:

${\frac {1}{3}}\displaystyle |a\cdot |b\times c||$

--CmaccompH89 19:05, 22 August 2006 (UTC)[reply]

That should be

{\frac {1}{3!}}\displaystyle |a\cdot (b\times c)|

Tom Duff 21:48, 12 September 2006 (UTC)[reply]

In order to remove the citation request I have relocated the associated paragraph to show a progression from triple scalar representation of volume to the formula using only the three edge lengths and three angles at one vertex. Frank M Jackson (talk) 18:36, 20 March 2008 (UTC)[reply]

law of sines

\angle OAB\cdot \angle OBC\cdot \angle OCA=\angle OAC\cdot \angle OCB\cdot \angle OBA.\,

Multiplying the angles themselves is so unusual that it ought to be expressed also in words. And if this is not multiplying the angles, then its meaning is obscure enough that it ought to be expressed also in words. —Tamfang 05:32, 3 May 2007 (UTC)[reply]

Setting d=0 by translating tetrahedron

Initially sentence is awkward: "If d = 0 is the origin of the coordinate system, then the formula becomes"

Simplification improves sentence, but removes insight into why we may set d=0: "If d = 0, then the formula becomes "

I have tried to keep clear sentence structure while retaining that insight (unfortunately making the sentence much more verbose): "If the tetrahedron has been translated so that vertex d is at the origin, then d = 0, and so"

Thoughts on the best way to present this? — Unsigned comment added by 98.203.237.75 (talk) 14:44, 17 January 2008 (UTC)[reply]

I agree. However, I would rather describe it as a reference frame choice, than a translation of the tetrahedron. Please see my edit. Paolo.dL (talk) 16:36, 17 January 2008 (UTC)[reply]

Ok, sounds good! —Preceding unsigned comment added by 98.203.237.75 (talk) 17:44, 17 January 2008 (UTC)[reply]

Solid angle and radii of spheres

I have added solid angle, radii of spheres associated with a regular tetrahedron and position of exsphere center into first table. Frank M Jackson (talk) 09:39, 19 March 2008 (UTC)[reply]

Practical applications

I miss some words about the use of tetrahedron-shaped nail constructs suitable for stopping cars... because no matter how many times they fall, one peak will always point up. Eh, I am sure someone will find better words than I do. —Preceding unsigned comment added by 129.240.72.130 (talk) 10:22, 11 April 2008 (UTC)[reply]

the general case

Had I written it, I'd prefer "general tetrahedron" to "generalized tetrahedron". Opinions? Rationale? —Tamfang (talk) 06:31, 27 May 2008 (UTC)[reply]

Yes, as far as I know, generalized has a different meaning. From Cambridge Dictionary:

generalized, UK USUALLY generalised

involving a lot of people, places or things:

He spoke of generalized corruption in the government.

Isolated showers will give way to more generalized rain later in the day.

NOTE: The opposite is localized.

The term is also used in statistics, meaning "(improperly) applied to a wider population". Paolo.dL (talk) 21:01, 28 May 2008 (UTC)[reply]

Second paragraph

I removed from the second paragraph the claim that tetrahedrons are the "second most common" type of pyramid. Perhaps it meant the second most common pyramid shape used for building classical monuments, which belongs at pyramid and not here (and would need a citation), or the second most commonly taught pyramid shape at school, which seems fairly unimportant and quite debatable. In a mathematical context (which this article is), "most common" is completely meaningless.

I replaced it by pointing out that tetrahedrons are not mearly pyramids but specifically are triangular pyramids (which is obvious of course, but only if you know it) Quietbritishjim (talk) 16:36, 15 August 2008 (UTC)[reply]

Excellent edit! Wasn't necessary to explain it though :-) When I saw the diff it was a pretty obvious enhancement. Tomeasy_{T C} 17:31, 15 August 2008 (UTC)[reply]

controversial formulae

These are in Coxeter Regular Polytopes, Table I(i):

dihedral angle	circumradius	midradius	inradius	surface	volume
acos(1/3)	${\sqrt {\frac {3}{2}}}l$	${\frac {l}{\sqrt {2}}}$	${\frac {l}{\sqrt {6}}}$	${\sqrt {3}}a^{2}$	${\frac {\sqrt {2}}{12}}a^{3}$

where l is half the edge length a. —Tamfang (talk) 19:53, 21 November 2008 (UTC)[reply]

What's your point? Supporting the current state of the article, objecting it, or providing a reference? Tomeasy_{T C} 01:29, 22 November 2008 (UTC)[reply]

The last. —Tamfang (talk) 03:08, 22 November 2008 (UTC)[reply]

You could do some very nice thing then: Add the reference to the article. Tomeasy_{T C} 10:42, 22 November 2008 (UTC)[reply]

Shall I help you adding the reference to the article? Tomeasy_{T C} 18:08, 24 November 2008 (UTC)[reply]

Sorry, I have just realized that you had put the reference already. Well done! 15:50, 25 November 2008 (UTC)

Newly discovered property of Regular Tetrahedron?

Some time ago I stumbled on a web site that states following: If a regular tetrahedron is intersected with a plane that is parallel to two opposite edges, anywhere between the edges, the resulting 2D shape is a rectangle (or square) with perimeter equal to twice the edge dimension. It is easily provable. I have not seen this in the literature about the tetrahedron. Is this a new theorem?-Drova (talk) 13:40, 4 February 2009 (UTC)[reply]

I read the article and I am convinced that the statement is true. I am a little bit reluctant to believe the claim that this property is only true for regular tetrahedra. This aside, I would welcome the inclusion of this property accorded by the visual proofs. Tomeasy_{T C} 15:25, 4 February 2009 (UTC)[reply]

Unfortunately, I have never contributed images to Wikipedia before. Can files in PDF or JPEG formats be used for such purpose. It would be better if some one with experience do that. It appears to me that tetrahedra with sides that are not equilateral triangles would be excluded from this rule, however I am not convinced either.Drova (talk) 16:44, 4 February 2009 (UTC)[reply]

Well, if you think of tetrahedra only, then a non-regular tetrahedra does not even have a unique edge dimension; thus, the statement does not apply by conception. However, one might think of complete different geometries, but I am not enough an expert to say something about this. Important for us is that the statement is a property of the regular tetrahedron.

With respect to images, all issues are about copyrights. So, you may not download one of the images from the linked website. If you have a copyright for an illustration (e.g., when you have made one yourself) uploading is no problem. I could also help you with that.

If you like, you can add a statement about the property in the article without illustration. Others would probably refine the wording. However, the addition of an illustration would greatly support understanding the statement. Tomeasy_{T C} 17:18, 4 February 2009 (UTC)[reply]

I disagree. I believe the statement must have been long known, but I think it should only be added to the article if appropriate sources can be found; otherwise, we're committing original research. Incidentally, the same property is true for any tetragonal disphenoid. —David Eppstein (talk) 17:52, 4 February 2009 (UTC)[reply]

As already mentioned, the statement does not apply to a disphenoid, because it does not have a unique edge length. However, one might reformulate and say that the perimeter of the rectangle is equal to twice the length of the opposing edges, which you probably meant. So, fine it is true for these tetrahedra as well. I think I even have a clear picture on how one can state this for an arbitrary tetraheadron. (The perimeter of the rectangle would be twice an interpolation of the lengths of the two opposing edges. The interpolation depends linearly on the distance of the plane and the respective opposing edge). Anyways... The original statement is still valid for the subject of the present article and worth to mention. Especially when illustrated, I find it adds value.

Personally, I do not see the OR problem as long as the derivation is purely logical and commonly comprehensible. However, I might be wrong and tried to see whether WP:OR says something specific about mathematical derivations, but it does not. So, I started a talk page section there, because I the question as to when a logical derivation becomes OR is anyway interesting. Tomeasy_{T C} 18:59, 4 February 2009 (UTC)[reply]

If the statement has been long known, we should be able to find an appropriate source that we could refer to. So far, I haven't been able to find one. I think it is an interesting statement, that would enhance the article. It is hard to believe that it is completely novel. In regard to tetragonal disphenoid, the Wiki article referenced above states that a regular tetrahedron is not considered to be a disphenoid.Drova (talk) 18:32, 4 February 2009 (UTC)[reply]

The disphenoid tetrahedron graphic I made, showing this tetrahedron inside of a cuboid shows clearly that cross sections from this orientation (along the y axis) will be rectangles. So this appears to be true at least for all digonal disphenoids, tetrahedra with 4 equal edges and two opposite (not necessarily) unequal edges. SockPuppetForTomruen (talk) 20:04, 4 February 2009 (UTC)[reply]

Perhaps a person with good imagination can see rectangles in the above disphenoid. Such person would also be able to imagine rectangles in almost any image of a tetrahedron. What interesting about the statement is that all such rectangular cross sections have constant perimeter equal to twice the edge distance. That can be easily proven for regular tetrahedron with edge E. If distance from the vertex to the point of intersection of the plane with the edge (measured along the edge) is a, then, from equilateral triangles, resulting rectangle would be (a)x(E-a). Perimeter of such rectangle is 2(E-a+a)=2E. Is there another proof, that is not based on equilateral triangles?-Drova (talk) 23:00, 4 February 2009 (UTC)[reply]

Regular tetrahedron coords?

I feel this would be useful addition to the article. It certainly would have helped me. I'm not sure where to put them however:

	x	y	z
A	0	0	0
B	1	0	0
C	0.5	sqrt(3)/2	0
D	0.5	sqrt(3)/6	sqrt(2/3)

--Skytopia (talk) 00:01, 16 February 2009 (UTC)[reply]

Or even simpler, alternated vertices of a cube! SockPuppetForTomruen (talk) 03:06, 16 February 2009 (UTC)[reply]

	x	y	z
A	1	1	1
B	1	-1	-1
C	-1	-1	1
D	-1	1	-1

Or indeed... that one. I'm amazed how those originals numbers can be simplified to ones and minus ones like that. Good stuff. I'm guessing there's not much point to having both coord sets in the article then? --Skytopia (talk) 14:27, 16 February 2009 (UTC)[reply]

Both are useful. The first comes from a dimensional series of regular simplexes. The second it just based on a special relation in 3D where the vertices of a cube contain two tetahedra. SockPuppetForTomruen (talk) 20:21, 16 February 2009 (UTC)[reply]

I added a section but noticed it is repeated under geometric relations. It is sort of a messy article, mixing regular and general tetrahedra. SockPuppetForTomruen (talk) 20:31, 16 February 2009 (UTC)[reply]

Okay, I'll leave it up to you and others here about adding the set I suggested as well. In any case, I like the addition you made to the start of the article - nice for them to get a mention early on.--Skytopia (talk) 05:53, 26 February 2009 (UTC)[reply]