text:
Talk:Tetrahedron

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

 * 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 and I'll send it back to you :-)

A cube can be divided into five tetrahedra. The central one is a regular tetrahedron whose edges are the diagonals of the cube's faces. The other four tetrahedra are not regular with three edges diagonals of the cube and the other threee 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)

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)

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)

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)

 * That should be $$\frac{1}{3!}\displaystyle|a \cdot (b \times c)|$$ Tom Duff 21:48, 12 September 2006 (UTC)

 * 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)

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)

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)

 * 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)

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

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)

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)

Note over 20 years later: This application is taken care of under the See Also section: Caltrop. How the Tetrahedron is being used pedagogically (practically?) is also taken up under Quadray_coordinates, sometimes called Caltrop Coordinates, which connects back to this page. Kirbyurner (talk) 21:44, 26 March 2021 (UTC)

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)

 * 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)

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)

 * Excellent edit! Wasn't necessary to explain it though :-) When I saw the diff it was a pretty obvious enhancement. T om ea s y T C 17:31, 15 August 2008 (UTC)

controversial formulae
These are in Coxeter Regular Polytopes, Table I(i):
 * What's your point? Supporting the current state of the article, objecting it, or providing a reference? T om ea s y T C 01:29, 22 November 2008 (UTC)

 * The last. —Tamfang (talk) 03:08, 22 November 2008 (UTC)
 * You could do some very nice thing then: Add the reference to the article. T om ea s y T C 10:42, 22 November 2008 (UTC)
 * Shall I help you adding the reference to the article? T om ea s y T C 18:08, 24 November 2008 (UTC)
 * 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)
 * 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. T om ea s y T C 15:25, 4 February 2009 (UTC)
 * If you unfold a tet into a linear net of four triangles, this rectangle is the straight line that transverses the net parallel to the top and bottom. If your line is drawn midway between top and bottom you get a square. Cloudswrest (talk)
 * An easy way to visualize this rectangle is to set the tetrahedron vertically on one of its edges. Assume the bottom edge goes left/right.   The opposite top edge is perpendicular, also level with the ground and goes from front to back.  Now slice the tet with a horizontal plane.   The intersection is this rectangle.  If the plane is nearer the ground the left/right edges will be the longer and the front/back edges will be the shorter.  As the plane rises vertically the edges change and reverse rolls.  At the midpoint in height you have a square.  This square can be thought of as the "equator" of the tetrahedron.  It traverses ALL faces of the tet similarly without any turning other than that induced by traversing the dihedral angle between faces.  The tetrahedron is the only regular polyhedron that admits a discrete Hopf fibration in three dimensions.  This is because this is a degenerate case with only ONE ring.  A second or more rings would collide or intersect. Cloudswrest (talk) 03:48, 8 October 2015 (UTC)

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)
 * 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. T om ea s y T C 17:18, 4 February 2009 (UTC)
 * 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)
 * 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. T om ea s y T C 18:59, 4 February 2009 (UTC)

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)

 * 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)

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)

 * Here's an external web site that illustrates this rather well http://www.matematicasvisuales.com/english/html/geometry/space/sectetra.html Cloudswrest (talk) 19:34, 8 October 2015 (UTC)

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:

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

Thanks for that, made a lot more sense than what's on the main page. The above is what's simplest for me (although I did swap the y and the z around in order to give a simple horizontal base).77.103.105.67 (talk) 20:23, 5 August 2019 (UTC)

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

 * 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)

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)

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)

 * 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)

One common formula for circumsphere, midsphere and insphere radius
I suggest in addition to formulas of circumsphere, midsphere and insphere radius of regular tetrahedron to add one simple common formula that covers all three of them and is applicable even in dimensions other than 3. The formula is: $$R=\sqrt\,a \,$$

where n is number of points at equal distance to each other (for regular triangle n=3, for regular tetrahedron n=4, etc.)

and k is number of points defining an entity the sphere “touches” (for vertex k=1, for edge k=2, for face k=3 and so on).

This is my first post to Wikipedia and I'm not native English speaker so forgive me if something is wrong :)

 * Coxeter Regular Polytopes Table I(iii) gives
 * $$R_j/l = \sqrt{\frac{2}{j+1}-\frac{2}{n+1}}$$
 * where j is the dimension of the tangent element (0 for a vertex...), n is the dimension of the space, and l is the half-length of an edge. That's a match.  —Tamfang (talk) 02:43, 22 June 2009 (UTC)

 * I suggest that you say that the radii of the circumsphere, midsphere and insphere for a regular tetrahedron are all related by the formula
 * $$ R_k=\sqrt{\frac{3-k}{8(k+1)}}$$
 * where k is dimension of the entity that the sphere “touches” (e.g. for vertices k=0, for edges k=1, for faces k=2).
 * Then make a reference to Coxeter formula saying that it can be generalized for higher dimensional Simplices and insert the Coxeter formula into the section on Simplices. Hope this helps. Frank M Jackson (talk) 18:07, 22 June 2009 (UTC)

Hi Tamfang

I didn’t know about Coxeter formula, I delivered it myself:) But I think already ancient Greeks knew it, any way to derive it one needs only Pythagoras theorem.

Hi Frank M Jackson

I’m fully aware that my definition of n and k are somehow weird, but defining them this way makes the formula extremely easy to remember - take 1/k-1/n divide by 2 and take square root (2 comes from the fact that unit n-simplex has edge sqrt(2)).

I understand that section Formulas for regular tetrahedron is quick reference, so this simple formula can be easily there, since IMHO it is easy to remember and useful (replaces three formulas that are more dificult to remember). As for yuor sugestion to include it in Simplices I'm not so sure. If one reads about simplices than he/she problably can get this formlula on the fly, so why it should be there?
 * Firstly it would be useful if you set yourself up with a proper user account rather than remain anonymous. This enables easier discussion between users and you can get help from Wikipedia administrators.
 * I believe that the table at the front of the article is a quick and useful reference and should not be altered. The fact that the circumsphere, midsphere and insphere are all related should be mentioned but in a separate place within the article. Quote the published formula as advised by Tamfang rather than your own interpretation and make a reference back to Coxeter. It avoids confusion.
 * The Coxeter formula relates to the more general Simplex so it may be worth a mention in the Wikipedia article on simplices as there is no reference to it there, at present. Frank M Jackson (talk) 18:44, 24 June 2009 (UTC)

 * I'd adjust the Coxeter formula to use the full edge length, conforming to the rest of the table.
 * $$R_j/a = \sqrt{\frac{1}{2(j+1)}-\frac{1}{2(n+1)}}$$
 * (someone check my algebra). —Tamfang (talk) 04:05, 29 June 2009 (UTC)

Minimum Volume property
It is said that a regular tetrahedron is the solid with the minimum volume for a given surface area, while a sphere has the maximum. See ‘Tetrahedral Hypothesis’, an outmoded geological speculation. This seems quite an important property, and maybe one of you clever chaps could say something about it? John Wheater (talk) 07:38, 1 July 2009 (UTC)

 * It's false for general solids, and even for general convex uniform polyhedra. Consider a uniform n-gonal prism (whose edges are equal): as n is increased with the surface area held constant, clearly the volume can be as small as you like. —Tamfang (talk) 17:06, 7 July 2009 (UTC)

 * Please don't go changing my spellings. I meant polyhedra (noun), not polyhedral (adjective). —Tamfang (talk) 20:09, 13 July 2009 (UTC)

 * I (no mathematician) am guessing that such ‘n-gonal prisms’ form a monotonically increasing series of forms, starting with tetrahedron and cube as 3-gonal and 6-gonal. Also that as n tends to infinity the form tends to that of a sphere.  Now, for tetrahedron, cube, and sphere, we may express the volume as k*A*sqrt(A), where k is constant for each form and A is the surface area, (and I guess this true of the other members?).  For tetrahedron, cube, and sphere, the constants are, I think: , <1/(6*sqrt(6))>, and <1/(6*sqrt(pi))>.  These evaluate to about 0.0227, 0.0680, and 0.0940.  So it looks to me that there is a flaw in your reasoning, and that  Holmes was  right in his 1944 geology book.  What do you think?, and could something be said? John Wheater (talk) 06:46, 9 July 2009 (UTC)

 * Tamfang is correct, and you are not. Consider the rectangular boxes (a very special case of the prisms mentioned by Tamfang) with side lengths x, 1/x, 1/x. As x goes to infinity, the surface area will be 4 + 2/x2, very close to 4. However the volume will be 1/x, very close to 0. By choosing x large enough you can make the volume far smaller than that of a regular tetrahedron with the same surface area. —David Eppstein (talk) 07:09, 9 July 2009 (UTC)
 * If we restrict ourselves to the Platonic solid then the tetrahedron is the one out of the five with min volume, but thats not a particularly amazing result. The Tetrahedal Hypothesis may be worth an article --Salix (talk): 08:04, 9 July 2009 (UTC)
 * I've created said article at Tetrahedal hypothesis.--Salix (talk): 16:05, 9 July 2009 (UTC)
 * Interesting, but shouldn't it be Tetrahedral Hypothesis or Tetrahedral hypothesis? Tom Ruen (talk) 19:19, 9 July 2009 (UTC)
 * Now at Tetrahedral hypothesis. —David Eppstein (talk) 20:22, 9 July 2009 (UTC)

 * David, what you say above is true, but I’m sorry to have caused you to waste your time pointing out the obvious fact that you can put two rectangles arbitrarily close together! I should have said that I was considering solids formed of polygons whose sides were equal, and, in my ignorance, I thought this was what Tamfang was discussing, and what I was exemplifying.  It seems though (thanks Salix) that not all such polygons can fit together in the right way (perils of extrapolating the soccer-ball pentagon!).  But is there some set of solids that includes the Platonics and the sphere?  Holmes says “Since, for a given surface area, a sphere is the regular figure with the greatest volume, while the tetrahedron is the regular figure with the smallest volume, it was thought that a contracting globe would tend to shrink to a tetrahedral form”.  And Qazi (2004), cited in Tetrahedral hypothesis, says this is a “basic principle of geometry”.  If there is a set of Volume/Area ratios for “regular figures”, and the sphere and tetrahedron are the end-points, that does seem interesting to me, but I’ll pipe down now if nobody else thinks so.  The Tetrahedral hypothesis itself is nothing but a historical curiosity, abandoned long before plate tectonics was accepted, but the idea of a sphere “contracting” to a tetrahedron is intriguing.
 * In this context "regular figure" must mean Platonic solid. Tamfang's example of a prism with all faces being regular polygons has much smaller volume, for a given surface area, than a regular tetrahedron. —David Eppstein (talk) 18:14, 10 July 2009 (UTC)
 * Yes, thanks, good. Now I know the meaning of  prism and Platonic solid I at last understand:  all the faces are regular polygons, but not the same regular polygon!  Tamfang, like you, was provoked into pointing out that two plates of arbitrarily large area can be placed close together!  If only I’d said ‘regular solid’…I note that a sphere is not actually a Platonic solid, but I guess an icosahedron is close, with a volume/area constant of 0.0856 (cf. tetrahedron 0.0227 and sphere .0940).  So this article is not the place for the ‘minimum volume’ property;  I might try adding something to Platonic solids and Tetrahedral hypothesis.  Many thanks to all for a marvellous bit of education, and goodbye (at last). John Wheater (talk) 10:17, 11 July 2009 (UTC)

 * "But is there some set of solids that includes the Platonics and the sphere?" Well, you can make a set of anythings, heh.  I guess you'd be most interested in the family of "geodesic" spheres, which approach a sphere as the number of pieces increases. —Tamfang (talk) 20:15, 13 July 2009 (UTC)

Tetrahedron is also a snub digonal dihedron/hosohedron
{3, 3} is also s{2, 2} I believe? Se Wythoff symbol. Shouldn't this be in the article? --116.14.72.74 (talk) 03:10, 29 July 2009 (UTC)

 * Yes! It's in the table on the right. Tom Ruen (talk) 03:46, 29 July 2009 (UTC)

convexity
Recently added:
 * The tetrahedron is the only convex polyhedron that has four faces.

Is there a non-convex polyhedron with four faces? —Tamfang (talk) 20:03, 15 August 2009 (UTC)

 * No, see the reference. Double sharp (talk) 05:36, 16 August 2009 (UTC)

 * This leads to a question about the smallest number of faces for a non-convex polyhedron. I can construct one with five faces, one of which is non convex and one with six faces all of which are convex. Not sure if its possible to construct a non-convex polyhedron with five convex faces.--Salix (talk): 09:21, 16 August 2009 (UTC)

 * I think not, but I haven't got a proof yet, and even if I did it would be original research. Not sure if there's a reference anywhere. I may do some Googling. Professor M. Fiendish 06:06, 22 August 2009 (UTC)

 * Can't seem to find it. Professor M. Fiendish 06:15, 22 August 2009 (UTC)

I asked this question later at the Mathematics reference desk: you need six convex faces to make a non-convex polyhedron, but only five if the faces don't have to be convex. Double sharp (talk) 15:41, 30 May 2019 (UTC)

Dual tetrahedron
This is my first Wikipedia edit so forgive me if this is not formatted correctly. There is a mistake near the top of this article where it says: If the original tetrahedron has edge length 1, its dual tetrahedron will also have edge length 1. I build polyhedral models and know that the Tetrahedron of edge length 1 will have a dual tetrahedron with edge length of 1/2. I could try a mathmatical proof but a simple model will show you that the 1 to 1 claim is incorrect. Please let me know if you implement this change. eric.higley@kla-tencor.com Thanks! Erichigley (talk) 16:13, 27 October 2009 (UTC)

 * Actually, the statement is misleading because the exact proportions of the dual depends on how you carry out the operation (i.e., the radius of the sphere of reciprocation). It is certainly possible to reciprocate a tetrahedron in such a way that its dual also has the same edge length, but this is not the only way to do it, as your models prove. This statement should probably be edited to not make these kinds of hidden assumptions.—Tetracube (talk) 17:06, 27 October 2009 (UTC)

 * OK I've now removed the line. TRhere might be a case for a more extensive treatment of duals lower down. --Salix (talk): 17:41, 27 October 2009 (UTC)

 * The hidden assumption, I think, is that the edges of the two polyhedra intersect; in other words, that the midspheres are equal. (See also canonical polyhedron. (Oh, darn.))  Do the other reg.poly. articles have analogous statements? —Tamfang (talk) 05:30, 12 November 2009 (UTC)

Actually, the dual of a tetrahedron is congruent to the original when inscribed in the same sphere. —The Doctahedron, 68.173.113.106 (talk) 22:44, 22 November 2011 (UTC)
 * Sure, but being inscribed in a sphere is not a self-dual statement: if a polyhedron is inscribed in a sphere, one would expect its dual to be circumscribed about the sphere. More specifically the polar reciprocation of a polyhedron inscribed in a sphere, with the polarity being defined with respect to the same sphere, gives a polyhedron circumscribed about the sphere. In contrast, the property of having a midsphere is self-dual: the polar of a polyhedron with a midsphere is a dual polyhedron with the same midsphere. For the other regular polyhedra, the edge length of the dual will be different than the edge length of the primal. —David Eppstein (talk) 22:53, 22 November 2011 (UTC)

dimensions
User:Quietbritishjim asserts, *generalized* tetrahedra (i.e. not 3D) have non-3D properties (and thus the heading "Three-dimensional properties of a generalized tetrahedron" is not redundant). Does this mean other simplex articles ought to have sections such as "Two-dimensional properties of general triangles" and "One-dimensional properties of general line-segments"? —Tamfang (talk) 15:45, 13 November 2009 (UTC)

 * Sorry I didn't reply to this at the time, I've only just seen this comment. I actually don't know anything about this section, except for the comment that you've quoted.  Someone has now removed "3D" from the title again, possibly making the same mistake, possibly because the section in fact also contains non-3D facts -- I don't have time to look properly at the moment.  One thing I can say though is why tetrahedra are particularly interesting when generalised: they are one of I think only three platonic solids that generalise to an arbitrary number of dimensions.  It's easy to construct the next one up, by just adding a point in the new dimension on the normal from the centre of the previous polytope.  I've no idea why three dimensional properties of such a body would be particularly interesting though! Quietbritishjim (talk) 00:14, 11 February 2010 (UTC)

 * A tetrahedron is a 3D figure; a general tetrahedron is not an N-dimensional simplex but a 3D solid bounded by four triangles which are not necessarily equilateral. What "non-3D facts" do you see here? —Tamfang (talk) 02:55, 16 February 2010 (UTC)

Tetrahedral Hypothesis
Thanks Salix for starting this article (see , above), and to all who helped my understanding. I've expanded it a bit, and hope you sympathise with the new presentation - see discussion there. --John Wheater (talk) 11:43, 10 February 2010 (UTC)

Dihedral Angle
I think that the dihedral angle is ,in fact, equal to arcsin{sqrt{3} / 3}, and arctan{sqrt{6} / 2} Would someone who knows how to edit that do so, under formulas, and the table? (A.glasswater (talk) 03:35, 10 August 2010 (UTC))
 * I've checked the formula and the ones in the article $$\arccos\left({1 \over 3}\right) = \arctan(2\sqrt{2}) \,$$ are correct. You can see this by considering a tetrahedron of edge length 2. The dihedral angle can be calculated from a triangle formed from the mid point of one edge and the opposite edge, this has sides 2, $$\sqrt{3}$$, $$\sqrt{3}$$. From the Law of cosines the we obtain $$\cos \theta=1/3$$.--Salix (talk): 06:41, 10 August 2010 (UTC)

This is an application of the so called Law of cosines for a trihedral angle. In fact if we call V,A,B,C the four points of a regular tetrahedron where A,B,C are the points of the base, given a good orientation, we have that <AVB = <BVC = <CVA = pi/3 since the tetrahedron is regular. Now the law of sines states cos(<AVB) = cos(<BVC)*cos(<CVA)+sin(<BVC)sin(<CVA)cos(a) where a is the dihedral angle of the opposite face <AVB. Since all angles are equal cos(pi/3)=cos(pi/3)^2+sin(pi/3)^2*cos(a) cos(a)=1/3 a=arcccos(1/3)

Lhrrwcc (talk) 19:47, 16 February 2011 (UTC)

Brain-numbing statements of syllogisms and worse

 * In the first paragraph:

"The tetrahedron is the only convex polyhedron that has four faces." A). It says that a tetrahedron has four faces. That is the definition of "tetrahedron". B). It is impossible to make any nonconvex polyhedron with just four faces. Therefore all all tetahedrons are automatically convex polyhedrons. If you assemble three suitable triangles into a pseudopolyhedron, then the size and the location of the fourth side are uniquely predetermined. It is impossible to place that fourth side to make the tetrahedron nonconvex. Therefore, stating that a tetrahedron is convex is a triviality. 98.67.97.108 (talk) 07:29, 13 August 2010 (UTC)

 * True. Meanwhile, please put new topics at the bottom, and don't correct other people's signed writings; they may have different style preferences. —Tamfang (talk) 19:04, 14 August 2010 (UTC)

Formula and Notation Ambiguity
It seems to me that there's a fair amount of ambiguity in the formulae given in this article. For example, the formula for the radius:
 * $$ R= \frac {|\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})|} {12V}$$

uses the absolute value symbols to mean the length of a vector, while the formula for V uses the absolute value symbols to mean absolute value:
 * $$ 6V= |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$$

Additionally, the notation $$\mathbf{x^2}$$ to notate the dot product of a vector with itself, as in $$\mathbf{a^2}(\mathbf{b} \times \mathbf{c})$$, is potentially ambiguous to many readers, especially when the proper notation is $$\mathbf{x} \cdot \mathbf{x}$$ or $$\left\|\mathbf{x}\right\|^2$$. In some texts and many programming languages, $$\mathbf{x^2}$$ is the vector whose individual values have been squared. nben (talk) 17:28, 8 September 2010 (UTC)

 * This overloading of |·| doesn't bother me; but I agree that a² is unfortunate. (In geometric algebra it means something distinctly other than a·a.) —Tamfang (talk) 00:09, 11 September 2010 (UTC)

 * Changing  a²    to   a·a   looks messy and could also confuse. I suggest that a simple comment is inserted before the start of the formulae in this section that says − " Nota bene, in the formulas below, the scalar   a²   represents the inner vector product   a·a " − Frank M Jackson (talk) 14:02, 11 September 2010 (UTC)

 * Now implemented.Frank M Jackson (talk) 13:00, 13 September 2010 (UTC)

 * Works for me. The absolute value overload does not bother me that much, except that the double-line notation is used in the absolute value article when talking about vector spaces.  I worry that readers will believe that the notation implies that the expression inside the absolute value brackets must ultimately be a scalar.
 * That said, I don't believe that in the formula for 6V it is correct to have the absolute value brackets at all, at least not when it is being applied to all formulas. E.g. If we use the expression for O, the circumcenter, in the article with the expression for V that I noted above (also in the article), we run into occasional problems due to the absolute value.  If,
 * $$\mathbf{a}=(-1, 0, 0),\;\mathbf{b}=(0,-1,0),\;\mathbf{c}=(0,0,-1)$$,
 * then the circumcenter is
 * $$\frac{\mathbf{a}^2(\mathbf{b}\times \mathbf{c}) + \mathbf{b}^2(\mathbf{c}\times \mathbf{a}) + \mathbf{c}^2(\mathbf{a} \times \mathbf{b})}{2|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|}

= \frac{(1, 1, 1)}{2} = (1/2, 1/2, 1/2)$$.
 * This is the same result we get if we use a, b, and c equal to the euclidean bases instead of their opposites. If the absolute value is removed from the formula for V, we get the correct result of (-1/2, -1/2, -1/2).  A proper derivation is long but results in the same conclusions. Of course, removing the absolute value invalidates the expression in the article for the radius, so either that needs to be modified or the equation for O needs to not use V. nben (talk) 15:43, 14 September 2010 (UTC)
 * Well spotted. I agree that the vector equations should not have volume repesented in absolute value. I will make the change accordingly.Frank M Jackson (talk) 16:31, 14 September 2010 (UTC)

Proof
Does Wikipedia have a proof that a tetrahedron cannot tile space?? The following proves a pentagon cannot tile a plane:

A regular pentagon has angles of 108 degrees. 360 / 108 = 3 1/3, which is not an integer, so a pentagon cannot tile a plane.

Any proof that a tetrahedron cannot tile space?? Georgia guy (talk) 18:45, 28 February 2011 (UTC)

 * Well, the dihedral angle of a regular tetrahedron is not a rational fraction of a circle. Many irregular tetrahedra do tile space, though.  (And regular tetrahedra can tile curved space, but I don't suppose that's what you were after.) —Tamfang (talk) 22:14, 28 February 2011 (UTC)

 * Agreed. dihedral angle is 70.53°, not a divisor of 360, so something else (like 2 tets+2 octs) are needed. And for irregular-tet example: disphenoid tetrahedral honeycomb Tom Ruen (talk) 10:01, 1 March 2011 (UTC)

 * It might be worth giving an explicit example of a non-regular tetrahedron that can tile space. Here's a simple one: take a cubic tiling, and divide each cube into six tetrahedra, by connecting its center to its faces with square pyramids. – OfficialURL (talk) 23:26, 26 January 2020 (UTC)
 * ✅ The example you mention is the Schläfli orthoscheme; I also added a mention of the Hill tetrahedron. —David Eppstein (talk) 23:31, 26 January 2020 (UTC)

 * With respect to the stated proof that a regular pentagon cannot tile a plane, an analogous proof that a tetrahedron cannot tile space is as follows. A regular tetrahedron has vertex solid angles of ArcCos(23/27).  4 π / ArcCos(23/27) = 22.79, to 2 decimal digits of precision, a number which is not an integer, and therefore a regular tetrahedron cannot tile space.  Karen.snowgirl (talk) 18:21, 7 September 2020 (UTC)

A link to similar objects
I am wondering the name for the similar object taking (0,0,0), (0,1,0), (0,0,1), (1,0,0)??? 174.91.223.216 (talk) 14:56, 23 March 2011 (UTC)

 * That's an irregular tetrahedron; it could be called a right isosceles tetrahedron. —Tamfang (talk) 17:53, 23 March 2011 (UTC)
 * Do any scalene tetrahedrons exist?? (This means no 4 faces are congruent.) Georgia guy (talk) 17:55, 23 March 2011 (UTC)

You might want to take a look at Schläfli orthoscheme. That's on a different type of tetrahedron (with vertices (0,0,0), (0,0,1), (0,1,1), (1,1,1)) but it discusses work of Fiedler and Coxeter on triangles like the one you mention in which there is a spanning tree of pairwise perpendicular edges. As for whether there are tetrahedra with no pair of congruent faces: yes. If you pick four random points (in any reasonable continuous probability distribution) then with probability one it will have this property. —David Eppstein (talk) 17:58, 23 March 2011 (UTC)

Degenerate form of antiprism and trapezohedron
Restored previous edit mentioning that a tetrahedron is a degenerate form of antiprism and of trapezohedron. These are not just random degenerate forms as implied by, but are special cases in these families of polyhedra where n = 2. These also highlight relationships of tetrahedra to octahedra and hexahedra respectively. 75.150.168.6 (talk) 04:33, 6 October 2013 (UTC)
 * I trust that TomRuen's subsequent edits have found a better home than the article lead for those thoughts. — Cheers, Steelpillow (Talk) 09:18, 6 October 2013 (UTC)

Incorrect definition of disphenoid
The page on disphenoids is inconsistent. Someone has been changing things to insist that a "digonal disphenoid" (two pairs of isoceles triangle faces) is a special case of a disphenoid (4 congruent faces). This is not true. Digonal disphenoids do not have 4 congruent faces in general. Here a disphenoid is incorrectly defined as a digonal disphenoid. 50.177.1.160 (talk) 21:29, 13 December 2013 (UTC)
 * How do you define a digonal disphenoid? The article says it has two pairs of congruent triangles. Some definitions say a disphenoid has 4 congruent triangles, but digonal diphenoid clearly contradicts that, so are you saying a digonal disphenoid is misnamed or misdefined? Tom Ruen (talk) 01:20, 14 December 2013 (UTC)
 * Linguistically, the two definitions are consistent. Consider for example elongating a cube parallel to one set of edges. The resulting figure is a cuboid, specifically an elongated cuboid: we add "-oid" to the root of "cube". Applying the same language rule to the sphenoid we would get a "digonal sphenoidoid" which is grammatically nonsensical - it is ether "like a wedge" or it is not, to say that it is "like a thing like a wedge" is superfluous. Linguistically we already have the "-oid" so there is no need to add it again. Thus, we find "digonal sphenoid" to be the correct form. What would probably be useful to help avoid this sort of ambiguity would be an accepted qualifier for the 4-congruent disphenoid, "equiangular disphenoid" or "equisphenal disphenoid," something like that, and a more general definition for the term "disphenoid". But it is likely that the mathematical community are in a mess over this sort of thing, in which case we can do no better than mirror that mess. — Cheers, Steelpillow (Talk) 13:12, 14 December 2013 (UTC)

Isosceles Tetrahedron
I notice that the definition given here for an isosceles tetrahedron is different to that at MathWorld. Here it says their faces are isosceles triangles, but MathWorld says it means that opposite edges have the same length. The article Disphenoid seems to agree with MathWorld. Auximines (talk) 16:54, 22 January 2014 (UTC)

 * I would be astonished if mathworld were correct. Clearly the property of equal-length opposite pairs applies to a tetrahedron having four congruent isosceles faces. But it is not exclusive because it applies to any tetrahedron having four congruent faces, even if they are scalene. I would not expect such a scalene-faced figure to be correctly described as an "isosceles tetrahedron." This is not the first suspicious material I have seen on Mathworld. Stay cool and try to reference a more reliable source. — Cheers, Steelpillow (Talk) 17:51, 22 January 2014 (UTC)

 * It looks like the equal-opposite edge definition is well-used, like this reference from John Leech (mathematician), which would define both the tetragonal and rhombic disphenoids as isosceles tetrahedra. Tom Ruen (talk) 23:04, 22 January 2014 (UTC)

 * Ah, of course, just because the tetrahedron is isosceles, it does not mean the face triangles are: in the tetrahedron the equal edges are not adjacent. Silly me. But I would now question whether an isosceles tetrahedron and disphenoid are synonyms. Is an isosceles tetrahedron with scalene faces really a disphenoid? — Cheers, Steelpillow (Talk) 10:34, 23 January 2014 (UTC)

Circumsphere radius / insphere radius ratio is 8
Really? --JWB (talk) 21:42, 16 August 2014 (UTC)

 * According to Coxeter (Regular Polytopes, Table I) and a bit of algebra, it's 3. —Tamfang (talk) 02:02, 17 August 2014 (UTC)

 * And the article does not say anywhere that it is 8. The section Formulas for a regular tetrahedron gives formulas from Coxeter for both radii, and their ratio is clearly 3, not 8. So there is no problem here. Dirac66 (talk) 02:07, 17 August 2014 (UTC)

 * It is indeed exactly three. Simply dissect a tetrahedron into tetrahedra and octahedra of one-third side length. The circumsphere of the innermost tetrahedron is just the insphere of the original. — Cheers, Steelpillow (Talk) 10:50, 17 August 2014 (UTC)

Ordinary

 * The tetrahedron is the simplest of all the ordinary convex polyhedra and the only one that has four faces.

What does "ordinary" mean here? And is "convex" necessary? —Tamfang (talk) 03:06, 17 August 2014 (UTC)

 * "Ordinary" means that we are not discussing twisted non-Euclidean forms such as the hemicube, dihedra and so forth. One could write "Euclidean" instead, but I think the meaning is clearer this way - calling them Euclidean begs the question as to what is so significant about Euclid and we will have visitors whose maths is not even that advanced. "Convex" is not necessary as long as we keep "ordinary" or "Euclidean", although it is true, and I think it could be removed. — Cheers, Steelpillow (Talk) 10:59, 17 August 2014 (UTC)
 * Then I thought, we could just say "elementary". To a lay person that implies basic junior school stuff, to a mathematician it implies a derivation from the principles in Euclid's Elements. Both are true. But is such a play on words cheating? — Cheers, Steelpillow (Talk) 11:03, 17 August 2014 (UTC)

 * Since the exact meaning of ordinary (or Euclidean or elementary) will not be obvious to all readers, it should be defined on the article page. Perhaps in a footnote at the end of the sentence which uses it. Dirac66 (talk) 13:22, 17 August 2014 (UTC)

 * "Ordinary" is not a defined mathematical term in this context. It is a word used in expressing definite ideas, just like the words "exact", "meaning" and "obvious" which you used in expressing your definite ideas. That is why I prefer it to "Euclidean" in the present context. — Cheers, Steelpillow (Talk) 16:21, 17 August 2014 (UTC)

 * The word "simplest" is entirely a matter of opinion, and opinion is not appropriate in Wikipedia articles. For instance, many people might say the cube is simpler since all of its angles are 90º. This sentence should be modified so it states only something that is factual, or else it should be deleted.

 * For instance, the sentence could state that four is the smallest possible number of vertices, and four is the smallest possible number of vertices, for any polyhedron. It also might make reference to the regular tetrahedron, which most people would say is simpler than the non-regular ones.67.55.241.99 (talk) 14:45, 24 August 2020 (UTC)

Meaning of demicube?
I clicked demicube in the "Fundamental convex regular and uniform polytopes in dimensions 2–10" box at the bottom of a page and was redirected here. There doesn't seem to be any explanation of what "demicube" means on this page. - Rainwarrior (talk) 09:41, 21 April 2016 (UTC)
 * See demihypercube for the n-dimensional case. When n < 5 it's not that interesting because it coincides with some of the other families (it's the digon in 2D, the tetrahedron in 3D, and the 16-cell in 4D). Double sharp (talk) 12:35, 21 April 2016 (UTC)
 * I included the term in the section with alternated cube, and relinked demicube to that section. Tom Ruen (talk) 13:22, 21 April 2016 (UTC)

3-simplex?
Surely the lead paragraph should mention the other synonym of the tetrahedron, that it's a 3-simplex (that is to say, the 3-dimensional analogue of a triangle)? I'm sure at least one good citation for this can be found, on Wolfram Mathematics or a similar site.

(Does Wikipedia have a "simplex" article? Generally, a "N-[dimensional] simplex" is the simplest possible N-dimensional shape (hence the name), and has N+1 vertices and N+1 faces, each face being a (N-1)-simplex.)

 * Technically the simplex is only the simplest, faced shape, if you define "simplest" as the smallest description allowing for the least amount of variance. There n-sphere is the simplest shape in n dimensions. User:Average  — Preceding unsigned comment added by 65.39.112.9 (talk) 02:56, 6 April 2020 (UTC)

2A02:C7D:419:2500:8D27:89E7:BBAE:3315 (talk) 06:48, 14 June 2017 (UTC)
 * It's in the second paragraph, with a link to simplex. Double sharp (talk) 06:58, 14 June 2017 (UTC)

Size and shape of points?
In the section "Regular tetrahedron" there is a sentence that states: "In a regular tetrahedron, not only are all its faces the same size and shape (congruent) but so are all its vertices and edges". Now, to my knowledge, vertices are points, and I wonder whether one can talk about the size and shape of points.Ekisbares (talk) 22:38, 16 January 2018 (UTC)
 * Good point. I've rephrased that part. —David Eppstein (talk) 22:40, 16 January 2018 (UTC)

A Commons file used on this page has been nominated for deletion
The following Wikimedia Commons file used on this page has been nominated for deletion: Participate in the deletion discussion at the. —Community Tech bot (talk) 17:55, 18 May 2019 (UTC)
 * Rhombicdodecahedron.jpg

Other considerations
It's not clear that the volume for the tetrahedron applies to all 4-faced pyramids or just tetrahedrons. It's very interesting if the former is the case. Also that the formula in this case only applies to pyramids whose apex lies above the base. User:Average64 — Preceding unsigned comment added by 65.39.112.9 (talk) 02:45, 6 April 2020 (UTC)

What is the "Face-vertex-edge angle "???
In the table of various geometrical quantities, one listing is for the "Face-vertex-edge angle".

But this is certainly not standard terminology in mathematics, and I am not sure what it refers to. I suggest that — whatever this means — it should be expressed much more clearly.67.55.241.99 (talk) 14:39, 24 August 2020 (UTC)

Seems like "Face-vertex-edge angle" should be changed to "Face-center-edge angle" where "face" is the center of a face, "center" is the center of the tetrahedron and "edge" is the midpoint of one of the edges of the "face". Tomldavis (talk) 19:34, 13 November 2021 (UTC)

What are "Plateau borders"
The table of values of geometrical quantities for the regular tetrahedron includes this sentence:

"Vertex-Center-Vertex angle, the angle between lines from the tetrahedron center to any two vertices. It is also the angle between Plateau borders at a vertex."

But nobody knows what "Plateau borders" are, and the linked Wikipedia article "Plateau's laws" does not mention the word "borders" anywhere in the article. Maybe this is just Too Much Information.67.55.241.99 (talk) 15:00, 24 August 2020 (UTC)

 * In the linked article "Plateau's laws", see section "Laws for soap films", points 3 and 4. Dirac66 (talk) 15:46, 24 August 2020 (UTC)

Distance from Centroid to Vertex
For a regular tetrahedron with unit edge lengths, it would be useful to have the distance from the centroid to a vertex. I think it's sqrt(3/8) = sqrt(6)/4, from the coordinate v4 for a regular tetrahedron inscribed in a unit sphere with its centroid at the center of the sphere. If that's correct, it should be added to the table.

In a regular tetrahedron with unit edge lengths, if one constructs spheres with radius 1/2 at each vertex, I think the radius of the sphere that can be inscribed in the space between those spheres, with its center at the centroid, is sqrt(6)/4 - 1/2, or about 11% of the edge length.

The ratio of the volume of the intersection of the spheres at the vertices with the tetrahedron, to the volume of the tetrahedron, is pi/sqrt(18), or about 74%, a result known to Kepler. This is the result for hexagonal closest packing of spheres, which Kepler conjectured was optimal.

Adding a sphere at the centroid with radius sqrt(6)/4-1/2 increases this ratio only a tiny bit, to about 74.64%.

This has implications for filling a space with spheres of different sizes, assuming optimal packing. I haven't worked out the radius for the spheres that can be inscribed in the four resulting gaps, or the next.... It would be nice to have a formula for the sequence of radii, so one could know the sizes of spheres necessary to fill space to a specified degree.

Van.snyder (talk) 01:53, 28 October 2020 (UTC)

 * The sqrt(3/8) in your first paragraph is already in the article. See section Angles and distances, 4th line of Table which has sqrt(3/8) a, and you have assumed unit edge length so a = 1. Dirac66 (talk) 00:42, 29 October 2020 (UTC)

Isogonic center
I find the following paragraph about the isogonic center to be blatantly false: "When all the solid angles at the vertices of a tetrahedron are smaller than π sr, O lies inside the tetrahedron, and because the sum of distances from O to the vertices is a minimum, O coincides with the geometric median, M, of the vertices." The isogonic center O is the point of minimum of OA+OB+OC+OD, it exists and it is unique since the sum of the distances from A,B,C,D is a convex function. By Lagrange multipliers, the point O is such that the sum of the normalized vectors OA,OB,OC,OD is zero. We may easily check that this is not true, in general, for the point M=(A+B+C+D)/4, which is the midpoint between the midpoint of AB and the midpoint of CD. If we slightly rotate CD around its midpoint, M stays in the same place, the sum of the normalized vectors MA,MB is the same, but the sum of the normalized vectors MC,MD changes, so M cannot be the isogonic center.

Also, M is a minimizer for $$PA^2+PB^2+PC^2+PD^2$$ (by the parallel axis theorem), hence it cannot be a minimizer for PA+PB+PC+PD too, unless we are in very peculiar situations.

--Elianto84 (talk) 22:03, 6 January 2021 (UTC)

Edges of a tetrahedron
How many edjes does it have 212.129.77.225 (talk) 19:38, 9 February 2022 (UTC)

 * Six. 2A00:23C7:548F:C01:74DF:CE53:6F62:D4DC (talk) 18:55, 7 February 2023 (UTC)

Application: Packaging
The now-rare tetrahedral Tetra Pak Classic carton shipped in hexagonal crates, and is the origin of the packaging multinational's name. There are supporting photos in the article about the company, including: scruss (talk) 16:23, 19 July 2022 (UTC)

multiple tetrahedral family that can be folded on a single A4 sheet
2^N multiple tetrahedral family that can be folded on a single rectangle sheet

Using a sheet of √1:√2 paper such as A4, you can fold a tetrahedron with four identical √3:√4 isosceles triangular faces into multiple pieces that are multiples of 2^N without cutting. Each of these tetrahedrons has four triangular paper faces, which do not overlap. That is, there are four times as many triangular faces as there are tetrahedra. In the case of a single sheet, groups of 2, 4, 8, and 16 tetrahedra of the same size are folded in various strange ways. Also, if you follow a certain rule (pattern), you can fold an infinite number of 2^N tetrahedra. 183.177.128.238 (talk) 05:20, 29 November 2023 (UTC)

title: Talk:Tetrahedron


text:
The Real McCoys

The Real McCoys is an American sitcom starring Walter Brennan, Richard Crenna, and Kathleen Nolan. Co-produced by Danny Thomas's Marterto Productions in association with Walter Brennan and Irving Pincus's Westgate Company, it was broadcast for six seasons: five by the ABC-TV network, from 1957 to 1962; and a final season by CBS, 1962–1963. Set in California's San Fernando Valley, it was filmed at Desilu studios in Hollywood.

Synopsis
The Real McCoys concerns the lives of a family originally from the Appalachian Mountains of West Virginia, from a fictional place named Smokey Corners. They relocated to California to live and work on a farm they inherited from Ben McCoy, brother to patriarch Amos McCoy. The family included Grandpa Amos (Walter Brennan), his grandson Luke (Richard Crenna), Luke's new wife Kate (Kathleen Nolan), Luke's teenage sister Tallahassie "Hassie" (Lydia Reed), and her 11-year-old brother "Little Luke" (Michael Winkelman). The double naming of the brothers is explained in the first full episode ("Californy, Here We Come"), when the elder Luke introduces Little Luke to Pepino Garcia (Tony Martinez) and says, "Well, you see, in the excitement of having him, Ma and Pa plum forgot they already had me." Only Crenna appeared in all 225 episodes.

The former West Virginians join the Grange farm association and hire Pepino when he informs them he was Ben's foreman. In the episode broadcast January 8, 1962, Pepino becomes an American citizen and assumes the surname "McCoy." The McMichaels, a brother and sister played by Andy Clyde and Madge Blake in 29 and 21 episodes, respectively, lived on the hill near the McCoys. Amos McCoy and George McMichael, both crotchety old men, sometimes quarreled, often about their games of checkers and horseshoes. Kate is friendly with the Flora McMichael, George's sister, and becomes involved with life in the community.

Although still in her 20s, Kate serves as a mother figure for Luke's younger siblings, Hassie and Little Luke. Many episodes have a moral theme consistent with the conservative opinions of Walter Brennan.

Jon Lormer was cast seven times as an actor for The Real McCoys during 1959 and 1960, six as the character Sam Watkins. Joan Blondell appeared three times near the end of the series as Aunt Win. Marjorie Bennett was cast three times as Amanda Comstock. Pat Buttram and Howard McNear also appeared three times; they were subsequently cast as Eustace Haney on CBS's Green Acres and as Floyd the Barber on CBS's The Andy Griffith Show. Olin Howland and Willard Waterman appeared five times each as Charley Perkins and Mac Maginnis, respectively.

Early in the series, Charles Lane, who often appeared in a character role on I Love Lucy, was cast twice as Harry Poulson, a fast-talking egg salesman; Hassie McCoy has an interest in Harry's son. During 1963, Jack Oakie appeared three times in the role of Uncle Rightly. Dick Elliott was cast twice as Doc Thornton, and Lurene Tuttle appeared twice as Gladys Purvis, the widowed mother of series character Kate McCoy, with Jay Novello in one of those appearances as Gladys' intended second husband, a retired photographer from Fresno.

Malcolm Cassell appeared several times as Hassie McCoy's boyfriend, Tommy. Edward Everett Horton (the narrator of Fractured Fairy Tales) played J. Luther Medwick, the grandfather of Hassie's other boyfriend, Jerry; Medwick and Amos soon clash. Verna Felton, a member of the December Bride cast, appeared once as Cousin Naomi Vesper. Jesse White, known later as the actor portraying the Maytag repairman for television commercials and subsequently a cast member of CBS's The Ann Sothern Show, portrayed a used car salesman named "San Fernando Harry" who clashes with Amos in "The New Car" (October 2, 1958). On June 1, 1961, Amos, Luke, and Kate return to West Virginia for the 100th-birthday gathering of "Grandmother McCoy", played by Jane Darwell. In one episode, Lee Van Cleef played an Air Force Security Guard; in another Tom Skerritt appeared as a letter carrier.

The episode "The Tycoon" (August 30, 1960) four years later coincidentally shared the title of Brennan's next ABC sitcom, The Tycoon, with his co-actor Van Williams. Barbara Stanwyck made a cameo appearance in the 1959 episode "The McCoys Go To Hollywood", which also features Dorothy Provine, and a glimpse of the Desilu Studios, where the series was filmed. In 1961, Fay Wray is featured in the episode "Theatre in the Barn", as herself. She volunteers to direct a local amateur production to raise money for the Grange.

Just before The Real McCoys ended as an ABC series, Nolan quit the series due to a contract dispute and was written out of the remaining scripts; her character died, but details were never given. Hassie left home to attend college, and Little Luke joined the United States Army. For the final season Hassie appeared only in the first episode; Little Luke was never seen again. Amos McCoy did not appear in the last eleven episodes; he was said to be back visiting family in West Virginia. Luke was a widower, and many of the stories concerned Grandpa trying to find him a new wife. This nearly succeeded when Luke met Louise Howard, portrayed by Janet De Gore, a widow with a young son, Greg, played by Butch Patrick, later of CBS's series The Munsters, a relationship which continued through the end of the series.

Home media
Infinity Entertainment released the first four seasons of The Real McCoys on DVD between 2007 and 2010.

On May 7, 2012, it was announced that Inception Media Group (IMG) had acquired the rights to the series. IMG subsequently re-released the first two seasons on DVD.

On June 4, 2012, IMG announced that it would be releasing a complete series set, featuring all 224 remastered episodes. Release of this set never happened, and the rights to the series' home videos later moved to SFM Entertainment, which released The Real McCoys: The Complete Series on August 29, 2017 in Region 1. This is a manufacture-on-demand (MOD) release, available exclusively from Amazon.com.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: The Real McCoys


text:
Talk:Masoud Barzani

Untitled
This page is a word-for-word copy of. I'm not sure if the text is copyrighted or not. -- Goatasaur

 * Yes, I replaced it with an external link ( 17:57 Mar 16, 2003 (UTC)

Requested move

 * The following discussion is an archived debate of the . Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

close early. — Khoikhoi 00:25, 21 September 2006 (UTC)

Mesud Barzani → Mesûd Barzanî – The correct spelling of his name in Kurdish is with the diacritics. I believe that not using them would essentially be misspelling his name. — Khoikhoi 05:12, 20 September 2006 (UTC)

Survey
Add "* Support" or "* Oppose" followed by a brief explanation, then sign your opinion with ~
 * Support. — Khoikhoi 05:12, 20 September 2006 (UTC)
 * Oppose. The article should be at Massoud Barzani. -   AjaxSmack    07:24, 20 September 2006 (UTC)
 * Oppose I agree with AjaxSmack- it should be at Massoud Barzani. This is in line with the policy WP:NAME which says article naming should give priority to what the majority of English speakers would most easily recognize I think this spelling is most commonly used by English-speakers, regardless of the Kurdish spelling. Just have a look at at all the redirects if you want evidence! AndrewRT - Talk 22:06, 20 September 2006 (UTC)
 * Ok...should I change it? — Khoikhoi 22:15, 20 September 2006 (UTC)
 * Oppose The usual spelling in English is Massoud Barzani. In Kurdish alphabet e is pronounced as a in cat, a likely source of confusion for English speakers, therefore it is better to stick with the English spelling.Heja Helweda 00:13, 21 September 2006 (UTC)

Discussion
Add any additional comments
 * Support The name should be Masoud Barzani, as this is the spelling the Kurdistan Region Presidency uses on its website, and President Barzani himslef signs English correspondece with this spelling. 159.255.160.88 (talk) 09:45, 16 February 2013 (UTC) KZibari
 * The above discussion is preserved as an archive of the debate. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

Copyvio deleted
I have deleted the copyvio text and restored a clean version. -- Aguerriero  ( talk ) 22:37, 16 October 2006 (UTC)

Move to Masoud Barzani
Both the spellings (Masoud and Massoud) are common in English texts. However, among these two, Masoud is more consistent with the original spelling. Also, in the official governmental website (www.krg.org), it's written as "Masoud". Jahangard 06:10, 3 March 2007 (UTC)

 * Under WP:NAME we should use the name most commonly used in english, not necessarily the name that is most "correct". A google search shows 1567 hits for Massoud in news vs 19 hits for Masoud. Web search shows 29900 hits for Masoud vs 147000 hits for Massoud. This shows that Massoud is used most of the time and hence I suggest it is kept here. By all means mention the alt spelling in the introduction text. AndrewRT(Talk) 12:12, 3 March 2007 (UTC)
 * I think when both spelings are common in English, the official spelling should take precedence. Jahangard 16:30, 3 March 2007 (UTC)
 * Even WP:NAME doesn't suggest using the number of Google search results as the only criterion for chosing the spelling. Jahangard 16:38, 3 March 2007 (UTC)

The official spelling should take precedence, just because their are more hits on an incorrect spelling does not mean it is right. It should be changed to Masoud.Csim125 (talk) 15:49, 8 March 2013 (UTC)

Corruption and nepotism
The brief information I have added on this subject has repeatedly been removed. As an encyclopaedia, Wikipedia, must of course include information that is less than flattering  about the subjects. Please do not remove again. If you have reservations about this information on Massoud Barzani, try rewriting instead.--Vindheim (talk) 18:08, 13 September 2008 (UTC)

Correct spelling of his name in English
The correct spelling for the name is Masoud Barzani, with one 's' in the first name. But the article title includes two occurances of 's'. His official biogrpahy at www.krp.org/MasoudBarzani, spells it with one 's'.

I have tried to 'move' the article to rename it, but the request is not allowed.

Mustafa was a moslem
Masoud Barzani's uncle Mahmoud, the elder brother of Mulla Mustafa Barzani, is reported to have become a Christian convert in 1931. However no such claims have  (afaIk) been made concerning the famous Mulla Mustafa.--Vindheim (talk) 21:27, 3 October 2008 (UTC)
 * Thank you. Mustafa Barzani's page should also be corrected accordingly. --CreazySuit (talk) 21:28, 3 October 2008 (UTC)

All Barzani Family in Iraq is Jewish not Muslim or Christian. You can look at Kurdish Jews — Preceding unsigned comment added by Eminonuk (talk • contribs) 14:30, 2 June 2011 (UTC)

Gulf wars
This is confusing. The first Gulf war was the Iran-Iraq war. The second Gulf war was 1990, the Third Gulf War 2003. Please reference the wars by year (Gulf war 1990). It is confusing to read something happened in the "first Gulf war" and then read about air-interdiction zones enforced by the USA and UK (who did not participate in the first Gulf war). --141.70.81.136 (talk) 12:35, 11 September 2011 (UTC)

Incorrect external link
Under the External Links section, please update the Kurdistan Region Presidency website link to: http://krp.org/english/pdisplay.aspx?sm=oYjJbpqKulY= — Preceding unsigned comment added by 159.255.160.88 (talk) 11:42, 11 February 2013 (UTC)

Some suggestions
There is a serious error in here -- the KDP-PUK civil war did not begin in "December 1994". I was in Hawler (Erbil) in July of '94 and saw it happening myself. It actually started in May 1994. The wiki entry on the civil war has the correct start date -- http://en.wikipedia.org/wiki/Iraqi_Kurdish_Civil_War

I would also like to offer a few suggestions:

1) Regarding spelling of the name, shouldn't we try to correct errors, no matter how popular, by indicating that the official spelling is Masoud? I've made this error (of two 's's instead of one) too many times in my own publications on the Kurds, and it's been pointed out to me that two s's indicate a stress that just doesn't exist in his name.

2) I suggest the following rewrite of the section on the "autonomous zone" -- as is, the description starts and stops at the wrong points in time, which will probably confuse people unfamiliar with the issue:

1st paragraph stays the same. After that, change to read: Just a few months after the creation of the autonomous zone, free elections (a first in Iraq) were held there in 1992. The vote split fairly evenly between the KDP and PUK. The two parties then agreed to divide all the posts of the new Kurdish government evenly between them, with a PUK vice-minister for every KDP minister and vice versa. In May 1994, however, fighting broke out between the Peshmerga of the PUK and of KDP (in a region subject to both international sanctions and Saddam Hussein's sanctions, resources were exceedingly scarce, and competition over what little was available probably helped spark the confrontation). Several armistice agreements were made and broken.

In the summer of 1996 Massoud Barzani called on the assistance of Saddam Hussein's regime to help him combat the PUK, which was receiving Iranian assistance. With the aid of the Iraqi army (some of whose Republican Guard units entered the Kurdish region for about 2 weeks' time), they drove the PUK over the border to Iran. The PUK eventually regrouped and with Iranian assistance retook Suleimani and parts of Hawler province. An end to the civil war was brokered in 1998 in the Washington Peace Accords, leaving the Kurdish zone divided between the two dominant parties; KDP in the Northwest and PUK in the Southeast.

From that time until 2003, there actually existed two parallel Kurdistan Regional Governments, with the KDP and PUK each running their own ministries headed by their own Minister of Interior, Prime Minister, Education Minister, and so forth. Gareth Stansfield [add citation to his 2003 book -- Iraqi Kurdistan: Political Development and Emergent Democracy, Routledge/Curzon, 2003] described the post-1998 status quo as actually beneficial, however, because it encouraged competition to deliver better governance in the two Kurdistan Regional Governments.

After 2003 the KDP and PUK embarked on an escalating series of efforts to reconcile their competing administrations, effectively creating a united Kurdish front in post-Saddam Iraq. Today all the ministries have been unified, as have Kurdistan's representation offices abroad (the KDP and PUK used to each run their own pseudo-embassies abroad). Although questions remain regarding the actual unification of especially the Ministry of Peshmerga (the equivalent of the Ministry of Defense in other countries), even this has officially been put under the command of the Kurdistan Regional Government and currently (2013) has a Minister from the PUK in charge, who answers directly to the Prime Minister (currently Nechirvan Barzani of the KDP).

3) The "criticism" section of the page looks problematic to me, given that Masoud Barzani is a living person. I just compared it to the page on Vladimir Putin (http://en.wikipedia.org/wiki/Vladimir_Putin) -- the critical language on Putin is much more careful (and the vast majority of the bio is much more full of praise for him, but that's a separate issue) and backed up by substantiated claims. On Masoud, I don't believe it's been proven that he personally owns much of anything.  I would thus suggest changing the following sentences: "Massoud Barzani and his relatives allegedly control a large number of commercial enterprises in Kurdistan-Iraq, with a gross value of several billion US dollars. The family is routinely accused of corruption and nepotism by some Kurdish media as well as international observers including Michael Rubin. President Barzani on several occasions has denied involvement in any commercial enterprises." The change I would suggest is: "Members of the Barzani family allegedly control a large number of commercial enterprises in Iraqi Kurdistan, with a gross value of several billion dollars, although no evidence of such ownership by Masoud Barzani himself exists.  While accusations of corruption against both the KDP-Barzanis and the PUK-Talabanis are often levied by both Kurdish sources and international observers like Michael Rubin [add the citation here], a lack of financial transparency in the region serves to both exacerbate the accusations and hamper efforts to find any evidence of malfeasance."

David Romano Thomas G. Strong Professor of Middle East Politics Missouri State University Author, The Kurdish Nationalist Movement (Cambridge University Press, 2006) — Preceding unsigned comment added by 173.202.176.147 (talk) 19:23, 15 February 2013 (UTC)

Hi, the page is protected but it should be updated in the referendum section that he is no longer president and that it is now Nechirvan Barzani Timrosen (talk) 11:37, 29 June 2021 (UTC)

Picture of Masoud Barzani
Masoud Barzani is known for his traditional clothing, it would be better to put a picture from him with his traditional clothing. --Alan Genco (talk) 15:55, 27 March 2013 (UTC)

Requested move 2013

 * The following discussion is an archived discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review. No further edits should be made to this section.

The result of the move request was: Moved. EdJohnston (talk) 21:29, 23 May 2013 (UTC)

– The first name in the title should be changed to Masoud for multiple reasons. The double s (as currently spelled incorrectly in the title) notates that Barzani's name should be pronounced with a "long s" sound in Kurdish/Arabic. However, that is not the correct way to pronounciate his name in the Kurdish/Arabic languages. When speaking his name in those languages it is spoken with a "short s." Writing the name in English as Massoud is an entirely different name in Kurdish/Arabic. For further reason that the name should be changed to Masoud, President Barzani's name English biography spells his first name as "Masoud" and not "Massoud" - http://krg.org/p/p.aspx?l=12&s=030000&r=316&p=226. Additionally, U.S. State Department use the spelling "Masoud" in the Secretary's published remarks and readouts. . Also,prominent U.S. news outlets spell his name "Masoud:" the Financial Times,, USA Today , and Foreign Policy Magazine Csim125 (talk) 21:09, 15 May 2013 (UTC)
 * Massoud Barzani → Masoud Barzani
 * The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page or in a move review. No further edits should be made to this section.

Semi-protected edit request on 9 September 2016
Halgurd (talk) 23:23, 9 September 2016 (UTC) Since 19th of August 2015, Barzani's presidency tenure has legally ended and is a controversial topic among various Kurdish political parties. While KDP still considers him the President of the Kurdistan region of Iraq, other parties such PUK and Goran dispute this.http://www.iraqoilreport.com/news/barzani-remains-in-office-after-term-expires-16206/ []
 * Red question icon with gradient background.svg Not done: it's not clear what changes you want to be made. Please mention the specific changes in a "change X to Y" format. KGirlTrucker81huh? what I'm been doing 17:39, 11 September 2016 (UTC)

PoB in the infobox
Responding to your edit summary, "it [Republic of Mahabad] was de facto where he was born; putting the historical state at the time is the standard on Wikipedia", I would like to emphasize that the city was Iranian territory at the time and Republic of Mahabad was an unrecognized state. You don't consider someone born in Raqqa in 2015, to be born in the Islamic State of Iraq and the Levant, do you? Pahlevun (talk) 18:16, 17 October 2017 (UTC)
 * Fine, change it back if you will, but link to Pahlavi dynasty and not the Islamic Republic of Iran. SpikeballUnion (talk) 18:19, 17 October 2017 (UTC)

Semi-protected edit request on 13 October 2019
The first line of the article says "Kurdistan Region Region" when it should only say "Kurdistan Region". Zucker42 (talk) 00:03, 13 October 2019 (UTC)

 * ✔ by the time I got here. –Deacon Vorbis (carbon • videos) 00:39, 13 October 2019 (UTC)

Nationality and citizenship
Pahlevun, how else would you like to describe the nationality? See maybe also the article Nationality where the nations who do not have a country are described. He sure has the Iraqi citizenship, but he also belongs to the Kurdish nation. Paradise Chronicle (talk) 20:10, 27 September 2021 (UTC)
 * Please note that in this context, nationality is a legal term and should not be with mistaken with ethnicity. For example, a citizen of American Samoa is not a U.S. citizen, but is a U.S. national while a person born in an incorporated territory would be both a citizen and national. Pahlevun (talk) 20:18, 27 September 2021 (UTC)
 * No in this context it is not a legal term, as stated in the article nationality. Then also, we refer to a wikipedia article, which is also not really recommendable. What I mean is that the Kurds are of a nation who share similar traditions like language, dances or the form they dress etc. Paradise Chronicle (talk) 20:43, 27 September 2021 (UTC)

His presidency
Masoud Barzani Became President after the KRG gained autonomy in 1992 he became president on August 31st 1992 185.100.23.36 (talk) 14:51, 29 November 2022 (UTC)

After him
Between November 1, 2017 and June 10, 2019 the office of President of Kurdistan Region was vacant. Aminabzz (talk) 15:00, 4 November 2023 (UTC)

title: Talk:Masoud Barzani