Talk:Tetrahedron

message: 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)
thread_title:  Changed CTB to CTA in the volume formula 

message: 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)
thread_title:  New stat table 

message: 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)
thread_title:  edit of irregular tetrahedron symmetries 

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

text: 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)
thread_title:  Proof for volume of tetrahedron 

message: :  \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)
thread_title:  law of sines 

message: 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)
replies: text: 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)
thread_title:  Setting d=0 by translating tetrahedron 

message: 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)
thread_title: Solid angle and radii of spheres

message: 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)
thread_title:  Practical applications 

message: 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)
replies: text: 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)
thread_title:  Second paragraph 

message: 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)
replies: text: 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)
replies: text:  The last. —Tamfang (talk) 03:08, 22 November 2008 (UTC)
replies: text:  You could do some very nice thing then: Add the reference to the article. Tomeasy T C 10:42, 22 November 2008 (UTC)
replies: text:  Shall I help you adding the reference to the article? Tomeasy T C 18:08, 24 November 2008 (UTC)

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

message: 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)
replies: text: 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)

text: 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)
thread_title:  Regular tetrahedron coords? 

message: {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)
replies: text: Yes! It's in the table on the right. Tom Ruen (talk) 03:46, 29 July 2009 (UTC)
thread_title:  [[Tetrahedron]] is also a [[snub]] [[digon|digonal]] [[dihedron]]/[[hosohedron]] 

message: 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)
replies: text: No, see the reference. Double sharp (talk) 05:36, 16 August 2009 (UTC)
replies: text: 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)
replies: text: 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)
replies: text: Can't seem to find it. Professor M. Fiendish 06:15, 22 August 2009 (UTC)
thread_title:  convexity 

message: 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)
replies: text: 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)
replies: text: 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)

text:  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)

text: 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)
thread_title: Dual tetrahedron

message: 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)
replies: text: 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)
replies: text:  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)
thread_title:  dimensions 

message: Thanks Salix for starting this article (see <Minimum Volume>, 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)
thread_title:   Tetrahedral Hypothesis 

message: 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)
replies: text: 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)
thread_title:  Dihedral Angle 

message: :::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)
replies: text: 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)
thread_title:  Brain-numbing statements of syllogisms and worse 

message: 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)
replies: text: 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)

text: 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)

text: 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)
replies: text:  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)

text: 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)
thread_title:  Proof 

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

message: 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)
replies: text: 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)
thread_title:  Degenerate form of antiprism and trapezohedron 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Incorrect definition of disphenoid 

message: 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)
replies: text: 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)

text: 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)
replies: text: 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)
thread_title:  Isosceles Tetrahedron 

message: Really? --JWB (talk) 21:42, 16 August 2014 (UTC)
replies: text: According to Coxeter (Regular Polytopes, Table I) and a bit of algebra, it's 3. —Tamfang (talk) 02:02, 17 August 2014 (UTC)
replies: text:  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)
replies: text: 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)
thread_title:  Circumsphere radius / insphere radius ratio is 8 

message: 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)
replies: text: 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)

text: 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)
thread_title:  Meaning of demicube? 

message: 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)
replies: text: It's in the second paragraph, with a link to simplex. Double sharp (talk) 06:58, 14 June 2017 (UTC)
thread_title:  3-simplex? 

message: 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)
replies: text: Good point. I've rephrased that part. —David Eppstein (talk) 22:40, 16 January 2018 (UTC)
thread_title:  Size and shape of points? 

message: 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)
thread_title:  A Commons file used on this page has been nominated for deletion 

message: 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)
thread_title:  Other considerations 

message: 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)
thread_title:  What is the "Face-vertex-edge angle "??? 

message: 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)
replies: text: 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)
thread_title:  What are "Plateau borders" 

message: 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)
replies: text: 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)
thread_title:  Distance from Centroid to Vertex 

message: 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)
thread_title:  Isogonic center 

message: How many edjes does it have 212.129.77.225 (talk) 19:38, 9 February 2022 (UTC)
replies: text: Six. 2A00:23C7:548F:C01:74DF:CE53:6F62:D4DC (talk) 18:55, 7 February 2023 (UTC)
thread_title:  Edges of a tetrahedron 

message: 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)
thread_title:  Application: Packaging 

message: 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)
thread_title:  multiple tetrahedral family that can be folded on a single A4 sheet