Talk:Hill tetrahedron

Originality?
How do these compare with Schl&auml;fli's orthoscheme, which Coxeter calls the "characteristic tetrahedron" of the cubic spacefilling? These were actually used in ancient China to calculate volumes for earthworks, along with a small family of related scissors-congruent polyhedra such as a certain square pyramid (3 of which tile a cube). As mentioned here -- Cheers, Steelpillow (Talk) 14:07, 15 December 2008 (UTC)
 * I don't know what is "Schl&auml;fli's orthoscheme". Judging by the name, $$Q(\alpha)$$ is unlikely to be one.  Of course, $$Q(\pi/2)$$ was known for centuries.  I see no problem with this.  On the other hand, there are quite a few references which refer to these as Hill tetrahedra.  If you know for a fact that $$Q(\pi/2)$$ is called Schl&auml;fli's orthoscheme or "characteristic tetrahedron", I would encourage you to add this info to the article. Igorpak (talk) 17:39, 15 December 2008 (UTC)
 * Well, I have no idea what $$Q(\pi/2)$$ means or what the associated figure looks like, and come to think of it I possibly misunderstand what "scissors-congruent" means. I cannot find anything intelligible on the web. Do you know of an online resource which gives basic information, understandable by the non-algebraic geometer? The cubic orthoscheme is a tetrahedron having three congruent, mutually orthogonal sides, into 48 of which the cube can be dissected - these may then be reassembled into other shapes. It is the topmost polyhedron in this diagram -- Cheers, Steelpillow (Talk) 12:50, 16 December 2008 (UTC)

Math symbols not displayed
This math symbol just looks like 3 boxes for me. (using Internet Explorer 7) Tom Ruen (talk) 18:42, 16 December 2008 (UTC)
 * $$ \cdots $$


 * Using Firefox 3 on Linux, it looks like three dots to me, as it should. Wikipedia is delivering the following code to the client browser:
 * I'd guess that IE is correctly interpreting the "\cdots" as three repeats of the same character, but the text renderer cannot identify the character passed to it by the mathml processor. I think this may be because the Wikipedia code is broken - I suspect it should be delivering the original content as typed in:
 * If so, then it's a wonder that Firefox gets it right. H'mm, I wonder which browser response is the more W3C compliant. -- Cheers, Steelpillow (Talk) 19:32, 16 December 2008 (UTC)
 * If so, then it's a wonder that Firefox gets it right. H'mm, I wonder which browser response is the more W3C compliant. -- Cheers, Steelpillow (Talk) 19:32, 16 December 2008 (UTC)
 * If so, then it's a wonder that Firefox gets it right. H'mm, I wonder which browser response is the more W3C compliant. -- Cheers, Steelpillow (Talk) 19:32, 16 December 2008 (UTC)


 * My assumption is that it was correctly substituting a DOT character, 3 of them, but that character isn't recognized by the font(s) available to IE. I'll gladly assume IE has failed me, AND I ought to do something about it, whatever that is (use FireFox, or download something from MS), but I figure the reality is I'm working on a half dozen different computers, AND largely "default systems" whatever that is, AND so will the average Internet user. SO I'd suggest "lowest common denominator" character sets be used. Otherwise Wikipedia becomes half-useless to perhaps 75% of Internet viewers! Tom Ruen (talk) 20:39, 16 December 2008 (UTC)


 * I checked Firefox and IE under VISTA and it looks fine in both cases. Mhym (talk) 04:37, 17 December 2008 (UTC)

Errors!
There are two errors in this article, much to my regret.

Construction:

$$Q$$ has 2 types of faces, one with edges 1, 1, and √2, the other with edges 1, √2, and √3. There are two faces of each type. Therefore no three faces can be congruent. Just three of the 6 dihedral angles are right.
 * Being right once does not prevent one from making errors. B.w.marx (talk) 11:21, 6 March 2010 (UTC)

Properties:

It probably should be mentioned that $$Q$$ is chiral, has a handedness. So to tile a cube one needs 3 left- and 3 right-handed copies of it. One CANNOT tile a REGULAR octahedron with any number of copies of $$Q$$, whatever their orientation. If one could, one could cube the tetrahedron, one octahedron and 4 tetrahedra were scissors congruent (equidissectable), which they are not.

Keep on improving, B.w.marx (talk) 08:55, 6 March 2010 (UTC)


 * You are correct, actually. In the future, when you see a mistake you might want to correct the wikipedia page yourself.  That's the point.  See WP:BB.  Mhym (talk) 10:56, 6 March 2010 (UTC)