Integer tetrahedra

[Richard Guy]

I'd like to underline what John has said.

When at school I was taught that for any problem
about a tetrahedron, you `put it in a box', i.e.,
draw parallel planes though each pair of opposite
edges, producing a parallel-epi-ped (so spelt
and pronounced).

If the box is rectangular, the diagonals of each
face are equal, so that pairs of opposite edges
are equal and there are only three different
edge-lengths and the faces are congruent.

If, in addition, the box is a square prism,
then the faces are isosceles and I've heard it
referred to as a `tetragonal bisphenoid' ---
was it in chap 5 of Coxeter's 1938 revision
of Rouse Ball's Mathematical Recreations &
Essays? --- in connexion with rotating rings
of tetrahedra?

While on the subject of tetrahedra. I'd like
to advertise a paper about the `orthocentric
tetrahedron' which has several surprising
properties and deserves to be better known:

H. Lob, Math. Gaz., 1935, p.102.

If you can't access that, you might be able to
find 

D.E.Rutherford, Vector Methods, Oliver & Boyd,
1939, 1943, pp.13-14,

where the properties are obtained by what I
like to call `wooden vectors' -- the good
old 3-D ones that we taught to engineers --
do they ever use them ?        R.

On Wed, 9 May 2001, John Conway wrote:

>    Hi Ed!  Nice to find someone interested in sphenoids
> (see below).  John Conway
> 
> On Tue, 8 May 2001, Ed Pegg Jr wrote:
> 
> > A few days ago, I asked for a tetrahedron with integer
> > sides, with an interior point an integer distance from
> > all vertices.  I discovered a spectacular result.
> >
> > A face of a skew tetrahedra has faces {a,b,c}.
> 
>     I presume you're referring to the kind of tetrahedron
> ("tetrahedra" is plural) whose faces are all congruent,
> and have edge-lengths  a,b,c ?  [This kind of tetrahedron
> is traditionally called a "sphenoid", by the way.]
> 
> > What is the radius of the Circumsphere?
> >
> >    Sqrt[(a^2 + b^2 + c^2)/8]
> >
> > Is this known? ( It's too nice a result for it to be mine. :-) )
> 
>    Sphenoids have been much studied, so more or less everything
> about them is "known", though maybe to only a few people.
> 
>    The great thing about sphenoids is that they can be obtained by
> taking alternate vertices of a rectangular box, or "cuboid", which
> usually gives easy proofs.
> 
>    In this case, take the vertices to be
> 
>      (x,y,z), (x,-y,-z), (-x,y,-z), (-x,-y,z),
> 
> so that the squared circumradius is  rr = xx+yy+zz, from which
> your formula easily follows using
> 
>     aa = 4(yy+zz),  bb = 4(zz+xx),  cc = 4(xx+yy).
> 
>    Since the sphenoid is what's left of the box after removing
> four "corner" tetrahedra of volume (2x.2y.2z)/6, its volume is
> 8xyz - 16xyz/3 = 8xyz/3, whose square is  (4xx.4yy.4zz)/9,
> easily seen to be
> 
>       bb+cc-aa cc+aa-bb aa+bb-cc
>       -------- -------- --------  / 9,
>          2        2         2
> 
> which I think is what you quoted.
> 
>     Regards,  JHC