Piero della Francesca's Tetrahedron Formula

The painter Piero della Francesca (who died in 1492) also studied 
mathematics, and one of his results leads to a 3-dimensional analogue
of Heron's formula for the volume of a general tetrahedron with edges
a,b,c,d,e,f, taken in opposite pairs (a,f), (b,e), (c,d).  Letting 
A,B,..,F denote the *squares* of these respective edge lengths, his 
formula was

 144 V^2  = - ABC - ADE - BDF - CEF + ACD + BCD + ABE + BCE
            + BDE + CDE + ABF + ACF + ADF + CDF + AEF + BEF
            - CCD - CDD - BBE - BEE - AAF - AFF

This polynomial for the volume of a tetrahedron in terms of the 
squares of the edge lengths doesn't factor, but I wonder what is the 
minimum number of "squared volume terms" necessary to express it.  
Certainly it be expressed in 5 terms,  as follows:

 144 V^2 = AF(-A+B+C+D+E-F)  +  BE(A-B+C+D-E+F)  +  CD(A+B-C-D+E+F)

              -  (A+F)(B+E)(C+D)/2   -  (A-F)(B-E)(C-D)/2

Can the formula be broken down into fewer than 5 terms, where each
"term" is a product of three linear functions of the squared sides?

Anyway, another interesting point is the fact that there are really
two distinct shapes (up to rotations and reflections), with distinct 
volumes, for any given set of six edge lengths and opposite pair 
assignments.  In other words, there are two distinct shapes for any 
given choice of "pairings" depending on how we choose to connect 
those pairs.  To illustrate, for the non-intersecting pairs {a,f}, 
{b,e}, and {c,d} we have the following two distinct tetrahedral 
graphs

     _________a_________      _________f_________
     \                 /      \                 /
     \ \             / /      \ \             / /
      \  \e       d/  /        \  \e       d/  / 
       \   \     /   /          \   \     /   /  
        \    \ /    /            \    \ /    /   
        c\    |    /b            c\    |    /b   
          \   |   /                \   |   /     
           \  |f /                  \  |a /      
            \ | /                    \ | /       
             \|/                      \|/        


The only change has been to transpose the edges a <-> f, but notice
that in the left-hand figure the edge "a" meets the edges c,e at 
one end and b,d at the other, whereas in the right-hand figure it
meets the edges b,c at one end and d,e at the other.

These two distinct configurations are can be seen algebraically
in the volume formulae.  Piero's formula corresponds to the arrangement
on the left, whereas the volume of the tetrahedron on the right would
be given by

 144 V^2  = - ABD - ACE - BCF - DEF + ACD + BCD + ABE + BCE
            + BDE + CDE + ABF + ACF + ADF + CDF + AEF + BEF
            - CCD - CDD - BBE - BEE - AAF - AFF

This is close to Piero's formula, but not exactly the same.  Where 
Piero had the four leading terms

                 - ABC - ADE - BDF - CEF

this formula has

                 - ABD - ACE - BCF - DEF

Actually both formulas are correct, but they represent the two
distinct tetrahedrons that can be constructed from six given edges 
with given pairings of non-intersecting edges (a,f), (b,e), (c,d).
This appears most clearly if we consider again Piero's volume 
formula expressed as a sum of five "squared volume" terms

 144 V^2 = AF[-(A+F)+(B+E)+(C+D)]  +  BE[(A+F)-(B+E)+(C+D)]

                   +  CD[(A+F)+(B+E)-(C+D)]

            -  (A+F)(B+E)(C+D)/2   -+  (A-F)(B-E)(C-D)/2

Notice that the first four terms are symmetrical in the non- 
intersecting pairs, but the fifth term is anti-symmetric, i.e.,
it reverses sign when we transpose any of the pairs.  That's the
reason I put a +- sign on that term.  With the - sign it gives
Piero's formula, whereas with the + sign it gives the volume for
the "conjugate" tetrahedron.  So this formula represents the two 
possible tetrahedron volumes (and shapes) for a given pairing of 
the edges.

Incidentally, it's interesting to note that the four terms in which
the two formulas differ

    Left-hand graph:    - ABC - ADE - BDF - CEF
    Right hand graph:   - ABD - ACE - BCF - DEF

are precisely the four trianglular faces of the respective tetrahedra.
In other words, the first tetrahedra has the four faces (a,b,c),
(a,d,e), (b,d,f) and (c,e,f), whereas the conjugate tetrahedron has
the faces (a,b,d), (a,c,e), (b,c,f), and (d,e,f).

Of course, we can easily express the volume of an arbitrary tetrahedron
by the usual formula in terms of the three edge vectors a,b,c emanating
from a single vertex

                      V  =  (1/6) a.(bxc)

where "." signifies the dot product and "x" is the cross product.  This
is easy to see if we recall that the area of the triangle with edges
b,c is  (bc/2) sin(A) where A is the angle between the edge vectors
b and c.  Also, the height of the pyramid on this base is  a*cos(q_a)
where q_a is the angle between the vector "a" and the perpendicular
to b and c.  Then, since the volume of any pyramid is 1/3 the base 
times the height, we have

              V = (abc/6) sin(A)cos(q_a)

                 = (abc/6) sin(A)sin(Q_a)

where Q_a is the angle between "a" and the plane of b and c.  Naturally 
this is consistent with the magnitudes of the dot and cross products.
By symmetry the above implies that

         sin(A)sin(Q_a) = sin(B)sin(Q_b) = sin(C)sin(Q_c)

for any three vectors a,b,c, because this is proportional to the
subtended volume, just as sin(A) is proportional to the subtended
area.  Hence, in a sense, this quantity sin(A)sin(Q_a) is the natural
3D analog of the 2D sine function.

Incidentally, suppose we want to express the standard formula 

              V = (abc/6)sin(A)sin(Q_A)

entirely in terms of the cosines of the angles A,B,C between the 
three edges a,b,c emanating from a point.  Consider unit vectors 
on the three edges, with the vertex at the origin of xyz coordinates 
and make one of the vectors [1,0,0] and another [cos(A),sin(A),0].  
Since the distance between two points on a unit sphere equals twice 
the sine of the half-angle between them, the unit vector [x,y,z] 
along the third edge must satisfy the conditions

               x^2 + y^2 + z^2 = 1

          [x-1]^2 + y^2 + z^2 = [2sin(B/2)]^2
 
       [x-cos(A)]^2 + [y-sin(A)]^2 + z^2 = [2sin(C/2)]^2

Substituting 1-x^2-y^2 for z^2 in the second equation gives 

         x  =  1 - 2sin(B/2)^2  =  cos(B)

and substituting this into the third equation gives

        y = [cos(C) - cos(A)cos(B)]/sin(A)

Substituting back into z^2 = 1-x^2-y^2, multiplying through by 
sin(A)^2, and noting that z = sin(Q_A), we have

  sin(A)^2 sin(Q_A)^2 

   = 1 - cos(A)^2 - cos(B)^2 - cos(C)^2 + 2cos(A)cos(B)cos(C)

This formula was called to my attention by Patrick Bell, who had
derived the equivalent of this independently based on the edge 
lengths of a tetrahedron alone, without involving trigonometric 
functions.  It's interesting that if we substitute the expressions
for the cosines in terms of the edge lengths (see the note on
Heron's Formula For Triangle Area) into the above expression 
we recover Piero's formula.  For more on tetrahedron volume formulas, 
see Heron's Formula for Tetrahedrons.