Piero della Francesca's Tetrahedron Formula
The painter Piero della Francesca (who died on Oct 12, 1492, the same
day Columbus sighted land on his first voyage to America) 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 doesn't factor, but it can be expressed as a sum of
five "squared volume 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?
It's interesting to note 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.
The entire expression is symmetrical under permutations of the
three pairs, so for a given set of six edge lengths (all satisfying
the triangle inequality with respect to each other) the number of
distinct pairings is 5*3*1 = 15, and for each of these there are
two distinct tetrahedra, so there are a maximum of 30 distinct
tetrahedra (with distinct volumes) that can be constructed from
six given edges.
Incidentally, 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 is well known - see for example, the discussion of rational
tetrahedra in Dickson's history of the theory of numbers, which refers to
a paper by K. Schwering in 1895. If we substitute the expressions for the
cosines in terms of the edge lengths (see the note on Heron's Formula)
into the above expression we recover Piero's formula. For more on
tetrahedron volume formulas, see Heron's Formula for Tetrahedrons,
and also the note on the Cayley-Menger determinant.
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