Heron's Formula For Tetrahedrons

The article Heron's Formula For Triangle Area gives a very direct
derivation of Heron's formula based on Pythagoras's Theorem for right
triangles. However, we might also observe that Heron's formula is
essentially equivalent to Pythagoras' Theorem for right tetrahedrons.
If B,C,D denote the areas of the three orthogonal faces of a right
tetrahedron, and A denotes the area of the "hypotenuse face", then
it's easy to show that

A^2 = B^2 + C^2 + D^2 (1)

This is essentially Heron's formula. Note that if a,b,c denote the
three orthogonal edges of the tetrahedron, then the areas B,C,D are
simply ab/2, ac/2, and bc/2. Furthermore, the edges of the hypotenuse
face d,e,f are directly related to a,b,c according to

d^2 = a^2 + b^2
e^2 = a^2 + c^2 (2)
f^2 = b^2 + c^2

so we can express the areas B,C,D in equation (1) in terms of d^2,
e^2, and f^2 to give Heron's formula explicitly.

Incidentally, it might be argued that this derivation does not
apply to obtuse triangles, because the "hypotenuse face" of a right
tetrahedron is necessarily acute (i.e., each of its angles must
be less than 90 degrees). However, the truth is that ANY triangle
can be the hypotenuse face of a right tetrahedron, provided the
orthogonal edge lengths and areas are allowed to be imaginary.

For ANY values of d,e,f we can solve equations (2) for the orthogonal
edges of the right tetrahedron whose hypotenuse is the triangle with
the edges lengths d, e, f. This gives

a^2 = (d^2 - e^2 + f^2)/2
b^2 = (d^2 + e^2 - f^2)/2
c^2 = (-d^2 + e^2 + f^2)/2

For example, suppose we want the area of a triangle with edge lengths
8, 5, and 5, which is an obtuse triangle. Substituting these into
the above equations gives

a = sqrt(32) b = sqrt(32) c = sqrt(-7)

so the "c" leg has imaginary length. Consequently, two of the three
orthogonal faces (those given by ac/2 and bc/2) are also imaginary.
However, these areas only appear _squared_ in the Pythagorean formula
for right tetrahedrons, so we're guaranteed to get a real area for
the hypotenuse face. As Hadamard said, "The shortest path to any
truth involving real quantities often passes through the complex

I honestly wouldn't be surprised if the ancient Greeks were aware of
the connection between the generalized Pythagorean theorem and Heron's
formula, but refrained from presenting it in that form because of
difficulties with interpreting the obtuse case. Recall that Descartes,
for one, believed the ancient Greeks had discovered most of their
theorems analytically by means of coordinate geometry and algebra, but
then concealed their methods, presenting them in synthetic form, so as
to make the results seem more daunting and impressive to the uninitiated.
It has always been doubtful that Heron's formula was discovered via
the throught process of Heron's proof, which is absurdly circuitous.
In any case, this is a nice example of how imaginary numbers can arise
naturally in dealing with questions of purely real quantities.

As for higher dimensional simplexes, there is no complete generalization
of Heron's formula giving the volume of a general tetrahedron in terms
of the areas of its faces, because the face areas don't uniquely determine
the volume (in contrast to the case of triangles, where the three
edge lengths determine the area). However, it IS possible to derive
a "Heron's formula" for tetrahedrons if we restrict ourselves to just
those that would fit as the "hypotenuse face" of a right 4D solid.
(Notice that EVERY triangle is the face of a right tetrahedron, which
explains why Heron's formula is complete for triangles).

To review, remember that Heron's formula for triangles is essentially
equivalent to Pythagoras' Theorem for right tetrahedrons. Let's let
A_xyo, A_xoz, and A_oyz denote the areas of the three orthogonal faces
of a right tetrahedron, and A_xyz denote the area of the "hypotenuse
face", so we have

(A_xyz)^2 = (A_xyo)^2 + (A_xoz)^2 + (A_oyz)^2 (3)

Now if we let L_x, L_y, L_z denote the three orthogonal edge lengths
of the tetrahedron, then the areas of its orthogonal faces are simply

A_xyo = (L_x)(L_y)/2
A_xoz = (L_x)(L_z)/2
A_oyz = (L_y)(L_z)/2

and so equation (3) can be re-written in the form

4(A_xyz)^2 = ((L_x)(L_y))^2 + ((L_x)(L_z))^2 + ((L_y)(L_z))^2 (4)

Furthermore, the three edges L1, L2, L3 of the hypotenuse face are
directly related to L_x, L_y, L_z by the 2D Pythagorean theorem

L1^2 = (L_x)^2 + (L_y)^2
L2^2 = (L_x)^2 + (L_z)^2 (5)
L3^2 = (L_y)^2 + (L_z)^2

Equations (5) are three linear equations in the three squared edge
lengths, so we can solve for these squared lengths in terms of L1,
L2, and L3, and then substitute these into equation (4) to give
the ordinary Heron's formula for triangles, as before.

Now, we can do the same thing for tetrahedrons based on the
generalized Pythagorean theorem for volumes of *right* 4D solids

(V_wxyz)^2 = (V_wxyo)^2 + (V_wxoz)^2 + (V_woyz)^2 + (V_oxyz)^2 (3')

If we let L_w, L_x, L_y, L_z, denote the orthogonal edge lengths of the
4D solid, then the volumes of the four orthogonal "faces" are simply

V_wxyo = (L_w)(L_x)(L_y)/6
V_wxoz = (L_w)(L_x)(L_z)/6
V_woyz = (L_w)(L_y)(L_z)/6
V_oxyz = (L_x)(L_y)(L_z)/6

so equation (3') can be rewritten as

36(V_wxyz)^2 = ((L_w)(L_x)(L_y))^2 + ((L_w)(L_x)(L_z))^2

+ ((L_w)(L_y)(L_z))^2 + ((L_x)(L_y)(L_z))^2 (4')

Furthermore, the four areas A1, A2, A3, A4 of the hypotenuse "face" are
directly related to L_x, L_y, L_z by the 3D Pythagorean theorem (4)

4(A1)^2 = ((L_w)(L_x))^2 + ((L_w)(L_y))^2 + ((L_x)(L_y))^2
4(A2)^2 = ((L_w)(L_x))^2 + ((L_w)(L_z))^2 + ((L_x)(L_z))^2 (5')
4(A3)^2 = ((L_w)(L_y))^2 + ((L_w)(L_z))^2 + ((L_y)(L_z))^2
4(A4)^2 = ((L_x)(L_y))^2 + ((L_x)(L_z))^2 + ((L_y)(L_z))^2

Thus, given the four face areas A1, A2, A3, A4, we have four equations
in the four unknowns L_2, L_x, L_y, L_z, so we can solve for these
values and then compute the volume of the tetrahedron using (4').

At this point people usually turn away from this approach, for two
reasons. First, everything we're doing is restricted to the "special"
tetrahedrons that can serve as the hypotenuse of a "right" 4D simplex,
so we're certainly not going to end up with a general formula
applicable to every tetrahedron (as is clear from the fact that we
have only four independent edge lengths here, whereas the general
tetrahedron has six). General formulas giving the volume in terms of
the edge lengths ARE available such as the one give by the Italian
painter Piero della Francesca. Of course, all such formulae can
be traced back to the well-known determinant expression for volumes.

The second reason that people usually give up on equations (5') is
that they are somewhat messy to solve, since they are non-linear in
the lengths. Still, we might decide to press on anyway. It turns
out (after extensive algebraic manipulation) that we can reduce (5')
to a single quartic in the square of any of the four edge lengths L_w,
L_x, L_y, or L_z. Arbitrarily selecting L_y, and letting A,B,C,D
denote 4 times the squares of the face areas (i.e., the left hand
sides of equations (5')), we can express the quartic in x = (L_y)^2
with coefficients that are functions of B and the elementary symmetric
polynomials of A,C,D

r = A+C+D s = AC+AD+CD t = ACD

In these terms the quartic for x = (L_y)^2 is

[12B] x^4
+ [3r^2 - 12s + 14Br - B^2] x^3
+ [2B(r^2 + 6s) - rB^2 - r^3 + 4rs - 36t] x^2
+ [2B(rs + 6t) - sB^2 + 4s^2 - sr^2 - 12rt] x
- [t(r-B)^2] = 0

Of course, the analagous quartics can be given for (L_w)^2, (L_x)^2,
and (L_z)^2, but once we have any one of them we can more easily compute
the others. For example, given L_y we can compute L_x from the relation
/ ((L_y)^2 + A)((L_y)^2 + D)
L_x = -L_y +- / ---------------------------
\/ ((L_y)^2 + C)

and the values of L_w and L_z follow easily, allowing us to compute
the volume using equation (4'). It would be nice if we could express
the volume as an explicit function of the face areas, but I don't
know if such a formula exists.

In the preceding discussion we developed a tetrahedral version of
Heron's formula for a restricted class of tetrahedra, namely those
that can serve as the hypotenuse of a "right" 4D simplex, but there
are other special classes of tetrahedra that possess interesting
volume formulas. The one that gives the closest analogue to Heron's
formula is the class of tetrahedra whose opposite edges lengths
are equal. Thus there are only three independent edge lengths, and
each face of the tetrahedron is identical. Letting (a,f), (b,e), and
(c,d) denote the pairs of opposite edge lengths, we can set a=f, b=e,
and c=d in the basic determinant expression for the volume, or
equivalently in Piero della Francesca's formula, and we find that
the resulting expression for the squared volume factors as

72 V^2 = (-a^2 + b^2 + c^2)(a^2 - b^2 + c^2)(a^2 + b^2 - c^2)

which is certainly reminiscent of Heron's formula for the area of
each face

16 A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)

This also shows that if each face is an identical right triangle, the
volume is zero, as it must be, since four such triangles connected
by their edges to give a tetrahedron necessarily all lie flat in the
same plane:
|\ /|
| \ / |
| \ / |
| \/ |
| /\ |
| / \ |
| / \ |

Obviously we can construct a regular tetrahedron with equilateral
triangles of the same area as these right triangles, and the volume
is V = a^3 / sqrt(72), which illustrates the fact that the face
areas of a tetrahedron do not in general determine it's volume.