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

plane".

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.

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