Heron’s formula for the area of a triangle with side lengths is
where is the semiperimeter. Today I’d like to try to prove this using as little geometry as possible.
If we assume that is a symmetric polynomial of degree in , there is only one polynomial it can be. This is because the area of a triangle satisfying any of is equal to zero. (Because area is a continuous function of the side lengths, any proposed area formula has to make sense in the degenerate cases.) This already implies that has to be divisible by (for example by the Nullstellensatz), and the remaining linear factor must be symmetric in , so we already know that
for some constant . But the area of a triangle with side lengths is , so
and as desired. No geometry necessary!
This is not a particularly deep argument, but during my math competition years when Heron’s formula was an important tool I never saw it proven in this way.
Justifying the assumption
is evidently homogeneous of degree in , and it is evidently symmetric, but is it evidently a polynomial? One approach is to let denote the altitude dropping down to side , and similarly for ; then
so the question is whether is a polynomial. Some handwaving about Stewart’s theorem suggests that the answer is yes, but I don’t want to do any computations to answer this question.
An answer much more in line with modern mathematics is that the square of the area of a parallelogram spanned by vectors is the determinant of its Gram matrix
The squares of the side lengths of the triangle are , and , so it does in fact follow that the determinant of the Gram matrix is a polynomial in the side lengths. Is this “obvious”? Arguably one can take this as following from the volume definition of the determinant, and then the question is why this definition is equal to the definition in terms of a sum over permutations. And that, of course, is a matter of exterior algebra. The Unapologetic Mathematician has written on determinants, although I’m not sure he discussed the volume definition in detail.