Heron’s formula for the area of a triangle with side lengths Image may be NSFW.
Clik here to view.
is
Image may be NSFW.
Clik here to view.
where Image may be NSFW.
Clik here to view.
is the semiperimeter. Today I’d like to try to prove this using as little geometry as possible.
The assumption
If we assume that Image may be NSFW.
Clik here to view.
is a symmetric polynomial of degree Image may be NSFW.
Clik here to view.
in Image may be NSFW.
Clik here to view., there is only one polynomial it can be. This is because the area of a triangle satisfying any of Image may be NSFW.
Clik here to view.
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 Image may be NSFW.
Clik here to view. has to be divisible by Image may be NSFW.
Clik here to view.
(for example by the Nullstellensatz), and the remaining linear factor must be symmetric in Image may be NSFW.
Clik here to view., so we already know that
Image may be NSFW.
Clik here to view.
for some constant Image may be NSFW.
Clik here to view.
. But the area of a triangle with side lengths Image may be NSFW.
Clik here to view.
is Image may be NSFW.
Clik here to view.
, so
Image may be NSFW.
Clik here to view.
and Image may be NSFW.
Clik here to view.
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
Image may be NSFW.
Clik here to view. is evidently homogeneous of degree Image may be NSFW.
Clik here to view. in Image may be NSFW.
Clik here to view., and it is evidently symmetric, but is it evidently a polynomial? One approach is to let Image may be NSFW.
Clik here to view.
denote the altitude dropping down to side Image may be NSFW.
Clik here to view.
, and similarly for Image may be NSFW.
Clik here to view.
; then
Image may be NSFW.
Clik here to view.
so the question is whether Image may be NSFW.
Clik here to view.
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 Image may be NSFW.
Clik here to view.
is the determinant of its Gram matrix
Image may be NSFW.
Clik here to view.![\mathbf{G}(\mathbf{v}, \mathbf{w})= \left[ \begin{array}{cc} \mathbf{v} \cdot \mathbf{v} & \mathbf{v} \cdot \mathbf{w} \\ \mathbf{w} \cdot \mathbf{v} & \mathbf{w} \cdot \mathbf{w} \end{array} \right]](http://s0.wp.com/latex.php?latex=%5Cmathbf%7BG%7D%28%5Cmathbf%7Bv%7D%2C+%5Cmathbf%7Bw%7D%29%3D+%5Cleft%5B+%5Cbegin%7Barray%7D%7Bcc%7D+%5Cmathbf%7Bv%7D+%5Ccdot+%5Cmathbf%7Bv%7D+%26+%5Cmathbf%7Bv%7D+%5Ccdot+%5Cmathbf%7Bw%7D+%5C%5C+%5Cmathbf%7Bw%7D+%5Ccdot+%5Cmathbf%7Bv%7D+%26+%5Cmathbf%7Bw%7D+%5Ccdot+%5Cmathbf%7Bw%7D+%5Cend%7Barray%7D+%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
.
The squares of the side lengths of the triangle Image may be NSFW.
Clik here to view.
are Image may be NSFW.
Clik here to view.
, and Image may be NSFW.
Clik here to view.
, 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.
