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## Heron’s formula

Heron’s formula for the area of a triangle with side lengths $a, b, c$ is

$K = \sqrt{s(s - a)(s - b)(s - c)}$

where $s = \frac{a + b + c}{2}$ is the semiperimeter. Today I’d like to try to prove this using as little geometry as possible.

The assumption

If we assume that $K^2$ is a symmetric polynomial of degree $4$ in $a, b, c$, there is only one polynomial it can be. This is because the area of a triangle satisfying any of $a + b = c, a + c = b, b + c = a$ 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 $K^2$ has to be divisible by $(a + b - c)(a - b + c)(-a + b + c)$ (for example by the Nullstellensatz), and the remaining linear factor must be symmetric in $a, b, c$, so we already know that

$K^2 = K_0(a + b + c)(a + b - c)(a - b + c)(-a + b + c)$

for some constant $K_0$. But the area of a triangle with side lengths $1, 1, \sqrt{2}$ is $\frac{1}{2}$, so

$\frac{1}{4} = K_0(2 + \sqrt{2})(2 - \sqrt{2})2 = 4K_0$

and $K_0 = \frac{1}{16}$ 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

$K^2$ is evidently homogeneous of degree $4$ in $a, b, c$, and it is evidently symmetric, but is it evidently a polynomial? One approach is to let $h_a$ denote the altitude dropping down to side $a$, and similarly for $b, c$; then

$\displaystyle K = \frac{ah_a}{2}$

so the question is whether $a^2 h_a^2$ 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 $\mathbf{v}, \mathbf{w}$ is the determinant of its Gram matrix

$\mathbf{G}(\mathbf{v}, \mathbf{w})= \left[ \begin{array}{cc} \mathbf{v} \cdot \mathbf{v} & \mathbf{v} \cdot \mathbf{w} \\ \mathbf{w} \cdot \mathbf{v} & \mathbf{w} \cdot \mathbf{w} \end{array} \right]$.

The squares of the side lengths of the triangle $0, \mathbf{v}, \mathbf{w}$ are $\mathbf{v} \cdot \mathbf{v}, \mathbf{w} \cdot \mathbf{w}$, and $(\mathbf{v} - \mathbf{w}) \cdot (\mathbf{v} - \mathbf{w})$, 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.

### 11 Responses

1. […] 本文譯自Annoying Precision, 並且處於自定版測試中. Share and Enjoy: […]

2. This works for the area of a cyclic quadrilateral as well…

3. […] Qiaochu Yuan mostra que é possível prová-la, sem usar métodos trigonométricos ou geométricos, no seu post Heron’s formula. […]

4. i my self vijay from india, i know so many proofs for herions formula, but this proof is quit different very nice.

5. The squared volume V^2 of a tetrahedron is likewise a polynomial function of its edge lengths, and the volume is zero if any of the faces are degenerate. Why does it not follow that V^2 has factors of the form (a+b-c)?

• A degenerate tetrahedron satisfies additional polynomial relationships among its side lengths: when one of the vertices degenerates to lie on one of the edges of a tetrahedron, the edge connecting the degenerate vertex to the opposite vertex becomes a cevian of a triangle, and there is a nontrivial polynomial relationship coming from Stewart’s theorem. This does not occur in the triangle case. (In other words, the correct fix should be that V^2 lies in the ideal generated by (a+b-c) and the relationship coming from Stewart’s theorem.)

• This was implicit, but I should say it explicitly: the reason we don’t need to worry about something similar happening in the triangle case is that the space of degenerate triangles satisfying a + b – c = 0 is Zariski dense in the line a + b – c = 0, and similarly for the other two degeneracy conditions. This doesn’t occur in the tetrahedron case because the relation coming from Stewart’s theorem puts us in a lower-dimension subvariety than we expect.

6. […] (HT: Annoying Precision) […]

7. One sometimes proves a result in number theory by aid of a little geometry, but here is a rare specimen of the reverse: a theorem in geometry (Heron’s formula) deduced with some help from number theory.

http://sci.tech-archive.net/Archive/sci.math/2006-11/msg05191.html

A typically eccentric LH effort. 😀

8. $G(v,w) = \left[ \begin{smallmatrix} v \\ w \end{smallmatrix} \right] \left[ \begin{smallmatrix} v & w \end{smallmatrix} \right]$, so if you believe that $\det$ is multiplicative and $\det \left[ \begin{smallmatrix} v & w \end{smallmatrix} \right]$ is the appropriate volume, then you’re done. It’s more or less clear that volume is an antisymmetric $n$-linear form; the $n$-linearity is obvious, and the antisymmetry follows from polarization, since the volume of a linearly-dependent set is $0$. Thus, if you believe that the space of $n$-linear antisymmetric forms is one-dimensional, you’re done: your argument does not require that determinant gives volume, but only that they are non-zero scalar multiples of each other. And of course, by $n$-linearity and antisymmetry, such a form is determined by its value on an ordered basis. I guess that’s how I’d finish the argument.

9. I didn’t use the geometric definition to begin with, because my treatment of linear algebra was purely, well, algebraic. But as I moved into multivariable integration, I defined parallelepipeds, the connection of signed volumes to determinants, and also the connection of unsigned volumes to absolute value of determinants.