TITLE: What does Heron's formula naturally deform?
QUESTION [11 upvotes]: Fixing three real numbers $a,b,c>0$ determines a triangle with
side-lengths $a,b,c$ (if admissible). Therefore, the area of a
triangle is a function in $a,b,c$. Due to the geometry of a
triangle, we know that the area is a symmetric function in
$a,b,c$. Indeed, Heron formula shows that
$$area(a,b,c) = \frac{1}{4} \sqrt{[(a+b+c)] \cdot [(a+b-c)(a-b+c)(-a+b+c)]}$$
What's interesting is that the radicand is the product of two
symmetric polynomials. The fact that it's a product (and
therefore not as "pure") motivates me to look at the following
expression:
$$f(a,b,c,d) = (a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)$$
Notice that $f$ is a symmetric polynomial, with $f(a,b,c,0)$
recovering the "impure" product.
Question: Does $f$ calculate anything? In other words, is
there any known combinatorial or geometrical meaning of $f$? I
will be glad to see if it deforms the area of a triangle in
some interesting way, but any interpretation is welcome.

REPLY [11 votes]: Another interpretation is that it is the radicand in brahmagupta's formula for the area for a cyclic quadrilateral.