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.

  • 3
    $\begingroup$ The expression you are interested in is exactly the radicand in the following formula for the volume of a tetrahedron: en.wikipedia.org/wiki/… It's worth noting that the a,b,c,d are not the edge lengths, which would have been quite convenient. Nonetheless, it is an interesting generalization of Heron's Formula. $\endgroup$ Dec 14, 2021 at 14:42
  • $\begingroup$ Following the citation, this result is due to William Kahan. You can find it on page 17/31 here. $\endgroup$ Dec 14, 2021 at 14:48
  • $\begingroup$ Fascinating! By the nature of that formula, one cannot take $d$ to zero alone. I'm still waiting to see if there's any other interpretation :) $\endgroup$
    – Student
    Dec 14, 2021 at 14:50

1 Answer 1


Another interpretation is that it is the radicand in brahmagupta's formula for the area for a cyclic quadrilateral.

  • $\begingroup$ This indeed deforms the area of the triangle by deforming the triangle to cyclic quadrilateral. Thank you! $\endgroup$
    – Student
    Dec 14, 2021 at 15:01

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