In his recent book, Love Triangle, Matt Parker playfully complains that Heron's formula is an "opaque formula, and I feel like you just chuck in the side-lengths, turn a series of arbitrary mathematical handles, and out pops the answer. It works, but it's not satisfying." Parker's remark got me thinking about whether there might be some kind of equidecomposition lurking beneath the surface.

Recall that Heron's formula says that the area of a triangle $ABC$ with sides $a$, $b$, $c$ (where side $a$ is opposite $A$ and so on; I abuse notation by using the same letter for the name of the side and its length) is $\sqrt{s(s-a)(s-b)(s-c)}$, where $s=(a+b+c)/2$ is the semi-perimeter. Now, if we let $r$ be the inradius of $ABC$, then it is easy to see that the area of $ABC$ is $rs$. Therefore, Heron's formula may be rewritten

$$\eqalignno{(s-a)(s-b)(s-c) &= r^2s.&(*)\cr}$$

Is there a nice way to dissect a (rectangular) cuboid with sides $s-a$, $s-b$, $s-c$ into finitely many pieces and reassemble them into a (rectangular) cuboid with sides $r$, $r$, $s$?

A couple of remarks:

• Every rectangular cuboid has Dehn invariant zero, so the desired scissors congruence certainly exists; the question is whether there is a "nice" dissection, ideally one that could form the basis for a (new?) proof of Heron's formula.
• The quantities $s-a$, $s-b$, $s-c$ have natural geometric interpretations; if $X$, $Y$, $Z$ denote the points of tangency of the incircle on sides $a$, $b$, $c$ respectively, then $s-a = AY = AZ$ and $s-b = BX = BZ$ and $s-c = CX = CY$.

In his recent book, Love Triangle, Matt Parker playfully complains that Heron's formula is an "opaque formula, and I feel like you just chuck in the side-lengths, turn a series of arbitrary mathematical handles, and out pops the answer. It works, but it's not satisfying." Parker's remark got me thinking about whether there might be some kind of equidecomposition lurking beneath the surface.

Recall that Heron's formula says that the area of a triangle $ABC$ with sides $a$, $b$, $c$ (where side $a$ is opposite $A$ and so on; I abuse notation by using the same letter for the name of the side and its length) is $\sqrt{s(s-a)(s-b)(s-c)}$, where $s=(a+b+c)/2$ is the semi-perimeter. Now, if we let $r$ be the inradius of $ABC$, then it is easy to see that the area of $ABC$ is $rs$. Therefore, Heron's formula may be rewritten

$$\eqalignno{(s-a)(s-b)(s-c) &= r^2s.&(*)\cr}$$

Is there a nice way to dissect a (rectangular) cuboid with sides $s-a$, $s-b$, $s-c$ into finitely many pieces and reassemble them into a (rectangular) cuboid with sides $r$, $r$, $s$?

A couple of remarks:

• Every rectangular cuboid has Dehn invariant zero, so the desired scissors congruence certainly exists; the question is whether there is a "nice" dissection, ideally one that could form the basis for a (new?) proof of Heron's formula.
• The quantities $s-a$, $s-b$, $s-c$ have natural geometric interpretations; if $X$, $Y$, $Z$ denote the points of tangency of the incircle on sides $a$, $b$, $c$ respectively, then $s-a = AY = AZ$ and $s-b = BX = BZ$ and $s-c = CX = CY$.
• EDIT Oct 15, 2024: I just learned of two papers that are soon to appear in Mathematics Magazine: "A Synthetic Geometry Proof of Heron’s Formula" by Colin Beveridge, and "An Astonishing Proof of Heron’s Formula" by Wagner Oliveira Costa Filho. But neither of these proofs involves a dissection of 3-dimensional solids.