Dissection proof of Heron's formula?

Question

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.

Answer 1

Heron's formula says the area of a triangle whose sides have lengths $a,b,c$ is $$ \frac14\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}. $$ It is true that this is an "opaque formula" with which you "just chuck in the side-lengths, turn a series of arbitrary mathematical handles, and out pops the answer," about which one should say "It works, but it's not satisfying."?

Notice that

• One must expect an expression that is homogeneous of degree $2$ as a function of $a,b,c.$
• $(a+b+c)$ should be a factor because the area is $0$ when $a=b=c=0.$
• $(a+b-c)$ should be a factor because when $a+b=c$ you have a degenerate triangle, whose area is $0.$
• The other factors must be there for the same reason.
• The number $1/4$ is seen to be right when one considers $a=b=c=1.$

This is not a dissection proof, but it addresses the first issue raised in the original posting.

Answer 2

Noam Elkies pointed out to me by email that there cannot exist a dissection (of the type I asked for) using bounded number of pieces. Consider a triangle with sides $T+1$, $T+1$, $2T$, so $s=2T+1$ and $r\approx \sqrt{T/2}.$ Then we are trying to dissect a cuboid $X$ with sides $T$, $T$, $1$ into a cuboid $Y$ with sides $\sqrt{T/2}$, $\sqrt{T/2}$, $2T+1$. It is then intuitively clear that since $X$ has thickness $1$, it is "too thin" relative to $Y$ for any dissection into $C$ pieces—where $C$ is a constant independent of $T$—to work. (A rigorous proof is a bit tricky; it's an old chestnut to show that a disc of diameter $D$ cannot be covered with fewer than $D$ rectangles of width $1$.)

On the other hand, it's still conceivable that by adding some auxiliary pieces $Z$, there might be a scissors congruence between $X+Z$ and $Y+Z$ using a bounded number of pieces.

Answer 3

Along with $S=rs$, there is a formula for the area $S=r_a(s-a)$, where $r_a$ is the corresponding exradius. Thus, $S=\sqrt{rs\cdot r_a(s-a)}$, and Heron's formula reads as $(s-b)(s-c)=rr_a$. This follows from the similarity of right triangles $BE_aX_{AB}$ and $IBI_c$, which in turn may be proved by dissection if you like.

Answer 4

I have been told that one of the reasons for the early popularity of Heron's formula was that it became a trade secret for Islamic lawyers, seeing that when a man died his land had to be divided equally among his sons. Triangulation and Heron's formula provided lawyers with a straightforward procedure for reckoning area.

Answer 5

Since this thread seems to have developed into a discussion of simple approaches to Heron, let me add my two cents. There is a very easy way to derive (even obtain from scratch) Heron´s formula and indeed to extend it to formulae in terms of the side lengths for ANY triangle quantity . If we denote the vertices by $A_1,A_2,A_3$ and the corresponding side lengths by $e_{12},e_{23},e_{31}$, then we can suppose that the former have coordinates $(0,0)$,$(e_{12},0)$ and $(p,q)$ for some $p,q$. One has equations $e_{23}^2=(p-e_{12})^2+q^2$ and $e_{31}^2=p^2+q^2$ which can be solved in a matter of seconds to obtain $p$ and $q$ in terms of the side lengths. The area is $\frac 12 q e_{12}$ which can now be expressed as a function of the side lengths. Any triangle quantity (by this I mean any real number associated with a triangle--trigonometric functions of the angles, altitudes, median lengths, etc.--which can be computed from the coordinates of the vertices) can be expressed in terms of the side lengths in a similar manner, namely by calculating it for the above triangle as a function of $p,q,e_{12}$ and then substituting for $p$ and $q$.

As a bonus, the same method can be used to express the volume of a tetrahedron, and indeed any other tetrahedral quantity, in terms of the side lengths (one supposes that the vertices are $(0,0,0)$,$(e_{12},0,0)$,$(p,q,0)$,$(r,s,t)$ and solves for $p,q,r,s,t$ in terms of the side lengths). In this case, the computations are rather more tedious but they can be done. If one is prepared to use Mathematica, then it is again a matter of seconds.

The same techniques cannot be applied to quadrilaterals since they are not rigid but one can use them for certain subclasses, notably for cyclic quadrilaterals (though there are some additional subtleties in this case).