A neat physical derivation of Heron's formula
Heron's formula gives the area $S$ of a triangle in terms of the lengths of its three sides $a,b,c$. It is a rather beautiful formula: it isn't complicated, and at the same time it is symmetric in $a,b,c$, fully reflecting the equal status of the three sides. Yet a formula with such symmetric beauty is usually derived through a process that is anything but symmetric — for example, the proof on Wikipedia — which always feels like a bit of a shame. The purpose of this post is to supply a symmetric derivation to make up for that. The title says "physical derivation," and the key word is "derivation" rather than "proof"; also, "physical" here doesn't mean we're drawing a physical analogy, but rather that the spirit and method of the derivation have a distinctly "physics flavor."
$$\sqrt{p(p-a)(p-b)(p-c)}$$
Before starting the derivation, let me offer one remark: Heron's formula seems to be the simplest of all possible formulas for computing a triangle's area from the lengths of its three sides.more
Basic assumptions
As I said, the spirit and method of this derivation have a "physics flavor" — but what does that mean exactly? In theoretical physics, physicists often start from a handful of very basic assumptions and derive a complete physical law from them, and only afterward check whether it agrees with experiment. Einstein and relativity are perhaps the pioneering example of this approach. Here we adopt a similar strategy: starting from some basic assumptions, we will construct Heron's formula directly.
Let the three side lengths of the triangle be $a,b,c$, and let the formula for the area in terms of the three side lengths be $S(a,b,c)$.
Basic assumptions
1. The area must have the dimension of length squared — this is probably the most basic principle of all.
2. $S(a,b,c)$ is symmetric in $a,b,c$. There's no reason to reject this, since we have no grounds to think any one side is special. Moreover, symmetry actually helps us find the correct formula.
3. The form of $S(a,b,c)$ should be as simple as possible. This one is imposed by hand: physicists want physical laws to be as simple as possible, so when deriving physical laws they add a simplicity principle — among all possible physical laws, take the simplest one. Whether this principle holds in mathematics is debatable, but for Heron's formula, it does indeed hold.
Known facts
The formula we derive must agree with some simple facts we already know — this is the requirement of consistency. Some facts we already know can be listed as follows:
4. When the sum of two side lengths equals the length of the third side, the area is 0, i.e., $S(a,b,a+b)=0$; in particular, when one of the sides is 0, the area is 0.
5. We already know the area formula for a right triangle, namely $S(a,b,\sqrt{a^2+b^2})=\frac{1}{2}ab$.
The derivation
Let's start from symmetry. There are many symmetric expressions that can be built from the three side lengths; the simplest is $abc$, but this obviously violates the first assumption, since its dimension is length cubed rather than length squared. The simplest fix is to consider
$$\left(abc\right)^{2/3}$$
This formula looks promising — it satisfies all three assumptions, and indeed the area is 0 when one of the sides is 0. But it doesn't fully satisfy fact 4, because when $a=1,b=1,c=2$, $\left(abc\right)^{2/3}$ is clearly nonzero, so this formula must be ruled out. Of course, there are many other candidates one could think up — for instance $ac+bc+ab$ also satisfies the three assumptions, but again it fails to give zero area when one side is 0, so it too is ruled out.
Fact 4 is a strong constraint, and to satisfy it we want a zero to appear precisely when the sum of two sides equals the third, and the simplest way to achieve this is $a+b-c$; then, imposing symmetry, we get
$$(a+b-c)(b+c-a)(c+a-b)$$
This expression now satisfies both symmetry and fact 4, but it fails the basic dimensional requirement. We could square it and then take a cube root, or multiply it by a length and take a square root. To decide which, we turn to fact 5.
This brings us to the most crucial — and most "physical" — part of the derivation. Fact 5 concerns right triangles, and we could simply substitute $c=\sqrt{a^2+b^2}$ directly, but because of the square root involved (a situation that occurs a lot in physics), this becomes cumbersome — even intractable. So instead of handling the full case, we consider only the infinitesimal case: suppose $b$ is infinitesimally small. Then
$$c=\sqrt{a^2+b^2}\approx a+\frac{b^2}{2a}$$
differs from $a$ only by a second-order infinitesimal, so to first-order accuracy, $c=a$. Meanwhile the area of the right triangle is $\frac{1}{2}ab$. In other words, for a triangle with sides $a,b,a$ where $b$ is infinitesimally small, the area is $\frac{1}{2}ab$ to first-order accuracy. Substituting $a,b,c=a$ into $(a+b-c)(b+c-a)(c+a-b)$ gives
$$b\times b\times (2a-b) \approx 2ab^2$$
We see that multiplying by $a$ and then taking a square root gets us to $ab$. However, multiplying by $a$ alone would break symmetry; the simplest symmetric fix is to multiply by $(a+b+c)$ instead. This gives us a candidate formula:
$$\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$$
The coefficient in front still needs adjusting, because substituting $a,b,c=a$ directly gives
$$\sqrt{(2a+b)\times b\times b\times (2a-b) }\approx \sqrt{4a^2 b^2}=2ab$$
which differs from $\frac{1}{2}ab$ by a factor of $\frac{1}{4}$. So a candidate formula that basically satisfies our three assumptions and two facts is
$$S(a,b,c)=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$$
And this is exactly the correct Heron's formula! At this point, we've "constructed" Heron's formula essentially out of thin air! Of course, strictly speaking, this is only a candidate — a highly plausible formula — and whether it's actually correct still needs to be proven. What follows is the proof, which is a matter of rigor and not particularly deep, so I won't include it here.
Why bother doing this?
Why go through all this trouble for a formula that is already so widely known? This process is neither a rigorous proof nor does it yield anything new — so what's the point? For the practically minded, at least for now, this exercise serves no practical purpose. And yet, it may still bring some unexpected benefits.
First, doing this — or more precisely, approaching the same problem through a different method — helps us understand the essence of the problem. By comparing different approaches, we come to see the strengths and weaknesses of each, and thereby grasp the key idea behind each one. Second, what's presented here is a simulated derivation process — or, put another way, a simulation of the process of discovering a new formula. Such a process matters a great deal when it comes to guiding physical discovery, and as someone who loves both mathematics and physics, I certainly hope this kind of thinking can bring some fresh vitality to mathematics as well — though, admittedly, this is only an experiment.
Translated automatically with claude-sonnet-5; all equations are reproduced verbatim from the source. Copyright remains with the original author.