By 苏剑林 | March 27, 2015
Heron's formula is used to calculate the area $S$ of a triangle given the lengths of its three sides $a, b, c$. It is a quite beautiful formula—not overly complex, and symmetric with respect to $a, b, c$, fully reflecting the equal status of the three sides. However, the derivation of such a symmetrically beautiful formula often involves an asymmetric process, such as the proof found on Wikipedia, which is somewhat of a shortcoming. The purpose of this article is to supplement this with a symmetric derivation. The title calls it a "physical derivation," but the key lies in "derivation" rather than "rigorous proof," and the "physics" here does not refer to a physical analogy, but rather to the idea and method of derivation, which has a distinct "physical flavor."
\[ \sqrt{p(p-a)(p-b)(p-c)} \]
Before beginning the derivation, I offer a comment: Heron's formula appears to be the simplest possible formula for calculating the area of a triangle from its three side lengths.
Basic Assumptions
As I mentioned, the thought process and method used here have a "physical flavor." What exactly does that mean? In theoretical physics, physicists often start from a few very basic assumptions to derive a complete physical law, and then verify whether it matches experimental results. Einstein and his theory of relativity are pioneers of this approach. We will now apply a similar logic, based on some basic assumptions, to "construct" Heron's formula directly.
Let the side lengths of the triangle be $a, b, c$, and the formula for the area be $S(a, b, c)$.
Basic Assumptions:
- Dimensional Analysis: Area must have the dimensions of length squared. This is perhaps the most fundamental principle.
- Symmetry: $S(a, b, c)$ must be symmetric with respect to $a, b, c$. We have no reason to reject this, as no side is more "special" than the others. Furthermore, symmetry actually helps us discover the correct formula.
- Simplicity: The form of $S(a, b, c)$ should be as concise as possible. This is a self-imposed principle; physicists hope that physical laws are as simple as possible, so they include the principle of parsimony in their derivations—choosing the simplest one among all possible candidates. Whether this principle holds in mathematics is debatable, but in the case of Heron's formula, it certainly does.
Known Facts:
Our derived formula must be consistent with some simple facts we already know; this is the requirement of compatibility. Some known facts include:
- When the sum of two side lengths equals the third side, the area is 0; i.e., $S(a, b, a+b) = 0$. In particular, if any side is 0, the area is 0.
- The formula for the area of a right-angled triangle is known: $S(a, b, \sqrt{a^2+b^2}) = \frac{1}{2}ab$.
Derivation Process
First, starting from symmetry, there are many symmetric expressions one can construct from the three sides. The simplest is $abc$, but this obviously fails the first assumption as its dimension is length cubed, not squared. An easy fix is to consider:
\[ (abc)^{2/3} \]
This formula is not bad; it satisfies the three assumptions, and the area is indeed 0 when any side is 0. However, it fails Fact 4. For instance, if $a=1, b=1, c=2$, then $(abc)^{2/3}$ is clearly not 0. So this formula is excluded. Many others could be considered; for example, $ac+bc+ab$ also satisfies the three assumptions, but it does not yield an area of 0 when one side is 0, so it is also excluded.
Fact 4 is a very strong condition. To satisfy Fact 4, we want a zero to appear when the sum of two sides equals the third. The simplest way to achieve this is with the term $a+b-c$, and per the requirement of symmetry, we should have:
\[ (a+b-c)(b+c-a)(c+a-b) \]
This expression satisfies both symmetry and Fact 4, but it does not satisfy the basic dimensional requirement (it has dimensions of $L^3$). We could square it and take the cube root, or multiply it by another "length" and take the square root. Which way to go will be determined by Fact 5.
This brings us to the most crucial part of this article—the most "physical" part. Fact 5 concerns the verification of a right-angled triangle. We could substitute $c = \sqrt{a^2+b^2}$ directly, but because of the square root (a situation common in physics), it becomes cumbersome (or even unworkable). Therefore, we don't look at the full case; we only consider the infinitesimal limit. Suppose $b$ is an infinitesimal quantity, so:
\[ c = \sqrt{a^{2}+b^{2}} \approx a + \frac{b^2}{2a} \]
This differs from $a$ only by a second-order infinitesimal. Thus, to first-order precision, $c = a$. The area of a right-angled triangle is $\frac{1}{2}ab$. In other words, for a triangle with sides $a, b, a$ where $b$ is infinitesimal, the area is $\frac{1}{2}ab$ to first-order precision. Substituting $a, b, c=a$ into $(a+b-c)(b+c-a)(c+a-b)$, we get:
\[ b \times b \times (2a-b) \approx 2ab^2 \]
We find that if we multiply this by another $a$ and then take the square root, we can obtain $ab$. However, multiplying by $a$ alone breaks symmetry. The simplest symmetric way is to multiply by $(a+b+c)$. Thus, we arrive at a candidate formula:
\[ \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)} \]
The leading coefficient needs adjustment. If we substitute $a, b, c=a$ directly, we get:
\[ \sqrt{(2a+b) \times b \times b \times (2a-b)} \approx \sqrt{4a^2 b^2} = 2ab \]
This differs from $\frac{1}{2}ab$ by a factor of $\frac{1}{4}$. Thus, a candidate formula that satisfies all three assumptions and both facts is:
\[ S(a,b,c) = \frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)} \]
This is exactly the correct Heron's formula! Up to this point, we have "vividly" constructed Heron's formula! Of course, from the perspective of formal logic, this is only a candidate, a highly probable formula; its correctness still requires a proof. The proof is a matter of formalization and is not advanced material, so it is not provided here.
Why Do This?
Why go to such lengths for a formula that is already widely known? This process is neither a rigorous proof nor does it yield anything new, so what is the meaning? For practitioners, at least for now, this task might seem meaningless. However, there could still be unexpected benefits.
First, doing this—or rather, using different methods to accomplish the same goal—helps us understand the essence of the matter. By comparing different approaches, we discover the strengths and weaknesses of each line of thought, thereby identifying the key elements of each. Secondly, this provides a simulated derivation process, or rather, a process of discovering new formulas. This kind of process holds significant meaning in guiding physical discovery. As an enthusiast of mathematics and physics, I naturally hope that this kind of thinking can bring some fresh energy to mathematics—though, of course, this is merely an attempt.