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# Hero's formula - a complex numbers proof.
Given an arbitrary triangle $T$ with sides $a,b,c$, Hero's formula states its area $A$ is given by
$$
A = \sqrt{s(s-a)(s-b)(s-c)}
$$
where $s = \frac{1}{2}(a+b+c)$ is its semi-perimeter. Elementary proofs using Pythagorean theorem or law of cosine are well known.
There is a nice proof using complex numbers (given by a then high school student ![[---images/---assets/---icons/exclaim-icon.svg]][^cite]), which goes as follows:
First draw the **incircle** $C$ of this triangle $T$ [^a] , and denote $r$ as the radius of $C$. Drawing the altitudes from the incenter to each sides cuts each side into two pieces, to which we label them as follows :
![[0 inbox/---files/Heros-formula-complex-numbers 2023-05-13 10.42.54.excalidraw.svg]]
%%[[0 inbox/---files/Heros-formula-complex-numbers 2023-05-13 10.42.54.excalidraw.md|đź–‹ Edit in Excalidraw]], and the [[0 inbox/---files/Heros-formula-complex-numbers 2023-05-13 10.42.54.excalidraw.dark.svg|dark exported image]]%%
Notice the complex number $r+ix$ has argument $\alpha$, $r+iy$ has argument $\beta$, and $r+iz$ has argument $\gamma$.[^why-1] And since $\alpha + \beta + \gamma = \pi$, the product of these three complex numbers has an argument of $\pi$.[^why-2] In other words its imaginary part is zero:
$$
{\frak Im}\left[(r+ix)(r+iy)(r+iz)\right] = 0
$$
So if we expand and consider the imaginary part, we get
$$
\begin{align*}
&{\frak Im}\left[(r+ix)(r+iy)(r+iz)\right]\\
&=xr^2+yr^2 + zr^2 -xyz \\
&=0
\end{align*}
$$
Namely $r = \sqrt{\frac{xyz}{x+y+z}}$ . Also note the area of the triangle is given by $A =r(x+y+z)$, whence the area is then $A = \sqrt{(x+y+z)xyz}$.
Finally, note the semi-perimeter $s = x+y+z$, and $x = s-c$, $y = s-a$, $z = s-b$, we have as claimed !
---
[^why-1]: Imagine putting these right triangle at the origin of the complex plane, then the complex number $r+ix$ will have an angle of $\alpha$, for example.
[^why-2]: Recall when doing complex multiplication, the angles add.
[^a]: Which exists by considering the intersection of the angle bisectors to find the incenter.
[^cite]: American Mathematical Monthly AMM Vol. 114, No. 10, 2007, p. 937.
#geometry #complex-numbers