A proposition of trigonometry, or complex numbers

Prop. If $0,1,z,w$ all lie on a circle centered at $1/2$ then $zw$ is the orthogonal projection of $0$ onto the line through $z$ and $w$.

Proof. If $z,w$ to lie on that circle, then $$(z-\frac12)(\overline{z}-\frac12)=\frac14,\text{ which simplifies to }z\overline{z}=\frac{z+\overline{z}}2$$ $$(w-\frac12)(\overline{w}-\frac12)=\frac14,\text{ which simplifies to }w\overline{w}=\frac{w+\overline{w}}2.$$ Introducing names $z=a+bi,w=c+di$ these equations turn into $$a^2+b^2=a,\quad c^2+d^2=c.$$ Two things need to be shown: $(1)$ $zw$ lies on the line through $z$ and $w$, and $(2)$ $zw$ is perpendicular to $z-w$.

$(1)$ Verify that the following determinant is zero. $$\begin{align*} \begin{vmatrix}1&a&b\\1&c&d\\1&ac-bd&ad+bc\end{vmatrix}&=acd+bc^2-acd+bd^2-a^2d-abc+abc-b^2d+ad-bc\\ &=b(c^2+d^2)-(a^2+b^2)d+ad-bc=bc-ad+ad-bc=0. \end{align*}$$ $(2)$ Verify that $zw/(z-w)$ is imaginary, or equivalently that $zw(\overline z-\overline w)$ is imaginary. $$\begin{align*} zw(\overline z-\overline w)+\overline{zw(\overline z-\overline w)}&=z\overline zw-zw\overline w+z\overline z\overline w-\overline zw\overline w=z\overline z(w+\overline w)-(z+\overline z)w\overline w\\ &=\frac{z+\overline z}2(w+\overline w)-(z+\overline z)\frac{w+\overline w}2=\frac12(z+\overline z)(w+\overline w)(1-1)=0. \end{align*}$$

Extremely rational cubic polynomials

By an extremely rational cubic polynomial, I mean a cubic polynomial $p$ such that both $p$ and $p'$ are products of rational linear factors. A parametrization will now be derived.

Seek to satisfy $p=(a+x)(b+x)(c+x),\ p'=3(d+x)(e+x)$ as polynomials.
$$\begin{align*} 0=p'(-e)&=ab+ac+bc-2(a+b+c)e+3e^2 \\ (3e-a-b-c)^2&=(a+b+c)^2-3(ab+ac+bc) \\ &=a^2+b^2+c^2-ab-ac-bc \end{align*}$$ We thus need to parametrize triples $(a,b,c)$ such that the right-hand side is a square.

By the cubic number system, I mean triples $(a,b,c)$ with the following operations.
$$\begin{align*} (a,b,c)+(d,e,f)&\equiv(a+d,b+e,c+f) \\ (a,b,c)(d,e,f)&\equiv(ad+bf+ce,ae+bd+cf,af+be+cd) \\ Q(a,b,c)&\equiv a^2+b^2+c^2-ab-ac-bc \end{align*}$$ This system is a close sibling of the usual complex number system. Its relevance to us is in the last definition. It is the expression we want to make into a square. Verify the identity $Q(zw)=Q(z)Q(w)$ to realize that our task can be fulfilled by setting $(a,b,c)\equiv(u,v,w)^2$. Normally, one would use fractions $a,b,c,u,v,w$ in this number system, but we can just as well use complex parameters $u,v,w$.

Here is the parametrization.
$$\begin{align*} p&=(a+x)(b+x)(c+x),\quad p'=3(d+x)(e+x),\quad p''=6(f+x) \\ a&=u^2+2vw,\quad b=w^2+2uv,\quad c=v^2+2uw\\ d,e&=\frac13((u+v+w)^2\pm Q(u,v,w)),\quad f=\frac13(u+v+w)^2\end{align*}$$
Here is an example.
$$\begin{align*} (u,v,w)&=(1,2,7) \\ (a,b,c)&=(1^2+2\times 2\times 7,\ 7^2+2\times 1\times 2,\ 2^2+2\times 1\times 7)=(29,53,18) \\ p&=(29+x)(53+x)(18+x)=27666+3013x+100x^2+x^3 \\ Q(u,v,w)&=1^2+2^2+7^2-1\times 2-1\times 7-2\times 7=31 \\ Q(a,b,c)&=29\times 29+53\times 53+18\times 18-29\times 53-29\times 18-53\times 18=961 \\ d,e&=\frac13({10}^2\pm 31)=\frac{131}3,23\\ p'&=3(\frac{131}3+x)(23+x)=(131+3x)(69+x)=3013+200x+3x^2\\ f&=\frac{100}3,\quad p''=6(\frac{100}3+x)=200+6x \end{align*}$$
An example involving complex parameters $u,v,w$ is best given visually and interactively.