Equilateral triangle in a circle Suppose you have a circle and consider three disjoint $60$ degree arcs $A,B,C$ in the circle. (i.e the arcs $A,B,C$ are separated by three arcs $x,y,z$ (with $x+y+z=180$ degrees and $x,y,z>0$)). Now take the chords on $x,y,z$. Call them $X,Y,Z$ respectively. Prove that the triangle that has its vertices at the midpoints of $X,Y,Z$ is an equilateral triangle.

 A: You have three points $x,y,z$ on the unit circle, and their images under a $60°$ rotation, which is multiplication by $\omega=e^{\frac{2i\pi}{6}}$. The three midpoints are
$$m=\frac{z+\omega y}{2},\quad, m'=\frac{x+\omega z}{2},\quad m''=\frac{y+\omega x}{2}$$ and the claim is that they form an equilateral triangle, i.e., the three sides
$$\begin{eqnarray}
m'-m&=\frac{x-z}{2}+\omega\,\frac{z-y}{2}\\
m''-m'&=\frac{y-x}{2}+\omega\,\frac{x-z}{2}\\
m-m''&=\frac{z-y}{2}+\omega\,\frac{y-x}{2}
\end{eqnarray}$$
have the same modulus. But they are of the form
$$\begin{eqnarray}
A+\omega B\\
C+\omega A\\
B+\omega C
\end{eqnarray}$$
with $A+B+C=0$, and therefore must have the same modulus. Indeed,
$$|A+\omega B|=|-(B+C)+\omega B|=|(1-\omega)B+C|=|B+\omega C|$$
since $1-\omega=\overline{\omega}=\omega^{-1}$ and $|\omega|=1$. The other equality is derived similarly. Hence
$$|m'-m|=|m''-m'|=|m-m''|$$
and the triangle is equilateral.
A: It can be done (even if it shouldn't be done!) by a mindless calculation.
If $\alpha$ is a cube root of $-1$, other than $-1$, then $1 - \alpha + \alpha^2 = 0.$
Let $u, v, w$ be any three complex numbers.
\begin{gather*}
0 = (1 - \alpha + \alpha^2)u = (1 - \alpha + \alpha^2)w \\
\therefore\
(-1 - \alpha^2)w + \alpha^2u + v = \alpha{u} + v - (\alpha{w} + u) \\
\text{i.e. }
\alpha^2[(\alpha - 1)w +(u - \alpha{v})] = (\alpha{u} + v) - (\alpha{w} + u) \\
\text{i.e. }
\alpha^2[(\alpha{w} + u) - (\alpha{v} + w)] = (\alpha{u} + v) - (\alpha{w} + u) \\
\end{gather*}
Dividing by 2, and taking $\alpha = (1 + \sqrt{3}i)/2$, so that the beginnings and ends of three 60-degree arcs of the unit circle in the complex plane are represented by $u$, $\alpha{u}$, $v$, $\alpha{v}$, $w$, $\alpha{w}$, we see that the complex numbers representing the oriented line segments joining the midpoints $(\alpha{w} + u)/2$, $(\alpha{u} + v)/2$, $(\alpha{v} + w)/2$, are each equal to $\alpha^2$ times the preceding one in cyclic order, i.e. they form an equilateral triangle.
Disjointness doesn't seem to play any part, so long as you take the 6 endpoints in the right order.
