CEMC Cayley, 2022

Solution 1

Since $A$, $B$, $C$, $D$, $E$, and $F$ are equally spaced around the circle,
moving from one point to the next corresponds to moving $\frac{1}{6}$ of the way around the
circle.

Therefore, moving from $A$ to $C$ corresponds to moving $\frac{2}{6}$ (or $\frac{1}{3}$) of the way around the
circle.

Since moving around the whole circle corresponds to moving through $360\degree$, then moving $\frac{1}{3}$ of the way around the circle
corresponds to moving through $\frac{1}{3} \times 360\degree = 120\degree$.

Thus, $\angle AOC = 120\degree$.

Solution 2

We join $A$, $B$, $C$, $D$, $E$, and $F$ to $O$.

Since $A$, $B$, $C$, $D$, $E$, and $F$ are equally spaced around the circle,
the angles made at the centre by consecutive points are equal. That is,
$$\angle AOB = \angle BOC = \angle COD = \angle DOE = \angle EOF = \angle FOA$$ Since these 6 angles form
a complete circle, the sum of their measures is $360\degree$.

Therefore, $$\angle AOB = \angle BOC = \angle COD = \angle DOE = \angle EOF = \angle FOA = \tfrac{1}{6} \times 360\degree = 60\degree$$ This means that $\angle AOC = \angle AOB + \angle BOC = 60\degree + 60\degree = 120\degree$.