, 2020

Solution:

Fix an orientation of the circle, and observe that the problem is equivalent to finding the number of ways to color ten equal arcs of the circle such that each arc is one of three different colors, and any two arcs which are separated by exactly one arc are of different colors. We can consider every other arc, so we are trying to color just five arcs so that no two adjacent arcs are of the same color. This is independent from the coloring of the other five arcs.

Let $a_{i}$ be the number of ways to color $i$ arcs in three colors so that no two adjacent arcs are the same color. Note that $a_{1}=3$ and $a_{2}=6$. We claim that $a_{i}+a_{i+1}=3 \cdot 2^{i}$ for $i \geq 2$. To prove this, observe that $a_{i}$ counts the number of ways to color $i+1$ points in a line so that no two adjacent points are the same color, and the first and $(i+1)$th points are the same color. Meanwhile, $a_{i+1}$ counts the number of ways to color $i+1$ points in a line so that no two adjacent points are the same color, and the first and $(i+1)$th points are different colors. Then $a_{i}+a_{i+1}$ is the number of ways to color $i+1$ points in a line so that no two adjacent points are the same color. There are clearly $3 \cdot 2^{i}$ ways to do this, as we pick the colors from left to right, with $3$ choices for the first color and $2$ for the rest. We then compute $a_{3}=6$, $a_{4}=18$, $a_{5}=30$. Then we can color the whole original circle by picking one of the $30$ possible colorings for each of the two sets of $5$ alternating arcs, for $30^{2}=900$ total.

Now, we must consider the rotational symmetry. If a configuration has no rotational symmetry, then we have counted it $10$ times. If a configuration has $180^{\circ}$ rotational symmetry, then we have counted it $5$ times. This occurs exactly when we have picked the same coloring from our $30$ for both choices, and in exactly one particular orientation, so there are $30$ such cases. Having $72^{\circ}$ or $36^{\circ}$ rotational symmetry is impossible, as arcs with exactly one arc between them must be different colors. Then after we correct for overcounting our answer is

$$\frac{900-30}{10}+\frac{30}{5}=93$$