2022 AMC 10A Problem 18

Step 1: describe the point by one number. Rotating about the origin and reflecting across the $y$-axis never change the distance from the origin. So the point always stays on the circle of radius $1$, and we only need its angle, measured counterclockwise from the positive $x$-axis. The start $(1, 0)$ has angle $0^\circ$.

Step 2: what each move does to the angle.
- Rotating $k$ degrees counterclockwise adds $k$: angle $A$ becomes $A + k$.
- Reflecting across the $y$-axis sends angle $A$ to $180 - A$. Examples: $0^\circ$ goes to $180^\circ$, $30^\circ$ goes to $150^\circ$, and $90^\circ$ stays put.

So $T_k$ (rotate, then reflect) turns angle $A$ into $180 - (A + k) = 180 - A - k$.

Step 3: try the first few steps.

The pattern: after an even number of steps $2m$, the angle is $-m$; after an odd number $2m + 1$, it is $179 - m$.

Step 4: see why the pattern holds. Do two steps in a row, $T_{2j-1}$ then $T_{2j}$, starting from angle $A$:
$$A \ \to\ 180 - A - (2j - 1) \ \to\ 180 - \big(180 - A - 2j + 1\big) - 2j = A - 1.$$
So each pair of steps just turns the point back by $1^\circ$.

Step 5: when is the point back at $(1,0)$? When the angle is a multiple of $360$.
- Even $n = 2m$: the angle $-m$ is first a multiple of $360$ at $m = 360$, so $n = 720$.
- Odd $n = 2m + 1$: the angle $179 - m$ is $0$ at $m = 179$, so $n = 359$. For smaller $m$ it is between $1$ and $179$, so it does not work sooner.

The smaller of the two is $359$.

The answer is $\boxed{\textbf{(A)}\ 359}$.

Rotations and reflections of a point on a circle only change its angle, and each one is a simple rule on that angle. Writing the rules down turns a geometry problem into arithmetic, and a small table exposes the slow backward turn hidden in each pair of steps. When the steps come in pairs, always check both even and odd stopping points.