2022 AMC 10A Problem 18

Every transformation preserves distance from the origin, so track the point by its angle $\theta$ measured from the positive $x$-axis; the starting point $(1,0)$ has $\theta = 0$.

A rotation by $k$ degrees sends $\theta \mapsto \theta + k$. Reflecting across the $y$-axis sends $(\cos\theta, \sin\theta)$ to $(-\cos\theta, \sin\theta)$, which is the point at angle $180 - \theta$. Therefore
$$T_k:\ \theta \longmapsto 180 - (\theta + k) = 180 - \theta - k.$$

Apply two consecutive transformations $T_{2j-1}$ then $T_{2j}$:
$$\theta \mapsto 180 - \theta - (2j-1) \mapsto 180 - \big(180 - \theta - (2j-1)\big) - 2j = \theta - 1.$$
So each pair of steps is simply a rotation by $-1$ degree.

Even $n = 2m$: the angle is $-m$, which is a multiple of $360$ first when $m = 360$, i.e. $n = 720$.

Odd $n = 2m + 1$: after $2m$ steps the angle is $-m$; then $T_{2m+1}$ gives $180 - (-m) - (2m+1) = 179 - m$. This is $0$ when $m = 179$, i.e. $n = 359$ (and no smaller odd $n$ works, since $179 - m$ is between $1$ and $179$ for $m < 179$).

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

Rotations and reflections of the unit circle act on a single number, the angle, by $\theta \mapsto \theta + k$ and $\theta \mapsto 180 - \theta$. Writing each $T_k$ as such a formula turns a geometric composition into arithmetic, and pairing steps reveals the hidden slow rotation. Always test both parities when the composition alternates between orientation-reversing and orientation-preserving maps.

• Checking only even n (which first works at n = 720) and missing that an odd number of transformations, ending on a reflection, already returns the point at n = 359.
• Using the wrong reflection formula ($\theta \mapsto -\theta$ is the reflection across the $x$-axis), which changes the net effect of each pair of steps.