2022 AMC 10A Problem 18

Since we are working with angles and reflections, working with polar coordinates would make this problem easier to deal with. Let $(r, \theta)$ be a polar coordinate. Rotating this by $k$ degrees counterclockwise maps the point to $(r, \theta + k^{\circ})$ and then reflecting it maps it to $(r, 180^{\circ} - \theta - k^{\circ}).$ Therefore, we have that $$T_k(r, \theta) = (r, 180^{\circ} - \theta - k^{\circ}).$$ From this, we can see that $$T_{k + 1}(T_k(r, \theta)) =$$$$T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) =$$$$(r, \theta - 1^{\circ}).$$ Now, let’s analyze what happens to the point $(1, 0^{\circ}).$ After $T_1,$ we get $(1, 179^{\circ}).$ After $T_2,$ we get $(1, -1^{\circ}).$ After $T_3,$ we get $(1, 178^{\circ}).$ After $T_4,$ we get $(1, -2^{\circ}).$ $$\vdots$$ After $T_{2m - 1},$ we get $(1, 180^{\circ} - m^{\circ}).$ After $T_{2m},$ we get $(1, -m^{\circ}).$ From this, we can see that the first time the angle is back to $0^{\circ}$ is after $T_{2(180)-1}=T_{359}.$ Therefore $n=359.$ Thus, A is the correct answer.