2022 AMC 12A Problem 18

Attempt Problem 18 of the 2022 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

Let TkT_k be the transformation of the coordinate plane that first rotates the plane kk degrees counterclockwise around the origin and then reflects the plane across the yy-axis. What is the least positive integer nn such that performing the sequence of transformations T1,T_1, T2,T_2, T3,T_3, …,\ldots, TnT_n returns the point (1,0)(1,0) back to itself?
359359
360360
719719
720720
721721
Answer: A
Concepts:transformationcasework
Small hint:
A point at angle θ\theta is sent by TkT_k to angle (180−k)−θ(180-k)-\theta
Big hint:
Track the angle of (1,0);(1,0); also consider odd n,n, where the net map is a reflection
Solution:
Rotating a point at angle θ\theta by k∘k^\circ gives θ+k,\theta+k, and reflecting across the yy-axis sends angle ϕ\phi to 180−ϕ.180-\phi. So TkT_k sends θ\theta to (180−k)−θ.(180-k)-\theta.
Starting from (1,0)(1,0) at angle 0,0, applying T1,T2,…T_1,T_2,\ldots gives angles 179,−1,178,−2,177,….179,-1,178,-2,177,\ldots. After an even number 2m2m of steps the angle is −m,-m, and after an odd number 2m+12m+1 it is 179−m.179-m.
For the point to return, the angle must be a multiple of 360∘.360^\circ. The even case needs m=360,m=360, i.e. n=720.n=720. The odd case needs 179−m=0,179-m=0, i.e. m=179m=179 and n=359,n=359, where the net reflection fixes (1,0).(1,0).
The least such nn is 359.359.
Thus, the correct answer is A.
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