2022 AMC 10A Problem 18

Below is the professionally curated solution for Problem 18 of the 2022 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:transformationpattern recognition

Difficulty rating: 1950

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,T2,T3,,TnT_1, T_2, T_3, \cdots, T_n returns the point (1,0)(1,0) back to itself?

359359

360360

719719

720720

721721

Solution:

Since we are working with angles and reflections, working with polar coordinates would make this problem easier to deal with.

Let (r,θ)(r, \theta) be a polar coordinate. Rotating this by kk degrees counterclockwise maps the point to (r,θ+k)(r, \theta + k^{\circ}) and then reflecting it maps it to (r,180θk.)(r, 180 - \theta - k^{\circ}.)

Therefore, we have that Tk(r,θ)=(r,180θk). T_k(r, \theta) = (r, 180 - \theta - k^{\circ}).

From this, we can see that Tk+1(Tk(r,θ))= T_{k + 1}(T_k(r, \theta)) = Tk+1(r,180θk)=T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) = (r,θ1).(r, \theta - 1^{\circ}).

Now, let's analyze what happens to the point (1,0).(1, 0^{\circ}).

After T1,T_1, we get (1,179).(1, 179^{\circ}).

After T2,T_2, we get (1,1).(1, -1^{\circ}).

After T3,T_3, we get (1,178).(1, 178^{\circ}).

After T4,T_4, we get (1,2).(1, -2^{\circ}).

\vdots

After T2n1,T_{2n - 1}, we get (1,180k).(1, 180^{\circ} - k^{\circ}).

After T2n,T_{2n}, we get (1,k).(1, -k^{\circ}).

From this, we can see that the first time the angle is back to 00^{\circ} is when n=180n = 180 and k=359.k = 359.

Thus, A is the correct answer.

Problem 18 in Other Years