2022 AMC 10A Problem 18

Question

Let

T_k

be the transformation of the coordinate plane that first rotates the plane

k

degrees counterclockwise around the origin and then reflects the plane across the

y

-axis. What is the least positive integer

n

such that performing the sequence of transformations

T_1,

T_2,

T_3,

\cdots,

T_n

returns the point

(1,0)

back to itself?

359

360

719

720

721

Track only the angle in polar coordinates

Two consecutive transformations simplify to a one-degree rotation

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.

Problem 18 in other years