2020 AMC 10B Problem 23

Attempt Problem 23 of the 2020 AMC 10B below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10B solutions, or check the answer key.

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

23.

Square ABCDABCD in the coordinate plane has vertices at the points A(1,1),B(1,1),C(1,1),A(1,1), B(-1,1), C(-1,-1), and D(1,1).D(1,-1). Consider the following four transformations:

L,L, a rotation of 9090^{\circ} counterclockwise around the origin;

R,R, a rotation of 9090^{\circ} clockwise around the origin;

H,H, a reflection across the xx-axis; and

V,V, a reflection across the yy-axis.

Each of these transformations maps the square onto itself, but the positions of the labeled vertices will change. For example, applying RR and then VV would send the vertex AA at (1,1)(1,1) to (1,1)(-1,-1) and would send the vertex BB at (1,1)(-1,1) to itself. How many sequences of 2020 transformations chosen from {L,R,H,V}\{L, R, H, V\} will send all of the labeled vertices back to their original positions? (For example, R,R,V,HR, R, V, H is one sequence of 44 transformations that will send the vertices back to their original positions.)

2372^{37}

32363\cdot 2^{36}

2382^{38}

32373\cdot2^{37}

2392^{39}

Answer: C
Concepts:transformationparity
Difficulty rating: 2060
Video solution:
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Written solution:

Each of L,R,H,VL,R,H,V moves every vertex to an adjacent corner of the square. Therefore after an odd number of transformations the labeling is in one of the four odd-parity states, and after an even number it is in one of the four even-parity states.

After any first 1919 transformations, the square is in an odd-parity state. From each odd-parity state, exactly one of L,R,H,VL,R,H,V sends the labeled vertices back to their original positions. Thus every sequence of the first 1919 transformations has exactly one valid final transformation.

There are 419=2384^{19}=2^{38} choices for the first 1919 transformations, so there are 2382^{38} valid sequences.

Thus, C is the correct answer.

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