2020 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 23 Hard

Square A B C D in the coordinate plane has vertices at the points A(1,1), B(-1,1), C(-1,-1), and D(1,-1). Consider the following four transformations: L, a rotation of 90^{\circ} counterclockwise around the origin; R, a rotation of 90^{\circ} clockwise around the origin; H, a reflection across the X-axis; and V, a reflection across the Y-axis. Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying R and then V would send the vertex A at (1,1) to (-1,-1) and would send the vertex B at (-1,1) to itself. How many sequences of 20 transtormations chosen from \{L, R, H, V\} will send all of the labeled vertices back to their original positions? (For example, R, R, V, H is one sequence of 4 transformations that will send the vertices back to their original positions.)(2020 AMC 10B, Question #23)

  • A.

    2^{37}

  • B.

    3 \cdot 2^{36}

  • C.

    2^{38}

  • D.

    3 \cdot 2^{37}

  • E.

    2^{39}

Answer:C

Solution 1: Let (+) denote counterclockwise/starting orientation and (-) denote clockwise orientation. Let 1,2,3, and 4 denote which quadrant A is in.

Realize that from any odd quadrant and any orientation, the 4 transformations result in some permutation of (2+, 2-, 4+, 4-). The same goes that from any even quadrant and any orientation, the 4 transformations result in some permutation of (1+, 1-, 3+, 3-). We start our first 19 moves by doing whatever we want, 4 choices each time. Since 19 is odd, we must end up on an even quadrant.

As said above, we know that exactly one of the four transformations will give us (1+), and we must use that transformation. Thus 4^{19}=(C) 2^{38}

Solution 2: Hopefully, someone will think of a better one, but here is an indirect answer, use only if you are really desperate. 20 moves can be made, and each move have 4 choices, so a total of 4^{20}=2^{40} moves. First, after the 20 moves, Point A can only be in first quadrant (1,1) or third quadrant (-1,-1). Only the one in the first quadrant works, so divide by 2 . Now, \text{C} must be in the opposite quadrant as A. B can be either in the second (-1,1) ) or fourth quadrant (1,-1), but we want it to be in the second quadrant, so divide by 2 again. Now as A and B satisfy the conditions, C and D will also be at their original spot. \frac{2^{40}}{2 \cdot 2}=2^{38}. The answer is C

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