2020 AMC 12B Problems/Problem 19

Revision as of 01:16, 8 February 2020 by Kinglogic (talk | contribs) (Solution)

Square $ABCD$ 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 $20$ transformations 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.)

$\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\  2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}$

Solution

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, 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. $\frac{2^{40}}{2\cdot2}=2^{38}$. The answer is $\boxed{C}$ ~Kinglogic

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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