2020 AMC 10B Problems/Problem 23

Revision as of 00:17, 8 February 2020 by Vvluo (talk | contribs) (Solution)

Problem

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

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 $4^{19}=\boxed{(C) 2^{38}}$

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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