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| − | Square <math>ABCD</math> in the coordinate plane has vertices at the points <math>A(1,1), B(-1,1), C(-1,-1),</math> and <math>D(1,-1).</math> Consider the following four transformations: <math>L,</math> a rotation of <math>90^{\circ}</math> counterclockwise around the origin; <math>R,</math> a rotation of <math>90^{\circ}</math> clockwise around the origin; <math>H,</math> a reflection across the <math>x</math>-axis; and <math>V,</math> a reflection across the <math>y</math>-axis.
| + | #REDIRECT [[2020 AMC 10B Problems/Problem 23]] |
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| − | Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying <math>R</math> and then <math>V</math> would send the vertex <math>A</math> at <math>(1,1)</math> to <math>(-1,-1)</math> and would send the vertex <math>B</math> at <math>(-1,1)</math> to itself. How many sequences of <math>20</math> transformations chosen from <math>\{L, R, H, V\}</math> will send all of the labeled vertices back to their original positions? (For example, <math>R, R, V, H</math> is one sequence of <math>4</math> transformations that will send the vertices back to their original positions.)
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| − | <math>\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}</math>
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| − | ==Solution==
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| − | Hopefully, someone will think of a better one, but here is an indirect answer, use only if you are really desperate.
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| − | <math>20</math> moves can be made, and each move have <math>4</math> choices, so a total of <math>4^{20}=2^{40}</math> moves. First, after the <math>20</math> moves, Point A can only be in first quadrant <math>(1,1)</math> or third quadrant <math>(-1,-1)</math>. Only the one in the first quadrant works, so divide by <math>2</math>. Now, C must be in the opposite quadrant as A. B can be either in the second (<math>(-1, 1)</math>) or fourth quadrant (<math>(1, -1)</math>) , but we want it to be in the second quadrant, so divide by <math>2</math> again. Now as A and B satisfy the conditions, C and D will also be at their original spot. <math>\frac{2^{40}}{2\cdot2}=2^{38}</math>. The answer is <math>\boxed{C}</math>
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| − | ~Kinglogic
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| − | {{AMC12 box|year=2020|ab=B|num-b=18|num-a=20}}
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| − | {{MAA Notice}}
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