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Difference between revisions of "2020 AMC 12B Problems/Problem 19"

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Since rotation <math>R</math> and rotation <math>L</math> cancels each other out, the difference between the numbers of them define the final position. The probability of the transformation returning the vertices to the orginal position given that there are even number of rotations is equal to the probability that
 
Since rotation <math>R</math> and rotation <math>L</math> cancels each other out, the difference between the numbers of them define the final position. The probability of the transformation returning the vertices to the orginal position given that there are even number of rotations is equal to the probability that
  
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<center><math>|n(R)-n(L)|\equiv0\pmod{4}</math></center>
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<center><math>|n(R)-n(L)|\equiv0\pmod{4}</math> when <math>|n(H)-n(V)|\equiv0\pmod{4}</math></center>
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or
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<center><math>|n(R)-n(L)|\equiv2\pmod{4}</math> when <math>|n(H)-n(V)|\equiv2\pmod{4}</math></center>
  
 
which is again, <math>\frac{1}{2}</math>.
 
which is again, <math>\frac{1}{2}</math>.

Revision as of 02:06, 8 February 2020

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

Solution 2

The total number of sequence is $4^{20}=2^{40}$.

Note that there can only be even number of reflections since they result in the same anti-clockwise orientation of the verices $A,B,C,D$. Therefore, the probability of having the same anti-clockwise orientation with the original arrangement after the transformation is $\frac{1}{2}$.

Next, even number of reflections mean that there must be even number of rotations since their sum is even. Even rotations result only in the original position or $180^{\circ}$ rotation of it.

Since rotation $R$ and rotation $L$ cancels each other out, the difference between the numbers of them define the final position. The probability of the transformation returning the vertices to the orginal position given that there are even number of rotations is equal to the probability that

$|n(R)-n(L)|\equiv0\pmod{4}$ when $|n(H)-n(V)|\equiv0\pmod{4}$

or

$|n(R)-n(L)|\equiv2\pmod{4}$ when $|n(H)-n(V)|\equiv2\pmod{4}$

which is again, $\frac{1}{2}$.

Therefore, $2^{40}\cdot\frac{1}{2}\cdot\frac{1}{2}=\boxed{\mathbf(C) 2^{38}}$ ~joshuamh111

2020 AMC 12B (Problems • Answer Key • Resources)
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Problem 18 		Followed by
Problem 20
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All AMC 12 Problems and Solutions

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