2020 AMC 10B Problems/Problem 23

The following problem is from both the 2020 AMC 10B #23 and 2020 AMC 12B #19, so both problems redirect to this page.

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


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, 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 3

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

Solution 4

Notice that any pair of two of these transformations either swaps the x and y-coordinates, negates the x and y-coordinates, swaps and negates the x and y-coordinates, or leaves the original unchanged. Furthermore, notice that for each of these results, if we apply another pair of transformations, one of these results will happen again, and with equal probability. Therefore, no matter what state after we apply the first $19$ pairs of transformations, there is a $\frac14$ chance the last pair of transformations will return the figure to its original position. Therefore, the answer is $\frac{4^{20}}4 = 4^{19} = \boxed{\textbf{(C) }2^{38}}$



Solution 5(Combinatorics)

So we have \[2\] Rotations and \[2\] reflections which make 4 isometries in total. We can start a sequence of isometries trying out different combinations. For example, rotating this square \[90\] degrees clockwise and \[90\] degrees counterclockwise is a sequence of \[1\] transformation away and \[1\] transformation back. Trying out bigger sequences, we eventually see that no matter which position the square is in, \[1\] of the above isometries will map the vertices of the square back to themeselves. Now we know that for \[19\] moves we can do whatever we want and for the \[20th\] move, we only have \[1\] option. So since each move has \[4\] options but the last one, the answer is \[4^{19} = \boxed{\textbf{(C) }2^{38}}\]

Video Solution

https://www.youtube.com/watch?v=XZs9DHg6cx0 ~ MathEx

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS