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2020 AMC 10B Problem 23

2020 AMC 10B Problem 23AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 4 of 5 · about 6 min · position P21-25

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:

$\quad\bullet\qquad$ $L,$ a rotation of $90^{\circ}$ counterclockwise around the origin;

$\quad\bullet\qquad$ $R,$ a rotation of $90^{\circ}$ clockwise around the origin;

$\quad\bullet\qquad$ $H,$ a reflection across the $x$ -axis; and

$\quad\bullet\qquad$ $V,$ 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.)

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

Answer

Key insight

Each move sends vertex A to an adjacent corner, so after 19 moves the square is one move from home, and exactly one 20th move finishes: 4^19.

Solution

Track where the label $A$ sits. Each of $L$ , $R$ , $H$ , $V$ moves whichever label is at a corner to an adjacent corner (a $90^\circ$ rotation or a reflection across a coordinate axis never sends a corner to itself or to the opposite corner). Color the corners like a checkerboard: $(1,1)$ and $(-1,-1)$ are "even," $(-1,1)$ and $(1,-1)$ are "odd." Every move flips the parity of $A$ 's corner.

A configuration of the labeled square is determined by where $A$ is and whether the square is flipped over (reflected) or not: $4 \times 2 = 8$ configurations. After $19$ moves, $A$ is at an odd corner, so the square is in one of the $4$ configurations with $A$ adjacent to home. Those four are exactly the results of applying a single $L$ , $R$ , $H$ or $V$ to the original square, and each is undone by exactly one of the four moves ( $L$ by $R$ , $R$ by $L$ , $H$ by $H$ , $V$ by $V$ ).

Therefore any sequence of $19$ moves can be completed in exactly one way, and no sequence of $19$ moves ends at the identity (wrong parity). The count is
$4^{19} = 2^{38}.$

The answer is $\boxed{\textbf{(C)}\ 2^{38}}$ .

Why this works

In "sequences of moves returning to start" problems, fix all but the last move and ask how many last moves work. Here a parity invariant guarantees that after an odd number of moves the square is never home but always exactly one move from home, and the four allowed moves are closed under inverses, so the completion is unique. This is the coset structure of the dihedral group in disguise, but the checkerboard argument needs no group theory.

Alternative approach

Let $f(n)$ be the number of $n$ -sequences ending at the identity and $g(n)$ the number ending at any fixed one-move-away configuration. By symmetry all four adjacent configurations are equally likely at each odd step, so $g(19) = 4^{19}/4 = 4^{18}$ each, and $f(20) = 4 \cdot g(19) = 4^{19}$ .

The trap

Trying to count sequences by how many of each letter appear (L, R must balance mod 4, etc.), which is a much harder computation.

Common mistakes

Trying to count sequences by how many of each letter appear (L, R must balance mod 4, etc.), which is a much harder computation.
Answering $4^{20}/8 = 2^{37}$ by assuming all $8$ configurations are equally likely after $20$ moves; parity makes half of them impossible.

Techniques

Map the objects to something easier to count · Use an invariant, parity, or coloring argument