AMC 10 · 2020 · #23

Grade 8 geometry-2d

transformations-compositionrotation-isometryreflection-symmetryparity pattern-recognitioncaseworksymmetry-argument ↑ Prerequisites: transformations-compositionreflection-symmetry

📏 Long solution 💡 3 insights

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:

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

Pick an answer.

(A)

$2^{37}$

(B)

$3\cdot 2^{36}$

(C)

$2^{38}$

(D)

$3\cdot 2^{37}$

(E)

$2^{39}$

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Toolkit + CCSS Solution

Understand

Restated: A unit-square labeled $ABCD$ undergoes a sequence of $20$ moves, each chosen from $\{L, R, H, V\}$ (90° rotations counterclockwise/clockwise and reflections across the $x$- and $y$-axes). How many of the $4^{20}$ sequences return every labeled vertex to its starting position?

Givens: Square $ABCD$ with vertices $A(1,1), B(-1,1), C(-1,-1), D(1,-1)$; Moves: $L$ (rot $90^\circ$ CCW), $R$ (rot $90^\circ$ CW), $H$ (reflect across $x$-axis), $V$ (reflect across $y$-axis); Sequence length: $20$; Total sequences: $4^{20}$; Answer choices: (A) $2^{37}$, (B) $3 \cdot 2^{36}$, (C) $2^{38}$, (D) $3 \cdot 2^{37}$, (E) $2^{39}$

Unknowns: Number of sequences whose net effect is the identity transformation

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #2 Make a Systematic List, #7 Identify Subproblems, #6 Guess and Check, #1 Draw a Diagram, #3 Eliminate Possibilities

Tool #9 (Easier Problem): work out the count for $n = 2$ moves, then $n = 4$, and look for the rule. Tool #5 (Pattern): the small cases reveal that exactly $\tfrac{1}{4}$ of all sequences of even length return home — a constant ratio regardless of $n$ (for $n \ge 2$). Tool #2 (Systematic List): enumerate the $4 \times 4 = 16$ length-$2$ sequences and check which give identity. Tool #7 (Subproblems): split the $20$ moves into $10$ pairs of consecutive moves; each pair has $4$ equiprobable effects on the square. Tool #6 (Guess and Check): the simple guess $4^{20}/4 = 4^{19} = 2^{38}$ is verified by both the pair-induction and the small-case computation. Tool #1 (Diagram) anchors the four moves as concrete permutations; Tool #3 (Eliminate) cross-checks the answer against the choice list.

Execute — Answer: C

#1 Draw a Diagram 8.G.A.1 Step 1

First, identify each move as a permutation of the four labeled vertices.
$L$ (CCW rotation by $90^\circ$): $A \to B \to C \to D \to A$.
$R$ (CW): $A \to D \to C \to B \to A$.
$H$ (reflect across $x$-axis): swaps $A \leftrightarrow D$ and $B \leftrightarrow C$.
$V$ (reflect across $y$-axis): swaps $A \leftrightarrow B$ and $C \leftrightarrow D$.
All four are symmetries of the labeled square that move every vertex to an adjacent labeled position.

$$L = (A B C D), \; R = (A D C B), \; H = (A D)(B C), \; V = (A B)(C D)$$

💡 Each move sends every labeled vertex to a neighbor — not to a diagonal.

#5 Look for a Pattern 4.G.A.3 Step 2

Crucial observation.
Each of the four moves sends labeled vertex $A$ to a neighbor of its current position — never to the diagonal corner.
So after ONE move, $A$ is at $B$ or $D$ (adjacent to start $(1,1)$); after TWO moves, $A$ is back at $A$ or at $C$ (diagonal).
Pattern: after an EVEN number of moves, $A$ is at $A$ or $C$; after ODD, at $B$ or $D$.
For the sequence to return $A$ home, the length must be even — and $20$ is even, so this is consistent.

$$A_{\text{after even}} \in \{A, C\}, \;\; A_{\text{after odd}} \in \{B, D\}$$

💡 Each move toggles between "diagonal class $\{A, C\}$" and "diagonal class $\{B, D\}$".

#2 Make a Systematic List 7.SP.C.8 Step 3

Work out $n = 2$ by Tool #2 (systematic list).
Enumerate the $16$ ordered pairs of moves and compose each.
By direct check, the pairs returning to identity are exactly: $LL$ gives rotation $180^\circ$ (not identity); $RR$ also $180^\circ$; $LR$ gives identity ✓; $RL$ gives identity ✓; $HH$ gives identity ✓; $VV$ gives identity ✓; mixed reflection pairs $HV, VH, LH, HL, LV, VL, RH, HR, RV, VR$ give $180^\circ$ rotation or another non-identity element.
Count: $LR, RL, HH, VV$ are exactly $4$.
So $4$ of $16$ return — ratio $1/4$.

$$N(2) = 4, \;\; \dfrac{N(2)}{4^2} = \dfrac{1}{4}$$

💡 Among $16$ length-$2$ sequences, exactly $LR, RL, HH, VV$ are identity.

#7 Identify Subproblems 7.SP.C.8 Step 4

Inductive step (Tool #7 subproblems + Tool #5 pattern).
Suppose for length $n \ge 2$ the count is $N(n) = 4^n / 4 = 4^{n-1}$.
Show $N(n+2) = 4^{n+1}$.
Append two arbitrary moves to a length-$n$ sequence: the final state after $n$ moves is some element $g \in $ orbit of $D_4$ on the labeled square; the next two moves act as a uniform distribution over $4$ specific elements of $D_4$ (one of which is identity).
So exactly $1/4$ of the $4^{n+2}$ length-$(n+2)$ extensions return home, giving $N(n+2) = 4^{n+1}$.

$$N(n+2) = 4 \cdot N(n), \;\; N(n) = 4^{n-1} \text{ for } n \ge 2 \text{ even}$$

💡 Last $2$ moves have a $\tfrac{1}{4}$ chance of canceling whatever the first $n$ did.

#9 Solve an Easier Related Problem 8.G.A.2 Step 5

Cleaner direct argument.
Define the "pair-effect" of two consecutive moves: any pair of moves gives one of four possible effects on the labeled square — the identity, the $180^\circ$ rotation, or one of the two diagonal reflections (across $y = x$ or $y = -x$).
For each fixed starting state, these four effects appear with exactly $4$ pairs each (out of $16$ pairs).
So after grouping the $20$ moves into $10$ pairs, each pair contributes uniformly from a 4-element subgroup of $D_4$.
The product of $10$ uniform elements from a group of order $4$ is the identity with probability exactly $1/4$ (for any $\ge 1$ pair).

$$P(\text{identity}) = \dfrac{1}{4}, \;\; N(20) = \dfrac{4^{20}}{4} = 4^{19} = 2^{38}$$

💡 Pairs of moves form a 4-element group; one pair in four is the identity in that group.

#6 Guess and Check 8.EE.A.1 Step 6

Compute: $4^{19} = (2^2)^{19} = 2^{38}$.
Match to choice $(C)$.

$$N(20) = 2^{38} \Rightarrow \textbf{(C)}$$

💡 $4^{19} = 2^{38}$ matches choice $(C)$ exactly.

#3 Eliminate Possibilities 7.SP.C.7 Step 7

Cross-check by Tool #3 (eliminate).
Total sequences $= 4^{20} = 2^{40}$.
The ratio of returning sequences is $1/4$, giving $2^{38}$ — choice $(C)$.
Choices $(A) = 2^{37}$ corresponds to ratio $1/8$ (would require $D_4$'s full order $8$ uniformly), and $(E) = 2^{39}$ to ratio $1/2$ — both inconsistent with the $4$-effect-per-pair count.

$$\text{Ratio} = \dfrac{1}{4} \Rightarrow N = 2^{40} / 4 = 2^{38}$$

💡 Among the 5 choices only $2^{38}$ matches a ratio of $1/4$.

[1] #1 8.G.A.1 First, identify each move as a permutation of the four labeled vertices. $L$ (CC

[2] #5 4.G.A.3 Crucial observation. Each of the four moves sends labeled vertex $A$ to a neighb

[3] #2 7.SP.C.8 Work out $n = 2$ by Tool #2 (systematic list). Enumerate the $16$ ordered pairs

[4] #7 7.SP.C.8 Inductive step (Tool #7 subproblems + Tool #5 pattern). Suppose for length $n \g

[5] #9 8.G.A.2 Cleaner direct argument. Define the "pair-effect" of two consecutive moves: any

[6] #6 8.EE.A.1 Compute: $4^{19} = (2^2)^{19} = 2^{38}$. Match to choice $(C)$.

[7] #3 7.SP.C.7 Cross-check by Tool #3 (eliminate). Total sequences $= 4^{20} = 2^{40}$. The rat

Review

Reasonableness: Sanity: total $4^{20} = 2^{40} \approx 10^{12}$, identity count $2^{38} \approx 2.75 \times 10^{11}$, ratio exactly $1/4$. Small case verification: $n = 2$ gives $4$ identity sequences ($LR, RL, HH, VV$); $4/16 = 1/4$ ✓. $n = 4$: enumeration (or pair-induction) gives $4^3 = 64$ identity sequences out of $256$; $64/256 = 1/4$ ✓. Choice $(C)$ confirmed.

Alternative: Tool #2 (Systematic List) extended: enumerate the $8$ elements of $D_4$, compute the transition matrix from one element to another after one move (each move sends a state to $4$ specific other states uniformly), and raise the matrix to the $20$-th power to read off the entry from identity to identity. Equivalent to a discrete-time random walk on $D_4$; the limiting probability after even $n \ge 2$ is exactly $1/4$ (uniform on the index-$2$ subgroup containing identity). The pair-grouping argument above is the shortcut.

CCSS standards used (min grade 8)

4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Tracking how each move sends a labeled vertex to an adjacent neighbor; even vs. odd parity of position.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Modeling each pair of moves as a uniform draw from $4$ possible effects and applying the $1/4$ ratio.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Enumerating the $16$ length-$2$ sequences to confirm exactly $4$ return to identity.)
8.EE.A.1 Know and apply the properties of integer exponents (Rewriting $4^{19} = 2^{38}$ to match choice $(C)$.)
8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Identifying each move ($L, R, H, V$) as a specific permutation of the labeled vertices.)
8.G.A.2 Understand that a two-dimensional figure is congruent to another using transformations (All four moves preserve the square; their composition lives in the symmetry group of the square.)

⭐ This AMC 10 problem only needs Grade 8 transformations: each pair of moves has exactly $4$ possible effects on the labeled square (one of which is the identity), so the ratio of identity sequences is $1/4$, giving $4^{20}/4 = 4^{19} = 2^{38}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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