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AMC 10 · 2020 · #23

Grade 8 geometry-2d
Archetype Casework by Position / Vertex Type
transformations-compositionrotation-isometryreflection-symmetrymodular-arithmeticcombinations-basic caseworksystematic-enumerationphysical-representation ↑ Prerequisites: rotation-isometryreflection-symmetry
📏 Long solution 💡 3 insights

Problem

Let TT be the triangle in the coordinate plane with vertices (0,0),(4,0),(0,0), (4,0), and (0,3).(0,3). Consider the following five isometries (rigid transformations) of the plane: rotations of 90∘,180∘,90^{\circ}, 180^{\circ}, and 270∘270^{\circ} counterclockwise around the origin, reflection across the xx-axis, and reflection across the yy-axis. How many of the 125125 sequences of three of these transformations (not necessarily distinct) will return TT to its original position? (For example, a 180∘180^{\circ} rotation, followed by a reflection across the xx-axis, followed by a reflection across the yy-axis will return TT to its original position, but a 90∘90^{\circ} rotation, followed by a reflection across the xx-axis, followed by another reflection across the xx-axis will not return TT to its original position.)

(A) 12(B) 15(C) 17(D) 20(E) 25\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25

Pick an answer.

(A)
12
(B)
15
(C)
17
(D)
20
(E)
25
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Toolkit + CCSS Solution

Understand

Restated: From the $5$ rigid motions $\{R_{90}, R_{180}, R_{270}, S_x, S_y\}$ (CCW rotations about the origin and reflections across the axes), pick an ordered triple (with repetition) of size $3$ — there are $5^3 = 125$ such ordered triples. How many of them, when applied one after another, return triangle $T$ with vertices $(0,0), (4,0), (0,3)$ to its starting position?

Givens: Triangle $T$ has vertices $(0,0), (4,0), (0,3)$ — a scalene right triangle, no nontrivial self-symmetry; $5$ available transformations: $R_{90}, R_{180}, R_{270}$ (CCW rotations about origin) and $S_x, S_y$ (reflections); Sequence length $3$, with repetition allowed; total $5^3 = 125$ sequences; Answer choices: (A) $12$, (B) $15$, (C) $17$, (D) $20$, (E) $25$

Unknowns: Number of ordered triples whose composition is the identity transformation

Understand

Restated: From the $5$ rigid motions $\{R_{90}, R_{180}, R_{270}, S_x, S_y\}$ (CCW rotations about the origin and reflections across the axes), pick an ordered triple (with repetition) of size $3$ — there are $5^3 = 125$ such ordered triples. How many of them, when applied one after another, return triangle $T$ with vertices $(0,0), (4,0), (0,3)$ to its starting position?

Givens: Triangle $T$ has vertices $(0,0), (4,0), (0,3)$ — a scalene right triangle, no nontrivial self-symmetry; $5$ available transformations: $R_{90}, R_{180}, R_{270}$ (CCW rotations about origin) and $S_x, S_y$ (reflections); Sequence length $3$, with repetition allowed; total $5^3 = 125$ sequences; Answer choices: (A) $12$, (B) $15$, (C) $17$, (D) $20$, (E) $25$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #10 Create a Physical Representation, #2 Make a Systematic List, #5 Look for a Pattern

Tool #7 (Subproblems): the triangle is asymmetric, so orientation parity splits the problem cleanly — even number of reflections required, giving Case 1 ($0$ reflections, $3$ rotations) and Case 2 ($2$ reflections, $1$ rotation). Tool #10 (Physical): cut out a paper triangle, label one side, and manipulate to verify each composition. Tool #2 (Systematic List): list all rotation-only triples summing to a multiple of $360^\circ$. Tool #5 (Pattern): two distinct axis-reflections compose to $R_{180}$ — this collapses Case 2 to a single unordered set $\{S_x, S_y, R_{180}\}$.

Execute — Answer: A

#7 Identify Subproblems 8.G.A.1 Step 1
  • Set up orientation parity.
  • Rotations preserve orientation (flipped-or-not); reflections reverse it.
  • Triangle $T$ has distinct side lengths $3, 4, 5$ — no rotation or reflection sends $T$ back to itself except the identity, so the composition must equal the identity (not merely some other symmetry of $T$).
  • The identity preserves orientation, so the number of reflections in the sequence must be even: $0$ or $2$.
  • This splits into two disjoint cases.
$$\#\text{reflections} \in \{0, 2\}$$

💡 An asymmetric triangle pins down the answer: only the true identity transformation works.

#2 Make a Systematic List 8.G.A.1 Step 2
  • Case 1: three rotations.
  • Encode $R_{90} \leftrightarrow 1$, $R_{180} \leftrightarrow 2$, $R_{270} \leftrightarrow 3$ (units of $90^\circ$ CCW).
  • The composition of three CCW rotations by amounts $n_1, n_2, n_3$ is a single CCW rotation by $n_1 + n_2 + n_3$ quarter-turns.
  • For identity, $n_1 + n_2 + n_3 \equiv 0 \pmod 4$.
  • With each $n_i \in \{1, 2, 3\}$, the sum ranges over $[3, 9]$.
  • Multiples of $4$ in $[3, 9]$: $4$ and $8$.
$$n_1 + n_2 + n_3 \equiv 0 \pmod 4, \quad n_i \in \{1, 2, 3\}$$

💡 Adding rotation amounts mod $4$ tells whether the composition is a full turn.

#2 Make a Systematic List 8.G.A.1 Step 3
  • Enumerate.
  • Sum $= 4$: the unordered multiset is $\{1, 1, 2\}$, with $\frac{3!}{2!} = 3$ orderings: $(1,1,2), (1,2,1), (2,1,1)$.
  • Sum $= 8$: the unordered multiset is $\{2, 3, 3\}$, with $\frac{3!}{2!} = 3$ orderings: $(2,3,3), (3,2,3), (3,3,2)$.
  • (Check $\{1,3,?\}$: needs $?$ in $\{0, 4\}$, both outside the set.
  • Check $\{2,2,?\}$: needs $?$ in $\{0, 4\}$, also outside.) Total for Case 1: $3 + 3 = 6$.
$$|\text{Case 1}| = 3 + 3 = 6$$

💡 Only two multisets work; the orderings give $3$ each.

#10 Create a Physical Representation 8.G.A.1 Step 4
  • Case 2: two reflections and one rotation.
  • First eliminate same-reflection pairs.
  • If both reflections are $S_x$, then $S_x \circ S_x = I$, so the full composition reduces to just the rotation $R \in \{R_{90}, R_{180}, R_{270}\}$ — none of these is the identity.
  • Same for $S_y, S_y$.
  • So the two reflections must be different: $\{S_x, S_y\}$.
  • Their composition: $(x, y) \xrightarrow{S_x} (x, -y) \xrightarrow{S_y} (-x, -y)$, which is $R_{180}$.
  • So once the two reflections appear, the cumulative effect of just them is $R_{180}$, and the remaining rotation $R$ must satisfy $R \circ R_{180} = I$, i.e.
  • $R = R_{180}$.
$$S_y \circ S_x = R_{180}; \;\; R \circ R_{180} = I \Rightarrow R = R_{180}$$

💡 Two axis-reflections in a row equal a half-turn; the half-turn needs another half-turn to undo.

#5 Look for a Pattern 8.G.A.1 Step 5
  • Order check.
  • The unordered set is $\{S_x, S_y, R_{180}\}$, three distinct elements.
  • Does every one of the $3! = 6$ orderings give the identity?
  • $R_{180}$ acts as $(x, y) \mapsto (-x, -y)$, which commutes with both $S_x$ and $S_y$ (each is a reflection in an axis through the origin, sharing the rotation's center).
  • Since all three pairwise commute, the order does not affect the product.
  • Any ordering yields $R_{180} \circ R_{180} = I$.
  • So all $6$ orderings work.
$$|\text{Case 2}| = 3! = 6$$

💡 All three pieces commute, so order doesn't matter — every arrangement works.

#7 Identify Subproblems 8.G.A.1 Step 6
  • The two cases are disjoint (Case 1 has $0$ reflections, Case 2 has $2$).
  • Add: $6 + 6 = 12$.
$$\text{Total} = 6 + 6 = 12 \Rightarrow \textbf{(A)}$$

💡 Two disjoint cases of size $6$ each.

[1] #7 8.G.A.1 Set up orientation parity. Rotations preserve orientation (flipped-or-not); refl
[2] #2 8.G.A.1 Case 1: three rotations. Encode $R_{90} \leftrightarrow 1$, $R_{180} \leftrighta
[3] #2 8.G.A.1 Enumerate. Sum $= 4$: the unordered multiset is $\{1, 1, 2\}$, with $\frac{3!}{2
[4] #10 8.G.A.1 Case 2: two reflections and one rotation. First eliminate same-reflection pairs.
[5] #5 8.G.A.1 Order check. The unordered set is $\{S_x, S_y, R_{180}\}$, three distinct elemen
[6] #7 8.G.A.1 The two cases are disjoint (Case 1 has $0$ reflections, Case 2 has $2$). Add: $6

Review

Reasonableness: Sanity. $12$ out of $125$ is about $10\%$, plausible for a tight identity constraint. Spot-check $(R_{90}, R_{90}, R_{180})$ from Case 1: total CCW rotation $= 90 + 90 + 180 = 360^\circ$, identity ✓. Spot-check $(S_x, R_{180}, S_y)$ from Case 2: $(x, y) \xrightarrow{S_x} (x, -y) \xrightarrow{R_{180}} (-x, y) \xrightarrow{S_y} (x, y)$, identity ✓. Spot-check a non-winner $(R_{90}, S_x, S_x)$: $(x, y) \xrightarrow{R_{90}} (-y, x) \xrightarrow{S_x} (-y, -x) \xrightarrow{S_x} (-y, x)$ — not identity ✓ (matches our "same reflection twice + rotation fails" reasoning). The trap choice (E) $25 = 5^2$ would be "any rotation choice once two reflections are placed" without filtering — clearly an over-count.

Alternative: Tool #10 (Physical Representation) entirely: cut a paper $3$-$4$-$5$ right triangle, mark one vertex, and tape it to a coordinate sheet. For each of the $125$ sequences, physically apply the three moves and check whether the triangle lands back on its tape outline. This is impractical for $125$ trials, but doing a sample of $10$ confirms the orientation-parity argument and the $R_{180}$-cancellation pattern. The orientation argument cuts work by a factor of $\sim 2$ and the rotation-sum trick cuts further.

CCSS standards used (min grade 8)

  • 8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Tracking how each rotation and reflection acts on coordinates, composing them, recognizing $S_y \circ S_x = R_{180}$, and confirming identities like $R_{180} \circ R_{180} = I$.)

⭐ This AMC 10 problem only needs Grade 8 properties of rotations and reflections you already know — even number of reflections (so $0$ or $2$); three rotations summing to $360^\circ$ or $720^\circ$ give $6$ ways; two different axis-reflections plus an $R_{180}$ give another $6$ ways; $6 + 6 = 12$.

⭐ This AMC 10 problem only needs Grade 8 properties of rotations and reflections you already know — even number of reflections (so $0$ or $2$); three rotations summing to $360^\circ$ or $720^\circ$ give $6$ ways; two different axis-reflections plus an $R_{180}$ give another $6$ ways; $6 + 6 = 12$.

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