AMC 10 · 2019 · #5

Grade 8 geometry-2d

Archetype Small-Domain Constrained Search

reflection-symmetrycoordinate-geometryslope-intercepttransformations-composition caseworksystematic-enumeration ↑ Prerequisites: coordinate-geometryslope-intercept

📏 Medium solution 💡 3 insights

Problem

Triangle $ABC$ lies in the first quadrant. Points $A$ , $B$ , and $C$ are reflected across the line $y=x$ to points $A'$ , $B'$ , and $C'$ , respectively. Assume that none of the vertices of the triangle lie on the line $y=x$ . Which of the following statements is not always true?

$\textbf{(A) }$ Triangle $A'B'C'$ lies in the first quadrant.

$\textbf{(B) }$ Triangles $ABC$ and $A'B'C'$ have the same area.

$\textbf{(C) }$ The slope of line $AA'$ is $-1$ .

$\textbf{(D) }$ The slopes of lines $AA'$ and $CC'$ are the same.

$\textbf{(E) }$ Lines $AB$ and $A'B'$ are perpendicular to each other.

Pick an answer.

(A)

Triangle $A'B'C'$ lies in the first quadrant.

(B)

Triangles $ABC$ and $A'B'C'$ have the same area.

(C)

The slope of line $AA'$ is $-1$.

(D)

The slopes of lines $AA'$ and $CC'$ are the same.

(E)

Lines $AB$ and $A'B'$ are perpendicular to each other.

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Toolkit + CCSS Solution

Understand

Restated: Triangle $ABC$ sits in the first quadrant. Each vertex is reflected across the line $y = x$ to get $A'$, $B'$, $C'$. No vertex lies on $y = x$. Of the five statements about $ABC$ and $A'B'C'$, find the one that is NOT always true.

Givens: Reflection rule across $y = x$: a point $(p, q)$ maps to $(q, p)$; All vertices $A$, $B$, $C$ are in the first quadrant; No vertex sits on the line $y = x$, so $p \ne q$ at each vertex; Answer choices: (A) $A'B'C'$ in Q1, (B) same area, (C) slope of $AA' = -1$, (D) slopes of $AA'$ and $CC'$ equal, (E) $AB \perp A'B'$

Unknowns: Which statement fails for some triangle

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem

Tool #1 (Diagram): sketch the line $y = x$ and a sample triangle with reflected image — this makes (A) and (B) visually obvious. Tool #3 (Eliminate): walk each statement; if it holds in general, cross it off. Tool #9 (Easier Problem): for (C) and (D), use two letters for a generic vertex $(p, q)$ and compute the slope formula directly; for (E), pick a concrete simple triangle and check whether $AB \perp A'B'$ — one failing example is enough.

Execute — Answer: E

#3 Eliminate Possibilities 8.G.A.3 Step 1

Check (A).
A point $(p, q)$ in the first quadrant has $p > 0$ and $q > 0$.
Its reflection $(q, p)$ also has both coordinates positive, so it stays in the first quadrant.
All three reflected vertices stay in Q1, so $A'B'C'$ is in Q1.
Statement (A) is always true — eliminate.

$$(p, q),\ p, q > 0 \;\Rightarrow\; (q, p),\ q, p > 0\;\checkmark$$

💡 Swapping two positive numbers leaves them positive.

#3 Eliminate Possibilities 8.G.A.1 Step 2

Check (B).
Reflection is a rigid motion — it preserves all distances, so the reflected triangle is congruent to the original, with the same side lengths and the same area.
Statement (B) is always true — eliminate.

$$\text{Reflection} \Rightarrow \text{congruent} \Rightarrow \text{same area}$$

💡 A mirror image is the same size — no stretching.

#9 Solve an Easier Related Problem 8.EE.B.6 Step 3

Check (C).
Let $A = (p, q)$ with $p \ne q$.
Then $A' = (q, p)$.
Slope of $AA'$ is $\dfrac{p - q}{q - p} = \dfrac{-(q - p)}{q - p} = -1$.
Statement (C) is always true — eliminate.

$$\text{slope}(AA') = \dfrac{p - q}{q - p} = -1\;\checkmark$$

💡 $AA'$ is perpendicular to $y = x$ (the mirror), so its slope is the negative reciprocal of $1$ — namely $-1$.

#9 Solve an Easier Related Problem 8.EE.B.6 Step 4

Check (D).
By the same calculation, slope of $CC'$ is also $-1$.
Both $AA'$ and $CC'$ have slope $-1$, so the slopes are equal.
Statement (D) is always true — eliminate.

$$\text{slope}(CC') = -1 = \text{slope}(AA')\;\checkmark$$

💡 Every 'point-to-its-reflection' segment crosses the mirror at $-1$ slope.

#3 Eliminate Possibilities 8.F.A.3 Step 5

Check (E) with a concrete triangle.
Pick $A = (1, 2)$ and $B = (3, 4)$, so $A' = (2, 1)$, $B' = (4, 3)$.
Slope of $AB = \dfrac{4 - 2}{3 - 1} = 1$.
Slope of $A'B' = \dfrac{3 - 1}{4 - 2} = 1$.
The two slopes are equal ($1$ and $1$), not negative reciprocals — so $AB$ and $A'B'$ are PARALLEL here, not perpendicular.
Statement (E) FAILS — this is the not-always-true one.

$$A = (1, 2), B = (3, 4):\ \text{slope}(AB) = 1,\ \text{slope}(A'B') = 1 \Rightarrow \text{parallel, not perpendicular} \;\Rightarrow\; \textbf{(E)}$$

💡 One counterexample is enough to break 'always true'.

[1] #3 8.G.A.3 Check (A). A point $(p, q)$ in the first quadrant has $p > 0$ and $q > 0$. Its r

[2] #3 8.G.A.1 Check (B). Reflection is a rigid motion — it preserves all distances, so the ref

[3] #9 8.EE.B.6 Check (C). Let $A = (p, q)$ with $p \ne q$. Then $A' = (q, p)$. Slope of $AA'$ i

[4] #9 8.EE.B.6 Check (D). By the same calculation, slope of $CC'$ is also $-1$. Both $AA'$ and

[5] #3 8.F.A.3 Check (E) with a concrete triangle. Pick $A = (1, 2)$ and $B = (3, 4)$, so $A' =

Review

Reasonableness: Verify (E) in general. Let $A = (p_1, q_1)$ and $B = (p_2, q_2)$. Slope of $AB$ is $m = \dfrac{q_2 - q_1}{p_2 - p_1}$. Slope of $A'B'$ between $(q_1, p_1)$ and $(q_2, p_2)$ is $\dfrac{p_2 - p_1}{q_2 - q_1} = \dfrac{1}{m}$. The product of slopes is $m \cdot \dfrac{1}{m} = 1$, NOT $-1$. So $AB$ and $A'B'$ are never perpendicular (perpendicular requires product $= -1$). The general analysis matches the concrete counterexample — (E) is the answer.

Alternative: Tool #1 (Diagram): draw $y = x$ as a mirror, then draw any triangle and its reflection. Visually, segments from each vertex to its mirror image are all perpendicular to $y = x$ (so they're parallel to each other and have slope $-1$), and the reflected triangle's sides are mirror images that are NOT perpendicular to the originals in general — confirming (A), (B), (C), (D) and breaking (E).

CCSS standards used (min grade 8)

8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Recognizing that reflection preserves distances and therefore area (statement B).)
8.G.A.3 Describe the effect of dilations, translations, rotations, and reflections on coordinates (Applying the rule $(p, q) \to (q, p)$ for reflection across $y = x$ — needed for statements A, C, D, E.)
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points (Computing slope of $AA'$ and $CC'$ as $\dfrac{p - q}{q - p} = -1$ (statements C, D).)
8.F.A.3 Interpret the equation y = mx + b as defining a linear function (Comparing slopes of $AB$ and $A'B'$ to check perpendicularity (negative reciprocals) — statement E.)

⭐ This AMC 10 problem only needs Grade 8 reflection and slope rules you already know — across $y = x$, points swap their coordinates. That makes (A), (B), (C), (D) all hold, but slopes of $AB$ and $A'B'$ multiply to $+1$ (not $-1$), so $AB$ and $A'B'$ are NOT always perpendicular. Answer: (E)!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 8)

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