AMC 10 · 2019 · #5
Grade 8 geometry-2dProblem
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?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $AB \perp A'B'$
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
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.
💡 Swapping two positive numbers leaves them positive.
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.
💡 A mirror image is the same size — no stretching.
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.
💡 $AA'$ is perpendicular to $y = x$ (the mirror), so its slope is the negative reciprocal of $1$ — namely $-1$.
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.
💡 Every 'point-to-its-reflection' segment crosses the mirror at $-1$ slope.
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.
💡 One counterexample is enough to break 'always true'.
8.G.A.3 Check (A). A point $(p, q)$ in the first quadrant has $p > 0$ and $q > 0$. Its r 8.G.A.1 Check (B). Reflection is a rigid motion — it preserves all distances, so the ref 8.EE.B.6 Check (C). Let $A = (p, q)$ with $p \ne q$. Then $A' = (q, p)$. Slope of $AA'$ i 8.EE.B.6 Check (D). By the same calculation, slope of $CC'$ is also $-1$. Both $AA'$ and 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.1Verify experimentally the properties of rotations, reflections, and translations (Recognizing that reflection preserves distances and therefore area (statement B).)8.G.A.3Describe 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.6Use 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.3Interpret 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)!
⭐ 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)!
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