AMC 10 · 2019 · #5

Grade 8 geometry-2d
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
reflection-symmetrycoordinate-geometryslope-intercepttransformations-composition caseworksystematic-enumeration ↑ Prerequisites: coordinate-geometryslope-intercept
📏 Medium solution 💡 3 insights
Problem
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.

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.

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

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 ⊥ A'B' — one failing example is enough.

1STEP 1

A point (p, q) with both coordinates positive reflects to (q, p), still positive — so A'B'C' stays in Q1. (A) is always true — eliminate.

(p, q), p, q ＞ 0 → (q, p), q, p ＞ 0 ✓
2STEP 2

Reflection is a rigid motion, so A'B'C' is congruent to ABC — same area. (B) is always true — eliminate.

Reflection → congruent → same area
3STEP 3

A = (p, q) gives A' = (q, p), so slope of AA' = p−qq−p\frac{p - q}{q - p} = -1. (C) is always true — eliminate.

slope(AA') = p−qq−p\frac{p - q}{q - p} = -1 ✓
4STEP 4

The same swap gives slope of CC' = -1 too, so AA' and CC' have equal slopes. (D) is always true — eliminate.

slope(CC') = -1 = slope(AA') ✓
5STEP 5

Take A = (1, 2), B = (3, 4): slope AB = 1 and slope A'B' = 1, so AB and A'B' are parallel, not perpendicular — (E) fails.

A = (1, 2), B = (3, 4): slope(AB) = 1, slope(A'B') = 1 → parallel, not perpendicular → (E)
Answer
Lines AB and A'B' are perpendicular to each other.
Verify (E) in general. Let A = (p₁, q₁) and B = (p₂, q₂). Slope of AB is m = q2−q1p2−p1\frac{q₂ - q₁}{p₂ - p₁}. Slope of A'B' between (q₁, p₁) and (q₂, p₂) is p2−p1q2−q1\frac{p₂ - p₁}{q₂ - q₁} = 1m\frac{1}{m}. The product of slopes is m · 1m\frac{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.
💡Key takeaway

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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