AMC 8 · 2019 · #21

Grade 6 geometry-2dalgebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

coordinate-geometryarea-trianglessystems-of-equationslinear-equations-two-var coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-trianglessystems-of-equations

📏 Medium solution 💡 2 insights

Problem

Three lines y = 5, y = 1 + x, and y = 1 - x cut out a single triangle in the coordinate plane. Find the area of that triangle.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The problem hands us three lines but no picture, and the question is geometric — exactly the setup that begs for Tool #1 (Draw a Diagram). A quick sketch on the coordinate plane reveals that two of the three vertices lie on the horizontal line y = 5, which makes that segment a perfect horizontal base. Tool #7 (Identify Subproblems) then splits the work into three clean pieces: (a) find the three vertices, (b) read off the base length and the height from the picture, (c) apply the triangle area formula. Breaking the problem this way avoids any need for a distance formula or coordinate-geometry algebra beyond solving simple one-step equations.

1STEP 1

Sketch the three lines: y = 5 is horizontal, and y = 1 + x and y = 1 - x meet at (0, 1), fanning up to each side at 45°.

Lines: y = 5, y = 1 + x, y = 1 - x

2STEP 2

Where y = 5 meets y = 1 + x, substituting gives x = 4, so the top-right vertex is (4, 5).

5 = 1 + x → x = 4 → (4, 5)

3STEP 3

Where y = 5 meets y = 1 - x, the same substitution gives x = -4, so the top-left vertex is (-4, 5).

5 = 1 - x → x = -4 → (-4, 5)

4STEP 4

Where the two slanted lines meet, 1 + x = 1 - x gives x = 0 and y = 1, so the bottom vertex is (0, 1).

1 + x = 1 - x → x = 0, y = 1 → (0, 1)

5STEP 5

The top vertices share y = 5, giving a horizontal base 8, and the drop down to (0, 1) gives height 4.

base = 4 - (-4) = 8, height = 5 - 1 = 4

6STEP 6

The area is half of base × height, so half of 8 × 4 = 16.

Area =

\frac{1}{2}

× 8 × 4 = 16 → (E)

Answer

The triangle is isosceles with a horizontal base of length 8 and a height of 4, so its area is half of an 8 × 4 = 32 rectangle, which gives 16. That matches choice (E). The answer is also plausible from the picture: the triangle fits snugly inside the rectangle with corners (± 4, 1) and (± 4, 5) — that rectangle has area 32, and the triangle is clearly half of it.

💡Key takeaway

This AMC 8 problem only needs Grade 6 coordinate-plane and triangle-area skills you already know!

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