AMC 8 · 2023 · #7

Grade 8 geometry-2d

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

coordinate-geometrylinear-equations-two-varslope-intercept coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryslope-intercept

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A small axis-aligned rectangle sits with one corner at (15, 3) and the opposite corner at (16, 5). One line is drawn through A(0,0) and B(3,1); another line is drawn through C(0,10) and D(2,9). We want to count how many points of the rectangle (its four sides and corners) lie on at least one of those two lines.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The picture is already on a grid, so the cleanest approach is to actually extend each line over to the rectangle and look (Tool #1). Because the two lines are independent, we can handle them as two separate small problems and add the answers (Tool #7): "does ℓ₁ hit the rectangle?" and "does ℓ₂ hit the rectangle?". Once we know the line's y-value at x = 15 and x = 16, we just compare with the rectangle's y-band [3, 5] — a very small case-check that immediately rules out four of the five choices (Tool #3).

1STEP 1

Sides parallel to the axes with opposite corners (15,3) and (16,5) make the rectangle exactly the band 15 ≤ x ≤ 16 and 3 ≤ y ≤ 5.

Rectangle = {(x,y) : 15 ≤ x ≤ 16, 3 ≤ y ≤ 5}

2STEP 2

From A(0,0) to B(3,1) the line rises 1 over 3, and through the origin that gives ℓ₁: y = $\frac{1}{3}$ x.

ℓ₁: y =

\frac{1}{3}

3STEP 3

Across the strip ℓ₁ runs from y=5 at x=15 up to $\frac{16}{3}$ ≈5.33 at x=16 — above the band except at one height, touching only the corner (15,5).

ℓ₁(15)=5, ℓ₁(16)=

\frac{16}{3}

≈ 5.33 → touches at (15,5) only

4STEP 4

Same for ℓ₂: y = $-\frac{1}{2}$ x + 10 gives 2.5 at x=15 and 2 at x=16 — the whole strip sits below the band [3,5], so no intersection.

ℓ₂: y=

-\frac{1}{2}

x+10; ℓ₂(15)=2.5, ℓ₂(16)=2

5STEP 5

ℓ₁ contributes the corner (15,5), ℓ₂ contributes none, and they never overlap, so the total boundary count lands on (B).

1 + 0 = 1 → (B)

Answer

Does 1 make sense? Line ℓ₁ has slope

\frac{1}{3}

starting from the origin, so at x=15 it should be at y=5 exactly — and the corner (15,5) of the rectangle sits at height 5, so a glancing touch at one corner is exactly what we expect. Line ℓ₂ starts high at (0,10) and drops by

\frac{1}{2}

per unit, so by x=15 it has fallen below 3 — it shoots under the rectangle entirely. One point total is the consistent answer.

💡Key takeaway

This AMC 8 problem only needs Grade 8 linear functions (the y = mx + b rule) you already know!

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