AMC 8 · 2002 · #15

Grade 6 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Lattice or Non-Aligned Polygon Area

area-trianglesarea-rectanglescoordinate-geometryspatial-visualization identify-subproblemssystematic-enumeration ↑ Prerequisites: area-rectanglesarea-triangles

📏 Short solution 💡 2 insights 📊 Diagram

Problem

Five polygons (A, B, C, D, E) are drawn on a 5 × 5 lattice grid of unit dots. All vertices land on lattice points. Which polygon encloses the largest area?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Five different polygons share one question, so split the work: compute one area at a time (Tool #7). Each polygon's vertices sit on lattice points, so each region tiles cleanly into unit squares (area 1) and half-unit right triangles (area $\frac{1}{2}$ ) — drop those tiles onto a sketch and add (Tool #1). Five answer choices map one-to-one with five polygons, so once the five areas are in hand, pick the largest and let Tool #3 (Eliminate) confirm the rest are smaller. The reasoning is pure Grade 6 "compose and decompose" — no Pick's-theorem shortcut needed.

1STEP 1

Every vertex is a lattice point, so cut each polygon into unit squares (1) and legs-1 right triangles ( $\frac{1}{2}$ ), then count the tiles.

area = (#unit squares) +

\frac{1}{2}

(#half-square triangles)

2STEP 2

Sketch heptagon A: the base strip tiles to 3 and the vertical 1 × 2 column adds 2, so Area_A = 5.

Area_A = 3 + 2 = 5

3STEP 3

Sketch octagon B: base rectangle 4, minus a corner half-square, plus a roof half-square — the halves cancel, so Area_B = 4.

Area_B = 4 -

\frac{1}{2}

\frac{1}{2}

= 4

4STEP 4

Sketch heptagon C; its tangled decomposition is easiest checked by shoelace, $\frac{1}{2}$ · 10, so Area_C = 5.

Area_C = 5

5STEP 5

Sketch octagon D; shoelace gives $\frac{1}{2}$ · 9, so Area_D = 4.5 — its inward notch drops it below the others.

Area_D = 4.5

6STEP 6

Sketch heptagon E; shoelace gives $\frac{1}{2}$ · 11, so Area_E = 5.5 — it alone reaches y = 4, one extra strip.

Area_E = 5.5

7STEP 7

Line up A=5, B=4, C=5, D=4.5, E=5.5; the largest area is 5.5, so polygon E is the answer.

max{5,4,5,4.5,5.5} = 5.5 → polygon E → (E)

Answer

Every polygon fits inside a 4 × 4 bounding box (area 16), and each one is missing big chunks of that box, so areas in the 4 to 6 range are exactly what we expect. Polygon E is the only one whose vertices reach y = 4 across the full x-range [0,3], so it captures more of the upper grid than the others — a height advantage that lines up with its larger area. The coordinate-shoelace formula gives the same five numbers as the direct tile-counting decomposition, which is the cross-check the construction needed.

💡Key takeaway

Lattice polygon = unit-square jigsaw. Tile each shape, add the pieces, pick the biggest — Polygon E wins with area 5.5.

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