AMC 8 · 2025 · #1

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-rectanglesarea-trianglesfraction-arithmeticpercentage identify-subproblemsarea-difference ↑ Prerequisites: area-rectanglesfraction-arithmetic

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

An eight-pointed star is drawn inside a 4 × 4 unit grid. What percent of the grid's area is shaded by the star?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The star is a single complicated shape, but it is built from simple pieces that already line up with the grid: one central 2 × 2 square and eight half-square triangles. Tool #7 (Identify Subproblems) cleanly splits the star into "central square area" + "8 point-triangle areas", each of which is one elementary area calculation. Tool #1 (Draw a Diagram) helps mark which unit cells are fully shaded, which are half-shaded, and which are empty, so we do not double-count. After computing the star area we form the ratio star/total, convert to percent, and use Tool #3 (Eliminate Possibilities) to confirm the match with choice (B).

1STEP 1

The 4 × 4 grid is 16 unit squares of area 1 each, so the whole board's area is 16.

4 × 4 = 16

2STEP 2

The middle 2 × 2 block sits entirely inside the star, a fully shaded square of area 4.

2 × 2 = 4

3STEP 3

Each of the 8 star points is a right triangle that is half a unit square, so together they add 8 × $\frac{1}{2}$ = 4.

8 ×

\frac{1}{2}

= 4

4STEP 4

Add the pieces: center square (4) plus the eight points (4) give a star area of 8.

4 + 4 = 8

5STEP 5

The star covers 8 of the 16 cells: $\frac{8}{16}$ = $\frac{1}{2}$ = 50%, matching choice (B).

\frac{8}{16}

\frac{1}{2}

= 50% → (B)

Answer

Cross-check by the complement: the unshaded region is made of 4 corner triangles (each a right triangle with legs 2 and 1, so area

\frac{1}{2}

· 2 · 1 = 1, total 4) plus 4 unit squares tucked next to the corners (total 4). The unshaded area is 4 + 4 = 8, so the shaded star area is 16 - 8 = 8 — the same 50%. Also, by symmetry the star looks like it covers "about half" of the grid, so 50% is the only believable choice; 40% would be too sparse and 60%/75%/80% would visibly engulf the corners.

💡Key takeaway

This AMC 8 problem only needs Grade 6 percent and ratio reasoning you already know — once you split the star into a center square and eight half-square points!

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