AMC 8 · 2022 · #1

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Rectilinear Frame / Box Border

area-rectanglesarea-trianglesspatial-visualization area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglesarea-triangles

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

The Math Team's logo is a multiplication-symbol shape drawn on a 1-inch grid. The shape is the polygon with vertices (1,2),(2,1),(3,2),(4,1),(5,2),(4,3),(5,4),(4,5),(3,4),(2,5),(1,4),(2,3). We need its area in square inches and must choose from 10, 12, 13, 14, 15.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The ×-shape is awkward to measure directly, but it is trapped inside a clean 4 × 4 bounding box. Tool #7 (Identify Subproblems) lets us split the problem into two easy pieces: (1) the area of the 4 × 4 box and (2) the area of the white triangular cut-outs around the logo. Tool #1 (Draw a Diagram) is useful for marking those cut-out triangles on the picture, and Tool #16 (Change Focus / Count the Complement) names the strategy of finding the shaded area indirectly by subtracting the unshaded area from the whole box.

1STEP 1

Fence the logo in the smallest grid square, corners (1,1) to (5,5); its area is 4 × 4 = 16 in².

Box area = 4 × 4 = 16 in²

2STEP 2

At each box corner the outline slices off a right triangle with legs 1, so its area is $\frac{1}{2}$ × 1 × 1 = $\frac{1}{2}$ in².

Corner triangle =

\frac{1}{2}

× 1 × 1 =

\frac{1}{2}

in²

3STEP 3

Mid-side, a bigger triangle (e.g. (2,1),(4,1),(3,2)) has base 2 and height 1, so its area is $\frac{1}{2}$ × 2 × 1 = 1 in².

Side triangle =

\frac{1}{2}

× 2 × 1 = 1 in²

4STEP 4

Add all the white space: four corner triangles plus four side triangles, 4 × $\frac{1}{2}$ + 4 × 1 = 6 in².

White = 4 ×

\frac{1}{2}

+ 4 × 1 = 2 + 4 = 6 in²

5STEP 5

Subtract the white space from the box: 16 - 6 = 10 in² → (A).

Logo area = 16 - 6 = 10 in² → (A)

Answer

Sanity check by symmetry: the logo has four identical "arms" plus a central diamond. The central diamond has vertices (2,3),(3,2),(4,3),(3,4) — a square of diagonal 2, so its area is

\frac{1}{2}

× 2 × 2 = 2 in². Each arm (for the right arm, vertices (3,2),(4,1),(5,2),(4,3)) is also a unit-diagonal square of area 2 in². Total = 2 + 4 × 2 = 10 in², matching answer (A). The result also fits inside a 16 in² box, which is the right order of magnitude.

💡Key takeaway

This AMC 8 problem only needs Grade 6 "area by cutting into triangles and rectangles" you already know!

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