AMC 8 · 2004 · #25

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Overlapping Translates

area-rectanglesarea-circlespythagorean-theoremspatial-visualizationreflection-symmetry area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglesarea-circlespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Two 4 × 4 squares are placed so they meet at right angles and each one cuts through the midpoints of the other's two intersected sides. A circle is drawn whose diameter is the segment between the two points where the square boundaries cross. Find the area of the shaded region — the union of the two squares with the circle removed.

Pick an answer.

(A)

$16-4\pi$

(B)

$16-2\pi$

(C)

$28-4\pi$

(D)

$28-2\pi$

(E)

$32-2\pi$

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shaded region is a compound shape, so Tool #7 (Identify Subproblems) splits it into three clean pieces: (a) the area of the union of the two squares, which itself splits into "two squares minus their overlap"; (b) the radius of the circle, which comes from the geometry of the overlap square; and (c) the area of the circle. Tool #1 (Draw a Diagram) is the unlock: marking the overlap as a 2 × 2 square and labeling its diagonal makes (a) and (b) visible at a glance. Each subproblem is then a single formula away.

1STEP 1

Bisecting the sides makes the shared central overlap a square of side 2 and area 4.

overlap side =

\frac{4}{2}

= 2, overlap area = 2 × 2 = 4

2STEP 2

By inclusion-exclusion the union of the two squares is 4² + 4² - 2² = 28.

union area = 4² + 4² - 2² = 16 + 16 - 4 = 28

3STEP 3

The diameter is the overlap's diagonal 2√(2), so the radius √(2) is half of it.

d = √(2² + 2²) = √(8) = 2√(2), r =

\frac{d}{2}

= √(2)

4STEP 4

The circle's area is π r² = π(√(2))² = 2π.

circle area = π r² = π (√(2))² = 2π

5STEP 5

Shaded area = union - circle = 28 - 2π, which is choice (D).

shaded = 28 - 2π → (D)

Answer

28-2π

Check magnitudes. Each square covers 16, so the union is between 16 (if they coincided) and 32 (if they were disjoint); our 28 sits right in that band. The circle is inscribed in the 2 × 2 overlap region's diagonal — its area 2π ≈ 6.28 is comfortably less than the overlap's 4 · π/2 ≈ a small fraction of the cross, leaving shaded ≈ 28 - 6.28 ≈ 21.7 ＞ 0. Eliminations: (A) 16 - 4π and (B) 16 - 2π start from a single square, ignoring the second. (C) 28 - 4π would need radius 2, which is the overlap's side, not the half-diagonal. (E) 32 - 2π forgets to subtract the overlap once. Only (D) is consistent.

💡Key takeaway

Break a tangled picture into pieces: the cross is two 4 × 4 squares minus their 2 × 2 overlap (28), the circle's radius is half the overlap's diagonal (√(2)), and subtracting the circle (2π) leaves 28 - 2π.

More like this