AMC 10 · 2020 · #14

Grade 7 geometry-2d

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

area-regular-hexagonarea-circlesequilateral-trianglesymmetry-argument identify-subproblemsarea-differencesymmetry-argument ↑ Prerequisites: area-regular-hexagonarea-circles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A regular hexagon has side length 2. Six semicircles lie inside it, one on each side, with each diameter equal to a side of the hexagon (so each semicircle has radius 1 and bulges into the hexagon). Find the area of the region that is inside the hexagon but outside all six semicircles.

Pick an answer.

(A)

$6\sqrt3 - 3\pi$

(B)

$\frac{9\sqrt3}{2} - 2\pi$

(C)

$\frac{3\sqrt3}{2} - \frac{\pi}{3}$

(D)

$3\sqrt3 - \pi$

(E)

$\frac{9\sqrt3}{2} - \pi$

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): subdivide the hexagon into a grid of small equilateral triangles of side 1 — 24 of them — so every region (shaded or white) is a clean union of these tiles or circular sectors. Tool #9 (Easier Problem): by 6-fold rotational symmetry, the shaded region is 6 congruent pieces; find the area of one piece and multiply. Tool #7 (Subproblems): each piece is (rhombus of 2 small triangles) minus (one 60° sector of radius 1). Tool #3 (Eliminate): the simplified expression 3√(3) - π matches exactly one answer choice.

1STEP 1

Long diagonals plus opposite-midpoint lines cut the side-2 hexagon into 24 congruent unit triangles (side 1).

hexagon = 24 small triangles of side 1

2STEP 2

By 6-fold rotational symmetry, the shaded region is 6 congruent pieces, one at each vertex — solve one, then times 6.

shaded = 6 · (one piece)

3STEP 3

Each vertex piece is a rhombus of 2 unit triangles minus a 60° sector of radius 1 at the vertex.

one piece = (rhombus of 2 unit triangles) - (60° sector of radius 1)

4STEP 4

Rhombus = 2 · $\frac{√(3)}{4}$ = $\frac{√(3)}{2}$ ; the 60° sector = π · $\frac{1²}{6}$ = $\frac{π}{6}$ .

rhombus =

\frac{√(3)}{2}

, sector =

\frac{π}{6}

5STEP 5

One piece = $\frac{√(3)}{2}$ - $\frac{π}{6}$ ; times 6 gives the shaded total 3√(3) - π.

shaded = 6 (

\frac{√(3)}{2}

\frac{π}{6}

) = 3√(3) - π

6STEP 6

Only choice (D) matches 3√(3) - π; the others miscount the √(3) or π coefficient.

3√(3) - π → (D)

Answer

3√3 - π

The hexagon's area is

\frac{3√(3)}{2}

· 4 = 6√(3) ≈ 10.39. The total semicircle area (with overlap) is 6 ·

\frac{π}{2}

= 3π ≈ 9.42 — so the white region cannot be that large; some semicircle area is double-counted. The shaded answer 3√(3) - π ≈ 5.196 - 3.14 ≈ 2.06 is positive and well below the hexagon area, exactly as expected for the leftover slivers near the vertices.

💡Key takeaway

This AMC 10 problem only needs Grade 7 area formulas you already know! Cut the side-2 hexagon into 24 small triangles of side 1. By 6-fold symmetry, the shaded region is 6 identical pieces near the vertices — each piece is a 2-triangle rhombus (area $\frac{√(3)}{2}$ ) minus a 60° sector of radius 1 (area $\frac{π}{6}$ ). Multiply by 6: 6( $\frac{√(3)}{2}$ - $\frac{π}{6}$ ) = 3√(3) - π, answer (D).

More like this